4U Revising Game (1 Viewer)

[Rhanoct]

start from this x^m

rule x^m/ x^m = 1

but x^m / x^m = x^m-m = x^0

therefore x^0 =1 xD

rofl

~i love you ron

 

[Kirjava]

offline? couldnt do it conics? big swing, no ding.

Spoiler

MP.NP = |(b/a)x0 - y0| / sqrt[(b/a)^2 + 1] . |-(b/a)x0 - y0| / sqrt[(b/a)^2 + 1]
= [(b/a)^2(x0)^2 - (y0)^2] / [(b/a)^2 + 1]
simplifying:
MP.NP = (b^2x0^2 - a^2y0^2) / (b^2 + a^2)

b^2x0^2 - a^2y0^2 = a^2b^2 (equ hyperbola)
.'. MP.NP = a^2b^2 / (b^2 + a^2)

Has there been a post for part iii yet? Might've missed it.

Anyway, Area of triangle PMN = 1/2*PN*PM = a^2b^2/2(b^2 + a^2)... Perhaps that was just trivial enough to go without saying, heh .

Praise be to whoever's idea this thread was by the way.

 

[ronnknee]

How about now?

Spoiler

(n+k-1)!
k!(n-1)!

k = 5
n = 14

= 18! ÷ 5!13!
= 8568

Haha honestly, I don't even know myself since I didn't bring up the question =p

 

[namburger]

Has there been a post for part iii yet? Might've missed it.

Anyway, Area of triangle PMN = 1/2*PN*PM = a^2b^2/2(b^2 + a^2)... Perhaps that was just trivial enough to go without saying, heh .

Praise be to whoever's idea this thread was by the way.

Sorry but your assuming that angle MPN is a rightangle.

 

[ronnknee]

While we wait:

If I_n = Int [tanⁿ x] dx where n >= 0, show that I_n = [tan^n-1x / (n - 1)] - I_n-2

 

[undalay]

While we wait:

If I_n = Int [tanⁿ x] dx where n >= 0, show that I_n = [tan^n-1x / (n - 1)] - I_n-2

int tanⁿx dx
= int tan^n-2x(sec²x -1) dx
= int tan^n-2xsec²x dx - I_n-2
= tan^n-1x / (n-1) - I_n-2

 

[ngogiathuan]

oh you want hard questions? thats easy:

(polynomials)
x^3 - x^2 -ax - b = 0 has 3 positive roots prove that x^3 -x^2 + bx + a = 0 has 1 positive root and 2 complex roots

prove that log_a(x) cannot be written as the quotient of 2 polynomials with real coefficients

find all polynomials with possibly complex coefficients satisfying p(x^2)-p(x)p(x+1) = 0

(conics)
given ellipse x^2/a^2 + y^2/b^2=1 a find the area of the elliptical sector define by the piece bounded by the ellipse and lines joining points (acos(t),bsin(t)) (acos(u),bsin(u)) to the focus S(ae,0)

(Mechanics)
A rocket accelerates itself by shooting out gases at the rate of m kilograms per sectond at a velocity of v relative to the rocket, if the mass of the rocket is M and initially it carries F kilograms of fuel and it's initial velocity is V, find the distance travelled in time T and it's terminal speed after all fuel has been burnt.

At any time, a particle at position (x,y) is subjected to acceleration of magnitude 10/(x^2 + y^2) towards the origin. initially it is at position (5,0) with velocity sqrt(3) upwards, describe the locus of such particle's path

(counting)
Given n colours, how many different ways can you colour the faces of a cube? two colouring are considered identical if one can rotate one into the other.

(inequality)
It is known that for a certain function f,
f(ua + (1-u)b) >= uf(a) + (1-u)f(b) for all u between 0 and 1 and all a,b.

prove that:

f(u_1 a_1 + u_2 a_2 + ... + u_n a_n) >= u_1 f(a_1) + u_2 f(a_2) + ... + u_n f(a_n)
where u_i are positive and u_i sum to 1.


