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Thankunext

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« Reply #4275 on: July 11, 2019, 02:39:05 pm »
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A particle is moving in a straight line so that its displacement x cm over time t seconds is given by x = t [square root (49-t^2)].
A) For how many seconds does the particle travel? Answer: 7s
B) How far does the particle move altogether? Answer: 49cm

LoneWolf

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« Reply #4276 on: July 11, 2019, 08:34:43 pm »
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It's just mathematical convention for the notation of the second derivative. dx^2 is literally the square of the differential. So dy^2/dx^2 would actually be (dy/dx)^2, which is the square of the first derivative. This is not the same as the second derivative, which is completely different. As a result, the notation d^2y/dx^2 is used to avoid confusion.

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Minivasili

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« Reply #4277 on: July 12, 2019, 11:32:26 am »
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Hey guys, this multi has me stuck i cant work it out. Help is much appreciated. Thanks

fun_jirachi

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« Reply #4278 on: July 12, 2019, 01:27:42 pm »
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A particle is moving in a straight line so that its displacement x cm over time t seconds is given by x = t [square root (49-t^2)].
A) For how many seconds does the particle travel? Answer: 7s
B) How far does the particle move altogether? Answer: 49cm

It's pretty clear from the square root function that t must be strictly less than or equal to 7, and since t cannot be negative, the particle is in motion between t=0 and t=7, hence it is in motion for 7 seconds.

For the second question, we find that:
$\frac{dx}{dt} = t \times \frac{-t}{\sqrt{49-t^2}} + \sqrt{49-t^2} \\ = \frac{49-2t^2}{\sqrt{49-t^2}} \\ \text{Hence, there is a turning point when} \ 49-2t^2=0 \\ ie. t=\frac{7\sqrt{2}}{2}$
Sub that value of t back into the original equation, and you'll find the particle travels there and back to the origin for a total of 49cm.

Hey guys, this multi has me stuck i cant work it out. Help is much appreciated. Thanks

As soon as you see an at least question for probability, your mind should instantly think 'how can I use the complement?' ie. (1-P(opposite of situation)).
Here, the complement of at least one movie ticket is going to be no movie tickets. The only way Darren can get no movie tickets is if he doesn't win at all ie. from both prizes, he doesn't win. The chance of this happening is 15/20 * 14/19, since the tickets are implied to not be replaced (and they never are in prize draws anyway.) Therefore, the chance of him winning at least one ticket is 1 minus this value, which is 17/38, hence the answer should be D.
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fiafishsauce

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« Reply #4279 on: July 13, 2019, 01:31:09 pm »
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Hi! For this question, I keep receiving 148 687.50 when the answers say $163 907.81. Where am I going wrong? Rachel starts working for a business at the beginning of 2005. If she retires at the end of 2034, how much superannuation will she have if$1000 is invested at the beginning of each year at 9.5% p.a.?

r1ckworthy

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« Reply #4280 on: July 13, 2019, 01:43:27 pm »
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Hi! For this question, I keep receiving 148 687.50 when the answers say $163 907.81. Where am I going wrong? Rachel starts working for a business at the beginning of 2005. If she retires at the end of 2034, how much superannuation will she have if$1000 is invested at the beginning of each year at 9.5% p.a.?

Hey! This question has a tiny little trick to it! Great job on finding $148 687.50. Everything you have done is correct except the total number of years. Look at the bit where the questions mentions the number of years: Quote If she retires at the end of 2034... That means that we must account for an extra year as well. This would mean that n=30, not n=29. We would get the answer$163 907.81 if we use n=30. I thought of this first but then dismissed it, and then got the same answer as you. When I used n=30, I got the correct answer.

Hopefully that helps!
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Shiv3n

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« Reply #4281 on: July 16, 2019, 07:58:43 pm »
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Hi guys, I'm having trouble with answering this question 16, quite unsure where to start even
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redpanda83

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« Reply #4282 on: July 16, 2019, 08:07:27 pm »
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Hi guys, I'm having trouble with answering this question 16, quite unsure where to start even
(a) pretty much try substituting those two quardinates into the equation - make two simultaneous equation and then solve. Try doing this for now!
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Shiv3n

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« Reply #4283 on: July 16, 2019, 09:11:35 pm »
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(a) pretty much try substituting those two quardinates into the equation - make two simultaneous equation and then solve. Try doing this for now!

Ahhh, yup after doing that, I was able to work thorugh the rest of the question with ease. Thank you!
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redpanda83

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« Reply #4284 on: July 16, 2019, 09:23:05 pm »
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Ahhh, yup after doing that, I was able to work thorugh the rest of the question with ease. Thank you!
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therese07

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« Reply #4285 on: July 17, 2019, 01:44:44 pm »
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Hi there!

I was just wondering to anyone who's doing CSSA or has done CSSA in the past, what trial papers would be the best to study for? I'm currently doing the James Ruse and the Baulkham Hills trial papers, but I was wondering is there any schools trial papers that may seem redundant to study for trials? as in its level of difficulty would never be in the CSSA?

Thank you!!
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emilyyyyyyy

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« Reply #4286 on: July 19, 2019, 11:16:27 am »
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A loan of $6000 over 5 years at 15% p.a. interest, charged monthly, is paid back in 5 annual instalments. (a) What is the amount of each instalment? The Smith family buys a car for$38 000, paying a 10% deposit and taking a loan out for the balance. if the loan is over 5yrs with interest of 1.5% monthly, find amount of each monthly repayment.
--> with this question, I took away the amount paid for the deposit and then answered how I normally would, but I still couldn't get the answer.

thanks!

david.wang28

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« Reply #4287 on: July 19, 2019, 01:11:53 pm »
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A loan of $6000 over 5 years at 15% p.a. interest, charged monthly, is paid back in 5 annual instalments. (a) What is the amount of each instalment? The Smith family buys a car for$38 000, paying a 10% deposit and taking a loan out for the balance. if the loan is over 5yrs with interest of 1.5% monthly, find amount of each monthly repayment.
--> with this question, I took away the amount paid for the deposit and then answered how I normally would, but I still couldn't get the answer.

thanks!
For a), make sure you divide 0.15 by 12, as the question states 'monthly' (12 months in 1 year). Let P = 6000, let M be instalments, and let An be amount owing (equals zero if all is paid back). Now An = P*1.0125^n - M(1.0125^n - 1)/0.0125, then do some simple algebraic manipulation and sub in An = 0, n = 60 months, P = 6000. You should get \$142.74.
For b), don't take away the amount paid for the deposit. Take P = 38000 - (38000*0.1) = 34200, and repeat the same process as for a), except for the fact that you are finding M. Have a go, and ask for more help if you need it
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RuiAce

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« Reply #4288 on: July 19, 2019, 04:21:21 pm »
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Hi there!

I was just wondering to anyone who's doing CSSA or has done CSSA in the past, what trial papers would be the best to study for? I'm currently doing the James Ruse and the Baulkham Hills trial papers, but I was wondering is there any schools trial papers that may seem redundant to study for trials? as in its level of difficulty would never be in the CSSA?

Thank you!!
CSSA past papers tend to have considerable difficulty compared to most past papers. Among selective papers, it's generally regarded that Sydney Grammar offers the hardest of the lot, so you can probably focus in that direction.

I doubt any paper would be "redundant" necessarily speaking, but it is interesting to note that despite having reasonable difficulty, James Ruse papers aren't actually the hardest

(Note that you can always start by asking your school if they have past papers they can offer you first. But presumably you've already done that)