Let the number be 'm' & 'n'
Hence
Product ; mn = 80
Sum ; m + n = 21
m = 21 - n
Substitute
(21 - n) n = 80
21n - n^2 = 80
n^2 - 21 n + 80 = 0 It is now in quadratic form to factor.
( n - 16)(n - 5) = 0
Hence the numbers are 5 & 16
It is 23 because 17+19+21+23 = 80
Let's denote the two numbers as x and y. We know that xy = 80 and x + y = 24. We can solve this system of equations by substituting one equation into the other. By substituting y = 24 - x into the first equation, we get x(24 - x) = 80. Simplifying this equation gives us a quadratic equation: x^2 - 24x + 80 = 0. Solving this quadratic equation will give us the values of x and y.
80 is.
The product of the two integers is -80.
The answer is -10 and -8. -8 + -10= -18 -8x-10 =80.
The numbers are: -10 and -8
23+21+19+17=80
It is 23 because 17+19+21+23 = 80
Let the number by 'm' & 'n' Hence mn = 80 (product) m + n = 21 (sum) Hence m = 21 -b Substitute (21 - n)n = 80 21m - n^(2) = 80 n^(2) - 21n + 80 = 0 Factor (n - 16)(n - 5) = 0 Hence n= 16 & m = 5
Let's denote the two numbers as x and y. We know that xy = 80 and x + y = 24. We can solve this system of equations by substituting one equation into the other. By substituting y = 24 - x into the first equation, we get x(24 - x) = 80. Simplifying this equation gives us a quadratic equation: x^2 - 24x + 80 = 0. Solving this quadratic equation will give us the values of x and y.
There's only one number that's equal to 80. It's 80.There's a good chance that there could be three numbers whose sum is 80, or threedifferent numbers whose joint product is 80, but the question doesn't ask for those.
41 & 39 = 1599. (40 x 40 = 1600 but I assume you meant different numbers)
-10 + -8 = -18 -10 x -8 = 80 The two numbers are therefore -10 and -8.
The sum of the first sixteen numbers of pi is 80.
8 + 10 = 188 x 10 = 80Therefore, the two numbers are 8 and ten
-73
80 as a product of prime numbers can be expressed as: 2x2x2x2x5