2u Mathematics Marathon v1.0 (2 Viewers)

[pLuvia]

Takes too long

 

[Riviet]

I'll probably use the ∫ when doing indefinites, as the large one is unnecessary, but the ones with boundaries i will use the large one.

 

[Riviet]

Evaluate ∫[4 to 0] root(1+x⁴ dx using Simpsons rule with 5 functions values

Spoiler

∫[4 to 0] root(1+x⁴ dx=h/3 . [y_o+y_n+4(y₁+y<sub>3[/sup]+...)+2(y2+y4+...)] where h=(b-a)/n and n is the number of sub-intervals

= 1/3 . [1+sqrt257+4(sqrt2+sqrt82)+2sqrt17]

= 22.39 (2dp)</sub>

<sub>Next Question:

If S and T are the roots of the equation x²-7x+2=0, without finding the roots, evaluate:

a) S+T

b) ST

c) 1/S + 1/T

d) S²+T²

e) S³+T³ [OPTIONAL]</sub>

 

[Mountain.Dew]

Next Question:

If S and T are the roots of the equation x²-7x+2=0, without finding the roots, evaluate:

a) S+T

b) ST

c) 1/S + 1/T

d) S²+T²

e) S³+T³ [OPTIONAL]

this one is simple. all u need to know is sum of roots and product of roots.

Essentially, all you need is that S + T = -b/a = - (-7)/1 = 7 (answer to (a))

and ST = c/a = 2 (answer to (b))

SO,

(c) 1/S + 1/T = (S+T)/ST=7/2 [merely sum of roots divided by product of roots.
(d) S^2 + T^2 = (S+T)^2 - 2ST = (7)^2 - 2(2) = 49 - 4 = 45
(e) this is a little bit tricky, at around 3U and 4U lvl.

with x²-7x+2=0, times both sides by x.

SO, we get x³-7x²+2x=0.....(J)

THEN, since S and T are roots, SUB into equation J (other root is x=0 BTW)

So, we have S³-7S²+2S=0 ....(K)
AND T³-7T²+2T=0....(L)

then, (K) + (L) yields --> S³ + T³ = 7(S² + T²) - 2(S+T)

so, S³ + T³ = 7(45) - 2(7) (using (d) and (a) )
= 301.

 

[YBK]

Mountain. Dew, you're posting up the next question

 

[Riviet]

Ah very clever Mountain Dew.

The model method for e) was actually S³+T³=(S+T)³-3ST(S+T) but yours was also interesting.

*waits for the next question from you*

 

[Mountain.Dew]

Ah very clever Mountain Dew.

The model method for e) was actually S³+T³=(S+T)³-3ST(S+T) but yours was also interesting.

*waits for the next question from you*

oh rats, forgot about that one line thing. that is much better method.

oh rats. i have to continue this grand marathon eh?

how about a nice juicy HSC question from 1992 eh?
everyone hates probability (save a few), so here goes:

HSC Mathematics Paper 1992 Question 8(b)

A box contain 12 chocolates of the same appearance. Four of the chocolates are hard, while the other 8 have soft centres. Kim eats three chocolates randomly selected from the box. Using a tree diagram or otherwise, find the probability that:
(i) the 1st chocolate Kim eats is hard
(ii) Kim eats three hard chocolates
(iii) Kim eats exactly one hard chocolate
And as two last additions by myself:
(iv) Kim eats no more than 3 hard chocolates.
(v) Kim eats a hard chocolate on his second pick.

 

[SoulSearcher]

A box contain 12 chocolates of the same appearance. Four of the chocolates are hard, while the other 8 have soft centres. Kim eats three chocolates randomly selected from the box. Using a tree diagram or otherwise, find the probability that:
(i) the 1st chocolate Kim eats is hard
(ii) Kim eats three hard chocolates
(iii) Kim eats exactly one hard chocolate
And as two last additions by myself:
(iv) Kim eats no more than 3 hard chocolates.
(v) Kim eats a hard chocolate on his second pick.

(i) P(1st chocolate is hard) = 4/12 = 1/3

(ii) P(3 hard chocolates are eaten) = 1/3 * 3/11 * 1/5
= 1/55

(iii) P(only 1 hard chocolate is eaten) = 3 * 1/3 * 8/11 * 7/10
= 168/330
= 28/55

(iv) I'm assuming that the 3 chocolates cannot be all soft otherwise the answer is 1, so

P(no more than 3 hard chocolates are eaten) = 1 - P(3 soft chocolates are eaten)
= 1 - (2/3 * 7/11 * 3/5)
= 1 - 14/55
= 41/55

(v) P(hard chocolate is eaten on second pick) = 2/3 * 4/11 + 1/3 * 3/11
= 8/33 + 1/11
= 1/3

 

[Riviet]

SoulSearcher, you have the honour to post the next question

(could you guys also put spoiler [ ]'s so others can do the questions to? thnx)

 

[sando]

sorry to interupt, but how do you do a spoiler???


my question will be coming soon

 

[Riviet]

Spoiler

quote this

 

[SoulSearcher]

yeah i'll make sure that i'll tag the answers next time i answer a question (if i can that is )

anyway, heres the next question:

A couple borrow $140 000 to buy a house. Interest is charged on the loan at 15% reducible, and the loan (including interest) is to be paid off by equal monthly instalments over 20 years, the first instalment to be made one month after time of purchase.

