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March 29, 2024, 04:41:47 pm

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AlphaZero

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Re: Mathematics Question Thread
« Reply #4020 on: February 18, 2019, 09:50:29 pm »
+3
Can someone please help with the working out of these? I tried using log but I think I'm doing something wrong  :'( :'(
1. Which term of the series 8 - 4 + 2 - ... is 1/128?
2. Which term of 54 + 18 + 6 + ... is 2/243?
3. Find the value of n if the nth term of the series -2 + 3/2 - 9/8 + ... is -81/128.

Indeed, using logarithms may bring a little trouble in Q1 and Q3 since the common ratio of the sequences are negative. But, we can avoid these if we just make some clever observations.

Question 1\begin{align*}\frac{1}{128}&=8\times \left(\frac{-1}{2}\right)^{n-1}\\
\implies \left(\frac{-1}{2}\right)^{n-1}&=\frac{1}{1024}\\
\implies \frac{(-1)^{n-1}}{2^{n-1}}&=\frac{1}{2^{10}}\tag{1}\end{align*} Now, let's look back at the original sequence. Clearly, the \(n\)th element of the sequence is positive only if \(n\) is odd. Notice that, if \(n\) is odd, then  \(n-1\) is even, and so equation \((1)\) becomes \[\frac{1}{2^{n-1}}=\frac{1}{2^{10}}.\]Thus, \(n=11\).

I'll let you try Q2 and Q3 on your own. If you're still having trouble, don't hesitate to ask again :)
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emilyyyyyyy

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Re: Mathematics Question Thread
« Reply #4021 on: February 20, 2019, 12:35:20 pm »
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Hi, with this question, i've drawn the graph and found the point of intersection. I've integrated what I think I have to, but I keep getting the wrong answer! How do I solve: Find the area enclosed between the curve y = x^3, the x-axis and the line y = -3x + 4.

Thanks!

RuiAce

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Re: Mathematics Question Thread
« Reply #4022 on: February 20, 2019, 01:41:06 pm »
0
Hi, with this question, i've drawn the graph and found the point of intersection. I've integrated what I think I have to, but I keep getting the wrong answer! How do I solve: Find the area enclosed between the curve y = x^3, the x-axis and the line y = -3x + 4.

Thanks!
Upon re-inspecting the question, because the region is bound also by the \(x\)-axis, it is the small triangle-like shape below both curves. This is a compound region, whose area is described by \( \int_0^1 x^3\,dx + \int_1^{4/3} (-3x+4)\,dx \).
« Last Edit: February 20, 2019, 01:44:22 pm by RuiAce »

emilyyyyyyy

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Re: Mathematics Question Thread
« Reply #4023 on: February 20, 2019, 02:08:56 pm »
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thank you!
I get that the intercept for the line y=-3x+4 is 4/3, but why do I use that instead of 1 when integrating the line, seeing as 1 is where the line and curve intersect?

AlphaZero

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Re: Mathematics Question Thread
« Reply #4024 on: February 20, 2019, 02:36:24 pm »
0
thank you!
I get that the intercept for the line y=-3x+4 is 4/3, but why do I use that instead of 1 when integrating the line, seeing as 1 is where the line and curve intersect?

Not too sure what you mean by using 4/3 instead of 1. We require both values. This image may help though.



We are trying to find the area of the red and blue regions. \[\text{The area of the red region is given by }\int_0^1 x^3\,dx.\] \[\text{The area of the blue region is given by }\int_1^{4/3} (-3x+4)\,dx.\]To find the total area, just add the two together.
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Kalliope.m

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Re: Mathematics Question Thread
« Reply #4025 on: February 20, 2019, 04:13:37 pm »
0
Hi, I’m struggling a bit with this question:

The first term of a geometric series is 3 and the common ratio is 4. Find the first term of the series that exceeds 300 000.

I got the inequality to find n, but I think I might be solving for it wrong. Could someone please help out? Thank you :)
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AlphaZero

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Re: Mathematics Question Thread
« Reply #4026 on: February 20, 2019, 04:49:56 pm »
+1
Hi, I’m struggling a bit with this question:

The first term of a geometric series is 3 and the common ratio is 4. Find the first term of the series that exceeds 300 000.

I got the inequality to find n, but I think I might be solving for it wrong. Could someone please help out? Thank you :)

So, this question asks us essentially to find the minimal value of \(n\) for which  \(3\times 4^{n-1}\geq300\,000\). There are a couple of ways to go about this. I've presented two methods below.

Method 1: (probably the better method)
First note that the sequence given by  \(T_n=3\times 4^{n-1}\) is clearly strictly increasing. So, one way we could go about this problem is by solving  \(3\times 4^{n-1}=300\,000\) and rounding our answer up to the nearest integer: \begin{align*}4^{n-1}&=300\,000\\
n-1&=\log_4(300\,000)\\
n&=1+\log_4(300\,000)\approx 9.30\ \ (\text{2DP})\end{align*} Thus, the first element to exceed \(300\,000\) is the \(10\)th.

Method 2: (trial and error)
You would find that \begin{align*}T_8&=49\,152<300\,000\\
T_9&=196\,608<300\,000\\
T_{10}&=786\,432>300\,000.\end{align*}Thus, our answer is the \(10\)th element.
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AlphaZero

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Re: Mathematics Question Thread
« Reply #4027 on: February 20, 2019, 10:42:25 pm »
+2
Can someone please help with these series questions? Sorry for asking a lot but I'm really stuck :'(
1. Lucia currently earns $25000. Her wage increases by 5% each year. Find her total earnings (before tax) in 6 years.
2. Write down an expression for the series 2 - 10 + 50 - ... + 2(-5[power k - 1]) in sigma notation and as a sum of n terms.
3. Find the sum of the first 10 terms of the series 3 + 7 + 13 + ... + [2(power n) + (2n - 1)] + ...

There is nothing wrong with asking a lot of questions :)

Question 1

Edit: I missed the word "total" in the question. Apologies.
\[S_6=\frac{\$25\,000(1.05^6-1)}{1.05-1}=\$170\,048.82\]

Question 2

This is a geometric series whose first term is \(2\) and common ratio is \(-5\).

So, the \(k\)th term of the series is given by  \(T_k=2\times(-5)^{n-1}\).

Therefore, the sum of the first \(n\) terms, \(S_n=T_1+T_2+\dots+T_n\) in sigma notation is just given by \[\sum_{k=1}^nT_k=\sum_{k=1}^n 2\!\times\!(-5)^{k-1}.\]
Question 3

In this question, \begin{align*}S=\sum_{k=1}^{10}\Big[2^k+2k-1\Big]&=\sum_{k=1}^{10}2^k+\sum_{k=1}^{10}(2k-1)\\
&=\underbrace{\sum_{k=1}^{10}2\!\times\! 2^{k-1}}_{\text{geometric series}}+\underbrace{\sum_{k=1}^{10}\big[1+2(k-1)\big]}_{\text{arithmetic series}}\end{align*}
The first series is just a geometric series with first term \(a=2\) and common ratio \(r=2\).

The second series is just an arithmetic series with first term \(a=1\) and common difference \(d=2\). Therefore, \begin{align*}S&=\frac{2(2^{10}-1)}{2-1}+\frac{10}{2}\big[2\!\times\! 1+2(10-1)\big]\\
&=2146\end{align*}
« Last Edit: February 21, 2019, 12:29:59 am by AlphaZero »
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RuiAce

