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polet [3.4K]
3 years ago
14

Find area of a parallelogram with a height of 8 inches and a base of 14 inches

Mathematics
1 answer:
12345 [234]3 years ago
4 0

Answer:

\boxed{ \bold{ \huge{ \boxed{ \sf{112 \:  {inches} \: ^{2} }}}}}

Step-by-step explanation:

Given,

Height of a parallelogram ( h ) = 8 inches

Base of a parallelogram ( b ) = 14 inches

Area of a parallelogram ( A ) = ?

<u>Finding </u><u>the</u><u> </u><u>area</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>parallelogram</u>

<u>\boxed{ \sf{area \: of \: parallelogram \:  =  \: base \:  \times  \: height}}</u>

⇒<u>\sf{area \: of \: parallelogram \:  =  \: 14  \: \times 8}</u>

⇒<u>\sf{area \: of \: parallelogram = 112 \:  {inches}^{2} }</u>

Hope I helped!

Best regards! :D

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Solve the following system of equations. Enter the x-coordinate of the solution. Round your answer to the nearest tenth.
Sergeu [11.5K]

Answer: <u>3.7</u>

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3 0
2 years ago
If sinA=√3-1/2√2,then prove that cos2A=√3/2 prove that
Ivan

Answer:

\boxed{\sf cos2A =\dfrac{\sqrt3}{2}}

Step-by-step explanation:

Here we are given that the value of sinA is √3-1/2√2 , and we need to prove that the value of cos2A is √3/2 .

<u>Given</u><u> </u><u>:</u><u>-</u>

• \sf\implies sinA =\dfrac{\sqrt3-1}{2\sqrt2}

<u>To</u><u> </u><u>Prove</u><u> </u><u>:</u><u>-</u><u> </u>

•\sf\implies cos2A =\dfrac{\sqrt3}{2}

<u>Proof </u><u>:</u><u>-</u><u> </u>

We know that ,

\sf\implies cos2A = 1 - 2sin^2A

Therefore , here substituting the value of sinA , we have ,

\sf\implies cos2A = 1 - 2\bigg( \dfrac{\sqrt3-1}{2\sqrt2}\bigg)^2

Simplify the whole square ,

\sf\implies cos2A = 1 -2\times \dfrac{ 3 +1-2\sqrt3}{8}

Add the numbers in numerator ,

\sf\implies cos2A =  1-2\times \dfrac{4-2\sqrt3}{8}

Multiply it by 2 ,

\sf\implies cos2A = 1 - \dfrac{ 4-2\sqrt3}{4}

Take out 2 common from the numerator ,

\sf\implies cos2A = 1-\dfrac{2(2-\sqrt3)}{4}

Simplify ,

\sf\implies cos2A =  1 -\dfrac{ 2-\sqrt3}{2}

Subtract the numbers ,

\sf\implies cos2A = \dfrac{ 2-2+\sqrt3}{2}

Simplify,

\sf\implies \boxed{\pink{\sf cos2A =\dfrac{\sqrt3}{2}} }

Hence Proved !

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Answer:

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Step-by-step explanation:

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