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3 years ago

Error error error error error

Mathematics

1 answer:

emmainna [20.7K]3 years ago

8 0

Error error error error error error error

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<img src="https://tex.z-dn.net/?f=%20%28%7B%20%7Bx%7D%5E%7B2%7D%20%20-%204%7D%29%5E%7B5%7D%20%28%20%7B4x%20-%205%7D%29%5E%7B4%7D

Makovka662 [10]

Let $u=x^2-4$ and $v=4x-5$ . By the product rule,

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=\dfrac{\mathrm d(u^5)}{\mathrm dx}v^4+u^5\dfrac{\mathrm d(v^4)}{\mathrm dx}$

By the power rule, we have $(u^5)'=5u^4$ and $(v^4)'=4v^3$ , but $u,v$ are functions of $x$ , so we also need to apply the chain rule:

$\dfrac{\mathrm d(u^5)}{\mathrm dx}=5u^4\dfrac{\mathrm du}{\mathrm dx}$

$\dfrac{\mathrm d(v^4)}{\mathrm dx}=4v^3\dfrac{\mathrm dv}{\mathrm dx}$

and we have

$\dfrac{\mathrm du}{\mathrm dx}=2x$

$\dfrac{\mathrm dv}{\mathrm dx}=4$

So we end up with

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=10xu^4v^4+16u^5v^3$

Replace $u,v$ to get everything in terms of $x$ :

$\dfrac{\mathrm d((x^2-4)^5(4x-5)^4)}{\mathrm dx}=10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3$

We can simplify this by factoring:

$10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3=2(x^2-4)^4(4x-5)^3\bigg(5x(4x-5)+8(x^2-4)\bigg)$

$=2(x^2-4)^4(4x-5)^3(28x^2-57)$

7 0

2 years ago

Please help me with this

Zanzabum

Answer:

−4y+32

Step-by-step explanation:

Let's solve for x.

−9

x+8

Step 1: Flip the equation.

−9

x+8=y

Step 2: Add -8 to both sides.

−9

x+8+−8=y+−8

−9

x=y−8

Step 3: Divide both sides by (-9)/4.

−9

y−8

−9

−4y+32

I think this is the right answer.

6 0

2 years ago

Your family needs a set of 4 tires. Which of the following deals would you prefer? Complete the explanation below to tell which

kherson [118]

Answer

Buy 2, get 2 free and/or 1/2 off

Step-by-step explanation:

OK, lets say that the tire price was 15. (l)= 15+15=30 (ll)= 15/45%= 33.33x4=133.32 (lll)= 15/2=7.5 7.5x4=30

6 0

2 years ago

What is the measurement for <8? A:108 B:98 C:252 D: 37.4

Verizon [17]

Answer:

c is a answer

yh i think

6 0

2 years ago

7x+14 2x2 + x-6 What is the GCF

fenix001 [56]

i will solve it first i need full form

6 0

2 years ago

Read 2 more answers