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Re: Mathematics Question Thread
« Reply #4028 on: February 20, 2019, 11:40:03 pm »
+5
Thank you so much for taking the time to answer my questions. :)
In regards to question 1, I kept getting $170047.82 (2dp) as my answer??
In regards to question 2,  how do you know that k=1 and how come you need to change from k to n? I don't know when you have to do that and which part you need to change.
Apologies, I just need some clarification.
\[ \text{His solution is slightly off for Q1.}\\ \text{We're interested in the }\textbf{total}\text{ earnings accumulated.}\\ \text{Not just the sixth period itself.} \]
\[ \text{Because the earnings in the }n\text{-year can be modelled by}\\ \text{a geometric sequence, with }a=25000\text{ and }r=1.05,\\ \text{what we're interested in is }S_6 = 25000\left( \frac{1.05^6-1}{0.05} \right) \]
Which evaluates to the given answer.

With Q2, that one has more than one correct answer in regards to what \(k\) we can start from. We can start from whatever value of \(k\) we like, so long as it works. Starting with \(k=1\), we have \( \sum_{k=1}^n 2(5)^{k-1} \) like he wrote. But we could, for example, start from \(k=0\) to obtain \( \sum_{k=0}^{n-1}2(-5)^k \). Or we could start from say \(k=2\), and obtain \( \sum_{k=2}^{n+1} 2(5)^{k-2}\).

Convention is typically to start with \(k=1\) though as far as the HSC course is concerned.

I am not sure what your second question is about with "change from k to n". EDIT: Looking again, I see that your expression in Q2 was in terms of \(k\) to begin with. So something like this would've been better, for example.
\[ \sum_{j=1}^k 2(5)^{j-1} \]
« Last Edit: February 20, 2019, 11:47:05 pm by RuiAce »

RuiAce

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Re: Mathematics Question Thread
« Reply #4029 on: February 21, 2019, 07:43:17 pm »
0
Can someone help with this question? I do not know how to write the equation or what formula to use. It is a Fibonacci problem which is that 1, 1, 2, 3, 5, 8, ... thing.
A man entered an orchard through 7 guarded gates and gathered a certain number of apples. As he left the orchard he gave the guard at the first gate half the apples he had and 1 more. He repeated this process for each of the remaining 6 guards and eventually left the orchard with 1 apple. How many apples did he gather? (He did not give away any half apples.)
Thank you to whoever will answer this question.
Where is this question from? Through recurrence relations I obtain an answer of 382 (which seemed to work when I backtracked as well), but apart from annuity like applications recurrences are not a part of the current HSC course.

RuiAce

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Re: Mathematics Question Thread
« Reply #4030 on: February 21, 2019, 08:07:52 pm »
+4
What really? The Fibonacci thing is not in the HSC course? It is in the Year 12 2u Grove Textbook in Chapter 7 - Series and my teacher showed us the pattern but not a lot of examples.
May I still have the working out because my teacher assigned that bit for homework. I will clarify with my teacher whether we will be tested on that.
Ok I tracked down the question in the textbook and it looks like it's a puzzle. (And it got tagged with a hilarious "This is a hard one!" in the MX1 version at least.)

I would be extremely concerned if this appeared in any HSC exam. Nevertheless, here's the full recursive approach.
\[ \text{Let }T_n\text{ be the amount of apples the man has left}\\ \textbf{after }\text{giving the }n\text{-th guard his apples.} \]
Note that the question can be done using 'before' instead of 'after'. You may like to experiment using 'before' instead.
\[ \text{We're essentially given that }T_7 = 1.\\ \text{Our aim is to backtrack to find }T_0\\ \text{which is the amount before even the 1st guard got his apples.} \]
\[ \text{The recurrence of interest is}\\ \boxed{T_n = \frac12 T_{n-1} - 1}.\\ \text{This results from the take-half, then give one more apple scenario}\\ \text{the question presented.} \]
\[ \text{The recurrence relation can be rearranged to}\\ \boxed{T_{n-1} = 2(T_n + 1)}.\\ \text{Once we're here, we can start backtracking.}\\ \begin{align*} T_6 &= 2(1 + 1) = 4\\ T_5 &= 2(4+1) = 10\\ T_4 &= 2(10+1) = 22\\ T_3 &= 2(22+1) = 46\\ T_2 &= 2(46+1) = 94\\ T_1 &= 2(94+1) = 190\\ T_0 &= 2(190+1) = 382\end{align*} \]
\[ \text{So he originally had 382 apples.} \]
Note that the problem was put in the geometric series section, so it had to somehow be related to geometric series. It turns out that one can actually prove that \(T_n = 384(2^{-n}) - 2\). This looks almost like a geometric series, except there is an extra \(-2\) hanging on the end interfering with the pattern. The formal derivation more or less uses first year uni techniques and is hence avoided, but there may be an intuition behind this formula that I haven't had the time to investigate yet.

