Question

1 out of 2 parts have been answered. the following model represents the total spent on education funding, in millions of dollars, in a state from 2002 to 2012, where the time t = 0 represents the year 2002: e(t) = 26 ln(2988 − 269t) 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 2007

Since \( t = 0 \) represents 2002, for 2007, \( t=2007 - 2002=5 \).

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

Given \( E(t)=26\ln(2988 - 269t) \), use the chain rule. The derivative of \( \ln(u) \) is \( \frac{1}{u}u' \), so:
\( E'(t)=26\times\frac{- 269}{2988 - 269t}=\frac{-26\times269}{2988 - 269t} \)

Step3: Substitute \( t = 5 \) into \( E'(t) \)

First, calculate the denominator: \( 2988-269\times5=2988 - 1345 = 1643 \)
Then, calculate the numerator: \( -26\times269=-7000 - 16\times269=-7000-4304=-11304 \) (or directly \( 26\times269 = 7000+16\times269=7000 + 4304 = 11304 \), so with the negative sign it's - 11304)
Now, \( E'(5)=\frac{-11304}{1643}\approx - 6.88 \). Rounding to the nearest integer, we get - 7. But since we are talking about the rate of change (the negative sign indicates a decrease), but the question says "the rate at which the total spent... was changing", and we take the absolute value? Wait, no, the derivative gives the rate. Wait, let's check the calculation again. Wait, \( 26\times269 \): 26*200=5200, 26*69 = 1794, so 5200+1794=6994. Oh! I made a mistake earlier. 26*269: 26*(270 - 1)=26*270-26=7020 - 26 = 6994. So numerator is - 6994. Then \( E'(5)=\frac{-6994}{1643}\approx - 4.26 \). Wait, no, wait 2988 - 269*5: 269*5=1345, 2988 - 1345=1643. Then 26*269=6994. So \( E'(t)=\frac{-26\times269}{2988 - 269t}=\frac{-6994}{1643}\approx - 4.26 \). Rounding to the nearest integer, - 4? Wait, no, maybe I messed up the derivative. Wait, the function is \( E(t)=26\ln(2988 - 269t) \). The derivative of \( \ln(u) \) is \( \frac{u'}{u} \), so \( u = 2988 - 269t \), \( u'=-269 \). So \( E'(t)=26\times\frac{-269}{2988 - 269t}=\frac{-26\times269}{2988 - 269t} \). Let's compute 26*269: 20*269=5380, 6*269=1614, so 5380+1614=6994. So \( E'(t)=\frac{-6994}{2988 - 269t} \). Now, for \( t = 5 \), denominator is 2988 - 1345=1643. So \( E'(5)=\frac{-6994}{1643}\approx - 4.26 \). Rounding to the nearest integer, - 4? Wait, but maybe the question is about the magnitude? Wait, no, the rate of change. Wait, maybe I made a mistake in \( t \). Wait, 2002 is \( t = 0 \), so 2007 is \( t = 5 \), correct. Wait, let's check the calculation of 2988 - 269*5: 269*5: 270*5=1350, minus 1*5=5, so 1350 - 5=1345. 2988 - 1345: 2988 - 1300=1688, 1688 - 45=1643. Correct. Then 6994 divided by 1643: 1643*4=6572, 6994 - 6572=422, so 4 + 422/1643≈4 + 0.257≈4.257, so - 4.257, which rounds to - 4. But maybe the problem expects the absolute value? Wait, no, the derivative is the rate of change. If the derivative is negative, it means the total spent is decreasing at that rate. But let's check again. Wait, maybe I messed up the derivative. Wait, the function is \( E(t)=26\ln(2988 - 269t) \). So derivative is \( E'(t)=26\times\frac{-269}{2988 - 269t} \). So that's correct. So with \( t = 5 \), we have \( E'(5)=\frac{-26\times269}{2988 - 269\times5}=\frac{-6994}{1643}\approx - 4.26 \), which rounds to - 4. But maybe the question has a typo, or I made a mistake. Wait, maybe \( t = 0 \) is 2002, so 2007 is \( t = 5 \), correct. Alternatively, maybe the function is \( E(t)=26\ln(2988 + 269t) \)? But the problem says 2988 - 269t. Hmm. Alternatively, maybe I miscalculated 26*269. Let's do 26*269: 26*(200 + 60 + 9)=26*200 + 26*60 + 26*9=5200 + 1560 + 234=5200+1560=6760+234=6994. Correct. 2988 - 269*5=2988 - 1345=1643. Correct. So 6994/1643≈4.26, so the rate is approximately - 4 million dollars per year (decreasing at 4 million dollars per year). But the question says "the rate at which the total spent on education funding was changing", so the answer is approximately - 4, but if we take the absolute value (rate of change in magnitude), maybe 4. Wait, but let's check with a calculator. 6994 divided by 1643: 1643*4=6572, 6994-6572=422, 422/1643≈0.257, so total is≈4.257, so - 4.257, which rounds to - 4. But maybe the problem expects positive? Wait, no, the derivative is negative, so the total spent is decreasing. But the question says "the rate at which... was changing", so the rate is - 4 (or 4 with a negative sign indicating decrease). But let's see, maybe I made a mistake in \( t \). Wait, 2002 is \( t = 0 \), 2003 is \( t = 1 \), ..., 2007 is \( t = 5 \), correct. So I think the answer is - 4, but rounding to the nearest integer, - 4. Wait, but maybe the problem has a different function. Wait, the user wrote \( E(t)=26\ln(2988 - 269t) \). So that's correct. So the rate of change is approximately - 4 million dollars per year. But let's check again. Wait, 26*269=6994, 2988 - 269*5=1643, 6994/1643≈4.26, so - 4.26, which rounds to - 4. So the answer is - 4. But maybe the question wants the absolute value, so 4. Wait, the problem says "the rate at which the total spent... was changing", so the rate can be negative (decrease) or positive (increase). Since the derivative is negative, it's a decrease. So the rate is approximately - 4 million dollars per year. But let's confirm with the calculation.

Answer:

\boxed{-4}