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help help help....test
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help help help....test
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jezza
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help help help....test
«
on:
21/01/2006 23:16:26 »
ok, so i got a maths test on monday, and i know that it is maths and not science, but i got this question this question that keeps coming up in past exams, i think it is to do with logorithums...it says:
solve 5x (thats 5 to the power x)=100
any ideas, i know its really simple but dont know how to do it
thanks
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jezza
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Re: help help help....test
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Reply #1 on:
22/01/2006 13:13:39 »
ok, so ive been thinking and can anyone verify that it means:
log100 divided by log5 = 2.86........?
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daveshorts
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Physics, Experiments
Re: help help help....test
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Reply #2 on:
22/01/2006 13:51:44 »
jSounds about right, you can check it by feeding 5^2.86 and seeing what it comes out as..
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Solvay_1927
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Re: help help help....test
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Reply #3 on:
22/01/2006 21:46:24 »
Jezza,
I'm probably too late to answer now, I guess. (I really should get on line more often.)
But your answer is correct, it can be explained as follows:
5^x = 100
Take logs of both sides:
log(5^x) = log(100)
But log(a^b) = b.log(a), so this equation becomes:
x.log(5)=log(100)
So you can divide both sides by log(5) to give:
x=log(100)/log(5)
GOOD LUCK with your exam tomorrow!
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jezza
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Re: help help help....test
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Reply #4 on:
22/01/2006 23:03:59 »
thanks chaps
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simeonie
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Re: help help help....test
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Reply #5 on:
23/01/2006 21:29:45 »
ya lost me at "ok"
----------------------
MY NAME IS NOT REALLY SIMEONIE IT IS SIMON!!!
«
Last Edit: 27/06/2009 14:40:36 by BenV
»
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MY NAME IS NOT REALLY SIMEONIE IT IS SIMON!!!
jezza
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Re: help help help....test
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Reply #6 on:
24/01/2006 12:19:46 »
ok, the rest went fine, thanks
but now tommorow i got geometry, which i suck at and i found this question that says:
find all angles
a
such that 0<
a
<360(degrees) and cos
a
=0.37
any ideas what so ever, i dont have a clue?
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rosy
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Re: help help help....test
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Reply #7 on:
24/01/2006 12:39:00 »
Sketch a cossine curve (you chould be able to do that, if not learn it!)
Then find the angle your calculator gives you for cos-1(0.37). Use the symmetry of the graph to find the other angle. There'll be 2 solutions. If you were looking for cos 2a there'd be 4 solutions.
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jezza
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Re: help help help....test
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Reply #8 on:
24/01/2006 17:15:55 »
thanks rosy, i just figured it out, i presume this is tht same as what your saying?
cos-1 0.37 = 68.28 + 360 (as its a cosine) = 428.28, which can then be checked by cos -428.28 = 0.37
so the answer being 428.28?
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rosy
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Re: help help help....test
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Reply #9 on:
24/01/2006 20:44:17 »
Well... you're on the right track, 428 will be *a* solution. But it's not between 0 and 360, is it?
The solutions you want will be 68 and 360 - 68 = 292 (like I said before, sketch the curve).
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Solvay_1927
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Re: help help help....test
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Reply #10 on:
24/01/2006 22:39:12 »
I'm too late again, I see. (I really should get on here more frequently.)
The general rule (as rosy has used) is that cos(x) = cos(360-x).
Same applies for sines too.
Good luck (again) Jezza.
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rosy
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Re: help help help....test
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Reply #11 on:
24/01/2006 22:54:22 »
Um, actually for sines it's sin(x)=360 + sin (x)... and also sin(x)=180 - sin(x)
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Solvay_1927
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Re: help help help....test
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Reply #12 on:
25/01/2006 00:47:28 »
Oops - buggeration - must remember to engage brain when posting things on here - what a twit.
Yes, sorry rosy (and sorry jezza).
sin(x) = - sin(360-x) [ and sin(x)= - sin(-x) ]
or
sin(x) = sin(180-x) [ and sin(x) = sin(360+x) ]
cos(x) = cos(360-x) [ and also cos(x) = cos(-x) and cos(x) = cos(360+x) ]
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jezza
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Re: help help help....test
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Reply #13 on:
25/01/2006 11:26:15 »
naa, its not to late, the exams at 4, but im kinda confused now, whats that stuff mean that you put in the brackets, is that to check?
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DoctorBeaver
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Re: help help help....test
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Reply #14 on:
25/01/2006 12:54:24 »
I can't even sine my name [
]
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Solvay_1927
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Re: help help help....test
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Reply #15 on:
25/01/2006 13:48:16 »
Jezza, the other formulae are just in case you get a different type of question.
E.g. if sin(45) = 0.707, what angle gives sin(x) = -0.707 ?
All of the the following answers are correct:
x = -45 (but maybe you haven't got on to the concept of negative angles, yet?)
x = 315 (i.e. 360-45)
x = 225 (i.e. 180+45)
And if sin(45) = 0.707,
then so is: sin(405) = 0.707 (cos 405 = 360+45)
and also: sin(135) = 0.707 (as 135 = 180-45)
If none of this looks familiar, then maybe you've not covered it in your studies yet, so it won't come up in your exams. In which case, you might be better to just forget you saw it (in case it just confuses you or makes you panic).
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jezza
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Re: help help help....test
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Reply #16 on:
26/01/2006 11:38:27 »
thanks, that was helpful, i think i had studied it, was just being a bit dim...although it didnt really help, beacuse it didnt come up on the exam....grrrr
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jezza
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Re: help help help....test
«
Reply #17 on:
26/01/2006 14:03:42 »
ok people, tomorrows test is the last one, and its on physics, but being me there are a few things im not quite sure on, so as im revising ill be contuelly adding to this thread with things im not to sure on.
firstly.....pendulms
how do you work out the maximum acceleration of a pendulm, i know how to work out the max velocity, but im stumped witht the max accleration?
also if you know the frequency, and the amplitude, how can you get the weight of the pendulm from that
thanks
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sharkeyandgeorge
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Re: help help help....test
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Reply #18 on:
27/01/2006 11:28:51 »
try this site
...sorry, you cannot view external links. To see them, please
REGISTER
or
LOGIN
hope it helps
"Defender of the Sea"
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Naked Science Forum
Re: help help help....test
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Reply #18 on:
27/01/2006 11:28:51 »
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