help-17

[vast shale]

Thanks so much omg

[vocal sleet]

@vast shale Has your question been resolved?

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[frank drift]

Prove that the angle bisecotor of an angle and the perpendicular bisector of the side opposite to that angle meet at the circumcircle of the triangle which we're talking about.

[glad owl]

So where are you stuck?

It's a very common result

Basically draw the angle bisector, let it interesting circumcircle at any point

Join the other 2 vertex of triangle

Now that new triangle is isosceles after doing basic angle chasing

Now it's easy right?

Then just draw the perpendicular

And prove that it's the perpendicular bisector of the base

Easyyy

[vocal sleet]

@frank drift Has your question been resolved?

[frank drift]

I contemplated the following:
DE = EB since AF is the perpendicular bisector of BD and the centre of the circle is A.
angle DAE = angle BAE
angle DCB = 2θ (let)
angle DCF = angle FCB = θ (since CF is the angle bisector of angle C)
angle DAB = 2 * 2θ = 4θ
angle DAE = 2θ
angle DBC = x (let)
angle ACB = x + 2θ - 90
angle FCA = θ - (x + 2θ - 90) = 90 - x - θ
angle CAF = 2θ + 2x
angle CFA = 180 - (2θ + 2x + 90 - x - θ) = 90 - θ - x = angle FCA
This implies AC = AF = radius = R
But the circle is the locus of every point at a fixed distance from a given point, which is here the central point A.
So F must lie on the circumcircle to satisfy that AF = AC = radius.
Done.
Is this proof of mine correct?

<@&286206848099549185>

[ripe parcel]

Riddle

[frank drift]

!help

[vocal sleet]

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

[glad owl]

L insufficient data

[ripe parcel]

ABC equilateral triangle

BE = AD

I’ve been thinking about it for a long time

Couldn’t solve it

So maybe it’s not possible with the given

[frank drift]

Bruh

go make your own channel

[ripe parcel]

Oh I’m supposed to make my own channel

My bad

[frank drift]

I think constructing another pair of intersecting lines might help (as shown in the figure)

[ripe parcel]

Don’t know how this server works lmao

Alr

[frank drift]

Can anyone see my proof and tell if it's correct?

<@&286206848099549185>

[mental egret]

Yes the proof looks correct

[glad owl]

Hm

[frank drift]

Okay, thanks.

[frank drift]

Don't worry, the question of theirs is solved.

.clsoe

.colseo

.cose

.lcose

.close

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[vocal sleet]

@lost basalt Has your question been resolved?

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[strange junco]

Who wanna grind with me on khan academy the pre-algebra section, i'm trying to pick up math.

[vocal sleet]

Please don't occupy multiple help channels.

[atomic jasper]

!1c

[vocal sleet]

Please stick to your channel.

[strange junco]

.close

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[atomic jasper]

#discussion

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[vocal sleet]

@limpid plover Has your question been resolved?

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

[limpid plover]

Do we need to use combinations?

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[weary harness]

Show that for this system there is an infinity of real solutions for x, y, z.

I honestly don't know how to solve this and the single Idea I had was to add them🤷‍♂️

If somebody can help me solve this problem I would really appreciate it!!!🙏

[vast shale]

My initial thought is think whether I can eliminate one variable

by doing one of the equations taking away another

if you end up with two equations that are the same that means there are infinitely many solutions

at least that’s how I did it when the equations are like ax+by+cz=0

[spiral turtle]

If we multiply through by xyz and add all three equations we get x^2 + y^2 + z^2 = xy + xz + yz + 7xyz

The fact that the left is a sphere is interesting, and it might be possible to do something with that?

I'm honestly not sure.

[weary harness]

yeah I also got that. the thing is that I have NO clue what to do with it

I stood on this problem for almost 2 hours today, so I would really appreciate a solution 😊

[bitter pilot]

😊 no

[spiral turtle]

Maybe a change of variables might be useful? a = 1/x, b = 1/y, c = 1/z

This would cut down a little bit in the degree of the expression.

ab - b^2 + bc = ac + 2b

And so on

[spiral turtle]

!noans

[vocal sleet]

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.

[spiral turtle]

Nothing personal, just against the server rules

[weary harness]

I'm learning really well when you show me how to actually solve it. I have problmes vizualizing hints that might or might not take me to the solution...

[spiral turtle]

Well, to be honest I don't see the solution either

I'm just trying stuff

[weary harness]

I also tried to replace y=z=k and x=t and I didn't get results...

[spiral turtle]

<@&268886789983436800>

Thx!

[weary harness]

huh?

[spiral turtle]

There was a spammer

ab - b^2 + bc = ac + 2b
ac + bc - c^2 = ab - 3c
-a^2 + ab + ac = bc - 6a

Each of these is a quadratic. From the third we can get

a^2 - (6 + b + c) + bc = 0

Or a = ((6+b+c) ± √((6+b+c)^2 - 4bc))/2

Which is gross, but I guess in principle can be substituted into the first and second equation.

@weary harness there is almost certainly a better way to do this

But I just don't see it

Remember though that 0 ≠ {a,b,c} due to the substitution

[weary harness]

yeah

I'll have to look at this tomorrow, because I'm way too tired rn.

should I close the thread?

[spiral turtle]

Up to you

It will time out automatically eventually if you aren't refreshing it

[weary harness]

.close

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[sudden delta]

i need help with understanding how integrating a^(x) dx gets you 1/ln(a) x a^(x), would be nice if sm1 could break it down for me

[oak magnet]

a^x = e^ln(a^x) = e^xln(a)

[sudden delta]

oh and then its just like integrating e^ax

ah i get it now tysm

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[slow olive]

I need help understanding Rotational Kinematics. I watched the lecture the professor provided us and am looking at kinematic equations but I am still unsure how I could solve these problems. This is one of the problems.

[slow olive]

This is Physics btw

possible kinematic equations

[edgy sapphire]

you'll also need v=wr

and come up with what both people have in common when on the same merry-go-round

[slow olive]

v=(omega)r right?

[vocal sleet]

@slow olive Has your question been resolved?

[heavy yoke]

both friends are on the same merry go round. what quantity might they both have in common based on that information

[slow olive]

Angular velocity since they are on the same merry go round?

[heavy yoke]

yes, the angular velocity is the same for the entire merry go round

[slow olive]

would the different between them be tangential velocity since they are on different points on the merry go round?

[heavy yoke]

yes

[slow olive]

for step one its just 8 seconds since they are on the same merry go round

[heavy yoke]

they do both have the same period, yes

[slow olive]

let me look at my notes to try to find the equation to find average linear speed

solved it. The friend was going at 4 m/s since they are twice as far from the center as you are

[vocal sleet]

@slow olive Has your question been resolved?

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[heavy palm]

number 38 is the one im doing

[heavy palm]

(2nd imnage

i should porb send seperate

look

https://cdn.discordapp.com/attachments/903463353417072641/1335420683567632538/image.png?ex=67a01aed&is=679ec96d&hm=f9a026cf39c4090de76510a0793d9382e28c8521816c43d41a6bca0263b710fa&

https://cdn.discordapp.com/attachments/903463353417072641/1335420683261317213/image.png?ex=67a01aed&is=679ec96d&hm=f98ba58abc5491ff31cae9cab75574fbe401af6fbda7b24abe7163173b7aebf9&

im unsure what my bounds would be

r,-r?

and how am i suppose to derive a formula for this

f(x) = ?

hint psl

[quiet echo]

Think about the 2D region you are integrating. Also note the definition of a torus.

[heavy yoke]

you have a formula for the surface area of a solid of revolution. think about how you might construct a torus as a solid of revolution

[heavy palm]

I know this is the area of a torus(4 × Pi^2 × R × r)

[quiet echo]

Okay. How would that relate here?

i.e., what graph (in 2D) should you be integrating over (via the definition of a torus)?

[heavy palm]

elipse?

[quiet echo]

Not necessarily.

[bitter pilot]

||how about a circle rotating around the y-axis||

[quiet echo]

Note the shape the outer ring has, if you take cross sections perpendicular to the shape.

[quiet echo]

I mean, I was going to lead them to that.

[bitter pilot]

mb

[quiet echo]

But we can still utilize it anyways.

It’s fine, it’s only one piece of the puzzle.

Try to draw (or graph in your mind) a 2D circle as described. Where would it’s center be, as well as it’s radius?

[heavy palm]

the origin

[quiet echo]

Hmm.

Well, it wouldn’t be the optimal place, but that works.

So what line would we be revolving it around? And what is the radius of the circle described?

[heavy palm]

y axis

[quiet echo]

Not necessarily. That’s if the circle is located somewhere else.

[heavy palm]

idk the radius of the circle being described but half of pie?

[heavy palm]

ik here its R+r

adn R-r

but thats all i know

[quiet echo]

The radius is arbitrary here.

[heavy palm]

wouldnt it be always revolving around the origin from where the circle is placed

placement of origin

i guess

.close

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[glad epoch]

is this correct?

[flat whale]

,w expand 3(x-3)-4(x-3)

[twin meteor]

Results provided by WolframAlpha

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

Requested by rie.mann

[steady plover]

you have missed a x in the 3rd step

[glad epoch]

wait where

3x(x-3)-4(x-3)?

[steady plover]

wait one second does the question say 6x^2 -26+24 or 6x^2-26x+24

[glad epoch]

first

[steady plover]

and what do you have to do with it?

