Free Math Worksheets

Build your own math worksheets from 21,000 problems for grades 3 to 12, from fractions to calculus. Every problem includes step-by-step solutions.

Build linear models from descriptions

Click problems to add them to your worksheet.

Difficulty: 1 – 5

5119118

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) \(19\,\text{in.}\)

5128778

A line \(g\) is parallel to the x-axis and passes through \(A(-4, -7)\). 1. Find the equation of line \(g\). 2. For each point, determine whether it lies above, below, or on line \(g\): \(B(-4, -5)\), \(C(10, -7)\), and \(D(0, -10)\). Justify each answer by comparing y-coordinates.

Hints

- What form does the equation of a horizontal line have? - A point's y-coordinate tells you its vertical position. - Compare each y-coordinate with \(-7\).

Solution

1. A line parallel to the x-axis is horizontal, so it has the form \(y = c\). Since it passes through \(A(-4, -7)\), its equation is \(y = -7\). 2. Point \(B\) has y-coordinate \(-5\), and \(-5 > -7\), so \(B\) is above the line. Point \(C\) has y-coordinate \(-7\), so it is on the line. Point \(D\) has y-coordinate \(-10\), and \(-10 < -7\), so \(D\) is below the line.

Answer

1) \(y = -7\) 2) \(B\) is above the line; \(C\) is on the line; \(D\) is below the line.

5130398

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.

5241978

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

5241998

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

5242018

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

5322208

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.

5332818

At the start of a \(15\)-minute observation, a rainwater tank already contains \(25\) gallons. A constant inflow adds \(5\) gallons per minute. The graph shows the tank's volume as a function of time. a) Use the graph to determine how many minutes pass before the tank contains \(60\) gallons. b) Write the function, including its domain for the observation interval, and find when the tank contains \(90\) gallons.

Figure for problem 533281

Hints

- The y-intercept is the starting volume. - The slope is the amount added each minute. - Use the stated observation interval for the domain. - Set the function equal to the target volume and solve for time.

Solution

1. The graph reaches \(60\) gallons at \(t=7\) minutes. 2. The initial value is \(25\), and the rate of change is \(5\) gallons per minute, so \(V(t)=5t+25\), with \(0\le t\le15\). Set \(V(t)=90\): \(90=5t+25\), so \(65=5t\) and \(t=13\).

Answer

a) \(7\) minutes b) \(V(t)=5t+25\), with \(0\le t\le15\); \(13\) minutes

5332828

A new candle is \(12\) inches tall. As it burns, its height decreases at a constant rate of \(0.75\) inch per hour. The graph shows candle height over time. a) Use the graph to find the candle's height after \(4\) hours. b) Write the function, including its contextual domain, and find how many hours pass before the candle is \(4.5\) inches tall.

Figure for problem 533282

Hints

- A decreasing height gives a negative slope. - The y-intercept is the initial candle height. - The contextual domain ends when the candle's height reaches \(0\). - Set the function equal to the target height and solve.

Solution

1. At \(t=4\), the graph shows a height of \(9\) inches. 2. The initial height is \(12\), and the rate of change is \(-0.75\) inch per hour, so \(h(t)=12-0.75t\). The candle burns from its initial height to \(0\) inches, so the contextual domain is \(0\le t\le16\). Set \(h(t)=4.5\): \(4.5=12-0.75t\), so \(-7.5=-0.75t\) and \(t=10\).

Answer

a) \(9\) inches b) \(h(t)=12-0.75t\), with \(0\le t\le16\); \(10\) hours

5332888

Two hikers, Alex and Bea, travel along a long straight road. Alex starts at mile marker \(0\) and walks at a constant speed toward higher mile markers. Bea starts at the same time at mile marker \(10\) and walks at a constant speed toward mile marker \(0\). The graph shows each hiker's position relative to Alex's starting point as a function of time \(t\), in hours. Write a function of the form \(s(t)=mt+b\) for each hiker on the displayed interval \(0\le t\le4\).

Figure for problem 533288

Hints

- The y-intercept is each hiker's starting position. - A positive slope means moving toward higher mile markers; a negative slope means moving toward lower mile markers. - Use another point on each line to calculate slope.

Solution

1. Alex's graph starts at \(0\) and passes through \((2, 6)\). Its slope is \(m=\frac{6-0}{2-0}=3\), so \(s_A(t)=3t\), with \(0\le t\le4\). 2. Bea's graph starts at \(10\) and passes through \((2, 6)\). Its slope is \(m=\frac{6-10}{2-0}=-2\), so \(s_B(t)=-2t+10\), with \(0\le t\le4\).

Answer

\(s_A(t)=3t\) and \(s_B(t)=-2t+10\), with \(0\le t\le4\) for both functions

5335288

Two water tanks are observed at the same time. The graph shows volume \(V\), in gallons, as a function of time \(t\), in minutes. Write an equation of the form \(V(t)=mt+b\) for each tank, including the displayed domain.

Figure for problem 533528

Hints

- The y-intercept is the initial volume. - The slope is the change in gallons per minute. - Use two clear points on each graph. - Read each displayed time interval from the graph.

Solution

1. Tank 1 starts with \(10\) gallons, so \(b=10\). After \(5\) minutes it contains \(30\) gallons, so \(m=\frac{30-10}{5-0}=4\). Therefore, \(V_1(t)=4t+10\), with \(0\le t\le12\). 2. Tank 2 starts with \(50\) gallons, so \(b=50\). It is empty after \(10\) minutes, so \(m=\frac{0-50}{10-0}=-5\). Therefore, \(V_2(t)=-5t+50\), with \(0\le t\le10\).

