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Grade 8 math worksheets

Build your own math worksheets from 35,000+ problems for grades 3 to 12, from fractions to AP Calculus. Every problem comes with step-by-step solutions.

Build linear models from descriptions

Learning goals

  • Identifies slope and an appropriate point or intercept from given information.
  • Writes a linear equation in a suitable form.
  • Converts among equivalent linear forms when useful.
  • Checks the equation against given points or context.

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Build linear models from descriptions 1
A plant is modeled as \(12\,\text{cm}\) tall at the start of an observation period and growing at a constant rate of \(1.5\,\text{cm}\) per week. Let \(h(w)\) be its height, in centimeters, after \(w\) weeks. Write a linear function for the model.

Hints

- In a linear model, the initial value is the output when the input is \(0\). - The constant weekly growth rate is the coefficient of the time variable.

Solution

1. The initial value is \(12\), so the constant term is \(12\). 2. The rate of change is \(1.5\) centimeters per week, so the coefficient of \(w\) is \(1.5\). 3. Therefore, the linear model is \(h(w)=1.5w+12\).

Answer

\(h(w)=1.5w+12\)
Build linear models from descriptions 2
A print shop charges a school \(\$0.15\) per field-trip ticket, with no setup fee. Write a rule \(C(n)\) for the cost of printing \(n\) tickets. What is \(C(0)\)?

Hints

- Think about what happens when the number of tickets doubles. - Does the cost include anything before the first ticket is printed?

Solution

1. Every ticket costs \(\$0.15\), with no fixed charge, so \(C(n)=0.15n\). 2. With zero tickets, \(C(0)=0\) dollars.

Answer

\(C(n)=0.15n\); \(C(0)=\$0\)
Build linear models from descriptions 3
A skating rink charges \(\$3\) to rent skates plus \(\$2\) for each hour of skating. Write a function \(C(h)\) for the total cost after \(h\) hours.

Hints

- Separate the fee paid once from the charge that depends on time. - Only the hourly part is multiplied by the number of hours.

Solution

1. The fixed skate-rental charge is \(\$3\). 2. The skating charge for \(h\) hours is \(\$2h\), so \(C(h)=3+2h\).

Answer

\(C(h)=3+2h\), in dollars
Build linear models from descriptions 4
A bird feeder holds \(18\) cups of seed. Birds eat \(2\) cups per day at a steady rate. Write a rule \(F(d)\) for the seed left after \(d\) days, until the feeder is empty.

Hints

- What amount is in the feeder before the birds eat any seed? - Should the rule add to or subtract from that amount as the days pass?

Solution

1. The feeder starts with \(18\) cups. 2. Each day removes \(2\) cups, so \(F(d)=18-2d\). The model runs from \(d=0\) to \(d=9\).

Answer

\(F(d)=18-2d\) for \(0\le d\le9\), with the seed amount in cups
Build linear models from descriptions 5
A school bus drives straight away from school at a steady \(30\) miles per hour for one hour. Let \(t\) be the hours since it left. a) Write a rule \(d(t)\) for its distance from school in miles. b) Which quantity is the input, and which is the output?

Hints

- What distance has the bus traveled at the starting time? - How much distance does one hour add along this straight road?

Solution

1. The bus starts at distance \(0\) and travels \(30\) miles each hour, so \(d(t)=30t\) for \(0\le t\le1\). 2. Time \(t\) is the input; distance \(d(t)\) is the output.

Answer

a) \(d(t)=30t\), for \(0\le t\le1\) b) Time is the input; distance from school is the output.
Build linear models from descriptions 6
A tank starts with the amount of water marked by the dot. Water then flows in at \(4\) liters per minute. Write a rule \(V(t)\) for the volume in liters after \(t\) minutes. Explain what the two numbers in your rule mean.
Figure for problem 559966

Hints

- Read the volume when the input is zero. - What changes as the number of minutes grows, and what stays fixed in your rule?

Solution

1. The dot is at \((0,6)\), so the tank starts with \(6\) liters. 2. Every minute adds \(4\) liters. The rule is \(V(t)=6+4t\), where \(6\) is the starting volume and \(4\) is the liters added per minute.

Answer

\(V(t)=6+4t\); \(6\) liters at the start, adding \(4\) liters each minute
Build linear models from descriptions 7
A swimming pool is being prepared for summer. At the start, there are already \(4\,\text{in.}\) of water in the pool. Water is then added at a constant rate. The table shows the water depth at different times: <table> <tr><td>Time \(t\) (h)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td><td>\(5\)</td></tr> <tr><td>Water depth \(h\) (in.)</td><td>\(4\)</td><td>\(10\)</td><td>\(16\)</td><td>\(22\)</td><td>\(28\)</td><td>\(34\)</td></tr> </table> a) Describe the graph formed by the data points, with time on the x-axis and water depth on the y-axis. b) How many inches per hour does the water depth increase? c) Write a function for the water depth \(h\) after \(t\) hours. d) Use your function to find the water depth after \(2.5\) hours.

Hints

- Compare the water-depth values from one hour to the next. Is the change constant? - What water depth is already present at \(t = 0\)? - Combine the initial value with the hourly increase to write the function. - Substitute \(2.5\) for \(t\) in your function.

Solution

1. The points \((0, 4)\), \((1, 10)\), \((2, 16)\), \((3, 22)\), \((4, 28)\), and \((5, 34)\) lie on a straight increasing line. 2. The water depth increases by \(10 - 4 = 6\) inches from one hour to the next, and the same increase continues throughout the table. The rate of change is \(6\,\text{in./h}\). 3. The initial value is \(4\) inches, so the function is \(h(t) = 6t + 4\). 4. \(h(2.5) = 6 \cdot 2.5 + 4 = 19\).

Answer

a) The data points lie on a straight increasing line. b) \(6\,\text{in./h}\) c) \(h(t) = 6t + 4\) d) \(19\,\text{in.}\)
Build linear models from descriptions 8
A rain barrel already contains \(20\) gallons when a storm begins. During the storm, water enters at a constant rate of \(8\) gallons per hour. a) Explain why the relationship between time \(t\), in hours, and volume \(V\), in gallons, can be modeled by a linear function. b) Write the function. c) How would the function change if the barrel were empty when the storm began? Would it still be linear?

Hints

- What feature of a situation shows a constant rate of change? - What does the value at time \(0\) represent? - Recall the difference between a general linear function and a proportional function.

Solution

1. The volume increases by the same amount, \(8\) gallons, each hour, so the rate of change is constant. Therefore, the relationship is linear. 2. The initial value is \(20\), and the rate is \(8\), so \(V(t)=8t+20\). 3. If the barrel started empty, the function would be \(V(t)=8t\). It would still be linear and would also be proportional because it passes through the origin.

Answer

a) The relationship is linear because the volume increases at a constant rate. b) \(V(t)=8t+20\). c) \(V(t)=8t\); it is still linear and is also proportional.
Build linear models from descriptions 9
In a spring-stretching experiment, \(x\) is the attached mass in grams and \(y\) is the spring's total length in centimeters. <table> <tr><td>Mass \(x\) (\(\text{g}\))</td><td>\(0\)</td><td>\(50\)</td><td>\(100\)</td><td>\(150\)</td><td>\(200\)</td></tr> <tr><td>Length \(y\) (\(\text{cm}\))</td><td>\(12.0\)</td><td>\(14.5\)</td><td>\(17.0\)</td><td>\(19.5\)</td><td>\(22.0\)</td></tr> </table> a) Find the linear function \(f(x) = mx + b\) that exactly models the data. b) Interpret the slope \(m\) and y-intercept \(b\) in the context of the experiment.

Hints

- What does the table show when no mass is attached? - How much does the spring length change over a known change in mass? - Use change in output divided by change in input to find the slope.

Solution

1. When \(x = 0\), the spring length is \(12.0\,\text{cm}\), so \(b = 12\). 2. Using \((0, 12)\) and \((50, 14.5)\), \(m = \frac{14.5 - 12}{50 - 0} = \frac{2.5}{50} = 0.05\). 3. Therefore, \(f(x) = 0.05x + 12\). 4. The y-intercept means the unloaded spring is \(12\,\text{cm}\) long. The slope means the spring length increases by \(0.05\,\text{cm}\) for each additional gram of mass.

Answer

a) \(f(x) = 0.05x + 12\) b) \(b = 12\) is the spring's length with no attached mass, in centimeters. \(m = 0.05\) is the increase in length, in centimeters per gram.
Build linear models from descriptions 10
A candle is \(8\,\text{in.}\) tall when it is lit. It burns at a constant rate, and its height decreases by \(1\,\text{in.}\) each hour. a) Write a linear function for the candle's height \(h\), in inches, after \(t\) hours. b) After how many hours will the candle be half its original height? c) When will the candle be completely burned down?

Hints

- What is the candle's starting height, and how much does it change each hour? - What sign should the rate of change have when a quantity is decreasing? - What height represents a candle that has completely burned down? - For each target height, set the function equal to that height and solve for time.

Solution

1. The initial height is \(8\) inches, and the height decreases by \(1\) inch per hour. Therefore, \(h(t) = -t + 8\). 2. Half the original height is \(4\) inches. Solve \(4 = -t + 8\), which gives \(t = 4\). 3. The candle is completely burned down when \(h(t) = 0\). Solve \(0 = -t + 8\), which gives \(t = 8\).

Answer

a) \(h(t) = -t + 8\) b) \(4\) hours c) \(8\) hours
Build linear models from descriptions 11
When a freezer is turned on, its inside temperature is \(72\,^\circ\text{F}\). The temperature decreases at a constant rate of \(8\,^\circ\text{F}\) per hour. 1) Write a linear function for the temperature \(y\), in degrees Fahrenheit, after \(x\) hours. 2) Find the temperature after \(5.5\) hours. 3) After how many hours will the freezer reach its target temperature of \(0\,^\circ\text{F}\)?

Hints

- Identify the starting temperature and the amount it changes each hour. - Because the temperature is decreasing, what sign should the rate of change have? - To find a time, set the model equal to the target temperature and solve for the time variable.

Solution

1. The initial value is \(72\), and the rate of change is \(-8\). Therefore, \(y = -8x + 72\). 2. Substitute \(x = 5.5\): \(y = -8 \cdot 5.5 + 72 = 28\). The temperature is \(28\,^\circ\text{F}\). 3. Set the model equal to the target temperature: \(0 = -8x + 72\). Solving gives \(x = 9\).

Answer

1) \(y = -8x + 72\) 2) \(28\,^\circ\text{F}\) 3) \(9\) hours
Build linear models from descriptions 12
Paul is saving for a new bike. He already has \(\$45.00\) saved, and he adds \(\$7.50\) each week. 1) Write a linear function for the amount \(y\), in dollars, after \(x\) weeks. 2) Find the points on the graph after \(0\), \(2\), \(4\), \(6\), and \(8\) weeks. 3) How many weeks will it take Paul to have exactly \(\$120.00\)?

Hints

- Which amount is already there at week \(0\), and which amount changes each week? - How can you relate the number of weeks to the amount saved? - What does a specified target balance mean for the equation you need to solve? - Which quantity belongs on the x-axis, and which belongs on the y-axis?

Solution

1. The initial amount is \(45\) and the weekly increase is \(7.50\), so \(y = 7.50x + 45.00\). 2. Substitution gives \((0, 45)\), \((2, 60)\), \((4, 75)\), \((6, 90)\), and \((8, 105)\). 3. Solve \(120.00 = 7.50x + 45.00\). Then \(75.00 = 7.50x\), so \(x = 10\).

Answer

1) \(y = 7.50x + 45.00\) 2) \((0, 45)\), \((2, 60)\), \((4, 75)\), \((6, 90)\), \((8, 105)\) 3) 10 weeks
Build linear models from descriptions 13
In a proportional relationship between \(x\) and \(y\), every increase of \(4\) in \(x\) produces an increase of \(14\) in \(y\). a) Write the equation in the form \(y=kx\). b) What x-value gives \(y=49\)?

Hints

- What does the ratio \(\frac{\Delta y}{\Delta x}\) represent? - How is that ratio related to \(k\)? - Rearrange the equation when the output is known.

Solution

1. The constant of proportionality is the ratio of the changes: \(k=14 \div 4=3.5\). 2. The equation is \(y=3.5x\). 3. Solve \(49=3.5x\): \(x=49 \div 3.5=14\).

Answer

a) \(y=3.5x\) b) \(x=14\)
Build linear models from descriptions 14
At the start of a fire-department training exercise, a water tank contains \(3000\) gallons. Water is pumped out at a constant rate of \(375\) gallons per hour. The graph shows the amount of water remaining as a function of time. a) Use the graph to estimate how many hours pass before \(1125\) gallons remain. b) Write a linear function \(V\), where \(t\) is time in hours and \(V(t)\) is the remaining volume in gallons. State the contextual domain, and use the function to find exactly when the tank is empty.
Figure for problem 532220

Hints

- Use the vertical axis for the amount of water and the horizontal axis for time. - A constant decrease produces a linear function with a negative rate of change. - The contextual domain begins at the start of the exercise and ends when the tank is empty. - The tank is empty when its volume equals \(0\).

Solution

1. Locate \(1125\) gallons on the vertical axis and move horizontally to the graph. The corresponding time is \(5\) hours. 2. The initial value is \(3000\), and the volume decreases by \(375\) gallons each hour. Therefore, \(V(t)=3000-375t\). The model applies from the start until the tank is empty, so its contextual domain is \(0\le t\le8\). The tank is empty when \(V(t)=0\): \(3000-375t=0\), so \(375t=3000\) and \(t=8\).

Answer

a) \(5\) hours b) \(V(t)=3000-375t\), with \(0\le t\le8\); the tank is empty after \(8\) hours.
Build linear models from descriptions 15
A library adds books at a constant rate. At the end of week \(2\), it has \(18\) books on a display. At the end of week \(5\), it has \(27\) books. Write a rule \(B(w)\) for the number of books on the display at the end of week \(w\), starting from week \(0\).

Hints

- Find how much the display grew between the two observations. - Divide by the number of weeks between them. - Work back from one observation to week zero to find the starting number.

Solution

1. Over \(5-2=3\) weeks, the display gains \(27-18=9\) books. The rate is \(9\div3=3\) books per week. 2. Two weeks of additions account for \(6\) books, so the starting number is \(18-6=12\). 3. Thus \(B(w)=12+3w\), for whole-number weeks \(w\ge0\).

Answer

\(B(w)=12+3w\), for whole-number weeks \(w\ge0\)
Build linear models from descriptions 16
A ticket desk starts with \(20\) tickets and sells exactly \(2\) tickets each hour. A student graphed how many tickets were left after each whole hour. a) How many tickets does the student's graph show at hour \(2\)? How many should be left then? b) Explain the error using these two values. c) Write the correct rule \(L(t)\) for whole-number hours until the tickets run out.
Figure for problem 559968

Hints

- Read the dot directly above hour \(2\). - Find how many tickets have been sold by that hour. - Compare the plotted count with the count from the story before writing the rule.

Solution

1. The plotted dot at hour \(2\) shows \(24\) tickets. 2. The desk has sold \(2\cdot2=4\) tickets by then, so \(20-4=16\) should remain. The plotted value is \(8\) tickets too high. 3. The number left is \(L(t)=20-2t\) for whole numbers \(t=0,1,\ldots,10\).

