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Elina [12.6K]
3 years ago
12

The tides around Cherokee Bay range between a low of 1 foot to a high of 5 feet. The tide is at its lowest point when time, t, i

s 0 and completes a full cycle over a 24 hour period. What is the amplitude, period, and midline of a function that would model this periodic phenomenon?

Mathematics
1 answer:
Eddi Din [679]3 years ago
6 0
Given:
Low tide height = 1 ft
High tide height = 5 ft
Tide period, T = 24  houts

Let the height of the tide be modeled by the expression
h(t) = K + A cos(bt)
Because the period is 24, therefore
b = (2π)/24 = π/12

That is,
h(t) = K + Acos[(πt)/12]

When r=0, h = 1, therefore
K + A cos(0) = 1, ot
 K + A = 1             (1)
When t = 12 (half cycle), h = 5, therefore
K + A cos(π) = 5, or
K - A = 5             (2)

Add (1) and (2):
2K = 6
K = 3
From(1), obtain
A = 1 - 3 = - 2

Answer:
The required function is h(t) = 3  - 2 cos[(πt)/12]
The amplitude is 2 feet
The period is 24 hours
The midline of the function is h = 3 feet
A graph of the function is shown below.

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Check the picture below.  so, that'd be the triangle's sides hmmm so let's use Heron's Area formula for it.

~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{15}) ~\hfill a=\sqrt{[ 15- 10]^2 + [ 15- 5]^2} \\\\\\ ~\hfill \boxed{a=\sqrt{125}} \\\\\\ (\stackrel{x_1}{15}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{30}~,~\stackrel{y_2}{9}) ~\hfill b=\sqrt{[ 30- 15]^2 + [ 9- 15]^2} \\\\\\ ~\hfill \boxed{b=\sqrt{261}}

(\stackrel{x_1}{30}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{5}) ~\hfill c=\sqrt{[ 10- 30]^2 + [ 5- 9]^2} \\\\\\ ~\hfill \boxed{c=\sqrt{416}} \\\\[-0.35em] ~\dotfill

\qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{125}\\ b=\sqrt{261}\\ c=\sqrt{416}\\ s\approx 23.87 \end{cases} \\\\\\ A\approx\sqrt{23.87(23.87-\sqrt{125})(23.87-\sqrt{261})(23.87-\sqrt{416})}\implies \boxed{A\approx 90}

6 0
2 years ago
Mikheal wanted to rewrite the conversion factor "1 yard = 0.914 meters" to create a conversion factor to convert meters to yards
Levart [38]
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7 0
3 years ago
Can someone help me with this question please?
blagie [28]

Answer:

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  b) h ≈ 723.11 m

Step-by-step explanation:

<h3>a)</h3>

The equation you want to solve is the model with the given values filled in.

  D(h) = 3.57√h . . . . model

  96 = 3.57√h . . . . . equation for seeing 96 km to the horizon

__

<h3>b)</h3>

We solve this equation by dividing by the coefficient of the root, then squaring both sides.

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Dustin would need to have an elevation of 723.11 meters above sea level to see 96 km to the horizon.

8 0
2 years ago
Explain why the graph above represents a function.
zloy xaker [14]
We do not see a graph above
7 0
2 years ago
A boat is located a sea level. A scuba diver is 80 feet along the surface of the water from the boat and 30 feet below thw water
Sauron [17]

Given :

A boat is located a sea level. A scuba diver is 80 feet along the surface of the water from the boat and 30 feet below the water surface.

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To Find :

The slope between the scuba diver and fish.

Solution :

Horizontal distance between diver and fish is 20 feet.

Fish is 10 feet below the scuba diver.

Slope between scuba diver and fish is :

tan\ \theta = \dfrac{10}{20}\\\\tan\ \theta = 0.5

Therefore, the slope between the scuba diver and fish is 0.5 .

6 0
2 years ago
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