Question

the following model represents the total spent on education funding, in millions of dollars, in a state from 2000 to 2010, where the time ( t = 0 ) represents the year 2000: ( e(t) = 35 ln(2857 - 251t) ) find the rate at which the total spent on education funding in that state was changing in the year 2007. (round your answer to the nearest integer.) the total spent on education funding was ? by ? million dollars per year in the year 2007.

Explanation:

Step1: Determine the value of \( t \) for the year 2007

Since \( t = 0 \) represents the year 2000, for the year 2007, \( t = 2007 - 2000 = 7 \).

Step2: Find the derivative of \( E(t) \)

The function is \( E(t)=35\ln(2857 - 251t) \). Using the chain rule, the derivative of \( \ln(u) \) is \( \frac{1}{u}u' \), where \( u = 2857 - 251t \) and \( u'=-251 \). So, \( E'(t)=35\times\frac{-251}{2857 - 251t} \).

Step3: Substitute \( t = 7 \) into \( E'(t) \)

First, calculate the denominator when \( t = 7 \): \( 2857-251\times7=2857 - 1757 = 1100 \).
Then, substitute into \( E'(t) \): \( E'(7)=35\times\frac{-251}{1100} \). But we are interested in the rate of change (the magnitude or the value, considering the sign for direction, but here we can compute the value: \( 35\times\frac{-251}{1100}=\frac{-8785}{1100}\approx - 7.986 \). Wait, maybe I made a mistake in the function. Wait, the original function is \( E(t) = 35\ln(2857 - 251t) \)? Wait, maybe it's a typo, maybe 2857? Wait, let's re - check. If \( t = 7 \), \( 2857-251\times7=2857 - 1757 = 1100 \). Then \( E'(t)=\frac{35\times(- 251)}{2857 - 251t} \). So \( E'(7)=\frac{35\times(-251)}{1100}=\frac{-8785}{1100}\approx - 7.986 \). But the question is about the rate at which the total spent on education funding was changing. The negative sign indicates a decrease. But let's check the calculation again. Wait, maybe the function is \( E(t)=35\ln(2857 + 251t) \)? Because if it's minus, at \( t = 7 \), the argument of the log is positive, but the rate is negative. But maybe I misread the function. Wait, the user's image shows \( E(t)=35\ln(2857 - 251t) \). Let's proceed. So \( E'(t)=\frac{35\times(-251)}{2857 - 251t} \). At \( t = 7 \), denominator is \( 2857-251\times7 = 2857 - 1757=1100 \). So \( E'(7)=\frac{-8785}{1100}\approx - 8 \) (rounded to the nearest integer). But the question also asks for the total spent in 2007. Let's calculate \( E(7)=35\ln(2857 - 251\times7)=35\ln(1100) \). \( \ln(1100)\approx7.003 \), so \( E(7)=35\times7.003\approx245.105\approx245 \) million dollars. Wait, but the first part is the rate of change. Let's re - do the derivative. The derivative of \( \ln(u) \) is \( \frac{u'}{u} \), so \( E(t)=35\ln(u) \), \( u = 2857 - 251t \), \( u'=-251 \), so \( E'(t)=35\times\frac{-251}{u}=35\times\frac{-251}{2857 - 251t} \). At \( t = 7 \), \( E'(7)=35\times\frac{-251}{2857 - 1757}=35\times\frac{-251}{1100}=\frac{-8785}{1100}\approx - 7.986\approx - 8 \). So the rate of change is approximately - 8 million dollars per year (the negative sign means the funding is decreasing at a rate of 8 million dollars per year), and the total spent in 2007 is \( E(7)=35\ln(1100)\approx35\times7.003\approx245 \) million dollars.

Answer:

The rate of change in 2007 is approximately \(-8\) million dollars per year, and the total spent in 2007 is approximately \(245\) million dollars. (If we consider the rate of change's magnitude or the total, based on the question's two parts: "Find the rate..." and "The total spent...")

For the rate of change (first part): \(\boxed{-8}\) (million dollars per year)

For the total spent (second part): \(\boxed{245}\) (million dollars)