help-39

[brave hawk]

becuase it says DIFFERENT arrangements

so here C1____C2 is the same as C2____C1

because they're both C's

[midnight haven]

so when it is said randomly

they arent the same?

[brave hawk]

in the earlier example, because there were 2 D's, that means a D is twice as likely to be picked for the given slot

so you would consider them different

its like if you have 2 red balls in a bag, and a bunch of other balls of other colors, you would consider the two red balls as different balls

[midnight haven]

isnt it supposed to be 7!/3!

where did 6 come from

nvm

thanks

.close

[pearl pond]

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[brisk pike]

Acceleration of a particle has a value 'a' for a time t. It is followed immediately by a retardation of magnitude 'a' for time t/2. Consider this as one cycle. Initial velocity of particle was zero. The displacement of the particle after n such cycles in succession is:

[brisk pike]

we were told to explicity do this with the help of a graph as a hw exercise

the correct answer is $\frac{n(3n+4)}{8}$ but I get $\frac{n(2n+5)}{8}$

[jolly parrot]

rak³en

[brisk pike]

now my maths is correct

i am sure of that

i need someones help with graphing this

the graph we were asked to make is a v-t one and then take its area

should i send mines

[pearl pond]

@brisk pike Has your question been resolved?

[pearl pond]

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[pearl pond]

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[stiff forge]

🕵️
The number of calories in 12 oz. of light beer is given for a sample of many popular light beers.

93, 94, 105, 64, 94, 102, 99, 112, 112, 109, 105, 114, 100, 133, 95, 89, 109, 119, 114, 70, 110, 72, 120, 124, 90

all data copied up there.

Steps used
Calculate sample mean
Calcilated xi for each value
Sum square deviations then sample variance was found from there got sample standard deviation (16.99284209833)
Standard error divided by 5
fgot margin of error
100.48- and plus the final number which was 8.46824698401

[plush bramble]

enter your calculation here with ,calc

,calc 1 + 2

[jolly parrot]

Result:

3

[stiff forge]

,calc 93, 94, 105, 64, 94, 102, 99, 112, 112, 109, 105, 114, 100, 133, 95, 89, 109, 119, 114, 70, 110, 72, 120, 124, 90

[jolly parrot]

The following error occured while calculating:
Error: Unexpected operator , (char 3)

[stiff forge]

not sure how exactly that works @plush bramble

[plush bramble]

Calculate sample mean

[stiff forge]

,calc sample mean 93, 94, 105, 64, 94, 102, 99, 112, 112, 109, 105, 114, 100, 133, 95, 89, 109, 119, 114, 70, 110, 72, 120, 124, 90

[jolly parrot]

The following error occured while calculating:
Error: Unexpected operator , (char 15)

[plush bramble]

this is not the formula for sample mean

use + for addition

and / for division

,calc (1 + 2) / 6

[jolly parrot]

Result:

0.5

[stiff forge]

can i keep it in the list format or no.

[plush bramble]

no idea what you're asking

,calc is like any other calculator

[plush bramble]

example here

[stiff forge]

,calc )93+94+105+64+94+102+99+112+112+109+105+114+100+133+95+89+109+119+114+70+110+72+120+124+90) / 25

[jolly parrot]

The following error occured while calculating:
Error: Value expected (char 1)

[stiff forge]

,calc (93+94+105+64+94+102+99+112+112+109+105+114+100+133+95+89+109+119+114+70+110+72+120+124+90) / 25

[jolly parrot]

Result:

101.92

[plush bramble]

and all the other calculations? sample std dev. etc

[pearl pond]

@stiff forge Has your question been resolved?

[stiff forge]

,calc (79.57+62.73+9.49+1440.17+62.73+0.01+8.17+101.61+101.61+62.73+9.49+146.89+3.69+965.97+47.05+166.93+62.73+308.35+146.89+1017.45+65.29+897.45+345.22+524.21+121.25)

[jolly parrot]

Result:

6757.68

[stiff forge]

,calc 6757.68/24

[jolly parrot]

Result:

281.57

[stiff forge]

,calc sqrt 281.57

[jolly parrot]

The following error occured while calculating:
Error: Unexpected type of argument in function multiplyScalar (expected: number or Complex or BigNumber or bigint or Fraction or Unit or string or boolean, actual: function, index: 0)

[stiff forge]

@plush bramble not sure how sqrt works on here

[plush bramble]

,calc sqrt(9)

[jolly parrot]

Result:

3

[plush bramble]

just like any other calculator

[stiff forge]

,calc sqrt(281.57)

[jolly parrot]

Result:

16.780047675737

[plush bramble]

,w sqrt(9)

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by rie.mann

[stiff forge]

,calc 16.780047675737/5

[jolly parrot]

Result:

3.3560095351474

[stiff forge]

,calc 2.4921593×3560095351474

[jolly parrot]

The following error occured while calculating:
Error: Undefined symbol ×3560095351474

[stiff forge]

,calc 2.4921593(3.3560095351474)

[jolly parrot]

Result:

8.3637103739063

[stiff forge]

,calc 100.48-8.3637103739063

[jolly parrot]

Result:

92.116289626094

[stiff forge]

,calc 100.48+8.3637103739063

[jolly parrot]

Result:

108.84371037391

[stiff forge]

92.12
108.84

also wrong

[stiff forge]

did all work on here. not sure where i made the mistake. let me know if you see anything off. I appreciate the help ❤️

[pearl pond]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[plush bramble]

.reopen

[pearl pond]

✅

[stiff forge]

.reopen

[plush bramble]

where's the sample standard deviation?