RuiAce

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Re: Mathematics Question Thread
« Reply #4031 on: February 21, 2019, 08:51:00 pm »
+4
Wow thank you so much for taking the time to type the working out for me. :D
I understand most of what is going on apart from how to get the "recurrence of interest" bit: Tn = 1/2 Tn -1 -1.
"_______ of interest" is just a fancy way of saying "the __________ that we are interested in". So in English, it's basically saying it's the correct recurrence relation for this particular question.

But as for why it works, it's simply just using what they gave us. We firstly give them half the number of apples we currently have, so therefore we still have the other half left - this is the \( \frac12 T_{n-1}\) bit. But then we need to give them one more apple. This leaves us with one more apple less that we have, and hence we have the trailing \(-1\) term in \( \frac12 T_{n-1} - 1 \).

Cherre Ho

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Re: Mathematics Question Thread
« Reply #4032 on: February 21, 2019, 09:20:34 pm »
+1
"_______ of interest" is just a fancy way of saying "the __________ that we are interested in". So in English, it's basically saying it's the correct recurrence relation for this particular question.

But as for why it works, it's simply just using what they gave us. We firstly give them half the number of apples we currently have, so therefore we still have the other half left - this is the \( \frac12 T_{n-1}\) bit. But then we need to give them one more apple. This leaves us with one more apple less that we have, and hence we have the trailing \(-1\) term in \( \frac12 T_{n-1} - 1 \).
Ahhhh I see. Yay I understand!! Thank you very much. You explain it very well and in great detail (thumbs up) :D

annabeljxde

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Re: Mathematics Question Thread
« Reply #4033 on: February 22, 2019, 10:29:12 pm »
0
I’ve been stuck on this question for days now! I just can’t wrap my head around what the question is asking...

A sector of a circle with a radius 5cm and an angle of π/3 subtended at the centre is cut out of cardboard. It is then curved around to form a cone. Find it’s exact surface area and volume.

Is the question asking us to use the SA and V formula for a cone or is the resulting cone empty (where the SA includes both the interior and exterior of the cone)?

Can anyone please help me? Thank you!
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RuiAce

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Re: Mathematics Question Thread
« Reply #4034 on: February 22, 2019, 10:34:39 pm »
+3
I’ve been stuck on this question for days now! I just can’t wrap my head around what the question is asking...

A sector of a circle with a radius 5cm and an angle of π/3 subtended at the centre is cut out of cardboard. It is then curved around to form a cone. Find it’s exact surface area and volume.

Is the question asking us to use the SA and V formula for a cone or is the resulting cone empty (where the SA includes both the interior and exterior of the cone)?

Can anyone please help me? Thank you!
That is a fair question, since they don't seem to tell you whether we should treat it as a 'sealed-off' cone or an empty cone. So why not try out both methods and see which one works?

(However, it is a bit tricky. Recall that the net of a cone looks like what's shown on this website. Note that the sector is wrapped around to form the apex of the cone. Therefore, that radius of 5 becomes the slant height of the cone. Furthermore, the arc length \( \ell = 5\times \frac\pi3\) becomes the circumference of the attached circle.)

Although having said that, this kind of ambiguity would not appear in the HSC. Where did this question originally come from?