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how would you find ^3square root of 38 if you know how I would put that on a calculator that's fine too

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do you have a scienrific calculator?
Type it on calculator

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Other answers:

okay please hold
i gots 3.36
just so everyone absolutely knows the problem looks like this |dw:1347856966969:dw|
Yes thats it
o kay now how would you find the value of the expression : if sin beta=0.45, find cos(pi/2-beta)
its the same
okay then i did it right, what bout if tan(pi/2-beta)=-5.32, find cot beta
tan(pi/2-beta)= cot beta = -5.32
okay cool then I think i got it so if sin (beta-pi/2)=0.73, to find cos-beta i would do cos(sin^-1-pi/2) right? or cos^-1(sin^-1-pi/2)
remember these identities: \(sin(\pi/2-\beta)=cos\beta \\ tan(\pi/2-\beta)=cot\beta \\ cos(\pi/2-\beta)=sin\beta \) when the ratio has angle \(\pi/2 - \beta\) , it becomes coratio.
so is it the same if its beta-pi/2 instead of pi/2-beta?
no, its only for pi/2-beta for beta-pi/2, u write it as -(pi/2-beta) and use sin(-x)=-sin x cos(-x)=cos x tan(-x)=-tan x
okay so to do if sin (beta-pi/2)=0.73, find cos-beta it would be cos(-pi/2-sin^-1(0.73)?
there is no need to do \(sin^{-1}\)here. \(sin(\beta-\pi/2)=sin(-(\pi/2-\beta))=-sin(\pi/2-\beta)=-cos\beta\)
so then how would i do the problem?
so cos beta =- sin (beta-pi/2)= -0.73
oh well IDK how i did it but I got that answer too. okay so last one of these is if cot(-beta)=7.89, find tan( beta-pi/2) would it kinda be like the previous one?
yes. use tan(beta-pi/2) = tan(-(pi/2-beta))=??
please hold
0.03? wait I think i did it wrong let me try again
um okay one last try
ok, from where are u getting those numbers ?? tan(beta-pi/2) = tan(-(pi/2-beta))=-tan(pi/2-beta)=-cot beta and cot(-beta)=-cot beta also so tan(beta-pi/2) = cot(-beta) =7.89 see whether u understand this ?
well idk what i put for beta i tried it as just 7.89 and cot(7.89) here i'll try it again
well i understand how it works but idk how to put it into my calc to where i get that answer

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