help-19

[bronze canyon]

how do we find avg speed given time and distance

then, we need to find those times and distances

plug them in -> problem solved

[onyx jay]

avg speed is distance/time i think

sooo

theres 2 hours between 10:30 and 12:30

[bronze canyon]

correct

total distance/total time = average speed

[onyx jay]

so

155 km - 105km is 50km

no wait

it says he stopped in the middle

so the distances to b city and a town should be equal where he is

[bronze canyon]

yes

[onyx jay]

i think he travels 25 km within the 2 hours

[bronze canyon]

mhm

[onyx jay]

since if we subtract 25 from 155 and add it to 105 they become equal (130km)

25 + 130 is

wait

he started his journey before 10:30

he js saw the road sign at 10:30

that makes the problem more complicated

[bronze canyon]

okay town B is 155 km away, town A is 105 km away at 10:30

we need to use this to find the distance between them

as we found earlier, its 50 km

so, midway between them is 25 km

well

it actually depends if A and B are in the same direction

thats actually very important info not given in the question

[onyx jay]

we can assume they are i think

[bronze canyon]

so hes going to B via A

A is therefore 105 km to the LEFT, B is 155 to the RIGHT

so we actually need to add them

[onyx jay]

ohhh

so thats 260 km

[bronze canyon]

now halve it

[onyx jay]

thats 130km

[bronze canyon]

now find the total time

[onyx jay]

so he travelled 130km?

[bronze canyon]

mhm

[onyx jay]

in 2 hours

[bronze canyon]

now find the average speed

[onyx jay]

so the average speed is

65 km/h?

[bronze canyon]

yes

[onyx jay]

tysm for ur help!

[bronze canyon]

yw

[bronze canyon]

dont forget to close this channel if you dont have anything else

[odd edge]

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[odd edge]

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[spare hill]

Hi, I've measured the dimensions of a plastic bottle and I want to optimize a shape that I create with the most efficient dimensions for such a bottle.

I want to do a single variable surface area optimization for a combined shape consisting of a cylinder and a spherical cap. Here's my idea:

Firstly, I would sum the two volume formulas and sum the area formulas. Then I would have an equation like this

V = (pi)(r^2)(h_1) +(1/6)(pi)(h_2)(3a^2+(h_2)^2)
A = 2pirh_1 +2pir^2 + pi(a^2+(h_2)^2)

Here, I have measured my volume to be 500cm^3 and I'm trying to optimize my radius. I've also measured the total height of the water bottle to be 21cm. Since I have an h_1 variable at 15cm and an "h_2" variable at 6cm, I'm wondering if I can write them in terms of h, which is 21cm.

Can I say that I want the new dimensions of the water bottle to be h_1 = h(15/21) and h_2 = h(6/21) ? That way I can say I want to preserve the ratio of the two heights while optimizing the shapes and make sure there are only 2 variables, r and h, so that I don't have to do multivariable calculus.

Please let me know if this made any sense at all. I am open to discussion!! Thanks to anyone who is willing to read my tangent here

[bronze canyon]

so, you want the radii of the sphere and cylinder to be the same, im assuming

yes we can do this

well hold on

doing this strictly wont preserve the optimization itself

because i dont think the optimization works out to be this precise ratio all the time

now, if you need this precise ratio, then this works
but if you need to optimize on V,A, i dont think it does

[odd edge]

@spare hill Has your question been resolved?

[spare hill]

It’s fine, I don’t exactly want the exact optimization

I just want to avoid single variable calculus

[bronze canyon]

then this is fine

[spare hill]

Awesome!!

[bronze canyon]

as long as youre close, itll hold approximately

[spare hill]

Got it

Sorry what do u mean as long as im close?

[bronze canyon]

yeah

[odd edge]

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[spare hill]

Thanks!!

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[errant plank]

Could sombody check my solving method? what I did was use the area of a sector formula for radians but my program is saying my awnser is wrong

[errant plank]

(625pi)/18

was my final awnser

[tidal matrix]

forgive my horrible drawing, but it seems you found the area of a sector with radius 10 inches (this shape isn't exactly a sector)

[errant plank]

uhhhh

I assumed that this was the blade

of 10 inches

ohhhh

wait

[tidal matrix]

as did i! but the shape isn't actually a sector

oop i think you're there

[errant plank]

I thought I was onto somthing too

I guess not

why does it say this is the blade then?

[tidal matrix]

it defo is

but we don't want the full sector, we only want the strip of the blade

so we have to subtract the area on the "inside," i.e. from the edge of the blade to the center

[errant plank]

oh

so

thats

okay

I know how to solve it now

thank you!

.close

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[tidal matrix]

i already drew this but i think you got it

A-B

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[errant plank]

wait I did this

[errant plank]

and got

(sorry for ping I forgot it would do that

[tidal matrix]

thats ok :)

[errant plank]

my camera is being difficult >: (

okay got it to work

I got the area of both of the sectors

(11250pi)/36

for the bigger one

and

(8000pi)/36 for the smaller one

[errant plank]

and then subtracted the smaller one from the bigger one

but its still saying its wrong : C

[tidal matrix]

so you did r=10 for the inside one, and r=30 for the outside one?

or are you doing 2 * 5 and 2 * 15

[errant plank]

Sector A was r = 10+5+15

and sector B is r = 15+5

[tidal matrix]

ah ok, i misread the 10 part from last time instead of 20

[errant plank]

okay

uhhh

[tidal matrix]

but in any case, i think you're putting the wiper on the end of the 15in arm. the question says it's mounted "5 inches from the pivot point," which makes me think it's supposed to be this

sorry, i suck at verbalizing the setup of word problems, but in this case we'd have radii of 20 and 10

[errant plank]

huhh

for some reason I though this was the pivot point

okay

that makes sense actually

[tidal matrix]

to be perfectly honest, i think that could be the pivot point too

but in that case, your radii would be 5 and 15, right?

[errant plank]

hmmm?

[errant plank]

in this case?

[tidal matrix]

no, in the case that you mentioned

[errant plank]

this one?

wouldn't it be 30 and 20?

[tidal matrix]

unless im misunderstanding what you mean, wouldn't it be like this?

[errant plank]

this was my case

[tidal matrix]

ohh i see what you're saying now

[tidal matrix]

i think that line is just the body of the car? like it doesnt actually divide anything

[tidal matrix]

anyways, i think this makes the most sense based on how a back wiper looks irl, so let's try 20 and 10?

[errant plank]

okay I'll try that

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[dreamy totem]

how do i do this omfg

[dreamy totem]

do i have to get it to ref

[wooden cypress]

That's certainly one way of doing it

[dreamy totem]

how tho...

so what i did

R2-R1

but now i have column 1 and 2 of the third row as non-zeros

[wooden cypress]

As in you set R2 to R2 - R1?

You can do a similar thing to R3 as well to get a 0 in the first column

[dreamy totem]

ya

[dreamy totem]

ok so

i can have 2 options right

1. subtract R3 by 3R1

[dreamy totem]

or subtract R3 by 3R2

does it matter which one i do??

[dreamy totem]

WHERE IS THIS FROM

[azure temple]

What

I created it for u

[wooden cypress]

Chatgpt I expect

Don't give out answers though, it's not helpful

[azure temple]

Bruh

[wooden cypress]

Nah it doesn't matter which one you do, you'll still get the same answer in the end

Whichever one gives you nicer numbers really

[azure temple]

Let's see this question

[dreamy totem]

oh wat

[azure temple]

Is the answer correct

[dreamy totem]

oh i suppose R1 then?

the 0 in the augmented column looks nice

wait lowkey

before that

can you

[wooden cypress]

Yeah probably 1 then

[dreamy totem]

tell me if my understanding is correct

AB i assume

matrix vector product>?

[wooden cypress]

Aka the matrix product yes

You just apply the formula

[dreamy totem]

omfg

what formuka

ok i got like

[wooden cypress]

Or equivalently apply them as linear transformations to the identity matrix

[dreamy totem]

ok well

theyre linear transformation

so i AB = BA irhgt

right

[wooden cypress]

from wikipedia

It's not as bad as it looks

And no, $AB\ne BA$ in general

[clever fjord]

depression

[quasi falcon]

guys what you on

cant we just like: det(A) = 0?

[wooden cypress]

It's not that simple unfortunately

All that means is that there isn't a unique solution

[quasi falcon]

Rank check

[wooden cypress]

There could be infinitely many

isn't necessarily* a unique solution

not sure about that actually

Point is it's not sufficient

[quasi falcon]

If Rank(A) =/= Rank(A|b) then the system is no solutions

[dreamy totem]

oh

[bronze canyon]

wait

unique solution exists if det(A) \neq 0

[wooden cypress]

Yes but I thought he was suggesting we find a value of k where det A becomes 0

[quasi falcon]

det(A) = 0 indicates the coefficient matrix is singular
aka either inf solutions or zero

[wooden cypress]

I've not actually come across that before

[quasi falcon]

but theres a pretty easy test for it

[wooden cypress]

But I think that does work too

[wooden cypress]

If you think about the matrices as linear transformations then it makes sense

[bronze canyon]

i mean, since there arent inf solutions, this does actually work

[wooden cypress]

For example, if you think about flipping then rotating a 2d image, vs rotating it then flipping it, you'll get different images

[quasi falcon]

yes, thats precisely what i meant

IF theres any, k has to lead to det(A) = 0

Which becomes a pretty easy equality

Well, as long as you know to get det(A) for 3x3

[wooden cypress]

So yeah that would narrow down the possible solutions

Yes that would be a good way of doing it then as well

[quasi falcon]

Doing triangulation is crazy work ngl

[lone elbow]

this augmented system she is doing is essentially Ax = b where A is a 3x3 matrix, so if the matrix A is invertible you can do x = A^-1 . b, but we want no solutions so we need that A is not invertible, then for the k you found find for which of these you get an inconsistent system

[wooden cypress]

You say that like it's a lot of work lol

I agree your way is probably quicker

[dreamy totem]

ohhh

okay that kinda makes sense'

[dreamy totem]

okay what does ^-1 mean

[bronze canyon]

AA^{-1} = I = A^{-1}A, where I is the identity matrix

the ^-1 is just the inverse matrix

[wooden cypress]

I'd honestly suggest you practice multiplying matrices together, inverting them, and understanding how they work before you worry tooo much about questions like this

[lusty sonnet]

The 4th question

[bronze canyon]

#❓how-to-get-help

[wooden cypress]

go to #help-30

[lusty sonnet]

Damn

[bronze canyon]

it is important to note that not every A has an inverse
specifically, exactly when for some nonzero vector x, Ax=0
then there is no inverse

[dreamy totem]

😭

ive never learnt that before

omfg

[bronze canyon]

lets just focus on your problem, and we can talk about everything else later

we can do it by rref, or through his det(A)=0, whichever you want

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[rich orchid]

hello guys how to be good in maths

[sacred birch]

I believe this is not topic for help channel, but on what level are you and why do you need math in the first place?

