368
The two numbers that have a difference of 6 and a product of 40 are 10 and 4. To find them, you can set up the equations: ( x - y = 6 ) and ( xy = 40 ). Solving these gives you ( x = 10 ) and ( y = 4 ) (or vice versa).
Yes.
I think 40...
As a product of its prime factors: 2*3*13 = 78
1,2,4,8,16,32
To find the factors of 560 where the middle term is -24, we can express the equation in the form of (x^2 + mx + n = 0), where (m) is the middle term and (n) is the product of the factors. The factors of 560 that also add up to -24 are -14 and -40, since (-14 + (-40) = -24) and (-14 \times -40 = 560). Thus, the required factors are -14 and -40.
368
easy break it up 26- 2,1,13, and itself 26 40- 4,1,10,5,8,20,2, and itself 40 24- 1,2,3,4,6,8,12, and itself 24
Its factors are: 1, 3, 9, 27 and they add up to 40
when both factors in a multiplication problem are rounded up to estimate the product, the estimate is an overestimate.
The four factors of 33 add up to 48.
The only number between 30 and 40 with distinct prime factors which add up to 12 is 35.
-46
8 and 5
They are 1 and 2
The four factors of 33 add up to 48.