Free Math Worksheets

- Substitute \(t=0\) into each function. - Recall that \(e^0=1\). - The sign of \(k\) in \(be^{kt}\) determines whether the function grows or decays. - Substitute \(t=20\) into \(f_1\).

1. At \(t=0\), \(e^0=1\). Therefore, \(f_1(0)=600\), \(f_2(0)=450\), and \(f_3(0)=600\). Function (2) has the smallest initial value, \(450\). 2. An exponential function of the form \(be^{kt}\) describes decay when \(k<0\). Therefore, functions (2) and (3) describe decay. 3. \(f_1(20)=600e^{0.05(20)}=600e\approx 1630.97\).

a) Function (2), with an initial value of \(450\) b) Functions (2) and (3), because their exponent coefficients are negative c) \(f_1(20)\approx 1630.97\)

- Compare the given equation directly with the general form. - What happens to \(e^{kx}\) as \(x\) increases when \(k\) is negative? - Use a calculator and check the third decimal place before rounding.

1. Comparing \(ce^{kx}\) with \(15.5e^{-0.4x}\) gives \(c=15.5\) and \(k=-0.4\). 2. Since \(k<0\), the factor \(e^{kx}\) decreases as \(x\) increases. The model represents exponential decay. 3. \(B(5)=15.5e^{-0.4(5)}=15.5e^{-2}\approx2.10\).

a) \(c=15.5\) and \(k=-0.4\) b) Exponential decay, because \(k<0\). c) \(B(5)\approx2.10\)

- Compare both the differences and the ratios between measurements over equal time intervals. - Use the initial value and the growth factor for each \(2\)-day interval. - Determine how many \(2\)-day growth intervals occur in \(7\) days.

1. For a linear model, equal time intervals should have equal differences. Here, \(600-400=200\), but \(900-600=300\), so the data are not linear. 2. For an exponential model, equal time intervals should have equal ratios. Here, \(\frac{600}{400}=1.5\) and \(\frac{900}{600}=1.5\), so an exponential model fits. 3. The initial value is \(400\), and the growth factor for every \(2\) days is \(1.5\). Thus, \(f(t)=400(1.5)^{t/2}\), where \(t\) is measured in days. Equivalently, \(f(t)\approx 400(1.2247)^t\). 4. After \(7\) days, \(f(7)=400(1.5)^{7/2}\approx 1653.41\).

a) Exponential, because the ratios for equal time intervals are constant while the differences are not b) \(f(t)=400(1.5)^{t/2}\), or approximately \(f(t)=400(1.2247)^t\) c) About \(1653.41\,\text{cm}^2\)

- Identify the population at \(t=0\). - Compare the population at year \(1\) with the initial population. - Use the general form of an exponential function. - Divide the year-1 value by the initial value to find the growth factor.

1. The initial value is \(40\), so \(f(t)=40b^t\). 2. Use the value after one year: \(48=40b\), so \(b=\frac{48}{40}=1.2\). 3. Therefore, \(f(t)=40(1.2)^t\).

\(f(t)=40(1.2)^t\)

- Identify the number of users at \(t=0\). - Translate “doubles” into a growth factor. - Convert half a year and one year to months before substituting for \(t\).

1. The initial value is \(N(0)=4500\), and monthly doubling gives a growth factor of \(2\). Therefore, \(N(t)=4500(2)^t\). 2. Half a year is \(6\) months: \(N(6)=4500\cdot2^6=4500\cdot64=288{,}000\). 3. One year is \(12\) months: \(N(12)=4500\cdot2^{12}=4500\cdot4096=18{,}432{,}000\).

The function is \(N(t)=4500(2)^t\). After half a year: \(288{,}000\) users After one year: \(18{,}432{,}000\) users

- Divide the elapsed time by the doubling time to find the number of doublings. - Rewrite large powers using multiples of \(10\) in the exponent. - Percent difference compares the difference with the stated reference value.

