The sum of x and y decreased by their product is (x + y)- xy.
1/x + 1/y = (y+x)/xy But y + x = sum = 150, and xy = product = 40 So sum of reciprocals = 150/40 = 3.75
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 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
1/x+1/y y/xy+x/xy x+y/xy=12/24 12/24=1/2 1/2 is the answer.
1/x + 1/y y/xy + x/xy (y+x)/xy= 10/20 10/20=1/2 1/2 is the answer.
6x - (6 + y)
5(x^2 + y)
3x over x + y
1/x + 1/y = (y+x)/xy But y + x = sum = 150, and xy = product = 40 So sum of reciprocals = 150/40 = 3.75
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 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
7+xy
x*(y+z) = x*y + x*z This is the distributive property of multiplication over addition.
The square of the sum of ( x ) and ( y ) is expressed mathematically as ( (x + y)^2 ). This can be expanded using the formula for the square of a binomial, resulting in ( x^2 + 2xy + y^2 ). Thus, the square of the sum of ( x ) and ( y ) captures both the individual squares of ( x ) and ( y ) as well as twice their product.
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
x = 10, y = 25
4xy - 2xy