help-19

[ivory grove]

they said it was "good passes" in the beginning message

[atomic void]

oh i dint see it

im so blind

[harsh bison]

I think I might have solved it to be 1.728 ?

[ivory grove]

@harsh bison Can you elaborate on what you mean by "trying to get the average of the numbers while attaching a value to them?"

[ivory grove]

no no it's fine

[atomic void]

@harsh bison , you need to find the weighted average

[ivory grove]

Oh, ic

[atomic void]

cause there are 4 kinds, perfect, good, avg, and fail.

[harsh bison]

yep! 3,2,1,0

[atomic void]

maybe, perfect can be weighted as 1, fail is obv 0, and blah blah

[atomic void]

oh

so u want to know how to make excel do that?

[harsh bison]

yeah I did it by hand for 1 but i have 36 players to do that since we combined stat programs and they wont accept data automatically

[atomic void]

owhp

well, the sum was 733?

my calculator says no...

[harsh bison]

that was total number of passes

[atomic void]

oh

but you want to find the avverage right?

[harsh bison]

yes please! it should be below a 1.8

[atomic void]

i mean weighted

[atomic void]

hold up

alright

i kind of fforgot how to excel tbh, but this is quite simple

can you zoom out so i can see the row and column?

[atomic void]

ok

[harsh bison]

it was also 719 total passe, no idea where i got 733

[atomic void]

in the "total" you want avg?

the weighted avg i mean

[harsh bison]

yes please

[atomic void]

i think u should change the name

"total" is confusing

[harsh bison]

kk done

[atomic void]

hold up

i need to re load

ok on E2

we type (3 *A2 + 2 *B2 + C2)

then we want to divide by the sum

waitt not 2

we type =(3 *A2 + 2 *B2 + C2) / SUM(A2 , B2 , C2, D2)

@harsh bison it should work

[harsh bison]

wow awesome thank you!!

[atomic void]

yes, now just pull it down

you know right?

so it applies the function to every row

well, you need to put the data first

!done

[odd edge]

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[harsh bison]

.close

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[edgy holly]

I tried the previous suggestion. Can't seem to telescope. Help.

[spice wren]

till where does the summation go to?

[edgy holly]

from 1 to n

[spice wren]

aight

wait, then the variable inside summation shouldn't be n

[edgy holly]

The problem is some guys in our group are using formulas like tan(a)/cos(2a) = tan (2a)-tan(a), without any prior discussion or exposure. From where are they pulling these formulas out?

They are so easy in hindsight

But to think that will be used, is hell.

[spice wren]

pretty sure it goes on like
cos(2a) = (1 - tan²(a) )/(1 + tan²(a) )

[edgy holly]

nahh, thats different bro 😭

It isn't used here

I am sure as, that part of the theory is after this question...

[spice wren]

nah I was like just sharing a proof to that tan(a)/cos(2a) expression

[edgy holly]

here I have to do something similar like sin(a)/cos(3a)= a telescope

[edgy holly]

ohh, I did differently

[spice wren]

that question looks kinda difficult 💀

[edgy holly]

its really not when we discover the cheat code formula

[edgy holly]

...

[spice wren]

yeah

im not that good in sequences and series 🥲

[edgy holly]

We are in the same boat brother...

[spice wren]

ayy bruv I almost solved it

the summation

[edgy holly]

wait a bit, my chemistry teacher called. Will be back in 10 min

[spice wren]

@edgy holly I could do this, hope it helps

[edgy holly]

merci beaucoup

[spice wren]

thx to u too, I got it revised

[edgy holly]

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[restive condor]

Okay all the channels I message on keep getting closed so I’m gonna try to open another one,
I am a grade 8 that can do grade 3 math, I have dyscalculia and dyslexia so it makes math quite difficult, I need continuous help in math as I try to understand, I don’t understand anything further than average addition and subtraction, it’s quite embarrassing.

[quasi sparrow]

#study-discussion

[restive condor]

Hm?

[quasi sparrow]

Your question is a how-to-learn question

Unless you're working on a concrete math problem, please ask in #study-discussion

[restive condor]

Okay

What does concrete mean

[quasi sparrow]

Any homework question you encounter

[restive condor]

Ah okay

I will go there then

Thank you

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[kindred hazel]

Idk what’s happening but I’ve tried 2.08 1.44 and .208

[tidal matrix]

do you have 540,000 or the square root of 540,000 in the top right?

i am also confused as to where the 2324m came from

[odd edge]

@kindred hazel Has your question been resolved?

[kindred hazel]

I took the square root of 540,000 to get 2324

I always just use my answer in the equation of the next step to save time

[tidal matrix]

ok, so we don't need to take the square root of 540,000

and if we did, that square root is ~ 734.85

[tidal matrix]

can you see why?

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[honest panther]

I don't understand this at all, can anybody be willing to help me and explain step by step?

[manic grail]

take the coordinates of each point

A , B , C , D , E

and check how the x and y changes

so for example

x of A goes from -4 to 3

so x becomes x+7

also y of A goes from -2 to 1

but since its using reflection

if you can tell by moving the whole shape 7 to the right

[honest panther]

-4 + 3?

[manic grail]

and reflecting for y axis it becomes the new shape

[manic grail]

no 4+3

so if you move the whole shape 7 to the right and 1 up

[honest panther]

do we js ignore the negative

[manic grail]

then flip it symmetrically to x axis it becomes the new one

[manic grail]

no you are not

to find distance of A' to A

its

3 - (-4)

for x going from a to b

the difference is b-a

just happens that a is negative so its addition since the - cancel out

[honest panther]

wdym reflecting from y axis

im confused

I get how we got 7, but I have no idea what your talking about atm

[manic grail]

ok so

draw the shape

if we move all the points 7 to the right

then move the shape all points up by 1

[idle grove]

there are some rules for reflections across lines and axis's

[manic grail]

now from this shape

if you draw it

you can tell that the points of the shape in comparison to the points of the A' ... shape have same x coordinate and opposite y ones

so if one point has -2 the other point on the above has + 2

[honest panther]

ooo i see now

[manic grail]

so its the first answer

[honest panther]

x axis?

can you explain why its x axis

[manic grail]

do you know which the x axis is ?

when it says reflexion towards to x axis

it means that all the x coordinates are the same

[honest panther]

yeah i do

[honest panther]

ohh alright

[manic grail]

but all the y coordinates become -y

so (x,y) -> (x, -y)

for every point

[honest panther]

ic

thanks

.close

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[teal marsh]

A puma runs at a speed of 8 km/h. It travels in a S65°E direction for 1 hour, and then continues in an E20°S direction for half an hour.

