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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.
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What is the square root of 100 x4 y4?
sqrt(100x4y4) = sqrt(100)*sqrt(x4)*sqrt(y4) = 10*x2*y2
Is x equals y4 a function?
Assuming the domain and range are suitably defined, then yes. If not, then no.
What is the answer to 2x equals y4?
x=y4 /2
What is the square root of 40 x12 y9?
√(40 x12y9) = 6.3246√(y)* x6 *y4 or 6.3246x6y4.5 ■
Y to the power of 2 times y to the power of 2 equals?
y4
What is the gcf of y4 y8 y9?
The GCF is y4
How do you solve y4-84 plus 5y2?
You cannot solve it since only one side of an equation is given. If the equation was y4 + 5y2 - 84 = 0 then y4 + 12y2 - 7y2 - 84 = 0 or y2*(y2 + 12) - 7*(y2 + 12) = 0 or (y2 - 7)*(y2 + 12) = 0 then y2 = 7 or y2 = - 12 y = +or- sqrt(7) and, if you are in the complex domain, also y = +or- i*sqrt(12) where i is the imaginary square root of -1.
What is 4x-y4 what is the value of y?
4x-y4 what = 0
Which equation has y4 as solution?
The simplest equation would be y4 = 0
How do you express x4 plus y4 if given x plus y equals a xy equals b?
(x + y)2 = x2 + 2xy + y2 So x2 + y2 = (x + y)2 - 2xy = a2 - 2b Then (x2 + y2)2 = x4 + 2x2y2 + y4 So x4 + y4 = (x2 + y2)2 - 2x2y2 = (a2 - 2b)2 - 2b2 = a4 - 4a2b + 4b2 - 2b2 = a4 - 4a2b + 2b2
How can you do x4-y4?
(x2 + y2)(x + y)(x - y) = x4 - y4.
What are the factors of x2 - y4?
-2
