help-49

[barren wharf]

Does this question make sense?

[shell wigeon]

"the sum of value of n"?

[lyric charm]

the sum of all possible values of n, by the looks of it

of which there is only a finite amount here

[shell wigeon]

I guess finite is one word for it

[barren wharf]

Ok. Thanks.

I have another question.

This is written below the triangle.

[midnight plank]

@barren wharf Has your question been resolved?

[barren wharf]

Help is needed.

[stray hare]

This is out of my pay grade

I’m still in secondary school math

[violet rivet]

I think the question is not valid

by focusing on triangles ABE and ACD we get that <B+<E=<C+<D=180-alpha

[zenith snow]

If it could be said that AED is isosceles, there could be a way

[violet rivet]

by focusing on triangles ABD and ACE we get that <B+<D=<C+<E=180-(alpha+beta)

summing both equations gives us:

(<B+<E)+(<B+<D)=(<C+<D)+(<C+<E)
2(<B)+<E+<D=2(<C)+<D+<E
2(<B)=2(<C)
<B=<C

so triangle ABC should be isosceles but AB≠AC

[shell wigeon]

You're taking <E = <BEA and <E = <CEA (same with <D), of course you're getting a contradiction

[violet rivet]

ohh mb

[midnight plank]

@barren wharf Has your question been resolved?

[shell wigeon]

I feel like the expected solution uses trigonometry in some hidden way, like a theorem that's proven using trigonometry or something

[barren wharf]

Yes, that is what I want.

[barren wharf]

Not necessarily, of course.

<@&286206848099549185>

[midnight plank]

@barren wharf Has your question been resolved?

[barren wharf]

<@&286206848099549185>

[tropic ice]

What is the problem?

[barren wharf]

!!!!!

[tropic ice]

I think you have to use the theorem of cosine for express AD in function of the angle alpha plus beta with the triangle ABD and then express AD in function of the angle alpha for the triangle ACD. In this way you can obtain the value of beta, becouse you know the value of cos alpha and of alpha by AC and AH (where H Is the projection of A on BC)

[barren wharf]

I think this uses trigonometry.

[tropic ice]

Yes and you only need two formulas:

1. a square + b square - abcosalpha = c square
2. cos alpha = side adjacent to the angle on the hypotenuse

[tropic ice]

When you have obtain the angle alpha plus beta, you had to use another time the theorem of cosine to obtein AD from the triangle ABD and whit AH find HD. Now you use another time the theorem of cosine on ACE and find AE, with AE and AH you find EH, and with EH and HD you find the ED and also the result.

[lyric charm]

you missed a 2 & also use ^ for exponents. a^2 + b^2 - 2 ab cos(gamma) = c^2

[gentle hornet]

isnt it clearly stated to NOT use trigo?

[tropic ice]

yes sorry

[tropic ice]

ah oh my god I didnt see that

[gentle hornet]

so in this question, we can consider there are 3 unknowns, ED, AE, AD

using stewarts theorem in ABD and AEC we can get 2 equations

we just need a third

for this question, i think we can use the property that for convex polygons all angles are less than 180 degrees

so (sum of total interior angles - 3 x 60) / (number of remaining angles) < 180

this will give maximum value for n, and the minimum can be found by observation

[barren wharf]

I think we can frame the solution as using similar triangles only. Because these formulas don't involve compound angles and only simple ratios.

[tropic ice]

I think Is a good idea but i not so simple

I m trying to think on it

[gentle hornet]

ive been trying this for 15 minutes but havent gotten anywhere

maybe try constructing angle bisectors or using isogonal conjugates?

[pearl hull]

Where is the OP?

[bold peak]

Pins?

[tropic ice]

I think I understand how it's done, but I need to rework it rigorously: I imagined that by setting BC fixed and perimeter constant, as an angle theta varies, the position of the intersection point P between BC and the bisector varies. A generic formula for the bisector theorem can therefore be derived by setting BP equal to x and CP equal to x-15. The relationship should allow me to arrive at the result, but I need to verify.

This method should work, I don t know if there Is an easyer one, but now I can try to express it in a good form

[midnight plank]

@barren wharf Has your question been resolved?

[tropic ice]

The reasoning I had previously proposed had proven too complicated: nevertheless, I did some research and found that it is easily solvable using a simple theorem that was needed to know. The isogonal theorem states that if two lines AD and AE starting from the vertex A of a triangle ABC and intersecting the opposite side BC at D and E respectively are isogonal, then the following relationship holds: BE.CD/CE.BD=AB^2 /AC^2. In your case by this you find the relationship between BE and CE. Take now BE = x and CE = y. You have 392x=507y and x+y=15, this implies that x=BE= 8,46, so ED=9-8,64=0,36. This implies p=9,8 approximately.

[tropic ice]

My original idea was to find a direct relationship between ED and side AB (as it varies) with a fixed base BC and a constant perimeter, but this proved too cumbersome and time-consuming. This problem instead requires only knowledge of the formula of the isogonal theorem, which however I find difficult to demonstrate non-trigonometrically.

[midnight plank]

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[tardy bloom]

Calculators are not allowed

[torpid garnet]

Is crying allowed?

[tardy bloom]

Depends, if you do it silently then yes

[shell wigeon]

Are those binomial coefficients?

[tardy bloom]

Yes

[shell wigeon]

Time to write some factorials I guess

Why do they sometimes use a dot and sometimes a cross?

[tardy bloom]

Idk

[tardy bloom]

Only one 101 gets factored out, rest is a mess

[shell wigeon]

!show

[midnight plank]

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

[hard umbra]

do PIE

consider 404 balls, 101 coloured red, blue, yellow, green

how many ways to select 1 of each colour

[tardy bloom]

(101C1)^4

[hard umbra]

but count it with PIE

[tardy bloom]

What's that

[hard umbra]

principle of inclusion exclusion

[tardy bloom]

I've heard that but what does it mean exactly

[hard umbra]

well consider the sets R = not selecting a red ball, B = not selecting a blue ball, etc

you can say that selecting one of each colour is the complement of their union

and then use PIE to calculate the size of their union

[tardy bloom]

I don't get this

[hard umbra]

what's confusing you

[tardy bloom]

I don't get what you're trying to say to do

[hard umbra]

you'll need to evaluate S through double counting argument

[hard umbra]

it's clear the number of ways to do this is 101^4

but you must now count it alternatively using PIE to get another expression for the same number

[tardy bloom]

What this double counting?

[hard umbra]

counting twice

[tardy bloom]

So counting the same thing in two different ways?

[hard umbra]

yes

[midnight plank]

@tardy bloom Has your question been resolved?

[midnight plank]

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[twilit field]

I'm trying to show $Ha={x \in G | a \equiv x mod H}$

[grand pond]

wai

[twilit field]

$a \equiv b mod H \implies ab^{-1} \in H$

[grand pond]

wai

[scenic wyvern]

$a \equiv b \mod{H} \implies ab^{-1} \in H$

[grand pond]

Hanako

[scenic wyvern]

tidying up

[twilit field]

Heres what I tried

$x \in Ha$ so $x=ta; t\in H$

[grand pond]

wai

[twilit field]

a isn't in H , so xt^{-1}=a doesn't work

neither does $t= xa^{-1}$

[grand pond]

wai

[twilit field]

sure it tells me $t= xa^{-1} \in H$

[grand pond]

wai

[twilit field]

oooh

so by that a x^{-1} \in H too

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[twilit field]

$Ha={x \in G \mid a\equiv x\mod{H}}$: Let $x \in Ha$, there then exists $t \in H: x=ta$. We thus have $xa^{-1} =t \implies a^{-1}x \in H$.Therefore $Ha \subseteq {x \in G \mid a \equiv x \mod{H}}$.\ Let $t \in {x \in G \mid a \equiv x \mod{H}}.$ We then have $at^{-1} \in H$ so $ta \in H$. But $ta$ is the general form of an element of $Ha$. Thus ${x \in G \mid \equiv x \mod{H}} \subseteq Ha.$
Therefore $Ha={x \in G \mid a\equiv x \mod{H}}$

[grand pond]

wai

[twilit field]

Can I have this proof checked

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

.close

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[vocal flame]

Would you say that for the first question -2 2 and 4 are continuous from those sides or neither

And for number two is that correct

I’m pretty sure I don’t include 0 but not sure

Same for 1

[vocal flame]

I switched the -2 and 2

[vocal flame]

But how is it hard brackets if -3 wouldn’t be continuous

[lyric charm]

first off do not say "-3 is/is not continuous"

continuity is not a property of numbers but a property of functions

you're asking "how is the first interval closed on the left if the function is not continuous at -3?"

[vocal flame]

Ya

[lyric charm]

and the answer to that is: your function is continuous there. it is simply undefined to the left of -3.

so there is no left-hand limit to check and therefore no possible disagreement with function value or right-hand limit

[vocal flame]

Even though it would fail the second rule of continuity

Ok

And 0 is included?

[lyric charm]

your function is left-continuous at 0, that's why i included it in the interval (-1,0] which is just to the left of 0

[vocal flame]

Do I need then still or can I just go (-1,1)(1,3]

.close

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[vocal flame]

If I end up with LDNE on the inside

Like with problem 13

Does that make the whole thing LDNE

Or can I say the values in the other one where they don’t exist

Like could I say 13 is -4

[dreamy lichen]

what does LDNE mean?

[vocal flame]

Or just LDNE and I’m crazy

[grim vector]

the g(x) limit does not exist so its good

[dreamy lichen]

Left doesnt exist?

[vocal flame]

Limit does not exist

[dreamy lichen]

f(g(x)) can exist even if g(x) doesnt

(in the limit)

[vocal flame]

Look at 13 though

It goes to 1 on f(x) first

[dreamy lichen]

examine both the RHS limit and LHS limit

so from RHS, f(x) goes to 1, that's right

so g(f(x)) goes to?

whats the limit of g(f(x)) as f(x) -> 1

[vocal flame]

Does not exist

From the right it’s 1

[dreamy lichen]

this is the same thing as limit of g(u) as u -> u

[vocal flame]

And left is -2

[dreamy lichen]

okay sure, but we are doing only the limit from the right now

[vocal flame]

Ok

[dreamy lichen]

and then we do the left and check whether they are equal

[vocal flame]

Oh

That makes sense

So 1 then gx makes 2

And left would be -2 makes 2

So 2

[dreamy lichen]

yep

yep

[vocal flame]

Alr thanks bro

[dreamy lichen]

np

notice that even if g is crazy discontinuous, f(g(x)) can still exist in the limit

(the best example is constant f)

[vocal flame]

Can I do the same with 12

[dreamy lichen]

sure

[vocal flame]

But don’t I get different answers

Nope I’m crazy

[dreamy lichen]

in case you got different answer for left and right limit, then the limit wouldnt exist

[vocal flame]

Ok

Thanks

[dreamy lichen]

np

[vocal flame]

.close

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[vocal flame]

I’m back and I need more help

[vocal flame]

I’ve done all the problems but these 2

.rclockwise 270

[gaunt nimbus]

,race

,rccw

[drowsy ferry]

Whaat.

[gaunt nimbus]

Substitute x = a

[vocal flame]

Ya but is that really it

[drowsy ferry]

Oh I was looking at 17.

[gaunt nimbus]

Yes

[vocal flame]

So just get ca^2 -2a^2

[gaunt nimbus]

What

[vocal flame]

Wait

That’s just weird bruh😭

[gaunt nimbus]

No

[ a^2 + a \cdot a + a^2 = c]

[grand pond]

k

[gaunt nimbus]

Innit

[vocal flame]

3a^2

[drowsy ferry]

You need $\lim_{x \rightarrow a} f(x) = f(a)$.

[grand pond]

Hardy

[vocal flame]

Yes

[gaunt nimbus]

Ya we did that

[vocal flame]

For the other one tho it’s a graph but like

No explanation

.rccw

[vocal flame]

.rccw

,rccw

[vocal flame]

All I can think to do is plot points

[drowsy ferry]

That's the one I was saying whaat to.

