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Makovka662 [10]

3 years ago

7

PLEASE HELP WITH #9! I really don't remember how to do this. THANKS.

1 answer:

AysviL [449]3 years ago

7 0

The answer is A because you want to get "y" by itself. When you add the 10 over, you get 6x+9=3y. Then, you get "u" by itself by dividing everything by 3. You end up getting 2x+3=y and the slope is the number in front of x and the y-intercept is the number not attached to a variable.

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Find the surface area of the regular hexagonal pyramid shown to the nearest square meter. a. 406 m2 b. 393 m2 c. 819 m2 d. 240 m

elena-s [515]

Answer is A. 406m2 so there is the answer to anyone who comes across this.

8 0

3 years ago

Javier rounded to the nearest half to estimate the oof 3 2/5 and -3 7/8. How do the estimate and the actual product compare?

Leya [2.2K]

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You have to add them then dived

Step-by-step explanation:

8 0

2 years ago

Can somebody help me with this question. Thank you!

Alex_Xolod [135]

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Step-by-step explanation:

3 0

2 years ago

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Differentiate with respect to X <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7Bcos2x%7D%7B1%20%2Bsin2x%20%7D%20

Mice21 [21]

Power and chain rule (where the power rule kicks in because $\sqrt x=x^{1/2}$ ):

$\left(\sqrt{\dfrac{\cos(2x)}{1+\sin(2x)}}\right)'=\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'$

Simplify the leading term as

$\dfrac1{2\sqrt{\frac{\cos(2x)}{1+\sin(2x)}}}=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}$

Quotient rule:

$\left(\dfrac{\cos(2x)}{1+\sin(2x)}\right)'=\dfrac{(1+\sin(2x))(\cos(2x))'-\cos(2x)(1+\sin(2x))'}{(1+\sin(2x))^2}$

Chain rule:

$(\cos(2x))'=-\sin(2x)(2x)'=-2\sin(2x)$

$(1+\sin(2x))'=\cos(2x)(2x)'=2\cos(2x)$

Put everything together and simplify:

$\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{(1+\sin(2x))(-2\sin(2x))-\cos(2x)(2\cos(2x))}{(1+\sin(2x))^2}$

$=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2\sin^2(2x)-2\cos^2(2x)}{(1+\sin(2x))^2}$

$=\dfrac{\sqrt{1+\sin(2x)}}{2\sqrt{\cos(2x)}}\dfrac{-2\sin(2x)-2}{(1+\sin(2x))^2}$

$=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac{\sin(2x)+1}{(1+\sin(2x))^2}$

$=-\dfrac{\sqrt{1+\sin(2x)}}{\sqrt{\cos(2x)}}\dfrac1{1+\sin(2x)}$

$=-\dfrac1{\sqrt{\cos(2x)}}\dfrac1{\sqrt{1+\sin(2x)}}$

$=\boxed{-\dfrac1{\sqrt{\cos(2x)(1+\sin(2x))}}}$

5 0

3 years ago

Find the value of x.

maxonik [38]

180 = 132 + x + x

180 - 132 = 2x

48 = 2x

x = 24⁰

5 0

2 years ago

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