I need help 5=6(1q-5)-19

2 answers:

Rudiy273 years ago

7 0

ok distribute the 6 by multiplying it by 1q and -5
5=6q-30-19
Now subtract 30-19
5=6q-11
add 11 to both sides
16=6q
divide both sides by 6
q= 16/6
simplify
16/2=8, 6/2=3
q=8/3

IRISSAK [1]3 years ago

4 0

So : 5 = 6(1q - 5) - 19.

You would distribute the number outside the parentheses to the numbers inside the parentheses. This would leave you with :

5 = 6q - 30 - 19
5 = 6q - 49

Add 49 on both sides, because you're trying to isolate q, and you need to rid all the numbers on that side. Adding 49 on both sides would leave you with :

54 = 6q

Divide both sides by 6, because again, inverse operations, and you need to isolate the q. Divide both sides by 6, and you get :

9 = q, which turns into q = 9

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The result of expanding the trigonometry expression $\sin^2(\theta) * (1 + \cos(\theta))$ is $cos^0(\theta) + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)$

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The expression is given as:

$\sin^2(\theta) * (1 + \cos(\theta))$

Express $\sin^2(\theta)$ as $1 - \cos^2(\theta)$ .

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$\sin^2(\theta) * (1 + \cos(\theta)) = (1- \cos^2(\theta)) * (1 + \cos(\theta))$

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$\sin^2(\theta) * (1 + \cos(\theta)) = 1 + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)$

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$\sin^2(\theta) * (1 + \cos(\theta)) = cos^0(\theta) + \cos(\theta) - \cos^2(\theta) - \cos^3(\theta)$

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