(21.5 % text{ of } 999)^{1/3} + (42 % text{ o | vedclass

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(B) To solve the expression $(21.5 \% 	ext{ of } 999)^{1/3} + (42 \% 	ext{ of } 601)^{1/2} + ? = 28$:$1$. Approximate $21.5 \% 	ext{ of } 999$ as $

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(B) To solve the expression $(21.5 \% 	ext{ of } 999)^{1/3} + (42 \% 	ext{ of } 601)^{1/2} + ? = 28$:$1$. Approximate $21.5 \% 	ext{ of } 999$ as $21.5 \% 	ext{ of } 1000$,which is $215$. The cube root of $215$ is approximately $6$ (since $6^3 = 216$).$2$. Approximate $42 \% 	ext{ of } 601$ as $42 \% 	ext{ of } 600$,which is $0.42 	imes 600 = 252$. The square root of $252$ is approximately $16$ (since $16^2 = 256$).$3$. The sum is approximately $6 + 16 = 22$.$4$. If the total is $28$,then $22 + ? = 28$,which gives $? = 6$. However,based on the provided options and standard approximation logic,the value of the expression itself is $22$.

$(21.5 \% \text{ of } 999)^{1/3} + (42 \% \text{ of } 601)^{1/2} + ? = 28$ (Assuming the equation equals $28$ for a standard evaluation)

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