For the diagram below, form an equation and then solve it to calculate the value of x
1 answer:
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Answer:
Option 3
Step-by-step explanation:
we are given the graph of a parabola
with vertex at (0,-1) and symmetrical about y axis and also open up.
Hence the equation of the graph would be transformation of
by a vertical shift of 1 unit down.
So new equation after transformation would be
Or
Hence option 3 is right choice.
Verify:
Put Solving the equation we get
In the graph x intercepts are the same as -1 and 1
Hence our answer is right.
Answer:
The horizontal asymptote can be described by the line y = 6
The vertical asymptote can be described by the line x = -2
Step-by-step explanation:
* <em>Lets the meaning of vertical and horizontal asymptotes</em>
- <u><em>Vertical asymptotes</em></u> are vertical lines which correspond to the zeroes
of the denominator of a rational function
- <u><em>A horizontal asymptote</em></u> is a y-value on a graph which a function
approaches but does not actually reach
- If the degree of the numerator is less than the degree of the
denominator, then there is a horizontal asymptote at y = 0
- If the degree of the numerator is greater than the degree of the
denominator, then there is no horizontal asymptote
- If the degree of the numerator is equal the degree of the denominator,
then there is a horizontal asymptote at y = leading coefficient of the
numerator ÷ leading coefficient of the denominator
* <em>Lets solve the problem</em>
∵
∵ The numerator is 6x
∵ The denominator is x + 2
∴ The numerator and the denominator have same degree
∵ The leading coefficient of the numerator is 6
∵ The leading coefficient of the denominator is 1
∴ There is a horizontal asymptote at y = 6/1
∴ <em>The horizontal asymptote can be described by the line y = 6</em>
- Put the denominator equal zero to find its zeroes
∵ The denominator is x + 2
∴ x + 2 = 0
- Subtract 2 from both sides
∴ x = -2
∴ <em>The vertical asymptote can be described by the line x = -2</em>
Answer:
The cosine function to model the height of a water particle above and below the mean water line is h = 2·cos((π/30)·t)
Step-by-step explanation:
The cosine function equation is given as follows h = d + a·cos(b(x - c))
Where:
= Amplitude
2·π/b = The period
c = The phase shift
d = The vertical shift
h = Height of the function
x = The time duration of motion of the wave, t
The given data are;
The amplitude = 2 feet
Time for the wave to pass the dock
The number of times the wave passes a point in each cycle = 2 times
Therefore;
The time for each complete cycle = 2 × 30 seconds = 60 seconds
The time for each complete cycle = Period = 2·π/b = 60
b = π/30 =
Taking the phase shift as zero, (moving wave) and the vertical shift as zero (movement about the mean water line), we have
h = 0 + 2·cos(π/30(t - 0)) = 2·cos((π/30)·t)
The cosine function is h = 2·cos((π/30)·t).