Extension One Revising Game (1 Viewer)

Dragonmaster262 · May 12, 2009

study-freak said:
insufficient information unless you mean 5^2x-5^x-20=0
and I guess you actually mean 5^(2x)-5^x-20=0?

That's what I mean.

addikaye03 · May 12, 2009

shaon0 said:
S= S {inf} = (a/(1-r))

S-S{n}
= (a/(1-r)) - (a(r^n-1)/r-1)
= (a/(1-r)) - (a(1-r^n)/r-1)
= (ar^n)/(1-r)

S-S{n} < 0.0001
(ar^n/1-r) < 0.0001
(6(1/2)^n/(0.5)) < 0.0001
(1/2)^n < (0.0001/12)
n < ln(0.0001/12)/ln(1/2)
n < 16.87...
Thus, n is 16.

I don't think my answer is right though.

i) is correct, thats the easy part. ii) basically correct, the number has to follow that : 12( 1/2)^n< 10^-4 and 2^n>120000...do you understand what i mean? Well done nontheless though.

shaon0 · May 17, 2009

addikaye03 said:
i) is correct, thats the easy part. ii) basically correct, the number has to follow that : 12( 1/2)^n< 10^-4 and 2^n>120000...do you understand what i mean? Well done nontheless though.

Sorry, I may have assumed something that i wasn't meant to.

ninetypercent · May 23, 2009

Find the second derivative of the following function with respect to x:

$x\sqrt{x + 2}$

kurt.physics · May 24, 2009

ninetypercent said:
Find the second derivative of the following function with respect to x:

$x\sqrt{x + 2}$

$$let$\,\, u = x\,\, $and$\,\, v = (x+2)^{\frac{1}{2}}\,\,$

$Then:$

$u' = 1$

$v' = 1 \cdot \frac{1}{2} \cdot (x+2)^{- \frac{1}{2} }$

$= \frac{1}{2} \cdot(x+2)^{- \frac{1}{2} }$

$\frac{d}{dx}\left( x\sqrt{x + 2} \right) = u \cdot v' + v \cdot u'$

$= x \cdot \frac{1}{2} \cdot(x+2)^{- \frac{1}{2} } + (x+2)^{\frac{1}{2}} \cdot 1$

$= \frac{x}{2 \sqrt{x+2} } + \sqrt{x+2}$

$\frac{d^2}{dx^2}\left( x\sqrt{x + 2} \right) = \frac{d}{dx} \left( \frac{d}{dx}\left( x\sqrt{x + 2} \right) \right)$

$= \frac{d}{dx}\left(x \cdot \frac{1}{2} \cdot(x+2)^{- \frac{1}{2} } + (x+2)^{\frac{1}{2}} \cdot 1 \right)$

$= \frac{d}{dx}\left( x \cdot \frac{1}{2} \cdot(x+2)^{- \frac{1}{2} } \right) + \frac{d}{dx} \left((x+2)^{\frac{1}{2}} \cdot 1 \right)$

$\frac{d}{dx} \left((x+2)^{\frac{1}{2}} \cdot 1 \right) = 1 \cdot \frac{1}{2} \cdot (x+2)^{- \frac{1}{2} }$

$\frac{d}{dx}\left( x \cdot \frac{1}{2} \cdot(x+2)^{- \frac{1}{2} } \right):$

$$let$\,\,u= x\,\,$and$\,\,v = \frac{1}{2} \cdot (x+2)^{- \frac{1}{2} }$

$u' = 1$

$v' = \frac{1}{2} \cdot - \frac{1}{2} \cdot 1 \cdot (x+2)^{- \frac{3}{2} }$

$= - \frac{1}{4} \cdot (x+2)^{- \frac{3}{2} }$

$\frac{d}{dx} = uv' + vu'$

$= x \cdot - \frac{1}{4} \cdot (x+2)^{- \frac{3}{2} } + \frac{1}{2} \cdot (x+2)^{- \frac{1}{2} } \cdot 1$

$= -\frac{x}{4 \cdot (x+2)^{\frac{3}{2}} }+ \frac{1}{2\cdot\sqrt{x+2}}$

$\frac{1}{2\sqrt{x+2}} - \frac{x}{4 \sqrt{(x+2)^{3}} }$

kurt.physics · May 24, 2009

addikaye03 said:
y'=[3x+4]/(2rt(x+2)

u'=3x+4 v'=2(x+2)^1/2

u''=3 v''=(x+2)^-1/2 (1)

y''=v'u''+u'v''

