Author Topic: Mathematics Question Thread (Read 1312253 times) Tweet Share

Dragomistress · « **Reply #2520 on:** September 02, 2017, 04:49:47 pm »

This will probably be a pain for you to answer but...
I need help with all questions except for question 12 a and b.

RuiAce · « **Reply #2521 on:** September 02, 2017, 04:54:46 pm »

Quote from: Dragomistress on September 02, 2017, 04:49:47 pm

This will probably be a pain for you to answer but...
I need help with all questions except for question 12 a and b.

$\text{Note that in 18, we have the product of roots:}\\ \alpha \beta = \frac{k}{k} = 1$
$\therefore \beta = \frac1\alpha\\ \text{and thus the roots are reciprocals}$
____________________________
$\text{From a relevant Pythagorean identity}\\ \begin{align*}3-2\cos^2 x - 3\sin x &=0\\ 3-2(1-\sin^2x)-3\sin x &=0\\ 2\sin^2 x - 3\sin x+1\\ (2\sin x - 1) (\sin x - 1)&=0\\ \sin x &=\frac12, 1\end{align*}$
$\text{The only solution to }\sin x = 1\text{ over }0^\circ \le x \le 360^\circ\text{ is }x=90^\circ$
$\text{Considering the first and second quadrants,}\\ \sin x = \frac12 \implies x = 30^\circ,150^\circ$
$\therefore \boxed{x=30^\circ,90^\circ,150^\circ}$

RuiAce · « **Reply #2522 on:** September 03, 2017, 11:13:58 am »

Quote from: Dragomistress on September 02, 2017, 04:49:47 pm

This will probably be a pain for you to answer but...
I need help with all questions except for question 12 a and b.

$\text{12c: If one root is double the other,}\\ \text{then the roots are }\alpha\text{ and }2\alpha$
$\text{It follows from the sum of roots that }\\ 3\alpha = -3 \implies \alpha = -1\\ \text{Hence the roots are actually }-1\text{ and }-2$
$\text{and hence from the product of roots, }(k-1) = 2 \implies k = 3$
_________________
$\text{If the roots are reciprocals, then they are }\alpha\text{ and }\frac1{\alpha}\\ \text{Immediately by the product of roots,}\\ \frac\alpha\alpha = k-1 \implies k = 2$
_________________
$\text{d) is a bit trickier, because you need to know what consecutive means}\\ \text{but if they're consecutive numbers}\\ \text{then the roots are really just }\alpha\text{ and }\alpha+1$
(Because consecutive numbers are numbers one after the other.)
$\text{From the sum of roots, }2\alpha + 1 = -3 \implies \alpha = -2\\ \text{so the roots are }-2\text{ and }-1$
As it so stands, this coincides with c).

15 follows a similar procedure to 14 and you should now give it a go using similar ideas, and post any progress if you remain stuck.

Kevinl123 · « **Reply #2523 on:** September 03, 2017, 11:07:27 pm »

Carbon-14 is a radioactive substance with a half-life of 5730 years. It is used to
determine the age of ancient objects. A Babylonian cloth fragment now has 40% of the
carbon-14 that it contained originally. How old is the fragment of cloth?

RuiAce · « **Reply #2524 on:** September 03, 2017, 11:12:09 pm »

Quote from: Kevinl123 on September 03, 2017, 11:07:27 pm

Carbon-14 is a radioactive substance with a half-life of 5730 years. It is used to
determine the age of ancient objects. A Babylonian cloth fragment now has 40% of the
carbon-14 that it contained originally. How old is the fragment of cloth?

$\text{These types of questions are standard and you are expected to know the usual procedures}$
$\text{Suppose the initial mass of C-14 is }M_0\\ \text{Then, we have }\boxed{M = M_0 e^{-kt}}$
$\text{The half life is 5730 years, so only 50 percent of the substance remains after }t=5730\\ \text{i.e. when }t=5730, M = \frac{M_0}2$
$\text{Hence}\\ \begin{align*}\frac{M_0}{2} &= M_0 e^{-5730k}\\ \frac12 &= e^{-5730k}\\ \ln \frac12 &= -5730k\\ -k &= \frac{1}{5730}\ln \frac12 \end{align*}$
___________________________________
$\text{So }\boxed{M=M_0e^{\frac{t}{5730}\ln \frac12 }}\\ \text{We want to find }t\text{ when }M=0.4M_0$
$\begin{align*}\frac{2}{5}M_0 &= M_0 e^{\frac{t}{5730}\ln \frac12 }\\ \frac25 &= e^{\frac{t}{5730}\ln \frac12 }\\ \ln \frac25 &= \frac{t}{5730}\ln \frac12 \end{align*}\\ \text{You should be able to finish it off from here}$

amberberrie · « **Reply #2525 on:** September 05, 2017, 08:51:16 pm »

Please help with this question!! No one I ask can seem to figure it out!

Much appreciated,
Amber

RuiAce · « **Reply #2526 on:** September 05, 2017, 09:12:45 pm »

Quote from: amberberrie on September 05, 2017, 08:51:16 pm

Please help with this question!! No one I ask can seem to figure it out!

