solve the triangle
given 3 sides, we can use the law of cosines to find an angle
\[\frac{c^2-a^2-b^2}{-2ab}=cos(C)\]
wats the answer
can u help with this one
lol .... the answer is what you get when you fill in the s bs and cs
the parabola one can be tricky; you have to go with the geometric definition
takin a quiz i really need the answers lol
x^2 + 4x -2y -2 = 0 x^2 +4x = 2y+2 x^2 +4x +4 = 2y+2+4 (x+2)^2 = 2(y+3) that looks better; thats in the form of x^2=4ay
if you simply want someone to give you answers to a quiz; thats against the anti cheating policy of openstudy
i just really need to pass this class bro
but if you need help finding the answers, I can step you thru them :) but working for free to get you a good grade just aint my style ... But I aint the only one here, so maybe we can get someone to produce answers :)
thanks
(x+2)^2 = 2(y+3) the vertex is defined by what is modifying the x and ^ ^ y parts of this equation ^ ^ just negate them can you give it a shot?
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