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mars1129 [50]
4 years ago
15

Which expression is equivalent to (x^-4y/x^-9y^5)^-2 ? Assume x ≠ 0, y ≠ 0

Mathematics
2 answers:
iVinArrow [24]4 years ago
8 0

<u>Question 1: Which expression is equivalent to (x^-4y/x^-9y^5)^-2 ? </u>

(\frac{x^{-4} y}{x^{-9} y^{5} } )^{-2} \\\\&#10;=(\frac{x^{-4}x^{9}}{ y^{5} y^{-1} } )^{-2} \\\\&#10;=(\frac{x^{9-4}}{ y^{5-1}} )^{-2} \\\\&#10;=(\frac{x^{5}}{ y^{4}} )^{-2} \\\\&#10;=(\frac{y^{4}}{x^{5}} )^{2} \\\\&#10;=\frac{y^{4*2}}{x^{5*2}} \\\\&#10;=\frac{y^{8}}{x^{10}}

Hence, option A is correct i.e. \frac{y^{8}}{x^{10}} .

<u>Question 2: Which expression is equivalent to 28p^9q^-5/12p^-6q^7 ?</u>

\frac{28 p^{9} q^{-5} }{12 p^{-6} q^{7} } \\\\&#10;=\frac{28 p^{9} p^{6} }{12 q^{7} q^{5} } \\\\&#10;=\frac{28 p^{9+6} }{12 q^{7+5} } \\\\&#10;=\frac{28 p^{15} }{12 q^{12} } \\\\&#10;=\frac{7 p^{15} }{3 q^{12} }

Hence, option B is correct i.e. \frac{7 p^{15} }{3 q^{12} } .

<u>Question 3: Which expression is equivalent to (5ab)^3/30a^-6b^-7 ? </u>

\frac{(5ab)^{3} }{30 a^{-6} b^{-7} } \\\\&#10;=\frac{125 a^{3} b^{3} }{30 a^{-6} b^{-7} } \\\\&#10;=\frac{125 a^{3} b^{3} a^{6} b^{7} }{30} \\\\&#10;=\frac{25 a^{3+6} b^{3+7} }{6} \\\\&#10;=\frac{25 a^{9} b^{10} }{6}

Hence, option D is correct i.e. \frac{25 a^{9} b^{10} }{6} .

Flauer [41]4 years ago
4 0
<h3><u>Answer:</u></h3>

Ques 1)

\dfrac{y^8}{x^10}

( option: A)

Ques 2)

\dfrac{7p^{15}}{3q^{12}}

(Option B)

Ques 3)

\dfrac{25a^9b^{10}}{6}

( Option: D)

<h3><u>Step-by-step explanation:</u></h3>

Ques 1)

(\dfrac{x^{-4}y}{x^{-9}y^5})^{-2}\\\\\\=(x^{-4+9}y^{1-5})^{-2}\\\\=(x^5y^{-4})^{-2}\\\\=x^{5\times (-2)}y^{-4\times (-2)}\\\\=x^{-10}y^8\\\\=\dfrac{y^8}{x^10}

( since, we know that:

\dfrac{x^m}{x^n}=x^{m-n}

Also,

(x^m)^n=x^{m\times n} )

Hence, option: A is correct.

Ques 2)

\dfrac{28p^9q^{-5}}{12p^{-6}q^7}\\\\\\=\dfrac{28}{12}\times (p^{9+6}q^{-5-7})\\\\=\dfrac{7}{3}\times (p^{15}q^{-12})\\\\=\dfrac{7p^{15}}{3q^{12}}

Hence, option: B is correct.

Ques 3)

=\dfrac{(5ab)^3}{30a^{-6}b^{-7}}\\\\=\dfrac{125a^3b^3}{30a^{-6}b^{-7}}\\\\=\dfrac{125}{30}\times (a^{3+6}b^{3+7})\\\\=\dfrac{25}{6}\times (a^9b^{10})\\\\=\dfrac{25a^9b^{10}}{6}

Hence, option: D is true.

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