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Linear vs exponential comparison

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5283079
Examine the two tables. For each table, decide whether the pattern is linear or exponential, and fill in the missing values. a) <table> <tr> <td>Time \(t\)</td> <td>\(0\)</td> <td>\(1\)</td> <td>\(2\)</td> <td>\(3\)</td> <td>\(4\)</td> </tr> <tr> <td>Amount \(A\)</td> <td>\(200\)</td> <td>\(240\)</td> <td></td> <td>\(345.6\)</td> <td></td> </tr> </table> b) <table> <tr> <td>Time \(t\)</td> <td>\(0\)</td> <td>\(1\)</td> <td>\(2\)</td> <td>\(3\)</td> <td>\(4\)</td> </tr> <tr> <td>Amount \(B\)</td> <td>\(200\)</td> <td>\(240\)</td> <td></td> <td></td> <td>\(360\)</td> </tr> </table>

Hints

- Test whether consecutive differences are constant. - Test whether consecutive ratios are constant. - Decide whether addition or multiplication takes one value to the next. - Use the known value at the end of each table to check the pattern.

Solution

1. In table a, \(\frac{240}{200}=1.2\). Also, \(240\cdot(1.2)^2=345.6\), so the constant growth factor is \(1.2\), and the pattern is exponential. 2. The missing values in table a are \(240\cdot1.2=288\) at \(t=2\) and \(345.6\cdot1.2=414.72\) at \(t=4\). 3. In table b, \(240-200=40\), and \(200+4\cdot40=360\). The constant difference is \(40\), so the pattern is linear. 4. The missing values in table b are \(240+40=280\) at \(t=2\) and \(280+40=320\) at \(t=3\).

Answer

a) Exponential; \(288\) at \(t=2\) and \(414.72\) at \(t=4\) b) Linear; \(280\) at \(t=2\) and \(320\) at \(t=3\)
5283089
Compare two savings models. Both begin with \(\$500\) at \(t=0\). Model 1 increases by \(\$25\) each year. Model 2 increases by \(4\%\) each year. Identify the type of growth in each model, and find the balance after \(5\) years.

Hints

- Distinguish between increasing by a fixed amount and increasing by a fixed percent. - A constant absolute change produces what type of model? - A change based on the current value produces what type of model? - Convert the percent increase to a growth factor.

Solution

1. Model 1 has a constant dollar increase, so it is linear. Its balance after \(5\) years is \(500+5\cdot25=625\). 2. Model 2 has a constant percent increase, so it is exponential with growth factor \(1.04\). Its balance after \(5\) years is \(500\cdot(1.04)^5\approx608.33\).

Answer

Model 1: linear growth; \(\$625.00\) after \(5\) years Model 2: exponential growth; \(\$608.33\) after \(5\) years
5288759
The table shows the number of views of a viral video during its first three weeks online. <table> <tr> <td>Week \(n\)</td> <td>0</td> <td>1</td> <td>2</td> <td>3</td> </tr> <tr> <td>Views \(A_n\)</td> <td>\(200\)</td> <td>\(300\)</td> <td>\(450\)</td> <td>\(675\)</td> </tr> </table> 1. Calculate the ratios of consecutive values to show that the data follow a recursive model of the form \(A_{n+1}=a\cdot A_n\), and identify \(a\). 2. An alternative linear model has the form \(L_{n+1}=L_n+d\). Use weeks \(0\) and \(1\) to find \(d\). 3. Use the linear model to predict week \(3\), and find its percent error relative to the actual week-3 value.

Hints

- Compare consecutive values using both division and subtraction. - A linear model has a constant additive change. - A growth factor is applied by multiplication. - Percent error compares the absolute error with the actual value.

Solution

1. The consecutive ratios are \(\frac{300}{200}=1.5\), \(\frac{450}{300}=1.5\), and \(\frac{675}{450}=1.5\). Therefore, the recursive exponential model has \(a=1.5\). 2. From weeks \(0\) and \(1\), \(d=300-200=100\). Thus, \(L_{n+1}=L_n+100\). 3. The linear prediction is \(L_3=200+3\cdot100=500\). The absolute error is \(675-500=175\), so the percent error is \(\frac{175}{675}\cdot100\%\approx25.9\%\).

