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Hoochie [10]
3 years ago
8

Any help with an explanation would be appreciated!

Mathematics
1 answer:
Lilit [14]3 years ago
5 0

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>

============================================================

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}\\\\

<h3>Answer:  -2/9</h3>

============================================================

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\\\\

<h3>Answer:  12</h3>
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Answer:

m\angle CDE=14^o

Step-by-step explanation:

we know that

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see the attached figure to better understand the problem

so

∠DEF=∠ECD+∠CDE

substitute the given values

(4x+5)^o=(x+9)^o+(x+8)^o

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Combine like terms

(4x+5)^o=(2x+17)^o

Group terms

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2x=12

x=6

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m\angle CDE=(x+8)^o

substitute the value of x

m\angle CDE=(6+8)=14^o

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TIA


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