log of the square root of 'y' = 1/2 sqrt(y)
y6
5x times the square root of y
2x square root 6x y squared
Yes, if x and y = 1 √1 + √1 = √1 + 1 1 + 1 = 1 + 1 QED
square root (y) / Square root (3) root (y) / 1.73
x=y
log of the square root of 'y' = 1/2 sqrt(y)
y6
5x times the square root of y
If x is a multiplication symbol, then square root of 36(4)y^2 = square root of(4^2)(2^2)(y^2) = (4)(2)(y) = 8y If it is a variable x, then square root of 36x4y^2 = square root of (4^2)(x)(2^2)(y^2) = (4)(2)(y)(square root of x)= 8y(square root of x)
Oh, dude, the square root of x plus the square root of y is just that - the square root of x plus the square root of y. It's like adding apples and oranges, you can't really simplify it further. So, like, that's your answer, no need to overcomplicate things, man.
2x square root 6x y squared
The number you are referring to is the square root of the second number. In mathematical terms, if you have a number "x" and it produces the square of another number "y" when multiplied by itself, then x is the square root of y. For example, if x * x = y, then x is the square root of y.
Algebraically if we have a number 'x^2' Then its square is (x^2)^2 = x^4 For the square root of x^2 = +/-x
Yes, if x and y = 1 √1 + √1 = √1 + 1 1 + 1 = 1 + 1 QED
The square root of a number x is one that, when multiplied by itself gives the value x. So if y * y = x then y is the square root of x. But, (-y)*(-y) also equals x. So -y is also a square root of x. Thus, given any positive number, x, there are two numbers, y and -y whose square is x. So both of them are square roots of x. One of y and -y must be greater than 0 and the other must be less than 0. The one that is less than 0 is the negative square root. As an example, 4 * 4 = 16 and (-4) * (-4) = 16 so both, 4 and -4 are square roots of 16. -4 is the negative square root while 4 is the positive square root (also called the principal square root).