Evaluate: $(3.2x^6y^3) \times (2.1x^2y^2)$ when $x = 1$ and $y = 0.5$.

Given:

$(3.2x^6y^3) \times (2.1x^2y^2)$

To do:

We have to evaluate $(3.2x^6y^3) \times (2.1x^2y^2)$ when $x = 1$ and $y = 0.5$.

Solution:

$(3.2x^6y^3) \times (2.1x^2y^2)=3.2 \times 2.1 \times x^{6+2} \times y^{3+2}$

$= 6.72x^8y^5$

$= 6.72 \times (1)^8 \times (0.5)^5$

$= 6.72 \times 1 \times 0.03125$

$= 0.21$

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Updated on: 10-Oct-2022

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