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| 27 Feb 2012 01:57 AM |
Your tasks: 1. Please. Tell me, if x+y = a , and x*y = b ,
then what would x and y be if
- a=6.7 - b=7.6
:O??????
2. what is pi in base 3.8?
3. what is the number in the middle of the 207th row of pascal's triangle? assuming:
row 1: 1 row 2: 1 1 |
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| 27 Feb 2012 02:09 AM |
Easy!
1. x is x and y is y. 2. It is pi in base 3.8. 3. It is the number in the middle of the 207th row of pascal's triangle, assuming:
row 1: 1 row 2: 1 1
That was easy!
btw, I don't even know what pascal's triangle is.. |
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| 27 Feb 2012 02:16 AM |
If you didn't know what it was, you wouldn't know it even had rows. LIAR! |
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| 27 Feb 2012 02:19 AM |
(By the way... with a calculator [I see decimals *twitch*], the first one should be super easy... Do you know how to substitute variables?) |
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| 27 Feb 2012 02:59 AM |
I have all three answers.
Once you solve a problem, put a new one up :D |
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| 27 Feb 2012 03:20 AM |
hint for #2: pssst! the 501st row's center number is 116744315788277682920934734762176619659230081180311446124100284957811112673608473715666417775521605376810865902709989580160037468226393900042796872256!!!! |
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| 27 Feb 2012 03:24 AM |
You can use my function to add 2 positive integer strings: function add(a,b) local A,B=a:reverse(),b:reverse() local m=math.min(#A,#B) local carry=0 local ans="" for d=1,m do local r=tonumber(A:sub(d,d))+tonumber(B:sub(d,d))+carry if r>=10 then carry=1 r=r-10 else carry=0 end ans=r..ans end local nA="" if #A==#B and carry==1 then nA="1" carry=0 elseif #A>#B then nA=A:sub(m+1) elseif #B>#A then nA=B:sub(m+1) end if carry==1 then local n9s=("x"..nA):match("x9*"):sub(2) local a0s=nA:sub(#n9s+1) nA=string.rep("0",#n9s)..(1+(tonumber(a0s:sub(1,1)) or 0))..a0s:sub(2) carry=0 end return nA:reverse()..ans end |
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stravant
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| Joined: 22 Oct 2007 |
| Total Posts: 2893 |
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| 27 Feb 2012 07:57 AM |
"assuming: row 1: 1 row 2: 1 1"
Easy, by extrapolating the patter that you gave us the answer is clearly just one, since the whole triangle is just ones. |
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eletrowiz
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| Joined: 08 Dec 2008 |
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| 27 Feb 2012 08:39 AM |
1) Learned this recently in math v.v x+y = 6.7 x*y = 7.6 Simultaneous Equations so x = 6.7 - y (6.7-y)*y = 7.6 -y^2+6.7y=7.6 -y^2+6.7y-7.6=0 (Now for the calculator to do the quadratic equation...) x = 1.45, 5.25 (rounded to 3 figures) y= (6.7-1.45),(6.7-5.25) y= 5.25, 1.45
x = 1.45, 5.25 y = 5.25, 1.45
:D Thank goodness for calculators. |
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stravant
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| Joined: 22 Oct 2007 |
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| 27 Feb 2012 08:41 AM |
"(Now for the calculator to do the quadratic equation...)"
You don't know how to do it by hand? That's not good: x = (-b ± sqrt(b^2 -4ac)) / 2a
Do they not teach you how to do that and prove why it works? |
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eletrowiz
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| 27 Feb 2012 08:44 AM |
| I nknow that way, but its hassle do to it as you have to go through all the formula and solve it, so I was lazy and typed it up into my calculator. Luckily they strangely give us all the formulas in our math IGCSEs so there is really no need to memorize them until A levels. |
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| 27 Feb 2012 09:45 AM |
@stravant
I clearly said it was Pascal's triangle, so for those who don't know,
row 3: 1 2 1
@ electrowiz
Das is gut! Have a problem to exchange? |
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| 27 Feb 2012 09:53 AM |
I got
x=5.25328663106742 y=1.44671336893258
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Oysi
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sdfgw
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| 27 Feb 2012 11:51 AM |
"Do they not teach you how to do that and prove why it works?"
not 'till college
"but that's the opposite way around, anyway. Which proves the the formula works"
NO YOU CAN'T SAY THAT STRAVANT IS HERE
ok I haven't read thread but ok
1) x + y = 6.7 x = 6.7 - y xy = 7.6 6.7y - y^2 = 7.6 y^2 - 6.7y + 7.6 = 0
i'm in a completing square kind of mood
(y-3.35)^2 = y^2 - 6.7y + 11.2225 (y-3.35)^2 - 11.2225 + 7.6 = 0 (y-3.35)^2 = 3.6225 y - 3.35 = plussyminus1.903 y = plussyminus1.903 + 3.35 y = 1.447, y = 5.253
x = 6.7 - y x = 5.253, x = 1.447
well, fancy that.
