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September 10, 2025, 01:52:48 am

Author Topic: Mathematics Question Thread  (Read 1626112 times)  Share 

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pikachu975

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Re: Mathematics Question Thread
« Reply #2610 on: September 27, 2017, 05:47:57 pm »
+1
How do we integrate y=ln(x-2) ?

This is 4 unit so there must be a previous part

set u = ln(x-2) and dv = dx
du = 1/(x-2) and v = x

so the integral = xln(x-2) - integral of x/(x-2)
= xln(x-2) - integral of (1 + 2/(x-2) )
= xln(x-2) - x - 2ln(x-2) + C

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Mathew587

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Re: Mathematics Question Thread
« Reply #2611 on: September 27, 2017, 05:56:46 pm »
+1
You don't have any techniques for doing this in 2U. Please post the full question.
From basic arithmetic/the calculator, 25 = 32.
oh its hsc 2008 q10 a)
why can't we just use dy/dx = f'x / fx?

This is 4 unit so there must be a previous part

set u = ln(x-2) and dv = dx
du = 1/(x-2) and v = x

so the integral = xln(x-2) - integral of x/(x-2)
= xln(x-2) - integral of (1 + 2/(x-2) )
= xln(x-2) - x - 2ln(x-2) + C

ahah i was gonna ask you personally XD
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Becky234

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Re: Mathematics Question Thread
« Reply #2612 on: September 27, 2017, 05:59:21 pm »
+1
You don't have any techniques for doing this in 2U. Please post the full question.
From basic arithmetic/the calculator, 25 = 32.
ok thanks

Hi!

(Image removed from quote.)

Hope this helps
thanks for ur help

Mod action: Posts merged. At times like this, please resort to the "modify" function at the top right corner of a post, to refrain from multiple-posting.
« Last Edit: September 27, 2017, 06:18:17 pm by RuiAce »

RuiAce

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Re: Mathematics Question Thread
« Reply #2613 on: September 27, 2017, 06:08:24 pm »
+7
This is 4 unit so there must be a previous part

set u = ln(x-2) and dv = dx
du = 1/(x-2) and v = x

so the integral = xln(x-2) - integral of x/(x-2)
= xln(x-2) - integral of (1 + 2/(x-2) )
= xln(x-2) - x - 2ln(x-2) + C
Not going to lie; I had already suspected what the actual question was. But without an example, I wasn't going to go into definite integral territory with finding areas.
oh its hsc 2008 q10 a)
why can't we just use dy/dx = f'x / fx?

ahah i was gonna ask you personally XD





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« Last Edit: September 27, 2017, 06:18:40 pm by RuiAce »

Mathew587

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Re: Mathematics Question Thread
« Reply #2614 on: September 27, 2017, 06:41:15 pm »
+2
Not going to lie; I had already suspected what the actual question was. But without an example, I wasn't going to go into definite integral territory with finding areas.

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well...
that was dumb by me :/
thanks for the help :)
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Becky234

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Re: Mathematics Question Thread
« Reply #2615 on: September 27, 2017, 07:08:39 pm »
+1
can someome pls solve this question
The number of members of a new social networking site doubles every day. On
Day 1 there were 27 members and on Day 2 there were 54 members.
(i) How many members were there on Day 12? 1
(ii) On which day was the number of members first greater than 10 million? 2
(iii) The site earns 0.5 cents per member per day. How much money did the
site earn in the first 12 days? Give your answer to the nearest dollar

Natasha.97

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Re: Mathematics Question Thread
« Reply #2616 on: September 27, 2017, 08:01:24 pm »
+7
can someome pls solve this question
The number of members of a new social networking site doubles every day. On Day 1 there were 27 members and on Day 2 there were 54 members.
(i) How many members were there on Day 12? 1
(ii) On which day was the number of members first greater than 10 million? 2
(iii) The site earns 0.5 cents per member per day. How much money did the
site earn in the first 12 days? Give your answer to the nearest dollar

Hi!

