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x + y = x, solve for X (pg. 9)
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| whiskers |
| quote: | Originally posted by moncster
7th: Algebra1
8th: Geometry
9th: Adv Algebra
10th: Precal HP
11th: Calc B/C AP
i r teh win |
sooo... that makes it 11th grade.
AFAIK, they don't teach ANYTHING about calculus in precalculus, especially delta epsilon definitions |
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| mezzir |
| quote: | Originally posted by moncster
7th: Algebra1
8th: Geometry
9th: Adv Algebra
10th: Precal HP
11th: Calc B/C AP
i r teh win |
eye c no pics!
you r teh stfu now! |
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| placebo |
I'm in College Prep Math.
The first day we learned how to add and subtract.
Seriously, I'll scan copies of my first test...its funny as . |
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| mezzir |
| quote: | | 8. Peter knocked up 6 girls in his gang. There are 27 girls in the gang. What percentage of the girls in the gang has Hector knocked up? |
trick question!!!
unless there are only two guys in the gang and 27 girls
but it'd be hard to push people around with only two guys and 27 girls to look after |
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| Fire999 |
whiskers: Since you got an A in your A Levels' Mathematics, I've got one question for you:
Explain and define in your own words on the mystery behind Fibonnaci Sequence. Decipher it in mathematical order if possible.
Thanks! I reckon you'll get an A on this too. |
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| DrUg_Tit0 |
| quote: | | Suppose 0.999 ... = x. Then 10 x = 9.999 ... = 9 + x, so x = 1. |
Yes, this is a pretty simple solution that shows 0.999... indeed is 1. Actually, the 0.999... exists only because of the inherent fault of the decimal number system (or all number systems, for that matter) which is it's unability to show all fractions as a finite sequence of numbers. As someone mentioned earlier, x=0.999... is a result of x=(1/9)*9 numerical equation. For all senses and purposes, we can clearly see that (1/9)*9 is infact 1, because the 9's cancel each other out. But if we attempt to solve the problem step by step in a decimal number system, we get a result that can't be displayed as a finite sequence of numbers, a sort of glitch in the matrix, if you will. The best thing it can done is to try to get as close to the actual result as it can, or in other words, infinite sequence of 9's.
But what happens if we look at that equation from a tertiary base rather than from a decimal one? In that case (1/9)*9 (10) = (1/100)*100 (3). So now with a tertiary number system, we get a result with a finite number of decimals. From that point on, we can easily conclude that (1/100)*100 (3) = 0.01*100 (3) = 1 (3) = 1 (10). So there, a simple indirect proof that an infinitely small amount is infact zero. |
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| whiskers |
| quote: | Originally posted by Fire999
whiskers: Since you got an A in your A Levels' Mathematics, I've got one question for you:
Explain and define in your own words on the mystery behind Fibonnaci Sequence. Decipher it in mathematical order if possible.
Thanks! I reckon you'll get an A on this too. |
i appreciate the sarcasm, but i never said i got an A in math, so i win :p
what is the mystery, anyway, what's to decipher? a(1)=1, a(2)=1, a(n)=a(n-1)+a(n-2), that's the sequence. |
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| jdjd |
| quote: | Originally posted by whiskers
i appreciate the sarcasm, but i never said i got an A in math, so i win :p
what is the mystery, anyway, what's to decipher? a(1)=1, a(2)=1, a(n)=a(n-1)+a(n-2), that's the sequence. |
derive the formula for the nth term and i will give you a cookie |
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| whiskers |
| quote: | Originally posted by jdjd
derive the formula for the nth term and i will give you a cookie |
i don't think it's possible to do so without recursion |
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| jdjd |
| quote: | Originally posted by whiskers
i don't think it's possible to do so without recursion |
ya its something like a-sub-n = (1 + sqrt(3))^(n+1)/2 + (1 - sqrt(3))^n/2 i dunno thats probably way off but you get the idea.. it was mentioned in our Classical alg 1, but we didnt derive it... |
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| trite |
| Here it is 1/sqrt(5)*((((1+sqrt(5))/2)^x)-(((1-sqrt(5))/2)^x)) |
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| itsTrueSonic |
that's it .. i have had it .. what i am going to do is i am going to take the equation
x + y = x
and take the antiderivative of that, then apply simpson's rule and l'hopital's rule to the formula, then use the sum of formations and apply it to the simple bell curve, and try to find the -p and +p values of the function, while trying to find the median of the curve, and then taking the slope of the curve and the slope function of the curve will only give us half the answer. then we apply the derivative of the slope curve and then apply the law of cosines to the formula, and evaluate it. after than, take the sine and tangent of the formula, add it together, and we will create a whole new f'n sound for trance!!!
this is i believe the whole point to the formula. the new sound for trance.
woo hoooooooooooooooooooooooo |
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