1. \(2^{10}=1024\). The percent difference from \(1000\) is \(\frac{1024-1000}{1000}(100\%)=2.4\%\). 2. Twenty years contains \(10\) doubling periods, so the factor is \(2^{10}\approx 10^3\). Sixty years contains \(30\) doubling periods, so \(2^{30}=(2^{10})^3\approx (10^3)^3=10^9\). 3. Forty years contains \(20\) doubling periods. Thus, the factor is \(2^{20}=(2^{10})^2\approx 10^6\), and the estimated performance is \(10^{16}\cdot 10^6=10^{22}=1 \times 10^{22}\) operations per second.

a) \(2^{10}=1024\); percent difference: \(2.4\%\) b) \(2^{10}\approx 10^3\) in \(20\) years; \(2^{30}\approx 10^9\) in \(60\) years c) \(1 \times 10^{22}\) operations per second

- Rewrite an exponent of \(0.5\) as a square root. - Substitute \(b=1\) and compare several function values. - How do you find a y-intercept? - Recall the zero-exponent rule.

1. \(f(0.5)=(-4)^{0.5}=\sqrt{-4}\), which is not a real number. To define \(b^x\) for every real \(x\), the base must satisfy \(b>0\). 2. If \(b=1\), then \(f(x)=1^x=1\) for every \(x\). This is a constant function and does not represent exponential growth or decay. 3. A y-intercept occurs at \(x=0\). For every allowed base, \(f(0)=b^0=1\). Therefore, every graph passes through \((0, 1)\).

1. \((-4)^{0.5}=\sqrt{-4}\) is not real, so \(b>0\) is required. 2. When \(b=1\), \(f(x)=1\), a constant function. 3. Since \(b^0=1\), all graphs pass through \((0, 1)\).

- First identify the population at \(t=0\). - To find a doubling time, locate twice the initial population on the vertical axis and move horizontally to the graph. - From a point on the graph, move vertically to read the corresponding time or population.

1. The graph starts at \(N(0)=100\). 2. The population doubles when \(N=200\). The graph reaches this value at about \(3.1\) hours. 3. The graph reaches \(N=400\) at about \(6.2\) hours. 4. At \(t=9\), the graph is near \(750\), so the population is about \(750\) bacteria.

a) About \(3.1\) hours b) About \(6.2\) hours c) About \(750\) bacteria

- Read the y-intercept of the graph. - Start at \(105\) on the y-axis, move horizontally to the graph, and then move down to the time axis. - Think about what happens to the cooling rate as the tea's temperature gets closer to room temperature.

1. At \(t=0\), the graph has a y-value of \(180\), so the initial temperature is \(180\,\text{°F}\). 2. The graph reaches approximately \(105\,\text{°F}\) at \(t=10\) minutes. 3. As the tea approaches room temperature, the temperature difference between the tea and the room becomes smaller, so the tea cools more slowly. In this model, \(70\,\text{°F}\) is a horizontal asymptote, so the tea approaches that temperature without falling below it.

a) \(180\,\text{°F}\) b) About \(10\) minutes c) The tea cools more slowly as its temperature approaches room temperature, and the model has a horizontal asymptote at \(70\,\text{°F}\).

- Evaluate each function at \(x=0\) to find its y-intercept. - Compare each base with \(1\). - For graphs with the same initial value, distinguish growth from decay.

1. Equations (a) and (b) have y-intercept \(3\), matching graphs \(p\) and \(q\). Equations (c) and (d) have y-intercept \(1.5\), matching graphs \(r\) and \(s\). 2. A base greater than \(1\) produces growth, while a base between \(0\) and \(1\) produces decay. 3. Therefore, (a) matches \(p\), (b) matches \(q\), (c) matches \(r\), and (d) matches \(s\).

(a) – \(p\) (b) – \(q\) (c) – \(r\) (d) – \(s\)

- Read the initial value at \(t=0\). - The one-hour growth factor is the ratio \(N(1)/N(0)\). - Use clearly marked grid points.

1. At \(t=0\), the graph gives \(N_0=2\) for culture \(h\) and \(N_0=8\) for culture \(k\). 2. At \(t=1\), the graph gives \(h(1)=3\) and \(k(1)=6\). 3. For culture \(h\), \(2a=3\), so \(a=1.5\). For culture \(k\), \(8a=6\), so \(a=0.75\).

