Question

Consider the function h=100sin[pi/240(t)] + 50. For what values of t is h>100 where 0 13 years ago

Answer 1

so let`s start it .......
you need to solve
100 > 100sin(pi/240 t) + 50
50 > 100sin(pi/240 t)
1/2 > sin(pi/240 t)

can you proceed next?

Answer 2

I have no idea... maybe take the sin of both sides?

Answer 3

no , we need to find t so think about inverse operations
pi/240 t = 2pi *n + pi/3
do you get it
if yes then find the value of t in terms of n

Answer 4

i don't know what you just did with n all i got told was 2kpi and that k was an integer

Answer 5

ok i you can take k instead of n it`s one and the same thing
pi/240 t = 2kpi + pi/3
fine now?

Answer 6

why is it pi/3 at the end?

Answer 7

opps sorry it should have been pi/6

Answer 8

ok I don't know what to do now....I know that i need to get t by itself but I have no clue

Answer 9

t = ( 2kpi + pi/3 ) (240/pi) = (2k + 1/3)*240
now find all values of k such that 0<t<350
those should get you the answer .....

Answer 10

Thanks

Answer 11

welcome!!!