sqrt(1/121) = ±1/11
±0.82218*i where i is the imaginary square root of -1.
It is: 1/11
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 1/100 is 1/10
To find the sum of the square roots of the digits in the number 121, we first identify its digits: 1, 2, and 1. The square roots of these digits are √1 = 1, √2 ≈ 1.41, and √1 = 1. Adding these together gives us 1 + 1.41 + 1 = 3.41. Thus, the sum of the square roots of the digits in 121 is approximately 3.41.
±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
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
the square root of (1/4) is 1/2
1/6 because 121/11=11-5=6=6/1(reciprocal)=1/6
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 1/100 is 1/10
To find the sum of the square roots of the digits in the number 121, we first identify its digits: 1, 2, and 1. The square roots of these digits are √1 = 1, √2 ≈ 1.41, and √1 = 1. Adding these together gives us 1 + 1.41 + 1 = 3.41. Thus, the sum of the square roots of the digits in 121 is approximately 3.41.
Yes, the square root of 121 is rational because it equals 11, which is an integer. Rational numbers are defined as numbers that can be expressed as the quotient of two integers, and since 11 can be represented as 11/1, it meets this criterion.
7 + sqrt(-121) => 7 + sqrt( -1 X 121) => 7 + sqrt(-1) X sqrt(121) => 7 + sqrt(-1) *+/-11 = > 7 +/= 11sqrt(-1) The sqrt(-1) is designated the letter 'i' ( small /lower case 'i'). Hence 7 +/- 11i NB If you can find the 'square root of '-1'' , then the mathemtical world. would like to hear from you.
The square root of 1 over 15 is a constant and so differentiating it gives 0.