[stiff forge]

@plush bramble

[plush bramble]

looks like you rounded here

where did those numbers come from

[stiff forge]

i found xi-100.48 for each value but i didnt show all work ucz that would take a while lol

so like 93 I did (93-100.48)^2=56.5504

[plush bramble]

why 100.48

[stiff forge]

oh wait i was using the wrong value

it should be 101.92

,calc 6677.84/24

[jolly parrot]

Result:

278.24333333333

[stiff forge]

,calc sqrt(278.24333333333)

[jolly parrot]

Result:

16.680627486199

[stiff forge]

,calc 16.680627486199/5

[jolly parrot]

Result:

3.3361254972398

[stiff forge]

,calc 2.4921593(3.3361254972398)

[jolly parrot]

Result:

8.3141561839133

[stiff forge]

,calc 100.48+8.3141561839133

[jolly parrot]

Result:

108.79415618391

[stiff forge]

,calc 100.48-8.3141561839133

[jolly parrot]

Result:

92.165843816087

[stiff forge]

108.79
92.16

wrong @plush bramble

[plush bramble]

can't follow your work at all

[plush bramble]

what is 100.48 here

[stiff forge]

after calc standard deviation
found standard erro (divided the16.68 number by 5)
used t value (2.4921593)

t value for margin of error

wait ac

,calc 101.92-8.3141561839133

[jolly parrot]

Result:

93.605843816087

[stiff forge]

,calc 101.92+8.3141561839133

[jolly parrot]

Result:

110.23415618391

[stiff forge]

110.23
93.61

got it

thank you @plush bramble lmao that was a process

[plush bramble]

:t

[pearl pond]

@stiff forge Has your question been resolved?

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[pearl pond]

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[silent rock]

Point A to point B
from A to B - 9km distance crossed, V speed, T time
B to A - 3.5km distance crossed extra (aka 12.5km), V + 1km/h speed, T + 15 mins time
my only idea is to make it something like (A to B) = (B to A) - 3.5km but i dont know how to do it afterwards and im not even sure if its the right way to go

[silent rock]

oh

forgot to mention but

unknown time since start of journey, but final time is 16;15

like you get to point B at 16:15 on the clock

[rain vessel]

We can start by noting that Xf= Xi+VT+1/2AT^(2)

[toxic lichen]

nope

[rain vessel]

?

[toxic lichen]

this is a trip there and back with constant speed on each leg. acceleration is zero.

at least if i understand correctly

[rain vessel]

Yeah, we can just set it to zero

[silent rock]

no accelelration mentioned yeah

no use of it

[rain vessel]

The formula still holds

[toxic lichen]

ok then why bother introducing a term that you're going to zero out anyway? what is the point?

occam's razor!

[rain vessel]

I like the formula

[toxic lichen]

ok well it's unfortunate but your formula isn't a magic bullet.

[rain vessel]

Gives more insight into real world scenarios when changing velocity actually requires acceleration

[toxic lichen]

can we see the original problem just for reference

[silent rock]

im sorry but is it okay if i close this and redo it around half an hour later, i have to go right now

[toxic lichen]

sure

[silent rock]

okay thanks

.close

[pearl pond]

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[pearl pond]

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[silent rock]

Point A to point B
from A to B - 9km distance crossed, V speed, T time
B to A - 3.5km distance crossed extra (aka 12.5km), V + 1km/h speed, T + 15 mins time
my only idea is to make it something like (A to B) = (B to A) - 3.5km but i dont know how to do it afterwards and im not even sure if its the right way to go
starting from point A at ??:??, and after making all the distance to B and all the way back to A its 16:15 on the clock, infact were looking for the time of starting

Heres a photo because people asked for it, its bulgarian language

[toxic lichen]

vt = 9
(v + 1)(t + 1/4) = 12.5

1/4 е 15-те минути, преведени в часове

най-прозрачният модел е с 2 уравнения

[silent rock]

do you know bulgarian or are you using a translator

[toxic lichen]

говоря български

[silent rock]

олеле

[toxic lichen]

толкова ли си личи на гугъл преводач?

[silent rock]

просто не разбрах какво имаш предвид по "прозрачен"

[toxic lichen]

лесен за разбиране

[silent rock]

аааа

[toxic lichen]

иначе би могъл да изразиш скоростта на първата пътека като 9/t

и тогава ще излезе едно рационално уравнение на една променлива t вместо система от 2

[silent rock]

нещо което си имах на предвит е
9 = 9/v * v = (v+1)*(12.5/v+1) -3.5
Обаче просто незнам дали е вярно и как да го реша

[toxic lichen]

това ми изглежда някак си сбъркано, ама не мога да кажа къде...

[silent rock]

можеби защото като ги съкратиш и се получава 9 = 9 = 12.5 - 3.5

for any English speakers we still haven't come to an answer

[toxic lichen]

i have to go soon, sorry

if you are patient enough you could catch me in the morning

[silent rock]

Okay no worries

I can try to see in the morning if anyone's going to help as Im going to go to bed now

.close

[pearl pond]

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[austere hare]

Hey there, can someone help me with this

Im confused on how to integrate i think because its below the like axis

[brave schooner]

below what axis?

oh the y i see it

[plush moss]

Calculate the volume of rotation of y=x^4/2 around x=0

[brave schooner]

should be around x=0 no?

[austere hare]

so would it be like

[plush moss]

No but it's a vertical revolution

[austere hare]

2pi (integral sign) x^5/2 dx

?

I just am struggling with this shell method or disc or whatever too

wait but that only would give the inner region too (assuming that thats the right equation its prob not)

[brave schooner]

shell method would be best here

[austere hare]

The way i understand shell method is its v = 2pi (integrate sign) x * f(x) dx

like x being height and f(x) being radius

i dont get how to apply that here tho

[brave schooner]

you could find the inner volume then the outer, split it up in 2 parts

[austere hare]

do i split up the outer volume into two parts too?

