- What is the value when \(t=0\)? - What growth factor represents a \(2\%\) increase? - Is a constant percent increase linear or exponential?

1. The initial value is \(N_0=400\). 2. A \(2\%\) increase gives a growth factor of \(1+\frac{2}{100}=1.02\). 3. Substituting into \(N(t)=N_0b^t\) gives \(N(t)=400(1.02)^t\), which is choice c).

c) \(N=400(1.02)^t\)

- Interpret the factor as the fraction of the previous amount that remains. - Compare the remaining percent with \(100\%\).

1. The factor \(0.92\) means \(92\%\) of the previous value remains. 2. Therefore, \(100\%-92\%=8\%\) is lost each time \(t\) increases by \(1\).

\(8\%\) decrease

- A growth factor includes the original \(100\%\) plus the increase. - Convert the percent to a decimal before adding.

1. Convert \(7\%\) to \(0.07\). 2. A \(7\%\) increase gives factor \(1+0.07=1.07\).

\(1.07\)

- Find the percent that remains after the decrease. - Write the remaining percent as a decimal.

1. A \(15\%\) decrease leaves \(85\%\) of the previous amount. 2. As a decimal, \(85\%=0.85\), so the decay factor is \(0.85\).

\(0.85\)

- Compare the multiplication factor with \(1\). - The difference from \(1\) can be written as a percentage.

1. The factor \(1.08\) is \(0.08\) greater than \(1\). 2. This means an \(8\%\) increase per week, so the audience is growing.

An \(8\%\) increase per week; the audience grows.

- Compare each after-amount with its own before-amount. - Calculate the mass lost from each tray. How could you compare these losses fairly?

1. Each after-to-before ratio is \(0.9\), so \(90\%\) remains. 2. Drying removes \(10\%\) of the mass from each tray. 3. The masses lost are \(2\), \(6\), and \(9\) grams; each is one-tenth of that tray's starting mass.

Each tray loses \(10\%\). The amounts lost are \(2\), \(6\), and \(9\) grams because the starting masses differ.

- How much water is in the tank before the pump runs? - Think about what fraction remains, not what fraction disappears. - Each new hour starts with the water left from the previous hour.

1. The starting amount is \(80\) gallons. 2. If one-fourth is removed, three-fourths remains each hour, giving a factor of \(0.75\). 3. The model is \(W(n)=80(0.75)^n\).

\(W(n)=80(0.75)^n\)

- Does the second increase apply to the original value or the new value? - A total multiplier combines the two step multipliers.

1. Each increase multiplies the current value by \(1.10\). 2. After both changes, the factor is \(1.10\cdot1.10=1.21\). 3. The overall increase is \(21\%\), not \(20\%\).

Total factor \(1.21\), for a \(21\%\) overall increase.

- Compare the first-year interest with the original balance. - Add the first-year interest to find the second year’s beginning balance. - Apply the same rate to the new balance.

1. For part a, the rate is \(\frac{18}{1200} = 0.015 = 1.5\%\). 2. The balance at the beginning of the second year is \(\$1200 + \$18 = \$1218\). 3. For part b, the second-year interest is \(\$1218 \cdot 0.015 = \$18.27\).

a) The annual interest rate is \(1.5\%\). b) The second-year interest is \(\$18.27\).

- Use the first-year interest and principal to find the rate. - For part b, the rate stays the same while the principal changes. - For part c, find the balance at the beginning of the second year.

1. For part a, the rate is \(\frac{75}{3000} = 0.025 = 2.5\%\). 2. For part b, the interest is \(\$4500 \cdot 0.025 = \$112.50\). 3. For part c, the balance at the beginning of the second year is \(\$3000 + \$75 = \$3075\). 4. The second-year interest is \(\$3075 \cdot 0.02 = \$61.50\).

a) The annual interest rate is \(2.5\%\). b) The interest would be \(\$112.50\). c) The second-year interest would be \(\$61.50\).

- What fraction of the machine's value remains after one year? - How would the same percentage change be applied again in the second year? - Put the starting value and yearly factor into an exponential model.

1. The starting value is \(\$45{,}000\). 2. A decrease of \(12\%\) leaves \(88\%\) of the previous value each year, giving a factor of \(0.88\). 3. The model is \(V(t)=45{,}000(0.88)^t\).

\(V(t)=45{,}000(0.88)^t\)

- Identify the initial value. - What percent of the value remains after an \(18\%\) decrease? - Convert that percent to a decimal factor. - Use the general form of an exponential function.

1. The initial value is \(45{,}000\). 2. After an \(18\%\) decrease, \(82\%\) of the previous value remains, so the yearly decay factor is \(0.82\). 3. Therefore, the model is \(V(t)=45{,}000(0.82)^t\).

\(V(t)=45{,}000(0.82)^t\)

- How is a percent increase converted to a growth factor? - How many times is the annual factor applied over several years? - A total factor compares the future value directly with the initial value.

1. The annual growth factor is \(b=1+0.024=1.024\). 2. After \(3\) years, the area is \(4500\cdot(1.024)^3\approx4831.84\,\text{ha}\). 3. After \(10\) years, the area is \(4500\cdot(1.024)^{10}\approx5704.43\,\text{ha}\). 4. The total factor after \(5\) years is \((1.024)^5\approx1.1259\).

a) \(b=1.024\) b) After \(3\) years: \(4831.84\,\text{ha}\); after \(10\) years: \(5704.43\,\text{ha}\) c) \(1.1259\)

- Use the population at \(t=0\) as the starting value. - Find the factor that takes the starting count to the next hour's count. - Convert that factor into a percent increase before using it for six hours.

1. From the first hour, \(1040=800a\), so \(a=\frac{1040}{800}=1.3\). 2. The factor \(1.3=1+0.30\) means the population increases by \(30\%\) per hour. 3. After six hours, \(N(6)=800(1.3)^6\approx3861.45\), or about \(3861\) bacteria.

a) \(a=1.3\) b) \(30\%\) c) About \(3861\) bacteria

- A growth factor for a percent increase is greater than \(1\). - First find how many years have passed since 2020. - Use an exponent of \(10\) for the 2030 prediction. - Round counts of plants only after completing the calculation.

