Question

Evaluate (3.2x⁶y³) × (2.1x²y²) when x = 1 and y = 0.5

Ans

First multiply the expressions and then substitute the values for the variables.

To multiply algebric experssions use the commutative and the associative laws along with the law of indices, $a^m\times a^n=a^{m+n}$.
We have,

$$\begin{gathered} \left(3.2x^6{y}^3\right)\times \left(2.1x^2{y}^2\right) \\ =\left(3.2\times 2.1\right)\times \left(x^6\times x^2\right)\times \left(y^3\times y^2\right) \\ =6.72x^8{y}^5 \end{gathered}$$
Hence, $\left(3.2x^6{y}^3\right)\times \left(2.1x^2{y}^2\right)=6.72x^8{y}^5$

Now, substitute 1 for x and 0.5 for y in the result.

$$\begin{gathered} 6.72x^8{y}^5 \\ =6.72{\left(1\right)}^8{\left(0.5\right)}^5 \\ =6.72\times 1\times 0.03125 \\ =0.21 \end{gathered}$$

Hence, the answer is $0.21$.

Answer 1

To multiply algebraic expressions, we use commutative and associative laws along with the laws of indices, i.e.,​ $a^m\times a^n=a^{m+n}$.

We have:

$$\begin{gathered} \left(3.2x^6{y}^3\right)\times \left(2.1x^2{y}^2\right) \\ =\left(3.2\times 2.1\right)\times \left(x^6\times x^2\right)\times \left(y^3\times y^2\right) \\ =\left(3.2\times 2.1\right)\times \left(x^{6+2}\right)\times \left(y^{3+2}\right) \\ =6.72x^8{y}^5 \end{gathered}$$

$\therefore$ $\left(3.2x^6{y}^3\right)\times \left(2.1x^2{y}^2\right)=6.72x^8{y}^5$

Substituting x = 1 and y = 0.5 in the result, we get:

$$\begin{gathered} 6.72x^8{y}^5 \\ =6.72{\left(1\right)}^8{\left(0.5\right)}^5 \\ =6.72\times 1\times 0.03125 \\ =0.21 \end{gathered}$$

Thus, the answer is $0.21$.