[brave schooner]

no, just integrate the inner volume and then the outer

[austere hare]

uhh okay i ll try

[brave schooner]

with $2\pi \int_{a}^{b}x\cdot f(x) dx$

[austere hare]

alr

thans

thanks

x \cdot f(x)

yea

[brave schooner]

cdot yeah lol

rusty latex

[austere hare]

lmao youre good

[jolly parrot]

caspar

[brave schooner]

now im done

[austere hare]

okay and that would be the inner region correct?

so for the outer i would then do

[brave schooner]

same for both

[austere hare]

ah okay

thanks man

[brave schooner]

just w a = -1 cause its below the y axis

[austere hare]

just to confirm

when it says volume of the glass

its including the shaded region right

cuz the wording is like confusing there

or is it just what the glass can hold

[buoyant vapor]

.repoen

.reopen

[brave schooner]

part a) is the volume of the physical glass, while part b) is the volume wich the glass holds

[austere hare]

ohh okay

thanks

wait im like

trying B right now

and i got 32pi/3 by doing shell method and washer method or whatever

but its wrong

[brave schooner]

y from 0 to 7+1

or

[austere hare]

i still got it wrong

[brave schooner]

hmm

[austere hare]

I only have 1 more attempt for A and 3 more for B

[brave schooner]

your function is wrong no?

[austere hare]

how

[brave schooner]

wait

[austere hare]

wait

i forgot

to multiply by another y right

should i try this

that was wrong

[brave schooner]

use disc integration around y

[austere hare]

like this?

[brave schooner]

i think so yeah

any luck?

[pearl pond]

@austere hare Has your question been resolved?

[austere hare]

sry i didnt see that

ill try

YES

ok i got (b)

only one attempt left on (a) though so i guess ill wait till someone i know gets the right answert

[pearl pond]

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[queen gazelle]

is there a faster way/is this correct

[queen gazelle]

and is there a way to avoid hyperbolic trig

[versed mica]

,w ln^3(sec(x) + tan(x)) sec(x)tan(x)

[jolly parrot]

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[versed mica]

just why bro

like why

[queen gazelle]

🐺

[pearl pond]

@queen gazelle Has your question been resolved?

[pearl pond]

@queen gazelle Has your question been resolved?

[pearl pond]

@queen gazelle Has your question been resolved?

[pearl pond]

@queen gazelle Has your question been resolved?

[pearl pond]

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[midnight haven]

integral of sin2x and 2sinxcosx can be true or false

how did this happen?!?!

[tropic saddle]

your sentence makes no sense

are you asking why they are equal?

sin(2x)=2sin(x)cos(x) is a trig identity

[midnight haven]

yes

BUT

if you use u sub on the 2nd integral

it's not equal

it's gonna be sin²x + C

[tropic saddle]

go on desmos

plot the functions

they differ by a constant

[echo anvil]

So u r saying if you integrated sin(2x)
Then integrate 2sinx cosx
U'll get diff answers ??

[midnight haven]

yeah

so it really is the job for +C this time?

[echo anvil]

Yeah for sure

[midnight haven]

i wonder which constant value should it be to make it equal

[echo anvil]

That's why teachers Keep saying
DO NOT FORGOT THE C

[echo anvil]

Any interval

[midnight haven]

my calculator did "calculation timeout" when i equated both

and it's also -1/2cos2x ≠ sin²x

+C on both sides

but idk which constant value to make it equal

will linear alg help on this

[echo anvil]

I see

(c) does not work this way

[tropic saddle]

this is just another trig identity

[midnight haven]

yeah it is trig identity

[tropic saddle]

do you know a trig identity with a square of a trig function in it

[echo anvil]

If u want to make them equals u have to solve both integration for the same interval

[midnight haven]

which one

[midnight haven]

like this?

int(sin2x)dx = int(2sinxcosx)dx

idk what u mean

[tropic saddle]

well I'm if you know any identity like that

[echo anvil]

Put a value for x here

[midnight haven]

ohhhhh

bounds

[echo anvil]

Let's say π

[echo anvil]

Yeah whatever u call it in English

[midnight haven]

alr ill try pi

yeah the same

[tropic saddle]

no, plug the x=pi into the sin^2(x) and the -1/2cos(2x)

[echo anvil]

Try 10 and tell me what u got

[tropic saddle]

those integrals are the same

the +C cancels when you do F(b)-F(a)

[echo anvil]

I'm not quite sure what u r trying to do

[midnight haven]

oh right wait

i think i get denascite

holup

[echo anvil]

He doesn't have a given value for x

@midnight haven what the question want 😭

[midnight haven]

whar

[midnight haven]

this is the question

it differs by a constant

[tropic saddle]

the trig identity is sin^2(x)=1/2-1/2cos(2x)

the constant is 1/2

[midnight haven]

OK THANK YOU

ANSWER IS CORRECT

.close

[pearl pond]

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[midnight haven]

now i know how to find constants to equate identities

[echo anvil]

This might be helpful

[midnight haven]

HOLY CRAP THANK YOU

[echo anvil]

Couldn't find a pic in English so i just took a pic of my old books

[echo anvil]

Np

[tropic saddle]

there is also just the wikipedia article

[echo anvil]

Too lazy 🦥

Lol

[pearl pond]

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[midnight haven]

I tried to extend the side AD and BC to make equilateral triangle but i still cannot find the DC(((

[steep wagon]

you're in the right direction

Let's denote that point E(where the lines meet)

[toxic lichen]

see if you can name any other angles on the picture

[steep wagon]

write down some angles

[midnight haven]

So DC is median bisector and altitude

[steep wagon]

you didn't write what I was talking about

write down all angles that you can conclude with the current diagram

[midnight haven]

Okay

What about now is it right?

[steep wagon]

yes

do we know any relation about DE and EC?

[midnight haven]

7/4

[steep wagon]

Idk how did you get this

[midnight haven]

Oh wait nope

Yes 1/2

Cause it is x/2x

[steep wagon]

ok, fair, now find another relation and crush the problem

[echo anvil]

Are u Arabian if u don't mind me asking

[steep wagon]

no, I'm not

[echo anvil]

Oh, so just Muslim ig
Welcome brother 🤍

[midnight haven]

lol I did not need any relation

Thanks a lot for helping me

[pearl pond]

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

[midnight haven]

.close

[pearl pond]

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[pearl pond]

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[inland ivy]

How can the closure of a set in Euclidean space be different than the closure in a subset?

[inland ivy]

If $V\subset \bR^n$ and $A\subset V$ then why isn't $\bar{A}^V = \bar{A}$?

[jolly parrot]

kheer257

[inland ivy]

the definition for a cluster point also ends up being the same

[toxic lichen]

do you want an example

V = [0,2)
A = [1,2)

closure of A in V is A itself but in R it's [1,2]

[inland ivy]

hmm

but why isn't 2 a cluster point of A in V?

[toxic lichen]

cause it isn't in V

[inland ivy]

ahhhh

okay that was a really dumb question

thanks

[toxic lichen]

eh

[inland ivy]

so then cl_V(A) = V \cap cl(A)?

[toxic lichen]

better to be a fool now than to be a bigger fool later

[toxic lichen]

i.... think so?

but i am not 100% on that

[inland ivy]

any cluster point of A that lies in V is a cluster point of A in V, and vice versa

[pearl pond]

@inland ivy Has your question been resolved?

[pearl pond]

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[willow geode]

I need help with this problem

[craggy magnet]

You're solving for x ?

[willow geode]

yes

[craggy magnet]

Note f(x) the left hand side

Study the behaviour of f

[willow geode]

I got f(x)-f(7)> -2 square root of 7

[craggy magnet]

For example where f is monotonous

[willow geode]

but i still can't solve for x

[craggy magnet]

Did you study f properties ?

For example you can start by looking at its variations

---

Ex : lets say you can prove it has a minimum. If this minimum happens to be greater than the left hand side then the inequation holds for all x > 0
That's why looking at the variations can help

btw are you solving for integers or for real numbers ?
Because the problem is way easier for integers

[pearl pond]

@willow geode Has your question been resolved?

[willow geode]

It has no minimum

[willow geode]

integers

[craggy magnet]

What do you know on f ?

[craggy magnet]

Why ?

[willow geode]

the derivative will be 5x^4 +1/(2 square root of x)

[craggy magnet]

Thus the derivative is....

positive for all x > 0

So what do you know about f ?

[willow geode]

it has no minimum

[craggy magnet]

You know more than that

If its derivative it positive on R+

[pearl pond]

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[surreal osprey]

I was wondering if somebody could help me understand whether Sn-D^1 is contractible or not? I was reading Hatcher 2.B, and he proves that all homology groups vanish, but does that imply it must be contractible? If not how can I show it's contractible?
the case of N=2 I can show S^2-D1 is homemorphic to R^2 but not generally

[pearl pond]

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[pearl pond]

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[pearl pond]

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[neon jolt]

hello

[neon jolt]

may anyone pls explain how do we detect the overflow here?

I dont understand the point of the xor at the end

[toxic lichen]

the claim seems to be that an overflow happens iff C3 xor C4

[neon jolt]

I mean what is the logic behind this ?

[toxic lichen]

trying to figure it out

[neon jolt]

alr

[toxic lichen]

in particular i can't seem to visualize how C3 could be 1 and C4=0

[neon jolt]

maybe in the case where no carry at the end?

[toxic lichen]

how can there be no carry though

B3=A3=0?

ig then there is just no overflow though

hm. i seem as confused as you are about this xor

[neon jolt]

I mean

this circuit can be adder or substractor right?

so in the case of adding we'll have an overflow

[toxic lichen]

sure

oh like

[neon jolt]

ig

[toxic lichen]

oh signed 4bit

huh.

[neon jolt]

yeah

[toxic lichen]

ahhhh

[neon jolt]

but in the case of substractor maybe we wont get overflow

[toxic lichen]

ok then C3=1, C4=0 means two negatives added to a positive?

which would be underflow

[neon jolt]

could be...

I still dont get the big pic tbh

oh hold up

" if two binary numbers are considered to be unsigned then C bit detects a carry after additon or borrow after substraction

if the numbers are signed then the V bit detects an overflow"

It is written on my notes

but I dont get what does that exactly means

[toxic lichen]

well C is just the final carry from the addition

that one is easy

[pearl pond]

@neon jolt Has your question been resolved?

[pearl pond]

@neon jolt Has your question been resolved?

[pearl pond]

@neon jolt Has your question been resolved?

[pearl pond]

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[warm forum]

oh you're her

e

for a fixed x, fn(x) -> 0 as n-> infty therefore its pointwise?

[pearl pond]

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[winged matrix]

the function will only make sense if x is bigger than or equal to -2, plug in -2 instead of x and complete the problem, and then try -3 and see the difference

[cursive wraith]

sqrt(a) is only well defined when a >= 0

so the thing that is input into the square root function, x+2, must be >= 0

which leads to x >= -2

[winged matrix]

try to visualise these problems

imagine x as a line

[pearl pond]

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[robust kraken]

i have to show that any permutation of S_n is written as a product of transpositions

[robust kraken]

by induction

so i did initialization for n=2, it works well, for the heredity i suppose this property true for n in N and i have to show that its true for n+1. The problem is that I don’t know how to go from n to n+1...

if someone has any ideas 🙏🏼

idk if i can do it without support and things like that ?

only with permutations and transpositions

[rough stream]

Let's say I give you a permutation on S4.

Can you give me a permutation on S3, such that you can add the 4th element in, and put it in the correct spot with a single transposition?

It might help to actually draw out examples

[robust kraken]

for example i take $\sigma= (3 \ 1 \ 4 \ 2)$ and you want me to describe $\sigma$ with an S3 element composed with a transposition?

idk if i understood

[jolly parrot]

tm

[rough stream]

Yeah basically. Assuming we can make anything in S3 with transpositions, that would prove we can make this with only transpositions as well

Cycle notation might actually make this harder. Consider that as a permutation:
1 2 3 4
4 3 1 2

[robust kraken]

$\tau = (2 \ 4)$ and $\sigma’=(1 \ 3 \ 2)$ ?

