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I can find no loopholes in that question Simon.
I see what you're saying, but I think clarification would be needed to ensure that that we can use that type of reasoning.
The only problem I can see is the cheif who can answer truthfully or falsely. It's possible that he would ansewer "true" if you asked him if 1+1=2, but then he lies to your question, and would answer "no".
You should stick with the door you picked originally. If you keep your first choice you have a 50% chance of winning, while if you switch doors you only have a 25% chance of winning. To picture why this is, below are all the possible configurations:
CCGG
CGCG
CGGC
GCCG
GCGC
GGCC
G is for goat, C is for car. Assume that you chose door #1, the choice on the far left (it doesn't matter which you choose, #1 is just used for an example). In the first 3 cases you'll win if you keep your choice, while you'll lose in the second 3 cases. This gives you a 50% chance of winning.
Now, if you decide to switch, you'll lose every time in the first 3 cases. In the second 3 cases you have a 50% chance of winning. Those 3 cases only make up half of the possibilities, so your total chance of winning is 50% * 50% = 25% chance of winning, so you're better off keeping your first choice.
P = Product of first and last terms = -12y
I think you mean -12y².
S = Sum (what does "Sum" refer to?) or middle term = 4y
I'm not completely certain, but I think it's the value that your 2 factors have to add up to, or "sum" to.
= 2y(2y + 3) - 1(2y + 3) [How did it go from this: "2y(2y + 3) - 1(2y + 3)" to this "(2y + 3)(2y - 1)"?????]
They're using the distributive property. It can be confusing at first, so use replacement variables to visualize what's going on.
Let's use u = 2y + 3. The the left side of the equation becomes 2yu - 1u. Using the distributive property you can see that this is equal to (2y - 1)u. Now, substitute 2y + 3 back in for u and you get (2y - 1)(2y + 3). The trick is to treat everything inside the parentheses as a single variable.
Question 2 I think this question is just IMPOSSIBLE
2a² + 11a² + 5
P = 10a²
S = 11a²
F = Factors of P that give S = x, y
x = ????
y = ????
There are 2 possible factor pairing that can give 10a²: 2a*5a and a*10a. Only one of these adds up to S, or 11a², and that is a*10a. Thus, we get an expression that looks like this:
2a² + 10a + a + 5 = 2a(a + 5) + 1(a +5) = (2a + 1)(a + 5).
I don't understand your question. Are you trying to find the base value of T, like T(1)? Or are you trying to find a non-recursive definition of T(n)?
If it's the first, the problem is easy. You know T(3) = 3 x T(2) x 2^2 = 3 x 4 x T(2) = 36, so T(2) = 3. Using this same method you can find that T(1) = 1/2.
If you need a non-recursive definition of T(n), try looking backwards. What I mean is to do something like this:
The pattern here is pretty easy to see. We have 3 raised to the power of n-1 at the start. We then have 2 raised to the power of n^2 - (sum from i = 1 to n of i). This is then multiplied by our base T(1), which we know from above is 1/2. This leaves us with
Let's check that this is right. For T(1) we get 3^0 x 2^(-1) = 1/2. For T(2) we get 3^1 x 2^0 = 3. For T(3) we get 3^2 x 2^2 = 36. And finally, T(4) = 3^3 x 2^5 = 864, all of which are what we expected.
x + y > 2, x - y < 2
x > 2 - y, x < 2 + y
->
2 - y < 2 + y
->
0 < 2y
->
y > 0
The 'solution' would be y > 0, 2 - y < x < 2 + y (alternatively, |x - 2| < y). I've never heard of such a thing mikau, would you mind posting your method?
The limit can only be found in relation to x, it is not a constant. Clearly the limit does not exist when x = kpi, since we have a divide by 0 error in that case. When x does not equal kpi the cos^{n+1}x term will drop, leaving you with
This can probably be simplified further, but I'm too lazy to deal with the absolute values, so I leave that to you. What we're left with is this:
Why not 6(a+1)(a-1)(a-1) ?? Just wondering?
It is John. 6(a+1)(a-1)(a-1) = 6(a-1)(a^2-1), which is what luca said.
It sounds like he's given a list of names of students, each of whom has a list of other students that they cannot live next to. The example would then be:
Bob
Judy
Bill
Jane
Bozo
Julie
Bob cannot live next to:
Judy
Bozo
Judy cannot live next to:
Bob
Bill cannot live next to:
Jane
Bozo
Julie
Jane cannot live next to:
Bill
Julie
Bozo cannot live next to:
Bob
Bill
Julie cannot live next to:
Bill
Jane
I've never heard of this particular problem, but at first glance it appears to be at least NP, if not NP-complete. Question for Ricky: what's the difference between NP and NP-complete? Is it that every problem in NP can be reduced in polynomial time to a problem in NP-complete, but not every problem in NP can be reduced to another problem in NP?
Here's some information that may be helpful. Check this Wikipedia link:
http://en.wikipedia.org/wiki/Infinity#Cardinality_of_the_continuum
This paragraph in particular is of interest:
Cardinal arithmetic can be used to show not only that the number of points in a real number line is equal to the number of points in any segment of that line, but that this is equal to the number of points on a plane and, indeed, in any finite-dimensional space. These results are highly counterintuitive, because they imply that there exist proper subsets of an infinite set S that have the same size as S.
A proper subset is a subset that has fewer elements than it's parent. For example, let A be a set containing {1, 2, 3, 4}, B containing {1, 2, 3}, and C containing {1, 2, 3, 4}. B is a subset of A because all of it's elements are contained in A. It is also a proper subset because it does not contain every element of A, namely 4. C is a subset of A because all of its members are contained in A. However, C is not a proper subset of A because there are no elements in A that are not in C.
Now, re-read the Wikipedia quote. There exists proper subsets of S (S being any infinite set, countable or otherwise) that have the same size as S. One such example would be A = {1, 2, 3, 4, 5, ...} and B = {2, 3, 4, 5, 6, ...}. B is clearly a proper subset of A, but it can also be shown that they have the same size. It's counterintuitive, but it's been rigorously proven. This is another Wikipedia page that may be helpful to study:
http://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel
Fixed. Use \lim for limits, not \limit, and use \rightarrow instead of \arrow to make the arrow.
Judging by the above responses you originally accidentally typed the denominator as 3x^2 - 2, when it should be 3x^3 - 2. Assuming that you now have the correct expression, follow LuisDog's advice again, but this time with x^3:
The limits of each of the terms with an x in the denominator is 0 as x approaches infinity, so they can be safely removed from consideration. We're left with
which has no x, so the limit is just the expression itself, -1/3.
The only way I can think of is to say something along the lines of "
" decreases faster than t increases, as t increases" (horribly worded I know). Is that what the question wants?
Yes, that's what it's asking. Technically it wants you to show that
but it's basically the same thing once the constants are removed. Just by observing you can see that
will approach 0, since t is growing linearly while e^-t is decreasing exponentionally, but that's not exactly rigorous. I can't think of any easy theorems you can use to prove this so I'd just go with Ricky's suggestion and make a proof using the definition of limits. Be sure to include C and k for completeness, they shouldn't make the problem that much more complicated.Use the rule log(a) + log(b) = log(ab). Applying this rule to your example you'd get log(x) + log(8) = log(8x) = 2. It should be fairly straightforward to solve from there.