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March 29, 2024, 02:51:36 am

Author Topic: 3U Maths Question Thread  (Read 1230294 times)  Share 

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Fatemah.S

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Re: 3U Maths Question Thread
« Reply #3840 on: January 19, 2019, 03:52:45 pm »
0
Some hints:
- The only volume formula you need is \(\boxed{V = A\ell}\), where \(\ell\) is the length associated with the prism and \(A\) is the area of the shape that produces the prism. Because we're talking about a water trough here, we can assume that the trough just takes the shape of a triangular prism.
- Note that you're only interested in the volume of water in the trough, so a cross section of the trough should just be the right-angled isosceles triangle itself. (You may wish to draw out a right-angled isosceles triangle, whose height (as in the height from the apex) is \(x\), to see what to do from here. Strong emphasis on that it is a right-angled isosceles triangle. Note that common sense suggests that the right angle should be at the very bottom of the diagram.)
Extra hint once you've given up with the previous two
Your length of the prism is just 200. The diagram is supposed to help you find the area of the triangular cross section. Some quick trigonometry or Pythagoras may be required to find the length of the base for \(A=\frac12bh\).

Thank you so much! Once I had the diagram, it all made sense. :)

david.wang28

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Re: 3U Maths Question Thread
« Reply #3841 on: January 21, 2019, 11:05:33 am »
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Hello,
I'm having trouble with question 7 in the link below. Can anyone help me with this question please(I have no idea how to do it)? Thanks :)
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kiwii

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Re: 3U Maths Question Thread
« Reply #3842 on: January 21, 2019, 04:30:43 pm »
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Hi. Can someone help me with this maths induction question please. Im having trouble proving the 3rd step with the inequality. Its making me confused. please help! :))
4^n > 3n + 7
for n > or = to 2

david.wang28

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Re: 3U Maths Question Thread
« Reply #3843 on: January 21, 2019, 04:59:37 pm »
+1
Hi. Can someone help me with this maths induction question please. Im having trouble proving the 3rd step with the inequality. Its making me confused. please help! :))
4^n > 3n + 7
for n > or = to 2
Here is my solution. For inequalities, you need to state your induction hypothesis (step 2), and use your algebraic skills to get the negative answer. Hopefully this is the answer :)
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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #3844 on: January 21, 2019, 07:03:43 pm »
+3
Here is my solution. For inequalities, you need to state your induction hypothesis (step 2), and use your algebraic skills to get the negative answer. Hopefully this is the answer :)

I think you might have forgotten to sub \(k+1\) in the middle there, it should be \(3(k+1)+7\) I think! The process ends up the same though, nice job ;D

david.wang28

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Re: 3U Maths Question Thread
« Reply #3845 on: January 21, 2019, 08:12:13 pm »
+1
I think you might have forgotten to sub \(k+1\) in the middle there, it should be \(3(k+1)+7\) I think! The process ends up the same though, nice job ;D
Sorry, my mistake there. Though I do focus more on the process rather than the final answer. Thanks for pointing out the mistake Jamon! :)
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RuiAce

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Re: 3U Maths Question Thread
« Reply #3846 on: January 22, 2019, 04:23:52 pm »
+3
Hello,
I'm having trouble with question 7 in the link below. Can anyone help me with this question please(I have no idea how to do it)? Thanks :)

\[ \text{So upon equating coefficients and also doing some rearranging and factorising we obtain}\\ \binom{n}{r}^2(-1)^r+ \binom{n}{r-1}\binom{n}{r+1} \left[ (-1)^{r-1}+(-1)^{r+1} \right] + \binom{n}{r-2}\binom{n}{r+2} \left[(-1)^{r-2}+(-1)^{r+2} \right] + \dots + \binom{n}{1}\binom{n}{2r-1}\left[ (-1)^1 + (-1)^{2r-1} \right] + \binom{n}{0} \binom{n}{2r} \left[(-1)^0 +(-1)^{2r} \right] = \binom{n}{r}(-1)^r\]
Note that this equation is really long - when writing it down you'll likely need multiple lines.
\[ \text{We now divide both sides by }(-1)^r\text{ to obtain}\\ \binom{n}{r}^2+ \binom{n}{r-1}\binom{n}{r+1} \left[ (-1)^{-1}+(-1)^{1} \right] + \binom{n}{r-2}\binom{n}{r+2} \left[(-1)^{-2}+(-1)^{2} \right] + \dots + \binom{n}{1}\binom{n}{2r-1}\left[ (-1)^{r-1} + (-1)^{-r+1} \right] + \binom{n}{0} \binom{n}{2r} \left[(-1)^{-r} +(-1)^{r} \right] = \binom{n}{r}\\ \text{and moving that first }\binom{n}{r}\text{ term to the RHS we have}\\ \boxed{\binom{n}{r-1}\binom{n}{r+1} \left[ (-1)^{-1}+(-1)^{1} \right] + \binom{n}{r-2}\binom{n}{r+2} \left[(-1)^{-2}+(-1)^{2} \right] + \dots + \binom{n}{1}\binom{n}{2r-1}\left[ (-1)^{r-1} + (-1)^{-r+1} \right] + \binom{n}{0} \binom{n}{2r} \left[(-1)^{-r} +(-1)^{r} \right] = \binom{n}{r}-\binom{n}{r}^2} \]
\[ \text{However since }(-1)^{-1} = (-1)^1,\text{ for any integer }k\text{ we also have }\boxed{(-1)^{-k} = (-1)^k}.\\ \text{This simplifies the LHS down considerably and gives us}\\ 2\left[ \binom{n}{r-1} \binom{n}{r+1} (-1)^1 + \binom{n}{r-2} \binom{n}{r+2} (-1)^2 + \dots + \binom{n}{1} \binom{n}{2r-1} (-1)^{r+1} +\binom{n}{0}\binom{n}{2r} (-1)^r \right] = \binom{n}{r} \left[1 - \binom{n}{r} \right] \]
You should be able to take it from here. Note that you need to divide \(-1\) on the LHS and RHS just once more.

david.wang28

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Re: 3U Maths Question Thread
« Reply #3847 on: January 22, 2019, 06:00:11 pm »
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(Image removed from quote.)
\[ \text{So upon equating coefficients and also doing some rearranging and factorising we obtain}\\ \binom{n}{r}^2(-1)^r+ \binom{n}{r-1}\binom{n}{r+1} \left[ (-1)^{r-1}+(-1)^{r+1} \right] + \binom{n}{r-2}\binom{n}{r+2} \left[(-1)^{r-2}+(-1)^{r+2} \right] + \dots + \binom{n}{1}\binom{n}{2r-1}\left[ (-1)^1 + (-1)^{2r-1} \right] + \binom{n}{0} \binom{n}{2r} \left[(-1)^0 +(-1)^{2r} \right] = \binom{n}{r}(-1)^r\]
Note that this equation is really long - when writing it down you'll likely need multiple lines.
\[ \text{We now divide both sides by }(-1)^r\text{ to obtain}\\ \binom{n}{r}^2+ \binom{n}{r-1}\binom{n}{r+1} \left[ (-1)^{-1}+(-1)^{1} \right] + \binom{n}{r-2}\binom{n}{r+2} \left[(-1)^{-2}+(-1)^{2} \right] + \dots + \binom{n}{1}\binom{n}{2r-1}\left[ (-1)^{r-1} + (-1)^{-r+1} \right] + \binom{n}{0} \binom{n}{2r} \left[(-1)^{-r} +(-1)^{r} \right] = \binom{n}{r}\\ \text{and moving that first }\binom{n}{r}\text{ term to the RHS we have}\\ \boxed{\binom{n}{r-1}\binom{n}{r+1} \left[ (-1)^{-1}+(-1)^{1} \right] + \binom{n}{r-2}\binom{n}{r+2} \left[(-1)^{-2}+(-1)^{2} \right] + \dots + \binom{n}{1}\binom{n}{2r-1}\left[ (-1)^{r-1} + (-1)^{-r+1} \right] + \binom{n}{0} \binom{n}{2r} \left[(-1)^{-r} +(-1)^{r} \right] = \binom{n}{r}-\binom{n}{r}^2} \]
\[ \text{However since }(-1)^{-1} = (-1)^1,\text{ for any integer }k\text{ we also have }\boxed{(-1)^{-k} = (-1)^k}.\\ \text{This simplifies the LHS down considerably and gives us}\\ 2\left[ \binom{n}{r-1} \binom{n}{r+1} (-1)^1 + \binom{n}{r-2} \binom{n}{r+2} (-1)^2 + \dots + \binom{n}{1} \binom{n}{2r-1} (-1)^{r+1} +\binom{n}{0}\binom{n}{2r} (-1)^r \right] = \binom{n}{r} \left[1 - \binom{n}{r} \right] \]
You should be able to take it from here. Note that you need to divide \(-1\) on the LHS and RHS just once more.
Wow, this is very impressive from you, everything was perfect; your working out, your thinking, the process. Thank you very much Rui! :)
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kiwii

