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Ksivusya [100]
3 years ago
13

Solve -15-2g+6g=1+6g

Mathematics
2 answers:
katrin [286]3 years ago
7 0
<span>-15-2g+6g=1+6g
Combine like terms
-2g=16
Divide by -2 on both sides
g=-8</span>
AleksAgata [21]3 years ago
7 0
<span>-15-2g+6g=1+6g

First: combine like terms
You'll get: -15+4g=1+6g
Next: You would subtract 4g on both sides to where you can still get all of your like terms together
You'll get: -15=1+2g
Then: You'll subtract 1 on both sides because you are still getting your like terms together
You'll get: -16=2g
Finally: You'll divide each side by 2, so your g would be alone
You'll get: -8=g <That would be your answer

HOPE THIS HELPS! ^_^</span>
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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 !

8 0
3 years ago
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