Culture \(h\): \(N_0=2\) thousand, \(a=1.5\) Culture \(k\): \(N_0=8\) thousand, \(a=0.75\)

- Use the annual-compounding model for part a) and the continuous-compounding model for part b). - Interest earned is the ending balance minus the initial investment. - For part c), set the one-year growth factors equal.

1. With annual compounding, \(K(5)=15{,}000(1.032)^5\approx \$17{,}558.59\). The interest earned is \(\$17{,}558.59-\$15{,}000=\$2558.59\). 2. With continuous compounding, \(K(5)=15{,}000e^{0.032(5)}\approx \$17{,}602.66\). The interest earned is \(\$17{,}602.66-\$15{,}000=\$2602.66\). 3. For equal one-year growth factors, \(1+i=e^{0.032}\). Thus, \(i=e^{0.032}-1\approx 0.0325175\), or about \(3.25\%\).

a) Ending balance: \(\$17{,}558.59\); interest: \(\$2558.59\) b) Ending balance: \(\$17{,}602.66\); interest: \(\$2602.66\) c) \(3.25\%\)

- Rewrite \(f(x+h)\) using exponent rules. - Identify the part equal to \(f(x)\) and the remaining multiplier. - Recall the meanings of negative and fractional exponents.

1. In general, \(f(x+h)=5(3)^{x+h}=5(3)^x(3)^h=f(x)(3)^h\). Therefore, the change factor is \(3^h\). 2. For \(h=1\), the factor is \(3^1=3\). 3. For \(h=4\), the factor is \(3^4=81\). 4. For \(h=-2\), the factor is \(3^{-2}=\frac{1}{9}\). 5. For \(h=0.5\), the factor is \(3^{0.5}=\sqrt{3}\approx1.732\).

a) \(3\) b) \(81\) c) \(\frac{1}{9}\) d) \(\sqrt{3}\approx1.732\)

- Use exponent rules to simplify ratios of powers. - Interpret a rational exponent such as \(1.5\). - Rewrite both sides of \(4^s=32\) using the same base. - Replace \(x\) with \(x+s\) and factor the expression.

1. At \(x=0\), \(\frac{f(1)}{f(0)}=\frac{4}{1}=4\). At \(x=2\), \(\frac{f(3)}{f(2)}=\frac{64}{16}=4\). The ratio is constant and equals the base. 2. The factor is \(4^{1.5}=4\sqrt{4}=8\). 3. Solve \(4^s=32\). Since \(4=2^2\) and \(32=2^5\), \(2^{2s}=2^5\), so \(2s=5\) and \(s=2.5\). 4. In general, \(f(x+s)=a^{x+s}=a^xa^s=f(x)a^s\), so the factor is \(a^s\).

1. The ratio is \(4\) in both cases. 2. \(8\) 3. \(s=2.5\) 4. \(a^s\)

- Use the point with \(x=0\) first. - Substitute the value of \(c\) before using the second point. - Solve the resulting equation involving a square or a reciprocal.

1. A point with \(x=0\) gives \(c\), because \(a^0=1\). 2. In a), \(c=7\). Using \((2, 63)\), \(63=7a^2\), so \(a^2=9\). Since \(a>0\), \(a=3\). 3. In b), \(c=200\). Using \((-1, 50)\), \(50=200a^{-1}=\frac{200}{a}\), so \(a=4\).

a) \(c=7\), \(a=3\) b) \(c=200\), \(a=4\)

- Use the exponent laws to express the growth factor over an increase of \(2\). - Consider what happens when every power has base \(1\). - Interpret an exponent of \(0.5\) as a square root. - Ask whether that square root is real for a negative base.

1. For part a, increasing \(x\) by \(2\) multiplies the output by \(a^2\). Thus, \(a^2=4\). Since \(a>0\), \(a=2\). 2. For part b, if \(a=1\), then \(f(x)=1^x=1\) for every \(x\). This is a constant function, so its graph is the horizontal line \(y=1\), not a growth or decay curve. 3. For part c, \((-2)^2=4\) is real, but \((-2)^{0.5}=\sqrt{-2}\) is not a real number. Therefore, a negative base does not give a real-valued exponential function with domain \(\mathbb{R}\).

a) \(a=2\) b) \(a=1\) gives the constant function \(f(x)=1\), whose graph is the horizontal line \(y=1\). c) \((-2)^2\) is real, but \((-2)^{0.5}\) is not real, so the function cannot have all real numbers as its domain.