Answer

Tank 1: \(V_1(t)=4t+10\), with \(0\le t\le12\) Tank 2: \(V_2(t)=-5t+50\), with \(0\le t\le10\)

5119138

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) \(y=100-8x\)

5119218

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.

5119248

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 \(10\,\text{GB}\).

5119548

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) \(15\,\text{cm}\)

5120438

Two different candles are lit at the same time. Candle A is initially \(30\,\text{cm}\) tall. After \(4\) hours, it is \(18\,\text{cm}\) tall. Candle B is initially \(20\,\text{cm}\) tall and burns down at a constant rate of \(1\,\text{cm/h}\). a) Find the height of Candle A after \(7\) hours. b) After how many hours will Candle A be completely burned down? c) When will the two candles have the same height, and what will that height be?

Hints

- How many centimeters does Candle A lose in one hour? - Can you write a formula for the height of each candle after \(t\) hours? - What equation represents the moment when the two heights are equal?

Solution

1. Candle A loses \(30 - 18 = 12\,\text{cm}\) in \(4\) hours, so its burn rate is \(12 \div 4 = 3\,\text{cm/h}\). 2. For part a, after \(7\) hours Candle A has height \(30 - 7 \cdot 3 = 9\,\text{cm}\). 3. For part b, solve \(30 - 3t = 0\). Then \(3t = 30\), so \(t = 10\). Candle A burns down completely after \(10\) hours. 4. For part c, model the heights by \(h_A(t) = 30 - 3t\) and \(h_B(t) = 20 - t\). Set them equal: \(30 - 3t = 20 - t\). 5. Solving gives \(10 = 2t\), so \(t = 5\). Then \(h_B(5) = 20 - 5 = 15\,\text{cm}\).

Answer

a) \(9\,\text{cm}\) b) \(10\) hours c) After \(5\) hours, both candles are \(15\,\text{cm}\) tall.

5120458

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.

5120518

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> a) Describe the trends in the two programs. b) Assume both memberships changed at constant rates from 2019 to 2022. In about what year were the memberships equal? Justify your answer. c) If the basketball trend from 2019 to 2022 continues at the same rate, how many members will the program have in 2025?

Hints

- Identify which sequence increases and which decreases. - Compare the membership difference in 2019 and 2022. - Extend the basketball change from one three-year interval to the next.

Solution

1. Tennis membership decreases from \(240\) to \(190\), while basketball membership increases from \(80\) to \(210\). 2. In 2019, tennis had \(205 - 175 = 30\) more members than basketball. By 2022, tennis had \(190 - 210 = -20\) more members, so the difference decreased by \(50\) members over \(3\) years. 3. Reaching a difference of \(0\) requires \(\frac{30}{50}\) of that three-year change. This takes \(\frac{30}{50} \cdot 3 = 1.8\) years after 2019, which is approximately 2021. 4. Basketball membership increased by \(210 - 175 = 35\) members from 2019 to 2022. Continuing the same three-year change gives \(210 + 35 = 245\) members in 2025.

Answer

a) Tennis membership decreases, while basketball membership increases. b) The memberships were approximately equal in 2021. c) \(245\) members

5120578

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.

5120598

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.

5124528

A class is ordering printed T-shirts. Two companies offer these prices. - InkWorks charges a one-time setup fee of \(\$24\) plus \(\$7\) per shirt. - TeeDirect charges \(\$10\) per shirt with no additional fee. a) Find the cost of \(5\) shirts and \(10\) shirts from each company. b) Starting with how many shirts is InkWorks less expensive than TeeDirect?

Hints

- Compare the additional cost for each shirt. - How much is saved per shirt with InkWorks? - How many per-shirt savings are needed to offset the setup fee?

Solution

1. For \(5\) shirts, InkWorks costs \(\$24+5 \cdot \$7=\$59\), and TeeDirect costs \(5 \cdot \$10=\$50\). 2. For \(10\) shirts, InkWorks costs \(\$24+10 \cdot \$7=\$94\), and TeeDirect costs \(\$100\). 3. Set the costs equal: \(24+7x=10x\). Then \(24=3x\), so \(x=8\). The costs are equal at \(8\) shirts, so InkWorks is less expensive starting at \(9\) shirts.

Answer

a) For \(5\) shirts: InkWorks \(\$59\), TeeDirect \(\$50\). For \(10\) shirts: InkWorks \(\$94\), TeeDirect \(\$100\). b) InkWorks is less expensive starting at \(9\) shirts; the costs are equal at \(8\) shirts.

5125848

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. a) Which stack grows faster? Briefly explain. b) After how many minutes will the stacks contain the same number of boxes? c) How many boxes will be in each stack at that time?

Hints

- Identify the amount added to each stack every minute. - Write a linear expression for the number of boxes in each stack after \(x\) minutes. - What must be true of the two expressions when the stacks contain the same number? - Compare the two rates to answer part a.

Solution

1. Stack A grows faster because its rate is \(5\,\frac{\text{boxes}}{\text{min}}\), compared with \(2\,\frac{\text{boxes}}{\text{min}}\) for Stack B. 2. After \(x\) minutes, Stack A has \(15 + 5x\) boxes and Stack B has \(45 + 2x\) boxes. 3. Set the expressions equal: \(15 + 5x = 45 + 2x\). 4. Subtract \(2x\) and then subtract \(15\): \(3x = 30\). 5. Divide by \(3\): \(x = 10\). 6. Substitute \(x = 10\): \(15 + 5 \cdot 10 = 65\).

Answer

a) Stack A grows faster because it receives 5 boxes per minute, while Stack B receives 2. b) The stacks contain the same number after 10 minutes. c) Each stack contains 65 boxes.

