Answer:
a. -3
b. -3
Step-by-step explanation:
Fill in the function arguments and simplify.
a.
![\dfrac{f(x)-f(a)}{x-a}=\dfrac{(-3x+9)-(-3a+9)}{x-a}=\dfrac{-3x+9+3a-9}{x-a}\\\\=\dfrac{-3(x-a)}{(x-a)}=-3](https://tex.z-dn.net/?f=%5Cdfrac%7Bf%28x%29-f%28a%29%7D%7Bx-a%7D%3D%5Cdfrac%7B%28-3x%2B9%29-%28-3a%2B9%29%7D%7Bx-a%7D%3D%5Cdfrac%7B-3x%2B9%2B3a-9%7D%7Bx-a%7D%5C%5C%5C%5C%3D%5Cdfrac%7B-3%28x-a%29%7D%7B%28x-a%29%7D%3D-3)
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b.
![\dfrac{f(x+h)-f(x)}{h}=\dfrac{(-3(x+h)+9)-(-3x+9)}{h}=\dfrac{-3x-3h+9+3x-9}{h}\\\\=\dfrac{-3h}{h}=-3](https://tex.z-dn.net/?f=%5Cdfrac%7Bf%28x%2Bh%29-f%28x%29%7D%7Bh%7D%3D%5Cdfrac%7B%28-3%28x%2Bh%29%2B9%29-%28-3x%2B9%29%7D%7Bh%7D%3D%5Cdfrac%7B-3x-3h%2B9%2B3x-9%7D%7Bh%7D%5C%5C%5C%5C%3D%5Cdfrac%7B-3h%7D%7Bh%7D%3D-3)
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<em>Comment on questions of this type</em>
Perhaps the point of this exercise is to help you understand that the average slope computed by the first expression approaches the instantaneous slope computed by the second expression. This happens when a approaches x and when h approaches zero.
In this problem, where the function is a straight line, the slope is a constant (-3), so the average slope and the instantaneous slope are the same everywhere (for all values of x, a, h).
Answer:
19
— or 0.76000
25
Step-by-step explanation:
It should take around 37 minutes.
Basically substitute y into the first equation: 2x+8(2)=20 and solve. Then plug in x to the second equation etc.
I'm not sure but it's either C or D