3u Mathematics Marathon V 1.1 (1 Viewer)

Riviet

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Next question:

If tan(A+B)=1 and tan(A-B)=1/7, find all possible values of tanA and tanB.
 

Sober

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pLuvia said:
(i) part is actually
8 book, can be arranged in 8! ways, since there are identical ones, then 8!/(4!x4!)=70
Good point, rather unthinking of me.
Riviet said:
Next question:

If tan(A+B)=1 and tan(A-B)=1/7, find all possible values of tanA and tanB.
(1) tan(A+B) = 1
(2) tan(A-B) = 1/7

Let a=tan(A) and b=tan(b)

(1) ( a + b ) / ( 1 - ab ) = 1
(2) ( a - b ) / ( 1 + ab) = 1/7

Simultaneously solving yields two solutions:

tan(A) = 1/2, tan(B) = 1/3
and
tan(A) = -2, tan(B) = -3
 

Sober

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pLuvia said:
Feel free to post the next question ;) Or anyone
Express the area of an anulus in terms of the largest interval that fits inside it.
 

Riviet

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Sober said:
Express the area of an anulus in terms of the largest interval that fits inside it.
Isn't it just the area of the circle that the annulus is inscribed within?
 

Sober

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Refer to the image. Interestingly, x does not uniquely define the anulus but it does define its area.

 
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Riviet

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As the annulus approaches the area of the circle, the longest interval approaches zero and as the annulus approaches an area of zero, the longest interval approaches the diameter of the circle.
 

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Riviet said:
As the annulus approaches the area of the circle, the longest interval approaches zero and as the annulus approaches an area of zero, the longest interval approaches the diameter of the circle.
I mean like area=f(x), solve f(x).
 

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I think i might have it?

The largest interval that fits within an anulus MUST be at a tangent to the inner circle. This means that the angle between the radius and the tangent is 90 degrees.

Also because the tangent is also a chord of the outer circle, and the line joining this chord is perpendicular to it, the radius of the inner circle bisects the line with x length.

View attachment 12744

Let the inner cricle have a radius of r1 and the outer circle hve a radius of r2.

Please see the diagram for more info.

A = πr22 - πr12
= π(r22 - r12 )

Now using pythagoras theorem,

r22 = r12 + (1/2x)2

Substituting the above equation into the first,

A = π(r12 + (1/2x)2 - r12)
= π(1/2x)2
=πx2/4
 
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D

doiyoubi

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no one seems to be posting a question so ive got a question

The speed vm/s of a particle moving on a straight line is given by:
v2=64-16x-8x2
where the displacement from a fixed point O is x
(a)Find an expression for the acceleration and show the motion is SHM
(b)Find the period of the motion
(c)Find the amplitude of the motion
 

Mountain.Dew

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doiyoubi said:
no one seems to be posting a question so ive got a question

The speed vm/s of a particle moving on a straight line is given by:
v2=64-16x-8x2
where the displacement from a fixed point O is x
(a)Find an expression for the acceleration and show the motion is SHM
(b)Find the period of the motion
(c)Find the amplitude of the motion
heres my 2 cents...
v2=64-16x-8x2

(a) v2=64-16x-8x2

(1/2)v2=32-8x-4x2

d (1/2)v2/dx=d(32-8x-4x2)/dx

knowing that d2x/dt2 = d (1/2)v2/dx, then

d2x/dt2 = -8-8x

a = -8(x+1) <== this is in the form a = -n^2(x-a), where n = sqrt8, a = -1.

Therefore, it is SHM.

(b) knowing that n = sqrt8, THEN the period, T = 2π/n = 2π/sqrt8 = 2sqrt8π/8 = πsqrt8/4 seconds

(c) the amplitude occurs when v = 0

so, back to original equation:

v2=64-16x-8x2, sub v = 0

0 = 64-16x-8x2
0 = 8 - 2x - x2
x2 + 2x - 8 = 0

(x - 4)(x + 2) = 0

therefore x = 4, -2 ==> the two extremes are 4 and -2, so the amplitude = 4 - (-2) / 2 = 3

therefore amplitude = 3

Alternatively,

knowing that the centre of motion is when x = -1, we know that 4 and - 2 are both 3 units away from -1. therefore, amplitude is 3.