[jolly parrot]

tm

[rough stream]

I'll transpose 4 and 2 here to get:
1 2 3 4
2 3 1 4

Which I think is what you were going for, yes

The first three can be placed with transpositions, and we got here with a single transposition, so our original can be made with transpositions

[robust kraken]

hm i see

[rough stream]

Hopefully it's clear that it didn't actually matter what element of S4 we started with

[robust kraken]

yeah but how we choose these permutations of S3

[spare lark]

Heredity = inductive step

[robust kraken]

aide moi au lieu de corriger mes fautes d’anglais mdr

[spare lark]

No kaynex is doing it already

[robust kraken]

i need to do 2 cases separately i think

if sigma in Sn+1 fixes n+1 and if it doesn’t

if $\sigma(n+1)=n+1$ : \
then $\sigma|_{{1,...,n}} = \tau_1\cdots\tau_r$ by hypothesis \
so $\sigma = \tau_1\cdots\tau_r$
\ if $\sigma(n+1)=m \neq n+1$ : idk how to do this case 😥

do we have to return to a fixed point case?

as in the first case

[jolly parrot]

tm

[robust kraken]

I had put k instead of n 💀

oh i think i have an idea

[rough stream]

Sorry I was pulled away haha

[robust kraken]

do a composition of something with sigma to get a fixed point by this thing

[spare lark]

Any permutations can be written as product of disjoint cycles

Idk if you know this thing

[robust kraken]

and then we use case 1

[robust kraken]

yes but im not at ease with cycles 😥😥

[spare lark]

Ok continue induction then since its your first idea

[robust kraken]

ok but with what we composes sigma 😭

to get a fixed point

[rough stream]

Let σ ∈ S_(n-1). By assumption, σ can be made with transpositions.

[robust kraken]

yes

[rough stream]

Wait no I want to go the other way

Let σ ∈ Sn.

Use a transposition to put element n in slot n. (Unless it is already there, in which case we can do nothing)

By assumption, the other n-1 elements can be placed with transpositions

[robust kraken]

$\tau = (m \ n)$ ?

[jolly parrot]

tm

[rough stream]

Idunno

[robust kraken]

😭

[rough stream]

It would be very brave to say such a transposition doesn't exist

[spare lark]

Lets say we not brave for this one

[rough stream]

If n is in slot m, then the transposition is (m n) yeah

[robust kraken]

and then we have $\tau \circ \sigma$ fixed n

[jolly parrot]

tm

[rough stream]

I guess you could start by making the case that element n is in slot m, which doesn't lose generality

[robust kraken]

bc $\tau \circ \sigma(n)=n$

[rough stream]

Other than the case where n is already fixed

[jolly parrot]

tm

[robust kraken]

that’s what i did no?

[rough stream]

Yeah!

[robust kraken]

okkkkkkk i see

damn permutations is not for beginners

thanks 🙏🏼😭

[rough stream]

Np, feel free to ask if you have anything else!

[robust kraken]

i think i’m good lol thx

.close

[pearl pond]

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[peak kayak]

anyone can help me please

[peak kayak]

<@&286206848099549185>

[foggy hound]

this is long did u start it?

you should start with working on one triangle like ABC as a base of the pyramid for example

[peak kayak]

ik😭

[broken fossil]

do you need vectors to solve this?

[foggy hound]

maybe

but i believe there are different ways to solve this

vectors and coordinates are the straightest one

you can try pythagoras, cosintan

[peak kayak]

icl idk how to do it

[pearl pond]

@peak kayak Has your question been resolved?

[pearl pond]

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[jovial kiln]

where did the g(x) come from

do you mean cf(x)g(x)

[pearl pond]

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[civic zephyr]

i need help w this

[jovial kiln]

okay do you know the coordinate of that intersection

[civic zephyr]

uh not exactly

we can estimate but

not good right

[jovial kiln]

you should find the intersection by equating f(x) and g(x)

[civic zephyr]

ohhso its like 0,1.686 approx

then what?

[jovial kiln]

,w -x^2-1.5x+9 = -2x+7

[jolly parrot]

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[jovial kiln]

okay good

now you integrate from 0 to 1.68614

[civic zephyr]

which equation

[jovial kiln]

which curve is at the top here?

[civic zephyr]

top is the x^2

right?

[jovial kiln]

yes

on the interval of interest

now we do top - bottom

[civic zephyr]

ohhh

[jovial kiln]

so f(x)-g(x)

[civic zephyr]

ok

is that2.487?

[jovial kiln]

,w integrate (-x^2-1.5x+9 +2x - 7) from 0 to 1.68614

[jolly parrot]

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[civic zephyr]

ohh so 2.485?

[jovial kiln]

yes

maybe you made an approximation error

[civic zephyr]

yea prolly rounded to soon

can we do 1 more?

[jovial kiln]

first you need to tackle each region one by one

try finding the intersection point

[civic zephyr]

so set them equal to eachother?

0.654?

[jovial kiln]

,w -2x^2+8 = 1.75x+6

[jolly parrot]

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[jovial kiln]

yes

now integrate from 0 to 0.654

which curve is at the top?

in that interval

[civic zephyr]

wouldnt this be right-left

or wait

is it 2 sep curves

cause we hafta find area ssperaltey

so on the left x^2 is at the top

[jovial kiln]

we tackle this first

[civic zephyr]

ok

[jovial kiln]

yes x^2 is on top

[civic zephyr]

so -2x^2+8-1.75x-6

and i wud intergrate that

[jovial kiln]

yes

[civic zephyr]

that is 0.747?

[jovial kiln]

,w integrate -2x^2+8-1.75x-6 from 0 to 0.654

[jolly parrot]

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[jovial kiln]

yes

now we focus on this

[civic zephyr]

ok and for right side we itnergrate 1.75x+6+2x^2-8?

