Question

Find the quotient of the quantity 21 times x to the 2nd power times y to the 6th power plus 6 times x times y to the 3rd power minus 30 times x times y all over 3 times x times y. 21x2y6 + 6xy3 − 10 7xy5 + 2y2 + 10 7xy5 + 2y2 − 10 7xy5 + 2y − 10

Answer 1

$\cfrac{21x^2y^6+6xy^3-30xy}{3xy} = \cfrac{3xy(7xy^5+2y^2-10)}{3xy} =7xy^5+2y^2-10$

Answer 2

Answer:

Option C. $7xy^{5}+2y^{2}-10$ is the answer.

Step-by-step explanation:

The given fraction is $\frac{21x^{2}y^{6}+6xy^{3}-30xy}{3xy}$.

We have to find the quotient. As we know quotient is the net result of division of any fraction, so we will solve the fraction to get the value of quotient.

$\frac{3xy(7xy^{5}+2y^{2}-10)}{3xy}$

$7xy^{5}+2y^{2}-10$

Therefore the quotient is Option C. $7xy^{5}+2y^{2}-10$.