[rich orchid]

i suffer a lot in maths and i dont know why in exam i ma not able to solve question

[rich orchid]

i am in 10th grade india and i suffer a lot in quadratic language based question

[sacred birch]

I'm not sure, what exactly you learn, but generally I personally try to understand definitions fully, then theorems and later to solve problems until it gets comfortable enough.
Can you show example of what you cannot answer or solve?

[atomic hornet]

Alright, do you have any examples for helpers to walk you through?

[rich orchid]

wait

[rich orchid]

these type language based

[sacred birch]

this kind of problems assumes, that you will make a mathematical model.
At constant speed you have S=tv, where S is full path, t is time spent and v is velocity. I'm not sure, what they mean by increasing 5km each hour, like 59 min -> 10 km and 61 -> 15 km or continuously.
And other problems want from you to do something similar

[rich orchid]

hmm right but how to make command in maths acc to how many hours i need to do study maths daily and did you know about JEE

[sacred birch]

I saw JEE, I know that it's usually not too bad in terms of ideas but forces one to do a lot of work without mistakes, that is technically difficult. Also Jee has some books on web that explains every piece of every problem

the idea of problem 13 and similar is to model number the following way. a * 100 + 10 * b + c. and then you usually can solve that using algebra and this model in the form of equation

[rich orchid]

tell me bro acc to you how many hours i need to study maths

[sacred birch]

But sometimes it's too much and one making problem harder and need actually to just write down the number into equation

[sacred birch]

Sorry, I don't know an answer on that question. I believe you remember or can track your progress and answer on that question better than anyone else. Or maybe people can help you in school with this question, as they have needed statistics

[rich orchid]

thx bro

well your country

[sacred birch]

I'm not Indian

you are welcome, wish you good time and fast progress in math

[rich orchid]

so where are you from

[sacred birch]

Sorry, I don't like this kind of questions. I prefer people to guess that themselves

[rich orchid]

egypt

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[dense stone]

What is the tan(x + y) when there's 90° triangle ABC and BD == DE ,DE == EC ,EC == AC and AC == 2?

[wet ridge]

sending the original question is easier for us ty:)

[urban condor]

tan(x+y) = (tanx + tany)/(1 - tanx tany)

[wooden python]

you have to remember something: you have to include the diagram in a geometry question.
otherwise, nobody can tell what "angle x" or "angle y" mean.

but yes, that.

[mystic saffron]

which angles are called x and y.

[wooden python]

she sent the diagram already, you can see.

[dense stone]

The angle ABC and ADE

[mystic saffron]

Oh

[wet ridge]

tan x = 2/6 and tan y equals 1/2 use tan(x+y) to solve

[wet ridge]

the guy above sent you the formula

[mystic saffron]

Guys can yall help me

[wet ridge]

bo

[wooden python]

!occupied

[odd edge]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[wet ridge]

send your question in help-26/33/4

[wooden python]

go and claim your own channel @mystic saffron

[dense stone]

I think I solved it

[wet ridge]

it will be sent here or anyother channel

[wet ridge]

gg keep going!

[mystic saffron]

How

[wet ridge]

um

do you see 3 channels above this?

[strange aspen]

.refer back to this

[wet ridge]

https://discord.com/channels/268882317391429632/903486479064522752

[graceful elk]

!help

[odd edge]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[wet ridge]

post it there

[odd edge]

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[azure temple]

Hope everyone can understand the problem

[bitter folio]

putnam 2024

! status

[odd edge]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[copper quarry]

wheop woop wooop wop

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[odd edge]

@azure temple Has your question been resolved?

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[split summit]

how can i even tell most of these options

n is too large for me to really visualise or test the question

[frank rivet]

im not the one to help here but when yall solve it ping me im curious

[odd edge]

@split summit Has your question been resolved?

[split summit]

oh nvm

[split summit]

i solved it

basically if you convert the radians to degrees

you'd notice all the angles have a LCF of 6

so you can use that information to figure out the answer is C

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[turbid warren]

$\frac{1}{\cos\big(\sin^{-1}(x)\big)} = \frac{1}{+\sqrt{\cos^2\big(\sin^{-1}(x)\big)}}$

[turbid warren]

alr guys what im stuck on here is now

forget the letx

[woeful briar]

use cos^2 + sin^2 = 1

[turbid warren]

yea

but the bery last part

how have they gone from this to this

[woeful briar]

sin and sin^-1 are inverses

[turbid warren]

yea

nah this makes no sense

[urban condor]

wdym

[turbid warren]

i dont see how the x^2 has come from

like i know that there is an invrse there

[urban condor]

sin^2 (sin^-1 x) = ((sin(sin^-1 x))^2 = x^2

[turbid warren]

omds

yh im acc dumb

ty

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[devout steeple]

correct me if im wrong but this can be solved on desmos i jjs dont know how to

[narrow crypt]

Hi

[mystic nova]

k vs k

[narrow crypt]

Use the t= p^2n-1

Then compare the powers

[narrow crypt]

k

[devout steeple]

huh

[narrow crypt]

Ok u get p^(6/35)

And u know that it is equal to t

U also know that t=p^(2n-1)

So by comparing the exponents

U should be able to find n

[devout steeple]

oh okay tahnks

.close

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[cyan cape]

On a windy day, the position of a balloon changes from 82 m [N] to 13 m [S] away from
a child in 15 minutes. What is the average velocity of the balloon.

this is what i did
∆d[S] = 95m[S]

Vavg[S] = ∆d[S]/∆t
Vavg[S] = 95m[S]/900s
Vavg[S] = 0.1m/s [S]

here i rounded to 1 sigfigs because you convert minutes to seconds, but does the time value stay at 2 sigfigs

[warped glacier]

,calc 15 * 60

[clever fjord]

Result:

900

[cyan cape]

how

[warped glacier]

since the original 15 minutes was also to 2 sf

hmm

okay yes there is an argument that 900 is to 1 sf, I see

[cyan cape]

yeah

idk if its 2sf or 1sf

[warped glacier]

yes cause 50 seconds is nearly 1 minute

it's not to the nearest 5 seconds, that'd be too precise

[cyan cape]

what

it says 15 minutes

[warped glacier]

the issue is when you convert 15 minutes to seconds

[cyan cape]

so 15 min = 600s

[warped glacier]

since your unit is m/s

,calc 15 * 60

[clever fjord]

Result:

900

[cyan cape]

yeah sorry 900

but that doesnt change the sf value

[warped glacier]

I agree, so you take the least precision out of all of the values

which is 1 sf

so yes, 0.1 m/s south is correct

[cyan cape]

okay

man

physics is so confusing

[warped glacier]

it's cause physics is an experimental science

it's not like maths where everything can be exact

[cyan cape]

hmm makes sense

ok thank you guys

.close

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[warped glacier]

so like if I tell you my ruler is 20 cm

it can't be 20.000000000000000000000 cm exactly

I can measure my ruler with another ruler, and say that it's 20.0 cm, to the nearest mm

[warped glacier]

no worries!

[warped glacier]

but I can't know anything more than that

[cyan cape]

shouldnt we treat 15 minutes as an exact value

so 15min = inf sf

since the question is exactly 15 minutes

[odd edge]

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[quasi shard]

for theorem 12.8 why is he saying that for it to be linearly depedent that it must be k+1 vectors in the span. Is that because the first k are just the k vectors from the set s that we use to span v_n. The only issue I have with this is we know that both i+j and i-j should lbe in the span and you can see that the you can use them to make a non trivial representation of 0 with only k vectors.

[fair prism]

The theorem is not saying you need k+1 vectors. It is saying any set of k+1 vectors will be linear dependent. You can still have just 2 vectors that are linearly dependent

[quasi shard]

ok I see

thx

.sovled

[fair prism]

.close

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[smoky nacelle]

umm, i need help factoring this and i actually dont know where to start (i have expanded ts and have some attempt to group, i.e (x^3+1) but it doesn't seem to work, i will send a pic if u need)

[austere jolt]

no common factors are here, so the first thing coming to mind is to expand this and factor with another method

[atomic hornet]

[stuck cloak]

did you try with factor theorum?

[smoky nacelle]

what is that?

[wet ridge]

wat

factor thm for multivariable polynomial?

[stuck cloak]

yeah

[wet ridge]

wait

[stuck cloak]

thats stupid

[wet ridge]

assume x=y I guess

[stuck cloak]

what if we put x=y

[wet ridge]

ahh

[stuck cloak]

same wavelenght

[signal yacht]

ngl !xy

!xy

[odd edge]

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.

[smoky nacelle]

that's what is stated to me

the whole hw sheet is jst abt factoring tbh

[austere jolt]

show what you've tried I suppose

[smoky nacelle]

ts is what i’ve written in my actual work

[austere jolt]

would have appreciated centering the work, but let's see

[smoky nacelle]

i couldn't find my attempts

sry:(

[wet ridge]

I dont think you can factor it?

[austere jolt]

well

I think it's factorable

but it's going to be rather screwed up

you can factor this as a quadratic in y

[wet ridge]

youll get (y-1)^2 if x= y

[austere jolt]

now, don't bother trying to do any fancy tricks, I suggest you just throw this into the quad formula

it should work

[smoky nacelle]

the ax^2+bx+c i suppose?

[austere jolt]

yes

but first rearrange your expanded form into a quadratic in y

[smoky nacelle]

apologies but i dont i actually understand that

[austere jolt]

try this first

[smoky nacelle]

yes, i'm quite confused that part

[austere jolt]

you have expanded the expression, yes?

[smoky nacelle]

yes

[austere jolt]

do you know what a quadratic in a variable means?

[smoky nacelle]

hmm, i dont think i know tbh

[austere jolt]

when there are multiple variables, having a quadratic in one variable means to treat every other variables as constants/coefficients, and rearranging them to the quadratic form with the variable given as the target

for example, a quadratic in y would have ay^2 + by + c, but a, b, and c here can have other variables like x

so group your y^2 terms, y terms and constant terms (anything without y)

[smoky nacelle]

u mean ts?

[austere jolt]

yup! factor the y out of the middle pair though

remember, we want a quadratic to look like one so we don't accidentally toss y into the quad formula

[smoky nacelle]

oke, so what's next:)?

[austere jolt]

toss it into the quad formula blender where it belongs

(your roots will be in terms of x, and that's exactly the factored form you're looking for)

[smoky nacelle]

is this right (i searched for the quad formula as i’ve never learned abt it, and i also wanna know if the whole thing inside of the root is the factored form or is it any terms here that is the common factor)

[austere jolt]

that is exactly right

[smoky nacelle]

so...is the whole thing inside of the root or have been rooted part is the factored form?

[austere jolt]

as for your second question, you know how you express the factored form of quadratics as a(x - b)(x - c)?

[smoky nacelle]

yep (...orr maybe)

[austere jolt]

here you will have $(y - \alpha)(y - \beta)$, where $\alpha, \beta$ are your two roots

[clever fjord]

Céline

[austere jolt]

so, no, this will give you just $\alpha$ and $\beta$, but you'll still have to put it into this form

[clever fjord]

Céline

[austere jolt]

also, since this is tedious, i will give you one hint

one of the two roots is a quadratic in x

also, you should simplify the discrim before working further, otherwise you're gonna be stuck seeing a quartic under the square root that will haunt you in your dreams

[smoky nacelle]

do i need to root it out? or do i keep the whole thing in the root as alpha and beta?

