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ioda
3 years ago
12

What is the amplitude, period, and phase shift of f(x) = −3 cos(4x + π) + 6?

Mathematics
1 answer:
hammer [34]3 years ago
6 0
This is in the form of \sf acos(bx-c)+d.

Amplitude is equal to \sf |a|
Period is equal to \sf\dfrac{2\pi}{|b|}
Phase Shift is equal to \sf\dfrac{c}{b}

Plug in the values:

Amplitude:
\sf |a|\rightarrow |-3|\rightarrow\boxed{\sf 3}

Period:
\sf\dfrac{2\pi}{|4|}\rightarrow\dfrac{2\pi}{4}\rightarrow\boxed{\sf\dfrac{\pi}{2}}

Phase Shift:
\sf\dfrac{c}{b}\rightarrow\boxed{\sf -\dfrac{\pi}{4}}
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Find the midpoint of the segment with the following end points (10,5) and (6,9)
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Answer:

The mid-point between the endpoints (10,5) and (6,9) is:

  • \left(x,\:y\right)=\left(8,\:7\right)

Step-by-step explanation:

Let (x, y) be the mid-point

Given the points

  • (10,5)
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Using the formula to find the mid-point between the endpoints (10,5) and (6,9)

\left(x,\:y\right)=\left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)

Here:

\left(x_1,\:y_1\right)=\left(10,\:5\right),\:\left(x_2,\:y_2\right)=\left(6,\:9\right)

Thus,

\left(x,\:y\right)=\left(\frac{x_2+x_1}{2},\:\:\frac{y_2+y_1}{2}\right)

\left(x,\:y\right)=\left(\frac{6+10}{2},\:\frac{9+5}{2}\right)

\left(x,\:y\right)=\left(8,\:7\right)

Therefore, the mid-point between the endpoints (10,5) and (6,9) is:

  • \left(x,\:y\right)=\left(8,\:7\right)
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