help-10

[sonic pumice]

so do i need the coordinates of M?

[frozen wing]

Yes

[sonic pumice]

and then uhm the coordinates of A

[frozen wing]

We already know the coordinates of o,a,b,c,e,f yeah? We assumed a unit cube

[sonic pumice]

i think thats where im stuck

how do i get the coordinates

[frozen wing]

Ok

Ill draw a diagram and show you ok?

Gimme a min

[sonic pumice]

okok

[sonic pumice]

ohhh okok

[frozen wing]

Did you get it?

[sonic pumice]

how do you know that its a unit vector

[sonic pumice]

yes i think so

[frozen wing]

Assume for easier calculations

The ratios remain the same

[sonic pumice]

ok yes i understand

[frozen wing]

You can assume a 2x2x2 cube also

[sonic pumice]

yes

[frozen wing]

Ok ill give formula, yeah?

[sonic pumice]

yes

[sonic pumice]

okay

so now we have p

how do i prove its on the line

[frozen wing]

Yeah good question

Did you study equation of lines in 3d yet?

Like a+lambda r

[sonic pumice]

like you can make them equal and compare coefficients?

[frozen wing]

No starting point+ (some constant)*direction vector

[sonic pumice]

noo i dont know thst

[frozen wing]

Ok

Now to prove p is on the line fe, find distance fp and distance pe, we know that fe is 1 (we assumed), so fp+ep=1

[sonic pumice]

okay yes

i give up its too hard

[frozen wing]

If you try and practice, vectors is more of a scoring topic than others

[sonic pumice]

yeah youre right but i dont even know what to do next

[frozen wing]

You just have to keep in mid if some points are collinear, or they are divided in some ratio and stuff like that

[sonic pumice]

even with all the equations in front of me 😭

[frozen wing]

Wdym?

[sonic pumice]

have we proved it?

im confused

[frozen wing]

Yeah

[sonic pumice]

i cant even tell when the answer is in front of me 😭

[frozen wing]

Np, practice

People here are v helpful, just keep asking doubts

[sonic pumice]

thank you for your help

ill go practice some more

.close

[obtuse pebble]

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[obtuse pebble]

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[formal island]

Partial fractions 🎉

You can probably solve it without doing so anyways

[raven nest]

I'm fine doing it any way

[formal island]

u = x+1 and splitting the numerator or sum like that

[timid silo]

some fractions come anyway

[raven nest]

okay well how do I do it with partial fraction cuz I tried setting u=x+1 but that doesn't get the 2x out of the numerator

[viral blade]

what happened when you did this?

oh

[raven nest]

I just get 2x/u^2

[viral blade]

plug in u-1 for x then

[raven nest]

oh yea

I forgot I could do that

whoopsy daisy

brb

[formal island]

if ⁨u = x + 1⁩, then ⁨x = u - 1 ⁩
You can always re-organize the u-sub to your convenience

[neon vector]

I mean, what is 2x

[formal island]

x+x

[neon vector]

[raven nest]

dw I got it

I just forgot I could do the thing with x

then i did partial fractions

[timid silo]

do the thing with x

[raven nest]

I did

:3

I got 2ln|x+1|+(2/(x+1))

I tried setting 1+sqrt(x) = u

and x^4=u

neither worked out

[timid silo]

then

[raven nest]

then x^(3/2)dx=du

and I there's no x on top

[timid silo]

hm..

sqrt(x) = u

[raven nest]

okay

[timid silo]

this would do

[silver zinc]

-3/2

[raven nest]

no?

it is actually -1/2 tho

[silver zinc]

aftee differentiating

-1/2 * x^(-3/2)

[raven nest]

after differentiating, it goes from sqrt(x) to x^(-1/2)

because sqrt(x) is x^(1/2)

[timid silo]

where does 3/2 come form

[silver zinc]

oh mb im blind i mistook 1+sqrt(x) as 1/sqrt(x)

[raven nest]

anyways

du=1/sqrt(x) dx

not really quite sure what to do with that

[frank coral]

1/2sqrt(x)dx

[raven nest]

true

good point

nonetheless I still don't know how to use that

[frank coral]

What do you have now ?