[Typed it straight up on computer, thats why i had to do so many steps]

EDIT: Did i get it wrong?:S

Um, first you have a fraction, that is [3x+4]/[2rt(x+2)], then you use the product rule, where you are supposed to use the quotient rule.

addikaye03 · May 24, 2009

kurt.physics said:
Um, first you have a fraction, that is [3x+4]/[2rt(x+2)], then you use the product rule, where you are supposed to use the quotient rule.

i think i might type out these kinda Q using latex now, i never see the denominator, hence use product instead of quotient

gurmies · May 24, 2009

addikaye03 said:
i think i might type out these kinda Q using latex now, i never see the denominator, hence use product instead of quotient

I often do that as well - yields the same answer so xD

addikaye03 · May 24, 2009

gurmies said:
I often do that as well - yields the same answer so xD

i think im noob, gurmies lol

Question: If P(x)=ax^5+bx^4+cx^3+dx^2+ex+f, Write down P(-x)=-P(x).
i)If P(x) is an Even polynomial, show that a=c=e=0
ii) If P(x) is an Odd polynomial, show that b=d=f=0

Hence, state general results for the coefficients of odd and even powers of x when a polynomial is even or odd.

[Pretty unique little Q i thought]

kurt.physics · May 24, 2009

addikaye03 said:
i think im noob, gurmies lol

Question: If P(x)=ax^5+bx^4+cx^3+dx^2+ex+f, Write down P(-x)=-P(x).
i)If P(x) is an Even polynomial, show that a=c=e=0
ii) If P(x) is an Odd polynomial, show that b=d=f=0

Hence, state general results for the coefficients of odd and even powers of x when a polynomial is even or odd.

[Pretty unique little Q i thought]

$P(x) = ax^5+bx^4+cx^3+dx^2+ex+f$

$P(-x) = -ax^5+bx^4+-cx^3+dx^2 -ex+f$

$$For even functions,$\,\,P(x)=P(-x)$

$ie:$

$ax^5+bx^4+cx^3+dx^2+ex+f \equiv -ax^5+bx^4+-cx^3+dx^2 -ex+f$

$\therefore ax^5 \equiv -ax^5$

$\therefore cx^3 \equiv -cx^3$

$\therefore ex \equiv -ex$

$ax^5 \equiv -ax^5\,\,$if and only if$\,\, a = -a = 0$

$Similarly,$

$c = -c = 0$

$And:$

$e = -e = 0$

$\therefore a = c = e = 0\,\,$if P(x) is even$$

$P(-x) = -P(x)\,\,$for odd functions$$

$P(-x) = -ax^5+bx^4+-cx^3+dx^2 -ex+f$

$-P(x) = -ax^5-bx^4-cx^3-dx^2-ex-f$

$\therefore -ax^5+bx^4+-cx^3+dx^2 -ex+f \equiv -ax^5-bx^4-cx^3-dx^2-ex-f$

$\therefore bx^4 \equiv -bx^4$

$\therefore dx^2 \equiv -dx^2$

$\therefore f \equiv -f$

$bx^4 \equiv -bx^4\,\,$if and only if$\,\,b = -b = 0$

$Similarly:$

$d = -d = 0$

$And:$

$f = -f = 0$

$\therefore b = d = f = 0$

$When P(x) is even, then for odd powers, the co-efficients are zero$

$When P(x) is odd, then for even powers, the co-efficients are zero$

addikaye03 · May 24, 2009

kurt.physics said:
$P(x) = ax^5+bx^4+cx^3+dx^2+ex+f$

$P(-x) = -ax^5+bx^4+-cx^3+dx^2 -ex+f$

$$For even functions,$\,\,P(x)=P(-x)$

$ie:$

$ax^5+bx^4+cx^3+dx^2+ex+f \equiv -ax^5+bx^4+-cx^3+dx^2 -ex+f$

$\therefore ax^5 \equiv -ax^5$

$\therefore cx^3 \equiv -cx^3$

$\therefore ex \equiv -ex$

$ax^5 \equiv -ax^5\,\,$if and only if$\,\, a = -a = 0$

$Similarly,$

$c = -c = 0$

$And:$

$e = -e = 0$

$\therefore a = c = e = 0\,\,$if P(x) is even$$

$P(-x) = -P(x)\,\,$for odd functions$$

$P(-x) = -ax^5+bx^4+-cx^3+dx^2 -ex+f$

$-P(x) = -ax^5-bx^4-cx^3-dx^2-ex-f$

$\therefore -ax^5+bx^4+-cx^3+dx^2 -ex+f \equiv -ax^5-bx^4-cx^3-dx^2-ex-f$

$\therefore bx^4 \equiv -bx^4$

$\therefore dx^2 \equiv -dx^2$

$\therefore f \equiv -f$

$bx^4 \equiv -bx^4\,\,$if and only if$\,\,b = -b = 0$

$Similarly:$

$d = -d = 0$

$And:$

$f = -f = 0$

$\therefore b = d = f = 0$

$When P(x) is even, then for odd powers, the co-efficients are zero$

$When P(x) is odd, then for even powers, the co-efficients are zero$

nice work man, EXACTLY how i done it. Got a Q? i could post up another i suppose?

kurt.physics · May 24, 2009

addikaye03 said:
nice work man, EXACTLY how i done it. Got a Q? i could post up another i suppose?

Thanks. I dont have any questions... but yes it would be awesome if you could post another

addikaye03 · May 24, 2009

kurt.physics said:
Thanks. I dont have any questions... but yes it would be awesome if you could post another

no worries man

Question (Projectile): A golf ball is hit from a point O with an initial velocity V m/s so that it rises with an angle of 30* ( *=degrees) to the horizontal. Taking g=10ms/s show that x=1/2(Vrt(3t)), y=-5t^2+1/2(Vt)

a) in its decent, its clears a wall 11.25m high, and takes 4.5 sec from moment of projection to clear the wall. Calculate

i)value of V

ii) greatest hieght reached by ball

iii)horizontal distance from the wall to the point of impact of the ball with the ground

[that ought to keep you lot busy]

gurmies · May 24, 2009

$\ddot{x}=0 \\\\ \dot{x}=V\cos\alpha \\\\ x = Vt\cos\alpha \\\\ If \,\, \alpha=30^{\circ}, \,\, x = \frac{1}{2}\sqrt3Vt \\\\ \ddot{y}=-g \\\\ \dot{y}=V\sin\alpha - gt \\\\ y = Vt\sin\alpha - \frac{1}{2}gt^{2} \\\\ If \,\, \alpha = 30^{\circ}, g = 10, y = -5t^{2}+\frac{1}{2}Vt$

(A)

$11.25 = \frac{1}{2}.V(4.5)-5(4.5)^{2} \\\\ \frac{1}{2}V(4.5)=11.25+5(4.5)^{2} \\\\ V = \frac{2[11.25+5(4.5)^{2}]}{4.5} = 50 m/s$

(B)

$Greatest \,\, height \,\, when \,\, vertical \,\, velocity \,\, is \,\, equal \,\, to \,\, 0. \\\\ \dot{y}=0 \\\\ 25-10t=0 \\\\ t = 2.5s \\\\ y = 25(2.5)-5(2.5)^{2}=31.25 m$

(C)

$When \,\, y = 0, \,\, golf \,\, ball \,\, will \,\, hit \,\, the \,\, ground. \\\\ 25t-5t^{2}=0 \\\\ t(25-5t)=0 \\\\ t = 0, 5 \\\\ Since \,\, t = 0 \,\, represents \,\, initial \,\, position, \,\, we \,\, take \,\, t = 5. \\\\ x = 25t\sqrt3 = 125\sqrt3 m$

EDIT: Read question wrong, have to subtract that from wall distance - cbb to change now with LAGGY BoS

addikaye03 · May 24, 2009

gurmies said:
$\ddot{x}=0 \\\\ \dot{x}=V\cos\alpha \\\\ x = Vt\cos\alpha \\\\ If \,\, \alpha=30^{\circ}, \,\, x = \frac{1}{2}\sqrt3Vt \\\\ \ddot{y}=-g \\\\ \dot{y}=V\sin\alpha - gt \\\\ y = Vt\sin\alpha - \frac{1}{2}gt^{2} \\\\ If \,\, \alpha = 30^{\circ}, g = 10, y = -5t^{2}+\frac{1}{2}Vt$