[glad epoch]

factorise

[steady plover]

Won’t it just be 6x^2 - 2 then

[glad epoch]

huh

[steady plover]

can you explain what you have done from the 3rd step to the 4th step

[glad epoch]

which one is 3rd which one is 4th

[steady plover]

because your final answer is the factorisation of 6x^2 -26x+24

[glad epoch]

what

oh yeah

[steady plover]

That’s why I asked if it was 26x or just 26

[glad epoch]

im not too sure how to explain but i dissembled -13 because there are no common factors for it

[glad epoch]

im confused as well

thats what the teacher gave me

[glad epoch]

resulting in -9 + -4

3 is a factor of 9

3 is also a factor of 12

[steady plover]

This is fine but -13 has to be a coefficient of x

In your question the coefficient of x is 0

[glad epoch]

oh

how do i solve it then

[steady plover]

This basically

[glad epoch]

is it solvable

[steady plover]

if you can confirm this

[glad epoch]

its 26 im pretty sure

wait

let me check

[steady plover]

if it’s just 26 then you can just write 6x^2-2 and factor out a 2

[glad epoch]

wait what

how come

[steady plover]

if it was 26x what you have done is correct

[glad epoch]

where did the 2 come from

[steady plover]

Alternatively you can try expanding what you got and see if it matches your question

[steady plover]

-26+24

[glad epoch]

oh yeah

[glad epoch]

this is what the teacher showed though

and i got really confused

[steady plover]

I mean you can factorise 3x^2-1 further into two factors but I guess the teacher wants you to do it using 26x

<@&286206848099549185> can somebody else help with this dilemma

[glad epoch]

so it could be a mistake on the teacher’s part

[steady plover]

Yeah it mostly is should be 26x

you can confirm with them however

[steady plover]

then your answer is correct

[glad epoch]

i’ll ask my teacher when i see them

[glad epoch]

wait where did the -1 come from

[quiet echo]

Hm?

What is the problem here?

[quiet echo]

Huh?

[quiet echo]

He just factored out a 2 from the equation.

If you’re going to factor non-monics, use this neat trick; for ax^2+bx+c, when a is not 0 or 1, we can write it like x^2+bx+ac, factor by grouping that way, then bring the coefficient back.

[glad epoch]

that’s what i did

i think

[quiet echo]

Just don’t forget to write an x next time.

[glad epoch]

i did NOT forget to write an x

the teacher handed me

6x^2 - 26 + 24

[quiet echo]

They must’ve forgotten.

[glad epoch]

probably

ill ask 🕊️

.close

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[hot mauve]

this shouldnt be this hard but i cant find proper way of finding the median for a histogram

ive seen maybe the Median = L + (N/2 - F)/f * w, or the one to calculate all frequency by (average x interval) * (y, number of paitent) / patient count, total y)

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@hot mauve Has your question been resolved?

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[coarse marten]

why did this guy end his integration by parts after one segment. (the notations under D and I)

[sharp lynx]

because the new integral can be solved via power rule

no more partsing

[coarse marten]

how do you idnetify that?

[coarse marten]

what does that mean

[heavy yoke]

it's a product of two powers of x

[leaden ingot]

the new integral is easy to evaluate

[coarse marten]

oh

i see it

.close

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[vocal sleet]

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[warm rapids]

Find the average value of the function f(x) in the interval [1, 3] and determine the coordinates of the point at which it occurs.

[finite mist]

Well

[warm rapids]

So, I integrated it.

[finite mist]

Yep

[warm rapids]

And I got:

[finite mist]

Recall that

[warm rapids]

Yes.

[warm rapids]

So, let us do it together to verify.

[finite mist]

So that the avergae value

It askes

determine the coordinates of the point at which it occurs.

[finite mist]

So its literally f(x) = average value

Find x

[warm rapids]

yes, but I cannot tell if I did it correctly or not.

I did find X

but how can I know if I'm right?

[finite mist]

What do u mean

[finite mist]

it just x^2 - 2 = 19/3

Solve for x

[warm rapids]

This is what I got.

[finite mist]

U missing +- case

[warm rapids]

well, yeah.

sorry about that.

[finite mist]

Then

Since x belong to [1, 3]

U get rid of the minus case

[finite mist]

You still need to write down the +- case, then argue this. So u get full score

Else, as a teacher, i would deduct half the score for that question lol

[warm rapids]

Okay

okay, okay

hm, do I need to do something else?

[finite mist]

Not really

[warm rapids]

maybe replace the X with the X I found?

so I can determine the coordinates

I have X

[finite mist]

well u need to make the conclusion

[warm rapids]

but I need Y

[finite mist]

(x,y)

[finite mist]

y is the average value itself

[warm rapids]

Oh.

Thank you.

[finite mist]

lol

[warm rapids]

Buddy

not everyone is as sharp as you xD

sorry

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[waxen star]

can someone explain why the second equation = f(2i)f(-2i)

[waxen star]

this part

how did it suddenly become f(2i)f(-2i)

[dull depot]

those are the roots

[waxen star]

2i and -2i are the roots of (p^2+4)(q^2+4)(r^2+4)

not f(x)

???

wut

[quick vigil]

f(x) = a(x-x1)(x-x2)....
Where x1 ,x2 etc are roots

[woeful igloo]

p, q, r are the roots of f(x). Or in other words, f(x) = (x-p)(x-q)(x-r)

so you just replace x with 2i, and x with -2i, youd get the product

[waxen star]

yes but where did 2i and -2i come from

like how did (p^2+4)etc etc become f(2i)f(-2i)

[woeful igloo]

a^2 - b^2, where a = p, b = 2i

factorize (p^2+4) as a difference of squares

[waxen star]

yea

[woeful igloo]

sqrt(-4) = +/- 2i

[waxen star]

i got that but-

im confused here

how did this happen

[woeful igloo]

you just factorized p^2+4 as (p+2i)(p-2i), and so on

[waxen star]

yea

[woeful igloo]

now pick the terms (p+2i)(q+2i)(r+2i)

