(-4,6)
The x value changes bc since it is a reflection y cannot change and x becomes negative
The area of an equilateral triangle of side "s" is s^2*sqrt(3)/4. So the volume of the slices in your problem is
(x - x^2)^2 * sqrt(3)/4.
Integrating from x = 0 to x = 1, we have
[(1/3)x^3 - (1/2)x^4 + (1/5)x^5]*sqrt(3)/4
= (1/30)*sqrt(3)/4 = sqrt(3)/120 = about 0.0144.
Since this seems quite small, it makes sense to ask what the base area might be...integral from 0 to 1 of (x - x^2) dx = (1/2) - (1/3) = 1/6. Yes, OK, the max height of the triangles occurs where x - x^2 = 1/4, and most of the triangles are quite a bit shorter...
1) 3 times 15 is 45
6) because in order for it to even have a slope it has to be constant through out the whole thing.
Answer:
1 and -5
Step-by-step explanation:
![~~~~~x^3+3x^2-9x+5=0\\\\\implies x^3-x^2+4x^2-4x-5x+5=0\\\\\implies x^2(x-1)+4x(x-1)-5(x-1)=0\\\\\implies (x-1)(x^2 +4x -5) = 0\\\\\implies (x-1)(x^2+5x -x -5) = 0\\\\\implies (x-1)[x(x+5) -(x+5)] =0\\\\\implies (x-1)(x-1)(x+5)= 0\\\\\implies (x-1)^2(x+5) = 0\\\\\implies x = 1,~~ x = -5](https://tex.z-dn.net/?f=~~~~~x%5E3%2B3x%5E2-9x%2B5%3D0%5C%5C%5C%5C%5Cimplies%20x%5E3-x%5E2%2B4x%5E2-4x-5x%2B5%3D0%5C%5C%5C%5C%5Cimplies%20x%5E2%28x-1%29%2B4x%28x-1%29-5%28x-1%29%3D0%5C%5C%5C%5C%5Cimplies%20%28x-1%29%28x%5E2%20%2B4x%20-5%29%20%3D%200%5C%5C%5C%5C%5Cimplies%20%28x-1%29%28x%5E2%2B5x%20-x%20-5%29%20%3D%200%5C%5C%5C%5C%5Cimplies%20%28x-1%29%5Bx%28x%2B5%29%20-%28x%2B5%29%5D%20%3D0%5C%5C%5C%5C%5Cimplies%20%28x-1%29%28x-1%29%28x%2B5%29%3D%200%5C%5C%5C%5C%5Cimplies%20%28x-1%29%5E2%28x%2B5%29%20%3D%200%5C%5C%5C%5C%5Cimplies%20x%20%3D%201%2C~~%20x%20%3D%20-5)
Step-by-step explanation:
Slope: −2
y-intercept: (0,1)
Answer:
