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DanielleElmas [232]
4 years ago
12

Someone please help me serious answers only!!!!!!

Mathematics
1 answer:
Maru [420]4 years ago
5 0
Hello!

First of all, let's look at the x value 1. There is one point located at y value 2. We will plug our x value into each equation and see if it has an outcome of two.

A. 2≠-2
B.2≠1
C.2≠0
D.2=2
E.2≠4
F.2≠3

Therefore, D is the correct equation.
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4 years ago
What is the quotient?
exis [7]

The quotient is \frac{3(y-5)}{2}

Explanation:

The expression is \frac{2 y^{2}-6 y-20}{4 y+12} \div \frac{y^{2}+5 y+6}{3 y^{2}+18 y+27}

Now, reciprocal the term \frac{y^{2}+5 y+6}{3 y^{2}+18 y+27} and convert ÷ to ×

\frac{2 y^{2}-6 y-20}{4 y+12} \times \frac{3 y^{2}+18 y+27}{y^{2}+5 y+6}

Taking the common terms out, we get,

\frac{2\left(y^{2}-3 y-10\right)}{4(y+3)} \times \frac{3\left(y^{2}+6 y+9\right)}{y^{2}+5 y+6}

Factorizing each numerator and denominator, we get,

\frac{2(y-5)(y+2)}{4(y+3)} \times \frac{3(y+3)(y+3)}{(y+3)(y+2)}

Multiplying the terms, we have,

\frac{6(y-5)(y+2)(y+3)^{2}}{4(y+3)^{2}(y+2)}

Cancelling the common factors, we get,

\frac{3(y-5)}{2}

Thus, the quotient is \frac{3(y-5)}{2}

8 0
3 years ago
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