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3 years ago

PLEASE HELP ILL MARK BRAINLIEST

Mathematics

2 answers:

oee [108]3 years ago

8 0

Answer:

3*2n=6n

3*4=12

And that gets you to the 6n-12

And adding I will Mark you brainliest is a good way to get your answers (though you probably won’t get so much if you keep saying it but never do)

The answer is C

Assoli18 [71]3 years ago

5 0

Answer: (C)The third answer that talks about distributing the 3

Because once you use the distributive property ( by multiplying three times 2n-4) it will bring you to the second step which is to subtract on both sides

Hope you understand!!!!

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Please need help please

Marianna [84]

Answer:

D is the correct answer! Hope it hepls!

Step-by-step explanation:

Please mark brainliest!

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4 years ago

Hola buenos días ayuda

tangare [24]

Hi i don’t speak spanish but hi....?

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3 years ago

Read 2 more answers

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

5 0

2 years ago

Given that PQRS is a square, find x and y.

Dovator [93]

Every angle of a square equals 90 degrees.

2x - 4 = 90 3y + 6 = 90
2x = 90 + 4 3y = 90 - 6
2x = 94 3y = 84
x = 94/2 y = 84/3
x = 47 y = 28

so x = 47 and y = 28

4 0

3 years ago

Read 2 more answers

Omg plzzzzzzzzzzz HELPPPPPPPPPPP MEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

natta225 [31]

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers