Problem 1
We'll use the product rule to say
h(x) = f(x)*g(x)
h ' (x) = f ' (x)*g(x) + f(x)*g ' (x)
Then plug in x = 2 and use the table to fill in the rest
h ' (x) = f ' (x)*g(x) + f(x)*g ' (x)
h ' (2) = f ' (2)*g(2) + f(2)*g ' (2)
h ' (2) = 2*3 + 2*4
h ' (2) = 6 + 8
h ' (2) = 14
<h3>Answer: 14</h3>
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Problem 2
Now we'll use the quotient rule
![h(x) = \frac{f(x)}{g(x)}\\\\h'(x) = \frac{f'(x)*g(x)-f(x)*g'(x)}{(g(x))^2}\\\\h'(2) = \frac{f'(2)*g(2)-f(2)*g'(2)}{(g(2))^2}\\\\h'(2) = \frac{2*3-2*4}{(3)^2}\\\\h'(2) = \frac{6-8}{9}\\\\h'(2) = -\frac{2}{9}\\\\](https://tex.z-dn.net/?f=h%28x%29%20%3D%20%5Cfrac%7Bf%28x%29%7D%7Bg%28x%29%7D%5C%5C%5C%5Ch%27%28x%29%20%3D%20%5Cfrac%7Bf%27%28x%29%2Ag%28x%29-f%28x%29%2Ag%27%28x%29%7D%7B%28g%28x%29%29%5E2%7D%5C%5C%5C%5Ch%27%282%29%20%3D%20%5Cfrac%7Bf%27%282%29%2Ag%282%29-f%282%29%2Ag%27%282%29%7D%7B%28g%282%29%29%5E2%7D%5C%5C%5C%5Ch%27%282%29%20%3D%20%5Cfrac%7B2%2A3-2%2A4%7D%7B%283%29%5E2%7D%5C%5C%5C%5Ch%27%282%29%20%3D%20%5Cfrac%7B6-8%7D%7B9%7D%5C%5C%5C%5Ch%27%282%29%20%3D%20-%5Cfrac%7B2%7D%7B9%7D%5C%5C%5C%5C)
<h3>Answer: -2/9</h3>
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Problem 3
Use the chain rule
![h(x) = f(g(x))\\\\h'(x) = f'(g(x))*g'(x)\\\\h'(2) = f'(g(2))*g'(2)\\\\h'(2) = f'(3)*g'(2)\\\\h'(2) = 3*4\\\\h'(2) = 12\\\\](https://tex.z-dn.net/?f=h%28x%29%20%3D%20f%28g%28x%29%29%5C%5C%5C%5Ch%27%28x%29%20%3D%20f%27%28g%28x%29%29%2Ag%27%28x%29%5C%5C%5C%5Ch%27%282%29%20%3D%20f%27%28g%282%29%29%2Ag%27%282%29%5C%5C%5C%5Ch%27%282%29%20%3D%20f%27%283%29%2Ag%27%282%29%5C%5C%5C%5Ch%27%282%29%20%3D%203%2A4%5C%5C%5C%5Ch%27%282%29%20%3D%2012%5C%5C%5C%5C)
<h3>Answer: 12</h3>
2x - 3(x-1) = -1
2x -3x + 3 = -1
-1x + 3 = -1
-1x = -4
x = 4
y = 4 - 1
y = 3
Answer:
Converse of the alternate exterior angle theorem
Step-by-step explanation:
just took the test ;)
Divide 35423 by 15 and round the answer down to a whole number, then you get 2361. If you subtract 15*2361 from 35423 you get the remainder 8. So answer C is correct.
B. triangle ABC is isosceles. The angles add up 180 degrees.