5125928

Julian compares two monthly cell phone plans. The Basic plan costs \(\$12.00\) plus \(\$0.08\) per minute. The Flat plan costs \(\$22.00\), regardless of the number of minutes used. Starting at how many whole minutes is the Flat plan less expensive than the Basic plan? Write and solve an inequality.

Hints

- Write a cost expression for each plan. - Translate “less expensive” into an inequality. - Find the boundary where the plans have equal cost. - Use the next whole minute because the inequality is strict.

Solution

1. Let \(x\) be the number of minutes used. 2. The Basic plan costs \(12 + 0.08x\), and the Flat plan costs \(22\). 3. For the Flat plan to be less expensive, write \(22 < 12 + 0.08x\). 4. Subtract \(12\): \(10 < 0.08x\). 5. Divide by \(0.08\): \(125 < x\). 6. The least whole-number value greater than \(125\) is \(126\).

Answer

The Flat plan is less expensive starting at \(126\) minutes per month.

5126108

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 a) Find the total cost of each membership after \(4\) months. b) Starting with which full month is Power less expensive overall than Easy? Justify your answer with an inequality.

Hints

- Write a cost expression for each membership. - Include the one-time fee only in the Power plan. - Use a strict inequality for “less expensive.” - Interpret the decimal boundary using whole months.

Solution

1. After \(4\) months, Power costs \(45 + 4 \cdot 25 = 145\) dollars. 2. After \(4\) months, Easy costs \(4 \cdot 35 = 140\) dollars. 3. Let \(x\) be the number of months. Power is less expensive when \(45 + 25x < 35x\). 4. Subtract \(25x\): \(45 < 10x\). 5. Divide by \(10\): \(4.5 < x\). 6. The least whole-number month greater than \(4.5\) is \(5\).

Answer

a) After \(4\) months, Power costs \(\$145.00\), and Easy costs \(\$140.00\). b) Power is less expensive starting in month \(5\).

5126118

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 a) In March, the user talks for \(80\) minutes. Find that month's cost under each plan. For Smart, use one-twelfth of the annual fee as the monthly share. b) Starting at how many whole minutes per year is Smart less expensive overall than Basic?

Hints

- Keep monthly and annual costs separate. - Divide the annual fee by \(12\) for part a). - For part b), compare the two annual cost expressions. - Use the next whole minute after the strict boundary.

Solution

1. Smart's monthly share of the annual fee is \(114 \div 12 = 9.50\) dollars. 2. For \(80\) minutes, Smart costs \(9.50 + 80 \cdot 0.05 = 13.50\) dollars. 3. Basic costs \(80 \cdot 0.15 = 12.00\) dollars. 4. For the annual comparison, let \(x\) be the number of minutes. Smart is less expensive when \(114 + 0.05x < 0.15x\). 5. Subtract \(0.05x\): \(114 < 0.10x\), so \(x > 1140\). 6. The least whole-number value is \(1141\).

Answer

a) Smart costs \(\$13.50\), and Basic costs \(\$12.00\) for March. b) Smart is less expensive starting at \(1141\) minutes per year.

5126128

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 a) Find the monthly cost for each company when a customer drives \(40\) miles. b) For what monthly mileage is Company A less expensive than Company B?

Hints

- Write a cost expression for each company. - Use a strict inequality for “less expensive.” - Isolate the mileage variable after comparing the two costs.

Solution

1. At \(40\) miles, Company A costs \(15.00 + 0.20 \cdot 40 = 23.00\) dollars. 2. Company B costs \(0.45 \cdot 40 = 18.00\) dollars. 3. Let \(x\) be the number of miles. Company A is less expensive when \(15 + 0.20x < 0.45x\). 4. Subtract \(0.20x\): \(15 < 0.25x\). 5. Divide by \(0.25\): \(x > 60\).

Answer

a) Company A costs \(\$23.00\), and Company B costs \(\$18.00\). b) Company A is less expensive when the customer drives more than \(60\) miles per month.

5128128

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.

5128568

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.

5129228

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) \(93.75^\circ\).

5129628

A line \(g\) passes through \(P(-1, 7)\) and \(Q(2, -2)\). Determine whether \(R(0, 5)\) and \(S(1, 1)\) lie on line \(g\).

Hints

- First find the equation of the line through the two given points. - How can you find the slope from two points? - Once you know the slope, use one point to find the y-intercept. - Then substitute the x-coordinate of each test point.

Solution

1. The slope through \(P\) and \(Q\) is \(m = \frac{-2 - 7}{2 - (-1)} = -3\). 2. Using \(P(-1, 7)\), \(7 = -3(-1) + b\), so \(b = 4\). Thus \(g(x) = -3x + 4\). 3. \(g(0) = 4\), not \(5\), so \(R\) does not lie on the line. 4. \(g(1) = 1\), so \(S\) does lie on the line.

Answer

\(R(0, 5)\) does not lie on \(g\). \(S(1, 1)\) lies on \(g\).

5129638

An electric-vehicle charging plan has a monthly base fee of \(\$18.50\) plus \(\$0.80\) for each kilowatt-hour (\(\text{kWh}\)) used. a) A customer says a month with \(40\,\text{kWh}\) of charging cost exactly \(\$50.50\). Does that match the pricing model? b) How many kilowatt-hours were used if the monthly bill was exactly \(\$74.50\)?

Hints

- Which part of the cost is fixed, and which part depends on usage? - Write a linear model using the base fee and the cost per kilowatt-hour. - To test the customer's claim, substitute the stated usage. - When the total bill is known, set the model equal to that amount and solve for usage.