Answer

1. \(a=1.5\) 2. \(d=100\) 3. \(L_3=500\); percent error \(\approx25.9\%\)
5335039
The graph shows two lines and two exponential curves labeled \(p\), \(q\), \(r\), and \(s\), all passing through the point \(P(0, 1)\). Match each equation to its graph. Use the characteristics of linear and exponential growth. (a) \(y=1.5x+1\) (b) \(y=0.5x+1\) (c) \(y=(1.2)^x\) (d) \(y=2^x\)
Figure for problem 533503

Hints

- How can you distinguish a line from an exponential curve? - Compare the slopes of the two lines. - Compare how quickly the two curves become steeper. - Consider which functions grow fastest for large positive \(x\).

Solution

1. Graphs \(p\) and \(r\) are lines, so they represent the linear functions. Graphs \(q\) and \(s\) are curved, so they represent the exponential functions. 2. Graph \(p\) is steeper than graph \(r\). Since \(1.5>0.5\), graph \(p\) matches (a), and graph \(r\) matches (b). 3. Graph \(q\) grows faster than graph \(s\). Since \(2>1.2\), graph \(q\) matches (d), and graph \(s\) matches (c).

Answer

\(p\) \(\rightarrow\) (a) \(q\) \(\rightarrow\) (d) \(r\) \(\rightarrow\) (b) \(s\) \(\rightarrow\) (c)
5127879
Mr. Meyer invests \(\$2500\) for three years and compares two plans: Plan 1: The account pays \(3\%\) interest each year, and the interest is withdrawn at the end of each year. Plan 2: The account pays \(3\%\) interest each year, and the interest remains in the account and earns interest in later years. Find the total interest earned under each plan after three years. How much greater is the interest under one plan? Round the Plan 2 amount and the difference to the nearest cent.

Hints

- Decide whether the amount earning interest stays constant or changes each year. - Under Plan 2, use each year’s ending balance as the next year’s beginning balance. - Subtract the two total interest amounts to find the difference.

Solution

1. Under Plan 1, the yearly interest is \(\$2500 \cdot 0.03 = \$75\). Over three years, the total interest is \(\$75 \cdot 3 = \$225\). 2. Under Plan 2, the ending balance is \(\$2500(1.03)^3 = \$2731.8175 \approx \$2731.82\). 3. The total interest under Plan 2 is \(\$2731.82 - \$2500 = \$231.82\). 4. The difference is \(\$231.82 - \$225 = \$6.82\).

Answer

Plan 1 earns \(\$225\) in total interest. Plan 2 earns approximately \(\$231.82\) in total interest. Plan 2 earns \(\$6.82\) more.
5258449
The value of a collectible car increases from \(\$20{,}000\) to \(\$50{,}000\) over \(10\) years. a) Find the car's value after exactly \(5\) years if its value grows by a constant annual percent. b) Find the car's value after exactly \(5\) years if its value increases by a constant dollar amount each year. c) Compare the results. Which model gives the greater value halfway through the \(10\)-year period? Explain why a constant-percent model is often used for investments.

Hints

- Distinguish between adding the same dollar amount each year and multiplying by the same growth factor each year. - What is the linear model's value halfway between the starting and ending values? - For the exponential model, use the relationship between the \(10\)-year growth factor and the \(5\)-year growth factor. - Which type of average corresponds to each model?

Solution

1. For the exponential model, let \(V(t)=20{,}000b^t\). Since \(V(10)=50{,}000\), \(b^{10}=\frac{50{,}000}{20{,}000}=2.5\). Thus \(b^5=\sqrt{2.5}\), and \(V(5)=20{,}000\sqrt{2.5}\approx31{,}622.78\). 2. For the linear model, the total increase is \(\$30{,}000\) over \(10\) years, or \(\$3000\) per year. Therefore, \(L(5)=20{,}000+5\cdot3000=35{,}000\). 3. Since \(35{,}000>31{,}622.78\), the linear model gives the greater midpoint value. A constant-percent model is often used when each period's change is based on the current value, as with compound growth.