[vocal flame]

Fr bro

[drowsy ferry]

f(x) = 4 isn't even a property, unless they meant something like "for all x, f(x) = 4".

[vocal flame]

Which can’t because it’s also 3

[drowsy ferry]

I suppose that they could mean "there exists x such that f(x) = 4"

[vocal flame]

And 1

[drowsy ferry]

But it would be ridiculous to mean that and write what they did.

[vocal flame]

A kid in my grade sent this hw packet out to everyone

But like he’s dumb asf

He just went

And said impossible

Which is what I think I’m basically going to do

[drowsy ferry]

If each line with x in it is supposed to start with "there exists x such that" then it's doable.

[vocal flame]

Ya

[drowsy ferry]

Or alternatively, "for each of the following propeties, sketch a graph with that property"

But it's very much a case of guessing what the question is supposed to be because the question as written has no valid answer.

[vocal flame]

Would this fit?

[drowsy ferry]

Yeah, that sort of thing would work for the "exists x" on each line.

[vocal flame]

Ya

Alr

Thanks

.close

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[vivid terrace]

How would you long divide this

[spare beacon]

The other way around

Long Divide x^4 by x+1

[vivid terrace]

Okay that makes sense idk why i did that

[spare beacon]

Also you can do (x^4 - 1)+1 and split the integrals

[vivid terrace]

I was thinking of that but when i searched it up i was told that long division is better

Can you please show me? I think i had a similar idea

[spare beacon]

Seems bs

[spare beacon]

Sure

[vivid terrace]

Would be grateful

[spare beacon]

Btw you'll do exactly the same thing in long division under a different name

[vivid terrace]

Okay thank you so much, i’ll try it!

Im new to uni math we didn’t have this in high school so im trying to get familiar with it

On that note can you please show me how to long divide it? Ik the basics i just forgot how now

[spare beacon]

Sure

[spare beacon]

@vivid terrace

[vivid terrace]

Okay thank you so much its all clear now!

[vivid terrace]

Okay so i did the process correctly, now i wanna write my answer, i take what i got on top and then what

Isnt that supposed to be the final answer

[midnight plank]

@vivid terrace Has your question been resolved?

[spare beacon]

You'd have to integrate it

[midnight plank]

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[vivid terrace]

Okay

[vivid terrace]

Wait

[austere blade]

you opened another channel btw, the channel got automatically closed after a while

[austere blade]

you can type .close, to close it if you want to

[vivid terrace]

I didnt know how to open it 😭

[hushed mauve]

you open a channel by simply sending a message in an available channel

do you still need this channel?

[vivid terrace]

Yes just abit

[hushed mauve]

ok.

[vivid terrace]

I wanted to make sure of something
We expand it then we integrate normally right

[vivid terrace]

In any other channel? Then it will open this specific one

[hushed mauve]

if the channel is occupied then nothing happens of course

if the channel is available, you will occupy it until you're done

[vivid terrace]

Okay thank you so much

Im almost done just abit of a clarification

[hushed mauve]

sure

i mean channel's yours anyway

[vivid terrace]

Is this how you integrate it

[shell wigeon]

There should be a "dx" at the end of these lines but yes that's how you would do this one

[vivid terrace]

Theyre asking to integrate x^4/x+1

[shell wigeon]

x^4/(x+1)?

[vivid terrace]

Yes

So i did long division

I got that part, but how do i move on forward from that step

[shell wigeon]

Show your result

[vivid terrace]

This

[shell wigeon]

Oh ok, looks correct, but I'd like to make sure you understand what the result is exactly, can you write it out here?

[vivid terrace]

From what i got its x^5/5 + c but i think its wrong how i did it

[shell wigeon]

Oh that's why you did the thing in red

Ok that's not how it works

You divided x^4 by (x+1) and got (x^3-x^2+x-1), right?

[vivid terrace]

Yes the one on top

There is a remainder of 1

[shell wigeon]

Right

That remainder of 1 is really a 1/(x+1)

[vivid terrace]

Okay

[shell wigeon]

What you showed by doing this division is that (x^3-x^2+x-1 + 1/(x+1) ) * (x+1) = x^4

So your integral becomes the integral of (x^3-x^2+x-1+1/(x+1)), not of [(x^3-x^2+x-1)(x+1) + 1]

[vivid terrace]

Give me a second to comprehend

Im gonna write it down too

Why do we multiply at the end by (x+1)?

[shell wigeon]

Where

[vivid terrace]

Oh okay wait

[shell wigeon]

It's x^4/(x+1)

It's your divisor

[vivid terrace]

Right

Sorry sometimes i blackout the basics

[shell wigeon]

Yes if there's no remainder

[vivid terrace]

Okay okay, so bcus of the remainder we added it to our expression and multiplied it by (x+1)

[shell wigeon]

No no the remainder R just adds R/(x+1) to your result

I should've written x^4/(x+1) = x^3-x^2+x-1 + 1/(x+1)

[vivid terrace]

And so they asking for integration of x^4/(1+x) so we move it to the other side and integrate the left

Okay so yes its alone

[shell wigeon]

x^3-x^2+x-1 is your direct result, 1/(x+1) is the additional term you can't get rid of because it comes from the remainder

[vivid terrace]

So we integrate each alone

[shell wigeon]

Well now you integrate x^3-x^2+x-1 + 1/(x+1) instead of x^4/(x+1)

Since it's a sum, you can integrate each term separately

[vivid terrace]

Let me try and come back give me a second, ty for going through it with me

[shell wigeon]

$\int x^3-x^2+x-1+\frac{1}{x+1} ,dx$

Bot's broken

[vivid terrace]

Like this?

[vivid terrace]

No worries no worries

[shell wigeon]

No there is no (x+1) factor in the integral

[vivid terrace]

Okay so where does it go

[shell wigeon]

It's only x^3 - x^2 + x - 1 + 1/(x+1)

[vivid terrace]

OMG YES THE OTHER SIDEE

ISTGG ughh

Okay okay i got you im so sorry

It makes things easier then

[shell wigeon]

Kinda

[vivid terrace]

Let me try for the 192810 time

[shell wigeon]

Is ok

[vivid terrace]

This way

[shell wigeon]

Yes that looks good

[vivid terrace]

OMG IM SO HAPPY

i guess math is all about trying out till you get through right answer

Or ig its always like that for me

[shell wigeon]

No that's math I'm pretty sure

[vivid terrace]

Ty @shell wigeon and @spare beacon so so much, its my first time encountering such an integral and both of you helped me a ton!

I finished my enquiry

.close

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[grave kernel]

here, i was trying to simplify the LHS
the numerator should be just S
and intuitively, i was trying out with x1 = 1 and x2 = 2
summation of (S - x(i)) should be S?
but then the whole thing becomes 1
and then n/n-1 is, less than 1?
i dont think so, i made a mistake so, yeah

[grave kernel]

oh right summation S - x(i) isnt S

should the LHS be 1/n?

[lyric charm]

the numerator should be just S
wdym

[grave kernel]

oh wait yeah mb

thats not gonna happen

[past jay]

Hmm do you know the formula for summing natural numbers from 1 to n?

[grave kernel]

yeah n(n+1)/2

can i apply that here? i dont get it

[past jay]

Can you apply it to S?

[grave kernel]

im not getting it exactly

[past jay]

Isn't S the sum of the natural numbers from 1 to n?

[grave kernel]

but its the summation of x(i) and x(i) here are positive reals right

[past jay]

Ah mb

[grave kernel]

mhm yeah

[nova pike]

I know how to prove it. First do you understand the problem?

[grave kernel]

kinda..?

im having trouble figuring out the LHS exactly

[nova pike]

Okay should I walk it through for you?

[grave kernel]

yes please

[nova pike]

So we have S to be the sum of n positive real numbers, right? Let us just say that n = 3 for now and let S = a+b+c

[grave kernel]

wait

so this is kinda like nesbitt thingy that one

[nova pike]

Yep exactly!

[nova pike]

Nesbitt’s inequality is the case for n=3. The inequality you have is more like a generalisation for any n

[grave kernel]

so i should follow the same procedure for this one as well?

[nova pike]

Correct. Just try and see if it mathematically works out

[grave kernel]

i used AM GM for Nesbitts tho, going that way, that would be way too complicated tho, right?

[nova pike]

Probably, idk the AM-GM route for this one though

Oh wait I might know it, but it shouldn’t be that bad though

It might get a bit complicated in writing it

[grave kernel]

wait lemme try it out

okay i got it

thank you so much tho

.close

[midnight plank]

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[midnight plank]

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[undone heath]

To solve integral like this is it considered formalized sufficiently (rigorous enough for calculus) not to the standard of analysis but for computational calculus

[dawn dagger]

hi

[undone heath]

Hi

[undone heath]

I am still curious if I can use that cheat without having to prove it 🫣🫣

I mean it’s really powerful but I am unsure if it’s legit

[jaunty ivy]

which cheat did u use and where

[dawn dagger]

wonder too

[undone heath]

I uploaded a method I discovered on internet to connect a couple of identities

But those are minor details, is the solution overall okay? Like if a ta grades it they should count it fully correct right?

[jaunty ivy]

yeah the TA should give u full points

wait let me check

,w integrate sqrt(tanx)

[grand pond]

Results provided by WolframAlpha

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

Requested by leftover5270

[undone heath]

My method usually has domain mismatch

[jaunty ivy]

i see u seem to have a different method than WA

[undone heath]

Yes it’s similar to the integration using Euler identities

[undone heath]

So there are some domain issues but the result of integral isn’t unique the method works mostly fine (mostly)

[jaunty ivy]

domains dont really matter for indefinite integrals as long as it is defined at at least some domain so it is riemann integrable

if the method works then it is fine

[undone heath]

Yes, so usually one doesn’t need to justify rigorous for why the method can work and as long as computation works out it should be fine? 🫣 for cal course

[jaunty ivy]

for finding an antiderivative, u just need to justify the steps u make like substitutions (usually domains as well if it is a definite integral)

i can imagine the prof also asking why it is integrable

[undone heath]

The method is similar to this one but more practical so not that general I think “https://arxiv.org/pdf/2505.03754”

So the substitution is a bit different than usual

https://arxiv.org/pdf/2505.03754

[midnight plank]

@undone heath Has your question been resolved?

[midnight plank]

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[vivid terrace]

How to do the 2nd part of the question, sketching?

[vivid terrace]

Graphing the results i mean

[worldly canopy]

like on a numberline

or on x-y axis

xy axes I think

[shell wigeon]

Graph the two lines and specify which area satisfies the inequality

[vivid terrace]

How do we know the shape of the graph

[worldly canopy]

theres only one line

[worldly canopy]

x=10/3 is the line u want to draw

its just a vertical line

where x is 10/3 on each point on that line

then u shade the side where the x values are greater than 10/3

[vivid terrace]

Then i shade the part on the right bcus it must be greater

[worldly canopy]

yes but

for some specifications

they want u to draw specific types of lines

for strict and inclusive inequalities

usually one is a dashed line and one is a solid line, tho theres no set way for which is which

[vivid terrace]

Now i have 1 value only

1 line

Vertical on x = 10/3

[worldly canopy]

on the off chance that this is edexcel gcse maths, strict inequalities have a dashed line

otherwise look at ur spec

[worldly canopy]

yup one line

sounds all good

[vivid terrace]

Its a direct entry exam for math for my uni

[worldly canopy]

hm not sure then

[vivid terrace]

Finished gcs already

[shell wigeon]

I don't see how that's useful but I don't know what the question expects from you

[vivid terrace]

Okay

[vivid terrace]

Hopefully its just as we said

Seems easy enough

[shell wigeon]

I'd be doing something like that:

[worldly canopy]

some exam boards would penalise if the line is the wrong type (dashed/solid)

[vivid terrace]