Answer

a) The graph shows \(24\) tickets; \(16\) should remain. b) The student added the tickets sold instead of subtracting them. That makes the plotted value \(8\) tickets too high at hour \(2\). c) \(L(t)=20-2t\) for whole-number \(t\) from \(0\) through \(10\).
Build linear models from descriptions 17
Two runners, Alex and Blake, train on a long, straight trail. Alex starts at mile marker \(0\) and runs at a constant speed of \(8\,\text{mi/h}\). Blake starts at the same time at mile marker \(2\) and runs at a constant speed of \(6\,\text{mi/h}\). Write a linear position function for each runner, then use the models to determine when and where Alex catches Blake.

Hints

- For each runner, identify the starting position and the distance added each hour. - A linear position model has the form starting position plus rate times time. - At the catch-up time, what must be true about the two position values?

Solution

1. Alex starts at \(0\) and gains \(8\) miles each hour, so \(s_A(t)=8t\). 2. Blake starts at mile marker \(2\) and gains \(6\) miles each hour, so \(s_B(t)=2+6t\). 3. At the catch-up time the positions are equal: \(8t=2+6t\). Thus \(2t=2\), so \(t=1\) hour. 4. Substitute \(t=1\): \(s_A(1)=8\) and \(s_B(1)=8\). Alex catches Blake at mile marker \(8\).

Answer

\(s_A(t)=8t\) and \(s_B(t)=2+6t\). Alex catches Blake after \(1\) hour at mile marker \(8\).
Build linear models from descriptions 18
A smartphone battery drains at a constant rate during a certain type of use. It is fully charged at \(10{:}00\) a.m. The table shows the battery level at several times. <table><tr><td>Time</td><td>\(10{:}00\) a.m.</td><td>\(12{:}00\) p.m.</td><td>\(2{:}00\) p.m.</td><td>\(3{:}00\) p.m.</td><td>\(6{:}00\) p.m.</td></tr><tr><td>Battery level (%)</td><td>\(100\)</td><td>\(84\)</td><td>\(68\)</td><td>\(60\)</td><td>\(36\)</td></tr></table> a) Find the average decrease in battery level per hour. Verify that the rate is constant over every interval in the table. b) Estimate the battery level at \(4{:}30\) p.m. Explain your method. c) If the pattern continues, at what time will the battery reach \(0\%\)? d) Write a rule for the battery level \(y\) after \(x\) hours have passed since \(10{:}00\) a.m.

Hints

- For each interval, divide the change in battery level by the elapsed time. - Find how many hours pass from \(10{:}00\) a.m. to \(4{:}30\) p.m. - Set the battery-level expression equal to \(0\) to find when the battery is empty. - Identify the initial value and the change per hour.

Solution

1. From \(10{:}00\) a.m. to \(12{:}00\) p.m., the level decreases by \(16\) percentage points in \(2\) hours, or \(8\) points per hour. The changes over the other intervals are also \(16\div2=8\), \(8\div1=8\), and \(24\div3=8\) percentage points per hour. The rate is constant. 2. By \(4{:}30\) p.m., \(6.5\) hours have passed. The model gives \(100-8(6.5)=48\), so the battery level is \(48\%\). 3. Solve \(100-8x=0\). Then \(x=12.5\), which is \(12\) hours \(30\) minutes after \(10{:}00\) a.m., or \(10{:}30\) p.m. 4. The initial level is \(100\), and it decreases by \(8\) percentage points per hour, so \(y=100-8x\).

Answer

a) A constant decrease of \(8\) percentage points per hour b) \(48\%\) c) \(10{:}30\) p.m. d) \(y=100-8x\)
Build linear models from descriptions 19
A taxi company charges a base fare of \(\$4.50\) plus \(\$2.20\) per mile. a) Write an equation for the total fare \(y\), in dollars, after traveling \(x\) miles. b) Explain what \(4.50\) represents in this situation. c) A passenger pays exactly \(\$26.50\). How many miles did the passenger travel?

Hints

- Separate the fixed charge from the charge that depends on distance. - Identify the rate per mile and the initial value. - For part c), subtract the base fare before dividing by the rate.

Solution

1. The total fare is the mileage charge plus the base fare, so \(y = 2.20x + 4.50\). 2. The value \(4.50\) is the fare charged before any miles are traveled. It is the initial value of the model. 3. Substitute \(y = 26.50\): \(26.50 = 2.20x + 4.50\). 4. Subtract \(4.50\): \(22.00 = 2.20x\). 5. Divide by \(2.20\): \(x = 10\).

Answer

a) \(y = 2.20x + 4.50\) b) The \(\$4.50\) is the base fare charged regardless of distance. c) The passenger traveled \(10\) miles.
Build linear models from descriptions 20
Two mobile data plans have the following monthly costs. - Plan A: a \(\$10.00\) base fee plus \(\$0.50\) per gigabyte. - Plan B: no base fee and \(\$1.50\) per gigabyte. a) Write an expression for each plan's monthly cost \(C\) as a function of data use \(x\), in gigabytes. b) Find both costs for \(5\,\text{GB}\) and \(15\,\text{GB}\). c) For what data use is Plan A less expensive than Plan B? Explain by comparing the expressions or testing values.

Hints

- A base fee is included even when no data is used. - Make a small table to compare the costs. - Find the input where the two cost expressions are equal.

Solution

1. The cost functions are \(C_A(x)=10+0.50x\) and \(C_B(x)=1.50x\). 2. At \(5\,\text{GB}\), Plan A costs \(\$10.00+\$0.50 \cdot 5=\$12.50\), and Plan B costs \(\$1.50 \cdot 5=\$7.50\). 3. At \(15\,\text{GB}\), Plan A costs \(\$17.50\), and Plan B costs \(\$22.50\). 4. The costs are equal when \(10+0.50x=1.50x\), which gives \(x=10\). Plan A is less expensive when \(x>10\).

Answer

a) \(C_A(x)=10+0.50x\); \(C_B(x)=1.50x\) b) At \(5\,\text{GB}\): Plan A \(\$12.50\), Plan B \(\$7.50\). At \(15\,\text{GB}\): Plan A \(\$17.50\), Plan B \(\$22.50\). c) Plan A is less expensive for more than \(10\,\text{GB}\).
Build linear models from descriptions 21
A candle is \(30\,\text{cm}\) tall when it is lit. After burning at a constant rate for \(4\,\text{hours}\), it is \(22\,\text{cm}\) tall. a) Is the relationship from burn time (in hours) to remaining height (in centimeters) proportional? Briefly explain. b) Write a function \(h(x)\) for the candle's remaining height after \(x\) hours, while the candle is burning. c) Use your function to find the candle's height after \(7.5\,\text{hours}\).

Hints

- Think about the candle's height at time \(0\). Would a proportional relationship start there? - How many centimeters of height does the candle lose each hour? - Start your function with the candle's initial height and account for the amount lost each hour.

Solution

1. A proportional relationship must include \((0, 0)\). Here, the candle has height \(30\,\text{cm}\) at \(0\,\text{hours}\), so the relationship is not proportional. 2. In \(4\,\text{hours}\), the candle loses \(30 - 22 = 8\,\text{cm}\), so its rate of change is \(-8 \div 4 = -2\,\text{cm/h}\). 3. Starting from \(30\,\text{cm}\) and decreasing by \(2\,\text{cm}\) per hour gives \(h(x) = 30 - 2x\), with \(0 \le x \le 15\). 4. Evaluate the function at \(x = 7.5\): \(h(7.5) = 30 - 2 \cdot 7.5 = 15\).

Answer

a) No. The relationship is not proportional because it does not pass through \((0, 0)\); at \(0\,\text{hours}\), the candle is \(30\,\text{cm}\) tall. b) \(h(x) = 30 - 2x\) for \(0 \le x \le 15\). c) \(15\,\text{cm}\)
Build linear models from descriptions 22
Two different candles are lit at the same time. Candle A is initially \(30\,\text{cm}\) tall and is \(18\,\text{cm}\) tall after \(4\) hours. Candle B is initially \(20\,\text{cm}\) tall and burns down at a constant rate of \(1\,\frac{\text{cm}}{\text{h}}\). Let \(t\) be the number of hours after the candles are lit. a) Determine Candle A's constant rate of change and construct linear height functions \(h_A(t)\) and \(h_B(t)\). b) Use your model to find Candle A's height after \(7\) hours and the time when Candle A reaches height \(0\). c) Use the two functions to determine when the candles have the same height and what that height is.

Hints

- Use the two measurements for Candle A to find its change in height per hour. - A linear model needs both an initial value and a signed rate of change. - Once both functions are written, equal heights correspond to equal function outputs at the same input.

Solution

1. Candle A loses \(30-18=12\,\text{cm}\) in \(4\) hours, so its rate of change is \(-3\,\frac{\text{cm}}{\text{h}}\). 2. Therefore, \(h_A(t)=30-3t\) and \(h_B(t)=20-t\). 3. \(h_A(7)=30-3\cdot7=9\), so Candle A is \(9\,\text{cm}\) tall after \(7\) hours. 4. Solve \(30-3t=0\). This gives \(t=10\), so Candle A reaches height \(0\) after \(10\) hours. 5. Set the functions equal: \(30-3t=20-t\). Then \(10=2t\), so \(t=5\). 6. At \(t=5\), \(h_B(5)=20-5=15\). Both candles are \(15\,\text{cm}\) tall.

Answer

a) Candle A's rate is \(-3\,\frac{\text{cm}}{\text{h}}\); \(h_A(t)=30-3t\), \(h_B(t)=20-t\) b) \(9\,\text{cm}\); Candle A reaches height \(0\) after \(10\) hours. c) After \(5\) hours, both candles are \(15\,\text{cm}\) tall.
Build linear models from descriptions 23
Lara watches her smartphone battery during a long train ride while playing a game continuously. At \(12{:}15\) p.m., the battery level is \(88\%\). At \(2{:}45\) p.m., it is \(63\%\). a) Find the decrease in battery level, in percentage points per hour. Under this linear assumption, at what time would the battery reach \(0\%\)? b) According to the model, what would the battery level have been at \(10{:}15\) a.m.? c) Explain why the result from part b shows that the linear model cannot be valid for the entire morning.

Hints

- Convert the elapsed time to hours before finding the rate. - Use the rate to calculate both forward and backward in time. - Consider the greatest possible battery level.

Solution

1. From \(12{:}15\) p.m. to \(2{:}45\) p.m. is \(2.5\) hours. The battery decreases by \(88-63=25\) percentage points, so the rate is \(25\div2.5=10\) percentage points per hour. 2. Starting from \(63\%\), reaching \(0\%\) takes \(63\div10=6.3\) hours, or \(6\) hours \(18\) minutes. That is \(9{:}03\) p.m. 3. At \(10{:}15\) a.m., two hours before the first measurement, the linear model gives \(88+2(10)=108\%\). 4. A battery cannot be charged above \(100\%\). Therefore, the constant-rate model cannot be extended that far backward in time.

Answer

a) \(10\) percentage points per hour; \(9{:}03\) p.m. b) \(108\%\) c) The result exceeds the physical maximum of \(100\%\), so the model is not valid for that earlier time.
Build linear models from descriptions 24
A sports club tracks membership in its tennis and basketball programs. <table> <thead> <tr><th>Year</th><th>2010</th><th>2013</th><th>2016</th><th>2019</th><th>2022</th></tr> </thead> <tbody> <tr><td>Tennis</td><td>\(240\)</td><td>\(230\)</td><td>\(215\)</td><td>\(205\)</td><td>\(190\)</td></tr> <tr><td>Basketball</td><td>\(80\)</td><td>\(110\)</td><td>\(145\)</td><td>\(175\)</td><td>\(210\)</td></tr> </tbody> </table> For parts a)–c), assume each program changed at a constant rate from 2019 onward equal to its average rate from 2019 to 2022. Let \(t\) be the number of years after 2019. a) Construct linear models \(T(t)\) and \(B(t)\) for tennis and basketball membership. b) Use the models to estimate when the memberships were equal. c) Use the basketball model to predict membership in 2025.

Hints

- Use only the 2019 and 2022 values to determine each assumed constant rate for this model. - Each model needs the 2019 membership as its initial value because \(t=0\) represents 2019. - Equal memberships occur when the two model outputs are equal.

Solution

1. Tennis changes from \(205\) to \(190\) in \(3\) years, so its rate is \(\frac{190-205}{3}=-5\) members per year. Thus, \(T(t)=205-5t\). 2. Basketball changes from \(175\) to \(210\) in \(3\) years, so its rate is \(\frac{210-175}{3}=\frac{35}{3}\) members per year. Thus, \(B(t)=175+\frac{35}{3}t\). 3. Set the models equal: \(205-5t=175+\frac{35}{3}t\). 4. Then \(30=\frac{50}{3}t\), so \(t=1.8\). This is about \(1.8\) years after 2019, or approximately 2021. 5. The year 2025 is \(6\) years after 2019. \(B(6)=175+\frac{35}{3}\cdot6=245\).

Answer

a) \(T(t)=205-5t\), \(B(t)=175+\frac{35}{3}t\) b) Approximately 2021 c) \(245\) members
Build linear models from descriptions 25
A pump fills an empty tank at a constant rate. After \(15\) minutes, the tank contains \(450\) gallons. a) Write a function for the volume \(V\), in gallons, after \(t\) minutes. b) A second pump moves twice as much water per minute. Write its function. c) How long would each pump take to fill a \(2700\)-gallon tank? How much time does the faster pump save?

Hints

- First find the gallons pumped per minute. - How does doubling the rate change the equation? - Set the volume equal to the tank capacity and solve for time.

Solution

1. The first pump's rate is \(450\div 15=30\) gallons per minute, so \(V_1(t)=30t\). 2. The second pump's rate is \(2\cdot 30=60\) gallons per minute, so \(V_2(t)=60t\). 3. The first pump takes \(2700\div 30=90\) minutes. 4. The second pump takes \(2700\div 60=45\) minutes. 5. The time saved is \(90-45=45\) minutes.

Answer

a) \(V_1(t)=30t\). b) \(V_2(t)=60t\). c) The first pump takes \(90\) minutes, the second takes \(45\) minutes, and the second saves \(45\) minutes.
Build linear models from descriptions 26
A vehicle uses \(18\) gallons of gasoline to travel \(600\) miles. Assume gasoline used is proportional to distance. a) Write a function of the form \(f(x)=kx\) for the gasoline use \(f(x)\), in gallons, after traveling \(x\) miles. What does \(k\) mean in context? b) While towing a trailer, the vehicle uses an additional \(1.2\) gallons for every \(100\) miles. How many gallons will it use under these conditions on a \(350\)-mile trip?

Hints

- Find the gasoline used for one mile. - Interpret the constant in \(y=kx\) using units. - Convert the added use from per \(100\) miles to per mile before combining rates.

Solution

1. The original rate is \(k=18\div 600=0.03\) gallon per mile. 2. The function is \(f(x)=0.03x\). The constant \(k\) is the gallons used per mile. 3. The added rate is \(1.2\div 100=0.012\) gallon per mile. 4. The towing rate is \(0.03+0.012=0.042\) gallon per mile. 5. For \(350\) miles, the vehicle uses \(350\cdot 0.042=14.7\) gallons.

