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Mandarinka [93]
4 years ago
8

What is the measure of Angle y?

Mathematics
2 answers:
olga55 [171]4 years ago
7 0

Answer:

41

Step-by-step explanation:

180 - 139 = 41

____ [38]4 years ago
3 0

Answer: A: 41

Step-by-step explanation: The answer is a because the line equaled 180 degrees. So when the measurement of 139 is shown then you would subtract 139 from 180, and your sum is 41.

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Answer:

5 shares if only whole shares are allowed

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Step-by-step explanation:

112(0.80) / 16.20 = 5.530864

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Verify that:
Lelu [443]

Answer:

See Below.

Step-by-step explanation:

Problem 1)

We want to verify that:

\displaystyle \left(\cos(x)\right)\left(\cot(x)\right)=\csc(x)-\sin(x)

Note that cot(x) = cos(x) / sin(x). Hence:

\displaystyle \left(\cos(x)\right)\left(\frac{\cos(x)}{\sin(x)}\right)=\csc(x)-\sin(x)

Multiply:

\displaystyle \frac{\cos^2(x)}{\sin(x)}=\csc(x)-\sin(x)

Recall that Pythagorean Identity: sin²(x) + cos²(x) = 1 or cos²(x) = 1 - sin²(x). Substitute:

\displaystyle \frac{1-\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)

Split:

\displaystyle \frac{1}{\sin(x)}-\frac{\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)

Simplify:

\csc(x)-\sin(x)=\csc(x)-\sin(x)

Problem 2)

We want to verify that:

\displaystyle (\csc(x)-\cot(x))^2=\frac{1-\cos(x)}{1+\cos(x)}

Square:

\displaystyle \csc^2(x)-2\csc(x)\cot(x)+\cot^2(x)=\frac{1-\cos(x)}{1+\cos(x)}

Convert csc(x) to 1 / sin(x) and cot(x) to cos(x) / sin(x). Thus:

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

Factor out the sin²(x) from the denominator:

\displaystyle \frac{1}{\sin^2(x)}\left(1-2\cos(x)+\cos^2(x)\right)=\frac{1-\cos(x)}{1+\cos(x)}

Factor (perfect square trinomial):

\displaystyle \frac{1}{\sin^2(x)}\left((\cos(x)-1)^2\right)=\frac{1-\cos(x)}{1+\cos(x)}

Using the Pythagorean Identity, we know that sin²(x) = 1 - cos²(x). Hence:

\displaystyle \frac{(\cos(x)-1)^2}{1-\cos^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}

Factor (difference of two squares):

\displaystyle \frac{(\cos(x)-1)^2}{(1-\cos(x))(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}

Factor out a negative from the first factor in the denominator:

\displaystyle \frac{(\cos(x)-1)^2}{-(\cos(x)-1)(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}

Cancel:

\displaystyle \frac{\cos(x)-1}{-(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}

Distribute the negative into the numerator. Therefore:

\displaystyle \frac{1-\cos(x)}{1+\cos(x)}=\displaystyle \frac{1-\cos(x)}{1+\cos(x)}

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