Someone please tell me if this problem is humanly possible to get correct! I have a screenshot of it in the comments

Question

please.help.me · Answer

attachment

reemii · Answer

You only want the first time $t$, that is smaller than 24. Remove the “$+12n$” from your answer and that will be fine.

please.help.me · Answer

That did not give me the right answer either

daysiphoto · Answer

What subject is this?

please.help.me · Answer

Pre Calculus

please.help.me · Answer

Covers solving trigonometric equations that come from applications or word problems.

reemii · Answer

For (a):
$$h(6) = 53 + 50 \cdot \sin(\frac\pi{12}6 - \pi/2) = 53 + 50 \cdot \sin(\frac\pi2 - \pi/2 ) = 53 + 50\cdot \sin(0) = 53$$How is this wrong?

reemii · Answer

$$h(6) = 53 + 50 \cdot \sin(\frac\pi{12}6 - \pi/2)\
 = 53 + 50 \cdot \sin(\frac\pi2 - \pi/2 )\
 = 53 + 50\cdot \sin(0) = 53$$

reemii · Answer

For the last question, don't multiply by $\pi$ or anything, once you have the value of $t$.
So, first time at the top is $t=12$, second time is $t=12+24$, then $t=12+24+24$, etc.