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Mar 23 2018 12:46pm
2+2-1=???
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Mar 23 2018 07:57pm
Quote (BERRISWEET @ 24 Mar 2018 04:46)
2+2-1=???


221

sen F0rum ch3ese, ty. :P
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Apr 18 2018 03:36pm
Don't need it solved but perhaps it's a fun exercise for someone.

f(x) = x^3 - 11x^2 - 25x - 13
y = p*x + q
Line Y is the tangent of f(x) at A(a,f(a)) and intersects with f(x) at B (13,0). What is the sum of p+q?
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Jun 5 2018 01:56pm
Quote (Forg0tten @ 18 Apr 2018 23:36)
Don't need it solved but perhaps it's a fun exercise for someone.

f(x) = x^3 - 11x^2 - 25x - 13
y = p*x + q
Line Y is the tangent of f(x) at A(a,f(a)) and intersects with f(x) at B (13,0). What is the sum of p+q?


So let's see, we have 13p + q = 0, so p + q = - 12p
Let's find p. Since y is the equation associated with the tangent of f at (a,f(a)), we have that p = f'(a), so all we need is to find a.
f(a) + (13-a)f'(a) = 0 <=> a^3 - 11a^2 - 25a - 13 + (13 - a)(3a^2 - 22a -25) = 0
We already know that 13 is one of the roots of the third-degree polynomial, and it is possible either to proceed by identification (boring) or to simply guess that -1 is also a root, so a^3 - 11a^2 - 25a - 13 + (13 - a)(3a^2 - 22a -25) = -2(a - 13)(a + 1)(a - c) with -2*13c = -13(25 + 1) = -13*2*13 -> c = 13
So a^3 - 11a^2 - 25a - 13 + (13 - a)(3a^2 - 22a -25) = -2(a+1)(a - 13)^2
The only possible solutions are either -1 or 13.
Now it depends if "intersect" means simply share a point or cut across. I think the latter is what is meant, but anyway : In both cases -1 works and p = f'(-1) = 3 + 22 - 25 = 0 so p + q = 0. This is the only solution for the second case since at 13 f is convex (f''(13) = 6*13 - 22 = 78 - 22 = 56 > 0 so it does not "cut across" but stays under the curve
In the first case, there is also p = f'(13) = 3 * 169 -22 * 13 - 25 = 507 - 286 - 25 = 507 - 306 - 5 = 201 - 5 = 196 so p + q = - 12 * 196 = - 2280 - 72 = -2352

Geez, and this was supposed to be fun? I was already bored to tears and now this is the final nail in the coffin lmao




Oh and also,

Quote (jordando101 @ 10 Mar 2018 07:40)
So i just found this section of JSP, been on JSP for years. Also been a high school math teacher for years (undergrad in math, masters in computer science).

Willing to help and give good, detailed answers for fg!

Just shoot me a pm :)


Wait, you're asking for fg when you're a high school teacher (i.e. failed maths major) and all you can solve are these boring exercises? Haha, are you Jewish?



This post was edited by Hanako on Jun 5 2018 02:02pm
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Jun 6 2018 01:56pm
Your answer is incorrect, I can't really be arsed where it went in the wrong though. It's been so very long and I would have to relearn most of the things necessary to solve it, but I remember the answer :p
Would you like to try again before I spoil?
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Jun 6 2018 05:27pm
positive * negative = shako
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Jun 7 2018 03:04am
Please show the recursion pattern for the following integral:
Integral of dx/(x^2 + 1)^n, where for n = 1 it is equal to tan^-1(x) (arctan x).
It's a funny exercise to play with, without any help from googel ofc.
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Jun 7 2018 07:22am
Quote (Forg0tten @ 6 Jun 2018 21:56)
Your answer is incorrect, I can't really be arsed where it went in the wrong though. It's been so very long and I would have to relearn most of the things necessary to solve it, but I remember the answer :p
Would you like to try again before I spoil?


Your existence is incorrect, I can't really be arsed to find where it went in the wrong though (probably right at birth, even though you're now going to pretend you were trolling)
Now listen here, I'm always right, premed brainlet. As I said it depends on your definition of "intersect" for the green line but both of the lines I found definitely share the point B with the original polynomial curve and are tangent at a point (a,f(a)), which you did not bother to define so I could pick any a.
(and assuming Y is the plot line associated to y, which you did not define either)
(and assuming the tangent of f(x) at A is the tangent of f at A, else it is meaningless)
According to variations they are the only ones with these properties



Quote (Forg0tten @ 6 Jun 2018 21:56)
It's been so very long and I would have to relearn most of the things necessary to solve it


and you're 25? damn, med students really are bainlets. The meme is actually true. I hope you fail because having doctors this slow being surgeons and shit is really scary.
There is nothing to know to solve this problem (the only prerequisites are extreme boredom and patience to try and understand your ill-stated problem which was supposed to be entertaining)


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Jun 7 2018 08:01am
Quote (pzold @ 7 Jun 2018 11:04)
Please show the recursion pattern for the following integral:
Integral of dx/(x^2 + 1)^n, where for n = 1 it is equal to tan^-1(x) (arctan x).
It's a funny exercise to play with, without any help from googel ofc.


boring as fuck

for some reasons the image is rotated, but right click -> open image in a new tab does the trick

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Jun 7 2018 01:16pm
Here's a little one :

Suppose you have a square garden of side L, and you decide to plant 2 trees at random (meaning X-coordinate and Y-coordinate are both evenly chosen between 0 and L).

What is the average distance between the trees ?
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