1. The annual growth factor is \(b=1+0.054=1.054\), so the model is \(P(t)=1200(1.054)^t\), where \(t\) is years after 2020. 2. \(P(1)=1264.8\approx1265\), \(P(2)\approx1333.10\approx1333\), and \(P(3)\approx1405.09\approx1405\). 3. For 2030, \(t=10\). Then \(P(10)\approx2030.43\), or about \(2030\) plants. The \(10\)-year factor is \((1.054)^{10}\approx1.6920\). 4. For 2045, \(t=25\). Then \(P(25)\approx4468.86\), or about \(4469\) plants.

a) \(b=1.054\) b) 2021: about \(1265\); 2022: about \(1333\); 2023: about \(1405\) c) 2030: about \(2030\) plants; \(10\)-year factor: \(1.6920\) d) About \(4469\) plants

- What factor represents a decrease of \(6.2\%\)? - Identify the value at \(t=0\). - The power of the decay factor gives the fraction remaining.

1. The annual decay factor is \(1-0.062=0.938\). 2. With initial volume \(8400\,\text{m}^3\), the model is \(V(t)=8400(0.938)^t\). 3. The fraction remaining after \(8\) years is \((0.938)^8\approx0.5993\), so about \(59.93\%\) remains.

a) \(V(t)=8400(0.938)^t\) b) \(59.93\%\)

- Convert the percent increase to a growth factor. - Use the general form of an exponential function. - Substitute the given time into your function.

1. A \(22\%\) increase gives a weekly growth factor of \(1.22\). 2. The model is \(A(t)=0.5(1.22)^t\). 3. After \(5\) weeks, \(A(5)=0.5\cdot(1.22)^5\approx1.3514\), so the covered area is about \(1.35\,\text{m}^2\).

a) \(A(t)=0.5(1.22)^t\) b) About \(1.35\,\text{m}^2\)

- What must be true of consecutive ratios for an exponential model? - How is a growth factor related to a percent increase? - Use the exponential model to predict a value beyond the table.

1. The consecutive ratios are \(\frac{600}{400}=1.5\), \(\frac{900}{600}=1.5\), and \(\frac{1350}{900}=1.5\). Since the ratio is constant, the data follow an exponential model. 2. The hourly growth factor is \(r=1.5\). Because \(1.5-1=0.5\), the hourly percent increase is \(50\%\). 3. The model is \(N(t)=400(1.5)^t\). Thus \(N(6)=400\cdot(1.5)^6=4556.25\), which rounds to about \(4556\) people.

a) Yes. Each consecutive ratio is \(1.5\). b) \(r=1.5\); \(50\%\) increase per hour c) About \(4556\) people

- A factor above \(1\) represents growth; a factor below \(1\) represents decay. - The initial value is the coefficient outside the exponential power. - Use \(a=1\pm\frac{p}{100}\).

1. In row 1, a \(35\%\) increase gives \(a=1.35\), so \(f(x)=120(1.35)^x\). 2. In row 2, \(a=0.82\) means \(82\%\) remains, so the change is an \(18\%\) decrease. The function is \(f(x)=50(0.82)^x\). 3. In row 3, the function shows \(C=8\) and \(a=1.045\). This factor represents a \(4.5\%\) increase.

Row 1: \(a=1.35\); \(f(x)=120(1.35)^x\) Row 2: \(18\%\) decrease; \(f(x)=50(0.82)^x\) Row 3: \(4.5\%\) increase; \(a=1.045\); \(C=8\)

- The initial value is the function value at \(x=0\). - Compare each exponential factor with \(1\). - Evaluate the exponent before multiplying.

1. For \(f\), the initial value is \(400\). The factor \(1.015\) represents a \(1.5\%\) increase. 2. For \(g\), the initial value is \(25\). The factor \(0.6\) represents a \(40\%\) decrease. 3. For \(h\), the initial value is \(2\). The factor \(3\) represents a \(200\%\) increase. 4. \(f(2)=400\cdot(1.015)^2=412.09\).

a) \(f\): initial value \(400\), \(1.5\%\) increase; \(g\): initial value \(25\), \(40\%\) decrease; \(h\): initial value \(2\), \(200\%\) increase b) \(f(2)=412.09\)

- Identify the initial population. - Convert the percent decrease to a decay factor. - Raise the factor to the number of elapsed years.

1. The yearly decay factor is \(1-0.06=0.94\), so \(N(t)=450(0.94)^t\). 2. After \(5\) years, \(N(5)=450\cdot(0.94)^5\approx330.26\), or about \(330\) plants. 3. After \(10\) years, \(N(10)=450\cdot(0.94)^{10}\approx242.38\), or about \(242\) plants.

a) \(N(t)=450(0.94)^t\) b) After \(5\) years: about \(330\) plants; after \(10\) years: about \(242\) plants

- Find two plotted values exactly one hour apart. - Compare the later value with the earlier one using division. - How does the fraction remaining relate to the fraction lost?

1. The graph starts at \(c=20\,\text{mg/L}\) and shows \(16\,\text{mg/L}\) after one hour. 2. Thus \(a=\frac{16}{20}=0.8\). 3. Since \(80\%\) remains each hour, the concentration decreases by \(20\%\) per hour.

a) \(a=0.8\) b) A \(20\%\) decrease per hour

- Convert the percent decrease to the fraction that remains after one wash. - The given value is the result after the same factor has been applied three times. - Work backward from the final value rather than subtracting \(20\%\) three times.

1. A \(20\%\) decrease gives a remaining factor of \(0.8\) per wash. 2. If \(I_0\) is the initial index, then \(I_0(0.8)^3=51.2\). 3. Since \((0.8)^3=0.512\), \(I_0=\frac{51.2}{0.512}=100\).

