Evaluate: $(-8x^2y^6) \times (-20xy)$ for $x = 2.5$ and $y = 1$.

AcademicMathematicsNCERTClass 8

Given:

$(-8x^2y^6) \times (-20xy)$

To do:

We have to evaluate $(-8x^2y^6) \times (-20xy)$ when $x = 2.5, y = 1$.

Solution:

$(-8 x^{2} y^{6}) \times(-20 x y)=(-8) \times(-20) \times x^{2} \times x \times y^{6} \times y$

$=160 \times x^{2+1} \times y^{6+1}$

$=160 \times x^{3} \times y^{7}$

$=160 \times(2.5)^{3} \times(1)^{7}$

$=160 \times(\frac{5}{2})^{3} \times 1$

$=160 \times \frac{125}{8}$

$=20 \times 125$

$=2500$

raja
Updated on 10-Oct-2022 13:19:29

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