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March 29, 2024, 04:01:15 pm

Author Topic: 3U Maths Question Thread  (Read 1230530 times)  Share 

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anotherworld2b

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Re: 3U Maths Question Thread
« Reply #1140 on: January 05, 2017, 12:56:46 am »
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Hi i just wanted to check whether or not my answer is correct for this question. I got 127 as the answer but the answer for the question is apparently 2/125 but im not sure why

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #1141 on: January 05, 2017, 01:18:54 am »
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Hi i just wanted to check whether or not my answer is correct for this question. I got 127 as the answer but the answer for the question is apparently 2/125 but im not sure why

You are definitely correct! ;D the answer perhaps thought it was a division not an addition?

anotherworld2b

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Re: 3U Maths Question Thread
« Reply #1142 on: January 05, 2017, 01:58:16 am »
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thank you for the help :)
I have another quick question that i solved but i am not sure whether i am right or not

You are definitely correct! ;D the answer perhaps thought it was a division not an addition?

kiwiberry

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Re: 3U Maths Question Thread
« Reply #1143 on: January 05, 2017, 02:49:25 am »
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thank you for the help :)
I have another quick question that i solved but i am not sure whether i am right or not

hey!
I think you've forgotten to divide the 8^-2 instead of multiplying as it is also on the denominator :)
2^(4x-8) ÷ [(2^x) x (2^-6)]
= 2^(3x-2)
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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #1144 on: January 05, 2017, 10:22:08 am »
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hey!
I think you've forgotten to divide the 8^-2 instead of multiplying as it is also on the denominator :)
2^(4x-8) ÷ [(2^x) x (2^-6)]
= 2^(3x-2)

Yep, that's correct, thanks for that kiwiberry! ;D

anotherworld2b

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Re: 3U Maths Question Thread
« Reply #1145 on: January 05, 2017, 05:37:27 pm »
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Thank you very much fkr your help :)
hey!
I think you've forgotten to divide the 8^-2 instead of multiplying as it is also on the denominator :)
2^(4x-8) ÷ [(2^x) x (2^-6)]
= 2^(3x-2)

anotherworld2b

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Re: 3U Maths Question Thread
« Reply #1146 on: January 06, 2017, 12:47:13 pm »
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Hi i was wondering if i could get help with finding the equation of these exponential functions. I am able to find a simple translation but i do not know how to find a equarion with  more than one transformation at play.

jakesilove

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Re: 3U Maths Question Thread
« Reply #1147 on: January 06, 2017, 03:04:59 pm »
+1
Hi i was wondering if i could get help with finding the equation of these exponential functions. I am able to find a simple translation but i do not know how to find a equarion with  more than one transformation at play.

Definitely a pretty tough question! We will need to make some approximations, and work through each logically. First, we can assume that the function will look something like



Where a and b are constants. The graphs could actually be any number of functions (eg. there could be a constant before x, or before e), but I believe that this is the most standard type of shift that will be assessed. We can quickly find b; that's going to be the value that the function was shifted UP by. Clearly, it's going to be 2, giving us the function.



Now, we can plot a point to find the value of a. When x=0, y=5 (from the graph). Therefore




This gives us a function



Let's test another point, to check that this function is correct. We know that at x=-1, y=3.



Now, the right hand side equals 4.1, not 3, therefore we have found the wrong equation. Looks like we need to do this entire question properly, which is a bit more difficult. I'll leave the above for reference.

Our new function is going to look like the following



Now, when x=0, y=5. So,



When x=-1, y=3. So,



When x=1, y=11. So,



Also, when x approaches negative infinity, y approaches 2. So,



This last equation gives us d=2 (as e to the negative infinity is zero!).

Now, we need to have a crack at the simultaneous equations. I want to get rid of a, and so I'll divide the first equation by the second. Note that I will bring d (2) to the left hand side before dividing through, so I can easily cancel out the 'a' term.







Great! We managed to grab a value for b; our equation now looks like this;



Now, I want to KEEP the c term, so dividing isn't going to work. The two equations I think we could use are




(sorry I can't figure out how the number the equations; hopefully it's clear-ish!).

The problem is, if we multiply the terms we keep both a and c, and if we divide we lose both a and c. Using wolfram alpha, turns out that a=3, c=0. Can't directly see why (actually, it has to do with indice laws; trying expanding out the indice, and simplifying the e^log(3) etc!).

This gives us the final equation,



That was an absolute trekk and a half. I hope that you could follow that working; I really don't think this could possibly be examined, so I'm not going to go into more detail. Give the second one a go using a similar method, and post your working if you still need help!
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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #1148 on: January 06, 2017, 04:08:32 pm »
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Hi i was wondering if i could get help with finding the equation of these exponential functions. I am able to find a simple translation but i do not know how to find a equarion with  more than one transformation at play.

Just as a note on this question, it is WACE. They won't ask anything like this in the HSC, but the working done by Jake above is still really great to see for anyone wanting to really understand exponentials clearly ;D

anotherworld2b

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Re: 3U Maths Question Thread
« Reply #1149 on: January 06, 2017, 06:45:03 pm »
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thank you for your help :)
I will back after I try to digest and understand this question :D. It is definitely more difficult than I though :O

Definitely a pretty tough question! We will need to make some approximations, and work through each logically. First, we can assume that the function will look something like



Where a and b are constants. The graphs could actually be any number of functions (eg. there could be a constant before x, or before e), but I believe that this is the most standard type of shift that will be assessed. We can quickly find b; that's going to be the value that the function was shifted UP by. Clearly, it's going to be 2, giving us the function.



