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goldfiish [28.3K]
3 years ago
8

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:
LiRa [457]3 years ago
3 0
F ( x ) = a cos ( b ( x - c ) ) + d
Apmlitude: a = ( 12 - 52 ) / 2 = -40 / 2 = - 20
If the curve takes 6.25 hours from the low to the high tide them it will take 12.50 hours to complete a full cycle.
The period:   b = 2π/12.5 ;
c = 9.25; d = 12 + 20 = 32;
Answer:
f ( x ) = - 20 cos ( 2π/12.5 ·( x - 9.25 ) ) + 32 =
= - 20 cos ( 0.5024 ( x - 9.25 ) ) +32
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Answer:

On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4

Step-by-step explanation:

The given function is presented as follows;

f(x) = \dfrac{2}{x - 1} + 4

From the given function, we have;

When x = 1, the denominator of the fraction, \dfrac{2}{x - 1}, which is (x - 1) = 0, and the function becomes, \dfrac{2}{1 - 1} + 4 = \dfrac{2}{0} + 4 = \infty + 4 = \infty therefore, the function in undefined at x = 1, and the line x = 1 is a vertical asymptote

Also we have that in the given function, as <em>x</em> increases, the fraction \dfrac{2}{x - 1} tends to 0, therefore as x increases, we have;

\lim_  {x \to \infty}  \dfrac{2}{(x - 1)} \to 0, and \  \dfrac{2}{(x - 1)}  + 4 \to 4

Therefore, as x increases, f(x) → 4, and 4 is a horizontal asymptote of the function, forming a curve that opens up and to the right in quadrant 1

When -∞ < x < 1, we also have that as <em>x</em> becomes more negative, f(x) → 4. When x = 0, \dfrac{2}{0 - 1} + 4 = 2. When <em>x</em> approaches 1 from the left, f(x) tends to -∞, forming a curve that opens down and to the left

Therefore, the correct option is on a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4.

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find the scale, k, of H(3,4) if H'(9,12) after going through the transformation (x,y) ---&gt; (kx,ky) centered at the origin
stiks02 [169]

The value of scale k is 3

Step-by-step explanation:

In order to find the scale , we can either divide the coordinates of H' by the original point or compare both the points to find the original ordered pair.

Given

H(3,4)

H'(9,12)

The coordinates of H are multiplied with some integer k to form H'

So,

<u>For x-coordinate:</u>

3k = 9

k = 9/3 = 3

<u>For y-coordinate:</u>

4k = 12

k = 12/4 = 3

Hence,

The value of scale k is 3

Keywords: Coordinate geometry, scaling

Learn more about scaling at:

  • brainly.com/question/4771355
  • brainly.com/question/4786449

#LearnwithBrainly

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