If sqrt{x} + sqrt{441} = 0.02,then the value | vedclass

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Question

(A) Given the equation: $\sqrt{x} + \sqrt{441} = 0.02$We know that $\sqrt{441} = 21$.Substituting this value into the equation: $\sqrt{x} + 21 = 0.02$

Vedclass · Accepted Answer

$0.1764$

Vedclass · Answer

(A) Given the equation: $\sqrt{x} + \sqrt{441} = 0.02$We know that $\sqrt{441} = 21$.Substituting this value into the equation: $\sqrt{x} + 21 = 0.02$$\sqrt{x} = 0.02 - 21$$\sqrt{x} = -20.98$Since the square root of a real number cannot be negative,there is no real value for $x$ that satisfies this equation. However,if the equation was intended to be $\sqrt{x} 	imes \sqrt{441} = 0.02$:$\sqrt{x} 	imes 21 = 0.02$$\sqrt{x} = \frac{0.02}{21} \approx 0.000952$$x = (0.000952)^2 \approx 0.000000906$.Given the provided options,it appears there is a typo in the question. If the equation was $\sqrt{x} = 0.02 	imes \sqrt{441}$:$\sqrt{x} = 0.02 	imes 21 = 0.42$$x = (0.42)^2 = 0.1764$.

If $\sqrt{x} + \sqrt{441} = 0.02$,then the value of $x$ is

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