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kumpel [21]
3 years ago
12

Use the distributive property to remove the parentheses. 8(3 - y)

Mathematics
1 answer:
LiRa [457]3 years ago
3 0

Answer:

-8y+24

Step-by-step explanation:

8(3−y)

=(8)(3+−y)

=(8)(3)+(8)(−y)

=24−8y

=−8y+24

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For the composite function, identify an inside function and an outside function and write the derivative with respect to x of th
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Answer:

The inner function is h(x)=4x^2 + 8 and the outer function is g(x)=3x^5.

The derivative of the function is \frac{d}{dx}\left(3\left(4x^2+8\right)^5\right)=120x\left(4x^2+8\right)^4.

Step-by-step explanation:

A composite function can be written as g(h(x)), where h and g are basic functions.

For the function f(x)=3(4x^2+8)^5.

The inner function is the part we evaluate first. Frequently, we can identify the correct expression because it will appear within a grouping symbol one or more times in our composed function.

Here, we have 4x^2+8 inside parentheses. So h(x)=4x^2 + 8 is the inner function and the outer function is g(x)=3x^5.

The chain rule says:

\frac{d}{dx}[f(g(x))]=f'(g(x))g'(x)

It tells us how to differentiate composite functions.

The function f(x)=3(4x^2+8)^5 is the composition, g(h(x)), of

     outside function: g(x)=3x^5

     inside function: h(x)=4x^2 + 8

The derivative of this is computed as

\frac{d}{dx}\left(3\left(4x^2+8\right)^5\right)=3\frac{d}{dx}\left(\left(4x^2+8\right)^5\right)\\\\\mathrm{Apply\:the\:chain\:rule}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}\\f=u^5,\:\:u=\left(4x^2+8\right)\\\\3\frac{d}{du}\left(u^5\right)\frac{d}{dx}\left(4x^2+8\right)\\\\3\cdot \:5\left(4x^2+8\right)^4\cdot \:8x\\\\120x\left(4x^2+8\right)^4

The derivative of the function is \frac{d}{dx}\left(3\left(4x^2+8\right)^5\right)=120x\left(4x^2+8\right)^4.

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