[woeful igloo]

and compare with this

[waxen star]

oh okay

[woeful igloo]

same with the leftover

[waxen star]

okay

[woeful igloo]

one of these is same as -f(2i), other is same as -f(-2i)

[waxen star]

ok

.close

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[jagged cargo]

@cobalt crypt this is what i got so far

i tried somehow using the fact that Tv = 0, but then i can't make a relationship between v and T'v

[cobalt crypt]

you know that ker(T') (+) Y = X, and X is finite dimensional, so ker(T') is as well

so you can pick a basis of ker(T')

[jagged cargo]

yeah, but v is in ker(T), it's not like i can construct v by "extending the basis of ker(T') to ker(T)"

[cobalt crypt]

you cannot, but ker(T') is in direct sum with ker(S), so S will take a basis of ker(T') to linearly independent vectors in im(S)

let say the basis of ker(T') is v_1, ..., v_k, and so Sv = a_1 Sv_1 + ... + a_k Sv_k

then you can play the same trick again

define x = a_1 v_1 + ... + a_k v_k, w = v - x

wait does this work

[jagged cargo]

how?

[cobalt crypt]

okay yeah it does work

[cobalt crypt]

because S restricted to ker(T') is injective

[jagged cargo]

ah i see

[cobalt crypt]

consider S' : ker(S) (+) ker(T') -> W, then im(S') = { S'(u + v) = S'v | u \in ker(S), v \in ker(T') } = { S'v | v \in \ker(T') }

[jagged cargo]

isn't that effectively the same as S': ker(T') -> W?

[cobalt crypt]

well their image is the same, but their domains are not

[jagged cargo]

i see

[cobalt crypt]

but thats the whole point i guess, the ker(S) is not contributing to the image of S' at all, so we can rewrite Sv = linear combo of Sv_i

where you've chosen the v_i from ker(T') only

[jagged cargo]

can you elaborate on this? S|_ker(T') is injective, sure, but v is in ker(T), not in ker(T')

[cobalt crypt]

yeah

but we can test which part of v is in ker(S) using S

you kill part of v using S, and the part that isn't killed comes from ker(T'), over which S is injective

so you can invert Sv back to get the part of it which lives in ker(T')

[jagged cargo]

ohhhhh waiittttttt

previously i defined w = v - linear combo of e's, so v = w + linear combo of e's

if we apply S on it, then Sv = S(w + linear combo of e's) = S(linear combo of e's)

but like, the combo is from X, not ker(T')

[cobalt crypt]

[jagged cargo]

well the part i trimmed down from v using S is w, which, yes, comes from ker(S)

but what's left is from X, not ker(T')

X = ker(T') (+) Y btw, just to make sure we're on the same boat

[cobalt crypt]

yeah but i have no idea what you're talking about rn

[jagged cargo]

er, can you reiterate from this part again?

[cobalt crypt]

im S|_(ker(S) (+) ker(T')) = im S|_ker(T'), and S|_ker(T') is injective

pick a basis of ker(T'), v_1, ..., v_k

then for v \in ker(S) (+) ker(T'), Sv \in im S|_ker(T'), and therefore can be written as a linear combo of the image of the basis Sv_1, ..., Sv_k

[jagged cargo]

but we are proving ker(T) = ker(S) + ker(T'), why are you already assuming v in ker(S) + ker(T')

rn i'm doing ker(T) subset ker(S) + ker(T')

[cobalt crypt]

oh yeah thats a typo

take v \in ker(T)

hmm

okay maybe we have to do more then

[jagged cargo]

something cropped up, imma go rq

[cobalt crypt]

aight

[jagged cargo]

i need some time to think about this as well

[cobalt crypt]

glhf

[jagged cargo]

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[cobalt crypt]

i have a way out but its just getting more complicated

[jagged cargo]

[cobalt crypt]

i hate lack of infinite dimensional tools

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[kind light]

im noy rly sure how to translate the tangent requirement

heres a better image

[woeful igloo]

radical axis

[kind light]

i dont know any properties of it

[woeful igloo]

a point on radical axis has same power wrt both circles

[kind light]

all i know its a line so power one one is power on both

[woeful igloo]

yea, and what is power of a point outside the circle?

[kind light]

which would be line ap

[woeful igloo]

Like its geometric interpretation

[kind light]

what

oh ok it just clicked

[steep crater]

Oh lmao didnt see that

[kind light]

basically line AP is the radical axis, so power of any point on the radical axis say of D (call the intersection of AP and BC as D) is equal on both circles, so BD^2=CD^2 or BD=CD, which bisects BC

like this?

[steep crater]

Yea

[kind light]

ok

ty

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[outer gazelle]

simplify

[outer gazelle]

help

<@&286206848099549185>

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[vital sable]

which set is this

[vital sable]

is that like natural numbers + {0}?