It is known that f(x) = x^a where 0<1 and x>0 has the property.

hence derive that for 1/p + 1/q = 1, p,q positive

(a_1 b_1 + a_2 b_2 + ... + a_n b_n) <= (a_1^p + a_2^p + ...+ a_n^p)^(1/p) * (b_1^q + b_2^q + ... +b_n^q)^(1/q)

for positive a's and b's.

hence show that

(a_1^p + a_2^p + ...+ a_n^p)^(1/p) + (b_1^p + b_2^p + ...+ b_n^p)^(1/p) >=
((a_1+b_1)^p + (a_2+b_2)^p + ...+ (a_n+b_n)^p)^(1/p) for p>=1



(geometry)

Show that the altitudes of a triangle are concurrent
do the same for the medians
and the perpendicular bisector os sides.
now show the the 3 points where the lines meet are colinear.

ABC is a triangle. D,E lies on side BC, F,G lies on CA and H,I lies on AB.

angle BAD = angle CAE
angle CBF = angle ABG
angle ACH = andgle BCI

prove that if AD,BF and CH are concurrent, then so are AE BG and CI


I will post some more when you finish these ones

Here are some hard questions from affinity, from another topic. I post this again so everyone can have a look.
By the way, conics2008 u said that the conics and polynomial question here were "pretty straightforward". Im very much appreciated if you could post your working here.

 

[tommykins]

Haha bashing on conics2008 is fun, but I guess let's see if he can back up the amount of shit he spills.

 

[ronnknee]

int tanⁿx dx
= int tan^n-2x(sec²x -1) dx
= int tan^n-2xsec²x dx - I_n-2
= tan^n-1x / (n-1) - I_n-2

How can you just jump from the second last line to the last without working?

 

[WannaBang?]

Spoiler

S = 9
E = 5
N = 6
D = 7
M = 1
O = 0
R = 8
Y = 2

Win.

Very good.

 

[undalay]

How can you just jump from the second last line to the last without working?

CBF typing, its obvious. (sec^2 is the derivative of tan)

 

[conics2008]

3unitz,, maybe you should go back and read my posts. I knew it was P.D distance, I jst didn't know how to simplify it to make it look like what he asked..

soo that means i wont change my name.

hahahh LoL

 

[WannaBang?]

Irrelevant

LOL! Ronnknee has stamped my question as irrelevant! *Runs to the corner and cries*

 

[conics2008]

Haha bashing on conics2008 is fun, but I guess let's see if he can back up the amount of shit he spills.

Your one of the shit I spilled.

sorry if i hurt your feelings, no harm done =)

 

[undalay]

Heres onee:

Graph |x+y| = 2 labeling significant points.

 

[namburger]

CBF typing, its obvious. (sec^2 is the derivative of tan)

slack

= int tan^n-2xsec²x dx - I_n-2
=int tan^n-2x . [d(tan x)/dx] . [dx] - I_n-2{since diffrentiation of tan x with respect to x is sec^2 x}
= int tan^n-2x d(tan x) - I_n-2 treat tan x as a u

 

[conics2008]

Hey guys can you do this please.

A triangle has sides 3,4 & 5. A circle has been inscribed.

Find the area of the circle. Thanks

 

[ronnknee]

Heres onee:

Graph |x+y| = 2 labeling significant points.

Edit: Wrong solution my bad, it looked like the diamond equation


P(x) is a monic polynomial of degree 4 with interger coefficients and constant term 4. One zero is root 2, another zero is rational and the sum of the zeros is positive. Factorise P(x) over real numbers.

 

[kurt.physics]

P(x) is a monic polynomial of degree 4 with interger coefficients and constant term 4. One zero is root 2, another zero is rational and the sum of the zeros is positive. Factorise P(x) over real numbers.

Im not so sure this is right, my graphing program reakons its the one attached

This is wrong because, for example, at (-1,1) it is

/ -1+1/ doesnt equal 2

 

[conics2008]

sketch this... |x+y|=2

hence x+y=2 and -x-y=2


therefore its two parallel lines with x int = 2 and -2 and y int with 2 and -2


welldone for those two guys that used their graphing program...

Q:?

A triangle has sides 3,4 & 5. A circle has been inscribed.

Find the area of the circle