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Katarina [22]
3 years ago
10

Over a 24-hour period, the tide in a harbor can be modeled by one period of a sinusoidal function. The tide measures 4.35 ft at

midnight, rises to a high of 8.3 ft, falls to a low of 0.4 ft, and then rises to 4.35 ft by the next midnight.
What is the equation for the sine function f(x), where x represents time in hours since the beginning of the 24-hour period, that models the situation?

f(x)=
Mathematics
1 answer:
netineya [11]3 years ago
8 0

Answer:

  f(x) = 4.35 +3.95·sin(πx/12)

Step-by-step explanation:

For problems of this sort, a sine function is used that is of the form ...

  f(x) = A + Bsin(2πx/P)

where A is the average or middle value of the oscillation, B is the one-sided amplitude, P is the period in the same units as x.

It is rare that a tide function has a period (P) of 24 hours, but we'll use that value since the problem statement requires it. The value of A is the middle value of the oscillation, 4.35 ft in this problem. The value of B is the amplitude, given as 8.3 ft -4.35 ft = 3.95 ft. Putting these values into the form gives ...

  f(x) = 4.35 + 3.95·sin(2πx/24)

The argument of the sine function can be simplified to πx/12, as in the Answer, above.

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the height of the house is 408ft .

<u>Step-by-step explanation:</u>

Here we have , To estimate the height of a house Katie stood a certain distance from the house and determined that the angle of elevation to the top of the house was 32 degrees. Katie then moved 200 feet closer to the house along a level street and determined the angle of elevation was 42 degrees. We need to find  What is the height of the house . Let's find out:

Let  y is the unknown height of the house, and x is the unknown number of feet she is standing from the house.

Distance of house from point A( initial point ) = x ft

Distance of house from point B( when she traveled 200 ft towards street  = x-200 ft

Now , According to question these scenarios are of right angle triangle as

At point A

⇒ Tan32 =\frac{Perpendicular}{Base}= \frac{y}{x}

⇒ Tan32 = \frac{y}{x}

⇒ y=x(Tan32 )       ..................(1)

Also , At point B

⇒ Tan42 = \frac{y}{x-200}

⇒ y=(x-200)(Tan42)     ..............(2)

Equating both equations:

⇒ (x-200)(Tan42) = x(Tan32)

⇒ x(Tan42-Tan32)=Tan42(200)

⇒ x=\frac{Tan42(200)}{(Tan42-Tan32)}

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Putting  x=653ft in  y=x(Tan32 )  we get:

⇒  y=x(Tan32 )    

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⇒  y=408ft

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4 0
3 years ago
What values for q (0 ≤q≤2π)<br> satisfy the equation?<br><br> 22√sin q + 2 = 0
Vesna [10]
Answer:
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Explanation:
2√2 sin(q) + 2 = 0
2√2 sin(q) = -2
sin(q) = \frac{-2}{2 \sqrt{2} }
sin(q) = \frac{- \sqrt{2} }{2}

Now, we know that:
sin (45) = \frac{ \sqrt{2} }{2}

From the ASTC rule, we know that the sine function is negative in the third and fourth quadrant.
This means that:
either q = 90 + 45 = 135° which is equivalent to \frac{3 \pi }{4}
or q = 270 + 45 = 315° which is equivalent to \frac{7 \pi }{4}

Hope this helps :)
6 0
3 years ago
Read 2 more answers
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