[jovial kiln]

try doing it urself and ill verify ur answer

[civic zephyr]

thats 5.580?

[jovial kiln]

,w -2(2^2)+8

[jolly parrot]

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[jovial kiln]

,w integrate 1.75x+6+2x^2-8 from 0.654 to 2

[jolly parrot]

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[jovial kiln]

yes

[civic zephyr]

ok now 5.5806+0.747264

[jovial kiln]

yes

[civic zephyr]

wait but didnt we approximate the 0..654

wouldnt that hurt us tech

[jovial kiln]

thats why put ur final answer in 3 sig figs

[civic zephyr]

okk

so 6.328?

[jovial kiln]

i mean do u really need me to verify addition

[civic zephyr]

nah

alr yea its right

[jovial kiln]

okay good

[civic zephyr]

i didnt have a calc mb

can we do 1 more?

its a diff topic

[jovial kiln]

ugh

sure

[jovial kiln]

you need to recognize an equation for the base of the rectangle for each x

so an equation of B(x)

[civic zephyr]

waht

so for the base

its e^x-(4x+1)

[jovial kiln]

which one is the top equation

we want positive areas remember

[civic zephyr]

yop is 4x+1

[jovial kiln]

so we have 4x+1-e^x

[civic zephyr]

ohh

[jovial kiln]

now what is the height of the rectangle?

[civic zephyr]

1/2 base

[jovial kiln]

so the integral for volume is
∫A(x)dx
where A(x) is the area of the rectangle

now we look at the bounds

where is the point of intersection

[civic zephyr]

0.258 and 2.194
​

[jovial kiln]

,w e^x = 4x+1

[jolly parrot]

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[jovial kiln]

,w e^x=4x+1 solve x numerically

[jolly parrot]

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[civic zephyr]

oh shit

i prolly misteneted in graphing calc mb

[jovial kiln]

okay so integrate A(x) from 0 to 2.34

[civic zephyr]

ok

4.637?

[jovial kiln]

what was ur working

[civic zephyr]

?

itnergral of 0 to 3.336666

[jovial kiln]

what was your A(x)

[civic zephyr]

(4x+1-e^x)^2

dx

then took half of dat

[jovial kiln]

yes that is correct

[civic zephyr]

alr thanks

.close

[pearl pond]

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[peak kayak]

Solid A and solid B are similar. The ratio of the height of solid A to solid B is 2 : 5 The volume of solid A is 12cm^3 work out the volume of solid B.

[peak kayak]

can someone help me please

[simple jasper]

Is that the exact question

[peak kayak]

yea

nothing else

[spiral pivot]

@peak kayak

if you have similar shapes with similarity constant k, then if shape 1 has a length d, then shape 2 has a length kd.

If shape 1 has area d^2 then shape 2 has area k^2 d^2

What about volume?

[peak kayak]

ohhhh

it would be

d^3 and shape 2 k^3 d^3

[spiral pivot]

So d^3 = 12 cm^3

What is k^3?

[peak kayak]

ummm

tbh am not sure

[spiral pivot]

Well, shape A to shape B is a ratio of 2 : 5.

(what does this imply about k?)

[peak kayak]

k is bigger

[spiral pivot]

Can you find the value?

Hint,

We are given that the 2 : 5 is a ratio of heights.

So in this case the height of shape A is 2 while the height of shape B is 5.

(or more specifically shape A is 2c while shape B is 5c, but you will find that the unknown constant c cancels)

Can you set up an equation that you can solve for k using this information?

@peak kayak ^

[peak kayak]

ummm

let me think

wait is the scale factor 2/7

Idk

[spiral pivot]

So we have if the height of shape A is d, then the height of shape B is kd. We know d = 2, and we know kd = 5.

What is the value of k?

[peak kayak]

k = 5/2

[spiral pivot]

Ok, now what?

[peak kayak]

do i times 12 by 5/2

@spiral pivot

[spiral pivot]

@peak kayak

Nope

Think back

12 cm^3 is what?

[peak kayak]

the volume of solid A

[spiral pivot]

[fair creek]

weird problem, perhaps they're prisms

constant area maybe... ?

it wouldn't make much sense without constant area honestly

[peak kayak]

yea

[fair creek]

$V\propto h$ maybe?

[jolly parrot]

parabolicinsanity

[spiral pivot]

@peak kayak

If 12 cm^3 is the volume of shape A then 12 * k^3 is the volume of shape B

Because volume scales with k^3, (and area with k^2, and length with k)

@fair creek they just need to be similar shapes.

[fair creek]

oh yeah, they're similar 😔

[peak kayak]

so then how dyu find k

[fair creek]

ratios

[spiral pivot]

You already found k

[fair creek]

$\frac{V_2}{V_1}=\left(\frac{h_2}{h_1}\right)^3$ something like this 😔

[jolly parrot]

parabolicinsanity

[spiral pivot]

@peak kayak

I handheld you through the entire problem, but it doesn't seem like you understood the process at all. This is unfortunate because it means you probably didn't learn much of anything.

Can you review what was already written in this channel and see if it clicks?

[peak kayak]

alright

is just that its 3 am and am bare tired innit

but i need to get this finnished

[spiral pivot]

You found k here

[peak kayak]

so do u times 5/2 by 12

[spiral pivot]

No

k^3

[peak kayak]

but here u said 12 x k

[peak kayak]

ohhh

[fair creek]

cubed 😭

[peak kayak]

187.5

but thats too big compared to 12

[spiral pivot]

No it's not, volume grows rapidly

[fair creek]

it's (2.5)^3 times that's why it's big

[spiral pivot]

A number this large is expected

[peak kayak]

oh

[fair creek]

i think you should look more into scaling of 2d and 3d shapes

[spiral pivot]

Well

[peak kayak]

yea true

[spiral pivot]

,w 12 * (5/2)^3

[jolly parrot]

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[spiral pivot]

I can confirm that you have the correct answer

Good job!

