Well, assuming that x ∈ R and assuming that by "sum" you mean the sum of the area under curve and above the line y=0, and that the term you are giving is a function where f(x) = 10x - 62, then..
f(x) = 10x - 36
∴∫f(x)dx = 5x2 - 36x + C
Now, if you know the range in which you want this area, you can say:
a
∫f(x)dx = 5a2 - 36a - 5b2 + 36b
b
But I'm guessing that this isn't a calculus question at all, and that you should probably go do your own homework.
The operators are missing, the possible factorisations are:4x2 + 10x + 6 = 2(2x + 3)(x + 1)4x2 - 10x + 6 = 2(2x - 3)(x - 1)4x2 + 10x - 6 = 2(2x - 1)(x + 3)4x2 - 10x - 6 = 2(2x + 1)(x - 3)
-10x-6 = -12 -10x = -12+6 -10x = -6 x = 3/5
The sum of -8 and 6 is -2
6-3x = 5x-10x+10 6-3x = -5x+10 -3x+5x = 10-6 2x = 4 x = 2
2-10x = -8
12x - 6 = 10x +2 12x - 10x = 2 + 6 2x = 8 x = 4
6-3x = 5x-10x+2 4 = 8x-10x 4 = -2x x = -2
The operators are missing, the possible factorisations are:4x2 + 10x + 6 = 2(2x + 3)(x + 1)4x2 - 10x + 6 = 2(2x - 3)(x - 1)4x2 + 10x - 6 = 2(2x - 1)(x + 3)4x2 - 10x - 6 = 2(2x + 1)(x - 3)
- 10X - 5Y = - 6 add 10X to both sides - 5Y = 10X - 6 divide both sides by - 5 Y = - 2X + 6/5 slope(m) = - 2
The solution is 1/2 or 0.5. This is how: 2x+1=7-10x 2x=6-10x 12x=6 x=1/2
-10x-6 = -12 -10x = -12+6 -10x = -6 x = 3/5
I assume you mean factors for 4x2+10x+6.(2x+3)(2x+2).
the sum of -2 and 6 and 7 is (-2+6+7)
10x-7=4x+5 +7 +7 10x=4x+12 -4x -4x 6x=12 /6 /6 x=2
3.3333
The sum of -8 and 6 is -2
136