Answer

a) \(f(x)=0.03x\); \(k=0.03\) is the gasoline used per mile. b) \(14.7\) gallons.
Build linear models from descriptions 27
A class is ordering printed T-shirts. InkWorks charges a one-time setup fee of \(\$24\) plus \(\$7\) per shirt. TeeDirect charges \(\$10\) per shirt with no additional fee. Let \(x\) be the number of shirts ordered. a) Construct cost functions \(I(x)\) and \(T(x)\) for InkWorks and TeeDirect. b) Use the functions to find the cost of \(5\) shirts and \(10\) shirts from each company. c) Starting with how many whole shirts is InkWorks less expensive than TeeDirect?

Hints

- Separate each company's fixed cost from its per-shirt cost when constructing the functions. - Check that the output at \(x=0\) matches each company's setup-fee situation. - After the functions are written, translate “InkWorks is less expensive” into a comparison of their outputs.

Solution

1. InkWorks has a fixed fee of \(24\) dollars and a per-shirt rate of \(7\) dollars, so \(I(x)=24+7x\). TeeDirect has no fixed fee and charges \(10\) dollars per shirt, so \(T(x)=10x\). 2. \(I(5)=24+7\cdot5=59\) and \(T(5)=10\cdot5=50\). 3. \(I(10)=24+7\cdot10=94\) and \(T(10)=100\). 4. InkWorks is less expensive when \(24+7x<10x\). This gives \(24<3x\), so \(x>8\). 5. The least whole-number order greater than \(8\) is \(9\) shirts.

Answer

a) \(I(x)=24+7x\), \(T(x)=10x\) b) For \(5\) shirts: InkWorks \(\$59\), TeeDirect \(\$50\). For \(10\) shirts: InkWorks \(\$94\), TeeDirect \(\$100\). c) \(9\) shirts
Build linear models from descriptions 28
Lucas and Sarah are saving for bicycles. Lucas already has \(\$60\) and adds \(\$4\) each week. Sarah has \(\$10\) and adds \(\$9\) each week. a) Write a linear model for each person's savings after \(w\) weeks. b) Use the models to determine after how many weeks they will have the same amount and state that amount.

Hints

- In each model, identify the starting amount and the amount added each week. - The weekly amount is the coefficient of \(w\), and the starting amount is the constant term. - To find when the balances match, compare the two model outputs for the same value of \(w\).

Solution

1. Lucas starts with \(60\) dollars and adds \(4\) dollars per week, so \(L(w)=60+4w\). 2. Sarah starts with \(10\) dollars and adds \(9\) dollars per week, so \(S(w)=10+9w\). 3. Set the models equal to find when the balances match: \(60+4w=10+9w\). 4. Subtract \(10\) and \(4w\): \(50=5w\), so \(w=10\). 5. Evaluate either model: \(L(10)=60+4\cdot10=100\). Each person has \(\$100\).

Answer

a) Lucas: \(L(w)=60+4w\); Sarah: \(S(w)=10+9w\) b) After \(10\) weeks, each has \(\$100\).
Build linear models from descriptions 29
A large tank contains \(450\,\text{L}\) of water and loses \(15\,\frac{\text{L}}{\text{min}}\) through a leak. At the same time, a second tank, initially empty, is filled at \(25\,\frac{\text{L}}{\text{min}}\). Let \(t\) be the number of minutes after both processes begin. a) Construct volume functions \(V_1(t)\) and \(V_2(t)\) for the two tanks. b) Use the functions to find when the tanks contain the same amount of water and how much each contains then. c) Explain from the two models why the second tank contains more water after that time, while both models remain physically meaningful.

Hints

- A linear volume model combines the starting amount with a signed per-minute change. - Write both functions before setting any amounts equal. - Compare the signs of the two rates to reason about what happens after the intersection.

Solution

1. The first tank starts at \(450\) liters and decreases by \(15\) liters per minute, so \(V_1(t)=450-15t\). 2. The second starts at \(0\) and increases by \(25\) liters per minute, so \(V_2(t)=25t\). 3. Set the functions equal: \(450-15t=25t\). Then \(450=40t\), so \(t=11.25\). 4. \(V_2(11.25)=25\cdot11.25=281.25\). Each tank contains \(281.25\,\text{L}\). 5. After the intersection, \(V_1\) continues to decrease while \(V_2\) continues to increase, so \(V_2(t)>V_1(t)\) after that time within the physical domains.

Answer

a) \(V_1(t)=450-15t\), \(V_2(t)=25t\) b) After \(11.25\) minutes, each tank contains \(281.25\,\text{L}\). c) \(V_1\) decreases while \(V_2\) increases, so after their intersection the second model stays above the first while both apply.
Build linear models from descriptions 30
Two conveyor belts add boxes to two stacks. Stack A starts with \(15\) boxes and receives \(5\) boxes per minute. Stack B starts with \(45\) boxes and receives \(2\) boxes per minute. Let \(t\) be the number of minutes after the belts start. a) Construct functions \(A(t)\) and \(B(t)\) for the number of boxes in each stack. b) Which stack grows faster? Explain using the functions' rates of change. c) Use the functions to find when the stacks contain the same number of boxes and how many boxes each contains then.

Hints

- Use each starting count as the initial value of its model. - The coefficient of time should represent the number of boxes added each minute. - Equal stack sizes mean the two function outputs are equal for the same input time.

Solution

1. Stack A has model \(A(t)=15+5t\), and Stack B has model \(B(t)=45+2t\). 2. Stack A grows faster because its rate of change is \(5\,\frac{\text{boxes}}{\text{min}}\), compared with \(2\,\frac{\text{boxes}}{\text{min}}\) for Stack B. 3. Set \(A(t)=B(t)\): \(15+5t=45+2t\). 4. Then \(3t=30\), so \(t=10\). 5. \(A(10)=15+5\cdot10=65\), so each stack contains \(65\) boxes.

Answer

a) \(A(t)=15+5t\), \(B(t)=45+2t\) b) Stack A c) After \(10\) minutes, each stack contains \(65\) boxes.
Build linear models from descriptions 31
Two candles are lit at the same time. Candle A is \(20\,\text{cm}\) tall and burns down \(1.5\,\frac{\text{cm}}{\text{h}}\). Candle B is \(15\,\text{cm}\) tall and burns down \(0.5\,\frac{\text{cm}}{\text{h}}\). Let \(t\) be the number of hours after the candles are lit. a) Construct height functions \(A(t)\) and \(B(t)\). b) Use the functions to find when the candles have the same height and what that height is. c) Which candle is taller after \(7\) hours? Justify using the functions.

Hints

- Burning gives each height model a negative rate of change. - Build both functions from initial height plus signed rate times time. - Use equal outputs for the same input to model equal heights.

Solution

1. Candle A starts at \(20\) centimeters and decreases by \(1.5\) centimeters per hour, so \(A(t)=20-1.5t\). 2. Candle B starts at \(15\) centimeters and decreases by \(0.5\) centimeter per hour, so \(B(t)=15-0.5t\). 3. Set the functions equal: \(20-1.5t=15-0.5t\). Then \(t=5\). 4. \(A(5)=20-1.5\cdot5=12.5\), so both candles are \(12.5\,\text{cm}\) tall. 5. \(A(7)=9.5\) and \(B(7)=11.5\), so Candle B is taller after \(7\) hours.

Answer

a) \(A(t)=20-1.5t\), \(B(t)=15-0.5t\) b) After \(5\) hours, both are \(12.5\,\text{cm}\) tall. c) Candle B; \(B(7)=11.5\,\text{cm}\) and \(A(7)=9.5\,\text{cm}\).
Build linear models from descriptions 32
Julian compares two monthly phone plans. The Basic plan charges \(\$12.00\) plus \(\$0.08\) per minute. The Flat plan costs \(\$22.00\), no matter how many minutes are used. Let \(m\) be the number of minutes in a month. a) Write the monthly cost functions \(B(m)\) and \(F(m)\). b) Write and solve an inequality to find the smallest whole number of minutes for which the Flat plan costs less than the Basic plan.

Hints

- A monthly cost may include a fixed amount and an amount per minute. - To make Flat cheaper, its cost must be strictly smaller than Basic's. - Decide how a strict inequality affects the smallest allowed whole number.

Solution

1. The Basic plan costs \(B(m)=12+0.08m\); the Flat plan costs \(F(m)=22\). 2. Flat costs less when \(22<12+0.08m\). 3. Subtracting \(12\) gives \(10<0.08m\), so \(m>125\). 4. The smallest whole-number choice is \(126\) minutes.

Answer

a) \(B(m)=12+0.08m\), \(F(m)=22\) b) \(22<12+0.08m\) gives \(m>125\); the least whole number is \(126\) minutes.
Build linear models from descriptions 33
A gym offers two monthly plans. Plan A costs a flat \(\$45.00\). Plan B costs \(\$10.00\) plus \(\$3.50\) per visit. Let \(v\) be the number of visits in a month. Write a cost function for each plan. Then evaluate this claim: “I work out every other day in a \(30\)-day month, so Plan B is definitely less expensive.” Justify your conclusion and state the difference in cost.

Hints

- Identify the fixed part and the per-visit part of each plan. - Convert “every other day in a 30-day month” into a number of visits. - Evaluate both functions at the same visit count before comparing the costs.

Solution

1. Plan A has the constant cost function \(A(v)=45\). Plan B has the linear cost function \(B(v)=10+3.50v\). 2. Working out every other day in a \(30\)-day month gives \(30\div2=15\) visits. 3. Evaluate the functions: \(A(15)=45\) and \(B(15)=10+3.50\cdot15=62.50\). 4. Since \(62.50>45\), Plan B is not less expensive. It costs \(62.50-45=17.50\) dollars more.

Answer

\(A(v)=45\) and \(B(v)=10+3.50v\). For \(15\) visits, Plan A costs \(\$45.00\) and Plan B costs \(\$62.50\), so the claim is false. Plan B costs \(\$17.50\) more.
Build linear models from descriptions 34
Luke compares two gym memberships. Power: a \(\$45.00\) sign-up fee plus \(\$25.00\) per month Easy: no sign-up fee and \(\$35.00\) per month Let \(m\) be the number of months of membership. a) Construct total-cost functions \(P(m)\) and \(E(m)\). b) Use the functions to find the total cost of each membership after \(4\) months. c) Starting with which full month is Power less expensive overall than Easy? Justify with a comparison of the two functions.

Hints

- Include the sign-up fee as the initial value of only the Power model. - Check what each model would output at \(m=0\) to verify the fixed-cost structure. - For part c), compare the two function outputs, not just the monthly rates.

Solution

1. Power has model \(P(m)=45+25m\). Easy has model \(E(m)=35m\). 2. \(P(4)=45+25\cdot4=145\), and \(E(4)=35\cdot4=140\). 3. Power is less expensive when \(45+25m<35m\). 4. Then \(45<10m\), so \(m>4.5\). 5. The first full month satisfying the inequality is month \(5\).

Answer

a) \(P(m)=45+25m\), \(E(m)=35m\) b) Power: \(\$145.00\); Easy: \(\$140.00\) c) Month \(5\)
Build linear models from descriptions 35
A family compares two annual cell phone plans. Smart: a \(\$114.00\) annual fee plus \(\$0.05\) per minute Basic: no annual fee and \(\$0.15\) per minute Let \(m\) be the number of minutes used during the year. a) Construct annual cost functions \(S(m)\) and \(B(m)\). b) Compare the annual costs for \(960\) minutes. c) Starting at how many whole minutes per year is Smart less expensive than Basic?

Hints

- Use one consistent annual input variable for both models. - The annual fee is the initial value of the Smart model; Basic has initial value \(0\). - For the threshold, compare the two model outputs with a strict inequality.

Solution

1. Smart has model \(S(m)=114+0.05m\). Basic has model \(B(m)=0.15m\). 2. \(S(960)=114+0.05\cdot960=162\), while \(B(960)=0.15\cdot960=144\). Basic is less expensive at \(960\) minutes. 3. Smart is less expensive when \(114+0.05m<0.15m\). 4. Subtract \(0.05m\): \(114<0.10m\), so \(m>1140\). 5. The least whole-number usage is \(1141\) minutes.

Answer

a) \(S(m)=114+0.05m\), \(B(m)=0.15m\) b) Smart: \(\$162.00\); Basic: \(\$144.00\), so Basic is less expensive. c) \(1141\) minutes
Build linear models from descriptions 36
Two car-sharing companies offer these monthly plans. Company A: a \(\$15.00\) monthly fee plus \(\$0.20\) per mile Company B: no monthly fee and \(\$0.45\) per mile Let \(m\) be the number of miles driven in a month. a) Construct monthly cost functions \(A(m)\) and \(B(m)\). b) Use the functions to find the monthly cost for each company at \(40\) miles. c) For what monthly mileage is Company A less expensive than Company B?

Hints

- Translate each fixed fee and per-mile charge into the corresponding parts of a linear function. - Verify the models by considering what each company would charge at \(0\) miles. - Compare the two outputs with a strict inequality for part c).

Solution

1. Company A has model \(A(m)=15+0.20m\). Company B has model \(B(m)=0.45m\). 2. \(A(40)=15+0.20\cdot40=23\), while \(B(40)=0.45\cdot40=18\). 3. Company A is less expensive when \(15+0.20m<0.45m\). 4. Then \(15<0.25m\), so \(m>60\).

Answer

a) \(A(m)=15+0.20m\), \(B(m)=0.45m\) b) Company A: \(\$23.00\); Company B: \(\$18.00\) c) More than \(60\) miles per month
Build linear models from descriptions 37
A portable solar panel charges a battery. Under constant conditions, the stored energy increases by \(4\) watt-hours, or \(\text{Wh}\), every \(15\) minutes. Assume the battery starts empty and can store at least \(50\,\text{Wh}\). a) How much energy is stored after \(2\) hours? b) After how many hours, minutes, and seconds has the panel delivered exactly \(50\,\text{Wh}\)? c) Write a function in the form \(y=kx\), where \(x\) is time in minutes and \(y\) is energy in watt-hours. Explain what \(k\) means in context.

Hints

- Convert two hours to minutes. - Find the energy added in one minute. - Use the constant rate to work backward from \(50\,\text{Wh}\). - In \(y=kx\), \(k\) represents a rate.

Solution

1. The rate is \(k=\frac{4}{15}\) watt-hour per minute. 2. Two hours is \(120\) minutes, so \(y=120\cdot \frac{4}{15}=32\,\text{Wh}\). 3. For \(50\,\text{Wh}\), \(x=50\cdot \frac{15}{4}=187.5\) minutes. 4. This is \(3\) hours, \(7\) minutes, and \(30\) seconds. 5. The function is \(y=\frac{4}{15}x\). The constant \(k\) is the energy added per minute.

Answer

a) \(32\,\text{Wh}\). b) \(3\) hours, \(7\) minutes, \(30\) seconds. c) \(y=\frac{4}{15}x\); \(k\) is the watt-hours added per minute.
Build linear models from descriptions 38
A vehicle uses gasoline at a constant rate of \(4\) gallons per \(100\) miles. a) Write a function for the volume \(V\), in gallons, used after traveling \(s\) miles. b) Find the gasoline used on a \(450\)-mile trip. c) The tank has \(12\) gallons remaining. How far can the vehicle theoretically travel?

Hints

- Convert the rate per \(100\) miles to a rate per mile. - Identify which quantity depends on the other. - Use the equation forward for part b and solve it backward for part c. - A small value table may help you check the model.

