SOLUTION: The height of the ocean at the dock is modeled by the function, h= 3sin(pi*t/4) +5 where h is measured in feet and t is the time in hours. If t=0 refers to 12:00 a.m., what is the

Question 252605: The height of the ocean at the dock is modeled by the function, h= 3sin(pi*t/4) +5 where h is measured in feet and t is the time in hours. If t=0 refers to 12:00 a.m., what is the height of the ocean at 10:00 a.m.?
find the exact value of the vertical asymptotes for 0<=x<=pi for the function y=cot(3x)
if sin(x)= -3/5 with x in the quadrant 4, find the sec(x)
Thanks so much!!
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
I'll do the first two to get you going.

# 1

Since "t=0 refers to 12:00 a.m", this means that refers to 10:00 a.m. (just add 10 hours to 0)

Start with the given equation.

Plug in

Reduce.

Rearrange the terms

Subtract from the argument (this is valid because you'll end up on a coterminal angle).

Combine like terms.

Use the unit circle to evaluate the sine of to get 1.

Multiply

Add

So the height of the ocean at 10:00 a.m. is 8 feet.

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# 2

Remember that and . So this means that or in short,

To find the vertical asymptotes of , we'll set the denominator equal to zero and solve for 'x' (since division by zero is undefined).

Set the denominator equal to zero.

Take the arcsine of both sides.

or Evaluate the arcsine of 0 to get or . Don't forget to add on multiples of to each solution.

or Divide both sides by 3 to isolate 'x' in each case.

As a shortcut, you can condense the solution to where 'n' is an integer.

So if , where 'n' is an integer, then .

But since , this means that we're only going to look at the solutions (for n=0), (where n=1), (when n=2), and (when x=3). Note: any other solution is outside the interval .

So the four vertical asymptotes of are , , , and
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