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3 years ago

Find two numbers whose sum is 15 and the product is 36

2 answers:

8 0

The two numbers are 3 and 12.

I got these numbers by factoring 36.

1 * 36 sum : 37
2 * 18 sum : 20
3 * 12 sum : 15

It is easier to find the factor of a number and solve for the sum. Than to find the addends of a number and solve for the product.

There are 9 factors of 36 while 14 addends of 15.

Vedmedyk [2.9K]3 years ago

7 0

OK so if you're wanting 15 and 36 you will need to know what goes into each of them but if you just need 15 then let's try to find that one so, what does go into 15 well let's find out.

15:1×15,3×5
36:1×36,2×18,3×12,4×9,6²

so, based on what we have here there are a lot of answers between numbers 15&36 but one maybe two or more answers for what can go into 15 and 36.

3×12=36
12+3=15
so, the two numbers you need are 3 and 12.

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2 years ago

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2 years ago

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

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

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

3 0

3 years ago