\(100\)

- Which price does the store use when taking the second percentage? - Compare the combined multiplier with \(1\). - You could test your conclusion using any convenient starting price.

1. An increase of \(20\%\) multiplies by \(1.2\); a decrease of \(20\%\) multiplies by \(0.8\). 2. Their combined factor is \(1.2\cdot0.8=0.96\). 3. The final value is \(96\%\) of the original, a net \(4\%\) decrease.

No. The combined factor is \(0.96\), so the overall change is a \(4\%\) decrease.

- Read each amount from the graph before calculating a percentage. - Find the amount lost over the first hour and over the full two hours. - Which starting and ending readings belong to each question?

1. The plotted amounts are \(80\), \(40\), and \(20\) at hours \(0\), \(1\), and \(2\). 2. The first loss is \(40\) out of \(80\), or \(50\%\). 3. By hour \(2\), the loss is \(60\) out of the original \(80\), or \(75\%\).

a) \(50\%\) b) \(75\%\)

- Identify the starting depth and what fraction survives each day. - The same percentage applies to the new depth, not to the original \(40\) inches. - Round only after making the repeated-change calculation.

1. The bank keeps \(85\%\) of its previous depth each day, giving a factor of \(0.85\). 2. The model is \(S(d)=40(0.85)^d\). 3. After three days, \(S(3)=40(0.85)^3=24.565\approx24.6\) inches.

\(S(d)=40(0.85)^d\); after \(3\) days, approximately \(24.6\,\text{in.}\)

- Find the fraction remaining after one filter pass. - Combine the two passes before working backward. - The final amount is a part of an unknown initial amount.

1. Each pass leaves \(80\%\), so two passes leave a factor of \((0.8)^2=0.64\). 2. If \(M\) is the starting mass, \(0.64M=64\,\text{mg}\). 3. Dividing gives \(M=100\,\text{mg}\).

\(100\,\text{mg}\)

- Find the first year's interest and add it to the starting balance. - Use the ending balance from one year as the beginning balance for the next year. - Organize the two options in separate tables. - Pay attention to the different second-year interest rates.

1. For Option A: <table> <tr> <th>Year</th> <th>Beginning balance</th> <th>Interest rate</th> <th>Interest</th> <th>Ending balance</th> </tr> <tr><td>\(1\)</td><td>\(\$250.00\)</td><td>\(3\%\)</td><td>\(\$7.50\)</td><td>\(\$257.50\)</td></tr> <tr><td>\(2\)</td><td>\(\$257.50\)</td><td>\(3\%\)</td><td>\(\$7.73\)</td><td>\(\$265.23\)</td></tr> </table> 2. For Option B: <table> <tr> <th>Year</th> <th>Beginning balance</th> <th>Interest rate</th> <th>Interest</th> <th>Ending balance</th> </tr> <tr><td>\(1\)</td><td>\(\$250.00\)</td><td>\(2\%\)</td><td>\(\$5.00\)</td><td>\(\$255.00\)</td></tr> <tr><td>\(2\)</td><td>\(\$255.00\)</td><td>\(4\%\)</td><td>\(\$10.20\)</td><td>\(\$265.20\)</td></tr> </table> 3. Option A ends with \(\$265.23\), while Option B ends with \(\$265.20\). Therefore, Option A earns \(\$0.03\) more.

Option A: <table> <tr><th>Year</th><th>Beginning balance</th><th>Interest</th><th>Ending balance</th></tr> <tr><td>\(1\)</td><td>\(\$250.00\)</td><td>\(\$7.50\)</td><td>\(\$257.50\)</td></tr> <tr><td>\(2\)</td><td>\(\$257.50\)</td><td>\(\$7.73\)</td><td>\(\$265.23\)</td></tr> </table> Option B: <table> <tr><th>Year</th><th>Beginning balance</th><th>Interest</th><th>Ending balance</th></tr> <tr><td>\(1\)</td><td>\(\$250.00\)</td><td>\(\$5.00\)</td><td>\(\$255.00\)</td></tr> <tr><td>\(2\)</td><td>\(\$255.00\)</td><td>\(\$10.20\)</td><td>\(\$265.20\)</td></tr> </table> After two years, Option A has \(\$265.23\), Option B has \(\$265.20\), and Option A earns \(\$0.03\) more.

- Use columns for year, beginning balance, interest rate, interest, and ending balance. - The ending balance for one year becomes the beginning balance for the next year. - You can find the total interest by adding the yearly interest amounts or subtracting the original deposit from the final balance.

1. Compute each year's interest from that year's beginning balance: <table> <tr> <th>Year</th> <th>Beginning balance</th> <th>Interest rate</th> <th>Interest</th> <th>Ending balance</th> </tr> <tr><td>\(1\)</td><td>\(\$600.00\)</td><td>\(1.5\%\)</td><td>\(\$9.00\)</td><td>\(\$609.00\)</td></tr> <tr><td>\(2\)</td><td>\(\$609.00\)</td><td>\(2.5\%\)</td><td>\(\$15.23\)</td><td>\(\$624.23\)</td></tr> <tr><td>\(3\)</td><td>\(\$624.23\)</td><td>\(3.5\%\)</td><td>\(\$21.85\)</td><td>\(\$646.08\)</td></tr> </table> 2. The total interest is \(\$9.00+\$15.23+\$21.85=\$46.08\). This agrees with \(\$646.08-\$600.00=\$46.08\).

<table> <tr><th>Year</th><th>Beginning balance</th><th>Interest</th><th>Ending balance</th></tr> <tr><td>\(1\)</td><td>\(\$600.00\)</td><td>\(\$9.00\)</td><td>\(\$609.00\)</td></tr> <tr><td>\(2\)</td><td>\(\$609.00\)</td><td>\(\$15.23\)</td><td>\(\$624.23\)</td></tr> <tr><td>\(3\)</td><td>\(\$624.23\)</td><td>\(\$21.85\)</td><td>\(\$646.08\)</td></tr> </table> Ending balance: \(\$646.08\) Total interest: \(\$46.08\)

- Find the first year’s interest from the original deposit. - Use each year’s ending balance as the next year’s beginning balance. - Interest added to the account increases the amount that earns interest later. - Round money to the nearest cent at the end of each year.

