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cwrw238Best ResponseYou've already chosen the best response.2
can you continue from here?
 9 months ago

dontothabonbonBest ResponseYou've already chosen the best response.0
Thanks. I'm not sure what to do from here.
 9 months ago

cwrw238Best ResponseYou've already chosen the best response.2
tan x = opposite / adjacent = 1 / sqrt15 sec x = hypotenuse / adjacent = ?
 9 months ago

jdoe0001Best ResponseYou've already chosen the best response.1
http://faculty.wlc.edu/buelow/PRC/T2_1.jpg
 9 months ago

jdoe0001Best ResponseYou've already chosen the best response.1
dw:1372535888225:dw
 9 months ago

dontothabonbonBest ResponseYou've already chosen the best response.0
Alright now I see. Thanks to both of you.
 9 months ago

dontothabonbonBest ResponseYou've already chosen the best response.0
Would the hypotenuse become a negative when calculating sec? Or would it just be: sec x = 4/sqrt15?
 9 months ago

dontothabonbonBest ResponseYou've already chosen the best response.0
Or better yet, why did the 1 become negative when calculating the tangent?
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.0
Because your function is in the second quadrant where sine is positive and cosine is negative.
 9 months ago

cwrw238Best ResponseYou've already chosen the best response.2
the whole ratio became negative tangent : opposite = 1 , adjacent =  sqrt15 tan = 1 / sqrt15 = 1/sqrt15
 9 months ago

JhannybeanBest ResponseYou've already chosen the best response.0
\[\large \left(\frac{1}{\sqrt{15}}\right)\]
 9 months ago

dontothabonbonBest ResponseYou've already chosen the best response.0
So, sec = 4/sqrt15?
 9 months ago

dontothabonbonBest ResponseYou've already chosen the best response.0
ahh okay I see now thanks.
 9 months ago

cwrw238Best ResponseYou've already chosen the best response.2
x values to the left of yaxis are negative y values below x axis are also negative
 9 months ago
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