Answer

a) \(\$31{,}622.78\) b) \(\$35{,}000.00\) c) The linear model gives the greater value after \(5\) years. A constant-percent model represents compound growth because each increase is based on the current value.
5259269
An experiment records \(x_0=100\), \(x_1=150\), and \(x_2=225\) at times \(t=0\), \(t=1\), and \(t=2\) hours. a) Show numerically that the data follow an exponential pattern. Why do they not follow a linear pattern? b) What would \(x_2\) be if the change were linear while \(x_0\) and \(x_1\) stayed the same? c) Find the predicted value after \(5\) hours for both the linear and exponential models.

Hints

- Compare consecutive differences and consecutive ratios. - A linear pattern has a constant difference; an exponential pattern has a constant ratio. - Use \(x_n=x_0+nd\) for the linear model and \(x_n=x_0r^n\) for the exponential model.

Solution

1. The consecutive ratios are \(\frac{x_1}{x_0}=\frac{150}{100}=1.5\) and \(\frac{x_2}{x_1}=\frac{225}{150}=1.5\). The constant ratio shows exponential growth. The consecutive differences are \(150-100=50\) and \(225-150=75\), so the pattern is not linear. 2. A linear model based on \(x_0\) and \(x_1\) has common difference \(50\). Therefore, its next value would be \(x_2=150+50=200\). 3. For the linear model, \(x_5=100+5\cdot50=350\). For the exponential model, \(x_5=100\cdot(1.5)^5=759.375\).

Answer

a) The ratios are both \(1.5\), while the differences are \(50\) and \(75\). Therefore, the pattern is exponential, not linear. b) \(200\) c) Linear: \(350\); exponential: \(759.375\)
5281519
Two investment accounts begin with the same balance of \(\$800\). Account \(K_1\) grows linearly, and account \(K_2\) grows exponentially. After \(2\) years, both accounts have a balance of \(\$1800\). a) Write a function for each account's balance after \(t\) years. b) Determine which account has the greater balance after \(1\) year, and find the difference between the balances.

Hints

- Start with the general forms of a linear model and an exponential model. - Use the value after \(2\) years to determine each model's unknown rate. - Evaluate both functions at \(t=1\).

Solution

1. For the linear account, write \(K_1(t)=800+dt\). Since \(K_1(2)=1800\), \(800+2d=1800\), so \(d=500\). Thus \(K_1(t)=800+500t\). 2. For the exponential account, write \(K_2(t)=800b^t\). Since \(K_2(2)=1800\), \(800b^2=1800\), so \(b^2=2.25\) and \(b=1.5\). Thus \(K_2(t)=800(1.5)^t\). 3. After \(1\) year, \(K_1(1)=1300\) and \(K_2(1)=1200\). The linear account is greater by \(1300-1200=100\).

Answer

a) \(K_1(t)=800+500t\); \(K_2(t)=800(1.5)^t\) b) Account \(K_1\) has the greater balance after \(1\) year, by \(\$100\).
5283019
Compare two savings plans. Plan A starts with \(\$2000\) and increases by \(\$150\) each year. Plan B starts with \(\$1500\) and earns \(5\%\) interest each year, compounded annually. a) Write a function for the balance in each plan after \(t\) years. Identify each model as linear or exponential. b) Find the balance in each plan after \(10\) years. c) By testing whole-number values of \(t\) or making a table, determine the first full year when Plan B has a greater balance than Plan A.

Hints

- Decide which plan adds a fixed amount and which multiplies by a fixed growth factor. - Convert a \(5\%\) annual increase to a growth factor. - Continue a table of whole-year balances until the inequality reverses.