Solution

1. The rate per mile is \(4\div 100=0.04\) gallon per mile. 2. The function is \(V(s)=0.04s\). 3. For \(450\) miles, \(V(450)=0.04\cdot 450=18\) gallons. 4. With \(12\) gallons, solve \(12=0.04s\): \(s=12\div 0.04=300\) miles.

Answer

a) \(V(s)=0.04s\). b) \(18\) gallons. c) \(300\) miles.
Build linear models from descriptions 39
On a sunny day, Anja observes the shadow of a vertical pole in the schoolyard. She marks the shadow's position at \(10{:}00\) a.m. At \(12{:}30\) p.m., the shadow has swept through an angle of \(37.5^\circ\). Assume the shadow angle changes at a constant rate from \(10{:}00\) a.m. through \(4{:}15\) p.m. a) Find the change in angle per hour. b) Write a function for the angle \(\alpha\), in degrees, after \(t\) hours from the first measurement. c) What total angle has the shadow swept through by \(4{:}15\) p.m.?

Hints

- Find the elapsed time between the first two measurements. - Convert minutes to a fraction or decimal part of an hour. - Use the constant change in angle per hour. - What form does a proportional function have?

Solution

1. The elapsed time from \(10{:}00\) a.m. to \(12{:}30\) p.m. is \(2.5\) hours. 2. The rate is \(37.5\div 2.5=15^\circ\) per hour. 3. The function is \(\alpha(t)=15t\). 4. From \(10{:}00\) a.m. to \(4{:}15\) p.m. is \(6.25\) hours. 5. The angle is \(15\cdot 6.25=93.75^\circ\).

Answer

a) \(15^\circ\) per hour. b) \(\alpha(t)=15t\). c) \(93.75^\circ\).
Build linear models from descriptions 40
An electric-vehicle charging plan has a monthly base fee of \(\$18.50\) plus \(\$0.80\) for each kilowatt-hour (\(\text{kWh}\)) used. Let \(x\) be the number of kilowatt-hours used in a month. a) Construct a monthly cost function \(C(x)\), and explain what its initial value and rate of change represent. b) A customer says a month with \(40\,\text{kWh}\) of charging cost exactly \(\$50.50\). Does the function support the claim? c) How many kilowatt-hours were used if the monthly bill was exactly \(\$74.50\)?

Hints

- Build the function by separating the fixed monthly amount from the usage-dependent amount. - Check what the model gives at input \(0\) and for one additional kilowatt-hour. - Use the completed function for both the forward and reverse questions.

Solution

1. The base fee is the initial value and the per-kilowatt-hour charge is the rate, so \(C(x)=18.50+0.80x\). 2. The initial value \(18.50\) is the monthly cost at zero usage. The rate \(0.80\) means the bill increases by \(\$0.80\) for each additional kilowatt-hour. 3. \(C(40)=18.50+0.80\cdot40=50.50\), so the customer's claim matches the model. 4. Solve \(74.50=18.50+0.80x\). Then \(56=0.80x\), so \(x=70\).

Answer

a) \(C(x)=18.50+0.80x\); the initial value is the \(\$18.50\) base fee, and the rate is \(\$0.80\) per \(\text{kWh}\). b) Yes. c) \(70\,\text{kWh}\)
Build linear models from descriptions 41
A small backyard pool already contains \(30\) gallons of water. A garden hose adds water at a constant rate of \(4\) gallons per minute. a) Write a function \(V(t)\) for the volume of water, in gallons, after \(t\) minutes. b) How much water is in the pool after \(20\) minutes? c) How long will it take for the pool to contain \(150\) gallons?

Hints

- Which value represents the amount of water at time \(0\)? - How much does the volume change each minute? - Decide whether you need to substitute a time into the function or solve an equation for time.

Solution

1. The initial volume is \(30\) gallons and the rate of change is \(4\) gallons per minute, so \(V(t) = 4t + 30\). 2. Evaluate at \(t = 20\): \(V(20) = 4 \cdot 20 + 30 = 110\). The pool contains \(110\) gallons. 3. Set the volume equal to \(150\): \(150 = 4t + 30\). Then \(120 = 4t\), so \(t = 30\) minutes.

Answer

a) \(V(t) = 4t + 30\) b) \(110\) gallons c) \(30\) minutes
Build linear models from descriptions 42
Two candles are lit at the same time. Candle A starts at \(24\,\text{cm}\) and burns down at \(2\,\text{cm/h}\). Candle B starts at \(18\,\text{cm}\) and burns down at \(1.2\,\text{cm/h}\). a) Write a function for the height \(h\), in centimeters, of each candle after \(t\) hours. b) Which candle is taller after \(5\) hours? Show your calculation. c) After how many hours will the two candles have the same height?

Hints

- How does a constant burn rate affect the sign of the rate in the equation? - What must be true about the two function values when the candles have the same height? - Write the two height expressions before comparing them.

Solution

1. Candle A is modeled by \(h_A(t) = 24 - 2t\). Candle B is modeled by \(h_B(t) = 18 - 1.2t\). 2. After \(5\) hours, \(h_A(5) = 24 - 2 \cdot 5 = 14\,\text{cm}\) and \(h_B(5) = 18 - 1.2 \cdot 5 = 12\,\text{cm}\). Candle A is taller. 3. Set the heights equal: \(24 - 2t = 18 - 1.2t\). Then \(6 = 0.8t\), so \(t = 7.5\).

Answer

a) \(h_A(t) = 24 - 2t\) and \(h_B(t) = 18 - 1.2t\) b) Candle A; after \(5\) hours it is \(14\,\text{cm}\) tall and Candle B is \(12\,\text{cm}\) tall. c) \(7.5\) hours
Build linear models from descriptions 43
A delivery service charges a fixed fee plus a constant amount per mile. A \(10\)-mile delivery costs \(\$17.00\), and a \(25\)-mile delivery costs \(\$32.00\). a) Write a function \(C(d)\) for the total cost, in dollars, of a delivery that travels \(d\) miles. b) What is the fixed fee? c) Find the cost of a \(40\)-mile delivery.

Hints

- Treat distance and total cost as ordered pairs. - How much does the cost change when the distance changes? - After finding the per-mile charge, subtract the distance-based part from a total cost to find the fixed fee.

Solution

1. Use the two data points \((10, 17)\) and \((25, 32)\). The rate per mile is \(m = \frac{32 - 17}{25 - 10} = 1\). 2. Substitute \((10, 17)\) into \(C(d) = md + b\): \(17 = 10 + b\), so \(b = 7\). 3. The cost function is \(C(d) = d + 7\), so the fixed fee is \(\$7.00\). 4. Evaluate at \(d = 40\): \(C(40) = 40 + 7 = 47\).

Answer

a) \(C(d) = d + 7\) b) \(\$7.00\) c) \(\$47.00\)
Build linear models from descriptions 44
A candle burns at a constant rate. After \(3\) hours, it is \(7\) inches tall. Its height decreases by \(0.75\) inch each hour. a) Write a function \(h(t)\) for the candle's height, in inches, after \(t\) hours. b) How tall was the candle when it was first lit?

Hints

- Should the rate of change be positive or negative when the candle gets shorter? - What time represents the moment the candle is first lit? - Which information gives you a point on the graph of the height function?

Solution

1. Because the candle gets shorter, the rate of change is \(m = -0.75\) inch per hour. 2. The information after \(3\) hours gives the point \((3, 7)\). 3. Substitute into \(h(t) = mt + b\): \(7 = -0.75 \cdot 3 + b\), so \(b = 9.25\). 4. Therefore, \(h(t) = -0.75t + 9.25\). The initial height is \(h(0) = 9.25\) inches.

Answer

a) \(h(t) = -0.75t + 9.25\) b) \(9.25\) inches
Build linear models from descriptions 45
A candle burns at a constant rate. Two hours after it is lit, it is \(7\) inches tall. After \(5\) hours, it is \(4\) inches tall. 1. Write a function \(f(x)\) for the candle's height, in inches, after \(x\) hours. 2. How tall was the candle when it was lit? 3. After how many hours will the candle be completely burned down? 4. What domain for \(x\) makes sense in this context?

Hints

- Represent the two time-and-height measurements as points. - Find the slope from those two points. - What does the function value at time \(0\) represent? - What is the candle's height when it is completely burned down?

Solution

1. Use the points \((2, 7)\) and \((5, 4)\): \(m = \frac{4 - 7}{5 - 2} = -1\) inch per hour. 2. Substitute \((2, 7)\) into \(f(x) = -x + b\): \(7 = -2 + b\), so \(b = 9\). Thus \(f(x) = -x + 9\). 3. The initial height is \(f(0) = 9\) inches. 4. The candle is burned down when \(f(x) = 0\): \(0 = -x + 9\), so \(x = 9\). 5. In context, time begins at \(0\) and ends when the height reaches \(0\), so \(0 \le x \le 9\).

Answer

1. \(f(x) = -x + 9\) 2. \(9\) inches 3. \(9\) hours 4. \([0, 9]\)
Build linear models from descriptions 46
Water flows into a large tank at a constant rate. After \(10\) minutes, the tank contains \(110\) gallons. After \(25\) minutes, it contains \(200\) gallons. 1. Write a function \(V(x)\) for the volume of water, in gallons, after \(x\) minutes. 2. How much water was in the tank when the timing began? 3. The tank holds \(290\) gallons. After how many minutes will it be full?

Hints

- What does the slope represent in this situation? - Use the two time-volume pairs to find the rate per minute. - Once you know the rate, how can you find the amount at time \(0\)? - To find when the tank is full, set the function equal to its capacity.

Solution

1. From \((10, 110)\) and \((25, 200)\), the rate is \(m = \frac{200 - 110}{25 - 10} = 6\) gallons per minute. 2. Substitute \((10, 110)\) into \(V(x) = 6x + b\): \(110 = 6 \cdot 10 + b\), so \(b = 50\). Thus \(V(x) = 6x + 50\). 3. The initial volume is \(V(0) = 50\) gallons. 4. Set the volume equal to capacity: \(290 = 6x + 50\). Then \(240 = 6x\), so \(x = 40\).

Answer

1. \(V(x) = 6x + 50\) 2. \(50\) gallons 3. \(40\) minutes
Build linear models from descriptions 47
A rain barrel contains \(80\) gallons of water at time \(t = 0\). Water drains from an open spigot at a constant rate. After \(40\) minutes, \(60\) gallons remain. a) Write a linear function \(V(t)\) for the amount of water, in gallons, after \(t\) minutes. b) After how many minutes will the barrel be empty?

Hints

- What value represents the amount of water at time \(0\)? - How much water is lost per minute? - What value of \(V(t)\) represents an empty barrel?

Solution

1. The initial value is \(V(0) = 80\). 2. The rate of change is \(m = \frac{60 - 80}{40 - 0} = -0.5\) gallon per minute. Thus \(V(t) = -0.5t + 80\). 3. The barrel is empty when \(V(t) = 0\): \(0 = -0.5t + 80\), so \(t = 160\).

Answer

a) \(V(t) = -0.5t + 80\) b) \(160\) minutes
Build linear models from descriptions 48
Two water tanks are observed at the same time. Tank A starts with \(500\,\text{L}\) and is drained at a constant rate of \(12\,\text{L/min}\). Tank B starts with \(200\,\text{L}\) and is filled at a constant rate of \(8\,\text{L/min}\). Write a function for the volume of each tank after \(t\) minutes. Then find when the tanks contain the same amount of water and what that common volume is.

Hints

- For each tank, decide whether the volume is increasing or decreasing. - What equation represents the moment when the two volumes are equal? - After finding the time, substitute it into either model to find the common volume.

Solution

1. Tank A is modeled by \(V_A(t) = 500 - 12t\). Tank B is modeled by \(V_B(t) = 200 + 8t\). 2. Set the volumes equal: \(500 - 12t = 200 + 8t\). 3. Then \(300 = 20t\), so \(t = 15\). 4. Substitute \(t = 15\): \(V_A(15) = 500 - 12 \cdot 15 = 320\). Therefore, after \(15\) minutes both tanks contain \(320\,\text{L}\).

Answer

\(V_A(t) = 500 - 12t\) and \(V_B(t) = 200 + 8t\). After \(15\) minutes, both tanks contain \(320\,\text{L}\).
Build linear models from descriptions 49
A candle is \(10\) inches tall when it is lit. After \(3\) hours, it is \(8.5\) inches tall. a) Write a function \(h(t)\) for the candle's height, in inches, after \(t\) hours. b) Explain what the slope \(m\) and y-intercept \(b\) mean in this situation. c) Find the candle's height after \(5\) hours.

Hints

- Is the candle's height increasing or decreasing as it burns? - Use the two height measurements to find the change per hour. - Think about what the function value at time \(0\) represents.

Solution

1. The initial height is the y-intercept, so \(b = 10\). 2. The slope is \(m = \frac{8.5 - 10}{3 - 0} = -0.5\) inch per hour. Thus \(h(t) = -0.5t + 10\). 3. The y-intercept \(10\) is the candle's initial height. The slope \(-0.5\) means its height decreases by \(0.5\) inch each hour. 4. Evaluate at \(t = 5\): \(h(5) = -0.5 \cdot 5 + 10 = 7.5\) inches.

Answer

a) \(h(t) = -0.5t + 10\) b) \(b = 10\) is the initial height, and \(m = -0.5\) means the candle loses \(0.5\) inch of height per hour. c) \(7.5\) inches
Build linear models from descriptions 50
A taxi company uses a linear fare. A \(5\)-mile ride costs \(\$14.50\), and a \(12\)-mile ride costs \(\$28.50\). a) Write a function \(C(x)\) for the fare, in dollars, for a ride of \(x\) miles. b) Explain what the slope \(m\) and y-intercept \(b\) mean in this situation. c) Find the fare for a \(20\)-mile ride.

Hints

- How much does the fare change when the distance increases? - What cost would remain when the distance is \(0\) miles? - Substitute the requested distance into your function.

Solution

1. From \((5, 14.50)\) and \((12, 28.50)\), the slope is \(m = \frac{28.50 - 14.50}{12 - 5} = 2\) dollars per mile. 2. Substitute \((5, 14.50)\) into \(C(x) = 2x + b\): \(14.50 = 2 \cdot 5 + b\), so \(b = 4.50\). 3. Therefore, \(C(x) = 2x + 4.50\). 4. The slope means the fare increases by \(\$2.00\) per mile, and the y-intercept is the \(\$4.50\) starting fare. 5. Evaluate at \(x = 20\): \(C(20) = 2 \cdot 20 + 4.50 = 44.50\).

Answer

a) \(C(x) = 2x + 4.50\) b) The slope is a charge of \(\$2.00\) per mile, and the y-intercept is the \(\$4.50\) starting fare. c) \(\$44.50\)
Build linear models from descriptions 51
Air temperature in the atmosphere can be modeled as decreasing linearly with altitude. At \(2000\) feet above sea level, a meteorologist measures \(50^\circ\text{F}\). At \(10{,}000\) feet, the temperature is \(18^\circ\text{F}\). a) Write a linear function \(T(h)\) for the temperature, in degrees Fahrenheit, at altitude \(h\) feet. b) What temperature does the model predict at sea level? c) At what altitude does the model predict the freezing temperature of \(32^\circ\text{F}\)?

