To find three numbers that sum to 40 and have a product of 1400, we can set up the equations (x + y + z = 40) and (xyz = 1400). By trial and error or factoring, we can find that the numbers 10, 14, and 16 satisfy both conditions: (10 + 14 + 16 = 40) and (10 \times 14 \times 16 = 2240). However, upon checking the product again, it appears my initial multiplication was incorrect; the correct numbers that satisfy the conditions are 10, 10, and 20, as (10 + 10 + 20 = 40) and (10 \times 10 \times 20 = 2000). Therefore, the correct numbers are actually 10, 10, and 20.
8 and -5
The sum of 40 is simply 40 itself. The product of 400 typically requires two numbers to multiply together to yield 400. For example, 20 x 20 = 400. If you are asking for the sum of 40 and the result of multiplying two numbers to get 400, then it would be 40 + 400 = 440.
34
391
8 and 5
If the product of the two numbers is the sum times 24, then the product of the two numbers is 2400. 40 times 60 is 2400, and 40 plus 60 is 100. The two numbers are 40 and 60.
1/x + 1/y = (y+x)/xy But y + x = sum = 150, and xy = product = 40 So sum of reciprocals = 150/40 = 3.75
8 and -5
The sum of 40 is simply 40 itself. The product of 400 typically requires two numbers to multiply together to yield 400. For example, 20 x 20 = 400. If you are asking for the sum of 40 and the result of multiplying two numbers to get 400, then it would be 40 + 400 = 440.
37
34
20 and 20.
391
9 and -40 9 + -40 = -31 9 x -40 = -360
23 and 32
8 and 5
41 & 39 = 1599. (40 x 40 = 1600 but I assume you meant different numbers)