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Nastasia [14]
3 years ago
7

The area of a parallelogram is given by the formula A = bh . The area of the parallelogram on the right is 85 sq. units . Find i

ts base and height
Mathematics
1 answer:
vodka [1.7K]3 years ago
5 0

Answer:

Where's the image

Step-by-step explanation:

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1. Use the integer multiplication facts in their integer bubble to create six related integer division facts.

2.The quotient of any two integers (with a non zero divisor) will be a rational number. If and are integers, then - (p/q)= (---)=(----)

3. Mrs. McIntire, a seventh-grade math teacher, is grading papers. Three students gave the following responses to the same math problem: 1/-2 -(1/2) -1/2

4.On Mrs. McIntire’s answer key for the assignment, the correct answer is −0.5. Which student answer(s) is (are) correct? Explain.

Step-by-step explanation:

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Elaine had StartFraction 5 Over 8 EndFraction of a pizza left after lunch. She told 5 of her friends that they could share the r
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1. if csc β = 7/3 and cot β = - 2√10 / 3, Find sec β
slava [35]

Step-by-step explanation:

1.

\tan \beta  =  \frac{1}{ \cot \beta }  =  -  \frac{3}{2 \sqrt{10} }  =  -  \frac{3 \sqrt{10} }{20}

\csc \beta  \tan \beta  =  \frac{1}{ \cos \beta  }  =  \sec \beta

Therefore,

\sec \beta  = ( \frac{7}{3} )( -  \frac{3 \sqrt{10} }{20} ) =  -  \frac{7 \sqrt{10} }{20}

2.

\csc y =  \frac{1}{ \sin y}  =  -  \frac{ \sqrt{6} }{2}

=  >  \sin y =  -  \frac{ \sqrt{6} }{3}

Use the identity

\cos y =   \sqrt{1 -  \sin ^{2} y}    \:  \:  \: \:  \:  \:  \:  \:  \:  \:  \:  \\ =  \sqrt{1 -  {( -  \frac{ \sqrt{6} }{3}) }^{2} }  =  -  \frac{ \sqrt{3} }{3}

We chose the negative value of the cosine because of the condition where cot y > 0. Otherwise, choosing the positive root will yield a negative cotangent value. Now that we know the sine and cosine of y, we can now solve for the tangent:

\tan \beta  =  \frac{ \sin y}{ \cos y} =( -  \frac{ \sqrt{6} }{3} )( -  \frac{3}{ \sqrt{3} } ) =  \sqrt{2}

3. Recall that sec x = 1/cos x, therefore cos x = 5/6. Solving for sin x,

\sin x =   \sqrt{1 -  \cos ^{2} x} =  \sqrt{ \frac{11}{6} }

Solving for tan x:

\tan x =  \frac{ \sin x}{ \cos x}  =  (\frac{ \sqrt{11} }{ \sqrt{6} } )( \frac{6}{5} ) =  \frac{ \sqrt{66} }{5}

5 0
3 years ago
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