x+y = 1 so y = 1-x
xy = -56
Substitute,
x(1-x) = 56
or
-x2 + x -56 = 0
or you could multiply both sides by -1 and get
x2 - x + 56 = 0.
Solve that. You can use the quadratic formula with a = 1, b = -1, and c = 56
The numbers are 56 and 58.
the only two odd numbers are 57 and 59 which give a sum of 116
8 , and 7. x+y=15 >> x=-y+15 x*y=56 >> y*(-y+15)=56 , then solve for y , put y back into either equation for x.
just put 7x8 and u get 56 k
The numbers are 55, 56 and 57 OR 54, 56 and 58
The two numbers with the sum of 15 and the product of 56 are: 7 + 8 = 15 7 x 8 = 56
27 29
7 and 8
86
-56 and -1 -56+(-1)=-57 -56*(-1)=56
35, 21
The numbers are 4 and 14
seven and eight
They are: 7 and 8
Well if someone could help ME FIRST it would be GREAT :(
7 and 8
One number has to be 7, because that is a prime factor of 56. The other two numbers must total 8 and multiply to 8. Can't be done...