How do you come to this answer

[worldly canopy]

I dont think the original 2 lines are necessary

since the shaded region has been solved algebraically

[shell wigeon]

Then it's basically just stating an interval, no point in making it a graph

But who am I to question it

[vivid terrace]

They say graph the results in the question

[worldly canopy]

yeah pretty much, thats how I interpreted it

[vivid terrace]

Why is it results when its a result

I hope my point is going across

[worldly canopy]

maybe the "results" are all the points shaded

not sure

[vivid terrace]

Hmm

[shell wigeon]

I interpret it more like "the results of your work"

[vivid terrace]

Makes sense

Thanks everyone

[worldly canopy]

if these things come with a mark scheme or a written specification, I'd fs give it a read

[vivid terrace]

Wait lemme find it

Here

[worldly canopy]

Linear inequalities in one variable: Inequality symbols and relations, the meaning of
solution to an inequality, Solution of linear inequalities in one variable and the use of
inequality notation and interval notation to express the solution, graphical presentation of
solution on a number line. Solution of simple word problems involving inequalities.

It doesn't say anything about the type of line u need to use so I wouldn't worry

as for whether u need to draw the original 2 lines, I'm still unsure

I suppose better safe than sorry

[vivid terrace]

Okay yeah i got you

How to even get the 2 lines

[worldly canopy]

you'd draw the line like any other y=mx+c line, if its y>mx+c shade above the line, if its y<mx+c shade below

wait just draw the lines actually, they're not inequalities I just clocked

y = 1/3x + 2/3

y = 5/6x -1

[vivid terrace]

Im sorry i dont follow

How did we get these values

[worldly canopy]

from the original question

[shell wigeon]

Each side of the inequality

[worldly canopy]

yeah exactly

[shell wigeon]

The question is giving you two lines and asking "when is the first below the second"

[worldly canopy]

ur effectively shading where y=5/6x-1 "overtakes" the other

[worldly canopy]

exactlyyy

[vivid terrace]

You mean the left side snd the right side of the inequality given?

How do we make them into the points we use to draw the lines

[worldly canopy]

as for interval notation, you'd put $ \left(\frac{10}{3},\infty\right) $

$$ \left(\frac{10}{3},\infty\right) $$

[grand pond]

AnitaG

[vivid terrace]

Okay

[worldly canopy]

you just do "y="

to each side

[vivid terrace]

Oh just that

[worldly canopy]

and u get ur linear equation

[shell wigeon]

Have you worked with functions yet?

[vivid terrace]

Yes ive worked

On the basic level

[shell wigeon]

Well if you have

• f(x) = 1/3 x + 2/3
• g(x) = 5/6 x - 1
  These are two linear functions, their graphs are lines

The question is: for what x does f(x) < g(x)

[vivid terrace]

Okay yes i got the point now

[shell wigeon]

Here, f is the red line, g is the blue line

[vivid terrace]

The picture helped me a lot to understand

Thank you both nel and anitag

It turned out to be easier than i thought

[vivid terrace]

Also this saved me a ton

[tough inlet]

Hello.

[vivid terrace]

I think im done here, ty to you both

[vivid terrace]

Its better to use a free help channel if you have a question i’ll close this one

[tough inlet]

Oh ok.

[vivid terrace]

.close

[midnight plank]

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[midnight plank]

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[dreamy lichen]

Why is this true?

[dreamy lichen]

Why does g being irreducible guarantee existence of m

[lyric charm]

if such an $m$ existed it would be a root of $x^{k-1} + x^{k-2} + \dots + x + 1$, which divides $g(x)$.

fucking hell is texit really still down

whatever

my point should come across just as clearly

this obv assumes m ≠ 1 but you're not picking 1 for this

because 1 = 1^k

[dreamy lichen]

oh i get it now

thanks

i didnt realize that x^{k-1} + x^{k-2} + \dots + x + 1 would divide g

.close

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[tidal turret]

need help with 4)

oh, its not injective

take for example

f(1) = f(2) but 1 not equal 2

[shell wigeon]

You sure?

[tidal turret]

?

[shell wigeon]

What's f(1)?

[tidal turret]

n + 1 where n = 1

[shell wigeon]

Ok, and f(2)?

[tidal turret]

n/2 where n = 2

[shell wigeon]

1 + 1 = 2/2 ?

[tidal turret]

absolutely not

[shell wigeon]

So f(1) =/= f(2)

[tidal turret]

im stupid

[shell wigeon]

Is ok

Just take f(4)

[tidal turret]

f(4)=f(1)

[tidal turret]

ye, my immediate thought

[tidal turret]

this is the counterexample as to why it isn't injective

what about surjectivity dude?

[shell wigeon]

See if you can make all even numbers and all odd numbers with f(n)

[tidal turret]

dude I need help

[shell wigeon]

Take any even number p

Is there an n such that f(n) = p?

[tidal turret]

what is the definition of surj

[shell wigeon]

That for any element of the codomain, there is an element of the domain that maps to it

[tidal turret]

Let f : X -> Y be a function

Injectivity: f is said to be injective if ∀ x,y ∈ X . f(x) = f(y) ⟹ x = y
Surjectivity: f is said to be surjective if ∀ y ∈ Y . ∃ x ∈ X . y = f(x)
Bijectivity: f is said to be bijective if f is both injective and surjective

[shell wigeon]

In other words, if you input the whole domain into f, the output is the whole codomain

[tidal turret]

n = 2p

@shell wigeon

[shell wigeon]

Yes, no need to spam ping

Does that work for any p of the codomain?

[tidal turret]

ye

[shell wigeon]

So, for any p in N, there exists n in N such that f(n) = p

So you're done

[tidal turret]

no

we need a non even number

[shell wigeon]

I said even number here but I didn't realize it worked for odds too

2p is always even, so f(2p) = 2p/2 = p

[tidal turret]

https://tenor.com/view/dalek-explain-doctor-who-gif-5490200

[shell wigeon]

You're essentially mapping {2,4,6,8,...} into {1,2,3,4,...}

{2,4,6,8,...} is a subset of the domain

{1,2,3,4,...} is the codomain

You're covering the entire codomain, so f is surjective

[tidal turret]

p can be odd

[shell wigeon]

Yes but not 2p

[tidal turret]

wdym

[shell wigeon]

... I mean 2p is even

Not sure what's confusing you

The input is n = 2p

Since n is even, f gives you n/2 = 2p/2 = p, as output

That works for any p in N

[tidal turret]

.close

[midnight plank]

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[tidal turret]

let x be a natural number,
then f(2x) = 2x/x = x, and thus, we can get to any natural number in this manner

[midnight plank]

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[blissful trench]

<@&268886789983436800>

[rough geyser]

what is this

[blissful trench]

scam

[rough geyser]

is this propaganda

[midnight plank]

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[rough geyser]

oh right

[midnight plank]

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[fervent burrow]

I don't get the steps under that black line

So I found the x intercepts when y is 0

What do the 3 dots mean?

[shadow schooner]

it's a shorthand for therefore

[fervent burrow]

I see

[carmine sigil]

Much more rare, is three dots in an upside down triangle, which means because.

[distant elm]

Bonsoir j'ai besoin d'un mentor qui pourra m'aider en maths en privé

[carmine sigil]

but you'll see therefore reasonably often.

[shadow schooner]

!occupied

[midnight plank]

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).

[fervent burrow]

Kk

[fervent burrow]

So how is y=-12 when x was 0 too

[shadow schooner]

Just to structure things, up to and including the therefore they find the intercepts. Then after that the way I see it they restart from y = 2x^2+5x-12 and complete the square to get the turning point.

[shadow schooner]

well for this you can plug x=0 into 2x^2+5x-12

and see it's -12

2x0^2+5x0-12=0+0-12=-12

[fervent burrow]

Oh yeah got it

But what I'm not getting what they did do after I drew the black line

So we have 3 coordinates

Or intercepts?

[shadow schooner]

those are the 3 special point on the graph of y=2x^2+5x-12. At x=-4 and x=3/2, y=0

[fervent burrow]

Oh they completed the square

Ah I see

[shadow schooner]

they deduced this from the factored form (2x-3)(x+4)

[fervent burrow]

Thank you

.close

[midnight plank]

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[shadow schooner]

yeah it was kinda explained in the margin you just had to separate the arguments into chunks :kek:

[midnight plank]

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[lilac niche]

Hii!! can I get help with this problem? I’ve been stuck on it for 30 minutes now 💔

[prime hornet]

how're you feeling about LCMs now?

[lilac niche]

HIGHERRRRR

MY FAVVVV

[lilac niche]

feeling kinda pretty confident about it, but at the same time. I’m trying to understand more yk?

[prime hornet]

sure! do you want me to repeat what I said earlier about them?

[lilac niche]

yes

[prime hornet]

recall that the multiples of a number are all the numbers you get by repeatedly adding that number to itself. for instance, if the number is 5, then the multiples are 5, 10, 15, 20, 25, and so on

now if we have two numbers, a "common multiple" is any number that is a multiple of both numbers

here's an example:
the multiples of the number 2 are 2, 4, 6, 8, 10, 12 etc
the multiples of the number 3 are 3, 6, 9, 12, 15, 18, etc

the common multiples are the numbers that appear in both of these lists... which would be 6, 12, 18, etc in this case

the least common multiple is then the smallest common multiple between the two numbers. in our example, that would be 6

does this make sense?

we're going to need this to find a common denominator in just a moment

but I want to make sure that this concept is clear first

[lilac niche]

WAIT

I REMEMBER NOW LMAOAOAO

that is such a brain refresher

im basically looking for the common between the numbers that has the same number then crossing them out ?

[prime hornet]

yeah! to find the LCM, you can first find all the common multiples, and then cross out all of them but the smallest

[lilac niche]

HAAAAA I GET IT

[prime hornet]

if this is clear on you, then we can move onto the fractions

[lilac niche]

YES

I GET ITTT

[prime hornet]

so earlier, I mentioned that we can rewrite the number 4 as 4/1 to make our lives more convenient

oh, uh

[lilac niche]

?

[prime hornet]

the bot is down

[prime hornet]

okay, I'll avoid using the bot then

[prime hornet]

do you remember this?

[lilac niche]

4/1

[prime hornet]

yeah!

once we do that, the expression back in your problem is - 4/1 + (- 3/7), agreed?

[lilac niche]

agreed b

[prime hornet]

now, to add these fractions, we first need to find a common denominator

and that common denominator is going to be the LCM between the two denominators

let's do this step by step, just so we're clear on how to do this

first off: what are the two denominators in question?

[lilac niche]

denominators as in the bottom numbers right ?

[prime hornet]

(remember that if you have a fraction, the denominator is the number in the bottom)

yup!

I'm more or less asking you: what're the two bottom numbers in the expression - 4/1 + (- 3/7)?

[lilac niche]

oh. I was adding the fractions all together, I thought that’s what you meant.

[lilac niche]

3/7?

[prime hornet]

nono, we're taking this slow and steady

one step at a time

this is how we're going to add the fractions

[lilac niche]

OHHH

[prime hornet]

that's a single number, and not the denominators

[lilac niche]

wait now im confusing myself, am I finding what’s half

WAIT

OMG

[prime hornet]

let me draw, since @grand pond is down

@lilac niche we've got two fractions here; what are the two denominators/bottom numbers?

[lilac niche]

1 & 7?

[prime hornet]

exactly!

now our job is to find an LCM between these two numbers

[lilac niche]

there isn’t none ?

[prime hornet]

mhm, not quite

remember: list out the multiples of 1 and the multiples of 7 first

[lilac niche]

?

[lilac niche]

oh

[prime hornet]

can you list out the multiples of 1?

[lilac niche]

OH

[prime hornet]

those are the numbers you get by adding 1 to itself over and over again

[lilac niche]

1,2,3,4,5,6,7

[prime hornet]

exactly!

what about the multiples of 7?