[austere jolt]

you can clear the sq. root here thankfully

but first, expand and simplify the discrim

[smoky nacelle]

like simplifying the whole thing in the root? or jst root the the whole thing inside? (yes, ts is new to me and i have difficulty in adapting to smth new as hell)

[austere jolt]

what do you get for the thing under the root after expanding and combining like terms?

(that thing under the root is the discrim btw)

[smoky nacelle]

x^4-4x^3+4x^2

[austere jolt]

right, factor the most obvious thing out first

[hollow fulcrum]

x = y - 1 appears to be a factor

then we can use the factor theorem i think

[smoky nacelle]

idek what the hell on earth is that

[hollow fulcrum]

okay, so

did u notice that putting x as y-1 gives 0?

[austere jolt]

I'll give way to the new helper then

[hollow fulcrum]

The factor theorem basically states that,
if there is a polynomial (let degree to be 2):
ax² + bx + c
if putting x = n makes the above expression 0, then (x-n) is a factor

[smoky nacelle]

so i get $\sqrt{x^2\right(x-2)^2}$

[clever fjord]

Mei
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[austere jolt]

uh?

[hollow fulcrum]

u forgot the \left)

[smoky nacelle]

jst ignore the texit

[hollow fulcrum]

okay

[austere jolt]

did you mean $\sqrt{x^2 (x-2)^2}$?

[clever fjord]

Céline

[smoky nacelle]

i've done smth in the past and now it's like ts

[smoky nacelle]

ye

[austere jolt]

then that's correct, and now you can just root this

you're very close to the end

that's a full solution isn't it

!nosols

[odd edge]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[hollow fulcrum]

oh

sorry

[smoky nacelle]

there is smth i'm quite not understand here, but i forgor to tell: if i toss ts into the quad formula and it has the +/- thingy, then should i do +, - or both? (i've root them all out)

[austere jolt]

both

that's how you get the two roots

[hollow fulcrum]

you're supposed to take them as 2 factors

[austere jolt]

one +, one -

[smoky nacelle]

👍

aight, i got the 2 roots now, 1 is 2x^2-2x+2 and the other is 2x+2 (i've divided both by 2 and make them x^2-x+1-y and x-y+1 respectively, but i doubt if it would have any use or if it's the right way)

[austere jolt]

did you make them = 0 or something?

[smoky nacelle]

ehhh, yes

[austere jolt]

the roots of y should not have y in there

[smoky nacelle]

aight

[austere jolt]

y = {root}

also, when putting those roots as factors, y should be positive, not negative

[smoky nacelle]

noted

[austere jolt]

btw, you can just write the answer on paper and take a pic of that

would be easier for all involved, I believe

[smoky nacelle]

so now i divided both the 2x^2-2x+2 and the 2x+2 by 2 then put it here? or i will keep the roots?

[austere jolt]

no, put it in the factors

that's what you want, isn't it? you weren't told to find the roots of y, you were told to factor

so put it in the factors

so... we good?

if you're done, then remember to close the channel, and see you around

[smoky nacelle]

aighty then

[austere jolt]

unless you wanna confirm the factors

[smoky nacelle]

so u said the parentheses here are factored form of the quad equations, right?

then i wanna know if after i found the roots, i substitute it for alpha and beta respectively

and i get (y-x^2+x-1)(y-x-1) is right

[austere jolt]

yes. personally i would write it as (y - (x + 1))(y - (x^2 - x + 1)), but same

[smoky nacelle]

then tysm

you saved my deadline=))

.close

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[gloomy mason]

stuck

[manic sleet]

multiply (a) by (\frac{1-i}{1-i})

[gloomy mason]

did this

[clever fjord]

PajamaMamaLlama

[manic sleet]

and what did you get?

[gloomy mason]

wait

[manic sleet]

[gloomy mason]

lemme click a picture

its a real mess

can u wait for like 5 minutes please ?

[manic sleet]

[atomic hornet]

[gloomy mason]

i feel like i did some stupid mistake in calculation

[austere jolt]

what happened down here

[gloomy mason]

ya i knew

i just knew

wait

lemme just redo

why am i still getting the same denominator

[austere jolt]

oh wait you're rationalizing the denom, my bad

still, there should be an i^2 somewhere

that's the whole point of rationalizing the denom

(hint: the multiplication in the denominator should be a familiar form to you. what does it look like? remember what you've learnt about quadratics)

[odd edge]

@gloomy mason Has your question been resolved?

[odd edge]

@gloomy mason Has your question been resolved?

[odd edge]

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[outer wadi]

I got lost in the second part

[outer wadi]

it says prove alpha is an automorphism of Z_n

but didn't alpha come from Aut(Z_n)?

oh

nvm

.close

[odd edge]

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Use .reopen if this was a mistake.

[outer wadi]

I wasnt reading carefully

.reopen

[odd edge]

✅

[outer wadi]

ok kinda of stuck again lol

im not understanding the operation preservation part of the proof

This proof follows this example

Stuck on this part of the proof

im not really seeing how they split up beta

[woeful briar]

this part proves T: Aut(Zn) --> Zn is a homomorphism

[outer wadi]

Im not understanding the steps

for example when they split up beta(1)

into 1 + 1 + ... 1

[woeful briar]

beta(1) is some natural number mod n

[outer wadi]

I was thinking it was referencing this

[woeful briar]

like 4 = 1 + 1 + 1 + 1

thats what they did

writing beta(1) as a sum of beta(1) 1's

[outer wadi]

oh ok so beta(1) is just some number mod n, and they are summing 1, beta(1) many times to get beta(1)

and since alpha is an isomorphism we can split up the inside 1s into beta(1) many alpha(1)'s

ok so everything till alpha(1)beta(1) makes sense

let me try and explain what I think the justfification behind each step is

[odd edge]

@outer wadi Has your question been resolved?

