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sattari [20]
3 years ago
5

1) State the amplitude, period, phase shift, vertical shift and graph

Mathematics
1 answer:
bogdanovich [222]3 years ago
3 0

Answer:

See answers and explanations below (along with a visual graph)

Step-by-step explanation:

For the function y=acos(b(x+c))+d:

Amplitude: |a|

Period: \frac{2\pi}{|b|}

Phase shift: \frac{-c}{b}

Vertical shift: d

Therefore, for y=2cos(\frac{2}{3}(\theta+\pi))-1:

Amplitude: |a|=|2|=2, which is 2 units above and below the midline (see d)

Period: \frac{2\pi}{|b|}=\frac{2\pi}{|\frac{2}{3}|}=\frac{2\pi}{\frac{2}{3}}=3\pi, so the cycle will repeat every 3π units

Phase shift: \frac{-c}{b}=\frac{-\frac{2\pi}{3} }{\frac{2}{3} }=-\pi, or π units to the left

Vertical shift: d=-1, or 1 unit down

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Am I correct if not help plzzz
11Alexandr11 [23.1K]
Hey there!

You got the first part right where you added r to both sides. However, you need to make sure that you complete all actions on both sides, including canceling out parts of terms by division. The only thing you need to do is divide both sides by h instead of just the left. This will make your answer:

y =  \frac{b+r}{h}

Hope this helped you out! :-)
4 0
3 years ago
The line segment joining A(6, 3) to B(–1, –4) is doubled in length by having half its length added to each end. Find the coordin
mr Goodwill [35]

Answer:

  • (9.5, 6.5) and (-4.5, -7.5)

Step-by-step explanation:

Let the extended points be A' and B' and add the point M as midpoint of AB

<u>Coordinates of M are:</u>

  • ((6 - 1)/2, (3-4)/2) = (2.5, -0.5)

Now point A is midpoint of A'M and point B is midpoint of MB'

<u>Finding the coordinates using midpoint formula:</u>

  • A' = ((2*6 - 2.5),(2*3 - (-0.5)) = (9.5, 6.5)
  • B' = ((2*(-1) - 2.5), (2*(-4) - (-0.5)) = (-4.5, -7.5)

4 0
3 years ago
If CDE ~ GDF, find ED
qaws [65]

Answer:

10

Step-by-step explanation:

\triangle CDE \sim \triangle GDF.. (given) \\\\\therefore \frac{CD}{GD} =\frac{DE}{DF}.. (csst) \\\\\therefore  \frac{15}{x+3} =\frac{3x+1}{4}\\\\ \therefore   \: 15 \times 4 = (x + 3)(3x + 1) \\  \\ \therefore   \: 60 = 3 {x}^{2}  + x + 9x + 3 \\  \\ \therefore  3 {x}^{2}  + 10x + 3 - 60 = 0 \\ \therefore  3 {x}^{2}  + 10x  - 57 = 0 \\ \therefore  3 {x}^{2}  + 19x - 9x  - 57 = 0 \\ \therefore   \: x(3x + 19) - 3(3x + 19) = 0 \\\therefore   \:  (3x + 19)(x - 3) = 0 \\ \therefore   \: 3x + 19 = 0 \:  \: or \:  \: x - 3 = 0 \\  \therefore   \: x =  -  \frac{19}{3}  \:  \: or \:  \: x = 3 \\  \because \: x \: can \: not \: be \:  - ve \\ \therefore   \: x = 3 \\ ED = 3x + 1 = 3 \times 3 + 1  \\ \huge \red{ \boxed{ ED= 10}}

7 0
3 years ago
Need help as fast as possible will mark brainliest
MissTica

Answer:

Option C. is correct choice.

Step-by-step explanation:

20\pm 0.1\:^{\circ}\:C

Regards: Umer

6 0
3 years ago
Please help me out!!!!!
Gnoma [55]
The last answer: 2 or 0

Explanation:

When you graph this function, it crosses the x-axis once. In other words, that means there is only 1 real zero and 2 imaginary complex zeros. In addition, imaginary solutions always come in pairs, so there can’t be a odd number of them such as 3.
8 0
3 years ago
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