Question

can some one check my calculus one answers? They are pretty simple...
http://www.math.uh.edu/~pamb/1431EMCF%2010.pdf

Answer 1

1. b
2. b
3. a
4. b
5. 5
6. d
7. d
8. d
9. b
10. d

Answer 2

5. b **

Answer 3

1.d

Answer 4

2.d

Answer 5

no...sorry 2.c

Answer 6

3.a

Answer 7

I have that you got 2 and 6 wrong, I skipped the limits because I haven't done those in a long time

Answer 8

i dont know how to do #6

Answer 9

6.c

Answer 10

differentiate it

Answer 11

6 is C.\[f'(x)=h'(x)g'(h(x))\\f'(-3)=h'(-3)g'(h(-3))=5g'(2)=-15\]

Answer 12

f'(x)= g'(h(x)) h'(x)
now plug in x=-3 and put all the values...u'll get the answer

Answer 13

i grasp the second step but it is just deriving it that is hard for me.

Answer 14

thanks everyone!

Answer 15

\[\lim_{x\to0}\frac{\sin6x}{\sin4x}=\lim_{x\to0}\frac{\cos6x}{\cos4x}=1\]

Answer 16

6/4(lim x->0 sinx/sinx=1)

Answer 17

6/4(1)=3/2

Answer 18

i thought it was like this???

Answer 19

Nope, that would be the case for\[\lim_{x\to\infty}\frac{6\sin{x}}{4\sin{x}}\]