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Mathematics

1 answer:

Dennis_Churaev [7]2 years ago

6 0

Answer:

2.5, 5, 7, 8

Step-by-step explanation:

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2 years ago

What is the quotient 2y^-6y-20/4y+12 ÷ y+5y+6/3y^2+28y+27

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Answer with explanation:

$\rightarrow \frac{\frac{2y^2-6 y-20}{4 y+12}}{\frac{y^2+5 y+6}{3 y^2+28 y+27}}\\\\\rightarrow \frac{\frac{y^2-3y-10}{2 y+6}}{\frac{(y+2)(y+3)}{3 y^2+28 y+27}}\\\\\rightarrow \frac{\frac{(y-5)(y+2)}{2 (y+3)}}{\frac{(y+2)(y+3)}{3 y^2+28 y+27}}\\\\\rightarrow \frac{(y-5)(y+2)}{2 (y+3)}} \times {\frac{3 y^2+28 y+27}{(y+2)(y+3)}}\\\\ \rightarrow\frac{(y-5)\times(3 y^2+28 y+27)}{2 (y+3)^2}}$

→y²+5y+6

=y²+3 y+2 y+6

=y×(y+3)+2×(y+3)

=(y+2)(y+3)

→y² -3 y-10

=y² -5 y+2 y -10

=y×(y-5)+2×(y-5)

=(y+2)(y-5)

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Alexus [3.1K]

If you are saying:
(3/12) - (8/12)
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Following the rule of adding and subtracting fractions, the denominators will not change unless they are different.

So: (3/12) - (8/12) = -5/12

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