Between which two consecutive integers does the square root lie? 1, √120=
2, √56=
3, √19=
4, √240=
5, √80=
6, √27=
7, √199=
8, √38=
9, √102=
10, √11=
please help me

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:

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:

link

7

step-by-step explanation:

50-38=12

step-by-step explanation: