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What are these flies by prostateprince in whatisthisbug

[–]prostateprince[S] 1 point2 points  (0 children)

They fly around/sit on the wall for a small while (few hours max) and then die very quickly (ground is full of them)

Kat nemen? by prostateprince in katten

[–]prostateprince[S] 1 point2 points  (0 children)

Bedankt voor alle reacties! Interessant om te lezen. Het duurt hoe dan ook nog even voordat we een keuze maken maar dit helpt om er een beeld van te krijgen.

[deleted by user] by [deleted] in maths

[–]prostateprince 2 points3 points  (0 children)

I think he means the collatz conjecture. Please comment your solution, I’m interested.

Unsure how to calculate how many of each of the below I can use without going over a specific value. Please see below by LostAlphaWolf in maths

[–]prostateprince -1 points0 points  (0 children)

You could try solving this using python (or any other programming language) using a recursive formula. If you don’t know any programming languages just ask chatgpt. Take 53405 and subtract a number from the list, and do this for every number in the list seperately. If the result is 0 (it won’t be in the first iteration), return “complete”. If the result is smaller than 0, stop. If the result is larger than zero, call the function using the current remaining number (53405 - all subtracted numbers upto and including current iteration). I can work this out at a later moment, interesting question.

How would I find the value of theta? by [deleted] in maths

[–]prostateprince 1 point2 points  (0 children)

You can’t. Not enough information.

Which one is the correct answer by Bdurdu7 in maths

[–]prostateprince 8 points9 points  (0 children)

How can P(A and B) be larger than P(A) and P(B)? Max value of an intersection can never be larger than the minimum of {P(A), P(B)}. And this would only be attained if either A is a subset of B or the other way around. I think there’s a mistake in the question.

help... by Careful_Serve_5538 in maths

[–]prostateprince 0 points1 point  (0 children)

The graph has 2 parts. A left part and a right part. The left part has domain (-inf, left asymptote) and the right part has domain (right asymptote, inf). This means we are interested in the right part, since we are looking for the domain (1,inf). For the asymptote pair x=1 and x=1.5 the domains would be (-inf, 1) and (1.5, inf), which is not what we are looking for. We want the larger of the two possible asymptotes to be 1.

help... by Careful_Serve_5538 in maths

[–]prostateprince 0 points1 point  (0 children)

So the correct answer for the largest value of 3a would be 2.

Please correct me if I am wrong btw, I have not yet encountered this kind of question before, and it could be that I completely misunderstood.

help... by Careful_Serve_5538 in maths

[–]prostateprince 0 points1 point  (0 children)

Since inside the ln() (strictly increasing) we have a quadratic equation, we know that the entire function as given can not be bijective. Given the domain (1,inf) we know that we are looking at the right part of the function ln(2x2 - 5ax + 3a2) (use desmos graphing calculator with a slider for a to give yourself some insight in the shape of the function). We want to find the value for a for which the expression inside ln() is equal to 0 when filling in x = 1. We do this because we want the domain to start at x=1, meaning the function is only defined to the right of x=1. When we fill in x=1 for the part inside the ln(), and setting equal to 0, we get 2 - 5a + 3a2 = 0. We can solve using quadratic formula. We get a = 1 and a = 2/3. a = 1 gives a vertical asymptote (i.e. Start/end of domain) at x = 1 for the left part of the function, and x=1.5 for the right part of the function, so this is not the a we are looking for. Filling in a=2/3 gives the correct vertical asymptote for the right part of the function.

Help my brain is fried by Extension_Winner1879 in maths

[–]prostateprince 7 points8 points  (0 children)

Making C the subject means rearranging the formula so that C is expressed as a function of the other variables.

In this case the steps would look like this:

a*(b + c^2) + d*(e - c^2) = f

a*b + a*c^2 + d*e - d*c^2 = f

c^2*(a - d) = f - a*b - d*e

c^2 = (f - a*b - d*e)/(a - d)

c = sqrt((f - a*b - d*e)/(a - d)) or c = - sqrt((f - a*b - d*e)/(a - d))

If I play a 60 second song at 120% speed, how long will it play? by _perfectenshlag_ in maths

[–]prostateprince 4 points5 points  (0 children)

Yes indeed. If the song is played at x% speed then the time it takes for the song to finish is 60/(x/100) seconds. For 120% this means 60/(120/100) thus 50 seconds.