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adell [148]
3 years ago
7

Help please!!

Mathematics
2 answers:
sergiy2304 [10]3 years ago
6 0
<span>The cosine of pi/2 is 0 so the problem really says "find the angle between - pi/2 and pi/2 whose tan = 0. That would be the same as asking for the angle whose sin is 0. That would be 0.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
</span>
ElenaW [278]3 years ago
4 0

Answer:

The angle is 0°.

Step-by-step explanation:

Consider the provided trigonometric function.

tan^{-1}(cos\frac{\pi}{2})

The given interval [\frac{-\pi}{2}, \frac{\pi}{2}]

The value of cos\frac{\pi}{2}=0

Now substitute the respective value in the provided trigonometric function.

tan^{-1}(tan(0))=0 by inverse property of trigonometric function

Hence, the angle is 0°.

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For the group, it is 5(x+12)

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\Rightarrow 5(x+12)=110\\\Rightarrow x+12=22\\\Rightarrow x=\$\ 10

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7 0
4 years ago
Given that (7x-8)/(2x-1)(x-2) <br><br> Use Partial Fraction Decomposition to Find A and B.
koban [17]

We look for constants <em>a</em> and <em>b</em> such that

\displaystyle \frac{7x-8}{(2x-1)(x-2)} = \frac a{2x-1} + \frac b{x-2}

Rewrite all terms with a common denominator and set the numerators equal:

\displaystyle \frac{7x-8}{(2x-1)(x-2)} = \frac{a(x-2)}{(2x-1)(x-2)} + \frac{b(2x-1)}{(x-2)(2x-1)} \\\\ \frac{7x-8}{(2x-1)(x-2)} = \frac{a(x-2)+b(2x-1)}{(2x-1)(x-2)} \\\\ \implies 7x-8 = a(x-2) + b(2x-1) = (a+2b)x - 2a - b

Then

<em>a</em> + 2<em>b</em> = 7

-2<em>a</em> - <em>b</em> = -8

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2 (<em>a</em> + 2<em>b</em>) + (-2<em>a</em> - <em>b</em>) = 2(7) + (-8)

2<em>a</em> + 4<em>b</em> - 2<em>a</em> - <em>b</em> = 14 - 8

3<em>b</em> = 6

<em>b</em> = 2

Then

<em>a</em> + 2(2) = 7   ==>   <em>a</em> = 3

and so

\displaystyle \frac{7x-8}{(2x-1)(x-2)} = \frac 3{2x-1} + \frac 2{x-2}

7 0
3 years ago
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Can anyone help me with my Geometry question?
WITCHER [35]

Answer:

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Step-by-step explanation:

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7 0
3 years ago
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