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oobleck - ouch
Answers (6)
oops - go with Reiny. Do you see my mistake?
if tanx = 8/15 in QIII then sinx = -8/√161 cosx = -15/√161 so tanx/2 = (1-cosx)/sinx = (1 + 8/√161)/(-15/√161) = -(8+√161)/15 makes sense, since x/2 will be in QII where tangent is negative
the last is -1/7 But I'm sure you saw my typo ...
naturally, (60/16) * 12
Oops. My bad. Forgot to note that it's Cl2, not just Cl.
read twice before posting! Thanks, Damon.