1. Year 1 interest: \(\$4500.00 \cdot 0.02 = \$90.00\). The ending balance is \(\$4500.00 + \$90.00 = \$4590.00\). 2. Year 2 interest: \(\$4590.00 \cdot 0.02 = \$91.80\). The ending balance is \(\$4590.00 + \$91.80 = \$4681.80\). 3. Year 3 interest: \(\$4681.80 \cdot 0.02 = \$93.636 \approx \$93.64\). The ending balance is \(\$4681.80 + \$93.64 = \$4775.44\).

Year 1: Interest \(\$90.00\); balance \(\$4590.00\) Year 2: Interest \(\$91.80\); balance \(\$4681.80\) Year 3: Interest \(\$93.64\); balance \(\$4775.44\)

- 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.

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 \(\$231.82 - \$225 = \$6.82\).

Plan 1 earns \(\$225\) in total interest. Plan 2 earns approximately \(\$231.82\) in total interest. Plan 2 earns \(\$6.82\) more.

- Find how much interest Strategy A earns over all four years. - How much does Strategy B withdraw in one year at rate \(x\)? - Be sure to count cash withdrawn as well as the remaining principal.

1. Strategy A ends with \(2000(1.03)^4=2251.01762\) dollars, including \(251.01762\) dollars of interest. 2. With Strategy B, each annual withdrawal is \(2000\cdot\frac{x}{100}=20x\) dollars. 3. Four withdrawals total \(80x\) dollars. The account still holds \(2000\) dollars, so matching total money requires \(80x=251.01762\). 4. \(x=\frac{251.01762}{80}=3.13772025\), which rounds to \(3.14\%\).

\(x\approx3.14\%\) annually

- Use each year's ending balance as the next year's beginning balance. - Write each annual rate as a growth factor. - For part b, consider whether changing the order of factors changes a product.

1. For Option 1, the ending balance is \(\$4000(1.02)^3=\$4244.832\approx\$4244.83\). 2. For Option 2, the ending balance is \(\$4000\cdot1.01\cdot1.02\cdot1.03=\$4244.424\approx\$4244.42\). 3. Reversing the order of the rates in Option 2 would not change the product of the growth factors because multiplication is commutative. Therefore, the ending balance would be unchanged.

a) Option 1: \(\$4244.83\) Option 2: \(\$4244.42\) b) The ending balance would be the same because changing the order of the growth factors does not change their product.

- Compare the final population with the initial population to find the five-hour factor. - Think about how the hourly factor builds up over five equal intervals. - Convert the unrounded factor into a percent before rounding.

1. Using the observations, \(10{,}000=1200b^5\). 2. Thus \(b^5=\frac{10{,}000}{1200}=\frac{25}{3}\). 3. Taking the fifth root gives \(b=\left(\frac{25}{3}\right)^{1/5}\approx1.528\). 4. Using the unrounded factor, \((b-1)\cdot100\%\approx52.814\%\), or about \(52.8\%\) per hour.

a) \(b\approx1.528\) b) About \(52.8\%\) per hour

- Work out each balance after one year, then use it for the second year. - Compare the products of the two annual factors. - Is adding the annual rates the same as multiplying the factors?

1. Ava's balance is \(1000(1.04)^2=1081.60\) dollars. 2. Mateo's balance is \(1000\cdot1.02\cdot1.06=1081.20\) dollars. 3. Ava has \(\$0.40\) more. 4. Rates over successive years combine by multiplying their growth factors: \(1.04^2=1.0816\), while \(1.02\cdot1.06=1.0812\). Adding rates does not give the total change exactly.

Ava: \(\$1081.60\); Mateo: \(\$1081.20\). Ava has \(\$0.40\) more because the yearly growth factors are multiplied, not added.

- How many times is interest added in each plan over five years? - Find the factor for one interest period in each plan. - For the equivalent annual rate, combine two of Plan B's factors before converting to a percent.

1. Plan A has five annual increases: \(15{,}000(1.032)^5\approx\$17{,}558.59\). The interest is \(\$2558.59\). 2. Plan B has ten six-month increases: \(15{,}000(1.016)^{10}\approx\$17{,}580.38\). The interest is \(\$2580.38\). 3. Two six-month factors combine to give \((1.016)^2=1.032256\) in one year. The corresponding annual increase is \(3.2256\%\approx3.23\%\).

a) Balance \(\$17{,}558.59\); interest \(\$2558.59\) b) Balance \(\$17{,}580.38\); interest \(\$2580.38\) c) \(3.23\%\)

- What decay factor represents a \(12\%\) decrease? - What type of model represents a constant percent change? - Test consecutive full years near the threshold. - Identify the first year for which the value is below the limit.

1. The yearly decay factor is \(1-0.12=0.88\), so \(V(t)=54{,}000(0.88)^t\). 2. After \(5\) years, \(V(5)=54{,}000\cdot(0.88)^5\approx28{,}497.52\), which rounds to \(\$28{,}498\). 3. \(V(7)\approx22{,}068.48\), which is still above \(\$20{,}000\). \(V(8)\approx19{,}420.26\), which is below \(\$20{,}000\). Therefore, the value first falls below the threshold after \(8\) full years.

a) \(\$28{,}498\) b) \(8\) years

- Convert the percent increase to a growth factor. - Express a number of months as a fraction of a year. - The total growth factor depends on the elapsed time, not the initial value.