Hope this helps
Edit: Fixed part ii and iii
« Last Edit: September 27, 2017, 08:19:08 pm by Jess1113 »
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RuiAce

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Re: Mathematics Question Thread
« Reply #2617 on: September 27, 2017, 08:06:07 pm »
+4
Hi!
(Image removed from quote.)
Hope this helps
Fairly sure ii) is asking for \(T_n > 10^7\), not \(S_n\). Instead, iii) should be the one considering \( S_n\)

Becky234

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Re: Mathematics Question Thread
« Reply #2618 on: September 27, 2017, 08:27:14 pm »
+1
Fairly sure ii) is asking for \(T_n > 10^7\), not \(S_n\). Instead, iii) should be the one considering \( S_n\)
Thank you

Hi!
(Image removed from quote.)
Hope this helps
Edit: Fixed part ii and iii
thanks

Mod edit: Posts merged. You're welcome for the help, but you will need to resort to the modify button as chain-posting is against the forum rules. Further cases of this may result in a formal warning being issued.
« Last Edit: September 27, 2017, 08:33:29 pm by RuiAce »

Becky234

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Re: Mathematics Question Thread
« Reply #2619 on: September 27, 2017, 08:40:00 pm »
0
Can someone pls solve this question
Kim has three red shirts and two yellow shirts. On each of the three days,
Monday, Tuesday and Wednesday, she selects one shirt at random to wear. Kim
wears each shirt that she selects only once.
(i) What is the probability that Kim wears a red shirt on Monday? 1
(ii) What is the probability that Kim wears a shirt of the same colour on all
three days?
1
(iii) What is the probability that Kim does not wear a shirt of the same colour
on consecutive days?

Opengangs

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Re: Mathematics Question Thread
« Reply #2620 on: September 27, 2017, 08:48:49 pm »
+5
Can someone pls solve this question
Kim has three red shirts and two yellow shirts. On each of the three days,
Monday, Tuesday and Wednesday, she selects one shirt at random to wear. Kim
wears each shirt that she selects only once.
(i) What is the probability that Kim wears a red shirt on Monday? 1
(ii) What is the probability that Kim wears a shirt of the same colour on all
three days?
1
(iii) What is the probability that Kim does not wear a shirt of the same colour
on consecutive days?
(i) 3 red shirts, 5 shirts altogether. Thus, probability is 3/5

(ii) If she were to wear a red shirt on Monday, she'd have to wear a red shirt on Tuesday and Wednesday.
(3/5)(2/4)(1/3)
She can't wear a yellow shirt on the three consecutive days.
Thus, the probability is simply (3/5)(2/4)(1/3) = 1/10

(iii) P(not wear a shirt of the same colour) = P(RYR) + P(YRY) = (3/5)(2/4)(2/3) + (2/5)(3/4)(1/3) = 18/60 = 3/10

Becky234

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Re: Mathematics Question Thread
« Reply #2621 on: September 27, 2017, 09:28:35 pm »
+1
(i) 3 red shirts, 5 shirts altogether. Thus, probability is 3/5

(ii) If she were to wear a red shirt on Monday, she'd have to wear a red shirt on Tuesday and Wednesday.
(3/5)(2/4)(1/3)
She can't wear a yellow shirt on the three consecutive days.
Thus, the probability is simply (3/5)(2/4)(1/3) = 1/10

(iii) P(not wear a shirt of the same colour) = P(RYR) + P(YRY) = (3/5)(2/4)(2/3) + (2/5)(3/4)(1/3) = 18/60 = 3/10

Thank you :D

Can someone please help me with these two questions

Mod Edit: Post merge
« Last Edit: September 27, 2017, 09:51:42 pm by jamonwindeyer »

Opengangs

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Re: Mathematics Question Thread
« Reply #2622 on: September 27, 2017, 09:49:48 pm »
+4
Can someone please help me with these two questions









Fahim486

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Re: Mathematics Question Thread
« Reply #2623 on: September 28, 2017, 03:34:48 pm »
+1
Hey can someone help me out with these two questions. Thanks!

RuiAce

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Re: Mathematics Question Thread
« Reply #2624 on: September 28, 2017, 03:37:02 pm »
+1