[lilac niche]

7 14 21 28 35

[prime hornet]

great!

[lilac niche]

I forgot the rest idk

[prime hornet]

which number makes an appearance in both lists?

[lilac niche]

1!

[prime hornet]

that's okay! we won't need it

[prime hornet]

does 1 appear in the list

7 14 21 28 35?

the two lists are 1, 2, 3, 4, 5, 6, 7 and 7, 14, 21, 28, 35

which number is in both of them?

[lilac niche]

WAIT

7 SORRU

[prime hornet]

there you go

you got it!

so 7 is the LCM between 1 and 7, which shouldn't be too much of a surprise

with this in hand, we're ready to start rewriting our fractions

[lilac niche]

YAYAYAYAYAYAY

[prime hornet]

remember how we rewrote -4 as - 4/1 earlier, for convenience?

[lilac niche]

yes

[prime hornet]

well, we're going to do that again

we'd like to write - 4/1 as - (something)/7

let me draw it out

[lilac niche]

7?

[prime hornet]

it won't be 7, but let me first ask: do you know why we want to do this?

cause it might seem like a strange thing to want to rewrite this number again, right?

[lilac niche]

yes

[prime hornet]

but the reason is that we can only add fractions together when they have the same denominator

[prime hornet]

back here, we can see that the denominator of -3/7 is 7

so it's in our best interest to rewrite -4/1 to have a denominator of 7

because once we do that, we can simply add the numerators and call it a day

this is the pivotal reason why we wanted to find and LCM to begin with

[lilac niche]

OH

[prime hornet]

we need equal denominators to add

or else we're stuck

[lilac niche]

okay I understand, what do we do next ?

[lilac niche]

the ones I said?

[prime hornet]

the LCM is going to be the number we change our denominators to

in this case, the LCM of 1 and 7 is 7, so we want both of our fractions to have denominators of 7

but -3/7 already has a denominator of 7, so we don't have to do any work there

it's only -4/1 that needs to change

are you following so far?

[lilac niche]

I am actually

but I have a question

[prime hornet]

go ahead

[lilac niche]

both top numbers are changing because they are negatives?

[prime hornet]

no, the top number will only change as we need it to in order to change our fraction to have a bottom number of 7

[prime hornet]

that's what the ??? is here

we must find what the top number is going to change to

[lilac niche]

and then we are figuring out what the top number is being right?

[prime hornet]

the reason it changes at all is that we need to preserve equality: if we kept the top number as 4, then the equality would read -4/1 = -4/7

but that's clearly not an equation that holds

[prime hornet]

yes! we can get into that now, if you're ready

we're going to use a little trick from multiplication of fractions to do it

[lilac niche]

im ready :>

[prime hornet]

do you remember how to multiply fractions?

if I showed you this, could you tell me what it's equal to?

[lilac niche]

the butterfly method is how I was taught.

3 times 3

4 times 2

[prime hornet]

mm, that sounds a bit wrong

[lilac niche]

..

or not um

[prime hornet]

that's for when you have an equation to begin with and you want to simplify it, not really for multiplying fractions

but I'll tell you straight up: it's literally just "multiply the top numbers and multiply the bottom numbers"

[lilac niche]

Oh.

[lilac niche]

6/12?

[prime hornet]

exactly!

now, we're going to make use of this fact, as well as the fact that any number times 1 = itself

following so far?

if so, we can get right into it

[lilac niche]

?

that’s not the answer?

[prime hornet]

we're still in the process of doing this!

we're not done the problem yet

I'm taking this super duper slow, to illustrate the points clearly

[prime hornet]

are you ready to find ??? now?

[lilac niche]

yes

[prime hornet]

are you okay?

let me know if you're feeling overwhelmed

[lilac niche]

no I just really despise fractions.

im not a fan.

[prime hornet]

they're genuinely confusing at the start, to be fair

addition and subtraction of fractions in particular is smth students commonly find very confusing and difficult

you're doing great though, I promise

we're nearly done!!

[lilac niche]

I’m more of an multi & finding the slope person

[prime hornet]

fun c:

okay, let's find ??? now

[lilac niche]

okay okay

[prime hornet]

first things first, we're going to use the fact that any number times 1 = itself to get this

makes sense?

[lilac niche]

• 4/1 = - 4/1 x 1?

[prime hornet]

yup!

[lilac niche]

I know where the 4 came from

and I guess the 1?

[prime hornet]

we're conjuring up the 1 out of thin air

we're allowed to, because anything x 1 = itself

[lilac niche]

ohh

[prime hornet]

now, we use another special property of 1: any number divided by itself is equal to 1, so we can change 1 to 7/7

[lilac niche]

6/12 , 7/7

those two?

[lilac niche]

OH OKAY I SEE

[prime hornet]

nono, ignore 6/12

that was an unrelated thing that I wanted to use just to illustrate an idea

it has nothing to do with this problem

[prime hornet]

awesome!

now: can you multiply the two fractions?

[lilac niche]

28/7

[prime hornet]

amazing!

[lilac niche]

YAYYYYY

[prime hornet]

look at what we have now

we've successfully changed -4/1 to -28/7

so 28 = ???

we've done it!

[lilac niche]

HUH

WAIR

ITS NOT POSITIVE?

TUTOR 💔💔

[prime hornet]

the original number in your problem was -4, not 4 :p

[lilac niche]

oh

[prime hornet]

so we had to carry a negative sign everywhere

but with that out of the way, we can finally return to the original problem

[lilac niche]

so it’s -4/7 ?

or was it -7/28?

[prime hornet]

nono, -4/1 = -28/7

that's what we had just found out

[lilac niche]

i am beginning to get whelmed

[lilac niche]

OHHH

[prime hornet]

remember why we're doing this: we want to change -4/1 to have a denominator of 7 so that we can add our fractions together

[lilac niche]

so I enter the answer like 28 on top and 7 on the bottom?

[prime hornet]

this was the one and only reason we did all of this work to begin with

[prime hornet]

yes!!

[lilac niche]

what abt the negative..

