Question

Help please
http://prntscr.com/121os3

Answer 1

I'm going to use x in place of theta

Answer 2

cos(2x) = 2cos^2(x) - 1

cos(x+x) = 2cos^2(x) - 1

cos(x)cos(x)-sin(x)sin(x) = 2cos^2(x) - 1

cos^2(x)-sin^2(x) = 2cos^2(x) - 1

cos^2(x)-(1-cos^2(x)) = 2cos^2(x) - 1

cos^2(x)-1+cos^2(x) = 2cos^2(x) - 1

2cos^2(x) - 1 = 2cos^2(x) - 1

Answer 3

and I'm using those identities I gave you earlier

Answer 4

ok

Answer 5

http://prntscr.com/121qcq

Answer 6

what did you get

Answer 7

that's the function in general

Answer 8

what is the function in this case though

Answer 9

\[N = N _{0}e^{kt}\]

Answer 10

yes i know, but what is the function in this case

Answer 11

M=3500e^k(x)

Answer 12

ok closer, but what is the value of k

Answer 13

1500

Answer 14

you sure?

Answer 15

no

Answer 16

ok hold on

Answer 17

we know that

M = 3500e^(kx)

plug in (3,1500) to get

1500 = 3500*e^(k*3)

and solve for k

Answer 18

from where did we get 3

Answer 19

(3, 1500) is one point on the graph

Answer 20

ok for part b

Answer 21

once you find k, you plug in x = 9.5 to find M

Answer 22

K?

Answer 23

you first solve 1500 = 3500*e^(k*3) for k

Answer 24

what do you mean

Answer 25

k=0.76

Answer 26

@jim_thompson5910

Answer 27

let me check

Answer 28

that's not correct

Answer 29

then?

Answer 30

try solving 1500 = 3500*e^(k*3) for k again

Answer 31

1500 = 3500*e^(k*3)

1500/3500 = e^(3k)

3/7 = e^(3k)

e^(3k) = 3/7

...
...
...

k = ???

Answer 32

0.14

Answer 33

no

Answer 34

e^(3k) = 3/7

3k = ln(3/7)

k = ???

Answer 35

-0.282

Answer 36

better

Answer 37

M = 3500e^(kx)

turns into

M = 3500e^(-0.282x)

Answer 38

that's your equation, you then use that in part b and plug in x = 9.5 and evaluate

Answer 39

http://prntscr.com/121yhv