Answer:
Step-by-step explanation:
![{ \tt{\int\limits^2_1 {x^{2}-8x+8 } \, dx}} \\ \\ = { \tt{[ \frac{ {x}^{3} }{3} - 4 {x}^{2} + 8x ] {}^{2} _{1}}}](https://tex.z-dn.net/?f=%7B%20%5Ctt%7B%5Cint%5Climits%5E2_1%20%7Bx%5E%7B2%7D-8x%2B8%20%7D%20%5C%2C%20dx%7D%7D%20%5C%5C%20%20%5C%5C%20%3D%20%20%7B%20%5Ctt%7B%5B%20%5Cfrac%7B%20%7Bx%7D%5E%7B3%7D%20%7D%7B3%7D%20%20-%204%20%7Bx%7D%5E%7B2%7D%20%20%2B%208x%20%5D%20%7B%7D%5E%7B2%7D%20_%7B1%7D%7D%7D)
Substitute x with the limits:
You'll most likely have to use the quadratic formula to find your answer.
The vertex of the absolute value function defined by f(x) = Ix - 2I - 7 is (2,-7).
When you say comparing do you mean dividing or just comparing
1. Cross multiply
35x = 5(11)
35x = 55
Divide both sides by 35
x = 55/35
x = 11/7
2. (x - 2)/x = 3/8
Cross multiply
3x = 8(x - 2)
3x = 8x - 16
Subtract 8x from both sides
-5x = -16
divide both sides by -5
x = -16/-5
x = 16/5 OR 3 1/5
3. (a + 1)/(a - 1) = 5/6
cross multiply
6(a + 1) = 5(a - 1)
distribute
6a + 6 = 5a - 5
subtract 5a from both sides
a + 6 = -5
subtract 6 from both sides
a = -11
4. (1/3)x - 4 = (2/3)x + 6
multiply each term by 3 to clear the fractions
x - 12 = 2x + 18
subtract x from both sides
-12 = x + 18
subtract 18 from both sides
-30 = x
