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Sati [7]
1 year ago
8

Write a cosine function that Has a midline of 2 an amplitude of 3 and a period of 7pi/4

Mathematics
1 answer:
Alexxx [7]1 year ago
5 0

Given:

Amplitude of cosine function, A=3.

Period, T=7π/4.

Midline, D=2.

The time period can be expressed as:

T=\frac{2\pi}{B}

Put T=7π/4 to find the value of B.

\begin{gathered} \frac{7\pi}{4}=\frac{2\pi}{B} \\ B=\frac{4\times2}{7} \\ =\frac{8}{7} \end{gathered}

The general cosine function can be expressed as,

f(x)=A\cos (Bx)+D

Substitute B=8/7, A=3 and D=2 in above equation.

f(x)=3\cos (\frac{8}{7}x)+2

Therefore, the cosine function is,

f(x)=3\cos (\frac{8}{7}x)+2

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BC has a midpoint at M(6, 6). Point C is at (2, 10). Find the coordinates of point B.
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Answer:

The coordinates of B are (10,2)

Step-by-step explanation:

Hi there!

We know that BC has a midpoint M, with the coordinates (6,6), and the endpoint C, with coordinates (2, 10)

We want to find the coordinates of point B

The midpoint formula is (\frac{x_1 + x_2}{2}, \frac{y_1+ y_2}{2}), where (x_1, y_1) and (x_2 , y_2) are points. In this case, the point C has the values of (x_1, y_1), and B has the values of (x_2 , y_2)

We know that coordinates of M equal (\frac{x_1 + x_2}{2}, \frac{y_1+ y_2}{2})

In other words,

\frac{x_1 + x_2}{2} = 6\\ \frac{y_1+ y_2}{2}=6

Let's plug 2 for x_1 and 10 for y_1

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\frac{2 + x_2}{2} = 6\\ \frac{10+ y_2}{2}=6

Multiply both sides by 2

{2 + x_2} = 12\\{10+ y_2} =12

Subtract 2 from both sides in the first equation to find the value of x_2:

x_2=10

Now, for the second equation, subtract 10 from both sides to find the value of y_2

y_2=2

Now substitute these values for (x_2, y_2)

(x_2, y_2)=(10,2)

So the coordinates of point B are (10, 2)

Hope this helps!

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