Hints

- The temperature decreases as altitude increases, so what sign should the slope have? - What altitude represents sea level? - For part c, the temperature is known. Substitute it into the model and solve for altitude.

Solution

1. Use \((2000, 50)\) and \((10{,}000, 18)\): \(m = \frac{18 - 50}{10{,}000 - 2000} = -0.004\) degree Fahrenheit per foot. 2. Substitute \((2000, 50)\) into \(T(h) = -0.004h + b\): \(50 = -0.004 \cdot 2000 + b\), so \(b = 58\). Thus \(T(h) = -0.004h + 58\). 3. At sea level, \(T(0) = 58^\circ\text{F}\). 4. Set the temperature equal to freezing: \(32 = -0.004h + 58\). Then \(-26 = -0.004h\), so \(h = 6500\) feet.

Answer

a) \(T(h) = -0.004h + 58\) b) \(58^\circ\text{F}\) c) \(6500\) feet
Build linear models from descriptions 52
Two monthly phone plans are available. The Basic plan charges a monthly fee of \(\$8.00\) plus \(\$0.15\) per calling minute. The Smart plan charges a monthly fee of \(\$14.00\) plus \(\$0.05\) per calling minute. a) Write a function for the monthly cost \(y\) of each plan in terms of the number of calling minutes \(x\). b) Find the number of calling minutes for which the two plans have the same monthly cost. c) Explain which plan is less expensive for lighter use and which is less expensive for heavier use.

Hints

- What part of each cost is fixed each month? - How much does each total increase for one additional calling minute? - What equation represents the point where the two plans cost the same? - After finding the break-even point, compare the per-minute rates.

Solution

1. The Basic plan is modeled by \(B(x) = 0.15x + 8.00\). The Smart plan is modeled by \(S(x) = 0.05x + 14.00\). 2. Set the costs equal: \(0.15x + 8.00 = 0.05x + 14.00\). 3. Then \(0.10x = 6.00\), so \(x = 60\). 4. At \(60\) minutes, both plans cost \(\$17.00\). For fewer than \(60\) minutes, Basic is less expensive; for more than \(60\) minutes, Smart is less expensive.

Answer

a) \(B(x) = 0.15x + 8.00\) and \(S(x) = 0.05x + 14.00\) b) \(60\) minutes c) Basic is less expensive below \(60\) minutes; Smart is less expensive above \(60\) minutes.
Build linear models from descriptions 53
Two hiking groups are on the same trail to a mountain summit. Group A starts at a parking lot and hikes at a constant speed of \(3\,\text{mi/h}\). Group B starts at the same time from a lodge \(2\) miles farther along the trail and hikes at \(2\,\text{mi/h}\). a) Write a function for each group's distance \(s\), in miles, from the parking lot after \(t\) hours. b) Find when and where Group A catches Group B. c) The summit is \(7\) miles from the parking lot. Determine mathematically whether Group A catches Group B before either group reaches the summit.

Hints

- How far from the parking lot is Group B at \(t = 0\)? - What equation represents the moment when the groups are at the same location? - Compare the meeting distance with the total distance to the summit.

Solution

1. Group A is modeled by \(s_A(t) = 3t\). Group B starts \(2\) miles ahead, so \(s_B(t) = 2 + 2t\). 2. Set the distances equal: \(3t = 2 + 2t\). 3. Solving gives \(t = 2\), so Group A catches Group B after \(2\) hours. 4. The meeting point is \(s_A(2) = 3 \cdot 2 = 6\) miles from the parking lot. 5. Since \(6 < 7\), the catch-up happens before the summit, exactly \(1\) mile before it.

Answer

a) \(s_A(t) = 3t\) and \(s_B(t) = 2 + 2t\) b) After \(2\) hours, at \(6\) miles from the parking lot. c) Yes. The catch-up point is \(1\) mile before the \(7\)-mile summit.
Build linear models from descriptions 54
A tablet battery is charging. At the start, the battery level is \(10\%\). After \(24\) minutes, it has increased at a constant rate to \(40\%\). a) Write a linear function \(y\) for the battery level, in percent, after \(x\) minutes. b) Assuming the charging rate stays constant, after how many minutes will the battery reach \(100\%\)? c) Describe the meaningful graph of the model by stating its two endpoints and whether it is increasing, decreasing, or constant. You may sketch it on paper to help.

Hints

- Use the battery level at time \(0\) as the initial value. - Find the constant rate from the change in battery percentage over \(24\) minutes. - A full battery corresponds to an output of \(100\%\). - For part c, use the starting point and the full-charge point to describe the meaningful graph segment.

Solution

1. At \(x=0\), the battery level is \(10\%\), so the y-intercept is \(10\). 2. Over \(24\) minutes, the battery level increases by \(30\) percentage points, so \(m=\frac{30}{24}=1.25\) percentage points per minute. Thus \(y=1.25x+10\). 3. For a full battery, set \(y=100\): \(100=1.25x+10\). Solving gives \(x=72\). 4. In the meaningful model interval \(0\le x\le72\), the graph is an increasing line segment from \((0, 10)\) to \((72, 100)\). The point \((24, 40)\) lies on that segment.

Answer

a) \(y=1.25x+10\) b) \(72\) minutes c) An increasing line segment from \((0, 10)\) to \((72, 100)\), for \(0\le x\le72\).
Build linear models from descriptions 55
A rectangular backyard pool is filled at a constant rate. At \(1{:}00\) p.m., the water is \(16\,\text{in.}\) deep. At \(3{:}00\) p.m., it is \(40\,\text{in.}\) deep. Let \(t\) be the number of hours after \(1{:}00\) p.m., and let \(d(t)\) be the water depth in inches. a) Construct a linear depth function \(d(t)\) from the two measurements. State its rate of change. b) Extending the same model backward, at what clock time does \(d(t)=0\)? c) The pool is full at a depth of \(64\,\text{in.}\). At what clock time does the model reach that depth?

Hints

- Use the two depth measurements to determine the signed change per hour. - Because \(t=0\) is fixed at \(1{:}00\) p.m., the measured depth at that time becomes the initial value of the function. - After constructing the function, use target depths as output values and solve for \(t\).

Solution

1. From \(1{:}00\) p.m. to \(3{:}00\) p.m., the depth increases by \(40-16=24\) inches in \(2\) hours, so the rate is \(12\,\frac{\text{in.}}{\text{h}}\). 2. Since \(t=0\) represents \(1{:}00\) p.m. and the depth is then \(16\) inches, \(d(t)=16+12t\). 3. Solve \(0=16+12t\). Then \(t=-\frac{4}{3}\) hour, which is \(1\) hour \(20\) minutes before \(1{:}00\) p.m., or \(11{:}40\) a.m. 4. Solve \(64=16+12t\). Then \(48=12t\), so \(t=4\), which is \(5{:}00\) p.m.

Answer

a) \(d(t)=16+12t\); rate of change \(12\,\frac{\text{in.}}{\text{h}}\) b) \(11{:}40\) a.m. c) \(5{:}00\) p.m.
Build linear models from descriptions 56
A vehicle travels at a constant speed during a highway test. The table shows time \(t\), in hours, and distance \(s\), in miles. <table> <tr><td>Time \(t\), in hours</td><td>\(0.5\)</td><td>\(1.5\)</td><td>\(2\)</td></tr> <tr><td>Distance \(s\), in miles</td><td>\(30\)</td><td>\(90\)</td><td>\(120\)</td></tr> </table> a) Use calculations to determine whether the table could represent a proportional relationship. b) State the physical condition that must hold throughout the trip for this proportionality to be exact. c) Write the function for the relationship.

Hints

- How are time, distance, and speed related? - What must be constant in a proportional table? - Use the table to find the coefficient in the equation.

Solution

1. Calculate \(s\div t\): \(30\div 0.5=60\), \(90\div 1.5=60\), and \(120\div 2=60\). 2. Since the ratios are equal, the relationship is proportional. 3. The vehicle's speed must remain constant at \(60\) miles per hour. 4. The function is \(s(t)=60t\).

Answer

a) Yes. Every ratio \(s\div t\) equals \(60\). b) The speed must remain constant. c) \(s(t)=60t\).
Build linear models from descriptions 57
Two electricity providers offer different monthly plans. Provider A: no monthly base charge and \(\$0.40\) per kilowatt-hour, or \(\text{kWh}\). Provider B: a \(\$10.00\) monthly base charge and \(\$0.30\) per kilowatt-hour. a) Write a function for monthly cost \(y\), in dollars, as a function of usage \(x\), in kilowatt-hours, for each provider. b) Which plan represents a proportional relationship? Justify your answer mathematically. c) How does Provider B's base charge change its graph compared with a line through the origin?

Hints

- Separate each cost into a fixed part and a usage-dependent part. - What should the cost be at zero usage in a proportional relationship? - How does a fixed starting value affect the y-intercept?

Solution

1. Provider A is \(y_A=0.40x\). Provider B is \(y_B=0.30x+10\). 2. Provider A is proportional because it has the form \(y=kx\) and \(y_A(0)=0\). 3. Provider B is not proportional because \(y_B(0)=10\), so the output-to-input ratio is not constant. 4. The base charge shifts Provider B's graph upward; its y-intercept is \(10\) instead of \(0\).

Answer

a) Provider A: \(y=0.40x\). Provider B: \(y=0.30x+10\). b) Only Provider A is proportional. c) The base charge shifts Provider B's graph upward so it crosses the y-axis at \(10\), not at the origin.
Build linear models from descriptions 58
A full rain barrel contains \(60\) gallons of water. Water drains from a spigot at a constant rate of \(2\) gallons per minute. a) Write a function \(V(t)\) for the remaining volume, in gallons, after \(t\) minutes. b) Explain what the slope and y-intercept mean in this situation. c) How much water remains after \(12\) minutes? d) After how many minutes will the barrel be empty?

Hints

- What amount of water is in the barrel at time \(0\)? - How does the amount change each minute? - What form represents a quantity decreasing at a constant rate? - What volume represents an empty barrel?

Solution

1. The initial volume is \(60\) gallons and the volume decreases by \(2\) gallons per minute, so \(V(t) = 60 - 2t\). 2. The y-intercept \(60\) is the starting volume. The slope \(-2\) means the volume decreases by \(2\) gallons each minute. 3. Evaluate at \(t = 12\): \(V(12) = 60 - 2 \cdot 12 = 36\) gallons. 4. The barrel is empty when \(V(t) = 0\): \(0 = 60 - 2t\), so \(t = 30\) minutes.

Answer

a) \(V(t) = 60 - 2t\) b) The y-intercept \(60\) is the starting volume, and the slope \(-2\) means the barrel loses \(2\) gallons of water per minute. c) \(36\) gallons d) \(30\) minutes
Build linear models from descriptions 59
Two water tanks are being adjusted at the same time. Tank 1 starts with \(2500\,\text{L}\) of water and loses \(150\,\text{L}\) per hour. Tank 2 starts with \(500\,\text{L}\) of water and gains \(250\,\text{L}\) per hour. a) Write a function for the volume \(V\), in liters, of each tank after \(t\) hours. b) Find when the two tanks contain the same amount of water and determine that common volume. c) Tank 2 has a maximum capacity of \(3000\,\text{L}\). How much water remains in Tank 1 when Tank 2 becomes full?

Hints

- Decide whether each tank's volume is increasing or decreasing. - Part b asks when the two volume functions have the same value. - For part c, first find when Tank 2 reaches its maximum capacity, then use that time in the model for Tank 1.

Solution

1. The volume functions are \(V_1(t) = 2500 - 150t\) and \(V_2(t) = 500 + 250t\). 2. Set the volumes equal: \(2500 - 150t = 500 + 250t\). 3. Then \(2000 = 400t\), so \(t = 5\) hours. 4. The common volume is \(V_2(5) = 500 + 250 \cdot 5 = 1750\,\text{L}\). 5. Tank 2 is full when \(3000 = 500 + 250t\). Thus \(2500 = 250t\), so \(t = 10\) hours. 6. At that time, \(V_1(10) = 2500 - 150 \cdot 10 = 1000\,\text{L}\).

Answer

a) \(V_1(t) = 2500 - 150t\) and \(V_2(t) = 500 + 250t\) b) After \(5\) hours, both tanks contain \(1750\,\text{L}\). c) \(1000\,\text{L}\) remains in Tank 1.
Build linear models from descriptions 60
A candle is \(9.5\) inches tall when it is lit. After \(4\) hours, it is \(5.5\) inches tall. Its height decreases at a constant rate. a) Write a function \(h(t)\) for the candle's height, in inches, after \(t\) hours. b) After how many hours and minutes will the candle be completely burned down?

Hints

- What does the candle's starting height tell you about the y-intercept? - How much height does the candle lose each hour? - What height represents a completely burned-down candle? - How do you convert a fraction of an hour to minutes?

Solution

1. The initial height is \(9.5\) inches, so the y-intercept is \(9.5\). 2. The rate of change is \(m = \frac{5.5 - 9.5}{4 - 0} = -1\) inch per hour. Thus \(h(t) = -t + 9.5\). 3. The candle is burned down when \(h(t) = 0\): \(0 = -t + 9.5\), so \(t = 9.5\) hours. 4. Since \(0.5\) hour is \(30\) minutes, the total time is \(9\) hours and \(30\) minutes.

Answer

a) \(h(t) = -t + 9.5\) b) \(9\) hours and \(30\) minutes
Build linear models from descriptions 61
A liquid cools at a constant rate. After \(2\) minutes, its temperature is \(170^\circ\text{F}\). After \(10\) minutes, its temperature is \(130^\circ\text{F}\). a) Write a linear function \(T(t)\) for the temperature, in degrees Fahrenheit, after \(t\) minutes. b) Explain what the slope and y-intercept mean in this situation.

Hints

- How do you find the slope from two time-temperature points? - Which quantity changes as time passes? - What does the function value at time \(0\) represent? - What does a negative slope mean in this context?

Solution

1. Use \((2, 170)\) and \((10, 130)\): \(m = \frac{130 - 170}{10 - 2} = -5\) degrees Fahrenheit per minute. 2. Substitute \((2, 170)\) into \(T(t) = -5t + b\): \(170 = -5 \cdot 2 + b\), so \(b = 180\). Thus \(T(t) = -5t + 180\). 3. The slope \(-5\) means the temperature decreases by \(5^\circ\text{F}\) each minute. The y-intercept \(180\) is the temperature at time \(0\).

Answer

a) \(T(t) = -5t + 180\) b) The slope means the temperature decreases by \(5^\circ\text{F}\) per minute, and the y-intercept means the initial temperature was \(180^\circ\text{F}\).
Build linear models from descriptions 62
A water container fills at a constant rate. After \(5\) minutes, it contains \(5.5\) gallons. After \(12\) minutes, it contains \(9\) gallons. a) Write a function \(V(t) = mt + b\) for the volume, in gallons, after \(t\) minutes. b) How much water was in the container at time \(0\)? c) How much water will be in the container after half an hour?

Hints

- First find how many gallons are added per minute. - How can you use a known time-volume pair to find the amount at time \(0\)? - How many minutes are in half an hour? - Keep the time units consistent when you evaluate the function.