(i) Calculate the amount of each instalment to the nearest cent.
(ii) Calculate how much was still owed at the end of 10 years.

 

[insert-username]

A couple borrow $140 000 to buy a house. Interest is charged on the loan at 15% reducible, and the loan (including interest) is to be paid off by equal monthly instalments over 20 years, the first instalment to be made one month after time of purchase.

(i) Calculate the amount of each instalment to the nearest cent.
(ii) Calculate how much was still owed at the end of 10 years.

[I'm assuming this is 15% p.a. interest]

Spoiler

Monthly interest = 1.25%
Principal = $140 000
Let M be the monthly repayment.


i. Owe₁ = 140 000 x 1.0125 - M

Owe₂ = 140 000 x 1.0125² - M x 1.0125 - M

Owe₃ = 140 000 x 1.0125³ - M x 1.0125² - M x 1.0125 - M

Owe₂₄₀ = 140 000 x 1.0125²⁴⁰ - M(1 + 1.0125 + 1.0125² + ... + 1.0125²³⁹)

but Owe₂₄₀ = 0

Therefore 0 = 140 000 x 1.0125²⁴⁰ - M(1 + 1.0125 + 1.0125² + ... + 1.0125²³⁹)

Therefore M = (140 000 x 1.0125²⁴⁰) ÷ ([1.0125²⁴⁰ - 1]/[0.0125])

M = $1,843.51 (nearest cent)


ii. Owe₁₂₀ = 140 000 x 1.0125¹²⁰ - M(1 + 1.0125 + 1.0125² + ... + 1.0125¹¹⁹)

= 140 000 x 1.0125¹²⁰ - 1843.51(1 + 1.0125 + 1.0125² + ... + 1.0125¹¹⁹)

= 140 000 x 1.0125¹²⁰ - 1843.51([1.0125¹²⁰ - 1]/[0.0125])

= $114,264.45 (nearest cent)

How's that?


I_F

 

[Riviet]

I don't know, you tell me lol.

Feel free to post the next question when you are ready.

 

[insert-username]

Next question:

The formula for the surface area of a cylinder is given by S = 2πr(r + h), where r is the radius of its base and h is its height. Show that if a cylinder holds a volume of 54π m³, the surface area is given by the equation:

S = 2πr² + (108π)/r

Hence find the radius that gives the minimum surface area.


I_F

 

[Riviet]

Next question:

The formula for the surface area of a cylinder is given by S = 2πr(r + h), where r is the radius of its base and h is its height. Show that if a cylinder holds a volume of 54π m³, the surface area is given by the equation:

S = 2πr² + (108π)/r

Hence find the radius that gives the minimum surface area.


I_F

Spoiler

Volume of cylinder=πr²h=54π

r²h=54

h=54/r²

sub. h=54/r² into S = 2πr(r + h)

=> S=2πr(r+54/r²)

S=2πr²+108π/r [QED]

dS/dr=4πr-108π/r²=0

4π³=108π

r³=27

r=3

.: the radius that gives the minimum surface area is 3m.

Next Question:

Prove that the roots of the equation ax²+bx+c=0 are [-b+sqrt(b²-4ac)] / 2a by completing the square.

 

[sando]

do u just sub it into the quadratic formula?

 

[Riviet]

do u just sub it into the quadratic formula?

Read the question. You need to complete the square.

 

[insert-username]

Prove that the roots of the equation ax2+bx+c=0 are [-b+sqrt(b2-4ac)] / 2a by completing the square.

Spoiler

ax² + bx + c = 0.

ax² + bx = - c [Subtract c from both sides]

x² + bx/a = - c/a [Divide through by a]

x + bx/a + b²/4a² = -c/a + b²/4a² [Complete the square]

(x + b/2a)² = -c/a + b²/4a² [Factorise the perfect square]

(x + b/2a)² = (-4ac + b²) ÷ 4a² [Bring the RHS over a common denominator]

x + b/2a = ±√[-4ac + b² ÷ 4a²] [Take the positive/begative square roots of both sides]

x + b/2a = ±√[b² -4ac] ÷ 2a] [Simplify the RHS]

x = -b/2a ±√[(b² -4ac)] ÷ 2a] [Subtract -b/2a from both sides]

x = (-b ±√[b² -4ac]) ÷ 2a [Common denominator]

Next question:


The angle of elevation of a flagpole from a point a certain distance away from its base is 20 degrees. After walking 80m toward the flagpole, the angle of elevation is 75 degrees. Find the height of the flagpole to the nearest metre.


I_F

 

[Riviet]

I'll leave this one for someone else to do

Here's a hint if you need it:

Spoiler

Draw a diagram and label what is given in the question.