/sci/ - Science & Math

find a set [math] X [/math] such that [math] X^2=\mathbb{Q} [/math]

>>9739411
[math]r=\frac{ \left(\frac{6+8+10}{2}\right)^2 }{ (6+8+10) -1}[/math] MOTHAFUCKA
YOU CAN'T TOUCH DIS SHIT
I'M ON FIRE NIGGAAAAAAAA

>>9739418
What is X^2? Do you mean the Cartesian product?

>>9739418
[math]\pm \sqrt{\mathbb{Q}}[/math]

>>9739418
You need to be filtered out, retard.

>>9740012
18

>>9740196

>>9739411
I wanna say 75.4 but that seems too easy. I can draw it out and measure it, but that wouldnt be an academically accepted route for solving this question.

>>9740012
24

>>9740012
undetermined

>>9740208
cringe

>>9739411
~8,8?

>>9739451
brainlet here. can you explain that?

el bumpo

>>9740012
[eqn]A = \frac{1}{2} \sum^{n-1}_{i=0} (x_i~y_{i+1}-x_{i+1}~y_i)[/eqn]

>>9739451
Wrong.

>>9740716
No, he's right, see question 6:
http://www.cemc.uwaterloo.ca/contests/past_contests/2017/2017CSMCSolution.pdf

>>9740421

Find length AB in terms of the radius of the circles.

>>9740724
very nice, thank you

>mfw made a careless mistake and kept getting a 4th circle that intersects the other 3 (extended) circles at the origin

>>9740012
How would you even show that an answers wrong?

>>9740764
I think this one is impossible

>>9740803
That's brainlet talk.

let's say I am given a certain velocity-position function, how do I find the adequate velocity-time function.
for example:
V(x)=x^2
V(t)=?

>>9740012
A is the total area.
x is the blue area.

Clearly

52 > A/2.
32 + x > A/2.

36 < A/2.
16 + x < A/2.

This gives us bounds on x of 18 < x < 20.
assuming the answer is an natural number like every other area which is consistent with the file name as kindergartners are only taught "whole numbers" we have x = 19.

>>9739411
5yo childs solve this in ukrania every day

>>9740879
aborted fetuses in america came up with harder problem

>>9740826
vdv = ads nigga

>>9740809
>>9740764
not >>9740803 but I gave up and cheated https://www.desmos.com/calculator/elwmv1oxtg
and yet I still can't find a nice clean answer

>>9740764
[eqn] \frac{8}{29} \sqrt{19\sqrt{7} + 2} \approx 1.99441 [/eqn]
Jesus christ that was a bitch

>>9740996
The trick is to draw a line to the center of the the top middle circle (F), and then use trigonometric identities to find the (sine and cosine of) angles EOD and EOF, then find angle DOF using difference of angles formulae.
You can then draw FA and FB and use the cosine rule on triangles AOF and BOF to solve for OA and OB. The answer will be the difference.

It's simple, really.

>>9740764
[eqn] \frac{8}{29} \sqrt{19\sqrt{7} + 2} \cdot r \approx 1.99441 r[/eqn]
Jesus christ that was a bitch

>>9741027
>>9741033
Well done on adding the r.

(It's correct, too.)

>>9741035
nice

>>9740803
Since there are no values given for anything and assuming all of the circles are equivalent in terms of radius, AB = 1/3 * OE

>>9741038
Did you use this method >>9741028 or another?

find the area of the shaded part

question for Chinese elementary students

>>9741046
hurr

>>9740764
2r

>>9741045
I did it differently.

I set up the equation for the line:
[math] y = ax [/math]
The equation for the circle it passes through:
[math] (y-2)^2 + (x-3)^2 = 1 [/math]
And the equation for the circle it leans on:
[math] (y-2)^2 + (x-5)^2 = 1 [/math]

Then I determined [math] a [/math] by substituting the line equation into the equation of the circle it's leaning on and setting the discriminant of the resulting equation to 0 (to make sure there's exactly 1 intersection).
[math] a [/math] turned out to be [math] \frac{5+\sqrt{7}}{12} [/math], so the equation for the line is [math] y = \frac{5+\sqrt{7}}{12}x [/math].

Now that you have the equation for the line you can obviously just calculate the intersections by substituting it into the equation of the circle it passes through.
And then it's just [math] \sqrt{(x_1-x_0)^2 + (y_1-y_0)^2} [/math].
It becomes a real mess though so I had to use Wolfram Alpha.

Hint: not 1, brainlet.

>>9741114
9*1/9 = 1

>>9741128
( ) pls

>>9741130
((())) pls

>>9741114
>completed infinities exist

>>9741103
Less impressed now. The trig method can be done with pen and paper.

>>9741046
tired of this bait t b h

>>9739411
r = The jews did this. Look like I proved my IQ is way higher than yours.
t. /pol/

>>9741046
If you tilt your head to the left a bit, then the relevant lines can be represented by these equations:
[math] \sqrt{5^2-x^2} [/math] and [math] \sqrt{10^2-x^2} - \sqrt{5^2+5^2} [/math].
https://www.desmos.com/calculator/huawstesfd
I don't feel like explaining why if you're too brainlet too understand.

Finding the intersections isn't difficult and results in [math] x = \pm
\frac{5}{2}\sqrt{\frac{7}{2}}[/math].
Now we integrate the difference between these two equations from [math]
-\frac{5}{2}\sqrt{\frac{7}{2}}[/math] to [math]
\frac{5}{2}\sqrt{\frac{7}{2}}[/math] and multiply by 2 to find the total area:
[eqn] 50 \arcsin(\frac{\sqrt{14}}{4}) - 200 \arcsin(\frac{\sqrt{14}}{8}) + 25 \sqrt{7} \approx 29.27625 [/eqn]

>>9741177
go for it then, I'd like to see that

>>9741114
>Hint, not 1, brainlet

[math] \sum_{k=1}^\infty \frac{9}{10} = \frac{9/10}{1- 1/10}= \frac{9/10}{9/10}=1 [/math]

>>9741222
you proved /pol/ is full of retard

>>9741231
forgot the k in the sum

[math] \sum_{k=1}^{\infty} \frac{9}{10^k} =\frac{9/10}{1-1/10}=\frac{9/10}{9/10}=1 [/math]

>>9741046
If you tilt your head to the left a bit, then the relevant lines can be represented by these equations:
[math] \sqrt{5^2-x^2} [/math] and [math] \sqrt{10^2-x^2} - \sqrt{5^2+ 5^2} [/math] .
https://www.desmos.com/calculator/huawstesfd
I don't feel like explaining why if you're too brainlet too understand.

Finding the intersections isn't difficult and results in [math] x = \pm \frac{5}{2}\sqrt{\frac{7}{2}} [/math].
Now we integrate the difference between these two equations from [math] -\frac{5}{2}\sqrt{\frac{7}{2}}[/math] to [math] \frac{5}{2}\sqrt{\frac{7}{2}} [/math] and multiply by 2 to find the total area:
[eqn] 50 \arcsin(\frac{\sqrt{14}}{4}) - 200 \arcsin(\frac{\sqrt{14}}{8}) + 25 \sqrt{7} \approx 29.27625 [/eqn]

>>9741177
go for it then, I'd like to see that

>>9741046
If you tilt your head to the left a bit, then the relevant lines can be represented by these equations:
[math] \sqrt{5^2-x^2} [/math] and [math] \sqrt{10^2-x^2} - \sqrt{5^2+ 5^2} [/math] .
https://www.desmos.com/calculator/huawstesfd
I don't feel like explaining why if you're too brainlet too understand.

Finding the intersections isn't difficult and results in [math] x = \pm \frac{5}{2}\sqrt{\frac{7}{2}} [/math].
Now we integrate the difference between these two equations from [math] -\frac{5}{2}\sqrt{\frac{7}{2}} [/math] to [math] \frac{5}{2}\sqrt{\frac{7}{2}} [/math] and multiply by 2 to find the total area:
[eqn] 50 \arcsin(\frac{\sqrt{14}}{4}) - 200 \arcsin(\frac{\sqrt{14}}{8}) + 25 \sqrt{7} \approx 29.27625 [/eqn]

>>9741177
go for it then, I'd like to see that

>not making a graphic resolution
brainlets gtfo

>>9739418
Lol fagot
[math]\mathbb{Z}\times\mathbb{Z}\setminus\{0\} = \mathbb{Q}[/math]

I FUCKIN HATE MYSELF THIS IS WHY I CANT GO TO COLLEGE FUCK ME FUCK ME FUCK ME WHY CANT I FIGURE IT OUT REEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

>>9741249
obviously wrong

>>9741248
i did it and got somewhere around 12,36 and 12,47 probably yhe last one is more precise

>>9741253
boi you just got filtered

>>9739411
Close as I could get. Figure out the angles in blue and it's solved...

>>9741241
These were my original workings out from when I solved it originally, which I had to dig up to check your answer (and remember how I did it).

It's not tidy, and there's a lot of irrelevant stuff from some other problem I've already forgotten about. The answer ended up in the box on the right.

I could probably write it up neat if I was bothered enough.

>>9741433
oh nice, good job
that's definitely better than my method

>>9741433
were you writing with charcoal or something wtf

>>9741460
It's the patrician choice.

>>9741414
Seriously I've been trying to figure this out all afternoon. FUCK. This is why I can't ever do a math degree, I'm too goddamn retarded. I've flipped it around trying to figure it out but I can't do it without cheating and approximating. I cannot get a set of lines that do the radius and have the numbers match up. Fucking hell OP please provide an answer soon. I seriously want to fucking end my life, I am going to be a wagecuck forever. I was considering tyring to go to college for math someday but I am too much of a brainlet apparently, I give up, time to an hero in a few years when I get sick of my dead end job finally.

>>9741970
If you feel like cheating yourself, this genius has the exact answer: >>9739451

>>9741041
Do you have a reasoning for that other than AB looks like the diameter of one circle?

>>9740012
it's 28 you brainlets. 32 + 16 = 20 + X

>>9741421
You need to draw the triangle that the circle is the incenter of and then construct triangles to the equidistant tangent points

>>9741970
Have you ever made any attempt to actually study math on your own, or are you pussing out on pursuing a degree because you missed the piss on one fucking problem? The real brainlet filter is how you answer that question.

>>9742091
bingo, all other participants must kll themselves

>>9742113
what was the name of that rule/law, or as i have recently taken to calling it, rule plus law. I can do a proof but i know this has a name

>>9742116
>rule
>law
physicsfags gtfo

>>9742118
I'm doing math, computer /sci/ and physics so no

>>9742120
>doing math
>calling it a rule or law
wew lad, just drop out now

>>9742127
In B4 Go Back to /g/

it's meme on this meme

I'd just like to interject for a moment. What you’re referring to as Linux, is in fact, GNU/Linux, or as I’ve recently taken to calling it, GNU plus Linux. Linux is not an operating system unto itself, but rather another free component of a fully functioning GNU system made useful by the GNU corelibs, shell utilities and vital system components comprising a full OS as defined by POSIX.
Many computer users run a modified version of the GNU system every day, without realizing it. Through a peculiar turn of events, the version of GNU which is widely used today is often called “Linux”, and many of its users are not aware that it is basically the GNU system, developed by the GNU Project. There really is a Linux, and these people are using it, but it is just a part of the system they use.
Linux is the kernel: the program in the system that allocates the machine’s resources to the other programs that you run. The kernel is an essential part of an operating system, but useless by itself; it can only function in the context of a complete operating system. Linux is normally used in combination with the GNU operating system: the whole system is basically GNU with Linux added, or GNU/Linux. All the so-called “Linux” distributions are really distributions of GNU/Linux.

>>9742135
That's not the problem, /g/aywad. The problem is that no self respecting mathematics practitioner would call it a 'rule' and especially not a 'law'.

>>9742151
>self respecting
>posting on an image board
pick one
also I don't care what it's called, language doesn't matter

>>9742091
>>9742113
damn was about to type out something scathing and realized you guys are right

>>9739411
6.3

>>9741970
lol calm down my man, you don't need to be able to do stupid shit like this to be a math major

>>9742189
I can't prove that this answer is right, but I'm highly confident it is.

>>9741046
Wait wouldn't the eyelid shape area be 2 * (100 - 25*pi) (1/4 pi*r^2 where r is 10 for the area of the quarter-circle, 100 minus that for the area of the exclusion, and 2* that because there are two such identical areas)? Therefore the area of the eyelid should be 42.9204 cm2. The area of the grey circle would simply be pi r squared where r is 5 so it'd be 25 pi.

Wow then I just realized the eyelid tips, wow I am fucking retarded, fuck me. Yeah you probably do have to do some integral shit to figure it out.

>>9742194
Maybe not but it's just another example of how fucking retarded I am, I just looked at the solution and it uses some cartesian coordinates shit I didn't even think of because I thought that would be cheating, didn't realize it would make it so easy, also forgot the rule of tangents being perpindicular even though I spent the whole time trying to draw a straight line from the middle of the 10-wide semicircle and figure out the lengths with other right triangles.

>>9742107
>or are you pussing out on pursuing a degree because you missed the piss on one fucking problem?
I can't afford to go 50k in debt for something if I can't even do a basic geometry problem.

>>9742285
If you're using your performance on this, or any one problem, to determine your ability to succeed at getting a math degree, you're a fucking moron. If you can't do a problem, you look up an answer, study the fuck out of it until you understand how to get there and can replicate it yourself, then you pocket that information and move on. If you want to pursue a math degree, or anything of substance whatsoever, you're going to face a metric fuckton of problems. And big surprise, you're not going to know how to do every single one of them. You're gonna have your shortcomings and you either let those overwhelm you, or you learn from them and become a better person for it. Are you prepared & capable enough to gamble 50k on a degree right now? I don't know, that's for you to decide. But you're also not actually making that gamble right now; you have time. If you're not ready, stop sulking over a problem you couldn't solve and start studying.

You couldn't do a basic geometry problem. Awesome, that puts you in the same boat as the vast majority of math majors. Most modern curricula are, for better or for worse, severely lacking in rigorous material on Euclidean geometry. Problems like these are notoriously hard, not because they're genuinely hard problems, but because people aren't taught the tools needed to solve them. I have a math degree, it was a breeze to get, and I couldn't solve this. Unless you pursue research in something that directly uses Euclidean geometry, chances are very high that you can enroll in and complete a math degree, then go out into the real world and succeed in whatever job you do, without ever being able to solve a problem like this. Don't let one problem keep you from doing what you want to do with yourself, that's top tier brainlet shit right there.

>>9740879
T. Hohol

>>9741970
Read Euclid

>>9739411

the answer is 144/23

http://mathworld.wolfram.com/ApolloniusProblem.html

>>9742789

oops, I see I'm late to the party this guy already got it

>>9739451

How the fuck is that a brainlet filter nigga that question took me 20 minutes

>>9742819
Big deal it took me 5

>>9742091
How do you prove this? Why is this true?

>>9742346
>If you're using your performance on this, or any one problem, to determine your ability to succeed at getting a math degree,
How do I find out if I am intellectually capable of finishing one?

>>9739411
Not that I would know how to calculate this, but would a possible solution be to find the points where the three semicircles intersect with the larger circle, then measure the arcs to get circumference, and plug it into [math]C=2\pi r[/math] ?

>>9744185
you don't have to be a savant to be a math major, if you did good on the sat or act you're fine brother

>>9743276
It's been solved on here so many times yet I still don't have it memorized. Something about realizing some length invariant by drawing the meeting point as being the displaced version of a point right in the middle. You end up with that expression for the sums of the areas and boom.