Get some sleep now

Come back tomorrow and review this thread

[peak kayak]

tysm😭

[spiral pivot]

(you can find it in your ping notifications)

[peak kayak]

alright

.close

[pearl pond]

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[pearl pond]

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[mighty timber]

Why is this wrong?

[sharp vigil]

can you show your work?

[jovial kiln]

,w integrate (-10x+14)/((3x-5)(x+1))

yeah your working would help

[tardy reef]

The 3rd term can be split again, and merged with the 1st term

Maybe the autograder didnt like that

[jolly parrot]

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[tardy reef]

yea

the answer is almost correct

but you need to merge the 1st and 3rd terms

with log rules

[mighty timber]

I got it!

Thanks!

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[pearl pond]

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[latent quail]

These polynomials are independent to each others because u1 x p1(x) + u2 x p2(x) + u3 x p3(x) = 0 only exist when u1 = u2 = u3 = 0

[latent quail]

Is my explanation correct?

[tiny lintel]

Yup, but do you need to show it ?

[latent quail]

Not really, I just need a quick confirmation

What if I change p1(x) into x + x^2

[blissful cloak]

or another way to think about it is that these are the bases [1 0 0], [0 1 0], and [0 0 1] in P2

[latent quail]

Aren't these only applicable to vectors

[tiny lintel]

p1 p2 and p3 are vectors

[latent quail]

I don't get it, they look like polynomial to me

I'm referring to my original question

[blissful cloak]

then p1=p2+p3 ==> p1-p2-p3=0 which means (1,-1,-1) is in null of [p1 p2 p3] which means Null is not {0} which means a few other things similar statements about this set of polynomials

[blissful cloak]

a way to think about it

is in P2 (the set of a polynomials of degree 2)

they can be written as c+bx+ax^2, right?

[tiny lintel]

You are in the vector space of polynomials. So the vectors are polynomials

[latent quail]

Gimme a few minutes

[pearl pond]

@latent quail Has your question been resolved?

[sharp vigil]

what we're doing here is writing polynomials as coordinate vectors with respect to a "standard basis" {1, x, x^2}. coordinate vectors are nice because they allow us to do all the usual computations for vectors in R^n for more unusual vector spaces

[pearl pond]

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[latent quail]

yes

[latent quail]

makes sense

Thanks yall

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[prime token]

hey

[prime token]

suppose f, g are functions with graphs given

we're finding lim_(x->0) f(g(x))

but g(x) as x->0 doesnt exist

so does lim fg exist? why/why not

[sharp vigil]

well you might want to split up that limit into two sided limits, and analyze those limits separately

[inland ivy]

it definitely can

[prime token]

oh right i get it

its just felt weird that the answer i saw to a certain problem reduced directly from lim g that lim fg doesnt exist

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[still dirge]

How do I finish this?

[still dirge]

.solved

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[rapid wagon]

Is [a,a] considered a segment?

[toxic lichen]

by whom and in what context

[rapid wagon]

In general

Wdym by whom?

[toxic lichen]

what problem were you doing that led you to ask this

[grim ibex]

Can someone help me with trigo in #help-0

[toxic lichen]

!noadvert

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[rapid wagon]

It's not related to a problem

I just thought of it

[toxic lichen]

ok then like

it is an edge case

it behaves differently from segments with nonzero length

[rapid wagon]

I see

But by definition you can call it a segement right? Just an edge case

Ok so let's say I have a segment [a,b] then can I say that [a,a] is a subsegement

[versed mica]

it’s just a point

a <= x <= a —> x = a

[autumn fossil]

It's up to your book / professor

i can imagine them defining it to be either

it depends on the definition you're using

[rapid wagon]

Tbh i dont think its allowed to be subsegment with how we learn

[autumn fossil]

okay, then it's not

[versed mica]

who says subsegment

[rapid wagon]

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[placid junco]

by def in euclidean space i believe its bounded by two distinct endpoints.
But in some problems you can generalize it to include [a,a] as segment (if construction have edge cases as mentioned), but you need to mention it

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[midnight haven]

.

this is a math/business question

[midnight haven]

how does this work?

[midnight haven]

there is a mathematical equation for it but i cant figure it out

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[atomic oracle]

For a function:
f(x) = g(x) + h(x)
If g(x) in one-one and onto and h(x) is one - one and onto, can we say f(x) is one - one and onto?

[sharp vigil]

what if h(x) = -g(x)

[atomic oracle]

Then it would be one-one into?

[sharp vigil]

what would be the sum of g(x) + (-g(x))?

[atomic oracle]

f(x) = 0

[sharp vigil]

yes, is that one to one or onto?

[atomic oracle]

Its one to one but its not onto?

[sharp vigil]

is it one to one?

[toxic lichen]

wrong on both counts

ah shit i spoiled it

[atomic oracle]

Oh its many - one because
for every x it would be 0

Fair enough

So nothing can be said about f(x) on the basis of h(x) and g(x)?

We have to check the mapping of f(x) seperately?

[toxic lichen]

indeed

nothing can be said as-is

i will call XY on this though

!xy

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[atomic oracle]

I know how to solve this but i was just curious if there was a relation

[plush bramble]

lol

[toxic lichen]

neither x^2 nor -x^2/(1+x^2) are injective OR surjective on their own

so in fact you were quite deeply misguided

[atomic oracle]

Thats not what I meant 😅
I just wanted to know if there was a relation between the mapping of f(x) and g(x) and h(x)

[toxic lichen]

ok but your question does not have much to do with the problem at hand

thats what im saying

[atomic oracle]

Fair enough

Anyway thanks!

.close

[toxic lichen]

anyway no there is not really a way to go around analyzing f(x) on its own merits

[pearl pond]

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[spiral pivot]

There isn't a lot you can say about the sum of two functions regarding their injectivity or surjectivity. It might be a fun exercise to go through and make a list of all different possibilities of sums of functions.

