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The sum of two number is 10 and their product is 30 the sum of their reciprocals is?
Suppose the numbers are x and y. The sum of their reciprocals = 1/x + 1/y = y/xy + x/xy = (y+x)/xy = (x+y)/xy = 10/30 = 1/3
If the sum of two numbers is 7 and there product is 25 what is the sum of there reciprocals?
Suppose the two numbers are x and y. Then, the sum of THEIR reciprocals is 1/x + 1/y = y/xy + x/xy = (y + x)/xy = 7/25
Suppose y varies inversely with X use the information to find K and choose the equation of variation. X equals 7 and y equals 3?
xy = K so K = 21
How do you factor x-y plus xy-1?
(x-y) + (xy - 1) = (x - 1)(y + 1)
What is xy over y in simplest form?
Xy/y = x
The sum of two number is 10 and their product is 30 the sum of their reciprocals is?
Suppose the numbers are x and y. The sum of their reciprocals = 1/x + 1/y = y/xy + x/xy = (y+x)/xy = (x+y)/xy = 10/30 = 1/3
If the sum of two numbers is 7 and there product is 25 what is the sum of there reciprocals?
Suppose the two numbers are x and y. Then, the sum of THEIR reciprocals is 1/x + 1/y = y/xy + x/xy = (y + x)/xy = 7/25
The sum of two numbers is 10 and their product is 20 find the sum of the reciprocal of the number?
Suppose the numbers are x and y Then the sum of their reciprocals is 1/x + 1/y = y/xy + x/xy = (y+x)/xy = 10/20 = 1/2
How can you determine whether a function is even odd or neither?
Looking at the graph of the function can give you a good idea. However, to actually prove that it is even or odd may be more complicated. Using the definition of "even" and "odd", for an even function, you have to prove that f(x) = f(-x) for all values of "x"; and for an odd function, you have to prove that f(x) = -f(-x) for all values of "x".
What is the GCF of x squared y and xy squared?
The GCF is xy
X plus XY equals 8 solve for Y?
x+xy=8 xy=-x+8 y=-1+8/x
If xy equals x or yx equals x what is y?
xy = x ÷x y = 1
Why does an even number subtract to odd number is equal to odd number?
Suppose x is an odd number, then x leaves a remainder when divided by 2. That is, x = 2m+1 (for some integer m). Suppose y is an even number, then y is a multiple of 2 so suppose y = 2n (for some integer n). Then x-y = 2m+1 - 2n = 2m-2n + 1 =2(m-n) + 1 Since m and n are integers, then m+n is an integer so that the sum gives 1 more than a multiple of 2. And that is what an odd number is!
Robert's numbers is between 20 and 32 the sum of the digits is 10 what is the number how do i solve it?
Suppose the number, when written is XY. Then sum of digits = 10 implies that X+Y = 10 Also, 20 < XY < 32 implies that X = 2 or 3. Suppose X = 3 Then X+Y = 10 requires Y = 7. But that is not possible because the number cannot be greater than 32. So, X must be 2. And then, X+Y = 10 implies that Y = 8 and so the number is 28.
If x-3yy12then xy equal to?
If x - 3y = 12, we can solve for x in terms of y: x = 12 + 3y. Then, to find xy, we multiply x by y: xy = y(12+3y) = 12y + 3y^2.
What is XY square?
It is XY^2 or X*Y*Y
How would you prove that the sum of an even number and an odd number is always an odd number?
Suppose x is an even number and y is an odd number. Then x = 2*n for some integer n and y = 2*m + 1 for some integer m Therefore x + y = 2*n + 2*m + 1 = 2*(n + m) +1 Now, since n and m are integers, (n + m) is also an integer [by the closure of integers under addition]. Thus, x + y = 2*p + 1 where p = n + m is an integer. ie x + y is an odd integer.
