July 16, 2019, 08:03:34 pm

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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

#### DrewN20

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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.

Thank you Blyatman

#### 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.
Failing everything, but I'm still Flareon up.

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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: 4 minutes ago »
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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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