Math

QuestionFind the range where 31\sqrt{31} lies, rounded to one decimal place.

Studdy Solution

STEP 1

Assumptions1. We are asked to find two consecutive integers between which 31\sqrt{31} lies. . We are looking for an approximate value with one decimal place accuracy.

STEP 2

First, we need to identify two consecutive integers that could potentially contain 31\sqrt{31}. Since we know that 52=255^2 =25 and 62=366^2 =36, we can conclude that 31\sqrt{31} must lie between5 and6.

STEP 3

To confirm this, let's calculate the squares of5 and6.
52=255^2 =2562=366^2 =36

STEP 4

Since 25<31<3625 <31 <36, we can confirm that 31\sqrt{31} lies between and6.

STEP 5

Now we need to identify the decimal value. Since we are looking for one decimal place accuracy, let's check if 31\sqrt{31} is closer to5.5. We know that (5.5)2=30.25(5.5)^2 =30.25.

STEP 6

Since 30.25<3130.25 <31, we can conclude that 31\sqrt{31} is greater than5.5.

STEP 7

Next, let's check if 31\sqrt{31} is closer to5.6. We know that (5.6)2=31.36(5.6)^2 =31.36.

STEP 8

Since 31<31.3631 <31.36, we can conclude that 31\sqrt{31} is less than5.6.

STEP 9

Therefore, to one decimal place, 31\sqrt{31} must lie between5.5 and5.6.

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