SOLUTION: Sorry I had asked the same question before but by accidentally i wrote the wrong equation so i am rewriting the question. I am sorry! For a certain day, the depth of water, h, in

Question 1042217: Sorry I had asked the same question before but by accidentally i wrote the wrong equation so i am rewriting the question. I am sorry!
For a certain day, the depth of water, h, in metres in Tofino, B.C, at time t, in hours, is given by the formula:
h(t)=7.8+3.5sin[pi/6(t-3)], tE[0,24]. Assume t=0 represents midnight. Provide an algebraic solution to determine the time(s) of day, the water reaches the depth of 10.29m. (Give the answer in hours and minutes)
Also if you could please answer it step by step so I can fully understand how to answer the question.
Thank you
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
For a certain day, the depth of water, h, in metres in Tofino, B.C, at time t, in hours, is given by the formula:
h(t)=7.8+3.5sin[pi/6(t-3)], tE[0,24]. Assume t=0 represents midnight. Provide an algebraic solution to determine the time(s) of day, the water reaches the depth of 10.29m. (Give the answer in hours and minutes)
-----
Solve::
7.8 + 3.5*sin((pi/6)(t-3) = 10.29
-----
3.5*sin((pi/6)(t-3)) = 2.49
---
sin((pi/6)(t-3)) = 0.711
----
Take sin^-1 of both sides to get:
(pi/6)(t-3) = 0.791+-(2pi*n) or (pi/6)(t-3) = pi-0.7911 = 2.35+-(2pi*n)
-----9
t-3 =1.51+-(12) or (t-3) = 4.488+-(12)
Since you want t E [0,24] t = 16.488 + 3 = 19.488 = 19 hr 0.488sec
= 19hr +0.488*60min = 19hr 29min + 0.28(60)sec = 19hr 29min 17sec
------------
Cheers,
Stan H.
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