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q1) Carry out the following divisions 

ii) 

-36 y^3 \div 9 y^2

Answers (1)

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We have,
         -36y^{3} = -1 \times 2 \times 2 \times 3 \times 3 \times y \times y \times y
           9y^{2 }  = 3 \times 3 \times y \times y
Therefore,
                  \frac{-36y^{3}}{9y^{2}} = \frac {-1 \times 2 \times 2 \times 3 \times 3 \times y \times y \times y}{3 \times 3 \times y \times y} = -4y

Posted by

Gautam harsolia

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