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3 years ago

ALL MY POINTS PLZ BE RIGHT

Mathematics

2 answers:

shepuryov [24]3 years ago

7 0

Answer:

-xy^6 +2y -6

Step-by-step explanation:

(-6x^2 y^8+12xy^3-36xy^2)

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6xy^2

Divide each term by 6 in the numerator and denominator

(-6/6x^2 y^8+12/6xy^3-36/6xy^2)

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6/6xy^2

-x^2y^8 +2xy^3 -6xy^2

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xy^2

Divide each term by x in the numerator and denominator

-x^2/xy^8 +2x/xy^3 -6x/xy^2

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x/xy^2

-xy^8 +2y^3 -6y^2

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y^2

Divide each term by y^2 in the numerator and denominator

Remember when dividing, we subtract the exponents

-xy^8/y^2 +2y^3/y^2 -6y^2/y^2

---------------------------------

y^2 /y^2

-xy^6 +2y^1 -6

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-xy^6 +2y -6

Anastaziya [24]3 years ago

7 0

Answer: BBBBBBBBBBBBBBBBBBBBBB

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Divide the rational expressions and express in simplest form. When typing your answer for the numerator and denominator be sure

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Dividing by a fraction is equivalent to multiply by its reciprocal, then:

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Applying the quadratic formula to the second polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the third polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}$

Applying the quadratic formula to the fourth polynomial:

$\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}$

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