1. A \(6\%\) increase gives a yearly growth factor of \(1.06\), so \(W(t)=500(1.06)^t\). 2. The total growth factors are \((1.06)^2=1.1236\), \((1.06)^5\approx1.3382\), and \((1.06)^{10}\approx1.7908\). 3. Three months is \(\frac{3}{12}=0.25\) year, so \(W(0.25)=500\cdot(1.06)^{0.25}\approx507.34\). 4. After \(1.5\) years, \(W(1.5)=500\cdot(1.06)^{1.5}\approx545.67\).

a) \(W(t)=500(1.06)^t\). Factors: \(1.1236\) after \(2\) years, \(1.3382\) after \(5\) years, and \(1.7908\) after \(10\) years. b) After \(3\) months: \(\$507.34\); after \(1.5\) years: \(\$545.67\)

- A decay factor is less than \(1\). How do you find it from the percent decrease? - Raise the annual factor to the number of years. - First find the fraction of the original value that remains after \(3\) years.

1. The annual decay factor is \(b=1-0.15=0.85\). 2. After \(2\) years, the value is \(12{,}500\cdot(0.85)^2=\$9031.25\). 3. After \(5\) years, the value is \(12{,}500\cdot(0.85)^5\approx\$5546.32\). 4. After \(3\) years, the remaining-value factor is \((0.85)^3=0.614125\). Therefore, the total percent decrease is \((1-0.614125)\cdot100\%=38.5875\%\approx38.59\%\).

a) \(b=0.85\) b) After \(2\) years: \(\$9031.25\); after \(5\) years: \(\$5546.32\) c) \(38.59\%\)

- Divide each year's balance by the previous year's balance. - How does a growth factor greater than \(1\) relate to a percent increase? - Use compound growth to calculate many years at once.

1. The consecutive ratios are \(\frac{5125.00}{5000.00}=1.025\), \(\frac{5253.13}{5125.00}\approx1.025\), and \(\frac{5384.46}{5253.13}\approx1.025\). Thus the annual growth factor is \(b=1.025\). 2. Since \(1.025=1+0.025\), the annual interest rate is \(2.5\%\). 3. After \(10\) years, \(K_{10}=5000\cdot(1.025)^{10}\approx6400.42\).

a) Each ratio is approximately \(1.025\). b) \(2.5\%\) c) \(\$6400.42\)

- Decide whether the situation describes growth or decay. - Convert the percent decrease to a decimal factor. - Use a power for repeated yearly changes. - The multi-year factor is the annual factor raised to the number of years.

1. The annual decay factor is \(b=1-0.021=0.979\). 2. After \(4\) years, the annual use is \(850\cdot(0.979)^4\approx780.82\,\text{GWh}\). 3. The \(10\)-year factor is \((0.979)^{10}\approx0.8088\). Therefore, the annual use after \(10\) years is \(850\cdot(0.979)^{10}\approx687.46\,\text{GWh}\). 4. The total percent decrease is \((1-(0.979)^{10})\cdot100\%\approx19.12\%\).

a) \(b=0.979\) b) \(780.82\,\text{GWh}\) c) Factor: \(0.8088\); energy use: \(687.46\,\text{GWh}\) d) \(19.12\%\)

- Identify whether the unknown is the ending amount, starting amount, or percent rate. - Rearrange the same model for the requested unknown. - In part c), think about the one-year factor that produces the observed eight-year factor.

1. \(K_6=4000(1.035)^6\approx\$4917.02\). 2. \(K_0=\frac{8000}{(1.0225)^{10}}\approx\$6404.08\). 3. The growth factor in part c) satisfies \(b^8=\frac{4000}{2500}=1.6\), so \(b=(1.6)^{1/8}\approx1.0605106\). Thus \(p=100(b-1)\approx6.1\), giving a \(6.1\%\) annual rate.

a) \(K_6\approx\$4917.02\) b) \(K_0\approx\$6404.08\) c) \(p\approx6.1\), meaning \(6.1\%\) per year

- Convert the annual interest rate to a growth factor. - Separate the total balance from the increase during one specific year. - What total balance corresponds to a gain of more than \(\$500\)? - Check consecutive full years near that target.

1. The growth factor is \(1.0175\), so \(K_n=2500(1.0175)^n\). 2. \(K_5\approx2726.54\), and \(K_{10}\approx2973.61\). 3. The interest earned during year \(10\) is \(K_{10}-K_9\approx2973.61-2922.47=51.14\). 4. A gain greater than \(500\) requires \(K_n>3000\). Since \(K_{10}\approx2973.61<3000\) and \(K_{11}\approx3025.65>3000\), the account first exceeds the target after \(11\) full years.

a) After \(5\) years: \(\$2726.54\); after \(10\) years: \(\$2973.61\) b) \(\$51.14\) c) \(11\) years

- Write a compound-growth function for each account. - Compare the balances after \(5\) and \(15\) years. - A lower initial balance can eventually become greater if its growth factor is larger. - Check consecutive whole years near where the order changes.

1. The models are \(A(n)=4000(1.03)^n\) and \(B(n)=4500(1.02)^n\). 2. After \(5\) years, \(A(5)\approx4637.10\), and \(B(5)\approx4968.36\). 3. After \(15\) years, \(A(15)\approx6231.87\), and \(B(15)\approx6056.41\). Account A is greater. 4. Since \(\frac{A(n)}{B(n)}=\frac{4000}{4500}\left(\frac{1.03}{1.02}\right)^n\) increases with \(n\), it is enough to check consecutive years near the crossover. At \(n=12\), \(A(12)\approx5703.04<B(12)\approx5707.09\). At \(n=13\), \(A(13)\approx5874.13>B(13)\approx5821.23\). Therefore, Account A first becomes greater after \(13\) years.

a) Account A: \(\$4637.10\); Account B: \(\$4968.36\) b) Account A: \(\$6231.87\); Account B: \(\$6056.41\). Account A is greater. c) \(13\) years

- Convert each percent increase to a growth factor. - Use the exponential model to calculate each future value directly. - Round counts only after completing each calculation. - Subtract the two unrounded \(15\)-year predictions before rounding.

