posted by ba .
For a certain day,the depth of water,h, in metres in PEI, in hours is given by the formula:h(t) = 7.8 + sin3.5 (pi/6(t-3)), t E [0,24], assume t=0 represents midnight. Provide an algebraic solution to determine the time(s) of day, the water reaches the depth of 10.29 m( give your answer in hours and minutes)
Stacey/ba, please don't switch names
7.8 + sin3.5 (pi/6(t-3)) = 10.29
sin3.5 (pi/6(t-3)) = 2.49
not possible, sin(?) cannot be > 1
when you first posted this, you had
I will change it to
7.8 + 3.5sin (pi/6(t-3)) = 10.29
3.5sin (pi/6(t-3)) = 2.49
sin (pi/6(t-3)) = .71142..
(pi/6(t-3)) = .79152 or (pi/6(t-3)) = π - .79152 = 2.35..
if (π/6)(t-3) = .79152..
t-3 = 1.5117..
t = 4.5117..
find the other value of t, using the second part of above
Lol sorry but that's what I got 4.51 I just wasn't sure how to get the second value
could you please help in finding the second value of t. thank you