From its final position, it is seen by a control tower located 1 km away in a straight line.

Calculate the straight-line distance from the control tower to the refuge.

[odd edge]

@teal marsh Has your question been resolved?

[glacial furnace]

what do you need help with in the question?

[teal marsh]

I only need the distance from the tower to the first point from where it came out

[glacial furnace]

im not allowed to do the question for you but I can walk you through it

[teal marsh]

Np

[glacial furnace]

first lets make a diagram

you understand what s65E means?

[teal marsh]

Uhmmm kind of

[glacial furnace]

so you just take a compass like this

im drawing on ms paint with a mouse it may be a littl emessy

you start at the first letter, in this case it's S

then you go 65° towards the second letter, in this case it's E

so it looks like this

[teal marsh]

Kay

[glacial furnace]

so with that we can make this diagram for the puma

can you explain what it means by a control tower 1km away in a straight line?

is that what the question says?

[teal marsh]

Oh srry

1 sec

A puma escaped from a wildlife refuge, running at 8 km/h for one hour in a S65°E direction, then E20°S for half an hour at the same speed.
At that point, it was spotted by a control tower located 1 kilometer from the puma's location, so the animal was located exactly on the same straight line that connected the refuge to the control tower. Determine the straight-line distance from the control tower to the refuge.

That's the whole question

[glacial furnace]

wait this question is weird

[teal marsh]

YUP

[glacial furnace]

so if i draw a sraight line through point A (where the puma started) and point C (where the puma is right now) the control tower is on that line?

[teal marsh]

No, the pont C is within the line that joins the tower with the point A.

[glacial furnace]

here's an updated image that makes the 0.5 line shorter

could you copy it and draw where you think the control tower is?

[teal marsh]

Kay

1 sec

[glacial furnace]

is it like this?

[teal marsh]

Yup

[glacial furnace]

okay okay that makes send

sense*

so we're trying to find the distance from T to A

aka the red line

[teal marsh]

Yes

[glacial furnace]

the first thing to do is turn the lines that are measured in hours into distance

the puma runs at 8 km/h

so with logic you can assume in 1 hour it will run 8km

and in half of that time, it will run half the distance, so in 0.5 hours it will run 4km

[teal marsh]

Yup

[glacial furnace]

so we can draw the diagram like this

do you know what the Z pattern is?

[teal marsh]

No idea

lol

[glacial furnace]

ok I'll show you

if we have a line passing through two parralel lines, you can follow a Z pattern like this

the angles on the left are the same as the angles on the right but they are flipped

the red line is the z pttern

[teal marsh]

Ohhh

I understand

[glacial furnace]

so if i have an angle like this

i know this angle is also 50 degrees

going back to the diagram of th epuma, do you notice a Z pattern with the 65 degrees?

[teal marsh]

I think so

[glacial furnace]

can you explain where

[teal marsh]

From point B to A the angle would be 20?

[glacial furnace]

ok thats my bad its hard to talk about this with words

i'll just shwo you

[teal marsh]

Haha okey

[glacial furnace]

the 65 degree angle is on Z pattern i marcked with red

this means that this angle is also 65 degrees

[teal marsh]

Ohhh

It makes sense

[glacial furnace]

now we need to find the other side of the angle on point B

I'll make this part of the diagram bigger

we're looking for x

[teal marsh]

70?

[glacial furnace]

yes

i assume you did 90 - 20>

?

[teal marsh]

Yes

[glacial furnace]

ok greath

I'll make a new diagram this one is getting messy

[teal marsh]

Kay

[glacial furnace]

it's not very useful to have 65 and 70 degrees apart, so we can add them together to have one big angle there

just like this

so if we're trying to find the red line, that means we need to find this blue line

do you know what equation will give us this blue line?

[teal marsh]

Pythagoras?

[glacial furnace]

pythagoras' theorem only works with a right angled triangle. The angle we have is 135 degrees

[teal marsh]

The cosine theorem?

[glacial furnace]

yes thats right

[teal marsh]

lol

Kay

[glacial furnace]

so $c^2 = a^2 + b^2 - 2ab . cosC

so $c^2 = a^2 + b^2 - 2ab . cosC$

so $c^2 = a^2 + b^2 - 2abcosC$

[clever fjord]

CristianThe12

[glacial furnace]

is that right

its been a bit since ive used it

[teal marsh]

Yup

[glacial furnace]

ok so because the line we're trying to find is the longest of the lines (you can tell because the angle opposite to it is over 90) we're going to give it the variable c

the two other side lengths are a and b, it doesnt matter which one is which

[teal marsh]

Kay

[glacial furnace]

and the capital C is the angle opposite to the line we're trying to find, so it's 135

can you try to do that

[teal marsh]

Sure

It gives me 11.9

[glacial furnace]

lets find out if thats right

$c^2 = 4^2 + 8^2 - 2(4)(8)cos(135°)$

[clever fjord]

CristianThe12

[glacial furnace]

what is that r

igrnor that

simplifying a little we get $c^2 = 16 + 64 - (64)cos(135)$

[clever fjord]

CristianThe12

[glacial furnace]

$c^2 = 80 - 64cos(135)$

[clever fjord]

CristianThe12

[glacial furnace]

ok now we're going to find cos(135) and approximate it in the equation

it's approximately -0.7071

so $c^2 = 80 - 64(-0.7071)$

[clever fjord]

CristianThe12

[glacial furnace]

$c^2 = 131.65$

[clever fjord]

CristianThe12

[glacial furnace]

$c = 11.47$

[clever fjord]

CristianThe12

[glacial furnace]

you got 11.9?

