How do you divide $\frac { 13u ^ { 4} + 6u ^ { 3} + 4u ^ { 2} } { 2u ^ { 2} }$ ?

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How do you divide $\frac { 13u ^ { 4} + 6u ^ { 3} + 4u ^ { 2} } { 2u ^ { 2} }$ ?

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Answer 1 · 2017-06-06T18:16:17

$\frac{13}{2}u^2 + 3u + 2$

Explanation:

The denominator is only a single term so you can split the fractions up and divide them individually: $\frac{13u^4 + 6u^3 + 4u^2}{2u^2} = \frac{13u^4}{2u^2} + \frac{6u^3}{2u^2} + \frac{4u^2}{2u^2}$ , and the " $u$ "s cancel out to give: $\frac{13}{2}u^2 + 3u + 2$ . I hope I have explained it in enough detail.

Answer 2 · 2017-06-06T18:36:49

$13/2u^2+3u+2$

Explanation:

$"'split' up the terms in the numerator dividing each one by"$
$"the denominator"$

$rArr(13u^4+6u^3+4u^2)/(2u^2)$

$=(13u^4)/(2u^2)+(6u^3)/(2u^2)+(4u^2)/(2u^2)$

$=13/2u^2+3u+2$

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How do you divide $\frac { 13u ^ { 4} + 6u ^ { 3} + 4u ^ { 2} } { 2u ^ { 2} }$ ?

2 Answers

Explanation:

Explanation:

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Science

Math

Humanities

... and beyond

How do you divide \frac { 13u ^ { 4} + 6u ^ { 3} + 4u ^ { 2} } { 2u ^ { 2} }\frac { 13u ^ { 4} + 6u ^ { 3} + 4u ^ { 2} } { 2u ^ { 2} }?

2 Answers

Explanation:

Explanation:

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How do you divide $\frac { 13u ^ { 4} + 6u ^ { 3} + 4u ^ { 2} } { 2u ^ { 2} }$ ?