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LUCKY_DIMON [66]

2 years ago

7

6w^2 + 11w +8w^2 + 15w -2 and 14w^2 +26w-2 are the equivalent

1 answer:

Agata [3.3K]2 years ago

5 0

Answer:

6w^2 + 11w +8w^2 + 15w -2

Get same power near,

6w^2 + 8w^2 + 11<em>w</em> + 15<em>w</em> -2

w^2 (6 + 8) + <em>w(11</em> + 15) -2

14w^2 +26w-2

Yes equivalent

Step-by-step explanation:

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Why are the solutions to the proportions 50/x =10/20 and 10/50=20/x the same

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Answer:

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Step-by-step explanation:

Given

$u =$

$v =$

Required:

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$cos\theta = \frac{12-28}{|u|.|v|}$

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For a vector

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So;

$|u| = \sqrt{2^2 + 6^2}$

$|u| = \sqrt{4 + 36}$

$|u| = \sqrt{40}$

$|v| = \sqrt{4^2+(-7)^2}$

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$|v| = \sqrt{65}$

So:

$cos\theta = \frac{-16}{|u|.|v|}$

$cos\theta = \frac{-16}{\sqrt{40}*\sqrt{65}}$

$cos\theta = \frac{-16}{\sqrt{2600}}$

$cos\theta = \frac{-16}{\sqrt{100*26}}$

$cos\theta = \frac{-16}{10\sqrt{26}}$

$cos\theta = \frac{-8}{5\sqrt{26}}$

Take arccos of both sides

$\theta = cos^{-1}(\frac{-8}{5\sqrt{26}})$

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$\theta = cos^{-1}(\frac{-8}{25.495})$

$\theta = cos^{-1}(-0.31378701706)$

$\theta = 108.288386087$

<em></em> $\theta = 108.29$ <em> (approximated)</em>

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