Solution

1. Plan A has a constant yearly increase, so \(A(t)=2000+150t\) is linear. Plan B has a yearly growth factor of \(1.05\), so \(B(t)=1500(1.05)^t\) is exponential. 2. After \(10\) years, \(A(10)=2000+150\cdot10=3500\), and \(B(10)=1500\cdot(1.05)^{10}\approx2443.34\). 3. Testing whole-number years from \(t=0\) upward shows that Plan A is at least as large through \(t=30\). At \(t=30\), \(A(30)=6500\) and \(B(30)\approx6482.91\), so Plan A is still greater. At \(t=31\), \(A(31)=6650\) and \(B(31)\approx6807.06\), so Plan B is greater for the first time.

Answer

a) \(A(t)=2000+150t\), linear; \(B(t)=1500(1.05)^t\), exponential b) Plan A: \(\$3500.00\); Plan B: \(\$2443.34\) c) After \(31\) years
5283029
Two plant cultures are grown under different conditions. Culture 1 initially covers \(500\,\text{cm}^2\), and its area increases by \(200\,\text{cm}^2\) each week. Culture 2 initially covers \(50\,\text{cm}^2\), and its area increases by \(50\%\) each week. a) Complete the table. Round Culture 2's areas to the nearest whole square centimeter. | Week \(t\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | | :--- | :--- | :--- | :--- | :--- | :--- | :--- | | Culture 1 in \(\text{cm}^2\) | \(500\) | | | | | | | Culture 2 in \(\text{cm}^2\) | \(50\) | | | | | | b) Use the changes in area to explain which culture grows linearly and which grows exponentially. c) After how many full weeks does Culture 2 first cover a greater area than Culture 1?

Hints

- Determine whether each week's change is a fixed amount or a fixed percent. - Write a rule for each row before filling the table. - Continue the models beyond week \(5\) until Culture 2 becomes greater.

Solution

1. Culture 1 follows \(A(t)=500+200t\), giving \(500, 700, 900, 1100, 1300, 1500\). 2. Culture 2 follows \(B(t)=50(1.5)^t\). Its rounded values are \(50, 75, 113, 169, 253, 380\). 3. Culture 1 is linear because its consecutive differences are always \(200\). Culture 2 is exponential because each exact value is \(1.5\) times the preceding value. 4. Continuing whole-number weeks beyond week \(5\) shows that Culture 1 remains greater through week \(9\). At week \(9\), \(A(9)=2300\) and \(B(9)\approx1922.17\). At week \(10\), \(A(10)=2500\) and \(B(10)\approx2883.25\). Therefore, Culture 2 first exceeds Culture 1 after \(10\) full weeks.

Answer

a) Culture 1: \(500, 700, 900, 1100, 1300, 1500\) Culture 2: \(50, 75, 113, 169, 253, 380\) b) Culture 1 is linear because it has a constant difference. Culture 2 is exponential because it has a constant growth factor. c) \(10\) weeks
5283119
You can choose between two pay plans for a \(14\)-day summer job. Plan 1: You earn \(\$30.00\) on the first day. Each following day's pay is \(\$15.00\) more than the previous day's pay. Plan 2: You earn only \(\$0.02\) on the first day. Each following day's pay is twice the previous day's pay. a) Find the amount you would earn on day \(14\) under each plan. Which plan pays more on that day? b) Write a function \(f(n)\) for the daily pay under each plan, where \(n\) is the day number. c) Identify the linear model and the exponential model. Briefly justify your choices.

Hints

- Decide whether the pay changes by adding the same amount or multiplying by the same factor. - List the first three days under each plan to identify the pattern. - Use an exponent to represent repeated multiplication. - Remember that no increase or doubling has occurred yet on day \(1\).

Solution

1. Plan 1 has a constant daily increase of \(15\), so \(f_1(n)=30+15(n-1)=15n+15\). On day \(14\), \(f_1(14)=225\). 2. Plan 2 has a constant growth factor of \(2\), so \(f_2(n)=0.02(2)^{n-1}\). On day \(14\), \(f_2(14)=0.02\cdot2^{13}=163.84\). 3. Since \(225>163.84\), Plan 1 pays more on day \(14\). Plan 1 is linear because it has a constant difference, and Plan 2 is exponential because it has a constant ratio.

