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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE General & Further Mathematics => Topic started by: Rosie on May 10, 2008, 08:56:53 am

Title: Questions
Post by: Rosie on May 10, 2008, 08:56:53 am: 1. The following are general first-order difference equations. Solve:
t_n+1 = 1.5t_n-8, t₁=32

I'm a little confused about the meaning of a first-order difference equation. Are they just equations that have a t_n and t_n+1, that is, they differ by one step.
Title: Re: Questions
Post by: Collin Li on May 11, 2008, 01:19:35 am: A first-order difference equation is a difference equation that has the smallest step differing with the highest step by 1.

In this case, the highest step term is $n+1$ , while the smallest step term is $n$ , hence the difference equation is first order.

So yes, your definition is correct. I know nothing of the Further Mathematics course, so I'm not sure whether you're required to know what a "difference equation" is in general. If you don't know, ask!
Title: Re: Questions
Post by: Rosie on May 11, 2008, 08:32:36 am: Can anyone solve the question I have above, this is a number pattern question
Title: Re: Questions
Post by: Mao on May 11, 2008, 08:54:14 am: what am I solving for....?

am i writing the first 10 numbers of the sequence down, or what... =\

(its not even an equation!) :P
Title: Re: Questions
Post by: bigtick on May 11, 2008, 09:23:21 am: first order means t(n) to the power of 1.
solve means find t(n) given the difference equation.
solution is t(n)=32x1.5^(n-1)+(-8)(1.5^(n-1) -1)/(1.5 -1).
Title: Re: Questions
Post by: Rosie on May 11, 2008, 10:45:08 am: Yes, you have that question on the top and you have to use the rule/equation for solving an equation that is both an arithmetic and geometric sequence.
It is t_n=ar^n-1+d $\frac {(r^{n-1}-1)} {r-1}$

You have to solve using this equation by putting in all teh required values that are given in the equation but I can't seem to get the correct answer.
Title: Re: Questions
Post by: Mao on May 11, 2008, 10:59:06 am: a=32, r=1.5, d=-8, n=variable :)
Title: Re: Questions
Post by: Rosie on May 11, 2008, 11:18:40 am: Ok, I'll show you the steps I did to solve this

t_n=32*1.5^n-1-8 $\frac{1.5^{n-1}-1}{1.5-1}$
t_n=32*1.5^n-1-8 $\frac{1.5^{n-1}-1}{0.5}$
I'm not sure what to do next. When simplifying teh fraction, should I divide -1 by 0.5 or do I bring 0.5 to the -8 to simplify the equation.
Title: Re: Questions
Post by: Collin Li on May 11, 2008, 11:51:15 am: Quote from: Rosie on May 11, 2008, 11:18:40 am
I'm not sure what to do next. When simplifying teh fraction, should I divide -1 by 0.5 or do I bring 0.5 to the -8 to simplify the equation.

$8\times\frac{1.5^{n-1}-1}{0.5} = \frac{8}{0.5}\left(1.5^{n-1}-1\right) = 16\left(1.5^{n-1}-1\right)$

Now you can just expand it and collect like terms.
Title: Re: Questions
Post by: Rosie on May 11, 2008, 12:37:24 pm: Hey, I've got it now. What I was doing wrong, was multiplying those numbers together instead of dividing.
However, I'm not sure why you divide it.
Title: Re: Questions
Post by: Mao on May 11, 2008, 01:44:15 pm: because:

$a \cdot \frac{b}{c} = \frac{a}{1} \cdot \frac{b}{c} = \frac{a\cdot b}{c} = \frac{a}{c} \cdot \frac{b}{1} = (a \div c) \times b$
Title: Re: Questions
Post by: dcc on May 11, 2008, 01:53:51 pm: "Do not use a cannon to kill a mosquito." - Confucius

$a_{n + 1} = 1.5a_{n} - 8, a_{1} = 32$

Substituting $n \mapsto n + 1$ , we arrive at another form for this equation:

$a_{n + 2} = 2.5a_{n + 1} - 1.5a_{n}$

$\mbox{Let }A(x) = \sum\limits_{n=0}^{\infty} a_{n} x^n$

Now multiply our new equation by $x^n$ :

$a_{n + 2} \cdot x^n = 2.5a_{n +1} \cdot x^n - 1.5a_{n} \cdot x^n$

Now, this really defines an infinite set of equations, one for each possible value of n, so we can write this with a summation of all the different variations of n:

$\implies \sum\limits_{n=1}^{\infty} a_{n + 2} \cdot x^n = 2.5 \sum\limits_{n=1}^{\infty} a_{n + 1} \cdot x^n - 1.5 \sum\limits_{n=1}^{\infty} a_{n} \cdot x^n$

$\implies \sum\limits_{n=3}^{\infty} a_{n} \cdot x^{n - 2} = 2.5 \sum\limits_{n=2}^{\infty} a_{n} \cdot x^{n-1} - 1.5 \sum\limits_{n=1}^{\infty} a_{n} \cdot x^n$

First Consider:
$1.5 \sum\limits_{n=1}^{\infty} a_{n} \cdot x^n = 1.5 \sum\limits_{n=0}^{\infty} a_{n} \cdot x^n - 1.5 a_{0} \cdot x^{0} = 1.5 \sum\limits_{n=0}^{\infty} a_{n} \cdot x^n - 40 = 1.5A(x) - 40$

Next:
$2.5 \sum\limits_{n=2}^{\infty} a_{n} \cdot x^{n-1} = 2.5x^{-1} \sum\limits_{n=2}^{\infty} a_{n} \cdot x^{n} = 2.5x^{-1} (\sum\limits_{n=0}^{\infty} a_{n} \cdot x^{n} - a_{0}x^{0} - a_{1}x) = 2.5x^{-1} (A(x) - \dfrac{40}{1.5} - 32x)$

Then:

$\sum\limits_{n=3}^{\infty} a_{n} \cdot x^{n - 2} = x^{-2} \sum\limits_{n=3}^{\infty} a_{n} \cdot x^{n} = x^{-2} (\sum\limits_{n=0}^{\infty} a_{n} \cdot x^{n} - a_{0}x^0 - a_{1}x - a_{2}x^2) = x^{-2} (A(x) - \dfrac{40}{1.5} - 32x - 40x^2)$

Combining these:

$\implies x^{-2} (A(x) - \dfrac{40}{1.5} - 32x - 40x^2) = 2.5x^{-1} (A(x) - \dfrac{40}{1.5} - 32x) - (1.5A(x) - 40)$

$\implies A(x) = \dfrac{\dfrac{40}{1.5} - \dfrac{104x}{3}}{1.5x^2 + 1 - 2.5x}$

Now expanding using the method of partial fractions:

$\implies A(x) = \dfrac{80}{1 - 1.5x} - \dfrac{\frac{160}{3}}{1-x} + \dfrac{\frac{208}{3}}{1 - x} - \dfrac{\frac{208}{3}}{1 - 1.5x} = \dfrac{\frac{32}{3}}{1 - 1.5x} + \dfrac{16}{1 - x}$

Now these are variations on the inifinite geometric sequence (which have a nice little series we can work with):

$\implies A(x) = \dfrac{32}{3} \cdot \dfrac{1}{1 - 1.5x} + 16 \cdot \dfrac{1}{1 - x} = \frac{32}{3} \sum\limit_{n=0}^{\infty} (1.5x)^n + 16 \sum\limit_{n=0}^{\infty} (x)^n = \dfrac{32}{3}(\sum\limit_{n=0}^{\infty} (1.5^n + 1.5)x^n) = \dfrac{\sum\limit_{n=0}^{\infty} (1.5^n + 1.5)x^n)}{\frac{3}{32}}$

BUT REMEMBER:

$A(x) = \sum\limits_{n=0}^{\infty} a_{n} x^n = \dfrac{\sum\limit_{n=0}^{\infty} (1.5^n + 1.5)x^n)}{\frac{3}{32}}$

Comparing Coefficients, it becomes apparent:

$a_{n} = \dfrac{32}{3} (1.5^n + 1.5)$
Title: Re: Questions
Post by: Mao on May 11, 2008, 01:58:38 pm: lol, rob you have WAY~ too much time on your hands :P

(and btw, further people, ignore this post. it does not affect you AT ALL)
Title: Re: Questions
Post by: clinton_09 on May 11, 2008, 01:59:26 pm: wow who would have thought this is further maths
Title: Re: Questions
Post by: Mao on May 11, 2008, 02:01:46 pm: Quote from: clinton_09 on May 11, 2008, 01:59:26 pm
wow who would have thought this is further maths
dcc, Ahmad II - self_control. :P
Title: Re: Questions
Post by: droodles on May 11, 2008, 03:04:37 pm: LOL...
Title: Re: Questions
Post by: Ahmad on May 11, 2008, 10:09:52 pm: Since dcc stole my favourite method - I propose an elementary one.

$t_{n+1} = \frac{3}{2} t_n - 8, t_1 = 32$

Multiply both sides by $(\frac{2}{3})^{n+1}$

$(\frac{2}{3})^{n+1}t_{n+1} =( \frac{2}{3})^n t_n - 8 (\frac{2}{3})^{n+1}$

Rearranging, then decreasing n,

$(\frac{2}{3})^{n+1}t_{n+1} -( \frac{2}{3})^n t_n = - 8 (\frac{2}{3})^{n+1}$

$(\frac{2}{3})^{n}t_{n} -( \frac{2}{3})^{n-1} t_{n-1} = - 8 (\frac{2}{3})^{n}$

$(\frac{2}{3})^{n-1}t_{n-1} -( \frac{2}{3})^{n-2} t_{n-2} = - 8 (\frac{2}{3})^{n-1}$

...

$(\frac{2}{3})^{2}t_{2} -( \frac{2}{3})^{1} t_{1} = - 8 (\frac{2}{3})^{2}$

Summing up LHS, and RHS respectively, noting that most of the terms on the LHS cancel, and subbing in t_1, while summing the geometric series on the right,

$(\frac{2}{3})^{n+1}t_{n+1} - \frac{64}{3} = -3\cdot 3 \cdot (\frac{2}{3})^2 (1 - (\frac{2}{3})^{n})$

Changing n+1 to n, while cleaning up,

$t_n = \frac{32}{3}\left( (\frac{3}{2})^n + \frac{3}{2}\right)$
Title: Re: Questions
Post by: Glockmeister on May 11, 2008, 10:11:43 pm: except I can't read latex.
Title: Re: Questions
Post by: Ahmad on May 11, 2008, 10:17:28 pm: You don't need to, anymore. $\LaTeX$ on the forum broke for a short while, but it's fine now.
Title: Re: Questions
Post by: Mao on May 11, 2008, 10:21:55 pm: sorry dcc, but i like Ahmad's method more than yours :P

i only got lost once :P (as opposed to 20000000 times)
Title: Re: Questions
Post by: dcc on May 11, 2008, 10:28:39 pm: Quote from: Mao on May 11, 2008, 10:21:55 pm
sorry dcc, but i like Ahmad's method more than yours :P

i only got lost once :P (as opposed to 20000000 times)

It's actually Ahmad's method, but i stole it :P
Title: Re: Questions
Post by: Ahmad on May 11, 2008, 10:32:12 pm: Generating function are cool though Mao, google generatingfunctionology if you're any interested.

For simple recurrences it's kind of like using integration instead of the 1/2bh formula to find the area of a triangle, but then it's much more general, just like integration is.
Title: Re: Questions
Post by: Rosie on May 16, 2008, 11:55:26 am: Is anyone familiar with the Fibonacci sequence. I have a question.

In the fibonacci sequence, t₇=13, t₉=34 and t₁₀=55
Name the terms represented by:
t₁₁+2t₁₂+t₁₃
Title: Re: Questions
Post by: Mao on May 16, 2008, 06:04:22 pm: $F_n = F_{n-1} + F_{n-2}$

$t_{11} + 2t_{12} + t_{13} = (t_{11} + t_{12}) + (t_{12} + t_{13}) = t_{13} + t_{14} = t_{15}$
Title: Re: Questions
Post by: plato on May 19, 2008, 11:45:45 pm: Quote from: Rosie on May 16, 2008, 11:55:26 am
Is anyone familiar with the Fibonacci sequence. I have a question.

In the fibonacci sequence, t₇=13, t₉=34 and t₁₀=55
Name the terms represented by:
t₁₁+2t₁₂+t₁₃

While Mao has simplified your expression to the term $t_{15}$ as you had requested, you may then want this value of $t_{15}$ . Use your stated values of $t_{9}=34$ and $t_{10}=55$ (you could ignore the value of $t_{7}$ !)

If the sequence is related to a Fibonnaci sequence, then each term is the sum of the TWO terms before it. Therefore,
$t_{11} = t_{9}+t_{10} = 34+55=89$
$t_{12} = t_{10}+t_{11} = 55+89=144$
$t_{13} = t_{11}+t_{12} = 89+144=233$

and so $t_{11}+2t_{12}+t_{13} = 89 + 2\times144 + 233 = 610$
Title: Re: Questions
Post by: Rosie on May 20, 2008, 06:43:03 pm: Number Patterns
Q1. A gardener moves a load of sand by emptying barrow loads of sand at points along a straight line. The first drop-off point is 20 metres from where the sand was originally dumped. The remaining loads of sand are then dropped off at five-metre intervals along the line.

After 15 barrow loads, all the sand has been moved. How far in total would the gardener have to walk to move all the sand and then return to the point where the sand was originally dumped?

Q2. A car purchased on 1 June 2005 loses value at a depreciation rate of 20% per year. The original purchase price was $25 000. The value of the car in dollars on 1 June 2010 will be:
Title: Re: Questions
Post by: Odette on May 20, 2008, 07:56:56 pm: Q1.
t1= 20
d= 5
n= 16
Using the formula tn= a + (n-1)d
t16= 20 + (15 x 5)
= 20 + 75
= 95
*I'm not 100% sure have done these in a while lol... may have interpreted the question wrong.

Q2.
P= 25000
R= 0.8
n= 5

Using depreciation formula- A= PR^n
A= 25000 x 0.8 ^5
A= $ 8192
Hope that helps ^_^
Title: Re: Questions
Post by: /0 on May 20, 2008, 11:22:21 pm: Quote from: Odette on May 20, 2008, 07:56:56 pm
Q1.
t1= 20
d= 5
n= 16
Using the formula tn= a + (n-1)d
t16= 20 + (15 x 5)
= 20 + 75
= 95
*I'm not 100% sure have done these in a while lol... may have interpreted the question wrong.

I think it's a sum formula instead of a nth term formula, because it asks u to find the total distance travelled, i.e the total length of

---->
<----
------>
<------
--------->
<---------

etc... 15 times back n forth

So to find the total disatnce in one direction then a = 20, d = 5, n = 15

$S_{15}=\frac{15}{2}(2(20)+(15-1)5)=825 m$

But since we have only counted the total distance he walks one way, we need to multiply be 2 because he has to return to pick up more sand => total distance is 1650m
I think that's right...
Title: Re: Questions
Post by: Odette on May 21, 2008, 08:12:34 am: Quote from: DivideBy0 on May 20, 2008, 11:22:21 pm
Quote from: Odette on May 20, 2008, 07:56:56 pm
Q1.
t1= 20
d= 5
n= 16
Using the formula tn= a + (n-1)d
t16= 20 + (15 x 5)
= 20 + 75
= 95
*I'm not 100% sure have done these in a while lol... may have interpreted the question wrong.

I think it's a sum formula instead of a nth term formula, because it asks u to find the total distance travelled, i.e the total length of

---->
<----
------>
<------
--------->
<---------

etc... 15 times back n forth

So to find the total disatnce in one direction then a = 20, d = 5, n = 15

$S_{15}=\frac{15}{2}(2(20)+(15-1)5)=825 m$

But since we have only counted the total distance he walks one way, we need to multiply be 2 because he has to return to pick up more sand => total distance is 1650m
I think that's right...
Yep yep, sorry missed that part of the question lolz... you're right ^_^
Title: Re: Questions
Post by: Rosie on May 22, 2008, 06:08:19 pm: Q. In an arithmetic sequence, a=250 and d=26. The first term in this sequence to exceed 500 is:
Title: Re: Questions
Post by: costargh on May 22, 2008, 06:13:10 pm: 500= a + ( n -1) * d

500= 250 + (n - 1) * 26

500 = 250 + 26n - 26

500 = 224+ 26n

276= 26n

276/ 26 = n

n = 10.6154
The first term (n) to pass 500 is therefore 11
Title: Re: Questions
Post by: Rosie on June 22, 2008, 01:58:11 pm: Q. A square has an area of 169cm². What is the length of the diagonal?

similar triangles:
Q. A hill has a gradient of 1 in 20, i.e. for every 20m horizontally there is a 1 m increase in height. If you go 300m horizontally, how high up will you be?

thanx
Title: Re: Questions
Post by: clinton_09 on June 22, 2008, 02:04:29 pm: for the similar triangle question i think you would just divide 300 by 20 which will give you 15m
Title: Re: Questions
Post by: lwine on June 22, 2008, 02:06:14 pm: 1. The area of a square is the length of its side squared. Therefore, $x^2=169\Rightarrow x=13\text{ cm}$ This is the length of its side.

To find the diagonal, we can use pythagoras theorem: $\sqrt{13^2+13^2}=13\sqrt{2}\text{ cm}$

2. If there is a 1m rise for every 20 meters, and we travel for 300 meters, then the number of meters we rise is equivalent to the number of times we travel 20 meters. Therefore, 300/20 = 15 times, so we rise a total of 15 times 1 meter, which equals 15 meters.
Title: Re: Questions
Post by: jsimmo on June 22, 2008, 02:23:50 pm: Quote from: Rosie on June 22, 2008, 01:58:11 pm
Q. A square has an area of 169cm². What is the length of the diagonal?

similar triangles:
Q. A hill has a gradient of 1 in 20, i.e. for every 20m horizontally there is a 1 m increase in height. If you go 300m horizontally, how high up will you be?

thanx

It's easier just to square root 169

√169 = 13cm
Title: Re: Questions
Post by: bucket on June 22, 2008, 03:22:58 pm: er thats what he did lol.
Title: Re: Questions
Post by: jsimmo on June 22, 2008, 06:15:37 pm: he didnt explain it well if it was
Title: Re: Questions
Post by: Rosie on June 29, 2008, 11:44:35 am: Q. A man whose eye is 1.7m from the ground when standing 3.5m in front of a wall 3m high can just see the top of a tower that is 100m away from the wall. Find the height of the tower.
How would I make the draw the two similar triangles because I can't understand what the question is saying?

Q. A spotlight is at a height of 0.6m above ground level. A vertical post 1.1m high stands 3m away, and 5m further away there is a vertical wall. How high up the wall does the shadow reach?

thanks
Title: Re: Questions
Post by: cara.mel on June 29, 2008, 12:22:13 pm: First one:
If you draw it out, it looks like this. Always draw the question out. (tower is blue, wall is red, person is a lovely Stick Figure
(http://img.photobucket.com/albums/v711/happypuff/Towerquestion.jpg)

Answer below (try to work it out for yourself first :P)
1.3/3.5 = x/(100 + 3.5)
x = 103.5*1.3/3.5 = 38.4
This is the tower's height above eye level. The actual height of the tower = 38.4 + 1.7 = 40.1m

Second question is essentially the same with different numbers
Title: Re: Questions
Post by: ninwa on June 29, 2008, 08:01:03 pm: hey, has any1 in this forum been to tsfx further intense workshops last semester??????
how was??????????
i reallly need feedback? is it worth going? do they give you bound reference? good notes?
Title: Re: Questions
Post by: Rosie on July 18, 2008, 04:46:23 pm: Q. A and B are two positions on level ground. From an advertising balloon at a vertical height of 750m, A is observed in an easterly direction and B at a bearing 160 degrees. The angles of depression of A and B as viewed from the balloon are 40 degrees and 20 degrees respectively. Find the distance between A and B.

Can someone help me with this question. I'm finding it hard to draw a diagram.
Title: Re: Questions
Post by: Rosie on July 19, 2008, 07:23:31 pm: *bump
Title: Re: Questions
Post by: /0 on July 19, 2008, 07:43:35 pm: Quote from: Rosie on July 18, 2008, 04:46:23 pm
Q. A and B are two positions on level ground. From an advertising balloon at a vertical height of 750m, A is observed in an easterly direction and B at a bearing 160 degrees. The angles of depression of A and B as viewed from the balloon are 40 degrees and 20 degrees respectively. Find the distance between A and B.

Can someone help me with this question. I'm finding it hard to draw a diagram.

Let the balloon be at point O. Let the horizontal distance of points A and B from the balloon be $OA$ and $OB$ respectively.
From the information:

$\tan {40^{\circ}}=\frac{750}{OA} \Rightarrow OA = \frac{750}{\tan{40^{\circ}}}$

$\tan{20^{\circ} = \frac{750}{OB} \Rightarrow OB = \frac{750}{\tan{20^{\circ}}}$

Now, by the Law of cosines,

$AB^2 = OA^2+OB^2 -2(OA)(OB)\cos{(160^{\circ}-90^{\circ})}$

$\Rightarrow AB = \sqrt{\left(\frac{750}{\tan{40^{\circ}}}\right)^2+\left(\frac{750}{\tan{20^{\circ}}}\right)^2-2\left(\frac{750}{\tan{40^{\circ}}}\right)\left(\frac{750}{\tan{20^{\circ}}}\right)\cos{70^{\circ}}$

This comes out to $1945.5 \mbox{m}$ . You could probably use componentizing to solve the last bit instead of the cosine law... and there's probably an easier way to do this...
Title: Re: Questions
Post by: Rosie on September 15, 2008, 08:12:38 pm: Urgent question!!!

MATRICES
Q. How many of the following five sets of simultaneous linear equations have a unique solution?

1. 4x + 2y = 10
2x + y = 5

2. x = 0
x + y = 6

3. x - y = 3
x + y = 3

4. 2x + y = 5
2x + y = 10

5. x = 8
y = 2

Answer choices
A. 1
B. 2
C. 3
D. 4
E. 5

How do i work out the answer especially the ones that dont have an equation, like x = 0, x = 8 etc.
Title: Re: Questions
Post by: Mao on September 15, 2008, 08:23:41 pm: 2 and 5 are trick sets

for 2, if x=0, then y must be 6. distinct solution

for 5, x=8, y=2, distinct solution
Title: Re: Questions
Post by: Rosie on September 15, 2008, 08:29:35 pm: ok so your saying that if x and y have a number greater than 0, there will be a unique solution
Title: Re: Questions
Post by: shinny on September 15, 2008, 08:36:03 pm: Quote from: Rosie on September 15, 2008, 08:12:38 pm
Urgent question!!!

MATRICES
Q. How many of the following five sets of simultaneous linear equations have a unique solution?

1. 4x + 2y = 10
2x + y = 5

2. x = 0
x + y = 6

3. x - y = 3
x + y = 3

4. 2x + y = 5
2x + y = 10

5. x = 8
y = 2

Answer choices
A. 1
B. 2
C. 3
D. 4
E. 5

How do i work out the answer especially the ones that dont have an equation, like x = 0, x = 8 etc.

Well for 1, the two equations are actually the same if you multiply the second one by two (and hence there is only one equation for 2 variables), and thus there are an infinite number of solutions. For 2, Mao's proved that already. 3, since there's two different equations, there will be a unique solution (x=3, y=0). For 4, since you have the same equation equal to two different numbers, there cannot be any possible solution, and thus there are no unique solutions. For 5, you don't need to solve anything; it's already given. So thus, C: 3

EDIT: and no, thats not what he's saying. Simultaneous equations will have a unique solution when there are as many unique equations as there are variables. i.e. in this case, you need two unique equations to solve for two variables, x and y. A unique solution just means that there is only one set of x and corresponding y values that can satisfy both equations to be true. For example in one, since both equations mean that 2x+y=5, you could have x=5 and y=-5, or x=3 and y=-1 and so on; there are an infinite number of possibilities. However in 4, there are no values that would satisfy both equations as 2x+y can equal both 5 and 10, which makes no sense at all.
Title: Re: Questions
Post by: rayray_t on September 28, 2008, 11:29:26 pm: A simpler way to solve that problem is to just write out the 'left hand side' of the equation in matrix form and work out the determinant. Any determinant that is equal to 0 will not have a unique solution.
Title: Re: Questions
Post by: Rosie on October 23, 2008, 04:21:45 pm: I just did the 2007 exam 1 vcaa paper and i am really stuck on Q7 of NUMBER PATTERNS. The answer is E but i am really sure that it is B. Can anyone help.
Title: Re: Questions
Post by: Rosie on October 23, 2008, 05:21:52 pm: In addition to this question, i am having trouble with getting the answers to two other questions.