[prime hornet]

the negative just sticks around

instead of 28/7, it's - 28/7

there's nothing to worry about

the negative can go on the top number, bottom number, or in front of the whole fraction

it makes no difference

anyhow, all of that hard work we did was just to get to this point :p

since we found that - 4/1 = - 28/7, we at last have this equality

[lilac niche]

ohh

[prime hornet]

and now we can just add the top numbers together

[lilac niche]

like that?

[prime hornet]

there you go!!

you've done it

[lilac niche]

WAIT

TUTOR WIZARD 💔💔💔

[prime hornet]

didja get it right?

[lilac niche]

do I add those fractions?

tutor wizard ? 💔

[prime hornet]

sorry, I had to answer my door

I'm back now

okay, to add fractions with equal denominators, you just need to add their top numbers and combine their bottom numbers

do you want me to give an unrelated example, for illustrative purposes?

[lilac niche]

I used desmos for it

[prime hornet]

ehh, you should learn how to do it yourself too

[prime hornet]

see this

[lilac niche]

no like I know how to do it.

im saying I used desmos to get the equations correct.

[prime hornet]

ah? I'm not sure I follow

[lilac niche]

bc that’s what I used back in high school all the time.

[prime hornet]

fair enough, but I know you're smart and capable enough to do this on your own

you don't need Desmos for this

[lilac niche]

I think you misunderstood what I explain, I did it on paper. I used desmos to see if I did it correctly.

[prime hornet]

I see

in that case, did the website accept your answer as correct?

[lilac niche]

yes

I got it!

[prime hornet]

congratulations

you did it!!

I hope I was able to clarify a few things along the way

[lilac niche]

you did you did !

[prime hornet]

if you practice these sorts of questions enough, you should be able to do them within a time frame like 10 seconds :p

[prime hornet]

nah, I only guided you along :p

you filled in all of the details

[lilac niche]

what happened if there’s 3 fractions 💔

[prime hornet]

you have two options:

choose any of two of them first, add them together, and reduce it to the case of only 2 fractions; or
get an LCM between all three of them, adjust the numerators all at once, and then add

[lilac niche]

oh.

[prime hornet]

can you first simplify everything in the brackets? :o

[lilac niche]

EVERYTHING ?

[prime hornet]

just the stuff in the brackets!

it looks scary, but it's really not

[lilac niche]

I don’t understand the word simplify

mean like division ?

or

[prime hornet]

well... not quite, because fractions are already in fully simplified form

[lilac niche]

ADD???

[prime hornet]

I'm more referring to the 4 - 2 and 5 + 3 expressions in your brackets

like, just do the addition/subtraction :p

[lilac niche]

I despise math deeply.

[prime hornet]

hey, we're gonna get through this together 🤍

take some deep breaths, calm your nerves, and let's face this together

if you're not feeling it rn, we can also stop

[lilac niche]

that’s how I feel rn

[prime hornet]

I love Spivak

his books are really nice

[prime hornet]

would you like to stop for now?

I don't want to force you to continue if you're feeling bad about this

prioritize yourself first

[lilac niche]

it’s honestly kinda hard, especially when it fractions I forgot to do by hand

[cinder yoke]

Do u guys know how i can get help like a channel like this for something i really need help in (sorry for being here js asking)

[lilac niche]

#help-46

[prime hornet]

hey there! welcome to the mathcord!

so this is a help channel, but it's currently being occupied by cosmos

you should open your own, like #help-11, to get help

[cinder yoke]

mmmmmmmmm ok ty

[prime hornet]

rule of thumb: if there's a name attached to the channel, it's occupied

if not, then you're free to ask in it

[lilac niche]

wizard tutor, can we stop here? ☹️

I feel an headache lingering

[prime hornet]

funny you ask, because I was about to apologize for having to take my own leave soon

[lilac niche]

omg can I call u the wizard tutor. ur like an magical wizard 🙂‍↕️

[prime hornet]

I'm off to have dinner in just a few minutes, so this is good timing

[prime hornet]

get some water, take a break, and enjoy your evening!!

I'm off now!

[lilac niche]

ohh okay ! BYE WIZARD TUTOR!

[midnight plank]

@lilac niche Has your question been resolved?

Channel closed

Closed by @lilac niche

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[chrome swift]

help plss

[chrome swift]

guys i only have 10 more minutes

im sure F is false and D is false

and C is false

@lavish venture MY GOAT

[midnight plank]

@chrome swift Has your question been resolved?

[mortal falcon]

do you still want help with this

[frozen lake]

@chrome swift pls close the channel if ur done lol

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[tidal turret]

need help with vi)

[frozen lake]

what help

like what you need

[tidal turret]

i need help

with vi)

[frozen lake]

obv

where are you getting stuck

[tidal turret]

in vi)

[frozen lake]

ok so

we have to check is f is injective or surjective right

or both

so

a>0: f(a) =2a

this is injective on Z+,ie positive integers

this is because 2a gives even numbers which are distinct for distinct a

now a<=0: f(a)=1-2a

this is injective on Z<=0

ie non positive integers

because it always gives +ve odd numbers which are distinct for distinct a

[scenic wyvern]

might wanna guide OP through instead

[frozen lake]

ig

OP?

u understand what I said so far?

@tidal turret

?

uhh

u alive?

[midnight plank]

@tidal turret Has your question been resolved?

[frozen lake]

@scenic wyvern can you close channel, OP is ignoring or is disconnected

[scenic wyvern]

i'm not supposed to do that unless OP says so (or is trolling)

[frozen lake]

ok then

[midnight plank]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[molten bay]

O(a)=3

O(b)=4

Then o(ab)?

[molten bay]

langrage method?

o(hk)=o(h)o(k)/intersection of h and k

12?

[west iron]

not possible to determine

[molten bay]

Why?

[west iron]

depending on the group it could have different values

[molten bay]

Group is abelian

[west iron]

that would be important

[molten bay]

So??

[hard umbra]

so you should've mentioned it in your question rather than making people help you solve a seemingly under specified problem

[molten bay]

Yeah my mistake

I will post it original