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#76 Re: Help Me ! » Proof by induction? » 2012-05-22 03:29:36

Thanks a lot, though the formula is n(n-1)/2 but i can account for that in my proof.

I need help with one more proof.

Prove that GCD(ab,c) = GCD(a,c) * GCD(b,c)

#77 Help Me ! » Proof by induction? » 2012-05-21 14:19:33

careless25
Replies: 14

Here is the question: http://pastebin.com/8yrJxMKE
I am sure this has to be proved by induction.
I am guessing my basis step would be, for n rocks where n = 2, then k*m = 1 = (n(n-1))/2 = 1.
induction hypothesis: basis step holds true for n >= 2

Now I am stuck, because k*m could break apart easily into many different possibilities and if i try to keep it such that k decreases and m is always 1, this doesnt take into account all the possibilities.

Can someone give me a little push in the right direction?

Thanks!

#79 Re: Help Me ! » probability » 2011-12-02 07:46:28

If we assume without replacement, then there are 9*9*8*7=4536  possible 4 digit numbers.

I am wondering how you got 9*9 when you say without replacement.
enlighten me! big_smile

Edit: Nevermind, I see what assumption you made.

#80 Re: Help Me ! » Proofs and Logic Numerically Equivalent » 2011-12-02 02:37:42

this might help:
http://en.wikipedia.org/wiki/Complex_number#Formal_construction

#81 Re: Help Me ! » probability » 2011-12-02 02:29:07

Are we allowed to choose a number more than once? for eg can we make a 4 digit number: 9999?

And to get you started:

How do you check if a number is divisible by 3?
How many such combinations are there from your set?

#82 Re: Help Me ! » p and c » 2011-12-01 07:18:57

Arent there 20 7s in 1-99??

7, 17, 27, 37, 47, 57, 67, 77, 87, 97, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79

#83 Re: Help Me ! » Circuits but math problem » 2011-12-01 05:51:32

Hi,

I have posted the question here, if you wanted to follow it:
http://forum.allaboutcircuits.com/showthread.php?t=62776

C25

#84 Re: Help Me ! » Circuits but math problem » 2011-12-01 02:37:37

Hi,

i think you are over thinking this by a mile.

I'm very confused about the 8 mA and 2 mA symbols

The 8mA and 2mA are ideal current sources. (neglect internal resistance R=0)
12V is an ideal voltage source.

now what I need to do is find the voltage across the 1kOhm resistor by simplifying the circuit. Thevenins therorem: take any two nodes, you want to find the voltage across; the rest of the circuit can be simplified into a voltage source and an equivalent resistor.

For this question what I have to do is find Vo by simplifying the rest of the circuits into a voltage source and a equivalent resistor.

Edit: To simplfy the circuit I can use KCL, KVL and superpositioning.

Let me know if you cant solve it, I will post this in a Physics forum.

Thanks,

C25

#86 Re: Help Me ! » Circuits but math problem » 2011-11-30 13:55:00

OK heres a question I am having trouble with:

Circuit: https://docs.google.com/open?id=0By2PVt4Tdl6VZjNmMGRmODgtOWVmYy00MTUxLWExNjItZmQ4MTMxMzYyZDAy

I have to solve for

using Thevenin's Theorem.

I dont know how to break apart the circuit, I tried just removing the 1kOhm resistor and trying to find V(Open Circuit) and R(thevenin) but I couldnt solve for V(OC).

Thanks in advance,
C25

#87 Re: Help Me ! » limit » 2011-11-27 17:15:46

factor out x from the inner braces:



factor out x from the whole term:


apply the limit and evaluate:

#88 Re: Help Me ! » Circuits but math problem » 2011-11-27 14:59:30

Thanks for the explanation, I eventually figured out how to cancel the units. big_smile

I need help with circuit problems, will you be able to help or do you know any good forums where I can ask my questions?

Thanks,

C25

#89 Help Me ! » Circuits but math problem » 2011-11-26 07:49:38

careless25
Replies: 8

I am trying to solve this problem but the units dont cancel out, maybe someone can shed some light on where I am going wrong.
These are equations from a circuits problem, applying KCL(nodal analysis) on a circuit. Here are the equations:



Now when I try to solve for Ix, after substituting for V1 and V2 into the second equation, I get stuck here:

I cannot cancel out the Ohm from the denominator of the first term. If I ignore it I get the right answer but I was just wondering if there was something I am missing here.

Thanks,

C25

#90 Re: Help Me ! » Perms and Combs! and some induction.. » 2011-06-04 07:32:42

Wow guys! Sorry for the confusion...this was how the problem was stated to us and it was very confusing!!!
Thanks for the help! I appreciate it, was busy for the last few days so I couldn't get back to you guys but looks like you figured it out.

#91 Re: Help Me ! » Cmpd Intrst » 2011-05-30 13:58:03

Its not the "=" thats messes things up, its the order of operations, remember PEDMAS/BEDMAS?

In the first case you are doing division first 0.088/2 and the following the PEDMAS rules you add 1 which gives you 1.044

In the second case you are adding first and then dividing. thats why your answers differ

#92 Re: Help Me ! » Perms and Combs! and some induction.. » 2011-05-30 05:37:49

The answer i posted is correct...i verified with Teachers Assistant.
You consider the third case i posted because you can have one card which of same suit at lets say card 2 and can also be the same value as card 1. So yea three cases, add them together and divide by total possibilities.

#93 Re: Help Me ! » Perms and Combs! and some induction.. » 2011-05-30 01:20:49

3 cases:
1. diff value, same suit as one card


2. diff suit, same value as one card

3. same suit, same value with another card

Final answer:

make sense???

#94 Re: Help Me ! » Perms and Combs! and some induction.. » 2011-05-30 00:12:34

That was stated in the question. (I did not post it when i posted the question)

#95 Re: Help Me ! » Perms and Combs! and some induction.. » 2011-05-29 23:58:55

Oh! I thot

was wat your answer was.
Nvm that is actually correct and agrees with my answer for the probability of getting a badugi.

#96 Re: Help Me ! » Perms and Combs! and some induction.. » 2011-05-29 23:54:10

AHHH!! HAHAHA Bobbym i am pretty sure that is not the right answer....remember the hint: it is between 55% and 60%...ur answer is 6.3%

#97 Re: Help Me ! » Perms and Combs! and some induction.. » 2011-05-29 23:50:05

Hi gAr!

Its actually
second card: different sut different VALUE so 52 - 13 - 3 = 36
third card: 52-26 - 4 = 22

#98 Re: Help Me ! » Perms and Combs! and some induction.. » 2011-05-29 22:01:00

Hahaha I knew I was right about the balls and slots, I was just mixing up the variables


Where n is the number of letters and k is the number of slots. We learned the right side of the equation in class so I am more accustomed to that, I just have to figure out which variable is which.

#99 Re: Help Me ! » Perms and Combs! and some induction.. » 2011-05-29 21:30:12

In other words, you are one card away from a Badugi. This is fairly tricky, so be careful with the counting here. The correct answer is between 55% and 60%.

#100 Re: Help Me ! » Perms and Combs! and some induction.. » 2011-05-29 21:25:56

There is never a 3 card badugi, for a badugi you need 4 cards with 4 different suits and values.

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