[outer wadi]

1. $T(\alpha\beta) = (\alpha\beta)(1)$ by definition of T $\newline$
2. $(\alpha\beta)(1) = \alpha(\beta(1))$ Notation change (function composition) $\newline$
3. $\alpha(\beta(1)) = \alpha(1 + 1 + \cdots + 1)$ Because $\beta(1)$ is a natural number mod n which can be decomposed into a sum of $\beta(1)$ many ones. $\newline$
4. $\alpha(1 + 1 + \cdots + 1) = \alpha(1) + \alpha(1) + \cdots + \alpha(1)$ By ismorphism of $\alpha$.$\newline$
5. $\alpha(1) + \alpha(1) + \cdots + \alpha(1) = \alpha(1)\beta(1)$ Honestly not sure, I want to say notation change but I can't back that claim up $\newline$
6. $\alpha(1)\beta(1) = T(\alpha)T(\beta)$ by definition of T $\newline$

oof let me fix that latex

[clever fjord]

Branshi

[outer wadi]

could someone clarify step 5

I know that alpha(1) is also a natural number mod n

and we have the sum of that number beta(1) times

so we have alpha(1)beta(1) but is that changing the operation from addition to multiplication

are these operations equivalent in this case?

but what if beta(1) was like pi or something, how would we know if alpha(1)pi is the sum of alpha(1) pi times?I think im just confusing myself now, is this just a notation change?

[woeful briar]

step 4 what is important is that alpha is a homomorphism

beta(1) is an integer

we define multiplication by a natural number in a ring
n * a = a + a +.... + a, n times

in this case because we are working in Zn which is pretty much integers, this and regular multiplicaiton coincide

[odd edge]

@outer wadi Has your question been resolved?

[odd edge]

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[ember pelican]

For q13 why is 5x^2/x^1/2 = 4/2-1/2=3/2?

[raven cloud]

any one avialble ?

i need help with straight line equation please

[jagged maple]

.occupied

[raven cloud]

i dont understand

[jagged maple]

!occupied

[odd edge]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[signal oar]

Where does it say that?

[raven cloud]

Sorry

[odd edge]

@ember pelican Has your question been resolved?

[tidal matrix]

this is just properties of exponents. $\frac{x^n}{x^m} = x^{n-m},$ and similarly $x^n \cdot x^m = x^{n+m}$

[clever fjord]

χασιβ ♥

[tidal matrix]

and then they just change 2 to 4/2 to subtract the fractions easier

if that confuses you, you can also do $$x^2 \cdot x^{-\frac 12} = x^{2 + (-\frac 12)} = x^{2 - \frac 12} = x^{\frac 42 - \frac 12}$$

[ember pelican]

Ahokay thanks

[clever fjord]

χασιβ ♥

[tidal matrix]

np! for your reference :)

(these are the product and quotient rules)

[odd edge]

@ember pelican Has your question been resolved?

[odd edge]

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[unreal briar]

I need to simplify the equation 3-(m+1)(2m+3)+((-1)^m)(m+3)^2+ ((-1)^(m+1))(m+4)^2 into 3-(m+2)(2m+5) my head can’t get around solving these massive equations

[mortal trench]

$3-(m+1)(2m+3)+((-1)^m)(m+3)^2+ ((-1)^m+1)(m+4)^2$

[clever fjord]

Mr. Gamer 🇵🇸

[unreal briar]

It’s ((-1)^(m+1)) not ((-1)^m+1)) my mistake sorry

[mortal trench]

$3-(m+1)(2m+3)+(-1)^m(m+3)^2+ (-1)^{m+1}(m+4)^2$

[clever fjord]

Mr. Gamer 🇵🇸

[unreal briar]

Yes

[mortal trench]

Looks good?

[unreal briar]

Yes

[unreal briar]

Exactly the same

[mortal trench]

What is m?

Any info about that

[unreal briar]

It’s a positive integer

[deft smelt]

You can do something to make it easier

[unreal briar]

How?

[deft smelt]

Dyk (-1)^m+1 what does it look like?

[unreal briar]

Uh

[deft smelt]

There’s a rule that says a^b x a^n = a^b+n

Right?

Now reverse this rule

[unreal briar]

Let me try

[deft smelt]

okay

[unreal briar]

Wait it’s (-1)^(m+1)

[deft smelt]

It’s gonna be (-1)^m x (-1)

[unreal briar]

Yes

[deft smelt]

That’s like all I can help with

Because im not sure if the way im solving it in my head is correct

@mortal trench

[mortal trench]

Split into cases after you factor out the -1

Either m even or m odd

[unreal briar]

Will expanding every make it easier

[mortal trench]

No

[unreal briar]

$3-(m+1)(2m+3)+(-1)^m((m+3)^2-(m+4)^2)$

[clever fjord]

Ajarco

[unreal briar]

It seems the same

[mortal trench]

Difference of squares

[unreal briar]

OH dam

what next? $3-(m+1)(2m+3)+(-1)^m(-2m-7)$

[clever fjord]

Ajarco

[unreal briar]

crud somethings wrong with the orginal equation

.close

[odd edge]

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[deft smelt]

Hey

[deft smelt]

How do I put an index on the redical on calculator

[faint knot]

what calculator are you using

[deft smelt]

Uh

[faint knot]

here you go

[deft smelt]

Where is that

.