Now, we can plot a point to find the value of a. When x=0, y=5 (from the graph). Therefore




This gives us a function



Let's test another point, to check that this function is correct. We know that at x=-1, y=3.



Now, the right hand side equals 4.1, not 3, therefore we have found the wrong equation. Looks like we need to do this entire question properly, which is a bit more difficult. I'll leave the above for reference.

Our new function is going to look like the following



Now, when x=0, y=5. So,



When x=-1, y=3. So,



When x=1, y=11. So,



Also, when x approaches negative infinity, y approaches 2. So,



This last equation gives us d=2 (as e to the negative infinity is zero!).

Now, we need to have a crack at the simultaneous equations. I want to get rid of a, and so I'll divide the first equation by the second. Note that I will bring d (2) to the left hand side before dividing through, so I can easily cancel out the 'a' term.







Great! We managed to grab a value for b; our equation now looks like this;



Now, I want to KEEP the c term, so dividing isn't going to work. The two equations I think we could use are




(sorry I can't figure out how the number the equations; hopefully it's clear-ish!).

The problem is, if we multiply the terms we keep both a and c, and if we divide we lose both a and c. Using wolfram alpha, turns out that a=3, c=0. Can't directly see why (actually, it has to do with indice laws; trying expanding out the indice, and simplifying the e^log(3) etc!).

This gives us the final equation,



That was an absolute trekk and a half. I hope that you could follow that working; I really don't think this could possibly be examined, so I'm not going to go into more detail. Give the second one a go using a similar method, and post your working if you still need help!

shreya_ajoshi

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Re: 3U Maths Question Thread
« Reply #1150 on: January 06, 2017, 08:25:05 pm »
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Hi!
I need help with these Permutations and Combinations questions

6. A motorist travels through eight sets of traffic light,each of which is red or green. He is forced to stop at three sets of lights.
a) In how many ways could this happen?
b) What other number of red lights would give an identical answer to part a)

7. Find how many ways the letters of the word SOCKS can be arranged in a line:
a) Without restriction
b) So that the two S's are together
c) So that the two S's are separated by at least one other letter
d) So that K is somewhere to the left of C

Thanks :)

RuiAce

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Re: 3U Maths Question Thread
« Reply #1151 on: January 06, 2017, 08:46:00 pm »
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Hi!
I need help with these Permutations and Combinations questions

6. A motorist travels through eight sets of traffic light,each of which is red or green. He is forced to stop at three sets of lights.
a) In how many ways could this happen?
b) What other number of red lights would give an identical answer to part a)

7. Find how many ways the letters of the word SOCKS can be arranged in a line:
a) Without restriction
b) So that the two S's are together
c) So that the two S's are separated by at least one other letter
d) So that K is somewhere to the left of C

Thanks :)



_________________________________________________









« Last Edit: January 06, 2017, 08:48:57 pm by RuiAce »

shreya_ajoshi

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Re: 3U Maths Question Thread
« Reply #1152 on: January 06, 2017, 09:19:09 pm »
0



_________________________________________________











Thank you RuiAce!
For part c) of question7, your answer is different to the answer they provided. They said the answer is 36, however your answer is 18. Would this be the correct way to approach it:
(3! x 3) + (3! x 2) + (3! x 1) = 36. The 3! represents the letters excluding the two S's. And the number we are multiplying it by after the 3! represents the number of groups that are formed altogether  when the 2 S's are grouped together.
And for part d), the answers say "half of them." Your answer is correct, but i don't understand why the answers they provided say 'it would be half the possibilities of all when there are no restrictions'

RuiAce

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Re: 3U Maths Question Thread
« Reply #1153 on: January 06, 2017, 09:59:26 pm »
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Thank you RuiAce!
For part c) of question7, your answer is different to the answer they provided. They said the answer is 36, however your answer is 18. Would this be the correct way to approach it:
(3! x 3) + (3! x 2) + (3! x 1) = 36. The 3! represents the letters excluding the two S's. And the number we are multiplying it by after the 3! represents the number of groups that are formed altogether  when the 2 S's are grouped together.
And for part d), the answers say "half of them." Your answer is correct, but i don't understand why the answers they provided say 'it would be half the possibilities of all when there are no restrictions'
That approach to c) should be correct, because it's the second approach of just splitting up the cases.

I found out why mine was wrong though. Going back to not reading the question clearly. I did the number of ways of 'exactly' 1 letter apart, not at least 1 letter apart.

Fixing my working up for c), this is the easiest way of handling it.

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shreya_ajoshi

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Re: 3U Maths Question Thread
« Reply #1154 on: January 07, 2017, 12:28:52 am »
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That approach to c) should be correct, because it's the second approach of just splitting up the cases.

I found out why mine was wrong though. Going back to not reading the question clearly. I did the number of ways of 'exactly' 1 letter apart, not at least 1 letter apart.

Fixing my working up for c), this is the easiest way of handling it.

_______________________





Ahhhh, I understand now. Thank you so much :)