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Tresset [83]
3 years ago
8

Question 13 and 14 please! Answer and show work :)

Mathematics
1 answer:
kozerog [31]3 years ago
8 0

Answer:

Problem 13) f(x)=4\,sin(\frac{1}{2} x+\frac{2}{3}\pi )-2

Problem 14) f(x)=cotan(x+\frac{1}{3} \pi)+2

Step-by-step explanation:

Recall how transformations affect the graph of the sine function, and how such is conveyed into the parameters A, B, C, and D that could be included in the general form of the function:

f(x)=A\,sin(Bx+C)+D

where the Amplitude of the transformed sine function is the absolute value of the multiplicative parameter A:

Amplitude = |A|

The period is (which for sin(x) is 2\pi) is modified by the parameter B in the following manner:

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

Where the phase shift is introduced as:

Phase shift = -\frac{C}{B}.

and finally any vertical shift is included by the constant D (positive means shift upwards in D many units, and negative means shift downwards D units)

Therefore, to have a sine function with the requested characteristics, we work on the value of the parameters A, B, C, and D one at a time:

1) Amplitude = |A|=4 then we use parameter A = 4

f(x)=4\,sin(Bx+C)+D

2) Period 4\pi, then we work on the parameter B:

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

4\pi=\frac{2\pi}{B}\\B*4\pi=2\pi\\B=\frac{2\pi}{4\pi} \\B=\frac{1}{2} which transforms the function into:

f(x)=4\,sin(\frac{1}{2} x+C)+D

3) phase-shift = -\frac{4}{3} \pi

Then knowing that B=\frac{1}{2}, we work on the value of parameter C:

Phase shift = -\frac{C}{B}

-\frac{4}{3} \pi=-\frac{C}{B} \\-\frac{4}{3} \pi=-\frac{C}{ \frac{1}{2} }\\-\frac{4}{3}* \frac{1}{2} \pi=-C\\C=\frac{2}{3} \pi

Therefore the function gets transformed into:

f(x)=4\,sin(\frac{1}{2} x+\frac{2}{3}\pi )+D

4) and finally the vertical shift of negative two units, that gives us the value D = -2

The complete transformed function becomes:

f(x)=4\,sin(\frac{1}{2} x+\frac{2}{3}\pi )-2

Now for problem 14, recall that the cotangent function is the reciprocal of the tangent function, therefore, their periodicity is the same: \pi

since you are asked for a cotangent function of period \pi as well, there is no multiplication parameter "B" needed (so we keep it unchanged - equal to one). B = 1

Then for the phase-shift which we want it to be -\frac{1}{3} \pi, we set the condition:

-\frac{1}{3} \pi=-\frac{C}{B} \\-\frac{1}{3} \pi=-\frac{C}{1}\\-\frac{1}{3} \pi=-C\\C=\frac{1}{3} \pi

And insert such in the cotangent general form:

f(x)=cotan(x+\frac{1}{3} \pi)+D

and finally include the desired vertical shift of 2 units:

f(x)=cotan(x+\frac{1}{3} \pi)+2

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Kaylib’s eye-level height is 48 ft above sea level, and addison’s eye-level height is 85 and one-third ft above sea level. how m
GalinKa [24]

The addison see to the horizon at 2 root 2mi.

We have given that,Kaylib’s eye-level height is 48 ft above sea level, and addison’s eye-level height is 85 and one-third ft above sea level.

We have to find the how much farther can addison see to the horizon

<h3>Which equation we get from the given condition?</h3>

d=\sqrt{\frac{3h}{2} }

Where, we have

d- the distance they can see in thousands

h- their eye-level height in feet

For Kaylib

d=\sqrt{\frac{3\times 48}{2} }\\\\d=\sqrt{{3(24)} }\\\\\\d=\sqrt{72}\\\\d=\sqrt{36\times 2}\\\\\\d=6\sqrt{2}....(1)

For Addison h=85(1/3)

d=\sqrt{\frac{3\times 85\frac{1}{3} }{2} }\\d\sqrt{\frac{256}{2} } \\d=\sqrt{128} \\d=8\sqrt{2} .....(2)

Subtracting both distances we get

8\sqrt{2}-6\sqrt{2}  =2\sqrt{2}

Therefore, the addison see to the horizon at 2 root 2mi.

To learn more about the eye level visit:

brainly.com/question/1392973

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Helga [31]

Answer:

Table B

Step-by-step explanation:

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Only Table B is a function since all input numbers are different.

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