# How do you approximate #sqrt(17.02)#?

4.1255302699 to 9 Decimal Places

Let

If we start with

So we see that very rapidly the method converges to a solution 4.1255302699 to 9 Decimal Places

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(I used a calculator from this point on)

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If you wanted the answer using Newton's Method; here it is

as the product of some iterative expressions in a spreadsheet:

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One way to approximate √17.02 is to recognize that it falls between √16 (which is 4) and √25 (which is 5). Since 17.02 is closer to 16, we can start with 4.1. Squaring 4.1 gives 16.81, which is close to 17.02. So, a reasonable approximation for √17.02 is approximately 4.1.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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