How do i estimate the square root and round to the nearest integer?

The square root of a number is a number such that

So, if a number is a perfect square, its square root is an integer. For example, the square root of 25 is 5, because .

If a number is not a perfect square, its square root is not an integer, which means that it is an irrational number, included between two integers.

If you want to find these two integers, you need to find two (consecutive) integers such that the first is "not enough", whereas the second is "too much".

Here's an example, suppose we want to find the square root of 20. We try some integers:

So, when we square 4 is slightly less than 20, and when we square 5 is slightly more than 20. This means that there must exist a number between 4 and 5 that, when squared, is exactly 20.

So, for sure

In order to round to the nearest integer, we consider that 4^2=16 was 4 units away from the goal (20), while 5^2=25 was 5 units away from the goal. So, 4^2 is nearest to the goal than 5^2, and so we round

For the record, we have

so our approximation was correct.

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step-by-step explanation:

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?