Recent content by blueparukia

I can get the answer through trigonometric substitution, the problem is it isn't in my textbook so therefore not part of the syllabus. I can integrate it fine by myself if I know the rule. All I've got is some basic substitution stuff (chain rule) and standard integrals, but I can't find a...

Not quite understanding. Since this is a definite integral, if I let x=20sin(theta), do I then sub in 0 and 20 as theta? There doesn't seem to be any trigonmetric substitution in the textbook? Is there anyway that formula could be arranged to fit with the short table of standard integrals?

Homework Statement [PLAIN]http://www4a.wolframalpha.com/Calculate/MSP/MSP184619c56eg84971107d00001e292h4340heh8g4?MSPStoreType=image/gif&s=35&w=135&h=53 It is a definite integral between x=0 and x=20. I forget how to integrate with a square root. Do you just change it to a power of a...

Wow, pretty specific example there. Thanks.

Homework Statement A man was murdered and inspectors found his body temperature to be 88 degrees at 4:30AM. Climate control in the room was set to 76 degrees. Temperature of his body was taken again 2 hours later and found to be 85.8 degrees. The standard body temperature for human...

PC, more powerful and whatnot. Aerospace engineers will most likely require high end CAD software that is unavailable for MAC and will require unbelievably high end hardware to run, as well as an extensive multi-monitor set up - which is just easier to achieve on Windows than OS/X.

Oops, got a bit lost there. So 60-θ = 43.853 So -θ = 43.853-60 = -16.146 θ = 16.146? :D Thanks a load, you've been unbelievably helpful.

Not really sure I know how to solve it like that. First of all, radius of satellite orbit = 6400+1600 = 8000km (sin 120)/8000 = (sin (C-θ))/6400 Would mean: sin (C-θ) = 0.693 And C-θ = sin^-1(0.693) = 43.853 But I can't see how I can get either C or θ from that, since it would...

Well 180-120 is equal to 60. That would mean the remaining two angles are 60 degrees combined. So C would be...60-θ?

No idea what that means. Does that mean I have to solve θ first? Sorry, I sound stupid, but I'm tired of this question and it's been a while since I've done trig.

Hmm, double checking my answer (40.5 degrees) is most likely not right since I substituted the wrong sides. The problem now is that the sine rule to my knowledge won't work with both side a and angle A missing. So I'm thinking use my original diagram to solve for side a then subsitute that into...

Yes I think I got it! Thank you so much, I have an answer :D Thank you very, very much. Hoping the answer is ~40 degrees.

OK, I can't figure this out. I'm stumped, I'm frustrated, I can't answer the question, it's been 4 days and the trig is just standing in the way of what I need to know. I am not skilled enough at trig to be able to work this out. The sides can't be equal on a right angle triangle, so if I...

Huh, common sense. It strikes again. Thanks a lot.

No idea, that's what I was trying to work out with sum of all angles = 180 degrees. And there's no other trig I can think of that will solve it without knowing theta or the other lengths.