- Raise the one-step factor to a power for several steps. - Convert a factor below \(1\) to a percent decrease. - Use the power-of-a-power rule for part c.

1. When \(x\) increases by \(2\), the function value is multiplied by \((0.75)^2=0.5625\). 2. When \(x\) increases by \(3\), the factor is \((0.75)^3=0.421875\). The percent decrease is \((1-0.421875)\cdot100\%=57.8125\%\). 3. Since \(g(2x)=(0.75)^{2x}=((0.75)^x)^2\) and \(y=(0.75)^x\), \(g(2x)=y^2\).

a) \(0.5625\) b) \(57.8125\%\) decrease c) \(g(2x)=y^2\)

- Substitute \(x-4\) into the function rule. - Apply the quotient rule for powers with the same base. - Simplify until the expression no longer contains \(x\). - Interpret a negative exponent as a reciprocal.

1. Substitute the function rule: \(\frac{f(x-4)}{f(x)}=\frac{ca^{x-4}}{ca^x}\). Canceling \(c\) and subtracting exponents gives \(a^{(x-4)-x}=a^{-4}=\frac{1}{a^4}\). This expression does not contain \(x\), so the ratio is constant. 2. For \(a=0.8\), the ratio is \((0.8)^{-4}=\frac{1}{(0.8)^4}=\frac{1}{0.4096}=2.44140625\).

1. \(\frac{f(x-4)}{f(x)}=a^{-4}=\frac{1}{a^4}\) 2. \(2.44140625\)

- A backward time step uses the reciprocal of the forward factor. - You do not need to find the one-unit growth factor. - Write the relationship for a two-unit forward step first.

1. The given relationship is \(f(t+2)=4f(t)\). 2. Moving backward by \(2\) time units uses the reciprocal factor, so \(f(t-2)=\frac{1}{4}f(t)\).

\(f(t-2)=\frac{1}{4}f(t)\)

- Evaluate the function at \(x=0\). - Compare the base with \(1\). - Decide whether a positive multiple of \(3^x\) can be zero or negative. - Analyze the graph far to the left.

1. \(f(0)=1.5\cdot3^0=1.5\), so the y-intercept is \((0, 1.5)\). 2. Since the base \(3\) is greater than \(1\) and the leading factor is positive, the function is increasing. 3. Because \(3^x>0\) for every real \(x\), and \(1.5>0\), the range is \((0, \infty)\). 4. As \(x\to-\infty\), \(3^x\to0\), so \(f(x)\to0\). Thus \(y=0\) is a horizontal asymptote.

a) \((0, 1.5)\) b) Increasing c) \((0, \infty)\) d) \(f(x)\to0\); horizontal asymptote \(y=0\)

- Substitute \(x+\frac{1}{3}\) into the exponent. - Distribute \(-3\) carefully. - Rewrite a negative exponent as a reciprocal. - Compare the new function expression with the original one.

1. Substitute the new input: \(h\left(x+\frac{1}{3}\right)=5(2)^{-3\left(x+\frac{1}{3}\right)}\). 2. Simplify the exponent: \(-3\left(x+\frac{1}{3}\right)=-3x-1\). 3. Therefore, \(h\left(x+\frac{1}{3}\right)=5(2)^{-3x}(2)^{-1}=h(x)\left(\frac{1}{2}\right)\). 4. The multiplication factor is \(\frac{1}{2}\).

\(\frac{1}{2}\)

- 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.

- Substitute both points to create two equations. - Divide the equations to eliminate one parameter. - The base of an exponential function is positive. - Substitute each requested x-coordinate into the completed function.

1. The points give \(18=ca^2\) and \(162=ca^4\). 2. Divide the equations: \(\frac{162}{18}=a^2\), so \(a^2=9\). Since \(a>0\), \(a=3\). 3. Substitute into \(18=ca^2\): \(18=9c\), so \(c=2\). Thus \(f(x)=2(3)^x\). 4. \(y_C=f(0)=2\), and \(y_D=f(-1)=2\cdot3^{-1}=\frac{2}{3}\).