For instance, x^3 - x^2 is surjective but not injective, x^2 is neither injective nor surjective, but their sum is both

But if you change to x^3 -2x^2 the sum is still not injective.

[pearl pond]

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[warm forum]

??????

[pearl pond]

@warm forum Has your question been resolved?

[pearl pond]

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[azure ether]

<@&286206848099549185>

[grizzled knoll]

!show

[pearl pond]

Show your work, and if possible, explain where you are stuck.

[frozen lantern]

!15min

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[azure ether]

Idk how to do it

Just do 1 for an example then I will understand

[verbal whale]

Have you been taught implicit differentiation?

I mean, did the teacher explain what it is and how to do it?

[spiral pivot]

I will implicitly differentiate the equation with respect to x.

\begin{align*}
x^2 + 2xy - y^2 &= 4 \
\dv{x}(x^2 + 2xy - y^2) &= \dv{x}(4) \
\dv{x}(x^2) + 2 \dv{x}(xy) - \dv{x}(y^2) &= 0 \
2x + 2(y + x y') - 2yy' &= 0 \
2x + 2y &= -2xy' + 2yy' \
y' &= \frac{2x+2y}{-2x + 2y}
\end{align*}

[jolly parrot]

OmnipotentEntity

[spiral pivot]

@azure ether ^ here is your example

[azure ether]

Thx

[azure ether]

For 4th line how y square became 2yy’

I just don’t understand the y’

[raw anchor]

Since you're differentiating with respect to x you gotta add the dy/dx aka y'

Hence it becomes 2yy'

[azure ether]

Ok thx

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[shrewd rune]

isnt thats impossible

i mean to find the domain

i gotta

e^x +1=0

e^x=-1

ln(-1)=x

wrong one?

i mean

doesnt have an answer?

[placid junco]

this just means that domain is R

[shrewd rune]

huh why

ohhh

cause e^x never will be -1

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[warm elbow]

\textbf{(b) (10 points) Solve the dual $P*$ using either the Two-Phase Method or the Big M Method} Maximize $z = -7w_1 -12w_2 - 10w_3$\\ Subject to \\ $-3w_1 + 2w_2 + 4w_3 \leq 1$ \\ $w_1 - 4w_2 -3w_3 \leq -3$\\ $-2w_1 - 8w_3 \leq 2$\\ $w_1, w_2, w_3 \geq 0$ \\ \\ \textit{Standard Form}: Introduce slack variables $w_4, w_5, w_6$\\ $-3w_1 + 2w_2 + 4w_3 + w_4 = 1$ \\ $-w_1 + 4w_2 +3w_3 -w_5 =3$\\ $-2w_1 - 8w_3 + w_6 = 2$\\ $z + 7w_1 + 12w_2 + 10w_3 = 0$\\ $w_1, w_2, w_3, w_4, w_5, w_6 \geq 0$ \\ \\ \textit{Initial Simplex Tableau} \\ $ \begin{array}{ccccccc|c} w_1 & w_2 & w_3 & w_4 & w_5 & w_6 & Z &\\ -3 & 2 & 4 & 1 & 0 & 0 & 0 & 1\\ -1 & 4 & 3 & 0 & -1 & 0 & 0 & 3 \\ -2 & -8 & 0 & 0 & 0 & 1 & 0 & 2\\ \hline 7 & 12 & 10 & 0 & 0 & 0 & 1 & 0\\ \end{array} $ \\ \\ Initial BFS: $w_1=0, w_2 = 0, w_3 = 0, w_4 = 1, w_5 = -3, w_6 = 2 \implies$ Infeasible. Introduce artificial variable $x_7$ \\ \\ \textit{Phase I} \\ Minimize $M = w_7$\\ Subject to \\ $-3w_1 + 2w_2 + 4w_3 + w_4 = 1$ \\ $-w_1 + 4w_2 +3w_3 -w_5 + w_7 =3$\\ $-2w_1 - 8w_3 + w_6 = 2$\\ $z + 7w_1 + 12w_2 + 10w_3 = 0$\\ $w_1, w_2, w_3, w_4, w_5, w_6, w_7 \geq 0$ $ \\ \begin{array}{cccccccc|c} w_1 & w_2 & w_3 & w_4 & w_5 & w_6 & w_7 & M &\\ -3 & 2 & 4 & 1 & 0 & 0 & 0& 0 & 1\\ -1 & 4 & 3 & 0 & -1 & 0 & 1 & 0 & 3 \\ -2 & -8 & 0 & 0 & 0 & 1 & 0 & 0 & 2\\ \hline 0 & 0 & 0 & 0 & 0 & 0 & -1 & 1 & 0\\ \end{array}$\\ \\ Eliminate $w_7$ from Tableau $ \\ \begin{array}{cccccccc|c} w_1 & w_2 & w_3 & w_4 & w_5 & w_6 & w_7 & M &\\ -3 & 2 & 4 & 1 & 0 & 0 & 0& 0 & 1\\ -1 & 4 & 3 & 0 & -1 & 0 & 1 & 0 & 3 \\ -2 & -8 & 0 & 0 & 0 & 1 & 0 & 0 & 2\\ \hline -1 & 4 & 3 & 0 & -1 & 0 & 0 & 1 & 3\\ \end{array}$ \pagebreak \\ Introduce $w_2$ into the solution, Pivot around $a_{12}$

[jolly parrot]

toast

[warm elbow]

I'm getting some kind of cycling from this and i'm not 100% sure why

[rough forge]

dont you have a negative 3

in your second constraint

but then multiplying by (-1) you get a greater equal inequality

yea that doesnt work it's supposed to be less equal because the slack variables are positive

[warm elbow]

i uh multipled by -1

oh hm

wait

[warm elbow]

i introduced a surplus variable then a artificial variable

bc its >=

so ig it should introduce slack variables w4, w_6 and surplus vaeriable w5

then i introduced artifical w7

[rough forge]

that's definitely it