| By Kingdvl (Kingdvl) on Sunday, October 05, 2003 - 04:20 pm: Edit |
I took the 2c today and had some questions.
1) The perimeter of a triangle is 55. Two legs are 15, 20. Find the angle across from the shortest leg.
I tried using the law of cosines but kept getting something undefined as I got something like 1.4=cos(x)
2) A population increases 10 percent a day. At the end of 100 days, how much bigger is the population than at the first day. I tried y=pe^Rt, but that didn't work. Not sure how to do it.
Thanks
| By Jason817 (Jason817) on Sunday, October 05, 2003 - 04:34 pm: Edit |
1. 65.3 degrees. I probably did it wrong. Can someone confirm it (i dont want to post incorrect work).
2. dunno
| By Thelazyone (Thelazyone) on Sunday, October 05, 2003 - 04:46 pm: Edit |
1. 44.0
2. 13780.6 1.1^100
| By Thelazyone (Thelazyone) on Sunday, October 05, 2003 - 04:49 pm: Edit |
For 1:
225 = 800 - 800cosa
800 cosa = 575
cosa=23/32
arccos(23/32)=44.0486
| By Geniusash (Geniusash) on Sunday, October 05, 2003 - 06:36 pm: Edit |
(1.1^100)-1*100=percent increase
| By Mo222 (Mo222) on Sunday, October 05, 2003 - 07:53 pm: Edit |
2)
p = (original amt)(1+%increase)^t
original amt = x, at day 0
p =x at day 0
p = x(1+.10)^100 , at day 100
p = x1.1^100
THe question is a little confusing, do they want to know how many TIMES bigger or the DIFFERENCE?
anyway, the population is 13781 (1.1^100) TIMES bigger
| By Kingdvl (Kingdvl) on Sunday, October 05, 2003 - 08:11 pm: Edit |
Doh! Figured out what I did wrong on the first question. I wrote down all the useful equations into my calculator beforehand, and wrote c^2=a^2+b^2-abcosC instead of -2abCOsC.
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