[raven nest]

I've got (sqrt(x)/(2(1+u))) du

which feels

very very wrong

[silver zinc]

sqrt(x) was just u

[raven nest]

oh true

[frank coral]

The 2 is not in the right place but that’s not a real problem

[timid silo]

[frank coral]

It’s 2u/1+u

[raven nest]

k

so 2u/(1+u) du

[frank coral]

Yup

[silver zinc]

its now similar to the first problem in the channel

[raven nest]

I mean is it tho, I feel like it's different

[frank coral]

Then just take the 2 out of the integral as it’s constant , and now you should be able to solve

[raven nest]

no partial fractions

[raven nest]

ngl I got no clue what the integral of u/(u+1) would be

[frank coral]

$\frac u {1+u} = \frac {1+u-1}{1+u}$

[warm shale]

Starlord

[raven nest]

correct

[frank coral]

How does that help ?

[silver zinc]

i assumed this is how you did the partial fractions in the first problem so it is similar

[raven nest]

uhhhh

partial fractions

no

yeah I got nothing chief

[frank coral]

Well $\frac {1+u-1}{1+u} = 1 - \frac 1 {1+u}$

[warm shale]

Starlord

[raven nest]

oh

true

so like

u- integral of (1/(1+u))

[frank coral]

Which is equal to ?

[raven nest]

no dang clue buddy

uhmm

[frank coral]

What is the integral of 1/u ?

[raven nest]

ln|u|

[frank coral]

What is the integral of 1/1+u ?

[raven nest]

oh is it ln|u+1|

[frank coral]

Yup

[raven nest]

ohh

[frank coral]

!done

[obtuse pebble]

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

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[zenith crest]

is S^ n(x) just the nth successor of X?

starting from x

[shut lagoon]

Yes that's the idea.

[zenith crest]

ty

.solved

[obtuse pebble]

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[broken otter]

10 000 = 0,1 + 0,46 log x

[broken otter]

how can i solve for x

[hidden compass]

If I gave you this, would you be able to solve it?

10 000 = 0,1 + x

[broken otter]

yup

[hidden compass]

How? Solve that one

[broken otter]

9999,9 = x

[hidden compass]

Awesome

Let's say, instead, that I give you the following. Can you solve it now?

10 000 = 0,1 + 0,46 x

[broken otter]

9999,9 = 0,46x

[hidden compass]

Good, go on

[broken otter]

x = 21 7138,91

[hidden compass]

That number comes from where?

You did 9999,9 / 0,46 right?

[broken otter]

9999,9/0,46

[hidden compass]

Awesome

So we have x = 9999,9/0,46

[hidden compass]

If instead I gave you

10 000 = 0,1 + 0,46 t

What would t be equal to?

You would do the same steps, I believe, wouldn't you?

[broken otter]

t = 21 7138,91

[hidden compass]

Which is t = 9999,9/0,46, do you agree?

[broken otter]

yea

[hidden compass]

Perfect

[hidden compass]

So can you now see how to solve this?

[hidden compass]

If you put t = logx, you have this exact thing, do you see it?

[broken otter]

21 738,91 = log x

[hidden compass]

Yes

But 9999,9 / 0,46 is not equal to 21738,91

So leave that number as 9999,9/0,46

[broken otter]

im not sure what to do

[hidden compass]

Well, you have logx = something

So you can exponentiate and find x

Never done something similar?

Suppose logx = 3, then x = e³

Or x = 10³ = 1000 if log means logarithm in base 10

[broken otter]

oh its 10^(9999,9/0,46) = x

[hidden compass]

Exactly

That's the correct form of the result

[broken otter]

ahh oka

well i gotta find a specific number

so i gotta do 9999,9/0,46

^10

[hidden compass]

Sure, use your calculator to do it

Of course you can't do that by hand

[broken otter]

2,36

[hidden compass]

But wait, I suspect there's something wrong in the original equation (or even the question)

[hidden compass]

Huh?