Solution

1. Let \(x\) be the number of kilowatt-hours and \(C\) be the cost in dollars. The model is \(C(x) = 0.80x + 18.50\). 2. For \(40\,\text{kWh}\), \(C(40) = 0.80 \cdot 40 + 18.50 = 50.50\). The customer's statement matches the model. 3. For a \(\$74.50\) bill, solve \(74.50 = 0.80x + 18.50\). Subtracting \(18.50\) gives \(56 = 0.80x\), so \(x = 70\).

Answer

a) Yes. The model gives \(\$50.50\) for \(40\,\text{kWh}\). b) \(70\,\text{kWh}\)

5129658

A line \(g\) passes through \(P(4, -1)\) and has slope \(m = -0.75\). A student wants to find more points on the line using slope triangles. a) Starting at \(P\), the student moves 8 units left. How many units must the student move up or down to land on the line again? b) What is the vertical change when the student moves exactly 1 unit right? c) Find an equation for line \(g\).

Hints

- How are slope, horizontal change, and vertical change related? - What does a negative slope tell you about the direction of the vertical change? - How can you use the given point and slope to find the y-intercept?

Solution

1. For part a, \(\Delta x = -8\). Since \(\Delta y = m\Delta x\), \(\Delta y = -0.75 \cdot (-8) = 6\). The student moves 6 units up. 2. For part b, \(\Delta x = 1\), so \(\Delta y = -0.75 \cdot 1 = -0.75\). The student moves \(0.75\) unit down. 3. Write \(g(x) = -0.75x + b\). Using \(P(4, -1)\), \(-1 = -0.75(4) + b\), so \(b = 2\). Therefore, \(g(x) = -0.75x + 2\).

Answer

a) 6 units up b) \(\Delta y = -0.75\), so \(0.75\) unit down c) \(g(x) = -0.75x + 2\)

5129708

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

5129718

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

5129728

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) \(\$47.00\)

5130018

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

5130218

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. \([0, 9]\)

5130228

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

5130308

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

5130318

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 \(320\,\text{L}\).

5130428

A water tank fills at a constant rate. After \(10\) minutes, it contains \(350\) gallons. After \(25\) minutes, it contains \(800\) gallons. a) Write a linear function \(V(t)\) for the volume, in gallons, after \(t\) minutes. b) How much water was in the tank at \(t = 0\)? c) The tank holds \(1700\) gallons. After how many minutes will it be full?

Hints

- Treat time as the input and volume as the output. - Use the two measurements to find how many gallons are added each minute. - What does the function value at \(t = 0\) represent? - To find the fill time, set the function equal to the tank's capacity.

Solution

1. Use \((10, 350)\) and \((25, 800)\): \(m = \frac{800 - 350}{25 - 10} = 30\) gallons per minute. 2. Substitute \((10, 350)\) into \(V(t) = 30t + b\): \(350 = 30 \cdot 10 + b\), so \(b = 50\). 3. Therefore, \(V(t) = 30t + 50\), and the initial amount is \(50\) gallons. 4. Set the volume equal to capacity: \(1700 = 30t + 50\). Then \(1650 = 30t\), so \(t = 55\) minutes.

Answer

a) \(V(t) = 30t + 50\) b) \(50\) gallons c) \(55\) minutes

5130448

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

5130938

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.

5130958

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.

5130968

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) Sketch the graph of the function on an appropriate coordinate plane.

Hints

- What is the battery level at time \(0\)? - How many percentage points does the battery gain per minute? - What output value represents a fully charged battery? - Choose axis scales that include the starting point and the full-charge point.

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 \(0 \le x \le 72\), the graph is the line segment through \((0, 10)\), \((24, 40)\), and \((72, 100)\).

Answer

a) \(y = 1.25x + 10\) b) \(72\) minutes c) The graph is the line segment through \((0, 10)\), \((24, 40)\), and \((72, 100)\) for \(0 \le x \le 72\).

5130988

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. a) Find the increase in water depth per hour and explain how this rate appears on a linear graph. b) If filling had begun earlier at the same rate, at what time was the pool empty? c) The pool is \(64\,\text{in.}\) deep. At what time will it become full and begin to overflow?

Hints

- Divide the change in depth by the elapsed time. - Work backward from the \(16\)-inch measurement to a depth of \(0\). - Find the remaining depth from \(40\) inches to the top.

Solution

1. In \(2\) hours, the depth increases by \(40-16=24\) inches. The rate is \(24\div2=12\) inches per hour, which is the slope of the line. 2. At \(1{:}00\) p.m., the pool contains \(16\) inches of water. Reversing at \(12\) inches per hour takes \(16\div12=\frac{4}{3}\) hour, or \(1\) hour \(20\) minutes. The pool was empty at \(11{:}40\) a.m. 3. From \(40\) inches to \(64\) inches is another \(24\) inches, requiring \(24\div12=2\) hours. The pool becomes full at \(5{:}00\) p.m.

Answer

a) \(12\,\text{in.}\) per hour; this is the slope. b) \(11{:}40\) a.m. c) \(5{:}00\) p.m.

5131158

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) \(s(t)=60t\).

5131738

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.

5131928

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

5131948

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

5132108

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.

5135178

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

5137588

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.

5137938

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) \(x = 5.5\)

5138188

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. a) How many visits in a month make the two plans cost exactly the same amount? b) A student plans to climb 5 times next month. Which plan costs less, and how much money does the student save? c) Starting at how many visits per month is the Summit plan less expensive overall? Briefly explain.

Hints

- Write an equation for the monthly cost of each gym. - What equation represents the point at which both plans cost the same amount? - For 5 visits, substitute \(5\) into both cost equations and compare the results.