1) I already did 1 so no
2) I can see how fun could be poked at those 1s but ok
C = ... god I know this
um
C = n!/(r!(n-r)!)
yes
so
C = 206!/(103!(206-103)!) = 206!/(103!*103!) = 206!/103!^2
i'm going to stop here but an online calculator told me that C = 5.71029445e+60 also I probably got the formula wrong so yeah |
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SN0X
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| 27 Feb 2012 02:04 PM |
math.pi(3.8) -_-
Don't go blurting into the "Scripters" thread if you can't do that. |
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Oysi
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| 27 Feb 2012 05:00 PM |
Alright; #2 is solved! anyone got a new question? |
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| 27 Feb 2012 05:01 PM |
| #1*, whoops. Post your own problem when you solve one. |
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Varp
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| 27 Feb 2012 05:17 PM |
"i'm in a completing square kind of mood"
Meh. I figured out the quadratic formula really quickly because I avoided completing the square. Doing so is obscuring the true nature of the proof. The proof I like is:
ax^2 + bx + c = 0
We know that, if b is 0, then:
ax^2 - c = 0 x^2 - c/a = 0 (Divide by a) x^2 = c/a x = ±sqrt(c/a)
That's not so hard. What if b ISN'T 0? Well, we can fix that. We can translate the whole problem. We want to find some n and d such that:
u = x + n au^2 + d = ax^2 + bx + c
We can do this algebraically, but we can do better than that. So, the the variable b happens to be the slope at the y-intercept. Since the vertex, at -b/2a, has a slope of 0 (this can be proven fairly easily), what if we set:
u = x + b/2a u - b/2a = x
Basically, we're finding a parabola whose vertex is at u = 0, which makes the problem really easy to solve. From this point, we need to find what the variable d is. This is just the value of the function at the vertex, then negating it*:
-d = a(-b/2a)^2 + b(-b/2a) + c -d = ab^2/4a^2 - b^2/2a + c -d = b^2/4a - b^2/2a + c -d = c - b^2/4a -d = -(b^2 + 4ac)/4a d = (b^2 - 4ac)/4a
Then, we know that:
au^2 + d = 0 u = ±sqrt(((b^2 - 4ac)/4a)/a) u = ±sqrt((b^2 - 4ac)/4a^2) u = ±sqrt(b^2 - 4ac)/2a
Then, substituting in x for u:
x - b/2a = ±sqrt(b^2 - 4ac)/2a x = b/2a±sqrt(b^2 - 4ac)/2a x = (b±sqrt(b^2 - 4ac))/2a
*An alternate method of finding d:
Substituting:
au^2 - d = a(u - b/2a)^2 + b(u - b/2a) + c au^2 - d = au^2 - bu + b^2/4a + bu - b^2/2a + c au^2 - d = au^2 + b^2/4a - b^2/2a + c -d = b^2/4a - 2b^2/4a + c -d = -b^2/4a + c d = b^2/4a - c d = (b^2-4ac)/4a
(Also, a similar transformation can be used to find a cubic or quartic formula, but those are a lot more involved than the quadratic formula) |
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| 27 Feb 2012 05:30 PM |
| I've always wondered about the quadratic forumula. Why 4? It seems that a constant in a formula like that is rather awkward. |
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Varp
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| Joined: 18 Nov 2009 |
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| 27 Feb 2012 05:36 PM |
"I've always wondered about the quadratic forumula. Why 4? It seems that a constant in a formula like that is rather awkward."
The answer to that depends on how you derive the quadratic formula. I mean, at least in what I did, it comes from the fact that the vertex is at x = -b/2a (the 2 there comes from the fact that the 2nd derivative of a quadratic is 2!a = 2a) and that this gets squared, giving b^2/4a^2. So, it's there because:
4 = 2!^2
But there's no real answer to that, since the quadratic formula isn't part of a series of formulas (the solution to ax + b = 0 isn't really related, and neither is the solution to ax^3 + bx^2 + cx + d = 0 or a quartic. It's REALLY REALLY not related to the solution to a quintic, since that one can't be expressed with radicals) |
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| 27 Feb 2012 05:40 PM |
| Trololol. Anyone have answers? |
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pauljkl
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| 27 Feb 2012 05:47 PM |
| Lrn2nCr on your calculator for the pascals triangle |
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| 27 Feb 2012 05:51 PM |
| Calculator's can't handle it. The math is too damn high. Use my add(a,b) function if you like, but that's boring. |
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