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Q: What is the square root of 1 over 121?

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Â±0.82218*i where i is the imaginary square root of -1.

It is: 1/11

1 square root of 144 =12 square root of 121=11so 12-11=1

1/3

the square root of (1/4) is 1/2

1/2

The square root of 1/100 is 1/10

It is i/3 where i is the imaginary number which is the square root of -1.

Square root of 25 = +or- 5 Square root of -36 = +or- 6i where i is the imaginary number such that i^2=-1 Square root of 121 = +or-11 So the 8 possible answers are: -16-6i, -16+6i, -6-6i, -6+6i, 6-6i, 6+6i, 16-6i and 16+6i

1/6 because 121/11=11-5=6=6/1(reciprocal)=1/6

1 over 6

The new square has an area of 121, so the length of a side is the square root of 121, or 11. So the length of the side of the old square was 10.

The square root of 1 over 15 is a constant and so differentiating it gives 0.

+/- 3i/2 where i is the imaginary square root of -1.

Since the square root of 1 is ±1, sqrt(1)/6 = ±1/6 = ±0.166...

4 over square root of 16 simplifies to 1 !

1/10

1/12

1/3

1/8

It is: 4/1 or simply as 4

The square roots of 121 are -11 and +11.

-1/9

The square root of 1 is also 1, so then you just take the square root of that, which is, again, 1.

√15/√30 = √(15/30) = √(1/2)