1. The growth factors are \(1.06\) and \(1.035\). 2. In the \(6\%\) scenario, the predictions are \(1200\cdot(1.06)^5\approx1606\), \(1200\cdot(1.06)^{10}\approx2149\), and \(1200\cdot(1.06)^{15}\approx2876\). 3. In the \(3.5\%\) scenario, the predictions are \(1200\cdot(1.035)^5\approx1425\), \(1200\cdot(1.035)^{10}\approx1693\), and \(1200\cdot(1.035)^{15}\approx2010\). 4. Using unrounded values, the difference after \(15\) years is \(1200\cdot(1.06)^{15}-1200\cdot(1.035)^{15}\approx865.45\), or about \(865\) birds.

a) About \(1606\), \(2149\), and \(2876\) birds b) About \(1425\), \(1693\), and \(2010\) birds c) About \(865\) birds

- Use the compound-interest model. - Consider how a small difference in annual rates accumulates over a long period. - Subtract the two \(20\)-year balances.

1. Bank A has growth factor \(1.032\). Its balances are \(30{,}000\cdot(1.032)^{10}\approx41{,}107.23\) and \(30{,}000\cdot(1.032)^{20}\approx56{,}326.82\). 2. Bank B has growth factor \(1.021\). Its balances are \(30{,}000\cdot(1.021)^{10}\approx36{,}929.95\) and \(30{,}000\cdot(1.021)^{20}\approx45{,}460.70\). 3. After \(20\) years, the difference is \(56{,}326.82-45{,}460.70=10{,}866.12\).

a) After \(10\) years: \(\$41{,}107.23\); after \(20\) years: \(\$56{,}326.82\) b) After \(10\) years: \(\$36{,}929.95\); after \(20\) years: \(\$45{,}460.70\) c) \(\$10{,}866.12\)

- What percent remains after removing \(12\%\)? - Each decrease is based on the newly reduced value. - Compare repeated exponential decay with simply multiplying \(12\%\) by the number of meters.

1. After a \(12\%\) decrease, \(88\%\) remains, so the decay factor is \(b=1-0.12=0.88\). 2. The percent intensity at depth \(d\) is \(I(d)=100(0.88)^d\). 3. \(I(2)=77.44\%\), \(I(5)\approx52.77\%\), and \(I(10)\approx27.85\%\). 4. The diver's claim is false because the \(12\%\) decrease is applied to the remaining intensity at each meter, not repeatedly to the original amount. At \(5\,\text{m}\), about \(52.77\%\) remains, so the total decrease is about \(47.23\%\), not \(60\%\).

a) \(b=0.88\); the factor is the fraction that remains after each \(12\%\) decrease. b) \(2\,\text{m}\): \(77.44\%\); \(5\,\text{m}\): \(52.77\%\); \(10\,\text{m}\): \(27.85\%\) c) The claim is false. The total decrease after \(5\,\text{m}\) is \(47.23\%\).

- Convert the percent decrease to a decay factor. - Repeated filtering raises the one-layer factor to a power. - For part c, test whole-number exponents until the remaining fraction is below \(0.10\).

1. The decay factor for one layer is \(b=1-0.18=0.82\). 2. After \(4\) layers, the remaining fraction is \((0.82)^4\approx0.4521\), so about \(45.2\%\) remains. 3. The factors are \(0.82\) for \(1\) layer, \((0.82)^2=0.6724\) for \(2\) layers, and \((0.82)^5\approx0.3707\) for \(5\) layers. 4. Test consecutive whole numbers near the threshold: \((0.82)^{11}\approx0.1127>0.10\), while \((0.82)^{12}\approx0.0924<0.10\). Therefore, at least \(12\) layers are required.

a) \(45.2\%\) b) \(1\) layer: \(0.82\); \(2\) layers: \(0.6724\); \(5\) layers: \(0.3707\) c) \(12\) layers

- Find the factor that remains after a \(20\%\) decrease. - Raise that factor to a power for a multi-hour interval. - One-fourth of the starting amount corresponds to a remaining fraction of \(0.25\). - Test consecutive whole hours near the threshold.

1. The hourly decay factor is \(0.8\), so \(A(t)=120(0.8)^t\). 2. After \(3\) hours, \(A(3)=120\cdot(0.8)^3=61.44\,\text{mg}\). 3. The \(4\)-hour decay factor is \((0.8)^4=0.4096\). 4. One-fourth of the initial amount is \(30\,\text{mg}\). Since \(A(6)\approx31.46\,\text{mg}>30\,\text{mg}\) and \(A(7)\approx25.17\,\text{mg}<30\,\text{mg}\), the amount is first below one-fourth after \(7\) full hours.

a) \(61.44\,\text{mg}\) b) \(0.4096\) c) \(7\) hours

- Convert the daily percent decrease to a factor less than \(1\). - Raise the daily factor to a power for longer intervals. - On day two, the \(12\%\) decrease is applied to the already reduced concentration.

1. The daily decay factor is \(b=1-0.12=0.88\). 2. After \(5\) days, the concentration is \(250\cdot(0.88)^5\approx131.93\,\text{mg/L}\). 3. The one-week decay factor is \((0.88)^7\approx0.4087\). 4. After two days, the remaining fraction is \((0.88)^2=0.7744\). Therefore, the total decrease is \(1-0.7744=0.2256\), or \(22.56\%\), not \(24\%\).

a) \(131.93\,\text{mg/L}\) b) Daily factor: \(0.88\); one-week factor: \(0.4087\) c) The actual two-day decrease is \(22.56\%\), so the claim is false.

- What factor represents retaining \(80\%\) of a height? - How many times is the factor applied after a given number of bounces? - How many centimeters are in one meter? - Substitute the given power estimate for part b.

1. The height factor is \(0.8\), and the initial height is \(2.50\,\text{m}=250\,\text{cm}\). 2. After the 3rd bounce, the height is \(2.50\cdot(0.8)^3=1.28\,\text{m}=128\,\text{cm}\). 3. After the 4th bounce, the height is \(2.50\cdot(0.8)^4=1.024\,\text{m}=102.4\,\text{cm}\). 4. After the 6th bounce, the height is \(2.50\cdot(0.8)^6=0.65536\,\text{m}=65.536\,\text{cm}\). 5. Using the given estimate, the height after the 10th bounce is approximately \(2.50\cdot0.1=0.25\,\text{m}=25\,\text{cm}\).

a) 3rd bounce: \(1.28\,\text{m}\) or \(128\,\text{cm}\); 4th bounce: \(1.024\,\text{m}\) or \(102.4\,\text{cm}\); 6th bounce: \(0.65536\,\text{m}\) or \(65.536\,\text{cm}\) b) About \(0.25\,\text{m}\), or \(25\,\text{cm}\)

- Relate the daily growth factor to a percent increase. - Over two days, apply the daily factor twice. - A half-day interval corresponds to an exponent of \(0.5\).

1. Over a time increase of \(h\) days, the growth factor is \((1.15)^h\), so the percent change is \(((1.15)^h-1)\cdot100\%\). 2. For \(h=1\), the factor is \(1.15\), giving a \(15\%\) increase. 3. For \(h=2\), the factor is \((1.15)^2=1.3225\), giving a \(32.25\%\approx32.3\%\) increase. 4. For \(h=0.5\), the factor is \(\sqrt{1.15}\approx1.07238\), giving an increase of about \(7.2\%\).

a) \(15\%\) increase b) About \(32.3\%\) increase c) About \(7.2\%\) increase

- Relate the multiplicative factor to the percent change. - A factor less than \(1\) represents a decrease. - Use negative and fractional exponents for the last two changes. - Start from \(g(x+h)=g(x)(0.8)^h\).

1. For a change of \(h\) in the input, the change factor is \((0.8)^h\), and the percent change is \(((0.8)^h-1)\cdot100\%\). 2. For \(h=1\), the factor is \(0.8\), so the value decreases by \(20\%\). 3. For \(h=2\), the factor is \(0.64\), so the value decreases by \(36\%\). 4. For \(h=-1\), the factor is \((0.8)^{-1}=1.25\), so the value increases by \(25\%\). 5. For \(h=0.5\), the factor is \(\sqrt{0.8}\approx0.8944\), so the value decreases by about \(10.6\%\).

(1) \(20\%\) decrease (2) \(36\%\) decrease (3) \(25\%\) increase (4) About \(10.6\%\) decrease

- Convert the annual percent decrease to a multiplication factor. - Use a power for repeated yearly decay. - Distinguish between the percent remaining and the percent lost. - Each year's decrease is based on the new, smaller population.

1. A \(10\%\) decrease gives a decay factor of \(b=0.9\). 2. After \(5\) years, the remaining factor is \((0.9)^5=0.59049\), so \(59.049\%\approx59.05\%\) remains. 3. After \(10\) years, the factor is \((0.9)^{10}=0.3486784401\approx0.3487\). 4. The loss after \(5\) years is \(100\%-59.049\%=40.951\%\approx40.95\%\). The loss after \(10\) years is \(100\%-34.86784401\%=65.13215599\%\approx65.13\%\). Since \(65.13\%\) is not twice \(40.95\%\), the claim is false.

1. \(b=0.9\) 2. \(59.05\%\) 3. \(0.3487\) 4. The claim is false. The losses are \(40.95\%\) after \(5\) years and \(65.13\%\) after \(10\) years.

- Use the exponent rule for adding numbers in an exponent. - Convert a multiplication factor to a percent change. - An exponent of \(0.5\) represents a square root. - Decreasing \(x\) corresponds to a negative input change.

1. If the input changes by \(k\), then \(f(x+k)=(1.44)^kf(x)\). The change factor is \((1.44)^k\). 2. For \(k=0.5\), the factor is \(\sqrt{1.44}=1.2\), so the value increases by \(20\%\). 3. For \(k=2\), the factor is \((1.44)^2=2.0736\), so the value increases by \(107.36\%\). 4. For \(k=-1\), the factor is \((1.44)^{-1}\approx0.6944\), so the value decreases by about \(30.6\%\).

a) \(20\%\) increase b) \(107.36\%\) increase c) About \(30.6\%\) decrease

- First find the initial temperature difference. - A \(15\%\) decrease leaves \(85\%\). - Add the current difference to the room temperature. - Consider what happens after the tea has remained in the room for a very long time.

1. The initial temperature difference is \(80-20=60\,^\circ\text{C}\). The difference is multiplied by \(0.85\) every \(5\) minutes. 2. After \(n\) intervals, the tea temperature is \(T_n=20+60(0.85)^n\). 3. \(T_1=71.0\,^\circ\text{C}\), \(T_2=63.35\,^\circ\text{C}\), \(T_3\approx56.85\,^\circ\text{C}\), and \(T_4\approx51.32\,^\circ\text{C}\). 4. The temperature is bounded below by the room temperature and approaches \(20\,^\circ\text{C}\). 5. The absolute drop becomes smaller because each \(15\%\) decrease is taken from a smaller temperature difference.

a) After \(5\) minutes: \(71.0\,^\circ\text{C}\); after \(10\) minutes: \(63.35\,^\circ\text{C}\); after \(15\) minutes: about \(56.85\,^\circ\text{C}\); after \(20\) minutes: about \(51.32\,^\circ\text{C}\) b) The temperature approaches the lower bound \(20\,^\circ\text{C}\). c) The absolute decrease becomes smaller each interval because the remaining temperature difference becomes smaller.

- A constant percent change suggests an exponential model. - Use the initial and final areas to find the yearly growth factor. - Convert the growth factor to a percent rate only after finding the factor.

