Mathematics Question Thread

[julia_warren13]

This is because if we try to minimise \(\sin ax\), the minimum value will be -1. This can be proven via calculus, but is more easily deduced just because the range of \( \sin ax \) is \( -1 \le y \le 1\).

\begin{align*}1+0.7\sin \frac{\pi t}{6}&=0.3\\ \sin \frac{\pi t}{6}&=-0.7\\ \frac{\pi t}{6} &=\frac{3\pi}{2}\\ t&= 9\end{align*}

Thanks Rui but I still don't quite understand how to get h=0.3 at low tide

[RuiAce]

Thanks Rui but I still don't quite understand how to get h=0.3 at low tide

This is because h is the height of the tide itself. Logically speaking, the low tide should be when the height of the tide is minimised.


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[sidzeman]

Hey I know this has been covered before but,
Why is the rationalisation of this root (n+1)− root n     and not root n - root (n+1)?

[RuiAce]

Hey I know this has been covered before but,
Why is the rationalisation of this root (n+1)− root n     and not root n - root (n+1)?

If you use that other one it will still work. They just chose to replace \( \sqrt{n}+\sqrt{n+1} \) with \( \sqrt{n+1} + \sqrt{n} \) first instead, which is still perfectly allowed.

[biffi023]

hey! bit of a dumb one sorry.. but how do you calculate something that has 'cos^2(theta)' or something like that?? as in to put it in the calculator i am guessing you have to do something with it cos u cant put a cos^2 or sin^2 etc in??
sorry very basic but hmm.. sadly i can hardly even do the basic stuff 
thanks!

[RuiAce]

hey! bit of a dumb one sorry.. but how do you calculate something that has 'cos^2(theta)' or something like that?? as in to put it in the calculator i am guessing you have to do something with it cos u cant put a cos^2 or sin^2 etc in??
sorry very basic but hmm.. sadly i can hardly even do the basic stuff 
thanks!

So you type cos of whatever number you have, and then put that in brackets before putting a square around it.

[hansolo9]

Hello
I'm stuck on part 2 and 3

thank you 

[RuiAce]

Hello
I'm stuck on part 2 and 3

thank you 

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\begin{align*}\frac94 CX^2 &=  \frac14 AB^2 + AB^2\\ \frac94 CX^2 &= \frac54 AB^2\\ 9 CX^2 &= 5 AB^2\end{align*}

[caitlinlddouglas]

Hey could someone please help me with this question? For
iii) what does it mean by doing d/V? I get that the displacement is 6units to the right (it cancels out between t=4 and t=6) but how do you then go on to workout when it returns to the origin?
Iv) what happens to the graph of the displacement if there is a point of inflexion on the velocity graph? The solution is indicated that it was just concave up?
Thanks heaps!!:)

[sidzeman]

For part iii, you get a result of n = 178.37 to repay the 346 095. The answers then round this down to 178, but shouldn't it be rounded up to 179? As is takes 178 AND a bit to repay the loan, surely just 178 payments would leave some amount still owing

[hobocop]

Hi, could I get some help with these two please

[RuiAce]

Hey could someone please help me with this question? For
iii) what does it mean by doing d/V? I get that the displacement is 6units to the right (it cancels out between t=4 and t=6) but how do you then go on to workout when it returns to the origin?
Iv) what happens to the graph of the displacement if there is a point of inflexion on the velocity graph? The solution is indicated that it was just concave up?
Thanks heaps!!:)

Some comprehensive discussion was given somewhere around here for part iii). Maybe the later bit of it will be useful to you.

As for iv), If the derivative graph has a point of inflexion then you actually DO NOT KNOW what happens to the original graph. So they probably just drew it concave up because it is meant to concave up.

(Remember: The second derivative of the derivative is the third derivative of the original thing. And you've never touched on applications of the third derivative at all.)

[RuiAce]

For part iii, you get a result of n = 178.37 to repay the 346 095. The answers then round this down to 178, but shouldn't it be rounded up to 179? As is takes 178 AND a bit to repay the loan, surely just 178 payments would leave some amount still owing

The HSC has left quite some contradictory evidence with this stuff. Technically, you should round up like you said so.

[RuiAce]

Hi, could I get some help with these two please

Please consult the compilation for the first one. You may expand on your question if the advice given was too brief though.

[caitlinlddouglas]

Some comprehensive discussion was given somewhere around here for part iii). Maybe the later bit of it will be useful to you.

As for iv), If the derivative graph has a point of inflexion then you actually DO NOT KNOW what happens to the original graph. So they probably just drew it concave up because it is meant to concave up.

(Remember: The second derivative of the derivative is the third derivative of the original thing. And you've never touched on applications of the third derivative at all.)

Hi thanks for that I can't see the discussion for iii) though?