[cyan talon]

most likely yes

[vital sable]

they could have just written whole numbers

bruh

thank you

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[vital sable]

.reopen

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✅

[vital sable]

wait

do you know how I would type that in latex

[cyan talon]

$\mathbb{N}_0$

[twin meteor]

aPlatypus

[vital sable]

thank you

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[silver dew]

could someone explain this to me??

[visual marsh]

what do u need to do?

[silver dew]

so it is a revision sheet for my exam tomorrow and this is the question i dont understand

it says 'Work out'

[visual marsh]

simply do what they asked?

1+144

by

2x9 - 13

[silver dew]

no i didnt understand the bottom like what to do in the division part im kinda dumb gng💀

[visual marsh]

simplify the denominator

and numerator

then divide 😭

[silver dew]

thanks

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[granite prairie]

how is this solved?

[finite mist]

Which section?

a, b, c, or d

@granite prairie

[granite prairie]

all

of the

m

but

i feel like if a is solved

i can solve the rest

AH WAIT

hold on

nvermind

wait a minute yea

no nevermind

again

i dont know what to do here

they arent considered similair shapes correct?

so i cant do small/large = small/large

[finite mist]

They are similar shape though

They both right triangle

Oh

[granite prairie]

what

sh

ah

i see

[finite mist]

Called the top point T for example

[granite prairie]

ahh

that makes alot more sense

so 3/7 = x/8

[finite mist]

Nope

[granite prairie]

guh

how come

[finite mist]

CX / OY

= TC / TO

[granite prairie]

whats cx and oy

what are

all the variabls

[finite mist]

We already know CX, OY, and CO

TC is the only variable

So solve for TC

And u get TO

which is the height of the original cone itself

[granite prairie]

ah

i see

i keep getting

13.25 as the total height

but its 18.7

3/7 = x (TC) / 8+x

[finite mist]

huh?

The height should be 14 though

[granite prairie]

that

is incorrect

[finite mist]

Thats literally your triangle

[granite prairie]

well

thats not what the mark scheme says

let me chec

maybe

its wrong

[granite prairie]

chat gpt says you're right as well

[finite mist]

I feel so humiliated that u trust chatgpt more than me

😭

[granite prairie]

no no no no

that isnt it

i just lie to have

more then one source

your answer made the most sense but its essentialy a 1v1 between you and the mark scheme

you get me?

[finite mist]

The mark scheme makes less sense cause u see, its a triangle and almost all given side are integer

So the teacher must select those number so that the final result will be a whole number

Usually...

[granite prairie]

that's very true actually

yea you are

very correct all in all

alright thank you

the rest should be

easy to solve

i assume

[finite mist]

Do u know how to solve for b

[granite prairie]

i hope

[granite prairie]

solve for the area of coxy

the shape

and then multiply by circumfrence?

its just a trapezoid right?

[finite mist]

No the volume

There's a quick trick to solve for these

[granite prairie]

which is?

[finite mist]

The cone volume = 1/3 of the cyclinder

[granite prairie]

ah

so

hold on wait

how do we know that

wait hold on what cylinder

ggggg

[finite mist]

The volume of the yellow cylinder = 3 * volume of the cone

[granite prairie]

gonna resend this so i dont have to scroll up to find it

AHHH

i see

well

OH

OHHHHHHHHH

that's so cool

[finite mist]

And its easier to find the cylinder volume

[granite prairie]

well total height is 14

[finite mist]

Cause its just the area of the circle * height

[granite prairie]

is it?

oh

cylinder

though toyu meant the cone

total area of the cylinder is 307.916

2/3 of that is

205.277333333

ccorrect?

[finite mist]

I mean the volume

for b

[granite prairie]

shit

wait

[finite mist]

So the frustum could be calculate by Volume of red - Volume of blue

[granite prairie]

i am doing things way too quickly right now

my apologies

yea

total area is 205.277333333

then

we find the area of the blue

volume*

which is just

holdon

we could do the same trick

area of circle * height * 2 all over 3

hold on i calculated it wrong

to begin with

[finite mist]

Its 1/3

Cone = 1/3 * Cylinder

[granite prairie]

ah

[finite mist]

Then we get the smaller cone

The one the frustum "delete" from the big one to the smaller one

Then u get the frustum volume

[granite prairie]

yes sir

so total area is 307.876666667and

the cone

the smaller cone

37.704

so answer is

270.172666667

i assume i

rushed and did something swrong so

let me revise my answer

yes sir i did

167.573333333

i think all is

correct and well

thank you very much for the help

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[hollow rivet]

hi i dont quite understand how this is possible

[hollow rivet]

i get the the first one is 1 out of 14 then 1 out of 13

but the second one, if 1 is already chosen, shouldn't there only be 13 choices?

or is it because BOB isn't guaranteed to be picked?

[lilac pebble]

bob may not serve as an officer at all

[vast shale]

its saying bob will only be an officer if he‘s president

so he cant be president

it took me forever to understand what this is saying

[hollow rivet]

gotcha

so lets say you have 6 slots and 26 things you can fit in them in any order

would the possibilities be 6^26 or 26^6?