[broken otter]

9999,9/0,46 = 21 738,91

oh idid a mistake

nvm

[hidden compass]

9999,9/0,46 ≠ 21 738,91

9999,9/0,46 ≈ 21 738,91

[broken otter]

when i do 10^21738,91 it doesnt work

[hidden compass]

Yeah because it's a huge number

[broken otter]

well that dont even make sense cause i gotta find a number lol

[hidden compass]

Much greater than the number of atoms in the whole universe, which is around 10⁸⁰

[broken otter]

no way

[tardy epoch]

why

[hidden compass]

#help-10 message. That's why I wrote this

[broken otter]

because it says for the average speed

[tardy epoch]

!original

[obtuse pebble]

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

[broken otter]

oh alr

[hidden compass]

Where did you get this from?

[broken otter]

from the text

[hidden compass]

Can you send the photo?

[broken otter]

yeah

[hidden compass]

🙈 🙈 🙈 🙈

[broken otter]

to put it in simple

[hidden compass]

10000 is the number of inhabitants, so x

[tardy epoch]

amazing

[hidden compass]

Not the velocity v(x)

[broken otter]

10 000 is x and v(x) is the speed

and like i said u have the same equation there

[hidden compass]

Not at all

[broken otter]

its different?

[hidden compass]

Of course!!

v(x) is the velocity, x is the number of inhabitants

They also have completely different unit of measure

[broken otter]

i wasnt aware of that

[hidden compass]

Are you serious??

So you're telling me you can't make the difference between 10000 people and 30 km/h ?

[broken otter]

10 000 people is x

[hidden compass]

Sure

[broken otter]

i already know it is

wait

i put 10 000 in v(x) by accident

[hidden compass]

That's the problem, indeed 😬 😅

[broken otter]

crap

[hidden compass]

Don't worry, it can happen sometimes 🫠 🤗

But next time you should realize sooner

[tardy epoch]

This is why you should alwayss

!original

[obtuse pebble]

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

[broken otter]

im just tired because i studied like 20 hours last week for my math exam just for my teacher to say today theres another exam tomorrow

[tardy epoch]

yea all the more reason to

!original

[obtuse pebble]

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

[broken otter]

ok ty guys v(x) = 1,85

.close

[obtuse pebble]

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[tardy epoch]

if people like you stopped doing the wrong problem, i'll stop using it

[obtuse pebble]

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[round shale]

Yo I need help with pre calc 11 dm

[obtuse pebble]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[neon vector]

Yo

waddup

[round shale]

Yo

I need help with pre calc 11

[tardy epoch]

!da2a

[obtuse pebble]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[daring sorrel]

I think he needs help with pre calc 11, not sure though

[round shale]

Unc

Pls help

I do

[tardy epoch]

[round shale]

How

Do I do that

[tardy epoch]

you ask math questions by either doing homework problems or reading your book until you get have a question

nobody knows what you need help with besides you so you need to put effort to articulate your question

[obtuse pebble]

@round shale Has your question been resolved?

[obtuse pebble]

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[sharp pulsar]

hi

[sharp pulsar]

okay so

im having trouble with B and C mostly C

for B, i did demorgan's law on the first part

and got not A + not B + not C + A notB

then not A + A notB to get notB

notB + notB to get notB

and im left with notA + notB + notC

or not(ABC)

for C im doing the same steps but im lost at
ABC notD + A notB + A notC + A notD + notA notB notC notD

ABCD' + AB' + AC' + AD' + A'B'C'D' how do i simplify from here

and just to make sure, the Kmap/TT for (b) would be 1's every where except at 111 right

apparently the answer for C is but I dont understand how still

how is the first term cancelled out

[obtuse pebble]

@sharp pulsar Has your question been resolved?