Alternatively,

simply get v2=64-16x-8x2 in the form: v = sqrt(a2 - x2). we realise that a2 = 9, so a = 3.
 
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Mountain.Dew

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now for my obligation to keep the ball rolling:

we are playing texas holdem. there are 3 player on the table, and a dealer.

the dealer deals out 2 cards to each player. after the inital round of betting, one card is discarded off the top of deck, followed by 3 'community cards' laid face up on the middle of the table.

now: the question:

1) what is the number of possible 2 card hands for:
(i) the 1st player?
(ii) the 2nd player?
(iii) the 3rd player?

2) what is the number of possible 3 'community cards' that could occur?

3) 2 more cards are to be laid down after more rounds of betting. before the 4th card is laid down, another card off the top of deck is discarded - then the 4th card is laid down. after another betting round, the same process applies for the 5th and final card.

(i) what are the possible 5 community cards that could occur?
(ii) how many possible poker hands can each player make with their inital hand and the community cards?

if u dont know the rules of texas holdem, please look here: http://www.texasholdem-poker.com/beginnersintro.php
 

SoulSearcher

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Mountain.Dew said:
now for my obligation to keep the ball rolling:

we are playing texas holdem. there are 3 player on the table, and a dealer.

the dealer deals out 2 cards to each player. after the inital round of betting, one card is discarded off the top of deck, followed by 3 'community cards' laid face up on the middle of the table.

now: the question:

1) what is the number of possible 2 card hands for:
(i) the 1st player?
(ii) the 2nd player?
(iii) the 3rd player?

2) what is the number of possible 3 'community cards' that could occur?

3) 2 more cards are to be laid down after more rounds of betting. before the 4th card is laid down, another card off the top of deck is discarded - then the 4th card is laid down. after another betting round, the same process applies for the 5th and final card.

(i) what are the possible 5 community cards that could occur?
(ii) how many possible poker hands can each player make with their inital hand and the community cards?

if u dont know the rules of texas holdem, please look here: http://www.texasholdem-poker.com/beginnersintro.php
I'll attempt, but I wouldn't be surprised if I'm well off :p
1) i) Number of possible hands for 1st player is 52C2 = 1326

ii) Number of possible hands for 2nd player is 50C2 = 1225

iii) Number of possible hands for 3rd player is 48C2 = 1128

2) Possible 3 community cards that can appear is 47C3 = 16215

3) i) Possible 5 community cards that can occur is 47C3 * 43C1 * 41C1

= 16215 * 43 * 41
= 28587045 (that doesn't sound right :p)

ii) Possible number of poker hands for 1st player is equal to 52C2 * 5C3 = 13260

Possible number of poker hands for 2nd player is equal to 50C2 * 5C3 = 12250

Possible number of poker hands for 3rd player is equal to 48C2 * 5C3 = 11280

I'm not too sure about question 3, the answers sound wrong :p
 
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Riviet

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I remember this one from Cambridge 3 unit, it works out nicely too. :p
 

Riviet

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SoulSearcher said:
Next Question:

Use the substitution u = x - 1/x to show that 1(√6+√2)/2 (1+x2)/(1+x4) dx = pi/4√2
let u=x-1/x
u2=x2-2+x-2
u2+2=x2+x-2
du=1+x-2dx
When x=(√6+√2)/2, u=(2+2√3)/(√6+√2)
When x=1, u=0
.'. 1(√6+√2)/2 (1+x2)/(1+x4) dx=1(√6+√2)/2 (1+x2)/x2(x2+x-2) dx
=0(2+2√3)/(√6+√2)du/(2+u2)
=(1/√2)[tan-1(u/√2)](2+2√3)/(√6+√2)0
=(1/√2).tan-1[(2+2√3)/(2√3+2)]
=pi/4√2, as required.

Next Question:
Use the substitution u=2x to find ∫ 4/(1+4x2) dx

 

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