[lilac pebble]

26^6

assuming you can reuse the things

[vast shale]

i think 26 * 25 * 24 * 23 * 22 * 21

[lilac pebble]

yes that's if you can't reuse them

[vast shale]

it says no person can do two jobs

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[vocal sleet]

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

[silver beacon]

everything

[devout kindle]

You can start by taking LCM of 4 & 3

LCM is the smallest positive integer that is divisible by both the numbers

so LCM would be 12

so every 12 day both swimming and band practice lessons would come together

and now you just have to count, how many times does it repeat

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[dull geode]

can i get an hint on how you are supposed to prove something like this

[dull geode]

i mean like i can prove normal open sets but like the bounds change right

so a,b and c,d

if a < c then the new interval would be either c, b or c,d if c,d is a total subset of a,b

and like the intersection will always be open cause it never includes

someone help

[neat rose]

here is an idea

consider I = (max(a,c), min(b,d))

[dull geode]

ohh thats smart

i mean so like if you do this wouldn't you just be proving that $(max(a,c), min(b,d))$ is an open set then

[twin meteor]

SushiMan

[neat rose]

yeah

[dull geode]

with the max and min you just minize the cases thts very smart

[neat rose]

rigorously consider $x \in I : \ \varepsilon = \min(x - \max(a,c); min(b,d) -x )$

[neat rose]

there might be something even more clever, but for now u can consider that and work ur way out

[dull geode]

yes thank you

[twin meteor]

Goëtia

[dull geode]

also do you think this is concrete

i guess all these things follow the same format for the intersection right

but the union is probably weird

cause youre proving each set seperately

[neat rose]

did u write this?

[neat rose]

not really, depends on the proof

there is always a clever way, but if the professor gives something try to learn from it

[dull geode]

yes i wrote this

[dull geode]

yeah its from a tutorial worksheet

[neat rose]

it contains some errors and unnecessary complexity

[dull geode]

where in particular

[neat rose]

let r = min(x-a, b-x)

consider y in (x-r, x+r)

x-r < y < x+r

a <= x-r (since x-a >= r)

[dull geode]

wait whats wrong with those

[neat rose]

then y in [a,b]

[dull geode]

i just sorta wrote it similar to the way my prof wrote the proof for this one

[neat rose]

u proved then (a,b) is open

[neat rose]

mhm wym?

[dull geode]

this is my prof proving that (0,1) is open

you think it'd be better to prove one of the cases like this

[neat rose]

look at ur definition of r in each case in the proof

[neat rose]

this is correct

[dull geode]

oh shoot i made it the same

[dull geode]

also this isnt my proof

but to make it more concrete do you think it'd be best to prove the other upperbound as well for each case or not nesscary

[neat rose]

formulate a question, i dont understand what you need?

[dull geode]

so in the proof, for case one you are prooving that x-r exists right, but would you have to prove that x+r also exist

[neat rose]

obviously yeah

thats the definition, so where is the problem?

[dull geode]

cause in this proof i didnt really do that for either case, so it'd probably be best to do so right

[neat rose]

this proof is invalid, because of the earlier reason i gave you, once you fix it , it will look like the proof ur professor gave

[dull geode]

ok ok alright thank you

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[obtuse thunder]

confused on this step

[silk osprey]

so you’re fine with how they got 1/4 sin^2(2x) right?

just confused how they got to the next step?

@obtuse thunder

[obtuse thunder]

Yes right

[silk osprey]

ok well recall that cos(2x) = 1- 2sin^2(x)

if you don’t remember that version then you can derive it from cos(2x) = cos^2(x) - sin^2(x)

<@&268886789983436800>

[oak magnet]

<@&268886789983436800>

[obtuse thunder]

oh man

[silk osprey]

bro

[oak magnet]

Get him out

Ty

[obtuse thunder]

ok yeah i see i think

[silk osprey]

ok so you see how they got it?

[obtuse thunder]

how does the 2x become 4x?

is the 2 squared?

[silk osprey]

well consider the fact that 4x = 2(2x)

so x -> 2x

we have cos(4x) = 1 - 2sin^2(2x)

remember x is just a variable

if it helps to see you can do some form of substitution like u = 2x

then cos(2u) = 1 - 2sin^2(u)

but u = 2x

so cos(4x) = 1 - 2sin^2(2x)

[obtuse thunder]

ah ok that makes sense its just rearranging the cos double angle formulae

thanks

[silk osprey]

you’re welcome

[obtuse thunder]

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[empty tiger]

So I'm trying to solve a two variable word problem and I found a YouTube guide to setup a chart for it but the missing information is different and idk how to change it to make it work

[empty tiger]

Idk why phone made it landscape the first time

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@empty tiger Has your question been resolved?

[reef agate]

Bc you faced your phone downwards before you rotated it

But it doesnt look like this is all the information- one sec

[cold hedge]

why do you think you need 2 variables here?

[reef agate]

Oh no im stupid

[empty tiger]

all the problems we've been doing this chapter have been. i asked chatgpt about it (ik bad idea for math cause its blatantly wrong a lot of the time) and it actually gets setup like this

[cold hedge]

yea this is the solution i was thinking off

of*

[empty tiger]

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[quick crag]

!status

[vocal sleet]

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

[trim pecan]

.close

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[hexed onyx]

Why does it write it as integral x^2-19 times the integral of dx?