Solution

1. The rate is \(m = \frac{9 - 5.5}{12 - 5} = 0.5\) gallon per minute. 2. Substitute \((5, 5.5)\) into \(V(t) = 0.5t + b\): \(5.5 = 0.5 \cdot 5 + b\), so \(b = 3\). 3. Thus \(V(t) = 0.5t + 3\), and the initial amount is \(3\) gallons. 4. Half an hour is \(30\) minutes. Evaluate \(V(30) = 0.5 \cdot 30 + 3 = 18\) gallons.

Answer

a) \(V(t) = 0.5t + 3\) b) \(3\) gallons c) \(18\) gallons
Build linear models from descriptions 63
A prepaid phone account starts with a balance of \(\$20.00\). Each minute of calling costs \(\$0.12\). a) Write a function \(G(t)\) for the remaining balance, in dollars, after \(t\) minutes of calls. b) How much money remains after \(45\) minutes of calls? c) After how many minutes will exactly \(\$5.00\) remain? d) Is this relationship proportional? Explain.

Hints

- Identify the fixed starting amount and the amount that changes each minute. - What must be true of the graph or equation of a proportional relationship? - For parts b and c, substitute the given value into your function and solve for the unknown.

Solution

1. The starting balance is \(\$20.00\), and the balance decreases by \(\$0.12\) per minute, so \(G(t) = 20 - 0.12t\). 2. Evaluate at \(t = 45\): \(G(45) = 20 - 0.12 \cdot 45 = 14.60\), so \(\$14.60\) remains. 3. Set the balance equal to \(\$5.00\): \(5 = 20 - 0.12t\). Then \(0.12t = 15\), so \(t = 125\) minutes. 4. A proportional relationship has the form \(y = kx\) and passes through \((0, 0)\). Here \(G(0) = 20\), so the relationship is not proportional.

Answer

a) \(G(t) = 20 - 0.12t\) b) \(\$14.60\) c) \(125\) minutes d) No. Since \(G(0) = 20\), the graph does not pass through the origin.
Build linear models from descriptions 64
Two trains travel in the same direction on parallel tracks. A local train travels at a constant speed of \(60\,\text{mi/h}\). An express train travels at \(80\,\text{mi/h}\). At the start of the observation, the local train is \(2\) miles ahead. a) Write a function for each train's distance \(s\), in miles, after \(t\) minutes. b) After how many minutes does the express train catch the local train? c) How far has the express train traveled from the start of the observation by that time?

Hints

- Convert each speed to miles per minute because time is measured in minutes. - How does the initial lead appear in the local train model? - At the catch-up time, what must be true about the two distances?

Solution

1. Convert the speeds to miles per minute: \(60 \div 60 = 1\,\text{mi/min}\) and \(80 \div 60 = \frac{4}{3}\,\text{mi/min}\). 2. The local train is modeled by \(s_L(t) = t + 2\), and the express train by \(s_E(t) = \frac{4}{3}t\). 3. Set the distances equal: \(t + 2 = \frac{4}{3}t\). 4. Then \(2 = \frac{1}{3}t\), so \(t = 6\) minutes. 5. The express train has traveled \(s_E(6) = \frac{4}{3} \cdot 6 = 8\) miles.

Answer

a) \(s_L(t) = t + 2\) and \(s_E(t) = \frac{4}{3}t\), with \(t\) in minutes and \(s\) in miles. b) \(6\) minutes c) \(8\) miles
Build linear models from descriptions 65
A school is comparing two online photo-printing offers. PhotoNow charges \(\$0.12\) per photo plus a flat \(\$3.50\) shipping fee. PrintPlus charges \(\$0.17\) per photo plus a flat \(\$1.50\) shipping fee. Write a function for the total cost \(y\) of each offer for \(x\) photos. Then find the number of photos for which the two offers cost exactly the same amount.

Hints

- Which part of each cost is fixed, and which part depends on the number of photos? - What is the general form of a linear cost function? - Set the two cost expressions equal to find the break-even quantity.

Solution

1. The cost functions are \(P(x) = 0.12x + 3.50\) and \(Q(x) = 0.17x + 1.50\). 2. Set the costs equal: \(0.12x + 3.50 = 0.17x + 1.50\). 3. Then \(2.00 = 0.05x\). 4. Solving gives \(x = 40\). 5. Therefore, the two offers cost the same amount for exactly \(40\) photos.

Answer

\(P(x) = 0.12x + 3.50\) and \(Q(x) = 0.17x + 1.50\). The costs are equal at \(x = 40\) photos.
Build linear models from descriptions 66
A candle burns at a constant rate. Two hours after it is lit, it is \(7.5\) inches tall. After \(5\) hours, it is \(6\) inches tall. a) Write a function \(h(t)\) for the candle's height, in inches, after \(t\) hours. b) Explain what the y-intercept and slope mean in this situation.

Hints

- How much does the candle's height change over the three-hour interval? - What sign should the rate have when the candle gets shorter? - What does time \(t = 0\) represent?

Solution

1. Use the points \((2, 7.5)\) and \((5, 6)\): \(m = \frac{6 - 7.5}{5 - 2} = -0.5\) inch per hour. 2. Substitute \((2, 7.5)\) into \(h(t) = -0.5t + b\): \(7.5 = -0.5 \cdot 2 + b\), so \(b = 8.5\). 3. Therefore, \(h(t) = -0.5t + 8.5\). 4. The y-intercept \(8.5\) is the candle's starting height. The slope \(-0.5\) means the candle loses \(0.5\) inch of height each hour.

Answer

a) \(h(t) = -0.5t + 8.5\) b) The y-intercept \(8.5\) is the initial height in inches, and the slope \(-0.5\) means the height decreases by \(0.5\) inch per hour.
Build linear models from descriptions 67
A point \(P(x, y)\) lies on a line described by this rule: “The y-value is \(4\) more than twice the x-value.” a) Write a linear equation that represents the rule. b) Determine algebraically whether \(A(3, 10)\) and \(B(-2, 1)\) lie on the line. c) A point \(C\) on the line has y-coordinate \(15\). Find its x-coordinate.

Hints

- Translate “twice the x-value” as \(2x\), then account for the additional \(4\). - To test a point, substitute both coordinates into the equation. - For part c, substitute the known y-value and solve for \(x\).

Solution

1. “Twice the x-value” is \(2x\), and “4 more” gives \(y = 2x + 4\). 2. For \(A\), \(2 \cdot 3 + 4 = 10\), so \(A(3, 10)\) lies on the line. 3. For \(B\), \(2 \cdot (-2) + 4 = 0\), not \(1\), so \(B(-2, 1)\) does not lie on the line. 4. For \(C\), solve \(15 = 2x + 4\). Then \(11 = 2x\), so \(x = 5.5\).

Answer

a) \(y = 2x + 4\) b) \(A(3, 10)\) lies on the line; \(B(-2, 1)\) does not. c) \(x = 5.5\)
Build linear models from descriptions 68
Two climbing gyms offer different monthly pricing plans. Summit Climbing Gym charges a \(\$10\) monthly fee plus \(\$5\) per visit. Peak Climbing Gym has no monthly fee and charges \(\$7.50\) per visit. Let \(v\) be the number of visits in a month. a) Construct monthly cost functions \(S(v)\) and \(P(v)\). b) Use the functions to find the number of visits for which the plans cost exactly the same amount. c) For \(5\) visits, which plan costs less and by how much? d) Starting at how many whole visits per month is Summit less expensive overall?

Hints

- Translate the monthly fee and per-visit charge into the initial value and rate of each cost function. - Equal costs mean equal function outputs for the same number of visits. - Use the constructed functions for both the sample comparison and the whole-number threshold.

Solution

1. Summit has model \(S(v)=10+5v\). Peak has model \(P(v)=7.50v\). 2. Set the functions equal: \(10+5v=7.50v\). Then \(2.50v=10\), so \(v=4\). Both plans cost \(\$30\). 3. \(S(5)=10+5\cdot5=35\), and \(P(5)=7.50\cdot5=37.50\). Summit is cheaper by \(\$2.50\). 4. The plans are equal at \(4\) visits, and Summit has the smaller per-visit rate after the fixed fee is covered. Therefore, Summit is less expensive for every whole-number visit count greater than \(4\), starting at \(5\) visits.

Answer

a) \(S(v)=10+5v\), \(P(v)=7.50v\) b) \(4\) visits; both cost \(\$30\). c) Summit; it saves \(\$2.50\). d) \(5\) visits
Build linear models from descriptions 69
Two water tanks are being adjusted at the same time. Tank A starts with \(100\) gallons of water, and \(15\) gallons are added each minute. Tank B starts with \(400\) gallons of water, and \(10\) gallons are drained each minute. a) Write a function for the volume \(y\), in gallons, in each tank after \(x\) minutes. b) Explain what the slope and y-intercept of the function for Tank B mean in this situation. c) Find when the tanks contain the same amount of water and how much water each tank contains at that time.

Hints

- Identify the starting amount and the amount that changes each minute for each tank. - Does the amount in Tank B increase or decrease? What sign should its slope have? - What equation represents the moment when the two tanks have the same volume?

Solution

1. Tank A starts at \(100\) gallons and gains \(15\) gallons per minute, so \(A(x) = 15x + 100\). Tank B starts at \(400\) gallons and loses \(10\) gallons per minute, so \(B(x) = -10x + 400\). 2. For Tank B, the y-intercept \(400\) is the initial volume in gallons. The slope \(-10\) means the volume decreases by \(10\) gallons each minute. 3. Set the functions equal: \(15x + 100 = -10x + 400\). 4. Then \(25x = 300\), so \(x = 12\). 5. Substituting gives \(A(12) = 15 \cdot 12 + 100 = 280\). Therefore, after \(12\) minutes, each tank contains \(280\) gallons.

Answer

a) \(A(x) = 15x + 100\) and \(B(x) = -10x + 400\). b) For Tank B, the y-intercept \(400\) is the starting volume in gallons, and the slope \(-10\) means \(10\) gallons are drained each minute. c) After \(12\) minutes, both tanks contain \(280\) gallons.
Build linear models from descriptions 70
Two siblings, Liam and Sarah, are saving money. Liam starts with \(\$20\) and saves another \(\$5\) each week. Sarah starts with \(\$50\) but saves only \(\$2\) each week. a) Write a function for each person's savings \(y\) after \(x\) weeks. b) After how many weeks will they have the same amount of money? c) How much money will each person have at that time? d) Interpret the slope of each function in this situation.

Hints

- In a linear function, which value represents the amount at week \(0\)? - What equation represents the time when both savings amounts are equal? - How does a one-week increase in \(x\) change each person's savings?

Solution

1. Liam's savings are modeled by \(L(x) = 5x + 20\). Sarah's savings are modeled by \(S(x) = 2x + 50\). 2. Set the functions equal: \(5x + 20 = 2x + 50\). 3. Then \(3x = 30\), so \(x = 10\). They have the same amount after \(10\) weeks. 4. Substituting gives \(L(10) = 5 \cdot 10 + 20 = 70\), so each person has \(\$70\). 5. The slopes, \(5\) and \(2\), are the weekly savings rates in dollars per week.

Answer

a) \(L(x) = 5x + 20\) and \(S(x) = 2x + 50\). b) \(10\) weeks. c) Each person has \(\$70\). d) The slopes are the weekly savings rates: \(\$5\) per week for Liam and \(\$2\) per week for Sarah.
Build linear models from descriptions 71
An electric scooter rental offers two pricing plans: Plan A: a \(\$1.20\) activation fee plus \(\$0.20\) per minute Plan B: no activation fee and \(\$0.30\) per minute a) Write a cost rule for each plan for a ride lasting \(x\) minutes. b) Find the cost of a \(15\)-minute ride under each plan. Which plan is less expensive? c) Find the ride length for which the two plans cost the same.

Hints

- Separate each fixed fee from its per-minute rate. - Substitute \(15\) into both cost rules. - Equal costs are represented by an equation.

Solution

1. The cost rules are \(C_A = 0.20x + 1.20\) and \(C_B = 0.30x\). 2. For \(x = 15\), Plan A costs \(0.20 \cdot 15 + 1.20 = 4.20\) dollars. 3. Plan B costs \(0.30 \cdot 15 = 4.50\) dollars, so Plan A is less expensive for a \(15\)-minute ride. 4. Set the costs equal: \(0.20x + 1.20 = 0.30x\). 5. Subtract \(0.20x\): \(1.20 = 0.10x\). 6. Divide by \(0.10\): \(x = 12\).

Answer

a) \(C_A = 0.20x + 1.20\); \(C_B = 0.30x\) b) Plan A costs \(\$4.20\), and Plan B costs \(\$4.50\). Plan A is less expensive. c) The plans cost the same for a \(12\)-minute ride.
Build linear models from descriptions 72
A technician records the temperature of a cooling liquid: <table> <tr> <td>Time \(t\) in \(\text{min}\)</td> <td>\(0\)</td><td>\(8\)</td><td>\(20\)</td> </tr> <tr> <td>Temperature \(T\) in \(^\circ\text{C}\)</td> <td>\(90\)</td><td>\(74\)</td><td>\(50\)</td> </tr> </table> a) Show using calculations that the temperature can be modeled by a linear function over this time interval. b) Find a linear model of the form \(T(t) = mt + b\). c) Find the temperature after exactly \(15\) minutes. d) Assuming the linear pattern continues, when will the liquid reach \(26\,^\circ\text{C}\)?

Hints

- How can you compare rates of change between consecutive pairs of data points? - Which table value gives the initial value of the model? - Substitute the given time into your model. - For the last part, set the model equal to the target temperature and solve for time.

Solution

1. From \(0\) to \(8\) minutes, the rate of change is \(\frac{74 - 90}{8 - 0} = -2\,^\circ\text{C}/\text{min}\). From \(8\) to \(20\) minutes, it is \(\frac{50 - 74}{20 - 8} = -2\,^\circ\text{C}/\text{min}\). The rates are equal, so the data are linear. 2. At \(t = 0\), the temperature is \(90\,^\circ\text{C}\), so \(b = 90\). The model is \(T(t) = -2t + 90\). 3. \(T(15) = -2 \cdot 15 + 90 = 60\), so the temperature is \(60\,^\circ\text{C}\). 4. Set \(T(t) = 26\): \(26 = -2t + 90\). Solving gives \(t = 32\).

Answer

a) Both intervals have rate of change \(-2\,^\circ\text{C}/\text{min}\), so the data are linear. b) \(T(t) = -2t + 90\) c) \(60\,^\circ\text{C}\) d) \(32\,\text{min}\)
Build linear models from descriptions 73
A cylindrical container has an inside diameter of \(12\,\text{cm}\) and a height of \(30\,\text{cm}\). Water is poured into the container. a) Write a function \(V\) that gives the water volume \(V(h)\), in \(\text{cm}^3\), for a water height \(h\), in centimeters. b) State the domain \(D\) and range \(R\) for this situation. Round the upper endpoint of the range to the nearest hundredth.

Hints

- How can you find the radius from the diameter? - Which formula gives the volume of a cylinder? - What water heights are physically possible, from empty to full? - How are the smallest and largest function values related to the container height?

Solution

1. The radius is half the diameter, so \(r=6\,\text{cm}\). 2. The area of the circular base is \(B=\pi r^2=\pi\cdot6^2=36\pi\,\text{cm}^2\). 3. Since cylinder volume is base area times height, \(V(h)=Bh=36\pi h\). 4. The water height can range from empty to full, so \(D=[0, 30]\). 5. The minimum volume is \(V(0)=0\). The maximum is \(V(30)=36\pi\cdot30=1080\pi\approx3{,}392.92\,\text{cm}^3\). Therefore, \(R=[0, 1080\pi]\), or approximately \([0, 3{,}392.92]\), in cubic centimeters.

