±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
It is i/3 where i is the imaginary number which is the square root of -1.
±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
1/6 because 121/11=11-5=6=6/1(reciprocal)=1/6
the square root of (1/4) is 1/2
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
Oh, what a happy little question! When you have the square root of ten over the square root of forty, you can simplify it by dividing the square roots. So, the answer is the square root of ten over the square root of forty simplifies to one over the square root of four, which simplifies further to one over two. Just a little math magic to brighten your day!
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.
It is i/3 where i is the imaginary number which is the square root of -1.