Author Topic: Mathematics Question Thread (Read 1613869 times) Tweet Share

Dragomistress · « **Reply #2580 on:** September 22, 2017, 06:57:59 am »

May I have these questions answered?

RuiAce · « **Reply #2581 on:** September 22, 2017, 09:52:20 am »

Quote from: Dragomistress on September 22, 2017, 06:57:59 am

May I have these questions answered?

In order, you would start with something along the lines of:

$ PR = 2\times \text{"distance from }y=1\text{"} $ so $ (x-3)^2+(y-2)^2 =4[ (x-x)^2 + (y-1)^2] $

$ \text{"distance from }x=0\text{"} = \text{"distance from }y=0\text{"} $ so $ (x-0)^2+(y-y)^2 = (x-x)^2+(y-0)^2 $

$ \frac{2}{5} PA = \frac{3}{5} PB $ so $ \frac{4}{25}[(x+6)^2+(y-5)^2] = \frac{9}{25}[(x-3)^2+(y+1)^2]$

These questions follow the exact same working and are repetitive so I will not clarify or complete the question any further. If you still have troubles, please indicate clearly where the issue is and/or post relevant working-out progress.

katnisschung · « **Reply #2582 on:** September 22, 2017, 12:01:57 pm »

http://educationstandards.nsw.edu.au/wps/wcm/connect/e8e5dd58-ab1f-482d-b849-221495c629c6/maths-hsc-exam-2010.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-e8e5dd58-ab1f-482d-b849-221495c629c6-lGhPb9T

for q 9 a) ii
of this paper my answer was 174.1431...... so money will run out on the 175th month therefore the answer is 174 because until that month there will be money remaining in the account correct? but i'm wrong which is annoying considering i'll lose a mark. can someone pls explain

EEEEEEP · « **Reply #2583 on:** September 22, 2017, 12:14:53 pm »

Quote from: katnisschung on September 22, 2017, 12:01:57 pm

http://educationstandards.nsw.edu.au/wps/wcm/connect/e8e5dd58-ab1f-482d-b849-221495c629c6/maths-hsc-exam-2010.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-e8e5dd58-ab1f-482d-b849-221495c629c6-lGhPb9T

for q 9 a) ii
of this paper my answer was 174.1431...... so money will run out on the 175th month therefore the answer is 174 because until that month there will be money remaining in the account correct? but i'm wrong which is annoying considering i'll lose a mark. can someone pls explain

In HSC maths, when it comes to exponentials, financial maths and all... they usually round up the END answer..,

So if its.. 1.56 days, they say 2 days.
IF its 174.1431 days (in this case), they round up to 175.

katnisschung · « **Reply #2584 on:** September 22, 2017, 12:22:03 pm »

http://educationstandards.nsw.edu.au/wps/wcm/connect/e8e5dd58-ab1f-482d-b849-221495c629c6/maths-hsc-exam-2010.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-e8e5dd58-ab1f-482d-b849-221495c629c6-lGhPb9T

for 9 b) iv) i can't quite tell from the marking guidelines if the graphs at x>4 is a straight line graph...considering the gradient from the given graph is a constant -3

RuiAce · « **Reply #2585 on:** September 22, 2017, 12:25:56 pm »

Quote from: katnisschung on September 22, 2017, 12:22:03 pm

http://educationstandards.nsw.edu.au/wps/wcm/connect/e8e5dd58-ab1f-482d-b849-221495c629c6/maths-hsc-exam-2010.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-e8e5dd58-ab1f-482d-b849-221495c629c6-lGhPb9T

for 9 b) iv) i can't quite tell from the marking guidelines if the graphs at x>4 is a straight line graph...considering the gradient from the given graph is a constant -3

It should most certainly be a straight line over the interval 4 < x < 6. The gradient of this line should be -3.

Quote from: katnisschung on September 22, 2017, 12:01:57 pm

http://educationstandards.nsw.edu.au/wps/wcm/connect/e8e5dd58-ab1f-482d-b849-221495c629c6/maths-hsc-exam-2010.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-e8e5dd58-ab1f-482d-b849-221495c629c6-lGhPb9T

for q 9 a) ii
of this paper my answer was 174.1431...... so money will run out on the 175th month therefore the answer is 174 because until that month there will be money remaining in the account correct? but i'm wrong which is annoying considering i'll lose a mark. can someone pls explain

Technically speaking, during the 175th month there is still money. It is only at the end of the 175th month where there won't be money anymore.

mary123987 · « **Reply #2586 on:** September 22, 2017, 12:31:39 pm »

Quote from: katnisschung on September 22, 2017, 12:01:57 pm

http://educationstandards.nsw.edu.au/wps/wcm/connect/e8e5dd58-ab1f-482d-b849-221495c629c6/maths-hsc-exam-2010.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-e8e5dd58-ab1f-482d-b849-221495c629c6-lGhPb9T

for q 9 a) ii
of this paper my answer was 174.1431...... so money will run out on the 175th month therefore the answer is 174 because until that month there will be money remaining in the account correct? but i'm wrong which is annoying considering i'll lose a mark. can someone pls explain

Hey , with respect to this question it is not right to assume that , what you rather should do is sub in the value (i.e n months ) in to the equation which you worked out in part (ii) 1 which is this formula : An = (P − 400 000) × 1.005n + 400 000.
now the question asks
For how many months after retirement will there be money left in
the account?
so what you need to do is sub in the value if you sub in 174.1431...... you get exactly 0
now consider rounding:
when tyou sub in 174 you will see you get $285.4307496
when you sub in 175 you will see you get $-1713.142097
so if you look at the question it states:Chris withdraws $2000 from the account at the end
of each month, without making any further deposits......
thus considering it is withdrawn at the end of 175 months not the beginning 175 is the answer.
Hope this helps

georgiia · « **Reply #2587 on:** September 23, 2017, 04:10:50 pm »

Could I please have help for these two questions? Thanks!!