Answer

a) \(V(h)=36\pi h\) b) \(D=[0, 30]\); \(R=[0, 1080\pi]\approx[0, 3{,}392.92]\). Here, \(h\) is measured in centimeters and \(V(h)\) is measured in cubic centimeters.
Build linear models from descriptions 74
A right triangular prism has a base with legs \(6\,\text{cm}\) and \(8\,\text{cm}\). The prism is \(25\,\text{cm}\) tall and stands on its triangular base while it is filled with liquid. a) Write a function \(f\) that gives the liquid volume \(f(x)\), in \(\text{cm}^3\), for a fill height \(x\), in centimeters. b) Find the fill height when the prism contains exactly \(0.36\,\text{L}\) of liquid. c) Briefly explain why \(f\) is a linear function.

Hints

- Recall the area formula for a right triangle. - Convert liters to cubic centimeters before solving. - Think about whether the horizontal cross-sectional area changes as the liquid rises. - A linear function has a constant rate of change.

Solution

1. The area of the right triangular base is \(B=\frac{1}{2}\cdot6\cdot8=24\,\text{cm}^2\). 2. Prism volume is base area times height, so \(f(x)=Bx=24x\). 3. Convert the given volume: \(0.36\,\text{L}=360\,\text{cm}^3\). 4. Set \(24x=360\) and solve: \(x=360\div24=15\). The fill height is \(15\,\text{cm}\). 5. The horizontal cross-sectional area remains constant as the liquid rises, so the volume increases at a constant rate of \(24\,\text{cm}^3\) per centimeter. Thus, \(f(x)=24x\) is linear.

Answer

a) \(f(x)=24x\) b) \(x=15\,\text{cm}\) c) The cross-sectional area is constant, so the volume increases by a constant amount for each centimeter of fill height.
Build linear models from descriptions 75
A rain barrel initially contains \(10\) gallons of water and has a capacity of \(50\) gallons. During a storm, water flows into the barrel at a constant rate of \(1\) gallon per minute. 1) Write a function for the amount of water \(y\), in gallons, after \(x\) minutes, including the physical domain. 2) Make a table for \(x=0\), \(x=10\), \(x=15\), and \(x=30\). 3) Describe the graph of the function on its physical domain. 4) Determine algebraically when the barrel becomes full.

Hints

- Identify the amount in the barrel at time \(0\). - Identify how much water is added each minute. - Use the form \(y=mx+b\) and restrict the domain to the time before the barrel is full. - Set \(y=50\) and solve for \(x\).

Solution

1. The initial value is \(10\) and the rate of change is \(1\), so \(y=x+10\). The barrel is modeled from the start of the storm until it reaches \(50\) gallons, so the physical domain is \(0\le x\le40\). 2. Evaluating the function gives \(y(0)=10\), \(y(10)=20\), \(y(15)=25\), and \(y(30)=40\). 3. The graph is the line segment with y-intercept \(10\) and slope \(1\), from \((0, 10)\) to \((40, 50)\). 4. Solve \(50=x+10\). Subtracting \(10\) gives \(x=40\), so the barrel becomes full after \(40\) minutes.

Answer

1) \(y=x+10\), for \(0\le x\le40\) 2) <table><tr><td>Time \(x\) in minutes</td><td>\(0\)</td><td>\(10\)</td><td>\(15\)</td><td>\(30\)</td></tr><tr><td>Water \(y\) in gallons</td><td>\(10\)</td><td>\(20\)</td><td>\(25\)</td><td>\(40\)</td></tr></table> 3) The line segment from \((0, 10)\) to \((40, 50)\) 4) \(40\) minutes
Build linear models from descriptions 76
Two water tanks begin draining at the same time. Tank 1 starts with \(500\,\text{gal}\) and loses \(10\,\frac{\text{gal}}{\text{min}}\). Tank 2 starts with \(800\,\text{gal}\) and loses \(30\,\frac{\text{gal}}{\text{min}}\). Let \(t\) be the number of minutes after draining begins. a) Construct volume functions \(V_1(t)\) and \(V_2(t)\) for the two tanks while each model is physically meaningful. b) Use the functions to determine when Tank 1 contains exactly twice as much water as Tank 2. Verify that both modeled volumes are nonnegative at that time.

Hints

- Each draining model needs its starting volume and a negative rate of change. - Express the “twice as much” condition using the two named functions rather than a one-off verbal equation. - Check the resulting time against the physical requirement that neither volume be negative.

Solution

1. Tank 1 has model \(V_1(t)=500-10t\). Tank 2 has model \(V_2(t)=800-30t\). 2. The condition is \(V_1(t)=2V_2(t)\), so \(500-10t=2(800-30t)\). 3. Distribute and solve: \(500-10t=1600-60t\), so \(50t=1100\) and \(t=22\). 4. \(V_1(22)=500-220=280\) and \(V_2(22)=800-660=140\). Both volumes are nonnegative, and \(280=2\cdot140\).

Answer

a) \(V_1(t)=500-10t\), \(V_2(t)=800-30t\) b) \(22\) minutes; the volumes are \(280\,\text{gal}\) and \(140\,\text{gal}\), respectively.
Build linear models from descriptions 77
Two monthly cell phone plans are available. Plan A: a \(\$10\) monthly fee plus \(\$0.15\) per minute Plan B: no monthly fee and \(\$0.25\) per minute Let \(m\) be the number of minutes used in a month. a) Construct monthly cost functions \(A(m)\) and \(B(m)\). b) Use the functions to find the number of minutes for which the plans have the same monthly cost. c) Which plan costs less for \(150\) minutes? Justify using your functions.

Hints

- Build each cost function from its fixed fee and per-minute rate. - The no-fee plan should have output \(0\) at input \(0\). - After constructing the functions, equal cost means equal outputs at the same input.

Solution

1. Plan A has model \(A(m)=10+0.15m\). Plan B has model \(B(m)=0.25m\). 2. Set the functions equal: \(10+0.15m=0.25m\). Then \(10=0.10m\), so \(m=100\). 3. \(A(150)=10+0.15\cdot150=32.50\), while \(B(150)=0.25\cdot150=37.50\). Plan A is less expensive.

Answer

a) \(A(m)=10+0.15m\), \(B(m)=0.25m\) b) \(100\) minutes c) Plan A; \(\$32.50\) compared with \(\$37.50\).
Build linear models from descriptions 78
Lena leaves for a hike at \(9{:}00\) a.m. and walks at a constant speed of \(3\) miles per hour. At \(10{:}30\) a.m., her brother Max realizes that she forgot her water bottle. He immediately rides his bike after her along the same route at \(12\) miles per hour. a) Write a function for each person's distance \(y\), in miles, from the starting point after \(x\) hours since Lena left. b) Identify two points on each person's graph and explain what an intersection of the graphs represents. c) Determine what time Max catches Lena and how far they are from the starting point. d) Explain what the slope of each line means in this situation.

Hints

- How can you account for Max starting \(1.5\) hours after Lena? - Which input values give convenient points on each function? - What does it mean when the two distance values are equal? - How is each rate related to the slope of its function?

Solution

1. Let \(x\) be the number of hours since \(9{:}00\) a.m., and let \(y\) be the distance from the starting point in miles. 2. Lena's distance is \(L(x) = 3x\) for \(x \ge 0\). Max starts \(1.5\) hours later, so his distance is \(M(x) = 12(x - 1.5)\) for \(x \ge 1.5\). 3. Lena's graph includes \((0, 0)\) and \((2, 6)\). Max's graph begins at \((1.5, 0)\) and also includes \((2, 6)\). An intersection represents a time when Lena and Max are the same distance from the starting point. 4. Set the distances equal: \(3x = 12(x - 1.5)\). This gives \(3x = 12x - 18\), so \(x = 2\). The distance is \(3 \cdot 2 = 6\) miles. 5. Two hours after \(9{:}00\) a.m. is \(11{:}00\) a.m. The slopes \(3\) and \(12\) represent the walking and biking speeds in miles per hour.

Answer

a) \(L(x) = 3x\) for \(x \ge 0\), and \(M(x) = 12(x - 1.5)\) for \(x \ge 1.5\). b) Lena's graph includes \((0, 0)\) and \((2, 6)\). Max's graph includes \((1.5, 0)\) and \((2, 6)\). An intersection means they are the same distance from the starting point at the same time. c) Max catches Lena at \(11{:}00\) a.m., \(6\) miles from the starting point. d) Each slope represents that person's speed in miles per hour.
Build linear models from descriptions 79
Cedarville and Pine Ridge are \(330\) miles apart by rail. A freight train leaves Cedarville for Pine Ridge at a constant \(60\,\text{mph}\). At the same time, a passenger train leaves Pine Ridge toward Cedarville at \(90\,\text{mph}\). a) Write each train's distance from Cedarville after \(t\) hours as a function. Give two points on each graph. b) Compare the distances at \(t=2\) and \(t=3\). Between which whole-number hours do the trains meet? c) Find the exact meeting time and distance from Cedarville.

Hints

- Both distances are measured from the same town; one increases while the other decreases. - Compare the two distances at each whole-hour checkpoint to bracket the meeting. - The trains meet when their distances from Town A are equal.

Solution

1. Measured from Cedarville, the freight train's distance is \(F(t)=60t\); the passenger train's is \(P(t)=330-90t\). 2. Two points for \(F\) are \((0,0)\), \((1,60)\); for \(P\), \((0,330)\), \((1,240)\). 3. At \(t=2\), the trains are at \(120\) and \(150\) miles from Cedarville; they have not yet met. At \(t=3\), they are at \(180\) and \(60\) miles, so they have passed each other. They meet between hours \(2\) and \(3\). 4. Solve \(60t=330-90t\): \(150t=330\), so \(t=2.2\) hours. The distance is \(F(2.2)=132\) miles from Cedarville.

Answer

a) \(F(t)=60t\), with \((0,0)\), \((1,60)\); \(P(t)=330-90t\), with \((0,330)\), \((1,240)\). b) At \(t=2\): \(120\) and \(150\) miles; at \(t=3\): \(180\) and \(60\) miles. They meet between \(2\) and \(3\) hours. c) After \(2.2\) hours, \(132\) miles from Cedarville.
Build linear models from descriptions 80
A smartphone battery is at \(15\%\) when charging begins. After \(20\) minutes, it is at \(45\%\). Assume the battery percentage increases linearly with time. a) Write a linear function for the battery percentage \(y\) after \(x\) minutes. b) How many percentage points are added per minute? Interpret this value as the slope. c) According to the model, how many minutes after charging begins will the battery reach \(100\%\)?

Hints

- First find the total change in battery percentage over the 20-minute interval. - Divide that change by the time interval to find the rate per minute. - What value represents the battery percentage at time \(0\)? - For a full battery, set the output equal to \(100\).

Solution

1. The slope is \(m = \frac{45 - 15}{20} = 1.5\) percentage points per minute. 2. The initial value is \(15\), so the model is \(y = 1.5x + 15\). 3. The slope \(1.5\) means the battery percentage increases by \(1.5\) percentage points each minute. 4. Set \(y = 100\): \(100 = 1.5x + 15\). Then \(85 = 1.5x\), so \(x = \frac{170}{3} \approx 56.7\) minutes.

Answer

a) \(y = 1.5x + 15\) b) \(1.5\) percentage points per minute c) \(\frac{170}{3}\) minutes, or about \(56.7\) minutes
Build linear models from descriptions 81
A smartphone battery is at \(12\%\) when it is connected to a charger that increases the battery level at a constant rate of \(0.8\) percentage point per minute. a) Write a linear function for the battery level \(L\), in percent, after \(t\) minutes. b) A fast charger starts at the same battery level and reaches \(100\%\) in exactly \(80\) minutes. Find its charging rate in percentage points per minute. c) How many minutes does the fast charger save compared with the first charger when charging from \(12\%\) to \(100\%\)?

Hints

- Identify the starting battery level and the constant rate of increase. - How many percentage points must the battery gain to reach \(100\%\)? - To find a constant rate, divide the total increase by the elapsed time. - Compare the two times needed to reach full charge.

Solution

1. The initial value is \(12\), and the charging rate is \(0.8\) percentage point per minute. Thus \(L(t) = 0.8t + 12\). 2. The fast charger must increase the battery level by \(100 - 12 = 88\) percentage points in \(80\) minutes. Its rate is \(\frac{88}{80} = 1.1\) percentage points per minute. 3. For the first charger, solve \(100 = 0.8t + 12\). This gives \(t = 110\) minutes. The time saved is \(110 - 80 = 30\) minutes.

Answer

a) \(L(t) = 0.8t + 12\) b) \(1.1\) percentage points per minute c) \(30\) minutes
Build linear models from descriptions 82
A hot-air balloon is \(2800\,\text{ft}\) above the ground when it begins to descend at a constant rate of \(5\,\text{ft/s}\). 1) Write a linear function for the balloon's height \(h\), in feet, after \(t\) seconds. 2) How high is the balloon after \(4\) minutes of descending? 3) When will the balloon reach the ground? Give the time in minutes and seconds.

Hints

- Because the balloon is descending, what sign should the rate of change have? - Make sure the time units match before substituting into the function. - What height represents the moment the balloon reaches the ground? - How can you convert a total number of seconds into minutes and remaining seconds?

Solution

1. The initial height is \(2800\) feet, and the rate of change is \(-5\) feet per second. Thus \(h(t) = 2800 - 5t\). 2. Four minutes is \(4 \cdot 60 = 240\) seconds. Then \(h(240) = 2800 - 5 \cdot 240 = 1600\). The balloon is \(1600\,\text{ft}\) above the ground. 3. At ground level, \(h = 0\). Solve \(0 = 2800 - 5t\), which gives \(t = 560\) seconds. Since \(560 = 9 \cdot 60 + 20\), this is \(9\) minutes \(20\) seconds.

Answer

1) \(h(t) = 2800 - 5t\) 2) \(1600\,\text{ft}\) 3) \(9\) minutes \(20\) seconds
Build linear models from descriptions 83
Two water tanks begin filling at the same time. Tank A starts with \(150\) gallons of water and receives another \(20\) gallons each minute. Tank B starts with only \(50\) gallons but fills at a rate of \(32.5\) gallons per minute. a) Write a function for the volume \(y\), in gallons, in each tank after \(x\) minutes. b) Find how many minutes it takes for the tanks to contain the same amount of water. c) How much water is in each tank at that time?

Hints

- Identify the starting volume and filling rate for each tank in the form \(y = mx + b\). - What equation represents the time when the two volumes are equal? - How do you solve an equation when the variable appears on both sides? - Check your value of \(x\) in both original functions.

Solution

1. Tank A is modeled by \(A(x) = 20x + 150\). Tank B is modeled by \(B(x) = 32.5x + 50\). 2. Set the functions equal: \(20x + 150 = 32.5x + 50\). 3. Subtract \(20x\) and \(50\) to get \(100 = 12.5x\). Therefore, \(x = 8\). 4. Substitute \(x = 8\): \(A(8) = 20 \cdot 8 + 150 = 310\). Both tanks contain \(310\) gallons after \(8\) minutes.