[teal marsh]

lol

Yes

[glacial furnace]

its very close but can you look at what I did and see what you did wrong

maybe i did something wrong

but just see what we did differently

[teal marsh]

I did it with quadratic formula

0.o

[glacial furnace]

lol

its pretty common to get those mixed up

cool that our numbers were so similar

lets hope my c = 11.47 is right

can we find out what the red line is now? (the distance from the tower tot he original point)

[teal marsh]

Just

Add 1

I guess

[glacial furnace]

thats exactly right

that means the answer to the question is 12.47km

does everything make sense?

[teal marsh]

Yes

Thank you so much bro

[glacial furnace]

no problem

good luck on your test if youre doing one

[teal marsh]

Ty

[glacial furnace]

.close

i think you gotta do that

[teal marsh]

.close

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[frail lintel]

why couldnt i cancel out the 1-x

[mystic nova]

You also have to cancel out 1-x in cos(1/1-x)) since we can't define cos(x) with x go to infinity

[frail lintel]

OHH

right

understood

.close

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[robust mortar]

hi, can anyone help me get an intuition for the functor-less def. of universal property (that is, a thing satisfies a universal property if it is a terminal [in this sense meaning initial or final] object of some category)

[odd edge]

@robust mortar Has your question been resolved?

[odd edge]

@robust mortar Has your question been resolved?

[low locust]

you will have more luck in the advanced channels

[robust mortar]

ok thank you

.close

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[celest violet]

how do you minus percentages with normal numbers

Nvm I got it

.close

[sharp oak]

@robust mortar
Let's say X is a terminal object.

X has a "universal property" in the sense that, for any object A, A has exactly one arrow into X.

The "for any object" and "exists exactly one arrow" are the important bits. While constructions may get more complicated, you still need those two facts

[robust mortar]

i think i understand the definition but i have trouble understanding what this "really means" for whatever property the category is based on

[sharp oak]

It's exactly the right information to set up this:

Let's say X and Y are both terminal objects. They're isomorphic.

This is always true of universal properties. Any two objects that have the same universal property are isomorphic objects

[robust mortar]

Oh okay wow that's significant

[sharp oak]

Trying to think up a nontrivial example haha. Maybe I'll grab one from a book

[robust mortar]

is something said to have a universal property only if each object has exactly one arrow going to it or does it include objects which have exactly one arrow going from it to each object

[sharp oak]

A more complicated example of a universal property is a product.

Let's say there's two objects X and Y, and we can find a "product object" X*Y, defined by this universal property:

For any object A, and any two arrows from A to X and A to Y, there exists exactly one arrow from A to X*Y

So in this case, there's a condition that the object A needs to follow. But if it does, then the "exactly one arrow" bit kicks in

As previously mentioned, if there's two objects that are a product of X and Y, then those two products are isomorphic

[robust mortar]

right okay

[sharp oak]

The neat thing here is we are making statements about the objects without needing any information about the objects. We are only using the arrows coming from the objects

[robust mortar]

I think i see

[sharp oak]

Like, a terminal object in Group is the group {e}.

We don't actually need to know what's in this group, we just need to know how other groups map to it

If there's another terminal object in Grp, it's isomorphic to {e}. In the language of group theory, there's exactly one group with one element

[robust mortar]

im not very familiar with groups yet b ut i think i get that

[sharp oak]

Haha I kind of ranted there, I guess I felt like talking about category theory atm. Feel free to ask if you have any questions

[robust mortar]

Might just have to take you up on that at some point

Thanks for explaining and giving examples, really did help

[odd edge]

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[rustic tartan]

Sup

[rustic tartan]

Ok, I want help with geometry.
I'm designing a bellows for a custom accordion.

I want to design it with a shape other than a rectangle or hexagon.

I don't know if this can be achieved. But it's an interesting puzzle to think about.

Best case scenario is that we're able to generalize a technique, that can maybe even be implemented into code, where we input a 2d shape and we get the 2d shapes necessary to cut the cardboard with to assemble the bellows.

[carmine idol]

hmm, what's the math problem?

[rustic tartan]

Do you think this is more of a physics thing?

[carmine idol]

by the way, if you know about the flexible polyhedra, you should be aware of the bellows conjecture

https://en.wikipedia.org/wiki/Flexible_polyhedron#Bellows_conjecture

[rustic tartan]

I'm not, I'll give it a read

[carmine idol]

basically, if the faces are rigid, you cannot change the volume

[rustic tartan]

Hmm...
I don't get it...

What I mean is like

Consider you want to make a bellows with a square cross section (with tapered corners)

The shape you need to cut the cardboard is the one in the image. 4 times, and then assemble.

That would result in a bellows like this

[carmine idol]

don't worry too much, just be aware the faces shouldn't be rigid

[rustic tartan]

Would it be possible to generalize a method to translate the cross section of the bellows to that zig zaggy pattern

[carmine idol]

lemme see if i understand, you want to figure out what shapes to cut to make any arbitrary polygon

maybe you can try relating the shape you cut to the length of a side and the adjacent angles

but there might be an issue with changing sides and changing angles that we might need to calculate

[rustic tartan]

Is there a timer on these question channels?
I think have a good way to do it
But I'll have to do it tomorrow because it's late now

The horizontal lines in the zigzag pattern are from the sides of the outer and inner polygons.
You add them in alternating patterns and that's half of it.

Now the vertical distance between these lines will result in the angle of the zigzag lines

There must be constraints on these angles

I believe there should be a relation between outer to inner area ratio and those angles

[desert jolt]

Probably take a screenshot of this

Before it’s lost forever

[carmine idol]

well, you can use the discord search feature to look for your own messages, or keep this on unread and search for something that pinged you

[odd edge]

@rustic tartan Has your question been resolved?

[odd edge]

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[shadow halo]

5 In p = root 6 In q

[latent scaffold]

Any context?

[shadow halo]

Express p in terms of q or euler

[latent scaffold]

Is this option 1 : $5\ln(p) = \sqrt{6}\ln(q)$ or option 2 : $5\ln(p) = \sqrt{6\ln(q)}$?

And what have you tried?

[clever fjord]

Azyrashacorki

[shadow halo]

Option 1

[carmine idol]

well probably find some way to isolate p

[shadow halo]

I got p=q^root 6 over 5

But on the answer sheet its p=q^6^1/10

[latent scaffold]

They are equivalent

[shadow halo]

Theyre the same?

[latent scaffold]

One sec just doing something I think they are but might be wrong

Yeah nvm I don't think they are I'll come back to it shortly

[odd edge]

@shadow halo Has your question been resolved?

[latent scaffold]

Yeah well what you did is correct regardless

It might depend on the format of the answer as the way you've typed is a bit ambiguous, but you did it right, so either they are equivalent, or the key is wrong, or you typed the problem wrong to begin with.

[odd edge]

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[fallen sparrow]

Give that abc = 60
And
1/a + 2/b + 3/c is 22/15
Find the value of a+b+c

[fallen sparrow]

I tried asking chatGPT but it use the substitution method instead of the algebraic ways to approach

[stray zinc]

Have you tried rewriting 1/a + 2/b + 3/c as a single fraction with denominator abc? What would the numerator look like?

[stray zinc]

yeah chatgpt is known for not being very elegant

[fallen sparrow]

bc+2ac+3ab over abc

[stray zinc]

Now, since you know
abc=60, what does that tell you about the numerator bc + 2ac 3ab when you multiply both sides by abc?

try setting up that equation and seeing what it simplifies to

[fallen sparrow]

So bc+2ac+3ab equals to 88?

And I don’t know how to proceed 😂

[stray zinc]

Maybe try expressing one variable in terms of the others from the product

and substitute into the sum bc + 2ac...

[fallen sparrow]

U mean try to write every factor in terms of a right?

[brittle plinth]

nah I think substitution is much better here

does a, b, c have to be integers?

[fallen sparrow]

Yes

[brittle plinth]

oof that's no good for substitutions

[odd edge]

@fallen sparrow Has your question been resolved?

[brittle plinth]

@fallen sparrow got an idea

since abc=60

one of those 3 numbers must be divisible by 3

[brittle plinth]

and you'll easily see that c can't be divisible by 3

so one of a or b must be 3 or 6

[restive kayak]

It seems to me that you cannot find the independent values of a, b and c; because you only have 2 equations, you would need a third one, you will have to perform tricks to find a + b + c

[brittle plinth]

a,b and c are integers so is that enough restrictions?

[restive kayak]

If you were given the restriction that they are integers then yes, but the guy in the problem didn't say they were positive integers.

[brittle plinth]

^

[restive kayak]

Oh

Sorry

[brittle plinth]

but if we do include negatives

[restive kayak]

I am blind

[brittle plinth]

either none or two numbers are negative

[restive kayak]

If we include negatives, two of the numbers must necessarily be negative, so that the product of the three equals 60.

[brittle plinth]

so it's easy to show that c must be positive and not equal to 1 (too big)

so c = 2 and 1/a + 2/b = -1/30 and ab = 30

2a + b = -1 but both a and b are negative

so impossible

all 3 numbers must be positive

[restive kayak]

Well the equations are fulfilled for the values 2; 3 and 10

a+b+c=15

[brittle plinth]

did you blind guess or?

[restive kayak]

Yea

[brittle plinth]

the only thing I know rn is this ^

[steep mantle]

trial and error isnt that bad since you can just factorize 60 🤷

[restive kayak]

To find the values formally, it would be necessary to restrict values and use modules.

Okay, I think we're going to have to prove at least one value.

If a=2 then bc=30; c=30/b and replacing in the 2nd equations we can operate a quadratic equation to find b and thus also find c

[brittle plinth]

my idea is to use AM-GM on 1/a + 2/b

we get 2√(c/30)+3/c<22/15

since c is not divisible by 3 then we can limit our options for c that way

[restive kayak]

Do you mean arithmetic mean and geometric mean?

[brittle plinth]

ye

the inequality

[restive kayak]

I felt like grabbing a pencil and paper and trying it.

but it's 1 in the morning in my country

😨

[odd edge]

@fallen sparrow Has your question been resolved?

[brittle plinth]

nvm it's better to restrict a

use AM-GM and you'll get a can be only 2 or 3

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[tardy knot]

I've seen people (usually on Math Stack Exchange) come up with insane AM-GM solutions with a bunch of numbers like this. Those solutions are neat and easily confirmed to be true. However, how does one come up with it?

[tardy knot]

For example here, I would never consider to use AM-GM for 54 numbers

For more context, the original problem was:
Suppose that x,y,z
are positive real numbers and x^5+y^5+z^5=3
Prove that {x^4\over y^3}+{y^4\over z^3}+{z^4\over x^3} \ge 3

[wooden python]

using \over instead of \frac

anyway this is baically weighted AM-GM.

[opaque eagle]

he's old school

[wooden python]

why's he on discord at all

discord is not old school

[tardy knot]

Please elaborate

[wooden python]

do you know what weighted AM-GM is in general

[tardy knot]

I just copied it from Math Stack Exchange, don't put the blame on me

[tardy knot]

Probably not

[wooden python]

ok in that case let me enlighten you

[tardy knot]

Please do

[opaque eagle]

I mean it's just the regular AM-GM but there are multiple copies of the same term

so there's a "weight" or a "frequency"

[wooden python]

let $w_1, \dots, w_n$ be positive numbers (``weights'') adding up to 1, and let $x_1, \dots, x_n$ be positive numbers whose mean(s) we're interested in. then $$\sum_{k=1}^n w_k x_k \geq \prod_{k=1}^n x_k^{w_k}.$$

[clever fjord]

Ann

[wooden python]

a variant of this is that if the weights don't sum to 1 but to some other total weight $W$ is $$\frac{1}{W} \sum_{k=1}^n w_kx_k \geq \paren{\prod_{k=1}^n x_k^{w_k}}^{1/W}$$

[clever fjord]

Ann

[wooden python]

regular AM-GM can then be recovered as a special case by setting each w_i to 1 (and thus W = n) @tardy knot

[tardy knot]

Uhm...

