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3 years ago

how to find the period, amplitude, and vertical shift of a sinusoidal graph. please explain in the simplest way possible, tyyyy

Mathematics

1 answer:

Kay [80]3 years ago

4 0

Answer:

Step-by-step explanation:

a) period is the length of one cycle ( one full squiggle ) If you are lucky and the graph is on exact numbers.. at a high point and to the next high point.. then just take that horizontal distance, that's one period also if it's not you could find the period, T, by using T = 2pi / k where k is a number that represents how fast , or how many times the wave cycles in one second or some other amount of time.

b) Amplitude is a number that goes right before the sinusoid, which is multiplied by it because, sin , cos, are at max 1 or at min -1

c) vertical shift is what is added , for upwards shift... before or after the sinusoid ie sin , cos, ect ect.

does this help?

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3 years ago

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Suppose N has a geometric distribution with parameter p. Derive a closed-form expression for E(N | N <= k), k = 1,2,... Check

vfiekz [6]

Answer:

P(X= k) = (1-p)^k-1.p

Step-by-step explanation:

Given that the number of trials is

N < = k, the geometric distribution gives the probability that there are k-1 trials that result in failure(F) before the success(S) at the kth trials.

Given p = success,

1 - p = failure

Hence the distribution is described as: Pr ( FFFF.....FS)

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Since N<=k

Pr (X =k) = p(1-p)^k-1, k= 1,2,...k

0, elsewhere

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Pr (Y= k) = p(1-p)^y......k= 0,1,2,3

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3 years ago

A triangular lot has sides of 215m, 185m, and 125m. Find the measures of the angles at its corners

Vlada [557]

Answer:

The measures of the angles at its corners are $59.1\°,35.4\°,85.5\°$

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

Find the measure of angle A

Applying the law of cosines

$185^{2}= 215^{2}+125^{2}-2(215)(125)cos(A)$

$2(215)(125)cos(A)= 215^{2}+125^{2}-185^{2}$

$cos(A)= [215^{2}+125^{2}-185^{2}]/(2(215)(125))$ $cos(A)=0.513953$

$A=arccos(0.513953)=59.1\°$

step 2

Find the measure of angle B

Applying the law of cosines

$125^{2}= 215^{2}+185^{2}-2(215)(185)cos(B)$

$2(215)(185)cos(B)= 215^{2}+185^{2}-125^{2}$

$cos(B)= [215^{2}+185^{2}-125^{2}]/(2(215)(185))$ $cos(B)=0.81489$

$B=arccos(0.81489)=35.4\°$

step 3

Find the measure of angle C

Applying the law of cosines

$215^{2}= 125^{2}+185^{2}-2(125)(185)cos(C)$

$2(125)(185)cos(C)= 125^{2}+185^{2}-215^{2}$

$cos(C)= [125^{2}+185^{2}-215^{2}]/(2(125)(185))$ $cos(C)=0.0784$

$C=arccos(0.0784)=85.5\°$

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3 years ago

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scZoUnD [109]

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