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Mathematics 30 Online
OpenStudy (anonymous):

Simplify: (21x^3y^4) / (-7xy) A. -3x^2y^3 B. -3xy C. 14x^2y^3 D. 14xy My answer: A PLease show me how to get the answer my teacher is a told brat.

OpenStudy (anonymous):

@pooja195

pooja195 (pooja195):

@Nnesha

OpenStudy (anonymous):

@Awolflover1 @zepdrix

Nnesha (nnesha):

\[\huge\rm \frac{ x^m }{ x^n } = x^{m-n}\] move exponent from bottom to the top of fraction (if bases are the same )

Nnesha (nnesha):

\[\large\rm \frac{21x^3y^4}{ -7xy} = -\frac{ 21 }{ 7 } \times \frac{ x^3 }{ x } \times \frac{ y^4 }{ y }\]

OpenStudy (anonymous):

that not right the / spouse to mean divid

Nnesha (nnesha):

yeah ik

Nnesha (nnesha):

what's not correct ??

OpenStudy (anonymous):

nothing I though you though that was a fraction

Nnesha (nnesha):

yep that's a fraction :-)

Nnesha (nnesha):

or you can factor x^3 remember exponent rule when you multiply same bases you should add their exponents \[\huge\rm x^m \times x^n = x^{m+n}\] so you can write \[\large\rm x^3 = x \times x \times x \]

Nnesha (nnesha):

\(\color{blue}{\text{Originally Posted by}}\) @Nnesha \[\large\rm \frac{21x^3y^4}{ -7xy} = -\frac{ 21 }{ 7 } \times \frac{ x^3 }{ x } \times \frac{ y^4 }{ y }\] \(\color{blue}{\text{End of Quote}}\) \[\large\rm -\frac{ 21 }{ 7 } \times \frac{ x \times x \times x }{ x } \times \frac{ y^4 }{ y }\] now cancel ot x

OpenStudy (anonymous):

so their is only 2 x left

Nnesha (nnesha):

^^yep x times x can be written as = ??? hint exponent rule x^m times x^n = x^{m+n}

OpenStudy (anonymous):

okay thanks

Nnesha (nnesha):

got it ??

Nnesha (nnesha):

so what's ur answer ??

OpenStudy (anonymous):

oops sorry -3x^2y^3

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