Evaluate following when x = 2, y = −1. (2xy)× ( x^2y 4 )× (x^2 )× (y^2 )

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 . (2xy)× ( x^2y 4 )× (x^2 )× (y^2 ) = (2×1 4 )× (x×x^2×x^2 )× (y×y×y^2) = (2×1 4 )× (x^ 1+2+2 )× (y^ 1+1+2 ) =1 2 x^5y^4 (2xy)× ( x^2y 4 )× (x^2 )× (y^2 )=1 2 x^5y^4 Substituting x = 2 and y = -1 in the result, we get: 1 2 x^5y^4 =1 2 (2)^5(−1)^4 =1 2 ×32×1 = 16 Thus, the answer is 16.

Evaluate following when x = 2, y = −1. (2xy)× ( x^2y 4 )× (x^2 )×…

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Question

Evaluate following when x = 2, y = −1.
$(2\text{xy})×\big(\frac{\text{x}^2\text{y}}{4}\big)×\big(\text{x}^2\big)×\big(\text{y}^2\big)$

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Answer

To multiply algebraic expressions, we use commutative and associative laws along with the law of indices, i.e. $a^m \times a^n=a^{m+n}$.
$(2\text{xy})×\big(\frac{\text{x}^2\text{y}}{4}\big)×\big(\text{x}^2\big)×\big(\text{y}^2\big)$
$=\big(2×\frac{1}{4}\big)×\big(\text{x}×\text{x}^2×\text{x}^2\big)×\big(\text{y}×\text{y}×\text{y}^2)$
$=\big(2×\frac{1}{4}\big)×\big(\text{x}^{1+2+2}\big)×\big(\text{y}^{1+1+2}\big)$
$=\frac{1}{2}\text{x}^5\text{y}^4$
$\therefore(2\text{xy})×\big(\frac{\text{x}^2\text{y}}{4}\big)×\big(\text{x}^2\big)×\big(\text{y}^2\big)=\frac{1}{2}\text{x}^5\text{y}^4$
Substituting x = 2 and y = -1 in the result, we get:
$\frac{1}{2}\text{x}^5\text{y}^4$
$=\frac{1}{2}(2)^5(−1)^4$
$=\frac{1}{2}×32×1$
$= 16$
Thus, the answer is $16.$