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Is the product of two perfect squares sometimes a perfect square?
yes..always a perfect square A perfect square is the product of an integer by itself. If you multiply a perfect square x² by another perfect square y² you get x²y² = x·x·y·y = x·y·x·y = (x·y)² which is a perfect square. Note that the product of two integers will also be an integer so x·y must be an integer because if x² and y² are perfect squares x must be an integer and y must be an integer and x·y is therefore a product of 2 integers.
What is the square root of x - y divided by square root of x y?
x=y
The sum of x and y decreased by their product?
The sum of x and y decreased by their product is (x + y)- xy.
A number that produce another number its square when multiplied by itself?
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.
How do you solve for x when it equals the square root of y to the fourth minus y?
If x equals the square root of ...., then you already have solved for x
Is the product of two perfect squares sometimes a perfect square?
yes..always a perfect square A perfect square is the product of an integer by itself. If you multiply a perfect square x² by another perfect square y² you get x²y² = x·x·y·y = x·y·x·y = (x·y)² which is a perfect square. Note that the product of two integers will also be an integer so x·y must be an integer because if x² and y² are perfect squares x must be an integer and y must be an integer and x·y is therefore a product of 2 integers.
How do you factor the special product?
It depends what the special product is. Common special products are: - perfect square trinomials ... x^2 + 2ax + a^2 = (x + a)^2 - difference of squares ... x^2 - y^2 = (x - y)(x + y)
How do you calculate r?
First, get the product of the summation of x squared and y squared and then find its square root. Divide the summation of x and y by the square root to get Pearson's r.
What you have learn in product of perfect square?
That the set of perfect squares is closed under multiplication. That is if x and y are any two perfect squares, then x*y is a perfect square.
What makes a product special?
In math, like algebra and calculus, a product is special when it is very common and worth knowing.Some examples area:(x + y) = ax + ay (Distibutive Law)(x + y)(x − y) = x2 − y2 (Difference of 2 squares)(x + y)2 = x2 + 2xy + y2 (Square of a sum)(x − y)2 = x2 − 2xy + y2 (Square of a difference)
Why is the product of two squared numbers multiplied together a square number?
Let's say you have any 2 non-zero whole numbers x and y.So x2 and y2 are square numbers.if you multiply them: x2 * y2 = (x*x)*(y*y).Since multiplication is commutative, rearrange to get: (x*y)*(x*y) = (x*y)2which is also a square number
What is the square root of x - y divided by square root of x y?
x=y
The sum of x and y decreased by their product?
The sum of x and y decreased by their product is (x + y)- xy.
How do you get the two numbers if product and quotient are given?
Suppose the two numbers are X and Y and you are given XY = A and X/Y = B. Then AB = XY*X/Y = X2 so that X = sqrt(AB) and A/B = XY / (X/Y) = XY*Y/X = Y2 so that Y = sqrt(A/B) So take the product and quotient that are given. Find THEIR product and quotient. Your two numbers are the square roots of these numbers. Depending on the context (or requirements of the question) you can either use only the principal square roots or select the necassary signs for the square roots. For example, given the product and quotient are 6 and 2/3, AB = 6*2/3 = 4 sand A/B = 6 / (2/3) = 6*3/2 = 9 The possible solutions are X = 2 and Y = 3 OR X = -2 and Y = -3
The difference between square and square roots?
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
Are the products of irrational numbers always irrational?
No. The product of conjugate pairs is always rational.So suppose sqrt(y) is the irrational square root of the rational number y. ThenThus [x + sqrt(y)]*[x - sqrt(y)] = x^2 + x*sqrt(y) - x*sqrt(y) - sqrt(y)*sqrt(y)= x^2 + y^2 which is rational.
Does x-y equals x square?
x-y=x² only if y = x - x²