RuiAce · « **Reply #2588 on:** September 23, 2017, 04:12:52 pm »

Quote from: georgiia on September 23, 2017, 04:10:50 pm

Could I please have help for these two questions? Thanks!!(Image removed from quote.)(Image removed from quote.)

You have already asked the first one
$\text{Whilst your limiting sum in itself is correct}\\ \text{To ensure that a limiting sum exists, we must check that the common ratio}\\ \text{satisfies }\boxed{|r| < 1}\text{ or equivalently }-1 < r < 1$
$\text{But this is true, as when }0^\circ < x < 45^\circ\\ \text{we have }0 < \tan x < 1\\ \text{which further implies }-1 < -\tan^2 x < 0$
$\text{So clearly }|r| = |-\tan^2 x| < 1\\ \text{and hence the limiting sum exists}$

georgiia · « **Reply #2589 on:** September 23, 2017, 04:25:21 pm »

Quote from: RuiAce on September 23, 2017, 04:12:52 pm

You have already asked the first one
$\text{Whilst your limiting sum in itself is correct}\\ \text{To ensure that a limiting sum exists, we must check that the common ratio}\\ \text{satisfies }\boxed{|r| < 1}\text{ or equivalently }-1 < r < 1$
$\text{But this is true, as when }0^\circ < x < 45^\circ\\ \text{we have }0 < \tan x < 1\\ \text{which further implies }-1 < -\tan^2 x < 0$
$\text{So clearly }|r| = |-\tan^2 x| < 1\\ \text{and hence the limiting sum exists}$

thank you!
Could I please also have help for this?

RuiAce · « **Reply #2590 on:** September 23, 2017, 04:54:04 pm »

Quote from: georgiia on September 23, 2017, 04:25:21 pm

thank you!
Could I please also have help for this? (Image removed from quote.)

Double check your graph. Over the domain $-\frac{\pi}{2} < x < \frac{\pi}{2} $, $y=\tan x$ only intersects $y=x$ at 0; never again.

georgiia · « **Reply #2591 on:** September 23, 2017, 04:58:49 pm »

Quote from: RuiAce on September 23, 2017, 04:54:04 pm

Double check your graph. Over the domain $-\frac{\pi}{2} < x < \frac{\pi}{2} $, $y=\tan x$ only intersects $y=x$ at 0; never again.

Im having trouble with graphing it. I know i can search it online but I want the process of finding it explained

RuiAce · « **Reply #2592 on:** September 23, 2017, 05:05:15 pm »

Quote from: georgiia on September 23, 2017, 04:58:49 pm

Im having trouble with graphing it. I know i can search it online but I want the process of finding it explained

I'm going to assume that you understand how to draw the shapes of the graphs separately with ease.
$\text{To understand why }\tan x \ge x\text{ over our domain of interest}\\ \text{we need to understand why }f(x)=\tan x\\ \textbf{increases faster}\text{ than }g(x) =<br />x$
$\text{The relevant derivatives are}\\ \begin{align*}f^\prime(x) &= \sec^2 x\\ g^\prime(x) &= 1\end{align*}\\ \text{and we want to examine why }f^\prime(x) \ge g^\prime(x)\\ \text{because if }f(x)\text{ increases faster than }g(x)\\ \text{and they both start from }(0,0)\\ \text{naturally one would stay above the other.}$
___________________________________
$\cos x\text{ has range }\boxed{-1\le \cos x \le 1}\text{ i.e. equivalently }\boxed{|\cos x |\le 1}\\ \text{so it follows that }\sec x\text{ has range } |\sec x | = \frac{1}{|\cos x|} \ge 1$
$\text{Since }\boxed{|\sec x| \ge 1}\text{, furthermore }\sec^2 x = |\sec x|^2 \ge 1$
$\text{So since }\sec^2 x \ge 1\\ f^\prime(x) \ge g^\prime(x)\text{ as required}\\ \text{and hence }\tan x\text{ increases faster than }x\\ \text{so }\tan x \ge x\text{ and hence the graph of }\tan x\\ \text{lies ENTIRELY above }y=x\text{ for our domain of interest}$

Nuwanda_g · « **Reply #2593 on:** September 23, 2017, 08:50:12 pm »

Hey guys,
I need help with 'iii.'

I dont get how the DF = FE-DE, became equal to EB-AE.

Thanks in advance

Mathew587 · « **Reply #2594 on:** September 23, 2017, 08:59:08 pm »

Quote from: Nuwanda_g on September 23, 2017, 08:50:12 pm

I dont get how the DF = FE-DE, became equal to EB-AE.
Thanks in advance

we know AE = DE = X (isoceles t)
we know EF = EB = Y (isos)

chuck that into your q;
DF = FE-DE, became equal to EB-AE...
DF = Y - X, ... Y-X
therefore EB-AE = FE-DE

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