 October 24, 2019, 12:25:49 am AuthorTopic: 3U Maths Question Thread  (Read 566728 times) Tweet Share

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diggity

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« Reply #4200 on: September 18, 2019, 08:57:56 am »
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Hullo, I need help with 2009 7) b & C). When I first did it, I was only able to do b) i), and going back I understand everything but C) i). So I was hoping someone could explain it, as the sample answers didn't help. Thank you!
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fun_jirachi

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« Reply #4201 on: September 18, 2019, 06:12:39 pm »
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Hey there!

Essentially we have that angle QSR = angle QTR since angles on the same arc are equal. Since T lies on the same plane as P, letting the distance from T to the wall be x1, we have that

Now clearly from the above function, as x increases, ϕ decreases (can be proven with a calculus approach or otherwise). Also, note that since x1 cannot be greater than x, ϕ cannot be less than θ. Essentially then, when S and T are the same point, ϕ=θ, which is the maximum for θ.

Hope this helps Failing everything, but I'm still Flareon up.

HSC 2018: Modern History  | 2U Maths 
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duncand

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« Reply #4202 on: September 27, 2019, 07:07:28 am »
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hi, I do not understand the answers to question 4 biii) in the 2011 hsc paper..
I am sure there is an easier alternative? b) In the diagram, the vertices of ABC lie on the circle with centre O. The point D lies on BC such that ABD is isosceles and ∠ABC = x.

(iii) Let M be the midpoint of AC and P the centre of the circle through A, C, D and O.
Show that P, M and O are collinear.
1

fun_jirachi

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« Reply #4203 on: September 27, 2019, 08:35:24 am »
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hi, I do not understand the answers to question 4 biii) in the 2011 hsc paper..
I am sure there is an easier alternative? The answer shows the easiest alternative: it's simply a theorem; albeit a relatively obscure one. Basically when you have two intersecting circles, the common chord ie. the chord between the two points has its midpoint collinear with the two centres.

Alternatively, you could prove the same result by considering that a line from the centres O and P to the midpoint of a chord on the circle M both meet at right angles, and hence must be part of the same line.

Hope this helps Failing everything, but I'm still Flareon up.

HSC 2018: Modern History  | 2U Maths 
HSC 2019: Physics | Chemistry | English Advanced | Maths Extension 1 | Maths Extension 2

mirakhiralla

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« Reply #4204 on: September 27, 2019, 06:41:29 pm »
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hey can someone pls do this q

fun_jirachi

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« Reply #4205 on: September 27, 2019, 07:12:01 pm »
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Hey there!

Next time, please indicate which part you're struggling with! Especially with exams coming up we really need to target missing links in your knowledge as opposed to doing specific questions for you; these links can be applied to infinitely many questions while this one question is only one of many Help us, help you! Part i) should be very familiar, perhaps one of the most common questions in projectile motion. Simply rearrange the horizontal displacement into a function of t in x, then substitute into the vertical displacement.

Part ii) needs you to substitute y=15, v=7sqrt10 and x=10 into the equation given in i). Note that since both pebbles pass through that point, both alpha and beta are solutions for theta in that equation, with alpha clearly being the smaller solution. Remember also that sec2x = 1 + tan2x, and you should be able to find a quadratic in tan theta, which you should then solve. The smaller solution will be tan alpha, as indicated, while the larger solution will be tan beta.

Part iii) takes a little more work. Draw up triangles with sides 2 and 1, such that tan alpha = 2 and sides 8 and 1, such that tan beta = 8. You'll find that for the pebble thrown at angle alpha, the time is

For the pebble thrown at angle beta, the time is
.
Since the pebble with angle beta was thrown first, the time difference is just (2) - (1), which is what is given in the question.

Hope this helps Failing everything, but I'm still Flareon up.

HSC 2018: Modern History  | 2U Maths 
HSC 2019: Physics | Chemistry | English Advanced | Maths Extension 1 | Maths Extension 2

mirakhiralla

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« Reply #4206 on: September 27, 2019, 07:42:30 pm »
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Hey there!