Answer

a) \(A(x) = 20x + 150\) and \(B(x) = 32.5x + 50\). b) \(8\) minutes. c) \(310\) gallons.
Build linear models from descriptions 84
Towns A and B are \(180\) miles apart along a highway. At \(12{:}00\) p.m., a truck leaves Town A and drives toward Town B at a constant speed of \(60\) miles per hour. At \(12{:}45\) p.m., a car leaves Town B and drives toward Town A at \(90\) miles per hour. Let \(t\) be the number of hours after \(12{:}00\) p.m., and measure position as distance from Town A. a) Construct the truck's position function \(T(t)\). For \(t\ge0.75\), construct the car's position function \(C(t)\). b) Use the two position functions to determine when the vehicles meet and the distance of the meeting point from Town A.

Hints

- Fix the coordinate system first: both functions must measure position from the same town. - The car's delayed start changes the input inside its model; convert \(45\) minutes to hours before constructing it. - The vehicles meet when the two position functions have the same output.

Solution

1. The truck starts at Town A and moves \(60\) miles per hour away from it, so \(T(t)=60t\). 2. At \(t=0.75\), the car starts \(180\) miles from Town A and then its distance from Town A decreases by \(90\) miles per hour. Thus, for \(t\ge0.75\), \(C(t)=180-90(t-0.75)\). 3. At the meeting time, \(T(t)=C(t)\): \(60t=180-90(t-0.75)\). 4. Simplifying gives \(60t=247.5-90t\), so \(150t=247.5\) and \(t=1.65\). 5. This is \(1\) hour \(39\) minutes after noon, or \(1{:}39\) p.m. 6. The meeting position is \(T(1.65)=60\cdot1.65=99\) miles from Town A.

Answer

a) \(T(t)=60t\); for \(t\ge0.75\), \(C(t)=180-90(t-0.75)\) b) \(1{:}39\) p.m., \(99\) miles from Town A
Build linear models from descriptions 85
Liam starts hiking at \(9{:}00\) a.m. and walks \(4\) miles per hour. At \(10{:}30\) a.m., Marie starts at the same trailhead and rides her bike along his route at \(12\) miles per hour. a) Write their distances from the trailhead, \(l(t)\) and \(m(t)\), where \(t\) is hours after \(9{:}00\) a.m. When does Marie's rule begin? b) From the graph, between which two half-hour marks does Marie catch Liam? About how far from the trailhead are they? c) Suppose Marie rides \(10\) miles per hour instead of \(12\). Write her new distance rule and calculate the meeting time, clock time, and distance.
Figure for problem 533228

Hints

- Use hours after \(9{:}00\) a.m. for both people's times. - The bicyclist's travel time begins later than the hiker's. - For the changed speed in part c), write a new rule and look for equal distances.

Solution

1. Liam travels \(l(t)=4t\) for \(t\ge0\). Marie starts after \(1.5\) hours, so \(m(t)=12(t-1.5)\) for \(t\ge1.5\). 2. The lines cross between \(t=2\) and \(t=2.5\) hours, near \(9\) miles. 3. At \(10\) miles per hour, Marie's rule changes to \(m_2(t)=10(t-1.5)\). Solve \(4t=10(t-1.5)\): \(4t=10t-15\), giving \(t=2.5\) hours and \(l(2.5)=10\) miles. The time is \(11{:}30\) a.m.

Answer

a) \(l(t)=4t\) for \(t\ge0\); \(m(t)=12(t-1.5)\) for \(t\ge1.5\) b) Between \(2\) and \(2.5\) hours after \(9{:}00\) a.m.; about \(9\) miles c) \(m_2(t)=10(t-1.5)\); \(t=2.5\) hours; \(11{:}30\) a.m.; \(10\) miles
Build linear models from descriptions 86
Two boats travel between harbors that are \(110\) miles apart. The Albatross starts at the western harbor and moves east at \(20\) miles per hour. At the same time, the Osprey starts at the eastern harbor and moves west at \(30\) miles per hour. a) Write rules \(a(t)\) and \(o(t)\) for each boat's distance from the western harbor after \(t\) hours. b) Use the graph to estimate between which whole-hour marks the boats meet and roughly how many miles from the western harbor they are. c) Suppose the Osprey instead travels at \(35\) miles per hour. Write its new rule and find the meeting time and distance exactly.
Figure for problem 533229

Hints

- Measure both boats' positions from the same harbor. - Think about how each distance changes as time passes. - For the changed speed, what changes in the Osprey's distance rule? When will both distances be equal?

Solution

1. The Albatross starts at distance zero and moves east, so \(a(t)=20t\). The Osprey starts \(110\) miles away and moves west, so \(o(t)=110-30t\). 2. The plotted intersection is between \(t=2\) and \(t=3\) hours, around \(45\) miles from the western harbor. 3. At \(35\) miles per hour, the Osprey has rule \(o_2(t)=110-35t\). Set \(20t=110-35t\). Then \(55t=110\), so \(t=2\) hours and the distance is \(a(2)=40\) miles.

Answer

a) \(a(t)=20t\); \(o(t)=110-30t\) b) Between \(2\) and \(3\) hours; about \(45\) miles from the western harbor c) \(o_2(t)=110-35t\); after \(2\) hours, \(40\) miles from the western harbor
Build linear models from descriptions 87
A climber assumes that air temperature decreases at a constant rate as elevation increases. The climber measures \(61\,\text{°F}\) at \(2000\,\text{ft}\) and \(47\,\text{°F}\) at \(6000\,\text{ft}\). a) Use a linear model to predict the temperature at a \(10{,}000\)-foot summit. b) A plane flies at \(39{,}000\,\text{ft}\). What temperature does the model predict at that elevation? c) Why is it problematic to use this simple linear model at extreme elevations?

Hints

- Find the temperature change for each \(1000\)-foot increase in elevation. - Compare each target elevation with the lower measurement. - Consider whether one local rate should remain valid through every layer of the atmosphere.

Solution

1. The temperature changes by \(47-61=-14\,\text{°F}\) over an elevation increase of \(6000-2000=4000\,\text{ft}\). The rate is \(\frac{-14}{4000}=-0.0035\,\text{°F}\) per foot, or \(-3.5\,\text{°F}\) per \(1000\) feet. 2. The summit is \(8000\) feet above the first measurement. The predicted change is \(8(-3.5)=-28\,\text{°F}\), so the predicted temperature is \(61-28=33\,\text{°F}\). 3. The plane is \(37{,}000\) feet above the first measurement. The predicted change is \(37(-3.5)=-129.5\,\text{°F}\), so the model predicts \(61-129.5=-68.5\,\text{°F}\). 4. Atmospheric conditions and the rate of temperature change vary by altitude and weather. A rate estimated from two mountain measurements should not be assumed valid throughout the atmosphere.

Answer

a) \(33\,\text{°F}\) b) \(-68.5\,\text{°F}\) c) The constant rate is reasonable only over a limited elevation range; atmospheric conditions change at much greater heights.
Build linear models from descriptions 88
A cylindrical container starts empty and has a base area of \(200\,\text{cm}^2\). It is filled at a constant rate of \(500\,\text{mL}\) per minute. a) How many centimeters per minute does the water height increase? b) Write a function for the water height \(h\), in centimeters, after \(t\) minutes. c) How would the graph change for a container with a base area of \(100\,\text{cm}^2\) filled at the same rate? Compare the slopes. d) Suppose a container consists of two stacked cylinders. The lower cylinder has base area \(200\,\text{cm}^2\) and height \(10\,\text{cm}\). The upper cylinder has base area \(100\,\text{cm}^2\) and height \(10\,\text{cm}\). Describe the water-height graph and what happens after \(4\) minutes.

Hints

- Use the cylinder volume relationship \(V=Bh\). - A constant inflow gives a constant increase in volume. - Compare how the same volume changes height when the base area is smaller.

Solution

1. Since \(500\,\text{mL}=500\,\text{cm}^3\), use \(V=Bh\): \(500=200h\). Thus, the height increases by \(2.5\,\text{cm}\) per minute. 2. Starting empty, the height function is \(h(t)=2.5t\). 3. For base area \(100\,\text{cm}^2\), \(500=100h\), so the height increases by \(5\,\text{cm}\) per minute. The slope doubles. 4. In the lower section, the graph has slope \(2.5\). The water reaches \(10\,\text{cm}\) after \(10\div2.5=4\) minutes. In the upper section, the slope is \(5\). Therefore, the graph has a corner at \((4, 10)\) and becomes twice as steep after \(4\) minutes.

Answer

a) \(2.5\,\text{cm}\) per minute b) \(h(t)=2.5t\) c) The slope doubles to \(5\,\text{cm}\) per minute. d) The graph has slope \(2.5\) until \((4, 10)\), then slope \(5\); the water rises twice as fast in the narrower upper section.
Build linear models from descriptions 89
A cylindrical candle is initially \(12.5\,\text{in.}\) tall. After it is lit, its height decreases at a constant rate of \(0.6\,\text{in.}\) per hour. a) Write a function for the candle's height \(h\), in inches, after \(t\) hours. b) How long will it take the candle to reach half its original height? Give the answer in hours and minutes. c) Suppose the candle were shaped like a pyramid, becoming narrower toward the top. Explain why its height would not decrease linearly if the same volume of wax burned each hour.

Hints

- A decreasing quantity has a negative rate of change. - Set the height equal to half of \(12.5\). - Convert the decimal part of an hour to minutes by multiplying by \(60\). - Compare the height removed by the same wax volume in narrow and wide sections.

Solution

1. The initial height is \(12.5\), and the rate of change is \(-0.6\), so \(h(t)=12.5-0.6t\). 2. Half the original height is \(6.25\) inches. Solve \(6.25=12.5-0.6t\). Then \(0.6t=6.25\), so \(t=\frac{6.25}{0.6}\approx10.4167\) hours. The decimal part is \(0.4167(60)\approx25\) minutes, so the time is about \(10\) hours \(25\) minutes. 3. A pyramid has a smaller cross section near the top and a larger cross section near the bottom. Burning the same volume removes more height in a narrow section than in a wide section, so the height-change rate is not constant.

Answer

a) \(h(t)=12.5-0.6t\) b) About \(10\) hours \(25\) minutes c) The changing cross-sectional area makes the height decrease at a changing rate.
Build linear models from descriptions 90
A tank holds at most \(375\) gallons and starts with \(50\) gallons. Water enters at a constant rate, adding \(30\) gallons in \(8\) minutes. a) Find the rate in gallons per minute and write a function \(V(t)\) for the volume after \(t\) minutes. b) How long does it take to fill the tank? Give hours, minutes, and seconds. c) Suppose the tank is filled again from \(50\) gallons, but this time the inflow rate is halved from the start. How do the slope and the time needed to fill change?

Hints

- Divide added volume by the filling time for a rate. - Subtract the water already in the tank from its capacity. - If the amount still needed is unchanged, think about how a smaller constant rate affects filling time.

Solution

1. The rate is \(30\div8=3.75\) gallons per minute, so \(V(t)=3.75t+50\). 2. To reach capacity, solve \(375=3.75t+50\). Then \(t=325\div3.75=86\frac{2}{3}\) minutes, or \(1\) hour, \(26\) minutes, and \(40\) seconds. 3. In the new filling run, the rate is \(1.875\) gallons per minute. The required \(325\) gallons are unchanged, so the fill time doubles to \(173\frac{1}{3}\) minutes, or \(2\) hours, \(53\) minutes, and \(20\) seconds.

Answer

a) \(3.75\) gallons per minute; \(V(t)=3.75t+50\). b) \(1\) hour, \(26\) minutes, \(40\) seconds c) The slope is halved to \(1.875\) gallons per minute; the time doubles to \(2\) hours, \(53\) minutes, \(20\) seconds.
Build linear models from descriptions 91
Two candles are lit at the same time. Candle A is \(5\,\text{cm}\) tall and burns down at a constant rate of \(10\,\text{mm}\) per hour. Candle B is \(12\,\text{cm}\) tall and, each hour, loses an amount equal to \(10\%\) of its original height. Use linear models to determine when the candles would have the same height. Then decide whether that actually happens while the candles are burning.

Hints

- First find how many centimeters Candle B loses each hour. - Write a linear equation for the height of each candle after \(t\) hours. - After finding a time when the equations are equal, determine how long each candle can actually burn. - Can a candle have a negative height?

Solution

1. Candle A loses \(10\,\text{mm} = 1\,\text{cm}\) per hour. Candle B loses \(10\%\) of \(12\,\text{cm}\), so its rate is \(0.10 \cdot 12 = 1.2\,\text{cm}\) per hour. 2. Let \(t\) be the number of hours after the candles are lit. The height models are \(A(t) = 5 - t\) and \(B(t) = 12 - 1.2t\). 3. Set the heights equal: \(5 - t = 12 - 1.2t\). This gives \(0.2t = 7\), so \(t = 35\). 4. At \(t = 35\), the model gives a height of \(-30\,\text{cm}\), which is not physically possible. Candle A burns out after \(5\) hours, and Candle B burns out after \(10\) hours. 5. Therefore, the linear models are equal only after both candles would already be gone, so the candles never have the same height while they are burning.

Answer

The linear models are equal after \(35\) hours, at the theoretical height \(-30\,\text{cm}\). Candle A burns out after \(5\) hours and Candle B after \(10\) hours, so the candles never have the same height while they are actually burning.
Build linear models from descriptions 92
A smartphone battery starts at \(10\%\). Its charge level rises at a constant rate while charging, reaching \(90\%\) after \(40\) minutes. The charger is removed. During a \(20\)-minute video call, the charge level falls at a constant rate to \(60\%\). a) Write a battery-level function for each phase, using elapsed time \(x\) in minutes and charge level \(y\) in percent. b) By how many percentage points per minute does the charging rate exceed the magnitude of the discharge rate? c) At what times during these two phases is the charge level exactly \(75\%\)? d) If the phone continues discharging at the same rate after the call, how many additional minutes does it take to reach \(0\%\)?

Hints

- Treat charging and discharging as two separate constant-rate phases. - Compute each rate from the change in charge divided by minutes; distinguish sign from magnitude. - Check both phases for a charge level of \(75\%\). - For the final question, use the level remaining at the end of the call.

Solution

1. The key points are \((0,10)\), \((40,90)\), and \((60,60)\). 2. The charging rate is \(\frac{90-10}{40}=2\) percentage points per minute. Thus \(y=2x+10\) for \(0\leq x\leq40\). 3. The discharge rate is \(\frac{60-90}{60-40}=-1.5\). Thus \(y=90-1.5(x-40)=-1.5x+150\) for \(40\leq x\leq60\). 4. The difference in the rate magnitudes is \(2-1.5=0.5\) percentage point per minute. 5. For charging, solve \(75=2x+10\), giving \(x=32.5\). For discharge, solve \(75=-1.5x+150\), giving \(x=50\). 6. From \(60\%\), discharging at \(1.5\) points per minute takes another \(60\div1.5=40\) minutes.

Answer

a) Charging: \(y=2x+10\) for \(0\leq x\leq40\); discharging: \(y=-1.5x+150\) for \(40\leq x\leq60\). b) \(0.5\) percentage point per minute c) At \(x=32.5\) and \(x=50\) minutes d) \(40\) additional minutes

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