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Re: 3U Maths Question Thread
« Reply #3848 on: January 23, 2019, 01:16:19 pm »
+1
Here is my solution. For inequalities, you need to state your induction hypothesis (step 2), and use your algebraic skills to get the negative answer. Hopefully this is the answer :)

Wow! Thank you so much David! The process made much more sense. :))

Rabi

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Re: 3U Maths Question Thread
« Reply #3849 on: January 24, 2019, 01:18:48 am »
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Quote from: Rabi on January 23, 2019, 07:40:56 pm
Hey I really need help with this question plz plz help.
"A water trough is 200 cm long and has a cross section of a right angled isosceles triangle. Show that when the depth of the water is x cm, the volume of water in the tank is 200x^2 cm^3. Water is poured in at a constant rate of 5 litres per minute. Find the rate at which the water level is rising when the depth is 30cm.

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Re: 3U Maths Question Thread
« Reply #3850 on: January 24, 2019, 01:32:53 am »
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Hey guys I have real troubles with this question I attacted it couple of times but it hit me harder so I need help.
" sand falling from a funnel forms a conical pile such that the height of the pile is one and a half times radius.
A). Show that the volume of the pile is given by V=pi r^3/2.
B). If the sand is falling at the rate of pi/10
meter cubed per minute . Find the rate at which the height is increasing when the pile of sand is 3m high.

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Re: 3U Maths Question Thread
« Reply #3851 on: January 24, 2019, 10:42:40 am »
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Quote from: Rabi on January 23, 2019, 07:40:56 pm
Hey I really need help with this question plz plz help.
"A water trough is 200 cm long and has a cross section of a right angled isosceles triangle. Show that when the depth of the water is x cm, the volume of water in the tank is 200x^2 cm^3. Water is poured in at a constant rate of 5 litres per minute. Find the rate at which the water level is rising when the depth is 30cm.

There are hints at the top of the page and at the bottom of the previous page to help you try this on your own! If you need more help with this question, tell us what you're struggling with and someone (probably Rui :P ) will address it.

Hey guys I have real troubles with this question I attacted it couple of times but it hit me harder so I need help.
" sand falling from a funnel forms a conical pile such that the height of the pile is one and a half times radius.
A). Show that the volume of the pile is given by V=pi r^3/2.
B). If the sand is falling at the rate of pi/10
meter cubed per minute . Find the rate at which the height is increasing when the pile of sand is 3m high.





Hope this helps :)
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r1ckworthy

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Re: 3U Maths Question Thread
« Reply #3852 on: January 24, 2019, 11:37:38 am »
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Hello all,
Hope all of you are enjoying the remaining days of the holidays!! (for all who are still in school)
I have a bit of trouble with the following question:

Prove by induction that is divisible by 5 for all odd integers.

I have proved it all the way upto n=K+1, but I am stuck with the following expression:
When I sub in my assumption, I get the following: And I don't know where to go from there.
Either I am on the right track or completely wrong. Please help!! :'( :'(
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meerae

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Re: 3U Maths Question Thread
« Reply #3853 on: January 24, 2019, 12:03:05 pm »
+1
Hello all,
Hope all of you are enjoying the remaining days of the holidays!! (for all who are still in school)
I have a bit of trouble with the following question:

Prove by induction that is divisible by 5 for all odd integers.

I have proved it all the way upto n=K+1, but I am stuck with the following expression:
When I sub in my assumption, I get the following: And I don't know where to go from there.
Either I am on the right track or completely wrong. Please help!! :'( :'(

Hi r1ckworthy!

unsure if I am right, but wouldn't you sub n=k+2 as k+1 would be even if k is odd? Try that...

Edit: I tried it, and it worked. Heres how I did it;

then sub in
and work from there.

Hope this helped!
meerae :)
« Last Edit: January 24, 2019, 12:10:55 pm by meerae »
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r1ckworthy

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Re: 3U Maths Question Thread
« Reply #3854 on: January 24, 2019, 12:14:33 pm »
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Hi r1ckworthy!

unsure if I am right, but wouldn't you sub n=k+2 as k+1 would be even if k is odd? Try that...

Edit: I tried it, and it worked. Heres how I did it;

then sub in
and work from there.

Hope this helped!
meerae :)

Thanks you very much!!!
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