Next time, please indicate which part you're struggling with! Especially with exams coming up we really need to target missing links in your knowledge as opposed to doing specific questions for you; these links can be applied to infinitely many questions while this one question is only one of many Help us, help you! Part i) should be very familiar, perhaps one of the most common questions in projectile motion. Simply rearrange the horizontal displacement into a function of t in x, then substitute into the vertical displacement.

Part ii) needs you to substitute y=15, v=7sqrt10 and x=10 into the equation given in i). Note that since both pebbles pass through that point, both alpha and beta are solutions for theta in that equation, with alpha clearly being the smaller solution. Remember also that sec2x = 1 + tan2x, and you should be able to find a quadratic in tan theta, which you should then solve. The smaller solution will be tan alpha, as indicated, while the larger solution will be tan beta.

Part iii) takes a little more work. Draw up triangles with sides 2 and 1, such that tan alpha = 2 and sides 8 and 1, such that tan beta = 8. You'll find that for the pebble thrown at angle alpha, the time is

For the pebble thrown at angle beta, the time is
.
Since the pebble with angle beta was thrown first, the time difference is just (2) - (1), which is what is given in the question.

Hope this helps really sorry for not clarifying, I needed ii and iii
thank you so much!!

spnmox

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« Reply #4207 on: October 01, 2019, 11:04:51 am »
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Hey guys, need help with this P+C question:
Find out how many arrangements are possible from the word STUDIO if the letter O is not first or last

RuiAce

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« Reply #4208 on: October 01, 2019, 11:20:55 am »
+1
Hey guys, need help with this P+C question:
Find out how many arrangements are possible from the word STUDIO if the letter O is not first or last
In a letter configuration _ _ _ _ _ _, simply start by arranging O in a place that's not the first or last. There are $4$ ways of doing this.

Then the remaining 5 letters are rearranged without further restriction, which is doable in $5!$ ways.

Which gives a total of $4\times 5!=480$ arrangements.  HSC tutoring by ATAR Notes - learn more!

spnmox

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« Reply #4209 on: October 01, 2019, 12:00:28 pm »
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thank you!! and two more:

In how many ways can a committee of three women and 4 girls be chosen from 7 women and 6 girls so that if the eldest woman is serving on the committee, then the youngest girl is not? answer = 375

In how many ways can 4 physics books and 3 maths books be arranged on a shelf if a selection is made from 6 different physics books and 5 different maths books? in how many of these arrangements are all the physics books next to each other? answer = 6C4 * 5C3 * 7! = 756000, 86400

RuiAce

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« Reply #4210 on: October 01, 2019, 12:35:36 pm »
+2
thank you!! and two more:

In how many ways can a committee of three women and 4 girls be chosen from 7 women and 6 girls so that if the eldest woman is serving on the committee, then the youngest girl is not? answer = 375

In how many ways can 4 physics books and 3 maths books be arranged on a shelf if a selection is made from 6 different physics books and 5 different maths books? in how many of these arrangements are all the physics books next to each other? answer = 6C4 * 5C3 * 7! = 756000, 86400

The first one can be down in two approaches. One approach is to go by cases. The other is to consider the complement - we cannot have both the eldest women and youngest girl on the committee.

Here is the complement approach. Without any restrictions, we simply choose from $\binom73\binom64=525$ ways. If both the eldest women and youngest girl are selected, then we need only 2 more women out of 6, and 3 more girls out of 5. Which is done in $\binom62 \binom53 =150$ ways.

Which works out, since $525-150=375$.

(You may want to figure out how the cases work to build more intuition. I just provide the answer for now: $\binom63\binom54 + \binom62\binom54 + \binom63\binom53=375$ ways.)
________________________________________________________________________________________

The first part is the classic 'select-and-then-arrange' scenario, where we first actually choose the books in question. Then we deal with the arrangement only after this is done.

Now for the second one, begin in the same way. We select 4 out of 6 physics books and 3 out of 5 maths books in $\binom64\binom53$ ways.

