2u Mathematics Marathon v1.0 (2 Viewers)

[Mountain.Dew]

either sando or soulsearcher can post up the next question

 

[SoulSearcher]

I'll let sando post up the next question

 

[sando]

ok then.


Q: A plant grows so that it increases its height each month by 0.2 of its previous month's height. If it grows to 3m, find its height in the first month

 

[SoulSearcher]

ok then.


Q: A plant grows so that it increases its height each month by 0.2 of its previous month's height. If it grows to 3m, find its height in the first month

Spoiler

I am confused about this question, so I'll just do what I can do

Since the plant grows by 20 percent of the previous month's height, the rate of growth is 1 + 0.2 = 1.2.

This is also a geometric series, so the value of r is equal to 1.2.

Therefore,
T_n = ar^n-1
T_n = 1.2^n-1 * a

Let T_n = 3, as it occurs in the geometric series 1.2^n-1 * a

Therefore
3 = 1.2^n-1 * a

and since a is equal to the first value of the series, then the value of a is

a = 3 / 1.2^n-1 where n is the amount of months for the plant to grow

therefore the height of the plant in its first month is equal to [ 3 / 1.2^n-1 ] metres

 

[Mountain.Dew]

mmmmmmm soul searcher, you might like to consider exponentials --> growth IE H = H(original) + e^ax kind of thing.

dunno if it will work though...im busy with uni stuff atm

 

[insert-username]

ok then.


Q: A plant grows so that it increases its height each month by 0.2 of its previous month's growth (the question doesn't work if it's 0.2 of the height - it'll just keep growing bigger and bigger, right past 3m - I_F). If it grows to 3m, find its height in the first month

Soulsearcher, you're not quite right. But since the question was a little badly worded, that's OK.

Spoiler

Let g = the plant's growth (it won't start at 0).

S_∞ = g

a/(1-r) = g

a/(0.8) = g

But g = 3 (final height) - a (initial height)

So a/0.8 = 3 - a

1.25a = 3 - a (division by 0.8 = multiplication by 1.25)

a + 1.25a = 3

a(1+1.25) = 3

a = 3/2.25

Therefore a = 1⅓ metres

Therefore g = 1⅔

To check, we can use the infinite sum formula to check the value of g

Does 1⅔ = 1⅓/0.8?

Yes, it does, therefore the plant's initial height is 1⅓ metres.

Next question:

Find the exact length of the line from the point (2,7) to the centre of the circle x² + 4x + y² - 6y - 3 = 0


I_F

 

[SoulSearcher]

Next question:

Find the exact length of the line from the point (2,7) to the centre of the circle x² + 4x + y² - 6y - 3 = 0


I_F

Ahh I get where I had gone wrong. That'll teach me to pay more attention to the question Anyways, I'll do this question

Spoiler

we need to convert this equation x² + 4x + y² - 6y - 3 = 0 into (x-h)² + (y-k)² = r²
where h and k are the x and y values of the centre respectively and r is the length of the radius.

therefore
x² + 4x + y² - 6y - 3 = 0
x² + 4x + y² - 6y = 3
x² + 4x + 4 + y² - 6y + 9 = 3 + 4 + 9
(x+2)² + (y-3)² = 16

therefore x = -2, and y = 3, and thus the co-ordinates of the centre of the circle is ( -2, 3 )

now using distance formula to calculate the length of line from the point ( 2, 7 ) to the point ( -2, 3 ), letting the point ( -2, 3 ) be ( x₁, y₁ ) and the point ( 2, 7 ) be ( x₂, y₂ )

D = √ [ ( x₂ - x₁ )² - ( y₂ - y₁ )² ]

D = √ [ ( 2 + 2 )² + ( 7 - 3 )² ]

D = √ [ 16 + 16 ]

D = √ 32

D = 4 √ 2 units

therefore the exact distance between ( -2, 3 ) and ( 2, 7 ) is 4 √ 2 units

Next Question:

A game is played in which two coloured dice are thrown once. The six faces of the blue die are numbered 4, 6, 8, 9, 10 and 12. The six faces of the pink die are numbered 2, 3, 5, 7, 11 and 13. The player wins if the number on the pink die is larger than the number on the blue die.

(i) Calculate the probability of the player winning a game
(ii) Calculate the probability that the player wins at least one in two successive games.

 

[Mountain.Dew]

Ahh I get where I had gone wrong. That'll teach me to pay more attention to the question Anyways, I'll do this question

Next Question:

A game is played in which two coloured dice are thrown once. The six faces of the blue die are numbered 4, 6, 8, 9, 10 and 12. The six faces of the pink die are numbered 2, 3, 5, 7, 11 and 13. The player wins if the number on the pink die is larger than the number on the blue die.

(i) Calculate the probability of the player winning a game
(ii) Calculate the probability that the player wins at least one in two successive games.

this might not be right, but here goes...my 2 cents

Spoiler

(i) consider each number on pink die --> 2,3,5,7,11,13...

now, consider if we rolled a 13 --> 1/6 chance --> there is a 6/6 chance player will win.
1/6 to roll a 11 --> there is 5/6 chance player will win, because there are 5 numbers on blue die less than 11.
1/6 to roll a 7 --> 2/6 chance to win.
1/6 to roll a 5 --> 1/6 chance to win.

and we do not consider 2 or 3...P(winning) = 0 for 2 and 3

So, P(winning) = 1/6[ 1 + 5/6 + 2/6 + 1/6]
= 1/36[6 + 5 + 2 + 1] =14/36 = 7/18

P(winning) = 7/18

(ii) P(W) = 7/18 therefore P(L)=11/18

P(win at least one game in 2 games) = WL + LW + WW = (7/18)(11/18) + (7/18)(7/18) + (11/18)(7/18) = 203/324

 

[Mountain.Dew]

okay, a bit of trig here:

PLC Ext 1 Trials 2003

Question 6
(a) given θ is actue.
(I) write sin(θ/2) in terms of cosθ
(ii) prove that tan(θ/2) = sinθ / (1 + cosθ)
(iii) if sinθ = 4/5, find the value of tan(θ/2)

hint for 2U people:

Spoiler

Consider double angle formulae --> sin2@ = 2sin@cos@
cos2@ = 1 - 2sin²@ = 2cos²@ - 1 = cos²@ - sin²@

 

[SoulSearcher]

No idea how to do this, for now

 

[Riviet]

Spoiler

Remember that the double angle formulae also apply for not just theta but also theta/2

 

[SoulSearcher]

okay, a bit of trig here:

PLC Ext 1 Trials 2003

Question 6
(a) given θ is actue.
(I) write sin(θ/2) in terms of cosθ
(ii) prove that tan(θ/2) = sinθ / (1 + cosθ)
(iii) if sinθ = 4/5, find the value of tan(θ/2)

hint for 2U people:

Spoiler

Consider double angle formulae --> sin2@ = 2sin@cos@
cos2@ = 1 - 2sin²@ = 2cos²@ - 1 = cos²@ - sin²@

Spoiler

6(a)(i)
sin θ = sin 2(θ/2), therefore
sin θ = 2 sin (θ/2) cos (θ/2), therefore
cos (θ/2) = sin θ / 2 sin (θ/2)

now cos θ = cos 2(θ/2), therefore
cos θ = cos² (θ/2) - sin² (θ/2)
cos θ = 2 cos² (θ/2) - 1
cos θ + 1 = 2 cos² (θ/2), therefore
cos² (θ/2) = { cos θ + 1 } / 2 and thus

cos (θ/2) = √ [ { cos θ + 1 } / 2 ]

now sub the value of cos (θ/2) into cos (θ/2) = sin θ / 2 sin (θ/2)
therefore
√ [ { cos θ + 1 } / 2 ] = sin θ / 2 sin (θ/2)
{ cos θ + 1 } / 2 = sin² θ / 4 sin² (θ/2)
cos θ + 1 = sin² θ / 2 sin² (θ/2)

therefore

2 sin² (θ/2) = sin² θ / 1 + cos θ
sin² (θ/2) = ( 1 + cos θ )( 1 - cos θ) / 2(1 + cos θ)
sin² (θ/2) = { 1 - cos θ } / 2

and therefore

sin (θ/2) = √ [ { 1 - cos θ } / 2 ]

6(a)(ii)
since tan θ = sin θ / cos θ
then tan (θ/2) = sin (θ/2) / cos (θ/2)
since sin (θ/2) = sin θ / 2 cos (θ/2)
then tan (θ/2) = { sin θ / 2 cos (θ/2) } / cos (θ/2)
tan (θ/2) = sin θ / 2 cos² (θ/2)
from above, cos θ + 1 = 2 cos² (θ/2), and thus

tan (θ/2) = sin θ / 1 + cos θ

6(a)(iii)
now sin θ = 4/5
therefore cos θ = √ [ 1 - sin² θ ]
cos θ = √ [ 1 - (4/5)² ]

cos θ = √ [ 1 - 16/25 ]
cos θ = √ [ 9/25]
cos θ = 3/5

therefore sub the values sin θ = 4/5 and cos θ = 3/5 into tan (θ/2) = sin θ / 1 + cos θ
tan (θ/2) = [ 4/5 ] / [ 1 + 3/5 ]
tan (θ/2) = [ 4/5 ] / [ 8/5 ]
tan (θ/2) = 4/8
tan (θ/2) = 1/2

At least I think thats right. Only just started the 3 unit course last week

 

[Mountain.Dew]

SoulSearcher, ur job isn't done yet *evil laugh*

please post up the next question *innocent smile*

 

[SoulSearcher]

Ok ok then. Sheesh

Next question:

The Numnber N of students logged onto a website at any time over a five hour period is approximated by the formula N = 175 + 18t² - t⁴

(a) What was the initial number of students logged onto the website?
(b) How many students were logged onto the website at the end of the five hours?
(c) What was the maximum number of students logged onto the website?
(d) When were the students logging onto the website most rapidly?

 

[Glen88]

(a) 175
(b) 0
(c) 256
(d) In the third hour

probably wrong

 

[pLuvia]

Please when you post please put it in spoilers [spoiler ]...[/spoiler] without the spaces.
Please post the next question

 

[Glen88]

Prove (d^2 y / dx^2) - 4(dy/dx) - 5y - 10 = 0 given that y = 3e^(5x) - 2

 

[Mountain.Dew]

Prove (d^2 y / dx^2) - 4(dy/dx) - 5y - 10 = 0 given that y = 3e^(5x) - 2

okay...heres my 2 cents.

Spoiler

y=3e^(5x) - 2

dy/dx = 15e^(5x),
d^2 y / dx^2 = 75e^(5x)

so: LHS = (d^2 y / dx^2) - 4(dy/dx) - 5y - 10 = 75e^(5x) - 4(15e^(5x)) - 5(3e^(5x) - 2) - 10
LHS = 75e^(5x) - 60e^(5x) - 15e^(5x) + 10 - 10 = 0 = RHS

therefore, (d^2 y / dx^2) - 4(dy/dx) - 5y - 10 = 0

Next question:

Solve ==> 2x/(x-2) > 1

 

[SoulSearcher]

I had an answer but was beaten by a minute

 

[drynxz]

Solve ==> 2x/(x-2) > 1

ill take a shot even thou i havent learnt it yet...

Spoiler

2x/(x-2) >1

(multiply both sides by (x-2)^2)
= 2x(x-2) > (x-2)^2
= 2x^2 - 4x >x^2-4x+4
= x^2 - 4 >0
= (x+2)(x-2)>0
*test point x=0
= (2)(-2)>0 doesnt hold true
therefore
= x > 2

pft maybe i got it wrong...i dunno

Next question:

Solve ==> solve simultaneously

x+2y-z = -5
2x-3y+4z = 28
4x+5y-3z = -10