Question
Evaluate $\left(3.2 x^6 y^3\right) \times\left(2.1 x^2 y^2\right)$ when $x = 1$ and $y = 0.5$.
✓
Answer
To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e. $a^m× a^n= a^{m+n}$.
We have:
$ \left(3.2 x^6 y^3\right) \times\left(2.1 x^2 y^2\right) $
$ =(3.2 \times 2.1) \times\left(x^6 \times x^2\right) \times\left(y^3 \times y^2\right) $
$ =(3.2 \times 2.1) \times\left(x^{6+2}\right) \times\left(y^{3+2}\right) $
$ =6.72 x^8 y^5$
Substituting $\mathrm{x}=1$ and $\mathrm{y}=0.5$ in the result, we get:
$ 6.72 x^8 y^5 $
$ =6.72(1)^8(0.5)^5 $
$ =6.72 \times 1 \times 0.03125 $
$ =0.21$
Thus, the answer is $0.21$.
We have:
$ \left(3.2 x^6 y^3\right) \times\left(2.1 x^2 y^2\right) $
$ =(3.2 \times 2.1) \times\left(x^6 \times x^2\right) \times\left(y^3 \times y^2\right) $
$ =(3.2 \times 2.1) \times\left(x^{6+2}\right) \times\left(y^{3+2}\right) $
$ =6.72 x^8 y^5$
Substituting $\mathrm{x}=1$ and $\mathrm{y}=0.5$ in the result, we get:
$ 6.72 x^8 y^5 $
$ =6.72(1)^8(0.5)^5 $
$ =6.72 \times 1 \times 0.03125 $
$ =0.21$
Thus, the answer is $0.21$.
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