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**Do the Math
5852 cr points
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26 / M / behind you...
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Posted 2/22/07
^no math here
Posted 2/22/07
^ there's an english here. without math, you'll never learn you insect. numbers are math, these numbers of pages are math, that age of urs is math, that crunchyroll points of urs are math. you will never know that i am the one who rule XD


ok here's mine, its just a distance problems so its not that very difficult to answer:

Mr. Honda and Mr. Toyota arrange to meet on the highway connection their hometowns. Mr. Honda drives at 45 kph and Mr. Toyota at 35 kph. They leave their houses which are 120 km apart at the same time. In how many hours will they meet?

:w00t: very very easy!!!
1087 cr points
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30 / M / Phillipines
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Posted 2/22/07
Find the solution wahehe
1. [x + y sin (y/x)]dx - x sin (y/x)dy = 0

clue use homogenous diff equation... hehe
Posted 2/22/07

ichi_ni240 wrote:

ok here's mine, its just a distance problems so its not that very difficult to answer:

Mr. Honda and Mr. Toyota arrange to meet on the highway connection their hometowns. Mr. Honda drives at 45 kph and Mr. Toyota at 35 kph. They leave their houses which are 120 km apart at the same time. In how many hours will they meet?

:w00t: very very easy!!!


That would be 1.5 hours, seeing as they are travelling at 80kmph in total and need to travel 120km in total

And;
[x + y sin (y/x)]dx - x sin (y/x)dy = 0
having done a page of working i have;
(x/y) = 2sin(x)
or
y = x/(2 sin(x))
PM me to say whether I'm correct

which is smaller;
-2 or +1?
1183 cr points
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30 / M / livin on the sun...
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Posted 2/22/07
+1

cosin 75 degrees
show me da solution very easy no calcu pls XD
1087 cr points
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30 / M / Phillipines
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Posted 2/22/07


RazieLSuphlatuS wrote:

+1

cosin 75 degrees
show me da solution very easy no calcu pls XD


0.26?



henz_lan wrote:


ichi_ni240 wrote:

ok here's mine, its just a distance problems so its not that very difficult to answer:

Mr. Honda and Mr. Toyota arrange to meet on the highway connection their hometowns. Mr. Honda drives at 45 kph and Mr. Toyota at 35 kph. They leave their houses which are 120 km apart at the same time. In how many hours will they meet?

:w00t: very very easy!!!


That would be 1.5 hours, seeing as they are travelling at 80kmph in total and need to travel 120km in total

And;
[x + y sin (y/x)]dx - x sin (y/x)dy = 0
having done a page of working i have;
(x/y) = 2sin(x)
or
y = x/(2 sin(x))
PM me to say whether I'm correct

which is smaller;
-2 or +1?



Wrong!!

nid to use this
y=vx
dy= xdx+xdv

use integration and differential after some solution you shoul get this answer
ln[x] + cos (x/y)=lnc

ill pm you the solution

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20 / M / Cali
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Posted 2/25/07
i think it's 1.1

here's mine 9.4 divide 6.9 =
Posted 2/25/07
oh god cant believe theres a maths thread.
seems i cant get away from all the maths whizzes flying around these days
424 cr points
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27 / F
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Posted 2/25/07
thank god its an easy one
1.362 (3 deci ap standard)

----> 10!=?
Posted 2/25/07
3628800



Find the particular solution which satisfies the given boundary condition in each of the following differential equation
a) x y (1 + x^2) dy/dx = 2 + y^2 ; y(1) = 0
b) dy/dx + y/x = cos 2x ; y(PI) = 0
d) dy/dx - y cotx = tan ^2 x ; y(PI/4) = 0


this solution would be so long...
Posted 2/28/07
what, can anyone answer mine
Posted 2/28/07

ichi_ni240 wrote:

what, can anyone answer mine


What r u a super nerd

Posted 2/28/07
a) x y (1 + x^2) dy/dx = 2 + y^2 ; y(1) = 0
[y/(2 + y^2)] dy = dx/[x(1 + x^2)]
Integral [y/(2 + y^2)] dy = Integral dx/[x(1 + x^2)]
(1/2) Ln (2 + y^2) = Integral dx/[x(1 + x^2)]
1/[x(1 + x^2)] = 1/x - x/(1 + x^2)
(1/2) Ln (2 + y^2) = Integral dx/[x(1 + x^2)] = Ln x - (1/2) Ln (1 + x^2) + c
(1/2) Ln (2 + y^2) = (1/2) [2 Ln x - Ln (1 + x^2)] + c
Ln (2 + y^2) = 2 Ln x - Ln (1 + x^2) + C
Ln (2 + y^2) = Ln [x^2 /(1 + x^2)] + C
e^[Ln (2 + y^2)] = e^[Ln [x^2 /(1 + x^2)] + C]
2 + y^2 = A x^2 /(1 + x^2)
y(1) = 0
2 + 0 = A 1/(1 + 1)
2 = A/2
A = 4
y^2 = 4x^2/(1 + x^2) - 2
y = sqrt(4x^2/(1 + x^2) - 2)


b) dy/dx + y/x = cos 2x ; y(PI) = 0
x dy/dx + y = x cos 2x
(d/dx) (xy) = x cos 2x
xy = integral (x cos 2x) dx
xy = (1/2) x sin (2x) - (1/2) Integral (sin 2x) dx
xy = (1/2) x sin 2x + (1/4) cos 2x + c
y(pi) = 0
0 = (1/2) pi * 0 + (1/4) (1) + c
c = -1/4
y = (1/2) sin 2x + (1/(4x)) cos 2x - 1/(4x)

d) dy/dx - y cotx = tan ^2 x ; y(PI/4) = 0
I think that - y cot x should be + y cot x
sin x dy/dx + y cos x = (sin x)^3 / (cos x)^2
(d/dx) (y sin x) = (1 - (cos x)^2) sin x / (cos x)^2
y sin x = Integral (sin x/ (cos x)^2 - sin x) dx
y sin x = 1/(cos x) + cos x + c
y(pi/4) = 0
0 = 1/(sqrt(2)/2) + sqrt(2)/2 + c
0 = sqrt(2) + sqrt(2)/2 + c
0 = (3/2) sqrt(2) + c
c = -(3/2) sqrt(2)
y = 1/( sin x cos x) + cot x - (3/2) sqrt(2) /sin x
y = csc x sec x + cot x - (3/2) sqrt(2) csc x

<_< nobody won't answer!


9! = ?
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Posted 2/28/07

ichi_ni240 wrote:


<_< nobody won't answer!


9! = ?


9 x 8 x 7 x 6 x 5 x 4 3 x 2


i needa start my Maths for Comp Science shit....
it's too hard !
424 cr points
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27 / F
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Posted 2/28/07
calculus is my worst subject right now

9!=362880

which step is wrong?
a=b plus a
2a=a+b minus 2 b
2a-2b=a-b factor
2(a-b)=(a-b) simplify
2=1
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