y = 36
Wiki User
∙ 9y agoYes, if x and y = 1 √1 + √1 = √1 + 1 1 + 1 = 1 + 1 QED
No.
y = square root of x y = x(1/2) y' = (1/2)x(1/2 - 1) y' = (1/2)x(-1/2) y' = (1/2)(square root of x)
A square root is taking a number X and determining which two same numbers times each other, or Y*Y equal X. If X equals 9, the square root of it is 3; that is 3 (Y) multiplied by 3 (Y) is 9.
log of the square root of 'y' = 1/2 sqrt(y)
If x equals the square root of ...., then you already have solved for x
Domian is x>-6 Range is y> or equal to 0
Yes, if x and y = 1 √1 + √1 = √1 + 1 1 + 1 = 1 + 1 QED
y = square root of x y = x(1/2) y' = (1/2)x(1/2 - 1) y' = (1/2)x(-1/2) y' = (1/2)(square root of x)
No.
A square root is taking a number X and determining which two same numbers times each other, or Y*Y equal X. If X equals 9, the square root of it is 3; that is 3 (Y) multiplied by 3 (Y) is 9.
square root (y) / Square root (3) root (y) / 1.73
y equals x plus 4 when y equals 0 then x equals 2i i is the square root of negative 1
The square root operation is not a function because for each value of y there can be 2 values of x - the principal square root and its negative. This can only be rectified by limiting the range of the opearation to the principal or positive square root. Furthermore, it also depends on the domain of the function. If y<4 then the square root is not defined within Real numbers. So, for y ≥ 4, x = +sqrt(y-4) is a function.
y=(8x).5 + (4x).5 = (2+2sqrt(2))x.5 y'=(1 + sqrt(2))/sqrt(x)
x=y
If: y = 8-x and y = x2+4x+2 Then: x2+4x+2 = 8-x => x2+5x-6 = 0 Solving the quadratic equation: x = 1 or x = -6 Points of intersection by substitution: (1, 7) and (-6, 14) Length of line is the square root of: (1--6)2+(7-14)2 = 7 times square root of 2