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ololo11 [35]
3 years ago
11

Estimate StartRoot 37 EndRoot 37 to the nearest tenth. Then locate StartRoot 37 EndRoot 37 on a number line.

Mathematics
1 answer:
Vesnalui [34]3 years ago
5 0

Estimating value of √37.

We know that

6^{2} = 36 and 7^{2} =49, so

6 < √37 < 7

If we take the average of 6 and 7, we get

\frac{6+7}{2}  = \frac{13}{2} = 6.5

Since, 6.5^{2} = 42.25

6 < √37 < 6.5

If we take the average of 6 and 6.5 , we get

\frac{6+6.5}{2}  = \frac{12.5}{2} = 6.25

Since, 6.25^{2} = 39.0625

6 < √37 < 6.25

If we take the average of 6 and 6.25 , we get

\frac{6+6.25}{2}  = \frac{12.25}{2} = 6.125

Since, 6.125^{2} = 37.515625

6 < √37 < 6.125

If we take the average of 6 and 6.125 , we get

\frac{6+6.125}{2}  = \frac{12.125}{2} = 6.0625

Since, 6.0625^{2} = 36.75390625

6.0625 < √37 < 6.125

If we take the average of 6.0625 and 6.125 , we get

\frac{6.0625+6.125}{2}  = \frac{12.125}{2} = 6.09375

Since, 6.09375^{2} = 37.1337890625

6.0625 < √37 < 6.09375

If we take the average of 6.0625 and 6.09375 , we get

\frac{6.0625+6.09375}{2}  = \frac{12.15625}{2} = 6.078125

Since, 6.078125^{2} = 36.943603515625

6.078125 < √37 < 6.09375

If we take the average of 6.078125 and 6.09375 , we get

\frac{6.078125+6.09375}{2}  = \frac{12.171875}{2} = 6.0859375

Since, 6.0859375^{2} = 37.03863525390625

Therefore,

√37 ≈ 6.0859375.

And if we round it to the nearest tenth, we get

√37 ≈ 6.1


Locating √37 on number line.

In order to locate √37 on number line first draw a line 0 to 6 on number line.

Then draw a perpendicular line segment of 1 unit on number 6 on number line.

Join the number 0 on the number line by the top point of perpendicular line segment on number 6 we drew in above step.

Finally, draw a curve by taking radius as Hypotenuse of the right trinagle form in the diagram shown.

The curve would cut the number line exactly at √37 on number line.

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