Answer

a) Plan 1: \(\$225.00\); Plan 2: \(\$163.84\). Plan 1 pays more on day \(14\). b) \(f_1(n)=15n+15\); \(f_2(n)=0.02(2)^{n-1}\) c) Plan 1 is linear because the daily difference is constant. Plan 2 is exponential because the daily growth factor is constant.
5283129
Two newly formed clubs have different membership patterns. Club A starts with \(200\) members and gains \(40\) members each month. Club B starts with \(50\) members, and its membership increases by \(15\%\) each month. a) Write a function \(M(t)\) for each club's membership \(t\) months after the clubs are formed. b) Find each club's membership after \(12\) months and after \(24\) months. Round Club B's membership to the nearest whole number. c) Explain which club has linear growth and which has exponential growth.

Hints

- Convert the \(15\%\) increase to a growth factor. - Distinguish between a fixed number of new members and a fixed percent increase. - Substitute each time value into the appropriate function.

Solution

1. Club A gains a constant number of members, so \(M_A(t)=200+40t\). 2. Club B has a monthly growth factor of \(1.15\), so \(M_B(t)=50(1.15)^t\). 3. After \(12\) months, \(M_A(12)=680\), and \(M_B(12)=50\cdot(1.15)^{12}\approx267.51\), or about \(268\) members. 4. After \(24\) months, \(M_A(24)=1160\), and \(M_B(24)=50\cdot(1.15)^{24}\approx1431.26\), or about \(1431\) members. 5. Club A is linear because its absolute monthly change is constant. Club B is exponential because its monthly percent change is constant.

Answer

a) \(M_A(t)=200+40t\); \(M_B(t)=50(1.15)^t\) b) After \(12\) months: Club A, \(680\); Club B, about \(268\). After \(24\) months: Club A, \(1160\); Club B, about \(1431\). c) Club A has linear growth, and Club B has exponential growth.
5283219
An initial investment of \(\$10{,}000\) is held for several years. Compare the two plans. Plan A: A fixed \(\$320\) is added at the end of each year. The added amounts do not earn additional interest. Plan B: The account earns \(3\%\) interest per year, compounded annually. a) Find the balance in each plan after \(1\), \(2\), and \(5\) years. b) Identify the linear model and the exponential model. Justify your answer using the yearly increases. c) Compare the balances after \(20\) years. Which plan gives the greater balance?

Hints

- Decide whether each plan adds the same dollar amount or applies a percent to the current balance. - Write a direct formula for the balance after \(t\) years. - Compare the yearly differences in the two models. - Consider what happens when the amount earning interest becomes larger each year.

Solution

1. Plan A follows \(A(t)=10{,}000+320t\). Its balances are \(A(1)=10{,}320\), \(A(2)=10{,}640\), and \(A(5)=11{,}600\). 2. Plan B follows \(B(t)=10{,}000(1.03)^t\). Its balances are \(B(1)=10{,}300\), \(B(2)=10{,}609\), and \(B(5)\approx11{,}592.74\). 3. Plan A is linear because the yearly increase is always \(\$320\). Plan B is exponential because the balance is multiplied by \(1.03\) each year, so its dollar increases become larger. 4. After \(20\) years, \(A(20)=16{,}400\), and \(B(20)=10{,}000\cdot(1.03)^{20}\approx18{,}061.11\). Plan B gives the greater balance.

Answer

a) Plan A: \(\$10{,}320.00\), \(\$10{,}640.00\), and \(\$11{,}600.00\). Plan B: \(\$10{,}300.00\), \(\$10{,}609.00\), and \(\$11{,}592.74\). b) Plan A is linear; Plan B is exponential. c) After \(20\) years, Plan A has \(\$16{,}400.00\), and Plan B has about \(\$18{,}061.11\). Plan B gives the greater balance.
5283229
Two biological cultures are observed under different conditions. At \(t=0\), each culture contains \(800\) organisms. Culture 1 increases each hour by \(15\%\) of its population from the previous hour. Culture 2 gains \(150\) organisms each hour. a) Write a function \(N(t)\) for each culture's population after \(t\) hours. b) Find both populations after \(3\) hours and after \(8\) hours. Round nonwhole populations to the nearest whole organism. c) Explain how the hourly increases in the two cultures change over time. Why will Culture 1 eventually exceed Culture 2, even though it grows more slowly at first?