[faint knot]

its on your calculator, suppose you look at it

[deft smelt]

I am..

[faint knot]

ngl youre gonna be cooked if you cant find a symbol on the thing in front of you

[deft smelt]

I am cooked

I have a quiz

I can barely work

[faint knot]

there arent very many radical symbols on this calculators

[deft smelt]

I found it

[faint knot]

very nice

[tacit stream]

To the left of ‘HEX’

[deft smelt]

It’s above x exponential thing

[faint knot]

dont spoil where it is

[deft smelt]

Sigh

Why cant I

[faint knot]

dont worry about it

it only took less than a minute

thats average

it just feels long because you didnt anticipate it to be a problem

thats alr

[deft smelt]

Do you think I’ll fail tomorrow

[faint knot]

if you look around this row

[faint knot]

you found the symbol

[deft smelt]

I did

[deft smelt]

I mean in general

I have redical and powers

[faint knot]

you found the symbol, you should be fine

[deft smelt]

I was working with ann

[faint knot]

as I was saying

[deft smelt]

Yes

[faint knot]

if you look around this row, youll see where the other radicals are

this is for square root

above that, cube root

similarly ^2 and ^3

[deft smelt]

Yes

I know these

[faint knot]

so this represents something more general

you have x^ a power

and also x being rooted to a particular power too

now because youc an put anything where the white square would be (it shows up as a blank for you to fill a number in when you press it)

you can actually use either symbol to do what you need

[deft smelt]

I know that

It’s not my first time using a calculator

I just usually

[faint knot]

so wdym youre gonna "fail?"

[deft smelt]

Work on index 2

[faint knot]

this is passing behavior bro

[deft smelt]

Not any other index

[faint knot]

its a button, you just found the button

[deft smelt]

My quiz isn’t about the calculator..

[deft smelt]

Yes.

[faint knot]

ok it doesnt sound like youre too interested in the actual button behavior

[deft smelt]

I’m not interested in math.in general

But that was all thank you forr helping me find it

[faint knot]

np and good luck

[deft smelt]

Thank you

.close

[odd edge]

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[deft smelt]

Hey again..

[deft smelt]

How to solve this

[wooden python]

aight lets set this aside for a bit and think about a simpler equation but which will have the same logic behind it

if i gave you this equation:

(x-19)(x-37) = 0

would you know what to do to solve it?

[deft smelt]

No..

They are both prime numbers

[wooden python]

do you know how to solve quadratic equations in factored form?

[deft smelt]

That I cant get anything out of

[wooden python]

19 and 37 are both prime but this is not relevant at all tbh.

[deft smelt]

Wdym not relevant

[deft smelt]

No I don’t

Well atleast I don’t understand what you mean

[wooden python]

...do you know what the word "relevant" means

what i meant is: it doesn't matter that the numbers in my equation are prime. it won't help you.

anyway ig give this a watch: https://youtu.be/qeByhTF8WEw?feature=shared

[deft smelt]

Yes

[jaunty hollow]

Am I free to join in?

[wooden python]

and also look up the words zero product property

bc thats really the key thing

cause this isnt about quadratics necessarily

but all equations that look like (something)(something) = 0

[mystic nova]

-# don't

[deft smelt]

I know what these are we call them second degree equations

[wooden python]

ok 2nd degree whatever.

[deft smelt]

Yes but

[wooden python]

degree matters a lot less than the idea of factoring.

[deft smelt]

What he’s solving aren’t prime numbers

[wooden python]

FORGET PRIMES

dear god

[deft smelt]

Okay

[wooden python]

i already said it doesnt matter that you got primes here. it doesnt. it doesnt!!

[deft smelt]

Okay..

[jaunty hollow]

@wooden python do you need me to help you?

[wooden python]

go watch the vid, go look up "zero product property" or read this thing i looked up for you just now: https://www.mathsisfun.com/algebra/zero-product-property.html

thats kinda all i can say atp

this is where you need to go, this is what you need to learn/revise

[deft smelt]

Okay im gonna close this till i watch and all so other people can ask

[deft smelt]

Okay thank youu

.close

[odd edge]

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[vivid inlet]

can someone help me factor nominator

[pastel steeple]

DEnominator or numerator?

[vivid inlet]

numerator

the upper

[pastel steeple]

well what do they have in common

[vivid inlet]

this is 3.1 limits

i only need to factor the upper

[pastel steeple]

great so what do they have in common

[vivid inlet]

beacuse the answer would be X to the power of 2 (x-2)

[vivid inlet]

wdym

[wooden python]

use ^ for exponents

[pastel steeple]

ok cool

[wooden python]

x^2

x^2(x-2) is the correct factorization

[vivid inlet]

thanks

[pastel steeple]

so uh whats the problem then

[vivid inlet]

but its to confusing

like im actully am trying to understand it for like an hour

i didnt answer it

[pastel steeple]

understanding how x^2 is factored out?

[vivid inlet]

no but like how we removed x^3 and then we still have x^2

[wooden python]

we didnt "remove" shit

(yet)

[pastel steeple]

x^3 = x * x^2

[vivid inlet]

https://tenor.com/view/i-understand-it-now-ok-i-understand-it-now-i-understand-it-gurm-bear-gif-4859748791707698675

thanks

calculus sucks ass

[odd edge]

@vivid inlet Has your question been resolved?

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[rapid junco]

7*(5-2)

[viscid shell]

21

[karmic arrow]

!noans

[odd edge]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[viscid shell]

mb

do u know abt priority of operations?

