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3 years ago

Find the exact value of cos(arcsin(one fourth)). For full credit, explain your reasoning.

1 answer:

5 0

We are asked to give the exact value of cos(arcsin(one fourth)). In this case, we shift first the setting to degrees since this involves angles. we determine first arc sin of one fourth equal to 14.48 degrees. then we take the cos of 14.48 degrees equal to 0.9682. Answer is 0.9682.

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Please hurry , help please??? find the values of the variables

MArishka [77]

Answer:

x=60

Step-by-step explanation:

The angles here are alternate interior angles, this means that they are congruent. Congruent means that they are equal to each other, so to solve just set them equal! (Your variable is x so you're solving for that)

$2x=120\\x=60$

Hope this helps! :)

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2 years ago

Identify the vertical asymptotes of <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%5E3-16x%7D%7B-3x%5E2%2B3x%2B18%7D" id="TexForm

navik [9.2K]

$\sf \frac{ {x}^{3} - 16x}{ { - 3x}^{2} + 3x + 18 }$

To find the vertical asymptotes set denominator as 0

$\sf - 3x ^{2} + 3x + 18 = 0$

Divide both sides of the equation by-3

$\sf \: {x}^{2} - x - 6 = 0$

Write -x as a difference

$\sf \: {x}^{2} + 2x - 3x - 6= 0$

Factor out x from the expression

$\sf \: x \times (x + 2) - 3x - 6 = 0$

Factor out -3 from the expression

$\sf \: x \times (x + 2) - 3(x + 2) = 0$

Factor out x+2 from the expression

$\sf(x + 2)(x - 3) = 0$

When the product of factors equals 0, at least one factor is 0

$\sf \: x + 2 = 0 \\ \sf \: x - 3 = 0$

Solve the equations for x

$\sf \: x = - 2 \\ \sf \: x = 3$

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