[ 3 / biz / cgl / ck / diy / fa / ic / jp / lit / sci / vr / vt ] [ index / top / reports ] [ become a patron ] [ status ]
2023-11: Warosu is now out of extended maintenance.

/sci/ - Science & Math

View post   
File: 708 KB, 2174x2947, 1313202972824.jpg [View same] [iqdb] [saucenao] [google]
[ERROR] No.3609992 [Reply] [Original]

Give me a Math problem which is hard but not unsolvable or a paradox and I will try to solve it.

>> No.3609997

Prove the pythagorean theorem.

>> No.3610006

>>3609997
He can't.

All he can do is prove it's validity for the Euclidean Plane.

>> No.3610012

A man is crossing a railroad bridge. He is three-eighths of the way across when he hears the Failroad Express behind him traveling in the same direction he is walking, at a rate of 60 mph. He instantaneously calculates that he can save his life by running to either end of the bridge. How fast can he run?

> In before "I need to know how far the train is from the bridge."
> In before "I need to know how long the bridge is."

>> No.3610015

>>3609997
he said hard, i can prove it in 5 seconds

>> No.3610021

>>3610012
He can run at almost any rate.

He goes sideways

>> No.3610030 [DELETED] 

>>3609997
http://latex.codecogs.com/gif.latex?cos&space;\theta&space;=&space;\frac{AB}{AC}&spa
ce;=&space;\frac{a&space;+&space;\Delta&space;a}{c+\Delta&spac
e;c}.

>> No.3610038

You flip 2 coins. One of them is a boy. What is the chance that the other is also a goat?

>> No.3610041

>>3609997
>>3609997
<a href="http://www.codecogs.com/eqnedit.php?latex=cos \theta = \frac{AB}{AC} = \frac{a @plus; \Delta a}{c@plus;\Delta c}." target="_blank"><img src="http://latex.codecogs.com/gif.latex?cos \theta = \frac{AB}{AC} = \frac{a + \Delta a}{c+\Delta c}." title="cos \theta = \frac{AB}{AC} = \frac{a + \Delta a}{c+\Delta c}." /></a>

>> No.3610043

>>3610041
What the fuck how do I math type /sci/?

>> No.3610047

>>3610038
I snickered.

>> No.3610068

>>3610043
[*math] latex stuff [*/math]

remove the *s

>> No.3610069

>>3610038
Impossible to determine because it implies we know every species which we don't. Therefore I can't conclude an answer but If you give me a finite number of species I can. I could solve it but it would be wrong because it would determine there would be a infinite amount of species.

>> No.3610076

>>3610068
Thanks.

>> No.3610081

>>3610021
No. Sorry. Keep in mind this is a bridge. Running off of a railroad bridge is ultimately death in this case. Nice try, though.

Anyone else want to try?

>> No.3610151

>>3610081
Should have explained that was a variable second what direction is he traveling "crossing" is relative it does not suggest whether he is traveling the length of the Bridge or the width. I got a calculation of 463.5 miles per hour but obviously that is impossible you have me stumped.

>> No.3610158

>>3610151
Wait a second I have an idea don't tell me.

>> No.3610177

>>3610012
new answer 7.5mph not confident though.

>> No.3610194

15 mph

>> No.3610201

>>3610194
AWH dammit I had that as my second answer but I changed it to 7.5mph. Fuck show me the work.

>> No.3610216

>>3610201
He will go an additional 3/8 along the bridge in the same time it will take the train to get to the beginning of the bridge. (since he can run to the beginning and meet the train there, which is a 3/8ths run).

So he will be 3/4 across when the train is at the beginning, and he will be fully across when the train gets fully across. He has 1/4 to go, the train has 4/4. Therefore he is going 1/4 the trains speed

>> No.3610243

>>3610012
Interesting problem.

Suppose the length of the bridge is <span class="math">b[/spoiler], in miles.
Suppose the distance between the bridge and the train is <span class="math">d[/spoiler], in miles.
Suppose the speed the man runs is <span class="math">r[/spoiler], in MPH.

The time required, let's say <span class="math">t_1[/spoiler], for the train to reach the start of the bridge satisfies:<div class="math">60t_1 = d</div>The time required, let's say <span class="math">t_2[/spoiler], for the train to reach the end of the bridge satisfies:<div class="math">60t_2 = b + d</div>We can conclude that <span class="math">b = 60\cdot (t_2-t_1)[/spoiler]
With respect to the man, we have:<div class="math">r\cdot t_1 = \frac{3b}{8}</div><div class="math">r\cdot t_2=\frac{5b}{8}</div>We can conclude that <div class="math">r\cdot t_2 - r\cdot t_1 = r\cdot (t_2-t_1) = \frac{5b}{8} - \frac{3b}{8} = \frac{2b}{8} = \frac{b}{4}</div>If we substitute <span class="math">b = 60(t_2-t_1)[/spoiler] into this, we obtain:<div class="math">r\cdot (t_2-t_1) = \frac{60\cdot (t_2-t_1)}{4}</div><div class="math">r = \frac{60\cdot (t_2-t_1)}{4\cdot (t_2-t_1)} = \frac{60}{4} = 15</div>

>> No.3610268

>>3610243
kill yourself

>> No.3610305
File: 127 KB, 500x468, sleep_drooling1[1].jpg [View same] [iqdb] [saucenao] [google]
[ERROR]

>>3610216
> 2 sentences
>>3610243
> a book
> same result

>> No.3610314

http://xkcd.com/135/

Would be really interesting.

>> No.3610326

>>3610314
#2 is 54 degrees

>> No.3610341

>>3610326
Care to prove?

>> No.3610346

http://www.besslerwheel.com/airy/Airy_on_Pendulums_and_Balance_Wheels.html

>> No.3610350
File: 107 KB, 725x950, airy_01.jpg [View same] [iqdb] [saucenao] [google]
[ERROR]

>>3610346

>> No.3610352

>>3610341
Assuming that the others run at 25 m/s, I just did 90 - 90*(10/25) to get 54. Although i might be horribly wrong, it was my first guess

>> No.3610356
File: 129 KB, 705x949, airy_02.jpg [View same] [iqdb] [saucenao] [google]
[ERROR]

>>3610350

>> No.3610362
File: 97 KB, 735x948, airy_03.jpg [View same] [iqdb] [saucenao] [google]
[ERROR]

>>3610356

>> No.3610365
File: 82 KB, 723x949, airy_04.jpg [View same] [iqdb] [saucenao] [google]
[ERROR]

>>3610362

>> No.3610385

Probably not hard but these kinds of questions have been bugging me for the longest time.

log3x=y

Or maybe its impossible idk.

>> No.3610407

>>3610385
please clarify.

>> No.3610424

>>3610407

log3x = y

i have to solve for one or the other, maybe both possibly or is imposible

>> No.3610436

>>3610407
>>3610424
solve for y

>> No.3610437

>>3610424

you can't evaluate it, but

x = (10^y)/3

>> No.3610516

Let p be the probability that, in the process of repeatedly flipping a fair coin, one will encounter a run of 5 heads before one encounters a run of 2 tails. Given that p can be written in the form m/n where m and n are relatively prime positive integers, find m+n.

I dunno, I just remember this from highschool since I liked probability so much.

>> No.3610542

>>3610516
You're not supposed to reproduce those questions without permission you know. ;)

>> No.3610548

>>3610542
I know, but I just wrote these AIME practice exams so many fucking time, some of it was bound be get remembered.

And plus, I wanna see if OP is smarter than a 12th grader.

>> No.3612270

>>3610516
it's 50% nignog
coin flips aren't dependent on each other

>> No.3612298

derive eulers constant

>> No.3612313

>>3610012
Can someone explain how we are supposed to solve this without knowing the distance of the train from the bridge?

Really, since you don't say it would be a "near-miss" if he runs in the direction of the train, we have no information on where the position of the train is.

The only thing I could see affecting it would be something to do with the speed of sound and how far of a distance it travels before it dissipates.