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Solve for w, where w is a real number.
Whole question is in the picture.​

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Sagot :

For solving for w, where w is a real number. The value of w would be equal to 100.

What is a perfect square?

Perfect squares are those integers whose square root is an integer.

Let x-a be the closest perfect square less than x, and let x+b be the closest perfect square more than x, then we get x-a < x < x+b (no perfect square in between x-a and x+b, except possibly x itself).

Then, we get:

[tex]\sqrt{x-a} < \sqrt{x} < \sqrt{x+b}[/tex]

Thus,[tex]\sqrt{x-a} and \sqrt{x+b}[/tex] are the closest integers, less than and more than the value of [tex]\sqrt{x}[/tex]. (assuming x is a non-negative value).

We have a w real number so,

[tex]\sqrt{w} = 10\\[/tex]

taking square both sides

[tex](\sqrt{w} )^2 = 10^2\\\\w = 10 \times 10\\\\w = 100[/tex]

Learn more about square root here:

brainly.com/question/7200235

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