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
3.5sin (pi/6(t-3))

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

To determine the time(s) of day when the water reaches a depth of 10.29 meters, we need to solve the equation h(t) = 10.29 for t within the given interval [0, 24].

The formula that represents the depth of water at a given time t is:
h(t) = 7.8 + sin(3.5(pi/6)(t-3))

Now, let's set h(t) to 10.29 and solve for t:

10.29 = 7.8 + sin(3.5(pi/6)(t-3))

To isolate the sine term, subtract 7.8 from both sides:

2.49 = sin(3.5(pi/6)(t-3))

Next, we need to find the inverse sine (arcsine) of both sides:

arcsin(2.49) = arcsin(sin(3.5(pi/6)(t-3)))

Since arcsin(sin(x)) = x only if x is within the valid range of arcsin (between -π/2 and π/2), we can simplify:

arcsin(2.49) = 3.5(pi/6)(t-3)

Now we solve for t:

t-3 = (arcsin(2.49) * (6/(3.5π))) + 3

t = (arcsin(2.49) * (6/(3.5π))) + 3

Using a calculator, evaluate the value of arcsin(2.49), multiply it by (6/(3.5π)), and add 3 to find the value of t. The resulting t will be in hours after midnight.

Please note that since the equation is a periodic function, there might be multiple values of t that satisfy the equation within the given interval [0, 24].