Hints

- Determine whether each culture adds a fixed amount or multiplies by a fixed factor. - Use the general forms \(y=mx+b\) and \(y=ab^x\). - Compare the absolute increase from one hour to the next in each model. - Use the values at \(3\) and \(8\) hours to observe how the comparison changes.

Solution

1. Culture 1 has exponential growth, so \(N_1(t)=800(1.15)^t\). Culture 2 has linear growth, so \(N_2(t)=800+150t\). 2. After \(3\) hours, \(N_1(3)=1216.7\approx1217\), and \(N_2(3)=1250\). 3. After \(8\) hours, \(N_1(8)\approx2447.22\approx2447\), and \(N_2(8)=2000\). 4. Culture 2 always gains \(150\) organisms per hour. Culture 1 gains \(15\%\) of an increasing population, so its absolute hourly increase becomes larger over time. A growing exponential model with factor greater than \(1\) eventually exceeds a linear model.

Answer

a) \(N_1(t)=800(1.15)^t\); \(N_2(t)=800+150t\) b) After \(3\) hours: Culture 1, about \(1217\); Culture 2, \(1250\). After \(8\) hours: Culture 1, about \(2447\); Culture 2, \(2000\). c) Culture 2 has a constant hourly increase. Culture 1's absolute hourly increase grows because \(15\%\) is taken from an increasingly large population.
5283519
An initial investment of \(\$12{,}000\) is placed in two different accounts. Account 1 has balances of \(\$13{,}440.00\) after \(2\) years, \(\$14{,}880.00\) after \(4\) years, and \(\$16{,}320.00\) after \(6\) years. Account 2 has balances of \(\$13{,}483.20\) after \(2\) years, \(\$15{,}149.72\) after \(4\) years, and \(\$17{,}022.23\) after \(6\) years. a) Determine whether each account follows a linear or exponential model. b) Find the yearly dollar increase for the linear model and the annual interest rate for the exponential model. c) Find the balance in each account after \(10\) years.

Hints

- Check whether equal time intervals produce equal differences or equal ratios. - Find the differences between consecutive balances. - Find the ratios between consecutive balances. - The data are given in \(2\)-year intervals, but part b asks for yearly growth.

Solution

1. For Account 1, the balance increases by \(1440\) every \(2\) years: \(13{,}440-12{,}000=1440\), \(14{,}880-13{,}440=1440\), and \(16{,}320-14{,}880=1440\). Therefore, it is linear, with a yearly increase of \(1440\div2=720\). 2. For Account 2, the consecutive \(2\)-year growth factors are \(\frac{13{,}483.20}{12{,}000}=1.1236\), \(\frac{15{,}149.72}{13{,}483.20}\approx1.1236\), and \(\frac{17{,}022.23}{15{,}149.72}\approx1.1236\). Thus the model is exponential. The yearly growth factor satisfies \(b^2=1.1236\), so \(b=1.06\), giving an annual interest rate of \(6\%\). 3. After \(10\) years, Account 1 has \(12{,}000+10\cdot720=19{,}200\). Account 2 has \(12{,}000\cdot(1.06)^{10}\approx21{,}490.17\).

Answer

a) Account 1 is linear; Account 2 is exponential. b) Account 1 increases by \(\$720.00\) per year. Account 2 earns \(6\%\) per year. c) Account 1: \(\$19{,}200.00\); Account 2: about \(\$21{,}490.17\)
5283659
Two candles, Candle A and Candle B, each begin with a height of \(25\,\text{cm}\). Candle A burns at a constant rate, so its height decreases by \(2\,\text{cm}\) each hour. Candle B's height decreases by \(10\%\) of its current height each hour. a) Find the height of each candle after \(5\) hours. b) Determine when Candle A reaches a height of \(0\,\text{cm}\). c) Identify the linear decay model and the exponential decay model. Justify your answer using the rates of change.