Q. A closed cubic box of side length 36cm is to contain a thin straight metal rod. The maximum possible length of the rod is closest to :
A. 36cm
B. 51cm
C. 62cm
D. 108cm
E. 216cm

i thought that the possible length of the rod can fit diagonal in the box and i get B. But the answer is c. any help
Title: Re: Questions
Post by: cara.mel on October 23, 2008, 05:31:40 pm: sqrt(36^2 + 36^2 + 36^2) = 62.4 => is C

It would be B if it was 2D (you were putting the rod in a square)

so for 3 dimensions you add all 3 sides (same for any number). I can visualise this in my head, but if you can't, don't worry about it :)
Title: Re: Questions
Post by: Rosie on October 25, 2008, 08:13:39 am: Quote from: Rosie on October 23, 2008, 04:21:45 pm
I just did the 2007 exam 1 vcaa paper and i am really stuck on Q7 of NUMBER PATTERNS. The answer is E but i am really sure that it is B. Can anyone help.

help needed with this question please
Title: Re: Questions
Post by: Odette. on October 25, 2008, 08:47:31 am: Quote from: Rosie on October 25, 2008, 08:13:39 am
Quote from: Rosie on October 23, 2008, 04:21:45 pm
I just did the 2007 exam 1 vcaa paper and i am really stuck on Q7 of NUMBER PATTERNS. The answer is E but i am really sure that it is B. Can anyone help.

help needed with this question please

Rosie, if it were B, the common ratio would have had to be between 0 and 1, however in the question it was stated that the common ratio was negative, hence E is the correct answer as it shows a graph that has a negative common ratio. If you need further explanation on this question, just ask.

Oh and another thing, you should have model graphs in your notes to help you, as in when r is >1, 0<r<1, -1< r<0 and r< -1
Title: Re: Questions
Post by: Rosie on November 02, 2008, 04:44:34 pm: can the Heron's formula only be used if we are given 3 sides of a triangle and not if we have found a side as this would lead to rounding errors. is that right?
Title: Re: Questions
Post by: Noblesse on November 02, 2008, 04:51:50 pm: Well I thought, usually if you have worked out a side and rounded it to the question specification (eg. to 2 decimal places). You are allowed to use that rounded answer in subsequent calculations. Anyone else heard this?

Generally, I prefer to use the 1/2*1side*2side*sin(x) instead, because that is all that is required to work out the other side, so you could use that instead. (unless of course you used SAA form to find the opposite side.)
Title: Re: Questions
Post by: vce01 on November 02, 2008, 09:30:57 pm: is it possible to graph a residual plot without finding the regression line guys? or do you HAVE to find it first before you can plot it? (just a tricky ques for me i wanted to clarify)
Title: Re: Questions
Post by: Noblesse on November 02, 2008, 09:36:03 pm: Fairly sure you have to find the regression line, although the calculator can do it without the regression and simply use the StatPlot function.
Title: Re: Questions
Post by: Mikey123 on November 02, 2008, 09:44:26 pm: Need a formular of effective interest guys.
Title: Re: Questions
Post by: Noblesse on November 02, 2008, 09:47:29 pm: (http://img508.imageshack.us/img508/5213/rese3.png)
Title: Re: Questions
Post by: vce01 on November 02, 2008, 09:49:12 pm: ^ thats from the essential book right? are you going to use that for your bound reference?
Title: Re: Questions
Post by: Mikey123 on November 02, 2008, 09:50:07 pm: Love ya
Title: Re: Questions
Post by: Noblesse on November 02, 2008, 09:52:26 pm: Quote from: vce01 on November 02, 2008, 09:49:12 pm
^ thats from the essential book right? are you going to use that for your bound reference?

Yeah it's from essential, but no, I made a bound reference book throughout the year - saved me a lot of hassle now! (more time for prac exams!)
Title: Re: Questions
Post by: Mikey123 on November 02, 2008, 09:58:14 pm: Okay anyone got a page with bearing formulas and hints?

I think my reference lacks that.
Title: Re: Questions
Post by: vce01 on November 02, 2008, 11:03:58 pm: neap exam 1 06 - q10 on core.

cant it be 1/y as well?
Title: Re: Questions
Post by: *Roxxii* on November 03, 2008, 08:54:04 am: Quote from: vce01 on November 02, 2008, 11:03:58 pm
neap exam 1 06 - q10 on core.

cant it be 1/y as well?

For data in that sort of shape, yes, either x^2, Log(y) or 1/y can be used to linearise the data.
But not all 3 give the same effect, some will linearise the data more than others.