Solution

1. Let \(x\) be the number of visits. The monthly cost functions are \(S(x) = 10 + 5x\) and \(P(x) = 7.50x\). 2. For part a, set the costs equal: \(10 + 5x = 7.50x\). Then \(2.50x = 10\), so \(x = 4\). Both plans cost \(\$30\). 3. For part b, \(S(5) = 10 + 5 \cdot 5 = 35\) and \(P(5) = 7.50 \cdot 5 = 37.50\). The Summit plan saves \(\$37.50 - \$35 = \$2.50\). 4. For part c, the plans have equal cost at 4 visits. Because Summit has the lower cost per additional visit, Summit is less expensive for every whole-number visit count greater than 4. Therefore, it is less expensive starting at 5 visits.

Answer

a) \(4\) visits; both plans cost \(\$30\). b) Summit Climbing Gym costs \(\$35\), compared with \(\$37.50\) at Peak Climbing Gym, so the student saves \(\$2.50\). c) Starting at \(5\) visits per month, the Summit plan is less expensive.

5152328

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.

5154828

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.

5223688

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.

5227128

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) \(32\,\text{min}\)

5231218

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.

5231228

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.

5237868

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

5239968

Two water tanks begin draining at the same time. Tank 1 starts with \(500\,\text{gal}\) and loses \(10\,\text{gal}\) per minute. Tank 2 starts with \(800\,\text{gal}\) and loses \(30\,\text{gal}\) per minute. After how many minutes will Tank 1 contain exactly twice as much water as Tank 2?

Hints

- Draining decreases each amount, so use subtraction in both expressions. - The entire Tank 2 expression must be multiplied by \(2\). - You can test your equation with a specific time to check that the expressions behave as expected.

Solution

1. After \(t\) minutes, Tank 1 contains \(500 - 10t\), and Tank 2 contains \(800 - 30t\). 2. Write the condition \(500 - 10t = 2(800 - 30t)\). 3. Distribute: \(500 - 10t = 1600 - 60t\). 4. Add \(60t\) and subtract \(500\): \(50t = 1100\). 5. Divide by \(50\): \(t = 22\).

Answer

Tank 1 will contain twice as much water as Tank 2 after \(22\) minutes.

5241368

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 a) Find the number of minutes for which the two plans have the same monthly cost. b) Which plan costs less for \(150\) minutes? Justify your answer.

Hints

- Write a cost rule for each plan. - Equal costs are represented by an equation. - For part b), substitute \(150\) into both rules and compare.

Solution

1. For \(x\) minutes, Plan A costs \(10 + 0.15x\), and Plan B costs \(0.25x\). 2. Set the costs equal: \(10 + 0.15x = 0.25x\). 3. Subtract \(0.15x\): \(10 = 0.10x\). 4. Divide by \(0.10\): \(x = 100\). 5. At \(150\) minutes, Plan A costs \(10 + 0.15 \cdot 150 = 32.50\) dollars. 6. Plan B costs \(0.25 \cdot 150 = 37.50\) dollars, so Plan A is less expensive.

Answer

a) The plans cost the same at \(100\) minutes. b) Plan A is less expensive at \(150\) minutes: \(\$32.50\) compared with \(\$37.50\).

5241818

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.

5241828

Towns \(A\) and \(B\) are \(300\) miles apart along a straight rail line. A freight train leaves Town \(A\) traveling toward Town \(B\) at a constant speed of \(60\) miles per hour. At the same time, a passenger train leaves Town \(B\) traveling toward Town \(A\) at \(90\) miles per hour. a) Write a function for each train's distance \(y\), in miles, from Town \(A\) after \(x\) hours. Identify two points on each function. b) Compare the trains' distances at whole-number times to estimate when and where they meet. c) Calculate the exact meeting time and distance to check your estimate.

Hints

- If a train is moving toward Town \(A\), what sign should its rate of change have when distance is measured from Town \(A\)? - What is the passenger train's starting distance from Town \(A\)? - At what whole-number time do the two functions have the same value? - Set the two distance expressions equal to check the estimate algebraically.

Solution

1. Let \(x\) be time in hours and \(y\) be distance from Town \(A\) in miles. The freight train is modeled by \(F(x) = 60x\). The passenger train starts \(300\) miles from Town \(A\) and moves toward it, so \(P(x) = 300 - 90x\). 2. The freight-train function includes \((0, 0)\) and \((1, 60)\). The passenger-train function includes \((0, 300)\) and \((1, 210)\). 3. At \(x=1\), the trains are at distances \(60\) miles and \(210\) miles from Town \(A\). At \(x=2\), both distances are \(120\) miles, so the comparison indicates that they meet after about \(2\) hours, \(120\) miles from Town \(A\). 4. To find the exact intersection, set the functions equal: \(60x = 300 - 90x\). 5. Then \(150x = 300\), so \(x = 2\) hours. The distance from Town \(A\) is \(F(2) = 60 \cdot 2 = 120\) miles.

Answer

a) \(F(x) = 60x\) and \(P(x) = 300 - 90x\). Two points on \(F\) are \((0, 0)\) and \((1, 60)\); two points on \(P\) are \((0, 300)\) and \((1, 210)\). b) Comparing whole-number times gives an estimated meeting point of \((2, 120)\). c) The exact values are \(2\) hours and \(120\) miles from Town \(A\).

5241938

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

5241988

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

5242028

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

5244118

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.

5262188

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) \(x=14\)

5268508

A smartphone battery drains at a constant rate while the phone is in use. At \(10{:}00\) a.m., the battery level is \(80\%\). At \(11{:}20\) a.m., it is \(60\%\). a) Find the battery loss in percentage points per hour. b) Assuming the phone has been used continuously at this rate, at what time was it unplugged at \(100\%\)? c) If the same rate continues, at what time will the battery reach \(0\%\)?

