What is the length of the hypotenuse of the triangle?

Mathematics

1 answer:

igor_vitrenko [27]2 years ago

8 0

Answer:

The answer is 37.

Step-by-step explanation:

#Hope it helps uh.......

You might be interested in

7/12+1/3 show your work and how you got your answer and you do all that and don’t scam me I will give branlist

Yanka [14]

Answer: 11/12

Step-by-step explanation:

7/12 and 1/3 don’t have the same denominator, so you can’t just add them right away. You have to give them the same bottom number. 1/3 = 4/12, so you can do it like this:

7/12 + 4/12 = 11/12

11/12 is in simplest form, so that should be correct! Hope this helps! :)

3 0

2 years ago

Read 2 more answers

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

alexira [117]

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

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