Qucik Question ...

[BadgerJelly]

Show what p is:

a^2p = b

p=?

[BadgerJelly]

Never mind. Was so easy it was confusing! haha! Really strange exercise.

[neuro]

so you are now switching from Husserl to algebra?

[Braininvat]

What's interesting about this. Does the answer make some sort of pun phrase, if you say it the right way?

"Be over a squared."

"Be over a to the second."

Wha?

[BadgerJelly]

so you are now switching from Husserl to algebra?

Haha! I still read him a little, but truth is no one here seems overly interested in tackling him in depth.

I will try again in the future to show why I believe his work is important for todays world and for science especially. I think others will beat me to it though, and probably already have somewhere?

He is a very slippery customer too in what I was presenting. Crisis was not a finished work and Husserl himself is regarded by many in philosophy as being quite ambiguous.

Biv -

I was just too dim witted to get my head around it. Cannot explain why, just a special kind of stupidity I have when looking at proofs.

[neuro]

What's interesting about this. Does the answer make some sort of pun phrase, if you say it the right way?

"Be over a squared."

"Be over a to the second."

Wha?

Not that it has much interest, actually, but I believe that the equation read a^{2p} = b, which solves as
2·p·log(a) = log(b), p=log(b)/(2·loga)

which however (pislogbiovertwologa) does not make up anything funny anyway

[BadgerJelly]

a=b therefore p=1/2

Not funny at all, sorry Biv :(

[Braininvat]

Ooops, misread your notation as meaning A²p=b. Which would give p = b/A².

This is why brackets are so important. :-)

[doogles]

Here' a new one that should be easy. I saw it somewhere a week ago and couldn't get my head around it properly as it kept coming back into my head. That was until I woke up in the middle of the night and in my semi-sleep state last night, worked it out. I think my answer is correct, but I would love someone to confirm my logic.

There are two painters. One can complete the painting of a shed in exactly 3 hours and the other in exactly 5 hours. This is a given. Now provided there is no rain, both remain uninjured and set out to do the job together until the shed is completely painted. They work uninterrupted, each working at the speed indicated above until the job is finished.

How long would they take?

[Braininvat]

Just express the surface area of the shed in units that are a multiple of 3 and 5. Say 30 square meters. One painter has a paint rate of 10 m/hr, the other of 6 m/hr. So their combined rate is 16 m/hr. Easy peasy. 1 and 7/8 hours.

[wolfhnd]

It doesn't matter what the surface area is.

[Braininvat]

It doesn't matter what the surface area is.

Of course it doesn't. My number of units was arbitrary. Just a math convenience. Hence the word "say" in front of "30 sq. meters."

[Lomax]

If it makes it easier to grasp, we can say that the surface area is 30x, where "x" just means "a 30th of the surface area". We can see that this will be true in all cases and that one painter is working at a rate of 10x per hour and the other is working at a rate of 6x per hour. Which produces the solution BiV gives.

[doogles]

Just express the surface area of the shed in units that are a multiple of 3 and 5. Say 30 square meters. One painter has a paint rate of 10 m/hr, the other of 6 m/hr. So their combined rate is 16 m/hr. Easy peasy. 1 and 7/8 hours.

Very clever Braininvat. I can't remember at all where I saw the question, but it has been cropping up in my mind the last week.

My first tendency was to say that if one did it in 3 and the other in 5, then the average time would 4. So together they would do it in 2 h. But something did not seem right about that. The problem has cropped up in my daydreaming a few times, and then last night, during a twilight sleep, I rationalised that one could paint 5/8ths of the area in the same time as the other did 3/8ths. Number one could paint the whole area of the shed in 180 mins, which meant he could paint 5/8ths of it in 112.5 minutes. As a check, it meant that the other could paint his share in 3/8ths of 300 minutes, which is 112.5 minutes or 1 and 7/8 h.

Area was also not relevant in the way I worked it out, but Lomax was more probably more systematic than intuitive in applying algebra.

Anyhow thank you all for playing.

[doogles]

While in mathematics and on the subjects of quick questions, I've had what I consider to be a very simple issue that I believe not only most people on the planet, but that every government in the world, got wrong at the end of the second millennium.

Given that there are side issues about the actual point in time of zero relative to the birth date of Christ and the adjustments made because of Julian, Gregorian and other factors, and agreeing that we've had things roughly in synch over the last couple of millennia, did we collectively get it wrong by celebrating the arrival of the 3rd millennium at the start instead of the end of the year 2000?

I think the answer to this question is very simple but is a classic example of how much of our collective mentality and personal thinking powers are frighteningly so close to that of a sheep.

[Lomax]

I recently turned 30 and several people congratulated me on entering my third decade. If only it were so.

[doogles]

Yes Lomax. It's another good example of the principle of what happened globally at midnight on 31st December 1999.

[mitchellmckain]

so you are now switching from Husserl to algebra?

He has been algebra questions for a while now.

What I don't understand is what he thinks is so strange about the question. Looks pretty standard to me -- one which is usually asked after you cover logarithms in class or in the reading.

[mitchellmckain]

If it makes it easier to grasp, we can say that the surface area is 30x, where "x" just means "a 30th of the surface area". We can see that this will be true in all cases and that one painter is working at a rate of 10x per hour and the other is working at a rate of 6x per hour. Which produces the solution BiV gives.

Sure... BUT...

Mastering the general algebraic method is more useful in the long run.

Use rate x time = amount, to write out an equation for the two finishing one shed in the same time.

(1/3) t + (1/5) t = 1 shed.

Multiply by 15 or get a common denominator and you quickly get t = 15/8 hours, same as Braininvat.

[neuro]

If it makes it easier to grasp, we can say that the surface area is 30x, where "x" just means "a 30th of the surface area". We can see that this will be true in all cases and that one painter is working at a rate of 10x per hour and the other is working at a rate of 6x per hour. Which produces the solution BiV gives.

if you prefer a more refined - though less straightforward - answer, the solution is
1/(1/3+1/5)
as it would be if you were summing up electric resistances in parallel
(in a sense, it is exactly your same answer for a unit surface area to be painted)