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\(16\) over the hypotenuse of the triangle with one side \(16\) and the other \(30\)

i.e. \[\frac{16}{\sqrt{16^2+30^2}}\] as a start

@PrincessHush nice picture!

Thanks! (: @satellite73

Doesn't theta in this case have one side on the x-axis?
|dw:1444266733897:dw|

cos(theta) = adjacent side over hypotenuse

g(x) = 4 sin(4x) − 3, what is the value of the y intercept?

cos(Θ) = x/r = x/sqrt(x^2 + y^2) = 16/sqrt[16^2 + (−30)^2] = 16/34 = 8/17.

so 8/17?

princess is correct, it would be 8/17

okay, thanks

rose, can you clarify the earlier comments for me? i don't quite understand what they were saying

You'll need to learn these. See attachment.

i haven't seen those since geometry 4 years ago 0.0 i forgot all about them

What is the 32nd term of the arithmetic sequence where a1 = −32 and a9 = −120?