Hints

- How many minutes pass between the two battery readings? - How many percentage points does the battery lose during that interval? - Find the loss per minute first, then convert it to a per-hour rate. - How long would a \(20\)-percentage-point drop take at that rate? - How long would a \(100\)-percentage-point drop take at the same rate?

Solution

1. From \(10{:}00\) a.m. to \(11{:}20\) a.m. is \(80\) minutes. The battery level decreases by \(20\) percentage points, so the rate is \(20 \div 80 = 0.25\) percentage point per minute, or \(0.25 \cdot 60 = 15\) percentage points per hour. 2. Going from \(100\%\) to \(80\%\) is a decrease of \(20\) percentage points, which takes \(20 \div 0.25 = 80\) minutes. Eighty minutes before \(10{:}00\) a.m. is \(8{:}40\) a.m. 3. Going from \(100\%\) to \(0\%\) takes \(100 \div 0.25 = 400\) minutes, or \(6\) hours \(40\) minutes. Starting at \(8{:}40\) a.m., the battery reaches \(0\%\) at \(3{:}20\) p.m.

Answer

a) \(15\) percentage points per hour b) \(8{:}40\) a.m. c) \(3{:}20\) p.m.

5268688

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. Find the time when the two vehicles meet and the distance of the meeting point from Town A.

Hints

- Picture the highway as a number line measured from Town A. Where is the car when it starts? - How does the car's distance from Town A change as it drives toward Town A? - Convert the \(45\)-minute delay to hours before writing the car's equation. - What must be true about the two position expressions at the meeting time?

Solution

1. Let \(t\) be the number of hours after \(12{:}00\) p.m., and let \(s\) be the distance from Town A in miles. 2. The truck's position is \(s = 60t\). The car starts \(0.75\) hour later, so for \(t \ge 0.75\), its position is \(s = 180 - 90(t - 0.75)\). 3. Set the positions equal: \(60t = 180 - 90(t - 0.75)\). 4. Simplifying gives \(60t = 247.5 - 90t\), so \(150t = 247.5\) and \(t = 1.65\). 5. Since \(0.65\) hour is \(39\) minutes, \(1.65\) hours after noon is \(1{:}39\) p.m. 6. The meeting point is \(60 \cdot 1.65 = 99\) miles from Town A.

Answer

The vehicles meet at \(1{:}39\) p.m., \(99\) miles from Town A.

5322158

Maya and Noah ride bicycles along the same route. Maya leaves the starting point at \(10{:}00\) a.m. Noah leaves the same point later but rides faster. The graph shows each rider's distance from the starting point as a function of time. a) Who rides faster? Find each rider's speed in miles per hour. b) Use the graph to determine when Noah catches Maya and how far they have traveled by then. c) Write an equation for each rider's distance \(y\), in miles, as a function of clock time \(x\), measured in hours. Solve the system to find the intersection, and compare it with your answer to part b).

Figure for problem 532215

Hints

- Find each speed by dividing distance traveled by elapsed time. - The intersection represents the time and distance at which the riders are together. - Use each rider's starting time and constant speed to write a linear equation. - Set the two distance expressions equal to solve for the meeting time.

Solution

1. Maya travels \(8\) miles from \(10{:}00\) a.m. to \(11{:}00\) a.m., so her speed is \(8\,\text{mph}\). Noah travels \(12\) miles from \(10{:}30\) a.m. to \(11{:}30\) a.m., so his speed is \(12\,\text{mph}\). Noah rides faster. 2. The graphs intersect at \((11.5, 12)\). Therefore, Noah catches Maya at \(11{:}30\) a.m., after each rider has traveled \(12\) miles. 3. Maya's equation is \(f(x)=8(x-10)=8x-80\) for \(x\ge10\). Noah's equation is \(g(x)=12(x-10.5)=12x-126\) for \(x\ge10.5\). Set the expressions equal: \(8x-80=12x-126\), so \(4x=46\) and \(x=11.5\). Then \(y=8(11.5-10)=12\). This matches the graph.

Answer

a) Noah; Maya rides at \(8\,\text{mph}\), and Noah rides at \(12\,\text{mph}\). b) \(11{:}30\) a.m.; \(12\) miles c) Maya: \(f(x)=8x-80\), \(x\ge10\) Noah: \(g(x)=12x-126\), \(x\ge10.5\) The intersection is \((11.5, 12)\).

5332028

A water tank is filled at a constant rate. The function \(V(t)\) gives the volume in gallons after \(t\) minutes. Two measurements, \(P_1\) and \(P_2\), are shown on the graph. a) Find the inflow rate, or slope, in gallons per minute. b) Find the volume already in the tank at \(t=0\). c) Write the function \(V(t)\). d) Find the volume after \(15\) minutes.

Figure for problem 533202

Hints

- Use change in volume divided by change in time to find the slope. - Substitute one point into \(V(t)=mt+b\) to find the initial value. - Evaluate the completed function at \(t=15\).

Solution

1. The points are \(P_1(2, 100)\) and \(P_2(8, 250)\). The slope is \(m=\frac{250-100}{8-2}=\frac{150}{6}=25\,\text{gal/min}\). 2. Substitute \(P_1\) into \(V(t)=25t+b\): \(100=25\cdot2+b\), so \(b=50\) gallons. 3. The function is \(V(t)=25t+50\). 4. \(V(15)=25\cdot15+50=425\), so the tank contains \(425\) gallons.

Answer

a) \(25\,\text{gal/min}\) b) \(50\) gallons c) \(V(t)=25t+50\) d) \(425\) gallons

5332138

Two cell phone plans have different monthly fees and different per-minute charges. Let \(x\) be the number of minutes used in a month. Use the graphs to write a cost function for Plan A and a cost function for Plan B, including the contextual domain. What does the intersection of the two lines mean in context?