1. Use \(A(n)=150q^n\). The data give \(215=150q^{12}\). 2. Thus, \(q=\left(\frac{215}{150}\right)^{1/12}\approx1.0304548\). 3. The annual percent increase is \((q-1)\cdot100\%\approx3.04548\%\approx3.05\%\).

About \(3.05\%\) per year

- Model the value with a constant yearly factor. - Use the initial and final values to solve for that factor. - A decay factor is less than \(1\); compare it with \(1\) to obtain the percent decrease.

1. Use \(V(n)=85{,}000q^n\). Then \(18{,}400=85{,}000q^7\). 2. Thus, \(q=\left(\frac{18{,}400}{85{,}000}\right)^{1/7}\approx0.8036316\). 3. The annual percent decrease is \((1-q)\cdot100\%\approx19.63684\%\approx19.64\%\).

About \(19.64\%\) per year

- Which graph stays constant, and what temperature does it show? - What quantity is decreasing by a percent: the tea's temperature or its difference from the room? - Convert the percent decrease into a factor for each minute. - Think about what happens to that difference over many minutes.

1. The room-temperature graph is at \(18\,\text{°C}\). The tea's curve starts at \(90\,\text{°C}\). 2. The initial difference is \(90-18=72\,\text{°C}\). A \(12\%\) decrease leaves a factor of \(0.88\), so \(T(x)=18+72(0.88)^x\). 3. \(T(5)=18+72(0.88)^5\approx56.00\,\text{°C}\), and \(T(10)=18+72(0.88)^{10}\approx38.05\,\text{°C}\). 4. As time passes, the positive difference \(72(0.88)^x\) gets smaller, so the tea approaches room temperature, \(18\,\text{°C}\).

a) Room: \(18\,\text{°C}\); tea initially: \(90\,\text{°C}\) b) \(T(x)=18+72(0.88)^x\) c) After \(5\) minutes: \(56.00\,\text{°C}\); after \(10\) minutes: \(38.05\,\text{°C}\) d) It approaches \(18\,\text{°C}\) as the temperature difference shrinks.

- For part a), apply the same growth factor once for each year. - For part b), find the balance after the first two years, then compare it with the goal. - For part c), first find the exact required rate, then check the two nearest tenth-percent rates to determine which one guarantees the goal.

1. For part a), the balance after three years is \(\$800.00(1.025)^3 = \$861.5125 \approx \$861.51\). Because \(\$861.51 > \$860.00\), the fixed rate is sufficient. 2. For part b), the balance after two years under the second account is \(\$800.00 \cdot 1.01 \cdot 1.02 = \$824.16\). The minimum third-year interest needed is \(\$860.00 - \$824.16 = \$35.84\). 3. For part c), the exact required rate is \(\frac{35.84}{824.16} \cdot 100\% \approx 4.3487\%\). A rate of \(4.3\%\) would give \(\$824.16 \cdot 1.043 \approx \$859.60\), which is below the goal. A rate of \(4.4\%\) gives \(\$824.16 \cdot 1.044 \approx \$860.42\), so the minimum rate in increments of \(0.1\%\) is \(4.4\%\).

a) Yes. The balance would be approximately \(\$861.51\). b) Mia must earn at least \(\$35.84\) in the third year. c) The minimum rate stated in increments of \(0.1\%\) is \(4.4\%\).

- The table uses two-year intervals, so convert the two-year factor to a one-year factor. - A negative time represents extrapolation into the past. - The percent remaining depends on the decay factor and elapsed time, not the initial value. - A depreciation model should give a value below the initial value for positive times.

1. Over two years, the value factor is \(\frac{768}{1200}=0.64\). Thus \(b^2=0.64\), so the annual factor is \(b=0.8\). 2. The initial value is \(1200\), so \(V(t)=1200(0.8)^t\). 3. \(V(6)=1200\cdot(0.8)^6\approx314.57\), so the value is about \(\$314.57\). 4. \(V(-1)=1200\cdot(0.8)^{-1}=1500\), so the modeled value one year earlier was \(\$1500\). 5. \(V(10)=1200\cdot(0.8)^{10}\approx128.85\). The remaining fraction is \((0.8)^{10}\approx0.1074\), or about \(10.74\%\).

a) \(b=0.8\); \(V(t)=1200(0.8)^t\) b) \(\$314.57\) c) \(\$1500\) d) \(\$128.85\), which is \(10.74\%\) of the original value

- Carefully compute the initial temperature difference. - Subtract the current difference from the room temperature. - Test consecutive whole-minute values near the threshold. - Distinguish mathematical equality from practical measurement.

1. The initial temperature difference is \(22-(-12)=34\,^\circ\text{C}\). The difference is multiplied by \(0.85\) each minute, so \(T_n=22-34(0.85)^n\). 2. \(T_0=-12.00\,^\circ\text{C}\), \(T_1=-6.90\,^\circ\text{C}\), \(T_2=-2.565\,^\circ\text{C}\approx-2.57\,^\circ\text{C}\), and \(T_3=1.11975\,^\circ\text{C}\approx1.12\,^\circ\text{C}\). 3. \(T_9\approx14.13\,^\circ\text{C}\), while \(T_{10}\approx15.31\,^\circ\text{C}\). Therefore, the temperature first exceeds \(15\,^\circ\text{C}\) after \(10\) full minutes. 4. The model assumes a constant room temperature and a constant relative decrease in the temperature difference. Real conditions vary, and measurement precision makes a very small difference practically indistinguishable from zero even though the model never reaches exactly \(22\,^\circ\text{C}\) at a finite time.

a) <table> <tr> <th>Time (min)</th> <th>Temperature (°C)</th> </tr> <tr><td>\(0\)</td><td>\(-12.00\)</td></tr> <tr><td>\(1\)</td><td>\(-6.90\)</td></tr> <tr><td>\(2\)</td><td>\(-2.57\)</td></tr> <tr><td>\(3\)</td><td>\(1.12\)</td></tr> </table> b) \(10\) minutes c) The model idealizes the conditions and approaches room temperature without reaching it exactly at a finite time.