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2 years ago

11

What does paragraph 26 reveal about Rachel's character?

1 answer:

4vir4ik [10]2 years ago

5 0

Answer:

your answer is A hope this helps

Step-by-step explanation:

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Phil gets a commission of 25% on all sales. He sold $1,500.00 worth of goods. What was the amount of his commission check?

Doss [256]

Answer:

the answer is 375.00 please leave a thanks id love that

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2 years ago

64.3 - 3 x 2 by the power of 3 divided by two

kolbaska11 [484]

If you meant:

64.3-3*(2³)/2

That would be 52.3 :)

8 0

2 years ago

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Help please thank you if u do!

dmitriy555 [2]

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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

2 years ago

What is the best approximation of the length of segment QS? (Note: cos 80° = 0.17)

matrenka [14]

Answer: $QS=11.8cm$

Step-by-step explanation:

1. You have the following information:

- The lenght of $RS$ is 2 centimeters.

- The angle m∠S=80°.

- Cos80°=0.17

2. Therefore, you can calculate the lenght of the segment $QS$ as following:

$cos80=\frac{adjacent}{hypotenuse}$

$adjacent=2cm\\hypotenuse=QS\\cos80=0.17$

3. Substitute values and solve for $QS$ :

$0.17=\frac{2cm}{QS}\\QS=\frac{2cm}{0.17} \\QS=11.76cm$

4. To the nearest tenth:

$QS=11.8cm$

6 0

2 years ago