30 degrees or pi/6
No.
The additive inverse of a number is the value that, when added to the original number, results in zero. For the square root of 52, which is approximately 7.21, the additive inverse would be -√52. Therefore, the additive inverse of the square root of 52 is -√52.
60
9 is the square ROOT of 81. Calculating a square root is the inverse operation of squaring a number.
yes
The inverse operation of taking the square root is to calculate the square.
Square root is the inverse operation of a square.
No.
XX or X*X, can be written as X squared. The inverse of a function "sort of cancels it out". I know the inverse of a square is the square root. Since we need the inverse of X squared, it's inverse is the square root of X. sqrt(x)
4
The square of the square root of 36. Which can also be stated as the square of 6.
The inverse operation is to take a square root.
The inverse operation of squaring a number is finding the square root of that number. In mathematical terms, if you square a number x, the result is x^2. The inverse operation would be taking the square root of x^2, which gives you the original number x. For example, if you square 3 (3^2 = 9), the square root of 9 is 3.
y = x2 where the domain is the set of real numbers does not have an inverse, because the square root function is a one-two-two mapping (except at 0). Any polynomial with more than one root, over the reals, has no inverse. y = 1/x has no inverse across 0. But it is possible to define the domain so that each of these functions has an inverse. For example y = x2 where x is non-negative has the square root function as its inverse.
For the modulus, first square both coordinates. Then add the results and square-root that sum. For the angle, divide the y-coordinate by the x-coordinate, then find the inverse tangent (tan-1) of that number.
60
The tangent is essentially the derivative of the function. The square-root is just what ever function that is takes two of that function to equal the tangent. If you need further help on this question just send me a message on my message board and id be glad to help you out.