Figure for problem 533213

Hints

- The y-intercept represents the monthly fee. - The slope represents the cost per minute. - The number of minutes used cannot be negative. - At an intersection, both plans have the same cost for the same number of minutes.

Solution

1. Plan A has a y-intercept of \(5\), representing a \(\$5\) monthly fee. Using the point \((100, 15)\), its slope is \(m=\frac{15-5}{100-0}=0.10\). Thus, \(A(x)=0.10x+5\), with \(x\ge0\). 2. Plan B has a y-intercept of \(10\), representing a \(\$10\) monthly fee. Using \((100, 15)\), its slope is \(m=\frac{15-10}{100-0}=0.05\). Thus, \(B(x)=0.05x+10\), with \(x\ge0\). 3. The lines intersect at \((100, 15)\). This means both plans cost \(\$15\) when \(100\) minutes are used.

Answer

Plan A: \(A(x)=0.10x+5\), \(x\ge0\) Plan B: \(B(x)=0.05x+10\), \(x\ge0\) At \(100\) minutes, both plans cost \(\$15\).

5332288

Liam leaves a trailhead at \(9{:}00\) a.m. and walks at a constant speed of \(4\) miles per hour. At \(10{:}30\) a.m., his sister Marie realizes that he forgot his lunch. She immediately rides her bike after him along the same route at \(12\) miles per hour. The time-distance graph shows each person's distance from the trailhead as a function of time \(t\), in hours since \(9{:}00\) a.m. The graph uses \(l\) for Liam and \(m\) for Marie. a) Use the graph to determine when and how far from the trailhead Marie catches Liam. b) Calculate the meeting time and distance exactly. What clock time do they meet?

Figure for problem 533228

Hints

- What does the intersection of the two lines represent in this situation? - How can Marie's \(1.5\)-hour delayed start be represented in her function? - Read the intersection coordinates from the graph, then check them using equations. - Convert the decimal part of the time in hours to minutes.

Solution

1. Liam's distance is \(l(t) = 4t\). Marie starts \(1.5\) hours later, so her distance is \(m(t) = 12(t - 1.5)\) for \(t \ge 1.5\). 2. The intersection of the graph represents the time and distance at which Marie catches Liam. It is at about \((2.25, 9)\). 3. Set the functions equal: \(4t = 12(t - 1.5)\). 4. Then \(4t = 12t - 18\), so \(8t = 18\) and \(t = 2.25\). 5. The distance is \(4 \cdot 2.25 = 9\) miles. Since \(0.25\) hour is \(15\) minutes, \(2.25\) hours after \(9{:}00\) a.m. is \(11{:}15\) a.m.

Answer

a) The graph shows that Marie catches Liam about \(2.25\) hours after Liam starts, about \(9\) miles from the trailhead. b) The exact meeting point is \((2.25, 9)\), so they meet at \(11{:}15\) a.m., \(9\) miles from the trailhead.

5332298

Two cargo boats, the Albatross and the Osprey, are traveling along a \(100\)-mile route between two harbors. The Albatross leaves Harbor A and travels toward Harbor B at a constant speed of \(20\) miles per hour. At the same time, the Osprey leaves Harbor B and travels directly toward Harbor A at \(30\) miles per hour. The graph shows each boat's distance from Harbor A as a function of travel time \(t\). The graph uses \(a\) for the Albatross and \(o\) for the Osprey. a) Use the graph to determine when the boats meet and how far the meeting point is from Harbor A. b) Confirm your result algebraically.

Figure for problem 533229

Hints

- When two objects move toward each other, how does the distance between them change? - What is the Osprey's starting distance from Harbor A? - What does a negative slope mean when the y-value is distance from Harbor A? - The intersection gives both the meeting time and the meeting distance.

Solution

1. The Albatross starts at Harbor A, so its distance from Harbor A is \(a(t) = 20t\). The Osprey starts \(100\) miles from Harbor A and moves toward it, so \(o(t) = 100 - 30t\). 2. The intersection on the graph is at about \((2, 40)\), so the boats appear to meet after \(2\) hours, \(40\) miles from Harbor A. 3. Set the functions equal: \(20t = 100 - 30t\). 4. Then \(50t = 100\), so \(t = 2\). 5. Substituting gives \(a(2) = 20 \cdot 2 = 40\). Therefore, the boats meet after \(2\) hours, \(40\) miles from Harbor A.

Answer

a) The graph shows that the boats meet after about \(2\) hours, about \(40\) miles from Harbor A. b) Algebra confirms the exact meeting time is \(2\) hours and the meeting point is \(40\) miles from Harbor A.

5120448

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.

5120668

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.

5125708

A large tank contains \(450\,\text{L}\) of water and loses \(15\,\text{L}\) per minute through a leak. At the same time, a second tank, initially empty, is filled at \(25\,\text{L}\) per minute. a) Find the time when the two tanks contain exactly the same amount of water. b) How much water is in each tank at that time? c) Explain without calculating why the second tank will contain more water after that time.

Hints

- Write one linear expression for each tank, paying attention to whether water is added or removed. - If the time includes a fractional minute, how can you convert the fractional part to seconds? - Imagine graphing both amounts. What happens after the two lines intersect? - Compare the directions in which the two amounts change.

Solution

1. After \(t\) minutes, the first tank contains \(450 - 15t\) liters and the second contains \(25t\) liters. 2. Set the amounts equal: \(450 - 15t = 25t\). 3. Add \(15t\) to both sides: \(450 = 40t\). Divide by \(40\): \(t = 11.25\). 4. Substitute \(t = 11.25\): \(25 \cdot 11.25 = 281.25\). Each tank contains \(281.25\,\text{L}\). 5. After the amounts are equal, the first amount continues to decrease while the second continues to increase, so the second amount stays greater.

