Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

do you have a scienrific calculator?
|dw:1347856598297:dw|
Type it on calculator

Not the answer you are looking for?

Search for more explanations.

Ask your own question

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
1.20?
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

Not the answer you are looking for?

Search for more explanations.

Ask your own question