For the question, it is asking which will be MOST likely to linearise the data.
From the graph, we can clearly see that the shape of the points resembles a parabola, not a hyperbolic function or a log function. Leaving you with the answer of C. x^2

Do you do methods by any chance?
I have only recently picked up that if you are aware of what each graph looks like, ie hyperbolas, logs, parabolas...
It is very useful in determining which transformation to use :)
Title: Re: Questions
Post by: Noblesse on November 03, 2008, 08:55:25 am: So, if the graph looks like a log graph then a log function will be the most likely to linear the graph the best? How would you know to log(x) or log(y)?
Title: Re: Questions
Post by: vce01 on November 03, 2008, 09:03:11 am: Quote from: *Roxxii* on November 03, 2008, 08:54:04 am
Quote from: vce01 on November 02, 2008, 11:03:58 pm
neap exam 1 06 - q10 on core.

cant it be 1/y as well?

For data in that sort of shape, yes, either x^2, Log(y) or 1/y can be used to linearise the data.
But not all 3 give the same effect, some will linearise the data more than others.

For the question, it is asking which will be MOST likely to linearise the data.
From the graph, we can clearly see that the shape of the points resembles a parabola, not a hyperbolic function or a log function. Leaving you with the answer of C. x^2

Do you do methods by any chance?
I have only recently picked up that if you are aware of what each graph looks like, ie hyperbolas, logs, parabolas...
It is very useful in determining which transformation to use :)

yeah i do methods, but not too good at it.

they wouldn't ask that type of question then though would they? further students not doing methods otherwise will be disadvantaged

another question...say for an example you're given a curve, like a parabola over the domain of R+ and a point of umm (4,16) is given. is it necessary for us to know how to find the rule connecting y and x and if so, how would you go about finding it?
Title: Re: Questions
Post by: *Roxxii* on November 03, 2008, 09:03:41 am: Quote from: Jamison on November 03, 2008, 08:55:25 am
So, if the graph looks like a log graph then a log function will be the most likely to linear the graph the best? How would you know to log(x) or log(y)?

Hmm, i have not yet established that :P (me= phail)
But we have a reference book, so there is no need to learn which transformations to use off by heart. yayyy :D

But i for a couple of the exams i did, i forgot to bring my reference book to the library so i actually worked out my answers using this method, without any reference material, and it worked for me lol. (or maybe i was just lucky...hmm!)
So yeah, ruled out a few questions first, but it was still a little bit of guesswork.

I might look into it a bit more after my exams when i have time :P
Title: Re: Questions
Post by: *Roxxii* on November 03, 2008, 09:15:23 am: Quote from: vce01 on November 03, 2008, 09:03:11 am
yeah i do methods, but not too good at it.

they wouldn't ask that type of question then though would they? further students not doing methods otherwise will be disadvantaged

another question...say for an example you're given a curve, like a parabola over the domain of R+ and a point of umm (4,16) is given. is it necessary for us to know how to find the rule connecting y and x and if so, how would you go about finding it?

This seems like Graphs and relations...hmm...
Usually, they will give you a question with like in your example, a parabola. (on the axis, x and y)
Then they will give a set of graphs with the axis labled y and x^2, with different points on it.
You will be asked which one is a correct representation of the given parabola.
Or they might do it the other way around.

Ok, i would first find the equation of the parabola, which the point of (4,16) lies on, giving you y=x^2
Then to find out which points lie on the graph with axis labeled y and x^2, simply delete the ^2 from your equation.
In this case ( u have given an easy example :P) The y=x^2 simply turns into y=x
You can then sub in certain the x-values into the equation, giving you the y-value...to find the co-ordinates that match on of the answers.

ie, if you were given 5 choices.
All with y and x^2 on the axis, but with the different points...
A. (2,4) B. (2,8) C. (2,2) D. (4,2) E. (4,16)

Your answer would be C. (2,2) as this point lies on the graph of y=x.

I hope this was what you were asking about, and that i havent written all this for nothing :P lolz
But this was a question is similar to a question that was pretty poorly handled on last years exam.
So i suggest people to learn this, as i am predicting another question of this sort in Graphs and relations today :P lolz
Title: Re: Questions
Post by: jsimmo on November 03, 2008, 10:09:36 pm: Can someone please tell me how the answer in NEAP 2005 (exam 2) question

1 bi. = (0,70) **I got (12,66)
1 bii. = (12,66) **I got (42.85, 42.85)

I think they stuffed up their answers? Agreed?