I'm dumbfounded here, sorry

Can you give an example?

[steep mantle]

if say $n=3$ then $\frac{w_1x_1+w_2x_2+w_3x_3}{w_1+w_2+w_3}\geq \sqrt[w_1+w_2+w_3]{x_1^{w_1}\cdot x_2^{w_2}\cdot x_3^{w_3}}$

[clever fjord]

skissue.in.a.teacup

[steep mantle]

this is basically the same as making like 2x+y=x+x+y>=3cbrt(x^2y)

[tardy knot]

I think i get the weighted AM-GM part, but i don't get how that yields those

How do you find the "coefficients" here

Like how many numbers to use

[steep mantle]

w_1 is 30, w_2 is 7, w_3 is 1, w_4 is 16

[wooden python]

the numerical coefficients on the left are the weights they use

the 30, 7, 1 and 16

[tardy knot]

But how does one find such values so that the right hand side end up with the ...x⁵ they need?

[wooden python]

wait are you sure this is the right original problem

i think we're missing some context here

[tardy knot]

Am I allowed to link the mathstackexchange solution?

[steep mantle]

this convenienyly works, but sometimes when it doesent you have to manipulate the weights abit, you can make a system of equations using the powers

[tardy knot]

Yes?

[wooden python]

you're not just allowed i'd really like to see it

[tardy knot]

They summed up the similar inequalities for y and z

And got this

[wooden python]

wow that is some fucked up ex-recto AM-GM

[tardy knot]

And are there tools for this?

If people are simply coming up with those by hand, I don't think inequalities are for me

[steep mantle]

i actually do t see what happend here lmao how??

[tardy knot]

[tardy knot]

Well they repeated this for y and z, then summed the 3 similar inequalities up

[steep mantle]

im not convinced thats the easiest method

[tardy knot]

I think someone else found smaller weights that worked but it was still kinda ridiculous

I guess I just need to get good, my skill issue

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[celest osprey]

How to solve relations such as R is defined by aRb a>=b

[atomic void]

Can you send a screenshot?

[desert jolt]

Do you the definitions for each type of relation

[mystic saffron]

B

[desert jolt]

@celest osprey

[mystic saffron]

Let there be 3 values a,b,c
(a,a)-reflexivw
(a,b)(b,c)Then (a,c) also-transitive
(a,b) Then (b,a) - symmetric
@desert jolt

[desert jolt]

No offense but i didn't ask you

[celest osprey]

Definition is R ={ aRb : a>=b}

[atomic void]

Yes

How about the properties?

[desert jolt]

No i mean reflexive, transitive and symmetric and equivalence

[celest osprey]

We use the definition to check for reflexivity, symmetricity and transitivity

[mystic saffron]

[atomic void]

Rudy was asking "what is transitivity, symmetricity, and transitivity? "

He wants you to type it out

[celest osprey]

It has already been typed out by Universe

[desert jolt]

Then what is the problem here

[celest osprey]

R is a set of real numbers

What exactly is the expression aRb

Is it a multiplied by Rb?

[desert jolt]

Its just notation

[celest osprey]

Of?

[sharp oak]

aRb is true, if a is related to b

Where "is related" needs to be supplied

[celest osprey]

a is given as greater than and equals to b

[desert jolt]

Yes

[celest osprey]

But what are a and b, real numbers?

[desert jolt]

Yes!

R is a relation in Real numbers

a and b have to be real so that they can be related via R

[celest osprey]

2>=1
1 is not >= 2 so it can't be symmetric

[desert jolt]

Their notation for relation R and set of real numbers R seems to be the same but its not

[desert jolt]

Correct

[celest osprey]

K thx

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[limpid cliff]

Hi, math question wrapped in a coding problem. I have a position and rotation transform that is changing over time. As an attempt to estimate the trajectory of the transform I take the positional and rotational velocity of the transform and get a dX and dW using the velocity as a linear approximation over some timespan dT.

Over several iterations, I need to apply dX and dW at each step of the trajectory (this is assuming that the rotational velocity and positional velocity (relative to the transform orientation) stay the same.

My question is this: I have 3 applications of these dX and dW that each produce convincing trajectories but are intuitively different in their intepretations.

The first is in the screenshot: apply the dW first and then the translation relative to the new rotation.

The second is to do a conventional transformation: i.e. future = future * delta where delta.position = dX and delta.rotation = dW.
(This would be applying the positional transformation first and then the rotational)

The third is to do a conventional transformation but flipped: future = delta * future.

As I said, all 3 produce similar and convincing results but only one of them can be 'physically correct'.

Ive also attached the transform multiplication method which is what i refer to when i say future * delta. Transforms store a position as a 3d vector and rotations as a quaternion.

[sudden marsh]

It's an artifact of finite time step. If you were to diminish it to infinitesimally small, that's when you'd get the 'real' trajectory. That said, that's not exactly practical.
If you want something that converges faster, I'd recommend you look at higher derivative methods like Runge-Kutta.

[limpid cliff]

Ok thanks, I'll have a look.

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[uncut seal]

i am not understanding constraints in a continuity question, especially the equal signs

[uncut seal]

here is the problem

the graph and the constraints

[wooden python]

are you not allowed to say |x| is continuous everywhere

[uncut seal]

continuous at a point

[wooden python]

continuous at every single point in R

[uncut seal]

f(x) = |x| at x=1 is not continuous

the question is: f(x) = |x| + |x-1| is continuous at x=0 and x=1 or not

i understand the graph but the constraints are not the same

please tell me why we're taking the equal sign

[clever plume]

modulus functions are continuous everywhere, but not differentiable at x=1, and 0

it is continuous since there is no break in the graph of the function

[uncut seal]

the answer is continuous but i am not understanding the constraints shown in the above picture

without the process, there will be less marks

how are we taking the constraints in the piecewise function?

[wooden python]

f(x) = |x| is not continuous at x=1?

what type of discontinuity is it?

removable, jump, asymptote?

does the limit of |x| as x->1 fail to exist? does it exist, but without being equal to |1|? why is it discontinuous?

[uncut seal]

let me take screenshot

@wooden python is the process ok?

[clever plume]

the constraints are simply there to define the function

using those constraints, you have to find the value of the function and its limiting value to find out whether it is continuous at the given points or not

[clever plume]

i'm sorry, it is incorrect

LHL as x approaches 1 is 1

[wooden python]

half of your x's do not look like x's and also your LHL value is wrong

[clever plume]

LHL=RHL so it is continuous at x=1

[clever plume]

lol that was rude

[wooden python]

at x=1 whether you approach from the left or from the right you're still governed by the condition x>=0 in BOTH cases @uncut seal

[wooden python]

rude how?

how do you think i should have said that?

[clever plume]

nvm

[clever plume]

i mean, you're not wrong

@uncut seal did you get where you went wrong?

[clever plume]

damn bro ghosted us

[uncut seal]

is it -(-x)

[clever plume]

no, it's just x since the constraint applies at x=0 not at x=1

[uncut seal]

understood

[uncut seal]

and the last thing, how are we taking constraints from this graph

[clever plume]

plot the graph as per the original function and expand the function as it changes for different values

the values at which it changes are basically the constraints

[paper hemlock]

|x| is continuous everywhere. But it's not conductive at x=0

[wooden python]

conductive?

did you mean differentiable?

we're not talking about differentiability here

[paper hemlock]

continuous function at x=x_0 plus continuous function at x=x_0, result is also continuous at x=x_0

[wooden python]

that graph is an overcomplication and you dont need it. the answer is simple: YOUR FUNCTION IS CONTINUOUS EVERYWHERE.

[uncut seal]

is this graph ok?

we are taking inequalities

but i dont understand the two different equal signs

[paper hemlock]

If the value of the left function is equal to the value of the right, the two different signs are allowed.

Take x=0 as an example. The value of -2x+1 at x=0 is 1, and it is also 1 on the right, so both signs can be allowed.

[uncut seal]

can you plz give ref so that i can look into?

@paper hemlock

[paper hemlock]

No need to ref. Function is actually a correspondence. As long as two functions yield the same value for the same input, they can be considered the same function.

[uncut seal]

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[brittle plinth]

let -a+b+c = x; a-b+c = y; a+b-c = z

then it's just am-gm

[frigid isle]

Am gm inequality

[eager yacht]

hmm

[frigid isle]

Clearly cause

A b c are > 0

[eager yacht]

I came up with this exercise

while studying a geometric problem

[frigid isle]

Triangle inequality may work?

But how would u get a product

[brittle plinth]

how would it even work here

[frigid isle]

.

[brittle plinth]

no you have to express a+b+c in terms of x,y and z

[fleet tapir]

Cosider x+y+z

[fleet tapir]

So a+b+c=x+y+z

Apply Am-Gm:
(x+y+z)/3>=?

[brittle plinth]

I think to make it more rigorous you have to show that at most one of 3 numbers x, y and z are negative

[fleet tapir]

Then you can apply am-gm

[eager yacht]

$$\mathrm{G}\leqslant \mathrm{A}\Longrightarrow \sqrt[3]{xyz}\leqslant \frac{x+\mathrm{y}+\mathrm{z}}{3}
\
\Longleftrightarrow xyz\leqslant \left( \frac{x+\mathrm{y}+\mathrm{z}}{3} \right) ^3=\frac{\left( a+b+c \right) ^3}{3^3},,\Box $$

[clever fjord]

AndyM

[brittle plinth]

yeah

[fleet tapir]

These are probably sides of a triangle

[eager yacht]

yes

[fleet tapir]

So then its easy

Have you heard of the triangle inequality

[eager yacht]

yes

[brittle plinth]

I find that it also works even if they arent 3 valid sides of a triangle

that'd make for a more interesting problem ig

[fleet tapir]

Would it? Atmost one of x,y,z can be negative(not hard to prove), and in that case the inequality is trivially true.

Oh i guess its more interesting than this

Ofcourse

[eager yacht]

When a,b,c are the sides of a triangle x,y,z are all positive, right?

cannot be maximal because it can be adjusted (increased
by equalizing its sides. We assume the theorem that says that equalizing two sides in a non-isosceles triangle increases its area.).
So the equilateral one E is the maximal because all its
sides are equalized.```

This is (in my own words) the proof for the maximality of the area of the equilateral triangle, presented here https://sci-hub.se/10.2307/30037526

I find it incomplete.

cannot be maximal because it can be adjusted (increased
by equalizing its sides).
So MAYBE the equilateral one E is the maximal.
It indeed IS because any adjustment we'd do to it
it will become T which cannot be maximal.
So E either is maximal 
or it is not (there exists no maximal triangle): impossible
because T either
1) has a base m=(a+b+c)/3 case in which it is obvious that it is smaller than E.
2) it has all three sides different than m. We still have to prove that in this case T is smaller than E.```

So to make this proof complete I used heron:

[brittle plinth]

ah nice

[fleet tapir]

Indeed, neat proof

Why is half of it typeset differently from the other half

[eager yacht]

Of course I'd love to hear that in fact this proof is complete in itself

Or to hear of alternative shorter proofs

[eager yacht]

the 1st half was typed some months back, the 2nd half was typed today in another software (AxMath)

I think another way to prove it is to show that the absolute difference between mₙ and mₙ₋₁ is strictly decreasing

d(m₁,m₀)>d(m₂,m₁)>d(m₃,m₂)>...>d(mₙ,mₙ₋₁)

This implies that mₙ approaches m=(a+b+c)/3

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[queen carbon]

hi i’m doing hw can someone explain to me why for question 4 b is a local max i assumed it was just r

[frigid isle]

Because

[autumn bolt]

see the y position of point b

values to the right of it are smaller

values to the left of it are smaller

[queen carbon]

oh i see they weren’t lined up to me

thank you

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[dreamy totem]

How do I solve this without a calculator? I don’t get why the person did 1/a.. is there another way to solve this question?

