What is the period of the function f(x)=sin(x/3) ?

Mathematics

1 answer:

vitfil [10]3 years ago

6 0

<h3>Answer: 6pi radians</h3>

(this is equivalent to 1080 degrees)

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Explanation:

f(x) = sin(x/3)

is the same as

f(x) = 1*sin( (1/3)(x-0) )+0

and that is in the form

f(x) = A*sin( B(x-C) )+D

The letters A,B,C,D are explained below

A = helps find the amplitude

B = 2pi/T, where T is the period

C = determines phase shift (aka left/right shifting)

D = determines vertical shift = midline

All we care about is the value of B as that is the only thing that is connected to the period T

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Compare f(x) = 1*sin( (1/3)(x-0) )+0 with f(x) = A*sin( B(x-C) )+D and we see that B = 1/3, so,

B = 2pi/T

1/3 = 2pi/T

1*T = 3*2pi ... cross multiply

T = 6pi

The period is 6pi radians. This is equivalent to 1080 degrees. To convert from radians to degrees, you multiply by (180/pi).

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A student repeatedly measures the mass of an object using a mechanical balance and gets the following values: 560 g, 562 g, 556

MrRissso [65]

Answer: 2.76 g

Step-by-step explanation:

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Then, $\overline{x}=\dfrac{\sum_{i=1}^{10} x_i}{n}\\\\\Rightarrow\ \overline{x}=\dfrac{560+562+556+558+560+556+559+561+565+563}{10}\\\\\Rightarrow\ \overline{x}=\dfrac{5600}{10}=560$

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Then, $\sigma=\sqrt{\dfrac{76}{10}}=\sqrt{7.6}=2.76$

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4 years ago

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