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:
Answer:
3
Step-by-step explanation:
f(x) =2x+3
Let x=0
f(0) =2*0+3
Multiply
f(0) =0+3
Add
f(0) =3
Answer:
3rd term: 24
10th term: 220
Step-by-step explanation:
<em>For the 3rd term</em>
2*3 + 2*(3^2) = 6 + 2*9 = 6 + 18 = 24
<em>For the 10th term</em>
2*10 + 2*(10^2) = 20 + 2*100 = 20 + 200 = 220
Answer:
X=7
Step-by-step explanation:
I am so sorry if this is not correct. I don´t know why there are 0s next to each number. I hope this helped. Again, I am sorry if this is incorrect
Answer:
x = -6/ -7/ -8/ -9/ -10/ -11...... every number less -5
Step-by-step explanation:
