Question

How would you estimate the square root the nearest integer to the nearest integer and the nearest tenth for this number?

Answer 1

SOLUTION:

Case: Square-root of numbers

Given:

$\sqrt{46}$

Required:

To find the square root to the nearest:

a) integer

b) tenth

Method:

a) integer.

We find the integer range for the square root of 46

$\begin{gathered} \sqrt{36}<\sqrt{46}<\sqrt{49} \\ 6x is a number closer to 7 than 6 because 46 is closer to 49

Hence;

[tex]\sqrt{46}\approx7(NearestInteger)$

b) tenth

We find the tenth range for the square root of 46

$\begin{gathered} \sqrt{44.89}<\sqrt{46}<\sqrt{46.24} \\ 6.7x is a number closer to 6.8 than 6.7 because 46 is closer to 46.24

Hence:

[tex]\sqrt{46}\approx6.8(NearestTenth)$

Final answers:

The square root of 46 is:

$\begin{gathered} \sqrt{46}\approx7(NearestInteger) \\ \sqrt{46}\approx6.8(NearestTenth) \end{gathered}$