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zalisa [80]
3 years ago
10

The water level varies from 12 inches at low tide to 52 inches at high tide. Low tide occurs at 9:15 a.m. and high tide occurs a

t 3:30 p.m. What is a cosine function that models the variation in inches above and below the water level as a function of time in hours since 9:15 a.m.?
Mathematics
1 answer:
balandron [24]3 years ago
5 0
52 inches - 12 inches = 40 inches
amplitude:  a = 40 inches / 2 = 20 

<span>f(x)=20cos(bx)+c</span>
the value of c is 32... since the centre of the has been moved up 32 units

the minimum amplitude =  32 - 20 = 12
the maximum amplitude = 32 + 20 = 52

<span>f(x)=20cos(bx)+32</span><span>
if the curve takes 6 1/4 hours from low to high tides (9:15 am to 3:30 pm)  then it will take 12 1/2 hours to complete a full cycle.

adjust the period by converting 12 1/2 hours to an angle measure.

360</span>°/12 = 30° 
30° / 12 = 15°

12 1/2 = 360° + 15° = 375°

f(x) = 20 cos(375°) + 32
f(x) = 20 * 0.97 + 32
f(x) = 19.4 + 32
f(x) = 51.4
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Answer:

If your solving for y...

The answer would be: y = -9/8x + 17/8 I believe.

For the second one... it would be : y = -1/7x + 8/7

Step-by-step explanation:

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If LR = 12 and PR = 4, find LP. Explain.
Softa [21]
<h2>Answer:</h2>

LP = 8 because LR + PR = LP according to the Segment Addition Postulate, and 8 + 4 = 12 using substitution

<h2>Step-by-step explanation:</h2>

From this problem, we know that:

LR = 12

PR = 4

So here we have a Line segment. Recall that a line segment has two endpoints, places where they end or stop and they are named after their endpoints, so the line segment here is LR whose measure is 12. Then, according to Segment Addition Postulate it is true that:

LP + PR = LR

By substituting LR = 12 and PR = 4, we have:

LP + 4 = 12

Subtracting 4 from both sides:

LP + 4 - 4 = 12 - 4

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3 years ago
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Which statement describes the inverse of m(x) = x2 – 17x?
stealth61 [152]

Answer:

The correct option is;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

Step-by-step explanation:

The given information is that m(x) = x² - 17·x

The above equation can be written in the form;

y = x² - 17·x

Therefore;

0 = x² - 17·x - y

From the general solution of a quadratic equation, 0 = a·x² + b·x + c we have;

x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}

By comparison to the equation,0 = x² - 17·x - y, we have;

a = 1, b = -17, and c = -y

Substituting the values of a, b and c into the formula for the general solution of a quadratic equation, we have;

x = \dfrac{-(-17)\pm \sqrt{(-17)^{2}-4\times (1) \times (-y)}}{2\times (1)} = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}

Which can be simplified as follows;

x =  \dfrac{17\pm \sqrt{289+4\cdot y}}{2}= \dfrac{17}{2} \pm \dfrac{1}{2}  \times \sqrt{289+4\cdot y}} = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +\dfrac{4\cdot y}{4} }}

And further simplified as follows;

x = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +y }} = \dfrac{17}{2} \pm \sqrt{y + \dfrac{289}{4} }}

Interchanging x and y in the function of the inverse, m⁻¹(x), we have;

m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

We note that the maximum or minimum point of the function, m(x) = x² - 17·x found by differentiating the function and equating the result to zero, gives;

m'(x) = 2·x - 17 = 0

x = 17/2

Similarly, the second derivative is taken to determine if the given point is a maximum or minimum point as follows;

m''(x) = 2 > 0, therefore, the point is a minimum point on the graph

Therefore, as x increases past the minimum point of 17/2, m⁻¹(x) increases to give;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }} to increase m⁻¹(x) above the minimum.

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