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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
Is the square root of 1 over 9 rational?
Yes and it is 1/3
What is the square root of 4 over 9?
The square root of the fraction 4/9 is either 2/3 or -2/3
Square root of 1 over 9?
sqrt(1/9) = 1/3
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.
Can 2 divided by 2 square root 3 be simplified?
Yes, the expression 2 divided by 2 square root 3 can be simplified. To simplify this expression, we need to rationalize the denominator. Multiplying both the numerator and the denominator by the conjugate of the denominator (2 square root 3), we get (2 * 2 square root 3) / (2 * 2 square root 3 * 2 square root 3). This simplifies to 4 square root 3 / 12, which further simplifies to square root 3 / 3.
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.