[somber portal]

probably to make it more basic and show that it's an immediate integral

[hexed onyx]

What does that mean

[somber portal]

integral of dx is x

[hexed onyx]

Oh

[somber portal]

integral of 19dx is 19x, or 19 times the integral of dx

since the integral of a constant times a function is the constant times the integral of the function

[hexed onyx]

So what did I do wrong here

Pearson is giving me whole number answers which I’m just not getting

[somber portal]

,rotate

[somber portal]

okay, first off, several notation errors

[hexed onyx]

Oh yeah no dx

Mb

[hexed onyx]

What’s wrong w 1/root42

[somber portal]

where does it come from?

[hexed onyx]

Isn’t is 1/b-a

For average value theorem

[somber portal]

wait, what are you exactly calculating?

[hexed onyx]

Average value of the function on the interval written

[somber portal]

oh, not the plain integral. okay then

[hexed onyx]

X^2-21 over 0 to root42

[somber portal]

you need to be a bit more careful with the notation. Also, what you get is, indeed, an integer

you're two steps off the solution

[hexed onyx]

Smfh bruh I just forgot to multiply it by root42

I multiplied 21*42 for part of my final answer 😐

Ik what to do

[somber portal]

good, now do it :)

[hexed onyx]

Have a good night thanks

I may be back 😊

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[high falcon]

Hi! If I have two sets, A=(1,2,2,3) and B=(2,3,4), the intersection of those sets is (2,3). The intersection operation does not include multiplicities. What if I wanted the result of my operation to be (2,2,3), including both times where an element from B showed up in A? Basically, I'm asking for A to be restricted by B, or for B to merely act as the set of elements that are "allowed" in A.

[high falcon]

But using multiplicities doesn't work because I want the elements of B to kind of be ignored in the result. Like, I also don't want the result to be (1,2,2,2,3,3,4)

(This is for a formal proof, trying to make this happen with standard notation)

It's so easy in like python but I have no clue how to go about it this way lol

[vocal sleet]

@high falcon Has your question been resolved?

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@high falcon Has your question been resolved?

[high falcon]

<@&286206848099549185>

I cry

[flat whale]

Show the original question

What you call set A is not actually a set

[high falcon]

I made the question up to explain what I am trying to do. The original statement I am trying to prove (using rigor - I know that it's pretty easy to solve straight up with the pigeonhole method) is "Show that if 7 integers are selected from the first 10 positive integers, there must be at least 2 pairs whose sum is 11."

my work in progress proof is as follows:
Let A = {1,2,3,4,5} and B = {6,7,8,9,10}
Let x ∈ A and y ∈ B
Let k = {(x, y) : x + y = 11}
∀x, ∃!y, x + y = 11
∴ |k| = 5
Let n = {a number of elements chosen at random from A ∪ B}
=> |n| - |k| ≤ |{(A x B) : A, B ∈ n} ∩ k| <-------- this basically calls on the pigeon hole principal
|n| = 7 => |{(A x B) : A, B ∈ n} ∩ k| ≥ 2
Thus, if 7 such integers are selected, at least 2 pairs will sum to 11.

There is one hole in my logic. The case where the two pairs whose sum is 11 are the same pair, i.e (1,10), (1,10). This creates a problem because
|{(A x B) : A, B ∈ n} ∩ k| will equal 1, not 2 because (1,10) will only appear once in the set {(A x B) : A, B ∈ n} ∩ k

That's where my question arises. I need a way to count both instances in the cases where the pairs are the same.

Also totally realizing I used () for A and B in the original quetion when obv I should have used {}. My b, sorry if that caused confusion!

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[sullen valve]

Is it wrong to find f-1(x) first?

[sullen valve]

I did this and it's not the right answer

^answer

Did I do something wrong in my algebra or is my approach inappropriate for the question

[cosmic cloud]

check the sign in the first one

after dividing by -2

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[pallid stream]

i dont get what the solution is doing? how do i solve this...

[vocal sleet]

Please don't occupy multiple help channels.

[leaden ingot]

it just uses mean value theorem

f'(c) = [f(4) - f(0)] / (4 - 0)

[agile imp]

Your problem gives as

f(x) = √(x)

Then, the given interval is [0,4] so,

f(0) = √(0) = 0
f(4) = √(4) = 2

[pallid stream]

ummm im not sure what to do next...

[agile imp]

Next you will solve for average rate of change

Just use this formula

[pallid stream]

so 2-0 / 4-0 = 4

like the answer?

[agile imp]

Use this formula

[agile imp]

Ye

After you solve that fraction u should solve for c

Which you should get 1/2

[pallid stream]

so that supposed coordinate is c and c=1 --> point is (1,1)?

[agile imp]

Ye

[pallid stream]

so to sum up the c value and the point that MVT guarantees will exist can be calculated using the equation (sent in the message)?

right?

[agile imp]

Yes absolutely

Wanna see my process?

