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No. If you square 1/6, you get 1/36, which is considerably smaller than 3. The square root of 3 is actually an irrational number - i.e. cannot be represented by a ratio of integers (fraction). The approximate value of the square root of three is 1.73205101, which lies between the values of 1 2/3 and 1 5/6
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What is the square root of -1 over 9?
It is i/3 where i is the imaginary number which is the square root of -1.
What is the square root of 48 over the square root of 3 simplified?
It is: 4/1 or simply as 4
Square root of 12 over the square root of 4?
2 root 3 over 2, so square root of 3
What is the answer to the square root of 15 over the square root of 3?
I think it is the square root of 5.
What is 1 over radical 54 in simplest form?
1 over (3 times the square root of 6)
What is the square root of -1 over 9?
It is i/3 where i is the imaginary number which is the square root of -1.
What is 1 divided by 1 plus the square root of 3?
root 3 - 1 all over 2
What is the square root of 48 over the square root of 3 simplified?
It is: 4/1 or simply as 4
Square root of 12 over the square root of 4?
2 root 3 over 2, so square root of 3
What is the square root of one ninth?
The square root of 1/9 is 1/3 because the square root of 1 is 1 and the square root of 9 is 3.
What is the square root of y over 3?
square root (y) / Square root (3) root (y) / 1.73
What is the square root of 4 over 9?
The square root of the fraction 4/9 is either 2/3 or -2/3
Is the square root of 1 over 9 rational?
Yes and it is 1/3
Square root of 1 over 9?
sqrt(1/9) = 1/3
Can 2 divided by 2 square root 3 be simplified?
Yes. Multiply both the bottom and top by root 3. Then you have 2 root 3 over two. 2 over 2 simplifies to 1, so the final answer is the square root of 3.
What is the answer to the square root of 15 over the square root of 3?
I think it is the square root of 5.
Find the two numbers whose product is 1 and whose sum is 1?
x + y = 1xy = 1y = 1 - xx(1 - x) = 1x - x^2 = 1-x^2 + x - 1 = 0 or multiplying all terms by -1;(-x^2)(-1) + (x)(-1) - (1)(-1) = 0x^2 - x + 1 = 0The roots are complex numbers. Use the quadratic formula and find them:a = 1, b = -1, and c = 1x = [-b + square root of (b^2 - (4)(a)(c)]/2a orx = [-b - square root of (b^2 - (4)(a)(c)]/2aSox = [-(-1) + square root of ((-1)^2 - (4)(1)(1)]/2(1)x = [1 + square root of (1 - 4]/2x = [1 + square root of (- 3)]/2 orx = [1 + square root of (-1 )(3)]/2; substitute (-1) = i^2;x = [1 + square root of (i^2 )(3)]/2x = [1 + (square root of 3)i]/2x = 1/2 + [i(square root of 3]/2 andx = 1/2 - [i(square root of 3)]/2Since we have two values for x, we will find also two values for yy = 1 - xy = 1 - [1/2 + (i(square root of 3))/2]y = 1 - 1/2 - [i(square root of 3)]/2y = 1/2 - [i(square root of 3)]/2 andy = 1 - [1/2 - (i(square root of 3))/2)]y = 1 - 1/2 + [i(square root of 3))/2]y = 1/2 + [i(square root of 3)]/2Thus, these numbers are:1. x = 1/2 + [i(square root of 3)]/2 and y = 1/2 - [i(square root of 3)]/22. x = 1/2 - [i(square root of 3)]/2 and y = 1/2 + [i(square root of 3)]/2Let's check this:x + y = 11/2 + [i(square root of 3)]/2 +1/2 - [i(square root of 3)]/2 = 1/2 + 1/2 = 1xy = 1[1/2 + [i(square root of 3)]/2] [1/2 - [i(square root of 3)]/2]= (1/2)(1/2) -(1/2)[i(square root of 3)]/2] + [i(square root of 3)]/2](1/2) - [i(square root of 3)]/2] [i(square root of 3)]/2]= 1/4 - [i(square root of 3)]/4 + [i(square root of 3)]/4 - (3i^2)/4; substitute ( i^2)=-1:= 1/4 - [(3)(-1)]/4= 1/4 + 3/4= 4/4=1In the same way we check and two other values of x and y.