Answer

a) The amounts are equal after \(11.25\) minutes, or 11 minutes 15 seconds. b) Each tank contains \(281.25\,\text{L}\). c) The first amount decreases while the second increases, so the second tank remains ahead after they are equal.

5125858

Two candles are lit at the same time. Candle A is \(20\,\text{cm}\) tall and burns down \(1.5\,\text{cm}\) per hour. Candle B is \(15\,\text{cm}\) tall and burns down \(0.5\,\text{cm}\) per hour. a) After how many hours will the candles have the same height? b) What will their height be at that time? c) Which candle will be taller after 7 hours? Justify your answer.

Hints

- Write a linear expression for each candle’s height after \(t\) hours. - Burning makes each candle shorter, so which operation should you use? - Once you find the time when the heights match, substitute it into either expression. - Compare the burn rates to predict what happens after the heights are equal.

Solution

1. After \(t\) hours, Candle A has height \(20 - 1.5t\), and Candle B has height \(15 - 0.5t\). 2. Set the heights equal: \(20 - 1.5t = 15 - 0.5t\). 3. Add \(1.5t\) to both sides: \(20 = 15 + t\). Subtract \(15\): \(t = 5\). 4. Substitute \(t = 5\): \(20 - 1.5 \cdot 5 = 12.5\). Both candles are \(12.5\,\text{cm}\) tall. 5. After 7 hours, Candle A is \(20 - 1.5 \cdot 7 = 9.5\,\text{cm}\), and Candle B is \(15 - 0.5 \cdot 7 = 11.5\,\text{cm}\). Candle B is taller.

Answer

a) The candles have the same height after 5 hours. b) Each candle is \(12.5\,\text{cm}\) tall then. c) Candle B is taller after 7 hours: \(11.5\,\text{cm}\) compared with \(9.5\,\text{cm}\) for Candle A.

5129788

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.

5129938

A rectangular water tank holds at most \(375\) gallons. It already contains \(50\) gallons when filling begins. An inlet adds \(30\) gallons in \(8\) minutes at a constant rate. a) Find the inflow rate in gallons per minute and write a function \(V(t)\) for the volume, in gallons, after \(t\) minutes. b) How long will it take to fill the tank completely? Give the time in hours, minutes, and seconds. c) Suppose a blockage cuts the inflow rate in half. Describe how the slope of the graph and the time needed to fill the tank would change.

Hints

- Divide the amount added by the elapsed time to find the rate per minute. - How many more gallons are needed to reach capacity? - In this context, what does the slope represent? - For the same amount of water, what happens to the required time when the rate is cut in half?

Solution

1. The inflow rate is \(30 \div 8 = 3.75\) gallons per minute. With an initial volume of \(50\) gallons, \(V(t) = 3.75t + 50\). 2. The tank is full when \(V(t) = 375\): \(375 = 3.75t + 50\). Thus \(325 = 3.75t\), so \(t = 86\frac{2}{3}\) minutes. 3. Since \(86\frac{2}{3}\) minutes is \(1\) hour plus \(26\frac{2}{3}\) minutes, the fill time is \(1\) hour, \(26\) minutes, and \(40\) seconds. 4. Halving the inflow rate halves the slope to \(1.875\) gallons per minute. The amount still needed is unchanged, so the fill time doubles to \(173\frac{1}{3}\) minutes.

Answer

a) The inflow rate is \(3.75\) gallons per minute, and \(V(t) = 3.75t + 50\). b) \(1\) hour, \(26\) minutes, and \(40\) seconds c) The slope is halved, and the fill time doubles.

5137798

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.

5142088

At the beginning of an observation, a smartphone battery is at \(10\%\). It is connected to a fast charger and reaches \(90\%\) after \(40\) minutes. The charger is then removed, and a \(20\)-minute video call lowers the battery level to \(60\%\). a) Write a function for battery level \(y\), in percent, as a function of elapsed time \(x\), in minutes, for each phase. b) Compare the charging rate with the magnitude of the discharging rate. By how many percentage points per minute do they differ? c) At what elapsed times was the battery level exactly \(75\%\)? d) If the phone continued to discharge at the same rate after the call, how many additional minutes would it take to reach \(0\%\)?

Hints

- Find the slope of each segment from its two endpoints. - Keep the sign of each slope separate from its magnitude. - Check both phases for a battery level of \(75\%\). - Divide the remaining battery level by the discharge rate.

Solution

1. The key points are \((0, 10)\), \((40, 90)\), and \((60, 60)\). 2. During charging, the slope is \(\frac{90-10}{40}=2\), so \(y=2x+10\) for \(0\le x\le40\). 3. During use, the slope is \(\frac{60-90}{60-40}=-1.5\), so \(y=90-1.5(x-40)=-1.5x+150\) for \(40\le x\le60\). 4. The charging rate is \(2\) percentage points per minute, and the discharging-rate magnitude is \(1.5\). They differ by \(0.5\) percentage point per minute. 5. During charging, solve \(75=2x+10\) to get \(x=32.5\). During use, solve \(75=-1.5x+150\) to get \(x=50\). 6. From \(60\%\), discharging at \(1.5\) points per minute takes \(60\div1.5=40\) additional minutes.

Answer

a) Charging: \(y=2x+10\), \(0\le x\le40\); use: \(y=-1.5x+150\), \(40\le x\le60\) b) \(0.5\) percentage point per minute c) At \(x=32.5\) minutes and \(x=50\) minutes d) \(40\) additional minutes

All problems may be used, copied and printed free of charge for school and tutoring, including paid tutoring. Commercial adaptations as well as publication or redistribution on the internet are not permitted.