We have a lot of square roots, and we want to find, for each one, two consecutive integers such that one is smaller than the square root and the other is larger.
The solutions are:
- 1) 10 and 11
- 2) 7 and 8
- 3) 4 and 5
- 4) 15 and 16
- 5) 8 and 9
- 6) 5 and 6
- 7) 14 and 15
- 8) 6 and 7
- 9) 10 and 11
- 10) 3 and 4
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To find the integers, we just need to find two integers such that one squared is smaller than the argument of the square root and the other squared is larger than the argument of the square root.
1) √120
Here we can use:
10*10 = 100
11*11 = 121
Then:
10 < √120 < 11
2) √56
Here we can use:
7*7 = 49
8*8 = 64
then:
7 < √56 < 8
3) √19
Here we can use:
4*4 = 16
5*5 = 25
then:
4 < √19 < 5
4) √240
Here we can use:
15*15 = 225
16*16 = 256
then:
15 < √240 < 16
5) √80
Here we can use:
8*8 = 64
9*9 = 81
then:
8 < √80 < 9
6) √27
Here we can use:
5*5 = 25
6*6 = 36
then:
5 < √27 < 6
7) √199
Here we can use:
14*14 = 196
15*15 = 225
then:
14 < √199 < 15
8) √38
Here we can use:
6*6 = 36
7*7 = 49
then:
6 < √38 < 7
9) √102
Here we can use:
10*10 = 100
11*11 = 121
then:
10 < √102 < 11
10) √11
Here we can use:
3*3 = 9
4*4 = 16
then:
3 < √11 < 4.
If you want to learn more, you can read:
https://brainly.com/question/10855879