[dreamy totem]

ping me pls

[orchid torrent]

The idea is that $\frac{\sin \theta}{\cos \theta}=\frac{-1}{1}$ does not tell you $\sin \theta=-1$ and $\cos \theta=1$ (aka you can't just directly compare the numerators and denominators). \ \ However, it does tell you that $\sin \theta$ and $\cos \theta$ are in a ratio of $-1:1$. They're letting $a$ be the value you need to divide $-1$ and $1$ by so that you can directly \text{"match"} the values in the numerator and denominator with the values of $\sin \theta$ and $\cos \theta$.

[clever fjord]

Civil Service Pigeon

[orchid torrent]

As a more concrete example, $$\frac{5}{7}=\frac{10}{14}$$ does not mean $5=10$ and $7=14$. However, if you divide the numerator and denominator of the right hand side's fraction by $2$, then you get $$\frac{5}{7}=\frac{5}{7}$$ at which point you can directly "compare" them since they're equal.

[clever fjord]

Civil Service Pigeon

[orchid torrent]

Except here, we don't know what we need to divide by, so we just call it a variable (in this case a)

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[vivid wyvern]

How do i find the coordinate of A

[pastel orbit]

what kind of triangle are you dealing with?

that might give you a clue

[vivid wyvern]

Equilateral

[gritty spire]

welcome to mathcord

[vivid wyvern]

I tried finding length of one side

[gritty spire]

try finding the height

using the 306090 triangle

[vivid wyvern]

Ohh ryt as in median

[pastel orbit]

a suggestion: try finding the coordinate of the red dot

that'll give you the x coordinate of A (and then hopefully you can see what to do from there)

[vivid wyvern]

Yeahh got it ,thanks

Yea

I found 2 values of y,while x is 0
How do i know which one belongs to the triangle

Okay no i figured it out

Thanks yall

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[long tinsel]

multiply the top and bottom with e^(-2x)

then sub the denominator

[mystic saffron]

Thanks a lot

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[mystic saffron]

hello

[narrow crypt]

hello the country of russia

[late sinew]

Hi, welcome to the server!
Read #❓how-to-get-help , #rules and #info .

[odd edge]

@mystic saffron Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

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[odd edge]

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[vivid wyvern]

How do i continue

[late sinew]

Differentiate

[wooden python]

take derivative of both sides

[late sinew]

Use product rule for lhs and chain rule+quotient rule for rhs

[wooden python]

you could also first apply sine to both sides to get y/x = sin(y(x+1))

might be a tad easier to deal with

[vivid wyvern]

I dun think i k them

[vivid wyvern]

Yeah lemme try

[late sinew]

,tex.diff rules

[late sinew]

product, quotient, chain ^^

[vivid wyvern]

Didnt know abt chain

[wooden python]

wait what

[late sinew]

u cant do this w/o chain

[wooden python]

how did you get to implicit differentiation without the chain rule

[late sinew]

[vivid wyvern]

I meam i didnt know it was callee chain rule

[late sinew]

o

ok

what do u call it then

[vivid wyvern]

Idk😃

[frigid isle]

U can use a shortcut for these

[vivid wyvern]

Yeah i would like to know that

[frigid isle]

u can use partial derivative

[vivid wyvern]

Yeah for that wont i need to separate y and x terms?

[frigid isle]

The method is a little confusing

Ur better off doing it normally

[vivid wyvern]

I was initially trying to do that

.close

[odd edge]

Channel closed

Closed by @vivid wyvern

Use .reopen if this was a mistake.

[late sinew]

?

Enlighten me

.reopen

[odd edge]

✅

[frigid isle]

Leave it

🤫

[late sinew]

Why

[vivid wyvern]

Cuz it's kinda tough to separate x and y terms maybe

[copper quarry]

leeave it exes are clutter

[late sinew]

Did u treat it as F(x,y)=0

ah

interesting

is it ok to assume it as a relation tho? @wanton bison

[frigid isle]

I’m saying this cause if u use it the expression might not match the result alot of the times

So it’s just better to use the core methods

[late sinew]

You can always bring it back in terms of the ans

[frigid isle]

Never worked for me

Cause the terms get clunky

When u take partial derivative

Also have to take it twice

And then divide

[late sinew]

[late sinew]

$\dv{y}{x}= -\frac{\pdv{F}{x}}{\pdv{F}{y}}$ ?

[late sinew]

damn too big

,tc alwayswide off

[clever fjord]

Transparent pixels will be added to your rendered LaTeX to improve previews.
Use the texw command to disable this for a single compile.

[frigid isle]

Yes

[late sinew]

alr

[frigid isle]

Very useful

[late sinew]

implicit definitely easier tho here

[frigid isle]

For some questions

[late sinew]

perhaps

[frigid isle]

Where they just ask u to find the value

[late sinew]

kk

.close

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[wanton bison]

Yes, you do that.

[wanton bison]

.solved

[odd edge]

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[waxen pivot]

Hello! I've been trying to figure out why and how formula above in the picture works but I can't figure it out. Can someone help me please.

[waxen pivot]

You can see it if you click on the picture

[vivid wyvern]

This is the dot product or scalar product

[waxen pivot]

Dot product

[vivid wyvern]

A and b r two vector and the angle between them is the angle fi

[vivid wyvern]

It is known as both dot and scalar

[waxen pivot]

O ok

[vivid wyvern]

What r u not understanding abt it?

[waxen pivot]

Why does it work?

I am trying to understand the proof but I don't get it

[vivid wyvern]

Can u share the full picture

[waxen pivot]

It' in dutch

[vivid wyvern]

I can translate

[atomic void]

do u know what dot product means geometrically?