Hints

- Decide whether each candle loses a fixed height or a fixed percent of its current height. - What factor represents a \(10\%\) decrease? - Which model has a decreasing absolute loss over time? - Write a function for each candle.

Solution

1. Candle A follows \(h_A(t)=25-2t\). After \(5\) hours, \(h_A(5)=25-2\cdot5=15\,\text{cm}\). 2. Candle B has decay factor \(1-0.10=0.9\), so \(h_B(t)=25(0.9)^t\). After \(5\) hours, \(h_B(5)=25\cdot(0.9)^5\approx14.76\,\text{cm}\). 3. Candle A reaches \(0\,\text{cm}\) when \(25-2t=0\). Solving gives \(t=12.5\) hours. 4. Candle A has linear decay because it loses a constant \(2\,\text{cm}\) per hour. Candle B has exponential decay because it retains the constant factor \(0.9\) each hour.

Answer

a) Candle A: \(15\,\text{cm}\); Candle B: about \(14.76\,\text{cm}\) b) \(12.5\) hours c) Candle A has linear decay because its hourly decrease is constant. Candle B has exponential decay because its hourly decay factor is constant.
5334999
Examine the three graphs labeled \(p\), \(q\), and \(r\). One graph represents a linear function, and the other two represent exponential functions. a) Which graph is linear? Find its equation. b) Match the exponential graphs to the functions \(f(x)=0.5(1.4)^x\) and \(g(x)=2(0.7)^x\). c) Use an example from the graph to explain the difference between linear decay and exponential decay.
Figure for problem 533499

Hints

- What geometric shape is the graph of a linear function? - Compare the y-intercepts with the initial values in the exponential functions. - Consider how each decreasing graph behaves as \(x\) becomes large.

Solution

1. Graph \(q\) is a line. It has y-intercept \(3\) and x-intercept \(6\), so its slope is \(\frac{0-3}{6-0}=-0.5\). Its equation is \(y=-0.5x+3\). 2. Graph \(p\) shows exponential decay and has y-intercept \(2\), so it matches \(g(x)=2(0.7)^x\). Graph \(r\) shows exponential growth and has y-intercept \(0.5\), so it matches \(f(x)=0.5(1.4)^x\). 3. The linear graph decreases by the same amount, \(0.5\), for each increase of \(1\) in \(x\). The exponential-decay graph decreases by the same percent, so it falls quickly at first and then levels off toward the x-axis.

Answer

a) Graph \(q\) is linear, with equation \(y=-0.5x+3\). b) Graph \(p\): \(g(x)=2(0.7)^x\); Graph \(r\): \(f(x)=0.5(1.4)^x\) c) Linear decay has a constant difference and appears as a line. Exponential decay has a constant ratio and approaches the x-axis without reaching it.
5350099
Let \(f(x)=2^x\). The three panels show the graphs of linear functions \(g_a\), \(g_b\), and \(g_c\) together with the graph of \(f\). Which linear graph intersects the graph of \(f\) at more than one point?
Figure for problem 535009

Hints

- Compare the exponential and linear function values at simple inputs. - A change in which graph is higher between two inputs indicates an intersection between them. - Check whether this change occurs in more than one interval.

Solution

1. In panel a), \(g_a(x)=x+2\). The graphs intersect at \(x=2\), since both values are \(4\). Also, at \(x=-2\), \(f(-2)=0.25>0=g_a(-2)\), while at \(x=0\), \(f(0)=1<2=g_a(0)\). By continuity, there is another intersection between \(-2\) and \(0\). 2. In panel b), \(g_b(x)=-x+1\). The functions intersect at \(x=0\), and the opposite monotonic behavior prevents a second intersection. 3. In panel c), \(g_c(x)=0.5x\). For \(x\le0\), \(g_c(x)\le0<f(x)\). For positive \(x\), the exponential graph remains above the line. Thus there is no intersection. 4. Therefore, only graph a) intersects \(f\) more than once.