Much appreciated,
Amber

$\text{Note that it is clear that the dice rolls and coin tosses are independent.}$
$\text{The probability of two heads and double six is simply}\\ \left(\frac12\right)^2 \times \left(\frac16\right)^2 = \frac1{144}$
Remark that this is also the probability of two tails and double six.
_____________________
$\text{The probability of two tails is still }\left(\frac12\right)^2\\ \text{As we are investigating dice rolls, we may simply list outcomes.}$
Favourable outcomes: (1,1), (2,2), (3,3), (4,4), (5,5)
Total possible outcomes: 36 of them
$\text{Hence our required probability is }\left(\frac12\right)^2 \times \frac5{36}=\frac5{144}$
Remark that this is also the probability of two heads and a double that's not double six
_____________________
$\text{Note that the only ways to score 13 are}\\ 5,3,3\\ 5,4,4$
$\text{The probability that }5,3,3,\text{ occurs}\\ \textbf{in that particular order}\\ \text{ is } \frac{1}{144}\times \frac5{144}\times \frac5{144}$
$\text{But we can have 5,3,3 or 3,5,3 or 3,3,5}\\ \text{i.e. three different ways of this occuring}\\ \text{So the required probability that this case occurs is }3\times \frac{25}{144^3}$
$\text{Furthermore, the probability that 5,4,4 occurs in that order is}\\ \frac{1}{144^3}\\ \text{But again, there are three possible ways this can occur}\\\text{So this case yields }\frac{3}{144^3}$
$\text{Hence our required probability is }\frac{75 + 3}{144^3}$
_____________________
$\text{Justin must score either 14 or 15.}$
$\text{The only way 15 can be attained is by scoring 5,5,5}\\ \text{This can only happen in one way, and has probability }\frac{1}{144^3}$
$\text{The only way 14 can be attained is by scoring 5,5,4 in any order}\\ \text{Note that once again this can only happen in one way.}\\ \text{The probability a particular way this happens is }\frac1{144}\times \frac1{144} \times \frac1{144} = \frac{1}{144^3}\\ \text{So the probability this case happens is }\frac{3}{144^3}$
$\text{Implying that when we regroup the cases,}\\ \text{Our required probability is }\frac4{144^3}$
It was also clear that this question came out of Maths in Focus (so therefore I went to check after I was done typing). Their answers for parts c) and d) look completely unrealistic.

arunasva · « **Reply #2527 on:** September 06, 2017, 01:07:45 am »

I'm like really bad in Probability. I thought practicing would solve it like anything else in life, however turns out I'm not even good with the concepts. Like what to multiply what to add, why multiply these fractions what kinda scenarios exist etc. Is there a really good guide on probability or advice for that topic ? Thanks

RuiAce · « **Reply #2528 on:** September 06, 2017, 08:47:53 am »

Quote from: arunasva on September 06, 2017, 01:07:45 am

I'm like really bad in Probability. I thought practicing would solve it like anything else in life, however turns out I'm not even good with the concepts. Like what to multiply what to add, why multiply these fractions what kinda scenarios exist etc. Is there a really good guide on probability or advice for that topic ? Thanks

And means Multiply
Or means Plus

i.e. When you're considering one single case and there's more than one factor involved, you're multiplying. But when you're regrouping seperate cases, you add.

arunasva · « **Reply #2529 on:** September 06, 2017, 10:16:08 pm »

Quote from: RuiAce on September 06, 2017, 08:47:53 am

And means Multiply
Or means Plus

i.e. When you're considering one single case and there's more than one factor involved, you're multiplying. But when you're regrouping seperate cases, you add.

makes more sense than anything ive read on that topic T.Thanks

hansolo9 · « **Reply #2530 on:** September 07, 2017, 12:55:06 pm »

Hello

May I have help with this question? (Part a)
Excuse the scribbling

thanks

mary123987 · « **Reply #2531 on:** September 07, 2017, 01:28:34 pm »

Hey guys I was just wondering if anyone knows what to do if their is a horizontal assymptote and the question asks you to sketch the derivative just for reference i'm referring to question 9c 2011 HSC 2 unit paper
regards Mary

pikachu975 · « **Reply #2532 on:** September 07, 2017, 01:35:33 pm »

Quote from: mary123987 on September 07, 2017, 01:28:34 pm

Hey guys I was just wondering if anyone knows what to do if their is a horizontal assymptote and the question asks you to sketch the derivative just for reference i'm referring to question 9c 2011 HSC 2 unit paper
regards Mary

At x = 1 f'(x) intercepts the x axis as f(x) has a stationary point
At x = 3 f'(x) has a stationary point since f(x) has a point of inflection
I think on f'(x) the horizontal asymptote would be y = 0 since the gradient approaches 0, and it would approach from the negative side

mary123987 · « **Reply #2533 on:** September 07, 2017, 02:03:17 pm »

oh right thankyou !!

left right gn · « **Reply #2534 on:** September 07, 2017, 05:06:04 pm »

Hey! can someone explain how to do this question??
I'm usually stumped when questions like these show up.

Search the forums now!

We have moved!

Author Topic: Mathematics Question Thread (Read 1312253 times) Tweet Share

Dragomistress

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

Kevinl123

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

amberberrie

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

arunasva

Re: Mathematics Question Thread

RuiAce

Re: Mathematics Question Thread

arunasva

Re: Mathematics Question Thread

hansolo9

Re: Mathematics Question Thread

mary123987

Re: Mathematics Question Thread

pikachu975

Re: Mathematics Question Thread

mary123987

Re: Mathematics Question Thread

left right gn

Re: Mathematics Question Thread

Recent Posts

User:
Password: