Ahnfeltia
3254011055738
★★★★
- Joined
- Oct 6, 2022
- Posts
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my ballsack is racist
Its Homer Simpson colored?my ballsack is racist
Nigga are you seriously asking me about my ballsack?Its Homer Simpson colored?
No, Your not Homer Simpson.Nigga are you seriously asking me about my ballsack?
the answer is "no" btw
A Turing machine is defined by the following scheme. Formulate an algorithm according to which this machine operates
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Is this computational complexity?Calculate the linear algorithmic complexity of the following Turing machine
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While computational complexity is more broad than Turing machines, the second question is indeed concerned with the computational complexity of a certain Turing machine.Is this computational complexity?
Correct .This is trivial by utilizing Mollweide's formula. Your ratio of cosines equals (a + c)/b = (b - k + b + k)/b = 2.
The problem didn't state it.What is the domain of F?
Yeah, I realized it was a silly question, hence why I tried to delete it in time.The problem didn't state it, but you can solve it solely using N
F(14,52) = F(14,38)*52/38 = F(14,24)*38/24*52/38 = F(14,10)*24/10*52/24 = F(10,14)*52/10 = F(10,4)*14/4*26/5 = F(4,10)*7*13/5 = F(4,6)*10/6*7*13/5 = F(4,2)*6/2*10/6*7*13/5 = F(2,4)*10/2*7*13/5 = F(2,2)*4/2*7*13 = 2*2*7*13 = 364F(x,x) = x.
F(x,y) = F(y,x).
(x + y)f(x, y) = yf(x, x + y).
Find F(14, 52)
CorrectF(14,52) = F(14,38)*52/38 = F(14,24)*38/24*52/38 = F(14,10)*24/10*52/24 = F(10,14)*52/10 = F(10,4)*14/4*26/5 = F(4,10)*7*13/5 = F(4,6)*10/6*7*13/5 = F(4,2)*6/2*10/6*7*13/5 = F(2,4)*10/2*7*13/5 = F(2,2)*4/2*7*13 = 2*2*7*13 = 364
How close is it?Once AI will be able to do math at the level of the best contemporary mathematicians, everything will be cope.
Very far away still, fortunately for me.How close is it?
S(n) = S(n-1) + Σ (i = 1 to n-1) i+(n/10) for n≥3Find S(n), given that S(1) = 0. S(2) = 1.2, S(3) = 1.2 + 1.3 + 2.3. S(4) = 1.2 + 1.3 + 1.4 + 2.3 + 2.4 + 3.4
What is that summation adding up?S(n) = S(n-1) + Σ (i = 1 to n-1) i+(n/10) for n≥3
Like for S(4) for example, the summation will be 1 + (4/10) + 2 + (4/10) + 3 + (4/10) = 1.4+2.4+3.4What is that summation adding up?
Also, it has to be as a function of n, not S(n).
. as in a multiplication sign, it's 1 times 2, 1 times 3 and so onLike for S(4) for example, the summation will be 1 + (4/10) + 2 + (4/10) + 3 + (4/10) = 1.4+2.4+3.4
Bruh I thought it was a decimal sign. You should've used the * sign for multplication.. as in a multiplication sign, it's 1 times 2, 1 times 3 and so on
Can you still solve it?Bruh I thought it was a decimal sign. You should've used the * sign for multplication.
S(n) = Σ (i = 1 to n-1) ( Σ (j= i+1 to n) i*j)Can you still solve it?
@trying to ascend ?S(n) = Σ (i = 1 to n-1) ( Σ (j= i+1 to n) i*j)
Is it a doule summation? Assuming it is, then it's not correct, provided that S(3) = 11 and yours gives 14.@trying to ascend ?
Wdym my summation does give me 11.Is it a doule summation? Assuming it is, then it's not correct, provided that S(3) = 11 and yours gives 14.
We will have sigma (upper equals 3 and lower 2) from (i)(i + 1) + Sigma (upper 3 and lower 3).
1.2 + 2.3 + 2.3.
You won't have pair of terms whose difference is higher than 1 in that summation.
Hint: The problem has a relation with arithmetic progressions
Was this what you wrote:Wdym my summation does give me 11.
(1 + 2 + ... + n)^2 = 1^2 + 2^2 + ... + n^2 + 2S(n). Since 1 + 2 + ... + n = n(n+1)/2 and 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6, one can easily solve for S(n) and find that S(n) = (n-1)n(n+1)(3n+2)/24.Find S(n), given that S(1) = 0. S(2) = 1.2, S(3) = 1.2 + 1.3 + 2.3. S(4) = 1.2 + 1.3 + 1.4 + 2.3 + 2.4 + 3.4
You have merely rewritten his function in terms of summation notation, which hardly counts as solving the problem.S(n) = Σ (i = 1 to n-1) ( Σ (j= i+1 to n) i*j)
Correct(1 + 2 + ... + n)^2 = 1^2 + 2^2 + ... + n^2 + 2S(n). Since 1 + 2 + ... + n = n(n+1)/2 and 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6, one can easily solve for S(n) and find that S(n) = (n-1)n(n+1)(3n+2)/24.
You have merely rewritten his function in terms of summation notation, which hardly counts as solving the problem.
@trying to ascend I concur. While using a "." for multiplication is fine when it comes to variables, using it this way between numbers is very confusing because virtually everyone will interpret it as a radix point. Moreover, I believe using a "." for multiplication is Indian notation, but Wikipedia suggests India uses the "." as the decimal separator as well, so I can't imagine using "." to indicate multiplication between numbers is common practice.Bruh I thought it was a decimal sign. You should've used the * sign for multplication.
Hint: power rule.f(x) = 2x^2 what is f’(x)
public PerChanceTest() {
super();
sentences.add("here is a salad vegi1 #, vegi2 # and herb #.");
sentences.add("how about this salad vegi1 #, vegi2 # and herb #. it goes well with tuna fish");
UniqueItemSizeLimitedPriorityQueue temp = new UniqueItemSizeLimitedPriorityQueue();
temp.setLimit(3);
wordToList.put("vegi1", temp);
temp = new UniqueItemSizeLimitedPriorityQueue();
temp.setLimit(3);
wordToList.put("vegi2", temp);
temp = new UniqueItemSizeLimitedPriorityQueue();
temp.setLimit(3);
wordToList.put("herb", temp);
}
If I understand you correctly, you have x different sentences, each with n variable slots and each slot permitting y options, correct? In that case, the total number of possible sentences is x*(y^n).if the amount of sentences is x, and each param list is marked as a1...an with y elements per list.
how many senstence options are there?
4xf(x) = 2x^2 what is f’(x)
Yes, it indeed looked confusing, though I was just copying the original problem@trying to ascend I concur. While using a "." for multiplication is fine when it comes to variables, using it this way between numbers is very confusing because virtually everyone will interpret it as a radix point. Moreover, I believe using a "." for multiplication is Indian notation, but Wikipedia suggests India uses the "." as the decimal separator as well, so I can't imagine using "." to indicate multiplication between numbers is common practice.
fair enough in that caseYes, it indeed looked confusing, though I was just copying the original problem
I thought that was okay.(1 + 2 + ... + n)^2 = 1^2 + 2^2 + ... + n^2 + 2S(n). Since 1 + 2 + ... + n = n(n+1)/2 and 1^2 + 2^2 + ... + n^2 = n(n+1)(2n+1)/6, one can easily solve for S(n) and find that S(n) = (n-1)n(n+1)(3n+2)/24.
You have merely rewritten his function in terms of summation notation, which hardly counts as solving the problem.
Generally not (although there technically are exceptions). Here's a rather extreme example: would you say that sum from j=2 to 4 of j is a valid solution to "find the value of 2 + 3 + 4"?I thought that was okay.
No but the upper limit for your example is bounded.Generally not (although there technically are exceptions). Here's a rather extreme example: would you say that sum from j=2 to 4 of j is a valid solution to "find the value of 2 + 3 + 4"?
Does any of the limits being a variable change anything significant? In that case you could argue that summation notation is more formal as opposed to the ellipses, but that's about it, no? At the end of the day you have merely recast the given expression in a more formal form without really doing any kind of simplification.No but the upper limit for your example is bounded.
that looks like a lot when it's put into a formula, maybe it could be used to write visual novel type of games, scriptsIf I understand you correctly, you have x different sentences, each with n variable slots and each slot permitting y options, correct? In that case, the total number of possible sentences is x*(y^n).
it is a lot, esp. for large nthat looks like a lot when it's put into a formula, maybe it could be used to write visual novel type of games, scripts
I take it you want the number of essentially distinct ones? Mirror images counting as the same figure etc.Consider 9 dots arranged as 3x3 what is the highest amount of closed figures that can be drawn by connecting the dots one by one in a single trial you can imagine this as drawing pattern lock in your phone i have achieved 10 closed figures so far
Wait i will upload an image to demonstrateI take it you want the number of essentially distinct ones? Mirror images counting as the same figure etc.
NotificationI take it you want the number of essentially distinct ones? Mirror images counting as the same figure etc.
Aha. I see what you mean now. 11 is possible.There are 10 closed figures in this image
I'm not convinced this is true. Take P(x) = ¼ + x. In that case your desideratum amounts to ¼ + ab ≤ (¼ + a)(¼ + b) for positive a & b, but this is evidently false for a = b = ¼.Let P(x) be a polynomial with positive real coefficients. Prove that
Sqrt(P(a)P(b) )≥ Sqrt(P( ab)),for all positive real numbers a and b.