Answer

Graph a)
5350109
The graph shows two models, \(f\) in blue and \(g\) in red. One model is linear, and the other is exponential. a) Which graph represents each type of growth? b) For what interval with \(x>0\) does the linear model have greater values than the exponential model?
Figure for problem 535010

Hints

- Which graph is a line, and which graph is curved? - Locate the points where the graphs intersect. - Between the intersection points, compare which graph lies higher.

Solution

1. The red graph \(g\) is a line, so it represents linear growth. The blue graph \(f\) is curved and becomes steeper, so it represents exponential growth. 2. The functions intersect at \(x=0\), because \(f(0)=3^0=1\) and \(g(0)=2\cdot0+1=1\). They also intersect at \(x=1\), because \(f(1)=3\) and \(g(1)=3\). 3. Between the intersection points, the line is above the exponential curve. For example, at \(x=0.5\), \(g(0.5)=2\), while \(f(0.5)=\sqrt{3}\approx1.732\). 4. Therefore, the linear model has greater values for \(0<x<1\).

Answer

a) The red graph \(g\) is linear, and the blue graph \(f\) is exponential. b) \((0, 1)\)
5127949
Two savers each begin with \(\$2000\) and keep the money invested for four years. Strategy A: The balance earns \(3\%\) per year, and the interest remains in the account and earns interest in later years. Strategy B: The original principal earns \(x\%\) per year, but the interest is withdrawn at the end of each year and does not earn additional interest. What annual rate \(x\) would make the total amount of money after four years exactly the same under both strategies? Round the rate to the nearest hundredth of a percent.

Hints

- Find the ending balance and total interest under Strategy A. - Under Strategy B, the same dollar amount of interest is paid each year. - Set the four-year interest earned under the two strategies equal.

Solution

1. Under Strategy A, the ending balance is \(\$2000(1.03)^4 = \$2251.01762\). 2. The total interest under Strategy A is \(\$2251.01762 - \$2000 = \$251.01762\). 3. Under Strategy B, the four equal annual interest payments total \(4\left(\$2000 \cdot \frac{x}{100}\right) = \$80x\). 4. Set the interest totals equal: \(80x = 251.01762\). 5. Solving gives \(x = \frac{251.01762}{80} = 3.13772025\), which rounds to \(3.14\%\).

Answer

Strategy B would need an annual interest rate of approximately \(3.14\%\).
5281529
Two microorganism populations are observed in a laboratory. At \(t=0\), each population contains \(100\) organisms. Population \(A\) grows linearly, and population \(B\) grows exponentially. After \(1\) hour, population \(B\) has exactly \(50\) more organisms than population \(A\). After \(2\) hours, population \(B\) is exactly twice the size of population \(A\). Write a function for each population, and find the size of each population after \(2\) hours.

Hints

- Write one equation from the relationship between the populations at each stated time. - Express the linear rate in terms of the exponential growth factor, then substitute. - In the second relationship, multiply the entire expression for population \(A\) by \(2\). - Factor the resulting quadratic equation.

Solution

1. Let \(A(t)=100+dt\) and \(B(t)=100r^t\). 2. The condition at \(t=1\) gives \(100r=100+d+50\), so \(d=100r-150\). 3. The condition at \(t=2\) gives \(100r^2=2(100+2d)\). Substituting \(d=100r-150\) gives \(100r^2=200+4(100r-150)\), so \(r^2-4r+4=0\). 4. Since \((r-2)^2=0\), \(r=2\). Then \(d=100\cdot2-150=50\). 5. The models are \(A(t)=100+50t\) and \(B(t)=100(2)^t\). Therefore, \(A(2)=200\) and \(B(2)=400\).

Answer

The functions are \(A(t)=100+50t\) and \(B(t)=100(2)^t\). After \(2\) hours, population \(A\) has \(200\) organisms and population \(B\) has \(400\) organisms.

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