[waxen pivot]

As I understand, it's the length of a vector?

[atomic void]

length of what vector?

[waxen pivot]

I don't know

[atomic void]

its literally in the picture here

dot product is a projection

[waxen pivot]

I will have to google that, projection has been a while

[atomic void]

right?

[vivid wyvern]

Do u know how to derive it?how to derive the abcostheta

[atomic void]

if you understand what a dot product is geometrically, it will be very easy to understand this

[waxen pivot]

What do you mean

[waxen pivot]

So it's the length of the projection of b on a, which is b.cos(fi), but why do you have to do .a too?

[atomic void]

wait, i will send a picture

[vivid wyvern]

Some sorta like this

[waxen pivot]

Oh yes I understand that, I can do that

[vivid wyvern]

In this sense dot product can be called the product of magnitude of a vector and it's projection on another vector

Am i wrong

[atomic void]

the diagram in your picture is not actually a dot product, the dot product is scaled by length of B too basically

i couldnt find a good picture

i recommend you to watch 3 blue 1 brown's video on it

[vivid wyvern]

Can u verify what i said?

[waxen pivot]

Can you send me a link

[atomic void]

wdym

[atomic void]

just search "dot product 3b1b"

on youtube

it will probably pop up

[vivid wyvern]

The pic and description

[waxen pivot]

Thx

[atomic void]

this is correct

[atomic void]

this is not very accurate

[vivid wyvern]

On the basis of angle?

[atomic void]

wdym

[vivid wyvern]

Like specifying angle will make it accurate?

[waxen pivot]

First 2 min and I get it

[atomic void]

uh, i meant ur definition is wrong

[atomic void]

yayyyy

!done

[odd edge]

If you are done with this channel, please mark your problem as solved by typing .close

[waxen pivot]

Thank you

It's the projection of b on a times a or the other way around

.close

[odd edge]

Channel closed

Closed by @waxen pivot

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[vivid wyvern]

Oh ok

[atomic void]

your definition = |a|^2 * cos(theta)

[vivid wyvern]

Yeah i made a mistake

[odd edge]

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[narrow schooner]

Find the volume of the solid formed when the area of the region bounded by the curves x^(2)-3y = 3 and x^(2)+ 4y = 0 is rotated about y-7=0

what should be my upperlimit and lowerlimit, the x or the y?

[narrow schooner]

I am really not sure, since I tried the X up and X down and got 82.1095 cubic units as the answer which I don't know if that is the correct answer.

I used washer method

if y-7=0, am i to use the X1 and X2?

but if its x+7=0, am i to use the Y1 and Y2?

[atomic void]

Wdym "upper limit or lower limit"?

Which variable u want to integrate with respect to?

[atomic void]

When using washer method :
If rotated about the x axis integrate with respect to x
Vice versa

[narrow schooner]

the integral

[narrow schooner]

yea

[atomic void]

Yeah

[atomic void]

Do u understand what i mean?

[narrow schooner]

so if x+7=0, i use the X?
and if y-7=0, i use the Y?

[atomic void]

Think about it. X + 7 = 0 is a vertical line right?

Basically the same as the y axis

Get it?

[narrow schooner]

I need ooga booga language

if X = Y
if Y = X

?

[atomic void]

go to desmos and graph x+7 = 0

is it a horizontal or vertical line?

[narrow schooner]

is a vertical line yea

so i use the Y?

[atomic void]

yes, integrate with respect to y

[narrow schooner]

Thank you, I have like trust issues with how I am solving with my maths

[atomic void]

lol

!done

[odd edge]

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[narrow schooner]

.close

[odd edge]

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[vivid wyvern]

What do we substitute here

[narrow crypt]

wtf is x^r

is that a number or variable?

[vivid wyvern]

x squared

x is variable

X³e^x^2

[narrow crypt]

u = x^2

then ibp?

[twin inlet]

yep

[vivid wyvern]

Uh i got it

Thnx

.close

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[odd edge]

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[hollow copper]

help me pls

[narrow crypt]

what do think

this means

this is interpretation problem

daddy chill

its sat. it aint that serious

[hollow copper]

its the equation

it saying it find the price of the small containers

i dont understand

[swift lake]

ow i thought of diophantine sorry lol

[vivid wyvern]

Answer is in the question

?

[swift lake]

4.51

[vivid wyvern]

Yea

[swift lake]

yeah

i think this is reading test

[vivid wyvern]

Yea could be

[hollow copper]

ohhh okay

thx sm

wait no

wait yes

thx

and

[narrow crypt]

what t would it be at the surface

[hollow copper]

what

500

[narrow crypt]

wonderful

[hollow copper]

is it 500

i was just guessing

[narrow crypt]

lets think about it

logically

what would the ans be

[hollow copper]

bro wat

[narrow crypt]

where did u get 500 from

[hollow copper]

cuz 500 is in the question lol

[narrow crypt]

😭

🥀

think abt

f(t) models no. of particles t mm under the surface

if we want the value at the surface

what should t be

[hollow copper]

isnt t 0.5

is in brackets and t is also in brackets

[narrow crypt]

gang

do u how functions work

[hollow copper]

like sorta

[narrow crypt]

so u substitute t with some number

and it outputs the thing u want right?

[hollow copper]

yeah

[narrow crypt]

so

what should we input here

so that f(t) spits out the no. of particles at the surface

[hollow copper]

uhm

500

[narrow crypt]

think abt it

t represents our position some milimeter under the surface

and we want the thing exactly on the surface

what should t be

[hollow copper]

idkkkk 😭

[narrow crypt]

ok

if i say that h represents my height above sea level

if im at sea level

what should h be

[hollow copper]

5'11

h + sea level

idk

[narrow crypt]

h = 0

right?

cuz im exactly at the sea level 😭

[hollow copper]

yeah

[narrow crypt]

using similar anaolgy

what should t be

[hollow copper]

t = 0

[narrow crypt]

ye