\(f(x)=2(3)^x\) \(y_C=2\), \(y_D=\frac{2}{3}\)

- Compare consecutive outputs using ratios rather than differences. - The output at \(x=0\) gives the initial value. - A decay factor is between \(0\) and \(1\). - Use the definition of a function to explain uniqueness of outputs.

1. Consecutive outputs have a constant ratio: \(\frac{6}{12}=\frac{3}{6}=\frac{1.5}{3}=0.5\). Therefore, \(g\) is exponential. 2. Since \(g(0)=6\), the initial value is \(6\), and the common ratio is \(0.5\). Thus, \(g(x)=6(0.5)^x\). 3. Verbally, the function begins at \(6\) when \(x=0\) and is multiplied by \(0.5\) whenever \(x\) increases by \(1\). Its graph decreases and approaches the x-axis without reaching it. 4. For every real value of \(x\), the expression \(6(0.5)^x\) produces exactly one real output, so the rule defines a function.

a) Exponential function: \(g(x)=6(0.5)^x\) b) The output is multiplied by \(0.5\) for each increase of \(1\) in \(x\); the graph decreases toward the x-axis. c) Each real input produces exactly one output.

- Use the base to determine whether a graph increases or decreases. - All four graphs share one point on the y-axis. - Evaluate the functions at \(x=1\) or \(x=-1\). - Compare the growth rates of the two increasing graphs. - Compare how quickly the two decreasing graphs fall.

1. Graphs \(p\) and \(q\) are increasing, so they correspond to bases greater than \(1\): \(3\) and \(1.5\). Graphs \(r\) and \(s\) are decreasing, so they correspond to bases between \(0\) and \(1\): \(0.5\) and \(0.2\). 2. Since \(3^x\) grows faster than \(1.5^x\) for positive \(x\), graph \(p\) is \(A\) and graph \(q\) is \(B\). At \(x=1\), their values are \(3\) and \(1.5\), respectively. 3. Since \(0.2^x\) decreases faster than \(0.5^x\) for positive \(x\), graph \(s\) is \(D\) and graph \(r\) is \(C\). At \(x=-1\), their values are \(5\) and \(2\), respectively.

Graph \(p\): \(A(x)=3^x\) Graph \(q\): \(B(x)=1.5^x\) Graph \(r\): \(C(x)=0.5^x\) Graph \(s\): \(D(x)=0.2^x\)

- For part c), use \(a^n b^n=(ab)^n\). - Relate \(g(x)\) to \(f(-x)\) when describing the symmetry.

1. For \(x\ge 0\), \(f\) increases and \(g\) decreases. Both pass through \((0,1)\). As complete functions, they are reflections across the y-axis because \(g(x)=f(-x)\). 2. At \(x=3\), \(f(3)=8\) and \(g(3)=0.125\). Thus, \(\frac{f(3)}{g(3)}=64\). 3. \(f(x)g(x)=2^x(0.5)^x=(2\cdot 0.5)^x=1\). The product is always \(1\).

a) For \(x\ge 0\), \(f\) increases and \(g\) decreases. The complete graphs are reflections across the y-axis. b) \(64\) c) \(f(x)g(x)=1\) for every \(x\).

- The parameter \(a\) is the function value at \(x=0\). - Choose a second point with coordinates that can be read exactly. - Substitute that point into \(y=ab^x\) and solve for \(b\).

1. For \(f\), the y-intercept is \((0, 5)\), so \(a=5\). The point \((1, 6)\) gives \(6=5b\), so \(b=\frac{6}{5}=1.2\). Therefore, \(f(x)=5(1.2)^x\). 2. For \(g\), the y-intercept is \((0, 2)\), so \(a=2\). The point \((1, 1)\) gives \(1=2b\), so \(b=0.5\). Therefore, \(g(x)=2(0.5)^x\).

\(f(x)=5(1.2)^x\) \(g(x)=2(0.5)^x\)

- Compare consecutive values using ratios. - The initial value is the value at \(t=0\). - Substitute \(t=4\) into the model.