>>9744185
What was your SAT score? Your intellect is not really the most important factor btw. An IQ of 115 should be sufficient, and you can reasonably approximate that from your SAT score if you've never had a proper IQ test.

Learning mathematics is most importantly about your ability to cope with frustration. You have to have the discipline and the drive to keep working on a problem when you are stuck and baffled and you hate everything and want to give up. You have to be willing to go to office hours and ask other students for help when you get stuck. Some problems will take hours, or days, or weeks to solve. You will be stuck. You will hate it. So you put it away and come back to it later with a fresh mind.

I'm currently working on a math undergrad with a second major in the social "sciences." I'm not a super genius but I work hard and get good grades, even in the upper division math courses. Of the people I've seen wash out of the mathematics major it is usually their work ethic and not their raw intelligence that is the limiting factor.

tl;dr stop whining and study faggot

>>9741249
Cringe

>>9744290
Not the anon your pertaining to, but thanks man -- I needed that.

>>9741046
2((1/4)arccos(sqrt{2}/4) + (sqrt{7}/8) -arccos(5/(4sqrt{2})))

>>9741114
9*(1/(1-0.1))=9*(10/9)=10

>>9741128
>>9741231
lol no retards

>>9744251
>>9744290
I got a 740 on my math SAT, which is lower than I would have liked.
>You have to be willing to go to office hours and ask other students for help when you get stuck.
RIP

>>9744290
>tfw 122 IQ based on wais-iv and got a 1670/2400 on my SAT, both times

>>9746575
>9*(1/(1-0.1))=9*(10/9)=10
Explain??

>>9746842
You'll be fine, but be prepared to work. I was used to everything being super easy all the time so it was a bit of an adjustment when I had classes that actually forced me to work hard to master the material.

>>9739889
Wrong.
[math]\pm\sqrt{\Q_{\geq 0}[/math]

>>9739889
>complex numbers are rationals
Wrong.

[math]\pm\sqrt{\Q_{\geq 0}}[/math]

>>9739889
>complex numbers are rational
Wrong.
[math]\pm\sqrt{\mathbb{Q}_{\geq 0}}[/math]

>>9741249
It's cells of a partition of that

Did I do anything wrong here?
t. cs dropout

>>9747447
well your answer is wrong, so probably.

>>9747447
It's wrong to assume that the lines that bisect the triangle sides and the points on the outer circle are also normal to the triangle sides

2+2sqrt2

>>9739411
You gave 3 side lengths and no base length for a figure with 2 sides and a base.

>>9749295
Wat

>>9740849
I like this explination best.
Thanks annon.

>>9749296
>In geometry, a base is a side of a polygon or a face of a polyhedron, particularly one oriented perpendicular to the direction in which height is measured, or on what is considered to be the "bottom" of the figure.

I guess its technically a side.im used to calling the base of a triangle a base.
Guess i cant weasle my way out of this one...

>>9749295
Retard

>>9741114
neck yourself, brainlet

>>9741222
OBSESSED

>>9740404

>>9749333
I dunno why he's having an existential crisis there, he should know he's 2D

>>9744290
That's a good motivational speech. Motivated me greatly. Thanks anon.

>>9740849
>Clearly
Not clear at all

>>9740764
Physicist here. Order of magnitude, I’d say about 10r.

>>9739411(OP)
About 6.587, easy. More precisely, squareroot (24+97/5)

>>9750627
So fucking wrong.

The correct answer is 144/23, if you want to give up, a link to the worked out answer is right here
>>9740724

>>9739411
6.260869565

>>9741482
it's hideous and nearly unreadable

>>9739451
Why -1 at the denominator?

>>9741970
>
You can calculate it with coordinates but there is probably a more elegant solution

>>9751133
Apparently its 144/23 or 6&6/23

>>9752025
Can confirm that the inner triangle with sides 3,4,5 allows due to the law of cosines to calculate the truncated radius which does indeed yield 144/23

>>9739418
?
\mathbb{N}\times\mathbb{N} = \mathbb{Q}