The four physics books $P_1$, $P_2$, $P_3$ and $P_4$ then need to be grouped together into something like $\boxed{P_1P_2P_3P_4}$. However, since no specific ordering of the physics books was required; just that they are grouped, we first count the number of groups - $4!$ ways.

With the physics books grouped, the problem now boils down to just arranging the four "objects" $M_1$, $M_2$, $M_3$ and $\boxed{physics}$. There are four objects here, and hence this is also doable in $4!$ ways.

Final answer: $\binom64\binom534!4!=86400$ ways.

Huh, weird... that's also the number of seconds in a day!
« Last Edit: October 01, 2019, 12:38:12 pm by RuiAce »  HSC tutoring by ATAR Notes - learn more!

spnmox

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« Reply #4211 on: October 01, 2019, 01:27:54 pm »
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The first one can be down in two approaches. One approach is to go by cases. The other is to consider the complement - we cannot have both the eldest women and youngest girl on the committee.

Here is the complement approach. Without any restrictions, we simply choose from $\binom73\binom64=525$ ways. If both the eldest women and youngest girl are selected, then we need only 2 more women out of 6, and 3 more girls out of 5. Which is done in $\binom62 \binom53 =150$ ways.

Which works out, since $525-150=375$.

(You may want to figure out how the cases work to build more intuition. I just provide the answer for now: $\binom63\binom54 + \binom62\binom54 + \binom63\binom53=375$ ways.)
________________________________________________________________________________________

The first part is the classic 'select-and-then-arrange' scenario, where we first actually choose the books in question. Then we deal with the arrangement only after this is done.

Now for the second one, begin in the same way. We select 4 out of 6 physics books and 3 out of 5 maths books in $\binom64\binom53$ ways.

The four physics books $P_1$, $P_2$, $P_3$ and $P_4$ then need to be grouped together into something like $\boxed{P_1P_2P_3P_4}$. However, since no specific ordering of the physics books was required; just that they are grouped, we first count the number of groups - $4!$ ways.

With the physics books grouped, the problem now boils down to just arranging the four "objects" $M_1$, $M_2$, $M_3$ and $\boxed{physics}$. There are four objects here, and hence this is also doable in $4!$ ways.

Final answer: $\binom64\binom534!4!=86400$ ways.

Huh, weird... that's also the number of seconds in a day!

thank you!!!

spnmox

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« Reply #4212 on: October 01, 2019, 01:54:12 pm »
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hey guys, another question:

Thirty people apply for a special housing package, but only 12 packages are available, and are allocated randomly. What is the probability that
a particular person applying will get a house? answer =  1/ 86 493 225

my question is, why is the probability not (29C11)/(30C12)? Wouldn't that be the probability of a particular person being chosen / randomly choosing people?

fun_jirachi

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« Reply #4213 on: October 01, 2019, 02:35:53 pm »
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Might be incorrect/not have the best explanation, but here's what I think:

This indicates the chance that when we pick a random person after prizes have been allocated, that they have a prize, which isn't what we want; instead, we want the chance that when we pick a person beforehand, they'll actually get a prize.
This chance is equal to the below expression (ie. picking all the prizewinners out of all the entrants)

« Last Edit: October 01, 2019, 02:55:40 pm by fun_jirachi »
Failing everything, but I'm still Flareon up.

HSC 2018: Modern History  | 2U Maths 
HSC 2019: Physics | Chemistry | English Advanced | Maths Extension 1 | Maths Extension 2

spnmox

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« Reply #4214 on: October 01, 2019, 08:03:28 pm »
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Might be incorrect/not have the best explanation, but here's what I think:

This indicates the chance that when we pick a random person after prizes have been allocated, that they have a prize, which isn't what we want; instead, we want the chance that when we pick a person beforehand, they'll actually get a prize.
This chance is equal to the below expression (ie. picking all the prizewinners out of all the entrants)

I sort of understand your explanation, but I don't get what you did in the fraction.