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VashaNatasha [74]
3 years ago
15

22% of people watch Netflix. 38% of people watch HBO. 10% of people watch both

Mathematics
1 answer:
timama [110]3 years ago
6 0

Answer:

40%

Step-by-step explanation:

Netflix:

22% - 10% = 12%

HBO:

38% - 10% = 28%

People who watch Netflix only: 12%

People who watch HBO only: 28%

People who watch both Netflix & HBO: 10%

People who watch HBO or Netflix: 12% + 28% + 10% = 40%

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Ksju [112]

Isolate the variable Y by multiplying 1.5 on both sides.

Y=1.6*1.5

Y=2.4

Hope this helps.

頑張って!

7 0
3 years ago
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Given equations A and B as 2/5x+y=12 and 5/2y-x=6, respectively, which expression will eliminate variable x?
balu736 [363]
The answer is C.) A+2/5B.
Adding equation A to 2/5B will result in canceling the x terms, i.e. 2/5x + 2/5(-x) = 0.
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3 years ago
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2,451 people came to watch the Easter parade. 745 of those people were adults. How many children came to the parade?​
8_murik_8 [283]

Answer:f

Step-by-step explanation:d

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3 years ago
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
Factorise this equation: 4x²-2x-5
Karolina [17]

Answer:

PEMDAS Answer:A

Step-by-step explanation:

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7 0
3 years ago
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