[rapid junco]

What's that

Order of operations?

[rapid junco]

I understand

You compute the expression inside the parantheses then multiply it by 7

Right?

[viscid shell]

yes basically

whats inside the parentheses

u always do first

[rapid junco]

Noted

[viscid shell]

then u can multiply and then u can add

[rapid junco]

Now the next question is use distributive property to expand the product

[viscid shell]

do u know abt this

using a*(b-c)=a*b-ca

[rapid junco]

I forgot it🙏

[rapid junco]

7(5-3) = 35-21 = 14

Is this what you are talking about?

<@&286206848099549185>

[vivid wyvern]

Do u know pedmas?

[rapid junco]

Yes I already solve that one

Now the examiner want me to use distributive property to expand the product

[vivid wyvern]

7(5-3)=7×5 -7×3

What is 7×3?

[rapid junco]

21

[vivid wyvern]

So isnt it supposed to be 35-21 and not 35-14

[rapid junco]

...

[rapid junco]

So, 14 is my answer?

[vivid wyvern]

U edite("...")

[vivid wyvern]

Yes.

[rapid junco]

Yes

[rapid junco]

Thanks tho

Should i post the similar problems that i couldn't solve?

[vivid wyvern]

Yes ,sure

[rapid junco]

(8-13)*(-3)

[vivid wyvern]

Also distributive?

[rapid junco]

First we have to evaluate the expression inside the paranthese then distributive

[vivid wyvern]

In that case ,do pedmas
What is 8-13?

[rapid junco]

-5

-5 * -3 = -15

[vivid wyvern]

(-5) * (-3)

[vivid wyvern]

Not 13

Do u know that (-)*(-)=(+)

[rapid junco]

New to me🙏

[vivid wyvern]

As in multiplying two negative s gives u positive

[rapid junco]

Do you mean we should treat subtraction as addition of a negative number?

[rapid junco]

Oh i understand

[vivid wyvern]

I didnt get ur phrasing

[vivid wyvern]

U did?

[rapid junco]

Yes

The answer= 15

[vivid wyvern]

Yes

Good

[rapid junco]

I have other problems ,if you are free🙏

[vivid wyvern]

Sure

[rapid junco]

I've been solving them since yesterday 😭😭

I cried and cried

[vivid wyvern]

issok drop ur questions

[rapid junco]

We have to distribute this first

[vivid wyvern]

Distribute?

That's the answer

[rapid junco]

Oh okay

[vivid wyvern]

Did u understand how u got 15?

[rapid junco]

I think that's all

[rapid junco]

Negative * negative= positive

[vivid wyvern]

Alright yea

[rapid junco]

(-)* (-3) = 15

Thanks ♥️

Bye

.close

[odd edge]

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[odd edge]

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[hot dome]

I got this problem from a video.
Find all primes of the form 2^(2^k) + 5.

[hot dome]

https://youtu.be/SyrJD1zZwmQ?si=t3DV_stNiK8mpYO_

I anticipate that the solution involves fermat primes but I don't know how to approach it.

[weary pelican]

It's simpler than that

what is 2^(2^k) + 5 mod [???]

[hot dome]

Mod 2?

It would be 5

[modest nacelle]

That won't help

[hot dome]

Bruh

How do I solve it 😭

I genuinly have no idea

[weary pelican]

you have to find some ??? such that 2^(2^k) + 5 mod [???] is 0 most of the time, so it's not prime because divisible by ???

mod 2 is definitely useless here, it's always 1

[hot dome]

mod 3?

If mod 2 is always 1, then mod 3 must be zero for most values of k

[modest nacelle]

Prove it.

[hot dome]

Or I can do it mod 4

x = 2^(2^k) + 5
= 2^(2^k) + 1 (mod 4)

It is true that if 2^n + 1 is prime, then n = 0 or 2^a

The converse is false, but I suspect that 2^(2^k) + 1 is prime for many values of k

[weary pelican]

if it didn't work mod 2, do you really think it will work mod 2 * something?

[weary pelican]

for an infinite amount of values of k, but it's false to say for "most values"

[odd edge]

@hot dome Has your question been resolved?

[hot dome]

I think that x mod 3 is always zero

Just figuring out how to prove it

Except for k = 0 of course

Just proved it with induction

Now what...

I think that's it

Thanks @weary pelican

.close

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[odd edge]

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[viscid shell]

i don't understand q2

[short terrace]

Do you know what equivalent norms are

[viscid shell]

ye

for all x in E

u can find c,d constants

[short terrace]

Well it's asking you to show that the norms corresponding to any two strictly positive functions are equivalent

[viscid shell]

yes

but i dont see how to majorize/minorize

[viscid shell]

i tried taking the supremum of Phi 1

and the lower limit of Phi1

but i tried it with some examples and it diddnt seem to work

i was also thinking of studying the function phi1/phi2

and find a lowerbound/upperbound

and i dont see what the hypothesis for them to be two strictly positive functions brings

[viscid shell]

oh wait i think it works

ye got it

.close

[odd edge]

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[dusky totem]

i cant understand permutation and combination

its confusing

[modest nacelle]

Wait, you meant to ask a problem?

[odd edge]

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[main cape]

I dont understand how to prove it using epsilon

[austere jolt]

lim as n tends to where though?

[main cape]

Infinity

[modest nacelle]

Which one?

[main cape]

The 1st one

Like when i tried it using epsilon i didnt get it

Like the freling as if u have proved something wasnt there like it felt off

[austere jolt]

for future helpers: it might be worth explaining or showing what you did