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user100 [1]
3 years ago
11

Find the missing side length

Mathematics
1 answer:
jolli1 [7]3 years ago
8 0
The missing side length would be 8
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Find the values of x that satisfy
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Setup 2 problems
2x - 7 < 15   and    2x - 7 > -15
2x < 22                  2x > -8
x < 11                    x > -4

Or you can write it  -4 < x < 11

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How can I do better on my test? my daily work is easy enough but I tend not to test well
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Find the missing length of the triangle pleeaasseee hurrryyy needdd urgenntttlllyylllyy
trasher [3.6K]

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1.2

Step-by-step explanation:

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Can someone help me on this/
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3 years ago
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Verify the identity. cotangent of x divided by quantity one plus cosecant of x equals quantity cosecant of x minus one divided b
Elenna [48]

Answer:

\frac{\cot x}{1+\csc x}=\frac{\csc x-1}{\cot x}

Step-by-step explanation:

We want to verify the identity:

\frac{\cot x}{1+\csc x}=\frac{\csc x-1}{\cot x}

Let us take the LHS and simplify to get the LHS.

Express everything in terms of the cosine and sine function.

\frac{\cot x}{1+\csc x}=\frac{\frac{\cos x}{\sin x} }{1+\frac{1}{\sin x} }

Collect LCM

\frac{\cot x}{1+\csc x}=\frac{\frac{\cos x}{\sin x} }{\frac{\sin x+1}{\sin x} }

We simplify the RHS to get:

\frac{\cot x}{1+\csc x}=\frac{\cos x}{\sin x+1}

We rationalize to get:

\frac{\cot x}{1+\csc x}=\frac{\cos x(\sin x-1)}{(\sin x+1)*(\sin x-1)}

We expand to get:

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

Factor negative one in the denominator:

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

Apply the Pythagoras Property to get:

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

Simplify to get:

\frac{\cot x}{1+\csc x}=\frac{-(\sin x-1)}{\cos x}

Or

\frac{\cot x}{1+\csc x}=\frac{1-\sin x}{\cos x}

Divide both the numerator and denominator by sin x

\frac{\cot x}{1+\csc x}=\frac{\frac{1}{\sin x}-\frac{\sin x}{\sin x}}{\frac{\cos x}{\sin x}}

This finally gives:

\frac{\cot x}{1+\csc x}=\frac{\csc x-1}{\cot x}

8 0
3 years ago
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