1. The ratios of consecutive values are \(\frac{30}{10}=3\), \(\frac{90}{30}=3\), and \(\frac{270}{90}=3\). The constant ratio shows exponential growth with factor \(b=3\). 2. The initial value is \(a=10\), so the model is \(y=10(3)^t\). 3. At \(t=4\), \(y=10\cdot3^4=810\).

a) \(b=3\) b) \(y=10(3)^t\) c) \(810\) users

- Determine whether each exponential function is increasing or decreasing. - Compare the order of the exponents with the order of the powers. - Find nearby powers of \(2\) that bracket \(0.2\).

1. Since \(0<0.6<1\), the function \(0.6^x\) is decreasing. Therefore, \(0.6^m<0.6^n\) implies \(m>n\). 2. Because \(5<7\) but \(a^5>a^7\), the exponential function must be decreasing. Thus \(0<a<1\). 3. \(2^{-3}=0.125\) and \(2^{-2}=0.25\). Since \(0.125<0.2<0.25\) and \(2^x\) is increasing, \(-3<k<-2\).

a) \(m>n\) b) \(0<a<1\) c) \(-3<k<-2\)

- Account for the negative initial temperature when finding the difference. - Use the value after \(10\) minutes to find either a 10-minute factor or a 1-minute factor. - The cake approaches room temperature from below. - For the table, use the fact that the difference is multiplied by \(0.8\) every \(10\) minutes.

1. The initial difference is \(22-(-18)=40\,^\circ\text{C}\). 2. If \(a\) is the decay factor per minute, then \(40a^{10}=32\). Thus \(a^{10}=0.8\), so \(a=0.8^{1/10}\approx0.9779\). The model is \(D(t)=40\left(0.8^{1/10}\right)^t\). 3. After \(30\) minutes, \(D(30)=40\cdot(0.8)^3=20.48\,^\circ\text{C}\). Since the cake is below room temperature, its temperature is \(22-20.48=1.52\,^\circ\text{C}\). 4. \(D(0)=40\), \(D(20)=40\cdot(0.8)^2=25.6\), \(D(40)=40\cdot(0.8)^4=16.384\), and \(D(60)=40\cdot(0.8)^6\approx10.49\), all in degrees Celsius.

a) \(40\,^\circ\text{C}\) b) \(a=0.8^{1/10}\approx0.9779\); \(D(t)=40\left(0.8^{1/10}\right)^t\) c) \(1.52\,^\circ\text{C}\) d) <table> <tr> <th>Time (min)</th> <th>Temperature difference (°C)</th> </tr> <tr> <td>\(0\)</td> <td>\(40\)</td> </tr> <tr> <td>\(20\)</td> <td>\(25.6\)</td> </tr> <tr> <td>\(40\)</td> <td>\(16.38\)</td> </tr> <tr> <td>\(60\)</td> <td>\(10.49\)</td> </tr> </table>

- 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. \(T_0=-12\,^\circ\text{C}\), \(T_1=-6.9\,^\circ\text{C}\), \(T_2=-2.565\,^\circ\text{C}\approx-2.57\,^\circ\text{C}\), and \(T_3\approx1.12\,^\circ\text{C}\). 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>\(-12\)</td></tr> <tr><td>\(1\)</td><td>\(-6.9\)</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.

- Write one equation for each given point. - Isolate one variable and substitute it into the other equation. - Apply the negative-exponent rule. - Use the completed function to find the missing outputs.

1. The two points give \(4=ca\) and \(32=ca^{-2}\). 2. From \(4=ca\), \(c=\frac{4}{a}\). Substitute into the second equation: \(32=\frac{4}{a}a^{-2}=4a^{-3}=\frac{4}{a^3}\). 3. Thus \(a^3=\frac{1}{8}\), so \(a=0.5\). Then \(c=\frac{4}{0.5}=8\). The function is \(y=8(0.5)^x\). 4. At \(x=2\), \(y=8\cdot(0.5)^2=2\). At \(x=-3\), \(y=8\cdot(0.5)^{-3}=64\).

\(a=0.5\), \(c=8\) \(R(2, 2)\), \(S(-3, 64)\)