Question

Hello, I need some calculas help and all help is appreciated

Answer 1

l

Answer 2

@kennyj123

Answer 3

a d c

Answer 4

are you totally sure?

Answer 5

yes

Answer 6

can u help me with some more plz?

Answer 7

\[\int\limits_{2}^{5}x^2~dx=\left[ \frac{ x^3 }{ 3 } \right]~from ~2~\to~5=\frac{ 1 }{3 }\left( 5^3-2^3 \right)=?\]

Answer 8

l

Answer 9

d e

Answer 10

you got a lot of questions

Answer 11

no just 10

Answer 12

can i get medal

Answer 13

\[\int\limits_{0}^{\frac{ \pi }{ 2 }}2\sin \left( 2x \right) dx=2\left[ \frac{ -\cos 2x }{2 }\right]~from~0~\to~\frac{ \pi }{2 }=-\left( \cos 0-\cos \pi \right)=\left( 1-(-1) \right)\]
=?

Answer 14

c

Answer 15

c for which one?

Answer 16

1-39,
2-2,
3-a=12, b=48

Answer 17

thnxs how bout the rest?

Answer 18

u=x^3
when x=2,u=2^3=8
when x=4,u=4^3=64

Answer 19

question 10