HSC Mathematics Marathon (3 Viewers)

JennaJameson · Aug 9, 2011

i found dis pr3tty intredasting

A haz mass 2m, B haz mass m, whole thing is rotated at constant angular velocity w, other deets r in teh diagram br8s

find teh tension in OA and OB

gogogogogo faggots

hscishard · Aug 9, 2011

Tension OB? Isn't it AB. Haven't done circular motion yet so idk

JennaJameson · Aug 10, 2011

yeh i meant AB you kent

hup · Sep 1, 2011

rawrence said:
New question, polynomials:

$\\\text{When the polynomial P(x) is divided by } (x^2+1) \text{ the remainder is } Ax+B\\\\\text{ i) Show that } A=\frac{P(i)-P(-i)}{2i} \text{ and }B=\frac{P(i)+P(-i)}{2}\\\\\text{ii) If P(x) is odd, find the remainder when P(x) is divided by }(x^2+1)$

first is easy and ii is 0

given A,B,C are angles in a triangle, show that
tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(A/2)tan(C/2) = 1

Drongoski · Sep 1, 2011

hup said:
first is easy and ii is 0

given A,B,C are angles in a triangle, show that
tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(A/2)tan(C/2) = 1

$\therefore \frac {A+B+C}{2} = 90$

$\therefore tan \frac {C}{2} = tan (90 - \frac {A} {2} - \frac {B}{2} ) = cot(\frac {A}{2} + \frac {B}{2}) = \frac {1}{tan (\frac {A}{2} + \frac{B}{2})} = \frac {1 - tan \frac {A}{2} tan \frac {B}{2}} {tan \frac {A}{2} + tan \frac {B}{2}}$

Now replace tan (C/2) in the LHS and simplify ending in $\frac {tan \frac {A}{2} + tan \frac {B}{2}}{tan \frac {A}{2} + tan \frac {B}{2} } = 1$

hup · Sep 1, 2011

Drongoski said:
$\therefore \frac {A+B+C}{2} = 90$

have you solved it or is that where you got up to

Drongoski · Sep 1, 2011

hup said:
have you solved it or is that where you got up to

Sorry. As a very slow typist, I key in the solution bit by bit.

hup · Sep 1, 2011

correct solution but mightve been easier just moving c/2 to the right side

evaluate

from (0 to pi/4)
∫(1-tanx)/(1+tanx) dx

thoth1 · Sep 1, 2011

hup said:
correct solution but mightve been easier just moving c/2 to the right side

evaluate

from (0 to pi/4)
∫(1-tanx)/(1+tanx) dx

log root(2)

hup · Sep 1, 2011

thoth1 said:
log root(2)

1/5+2/5^2+3/5^3 = ?

thoth1 · Sep 1, 2011

hup said:
1/5+2/5^2+3/5^3 = ?

wat. ill put up my solution in a second. wat did u do?

ohexploitable · Sep 1, 2011

thoth is correct, answer is $\log{\sqrt2}$

also, kate upton has saggy boobs

hup · Sep 1, 2011

thoth1 said:
wat. ill put up my solution in a second. wat did u do?

considered the series x + 2x^2 +3x^3+....
i also know of another way

thoth1 · Sep 1, 2011

ohexploitable said:
thoth is correct, answer is $\log{\sqrt2}$

also, kate upton has saggy boobs

tnk u for baking me up. and plz dont condescend kate upton.

hup · Sep 1, 2011

what did you get for answer

thoth1 · Sep 1, 2011

$\int_{0}^{\frac{\pi }{4}}\frac{1-\tan x}{1+\tan x}=\int_{0}^{\frac{\pi }{4}}\frac{1-\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}}=\int_{0}^{\frac{\pi }{4}}\frac{\cos x-\sin x}{\cos x +\sin x}=\left [ \ln (\cos x+\sin x) \right ]$

when you solve with limits pi/4 and 0 you get ln(root 2)

hup · Sep 1, 2011

thoth1 said:
$\int_{0}^{\frac{\pi }{4}}\frac{1-\tan x}{1+\tan x}=\int_{0}^{\frac{\pi }{4}}\frac{1-\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}}=\int_{0}^{\frac{\pi }{4}}\frac{\cos x-\sin x}{\cos x +\sin x}=\left [ \ln (\cos x+\sin x) \right ]$

when you solve with limits pi/4 and 0 you get ln(root 2)

i was talking about series q

thoth1 · Sep 1, 2011

which one is the series question?

ohexploitable · Sep 1, 2011

JennaJameson said:
i found dis pr3tty intredasting

A haz mass 2m, B haz mass m, whole thing is rotated at constant angular velocity w, other deets r in teh diagram br8s

find teh tension in OA and OB

gogogogogo faggots

let tension OA be T_A and OB, T_B

-----

at A:

$2mg+T_B\sin\beta=T_A\sin\alpha$
$4ma\omega^2=T_B\cos\alpha-T_B\cos\beta$

-----

at B:

$mg=T_B\sin\beta$
$3ma\omega^2=T_B\cos\beta$

-----

solving simultaneously,

$3mg=T_A\sin\alpha$
$7am\omega^2=T_A\cos\alpha$

$\tan\alpha=\frac{3g}{7a\omega^2}=\frac{3}{14}$

-----

draw a right angled triangle, $\sin\alpha=\frac{3}{\sqrt{205}}$

now it should be easy to solve for T_A and T_B

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