Question
Find product: $\big(\frac{1}{8}\text{x}^2\text{y}^4\big)×\big(\frac{1}{4}\text{x}^4\text{y}^2\big)×(\text{xy})×5$
✓
Answer
To multiply algebraic expressions, we can use commutative and associative laws along with the law of indices, $a^m \times a^n=a^{m+n}$.
We have:
$\big(\frac{1}{8}\text{x}^2\text{y}^4\big)×\big(\frac{1}{4}\text{x}^4\text{y}^2\big)×(\text{xy})×5$
$=\big(\frac{1}{8}×\frac{1}{4}×5\big)×\big(\text{x}^2×\text{x}^4×\text{x}\big)×\big(\text{y}^4×\text{y}^2×\text{y}\big)$
$=\big(\frac{1}{8}×\frac{1}{4}×5\big)×\big(\text{x}^{2+4+1}\big)×\big(\text{y}^{4+2+1}\big)$
$=\frac{5}{32}\text{x}^7\text{y}^7$
To verify the result, we substitute $x = 1$ and $y = 2$ in LHS; we get:
$\text{LHS}=\big(\frac{1}{8}\text{x}^2\text{y}^4\big)×\big(\frac{1}{4}\text{x}^4\text{y}^2\big)×(\text{xy})×5$
$=\big(\frac{1}{8}×(1)^{2}×(2)^{4}\big)×\big(\frac{1}{4}×(1)^{4}×(2)^{2}\big)×(1×2)×5$
$=\big(\frac{1}{8}×(1)^{2}×(2)^{4}\big)×\big(\frac{1}{4}×1×4\big)\times(1\times2)\times5$
$=2×1×2×5$
$=20$
Substituting $x = 1$ and $y = 2$ in $RHS$, we get:
$\text{RHS}=\frac{5}{32}\text{x}^7\text{y}^7$
$=\frac{5}{32}(1)^7(2)^7$
$=\frac{5}{32}\times1\times128$
$=20$
Because $LHS$ is equal to $RHS$, the result is correct.
Thus, the answer is $=\frac{5}{32}\text{x}^7\text{y}^7.$
We have:
$\big(\frac{1}{8}\text{x}^2\text{y}^4\big)×\big(\frac{1}{4}\text{x}^4\text{y}^2\big)×(\text{xy})×5$
$=\big(\frac{1}{8}×\frac{1}{4}×5\big)×\big(\text{x}^2×\text{x}^4×\text{x}\big)×\big(\text{y}^4×\text{y}^2×\text{y}\big)$
$=\big(\frac{1}{8}×\frac{1}{4}×5\big)×\big(\text{x}^{2+4+1}\big)×\big(\text{y}^{4+2+1}\big)$
$=\frac{5}{32}\text{x}^7\text{y}^7$
To verify the result, we substitute $x = 1$ and $y = 2$ in LHS; we get:
$\text{LHS}=\big(\frac{1}{8}\text{x}^2\text{y}^4\big)×\big(\frac{1}{4}\text{x}^4\text{y}^2\big)×(\text{xy})×5$
$=\big(\frac{1}{8}×(1)^{2}×(2)^{4}\big)×\big(\frac{1}{4}×(1)^{4}×(2)^{2}\big)×(1×2)×5$
$=\big(\frac{1}{8}×(1)^{2}×(2)^{4}\big)×\big(\frac{1}{4}×1×4\big)\times(1\times2)\times5$
$=2×1×2×5$
$=20$
Substituting $x = 1$ and $y = 2$ in $RHS$, we get:
$\text{RHS}=\frac{5}{32}\text{x}^7\text{y}^7$
$=\frac{5}{32}(1)^7(2)^7$
$=\frac{5}{32}\times1\times128$
$=20$
Because $LHS$ is equal to $RHS$, the result is correct.
Thus, the answer is $=\frac{5}{32}\text{x}^7\text{y}^7.$
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