[pallid stream]

yea sure

[agile imp]

Alr hold on

Wait

It won't send

There you go @pallid stream

Apologies for my penmanship

[pallid stream]

thanks for helping 😀

.close

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[trail delta]

Hey, how do I determine the factor rings in this example? I think, I got the maximal and prime ideals correct.

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[dense eagle]

i think it's easiest just to think about what it should be

i.e. if you have Z/12 / (3)

what are the elements of this field, and it should be kinda clear what the answer should be

[trail delta]

Is it like all m + (3) where m is in Z12?

[trail delta]

I really dont have any intuition behind quotient rings or groups

I know, they're like all additive cosets of the ideal, but i dont know how to work with this defn

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[crude locust]

dead server

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[trail delta]

.reopen

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✅

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[dense eagle]

the best way to think about quotient stuff is

G/H is "G with H killed"

so for example, Z[X]/(X^2) would be everything in Z[X]

but i treat X^2 = 0

unfortunately i have to dip now

btw if you don't get a response on uni mathematics like this, usually it's worth posting it in #groups-rings-fields instead

(you can still post it in the help channels, plenty of us who do uni mathematics still visit these help channels, but ur much more likely to get a faster response there)

[trail delta]

Okey, thanks

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[burnt lily]

why is this wrong?

[burnt lily]

i dont get it

oh wait i need to use the product rule

ye i was confusing the product and the sum rule

there are too many rules

[bitter pilot]

yes

[severe stone]

gelukt?

[burnt lily]

yep

[severe stone]

goudd

je kon ook gebruik maken van 2sin(x)cos(x) = sin(2x) 😉

dus 4 sin(x)cos(x) = 2 sin(2x)

[burnt lily]

ja maar die vergeet ik altijd

[severe stone]

probeer ze te onhouden gaat enorm veel tijd schelen met bepaalde vragen

[burnt lily]

ja er zijn zo ongelooflijk veel regels

[severe stone]

klopt maar de belangrijkste voor die zijn

cos²(x) + sin²(x)= 1

en dan de som regels

vanuit die kan je eigenlijk alle andere construeren

[burnt lily]

ja die ben ik ook weer al vergeten

ik ben echt slecht in deze dingen onthouden

lol

[severe stone]

hoe verder je in wiskunde komt

hoe meer je moet onthouden helaas

[burnt lily]

ja dat blijkt

ik vond wiskunde altijd wel leuk omdat het minder onthouden was en meer snappen

maar het wordt steeds meer onthouden

[severe stone]

klopt maar het wordt ook steeds meer snappen

dus dat scheelt

het zal alsnog altijd veel minder onthouden zijn dan andere vakken

[burnt lily]

fair

@severe stone wat is hier niet goed aan dan

dit klopt toch gewooon

of moet ik hierbij ook product regel done

ah ja

sorry voor de ping

[severe stone]

mknu maat

altijd product regel bij functies in de vorm van h(x)=f(x) * g(x)

en qoutient regel bij : h(x)= f(x)/g(x)

en je gebruikt kettingregel bij functies als h(x)= f(g(x))

[burnt lily]

hoe moet deze dan

er staat uitleg bij maar die snap ik neit

[severe stone]

aaah oke

ken je de product regel op deze manier?
(uv)'=u . v' + u'.v

[burnt lily]

ja

[severe stone]

oke u is x^4 en v is 2^x^4

je moet wel rekening houden van jouw functie v

[burnt lily]

maar ik weet die 2^x^4 niet

ja ik weet niet hoe ik 2^x^4 moet diff tho

[severe stone]

oke ja daar moet je rekening houden dat het een samengestelde functie is

[burnt lily]

alsin kettingregel?

[severe stone]

precies

want normaal heb je alleen 2^x, maar nu heb je h(x)=f(g(x)), f(x) is 2^x en g(x)=x^4

[burnt lily]

dat deed ik maar dat was fout

[severe stone]

kan je je werk laten zien?

[burnt lily]

moment dan reset ik de vraag ff dan kan ik het geformat doen

[severe stone]

sgoed

[severe stone]

aaah oke

laten we alleen op het 2e deel letten

en dan het deel binnen de haakjes

[burnt lily]

ja

[severe stone]

weet je wat de afgeleide is van h(x)=f(g(x))?

[burnt lily]

heb die haakjes alleen ff voor de duidelijkheid voor mezelf

[severe stone]

kan begrijpen als je dat niet helemaal kan aangezien het beetje abstract is

[burnt lily]

f'(x) * g'(x) dacht ik

[severe stone]

niet helaal

h'(x)=f'(g(x)) * g'(x)

[burnt lily]

ja dat bedoel ik

[burnt lily]

is dat niet wat ik hier doe

ik onthou altijd buitenste gediff * binnesnste gediff

die 4^x^2 lijkt mij dan 2*4^x^1 * [4^x]' te worden

[severe stone]

nee niet helemaal

oke wat is de afgeleide van 4^x

[burnt lily]

4

[severe stone]

nee nu heb je een exponentiele functie

ipv

x^4

[burnt lily]

oh

[severe stone]

dan zou het zijn 4x^3

[burnt lily]

4^x

ln(4) * 4 ^x

[severe stone]

precies

wat je anders kan doen

u=x²