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xz_007 [3.2K]
3 years ago
6

18. The yearly attendance at a local restaurant is 54,200 and grows continuously at a rate of 7.2% each year. What is the approx

imate attendance of the
restaurant in 21 years? (2 points)
136,150
58,239
245,840
233,394
Mathematics
1 answer:
Eva8 [605]3 years ago
8 0

Answer:

233,394

Step-by-step explanation:

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-1xf(-8)-4xg(4) pls help
KATRIN_1 [288]

Answer:

8f-16g

Step-by-step explanation:

-1×f(-8) -4×g(4)

-1×-8f -4×4g

8f -16g

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7is subtracted from the quotient of 48 divided by th<br>e sum of 5 &amp; difference of 11 and 8​
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Answer:

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3 years ago
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Rewrite the expression in terms of the given function 1/1-sinx - sinx/1+sinx
Feliz [49]
Your question seems a bit incomplete, but for starters you can write

\dfrac1{1-\sin x}-\dfrac{\sin x}{1+\sin x}=\dfrac{1+\sin x}{(1-\sin x)(1+\sin x)}-\dfrac{\sin x(1-\sin x)}{(1+\sin x)(1-\sin x)}=\dfrac{1+\sin x-\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}

Expanding where necessary, recalling that (1-\sin x)(1+\sin x)=1-\sin^2x=\cos^2x, you have

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

and you can stop there, or continue to rewrite in terms of the reciprocal functions,

\dfrac{1+\sin^2x}{\cos^2x}=\sec^2x+\tan^2x

Now, since 1+\tan^2x=\sec^2x, the final form could also take

\sec^2x+\tan^2x=\sec^2x+(\sec^2x-1)=2\sec^2x-1

or

\sec^2x+\tan^2x=(1+\tan^2x)+\tan^2x=1+2\tan^2x
7 0
3 years ago
Which expression has a value of 20?
vovangra [49]
<span>(16−8)×2+4(16−8)×2+4
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