(A)

$11.25 = \frac{1}{2}.V(4.5)-5(4.5)^{2} \\\\ \frac{1}{2}V(4.5)=11.25+5(4.5)^{2} \\\\ V = \frac{2[11.25+5(4.5)^{2}]}{4.5} = 50 m/s$

(B)

$Greatest \,\, height \,\, when \,\, vertical \,\, velocity \,\, is \,\, equal \,\, to \,\, 0. \\\\ \dot{y}=0 \\\\ 25-10t=0 \\\\ t = 2.5s \\\\ y = 25(2.5)-5(2.5)^{2}=31.25 m$

(C)

$When \,\, y = 0, \,\, golf \,\, ball \,\, will \,\, hit \,\, the \,\, ground. \\\\ 25t-5t^{2}=0 \\\\ t(25-5t)=0 \\\\ t = 0, 5 \\\\ Since \,\, t = 0 \,\, represents \,\, initial \,\, position, \,\, we \,\, take \,\, t = 5. \\\\ [B]x = 25t\sqrt3 = 125\sqrt3 m[/B]$

EDIT: Read question wrong, have to subtract that from wall distance - cbb to change now with LAGGY BoS

For other hes just means subtract the distance from wall to origin of projection ie. x=25rt3-225rt3/2=25rt3/2
<DIR></DIR>Well done nevertheless Gurmies!

addikaye03 · May 24, 2009

Question:

A deck of cards consists of a 4 groups each of 6 cards. Each group with a different colour and the cards number 1 to 6.

a) How many distinct 5-card subsets are possible from the deck If:
i)3 cards are marked with integer '5' and 2 are marked with '6'
ii)5 cards are all the same colour
iii) exactly 4 of the cards has the same integer on it
iv) the sum of the digits is greater than 27

[I found this difficult to think bout at first but once you think bout it right, its easy as]

Timothy.Siu · May 24, 2009

addikaye03 said:
Question:

A deck of cards consists of a 4 groups each of 6 cards. Each group with a different colour and the cards number 1 to 6.

a) How many distinct 5-card subsets are possible from the deck If:
i)3 cards are marked with integer '5' and 2 are marked with '6'
ii)5 cards are all the same colour
iii) exactly 4 of the cards has the same integer on it
iv) the sum of the digits is greater than 72

[I found this difficult to think bout at first but once you think bout it right, its easy as]

i)4C3 x 4C2
ii)6C5x4
iii)4C4x6x20?
iv)0?

addikaye03 · May 24, 2009

Timothy.Siu said:
i)4C3 x 4C2
ii)6C5x4
iii)4C4x6x20?
iv)0?

All correct cept the last part man, the last part is actually asking for the probability which is my error but the answer is (24+4+4)/24C5=4/5313 #... see if you can get that if not i'll explain, sorry again

azen · May 24, 2009

Help me with these questions ><" Thank you~ ^^

A cricket ball leaves the bowler's hand 2 metres above the ground with a velocity of 30m/s at an angle of 5 degrees below the horizontal. The equations of motion for the ball are

Acceleration (x double dot)=0, y DOUBLE DOT=-10

Take the origin to be the point where the ball leaves the bowler's hand.
i) Using caculusm prove that the coordinates of the ball at time t are given by
x=30tcos(5degrees), y=-30tsin(5degrees)-5t^2

ii) Find the time at which the ball strikes the ground
iii)caculate the angle at which the ball strikes the ground

addikaye03 · May 31, 2009

azen said:
Help me with these questions ><" Thank you~ ^^

A cricket ball leaves the bowler's hand 2 metres above the ground with a velocity of 30m/s at an angle of 5 degrees below the horizontal. The equations of motion for the ball are

Acceleration (x double dot)=0, y DOUBLE DOT=-10

Take the origin to be the point where the ball leaves the bowler's hand.
i) Using caculusm prove that the coordinates of the ball at time t are given by
x=30tcos(5degrees), y=-30tsin(5degrees)-5t^2

ii) Find the time at which the ball strikes the ground
iii)caculate the angle at which the ball strikes the ground

this Q has been answered like a billion times on here, have a look round...

new Q anyone?

Extension One Revising Game (1 Viewer)

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