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What is 3 times 3 square root 2?
3(3 square root of 2) = 9(square root of 2)
What is the relation between square of 2 and square of 3?
The square of 3 is 2.25 times the square of 2
Is a square 2 or 3 dimensional?
A square is 2, but a cube is 3.
How did you find the square of each number?
The square of a number is that number times itself. Square of 2 is 2*2=4 Square of 3 is 3*3=9
How do you do a fraction square?
If you want to square a fraction, you can square the numerator and the denominator separately. For example, the square of 2/3 is equal to 4/9.
What is the square root of 3 divided by 2 plus the square root of 3 divided by 2?
2 times the Square root of 3 + 4
Why is it called cubed and squared?
That is because a cube has 3 dimensions, and a square has 2.That is because a cube has 3 dimensions, and a square has 2.That is because a cube has 3 dimensions, and a square has 2.That is because a cube has 3 dimensions, and a square has 2.
What is Square root of 3 plus square root of 3?
sqrt(3) + sqrt(3) = 2*sqrt(3) NOT sqrt(3 + 2)
Square root 2 times square root 3 times square root 8?
square root 2 times square root 3 times square root 8
What is a fraction square?
It is the square of the numerator divided by the square of the denominator. Thus, for example, square(2/3) = square(2)/square(3) = 4/9
area of a box that is 3 ft wide 3 ft high and 3 ft long?
2*(3*3 + 3*3 + 3*3) = 2*27 = 54 square feet.2*(3*3 + 3*3 + 3*3) = 2*27 = 54 square feet.2*(3*3 + 3*3 + 3*3) = 2*27 = 54 square feet.2*(3*3 + 3*3 + 3*3) = 2*27 = 54 square feet.
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.
