# Write down the product of $-8x^2y^6$ and $-20xy$. Verify the product for $x = 2.5, y = 1$.

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Given:

$-8x^2y^6$ and $-20xy$

To do:

We have to write down the product of $-8x^2y^6$ and $-20xy$ and verify the product for $x = 2.5, y = 1$.

Solution:

The product of $-8x^2y^6$ and $-20xy$ is,

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

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

$= 160x^3y^7$

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

$= -8 \times (2.5)^2 \times (1)^6 \times (-20 \times 2.5 \times 1)$

$= -8 \times 6.25 \times 1 \times -20 \times 2.5$

$= (-50) \times (-50)$

$= 2500$

RHS $= 160 x^3y^7$

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

$= 160 \times 15.625 \times 1$

$=2500$

Therefore,

LHS $=$ RHS

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