Find 2 integers whose sum is 16 and the product is 80

No solution in real numbers

Step-by-step explanation:

a+b=16

a*b=80

Unfortunately it does not have solution

I think the question is mistyping or wrong

Step-by-step explanation:

A+B=16

A*B = -80

Convert to

A=16-B

Substitute

(16-B)*B=-80

16B-B^2+80=0

Change the sign to make it easier (multiply by negative number)

B^2 -16B-80

=(B - 4)(B + 20)

B1=-4 and B2=20

Then its match if the product is -80.

Or second option is, assuming that the sum is 18, not 16

The A would be 8 and the B would be 10

use this formula:

p(a\b) = p(a and b)/p(b).

take it from here.

.632

step-by-step explanation:

6 * 1/10 = 3/5

3*0.01 = 0.03

3/5 + 0.03 = .63

2*0.002 = .002

.63 + 0.002 = .632