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.

PreCalc & Trig Help, Picture Below!

I got my questions answered at in under 10 minutes. Go to now for free help!
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


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

1 Attachment
Using pythagoras' theorem u can find the length of the hypotenuse. a^2+b^2=c^2 Here a=2, b=3, c=? |dw:1362407511223:dw||dw:1362407540894:dw| Does that help? @Atkinsoha
I did all of that.. I'm stuck on maybe how to solve it, since I can't seem to come up with any of the answers that are given.

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

What is the value of hypotenuse?
(√13)? I hope..
yup right:)
nw what is \(\cos\theta\)
yup:) what is \(\cos^2\theta\)?
That's were I'm lost apparently. 3^2 = 9 and the (√13)^2 = 13?
WAIT. Is it A? 5/13? because then it would be 2(9/13)-1 which would be (18/13)-1 and one would be 13/13 so it would be 5/13? A?
HOLY BEJEBOUS! Thank you so much! I guess I just needed someone to break it down!
Can you help me with this one? I'm sure it's really easy, but I've been gone from school.. so haven't learned it :P
1 Attachment
let us say arcsinx=theta nd arccosx=beta so sin theta=x nd cos beta =x sin(arcsinx+arccosx)=sin(theta+beta) =sin theta*cos beta+sin beta * cos theta
cos theta=sqrt(1-sin^2theta) sin beta=sqrt(1-cos^2 beta) plug in the values and u will get the algebraic expression:)
Yeahh.. I don't understand at all. haha.
Really sorry. Hav to go nw. Please post this as a new ques. Hope some will sre help u.:) Really sorry

Not the answer you are looking for?

Search for more explanations.

Ask your own question