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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)
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!
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!
Edit: Nevermind, I see what assumption you made.
this might help:
http://en.wikipedia.org/wiki/Complex_number#Formal_construction
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?
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
Hi,
I have posted the question here, if you wanted to follow it:
http://forum.allaboutcircuits.com/showthread.php?t=62776
C25
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
Oops nice catch :$
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
factor out x from the inner braces:
Thanks for the explanation, I eventually figured out how to cancel the units.
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
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
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.
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
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.
3 cases:
1. diff value, same suit as one card
Final answer:
make sense???
That was stated in the question. (I did not post it when i posted the question)
Oh! I thot
was wat your answer was.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%
Hi gAr!
Its actually
second card: different sut different VALUE so 52 - 13 - 3 = 36
third card: 52-26 - 4 = 22
Hahaha I knew I was right about the balls and slots, I was just mixing up the variables
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%.
There is never a 3 card badugi, for a badugi you need 4 cards with 4 different suits and values.