[sharp pulsar]

pls bro im struggling 😭

[tame python]

makes sense, (b) is not(ABC)

[tame python]

along the line I used an absorption property

[tame python]

here, factor the A

then simplify BCD' + D'

(you need to justify that this simplifies to D', thers is a law called "absorption" you learned that lets you do this)

from there, you will have A(B' + C' + D') + A'B'C'D' which means A(BCD)' + A'B'C'D'

[sudden cloak]

Hello

[sharp pulsar]

ohhhh okay

thankyou i see

So i tried making a Kmap of this but i'm not sure I did it right

CD/AB
00 01 11 10
00 1 0 1 1
01 0 0 1 1
11 0 0 0 1
10 0 0 1 1

i boxed bottom 2, right column, top right four, and wrapped 0000 with 1000

CD rows and AB columns

[sharp pulsar]

anyone know how to make a kmap from this cuz mine is wrong fs because there isn't an A' from it's equation

[sharp pulsar]

.close

[obtuse pebble]

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[obtuse pebble]

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[balmy eagle]

Im struggling with 10(a), my teacher said that the answer should be more simple than what i got. Is there an error in the textbook?

[royal basin]

i can't make sense of the first line in your work

[winter cargo]

what simplified to give you 3 in the numerator

[balmy eagle]

I didn't show full work

[royal basin]

ok well then DO show full work lol

i have a suggestion to make your own life simpler though

take $x := \log_3(2)$ and express everything in the question in terms of $x$

[warm shale]

Ann

[winter cargo]

here you seemed to have simplified $-4 \log_3 ^2 2 + \log_3 ^2 4$ to $-3\log_3 ^2 2$ from line 1 to 2

all prior steps are correct though

[royal basin]

⁨\log⁩

[winter cargo]

oh k

[royal basin]

ok so yeah i concur with that

that seems to be the mistake

i will reiterate my suggestion: scrap all this, and
take $x := \log_3(2)$ and express everything in the question in terms of $x$

[warm shale]

Ann

[obtuse pebble]

@balmy eagle Has your question been resolved?

[obtuse pebble]

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[spice chasm]

how do you do part d without doing the whole formula

[warm shale]

Jash

[spice chasm]

i meant y bar

not x bar

and then for x bar i did x(x+2-x^2) inside the integrand

but i dont think ur meant to do that

[dim yacht]

given that it says most likely

I think you're just supposed to estimate

[spice chasm]

ye but how theyre all close to each other

[dim yacht]

put a dot where you think the center of mass would be

it does correspond to d

roughly

[spice chasm]

see when i did the formula i got like (.8, .95)

[dim yacht]

centroid*

[spice chasm]

,w integral of 3/20 (x+2-x^2)^2 from 0 to 2

wait

[warm shale]

Results provided by WolframAlpha

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

Requested by jashgg

[spice chasm]

ye i got that

the area of the region is 10/3

so 1/area=3/10 and then when u do 1/2f(x)^2 u get 3/20 f(x)^2

f(x)=x+2-x^2

[dim yacht]

I don't think you used the correct formula

[spice chasm]

this what ai did

idk why its like top^2-bottom^2

instead of just top - bottom all squared

[dim yacht]

I have no idea what AI is doing here tbh

[spice chasm]

its getting the right value of 1.8

[dim yacht]

but this is all you need to know to solve the problem

[spice chasm]

ye we learnt this in physics

how would you do integral of y dm

[dim yacht]

this could also give better insight

probably much better insight

[spice chasm]

idk how y dm turns into xf(x)dx and how x dm turns into 1/2f(x)^2 dx

[dim yacht]

yea I also don't remember the derivation xd

just know that these are the formulas

but also just estimating it should be somewhat clear that d is the correct answer

[spice chasm]

so u gotta do the top^2-bottom^2

ty

.close

[obtuse pebble]

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[zinc badge]

any tips on how to start this? it says i=0 for the first term of the sum btw the zero got a bit folded

[brazen gorge]

is this on a shirt

[zinc badge]

yes

[brazen gorge]

is this the full problem

!xy

[obtuse pebble]

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

[zinc badge]

I don't think the expected value or limit change anything tho, but I've never seen a problem like this before i don't do stats

[brazen gorge]

yeah the lim and E are the important part

[zinc badge]

oh okay

😭😭

[brazen gorge]

💀

well in the first part U ~ Bernoulli(1/2)

[zinc badge]

i never even seen something like this before I just wish i could do a similar problem before going into this one but idk what I'd even be searching for

[brazen gorge]

the max doesn't make much sense tho..

the sum is a random variable

[fossil crag]

you can define the max between random variables

R.Vs are functions before anything else

[zinc badge]

wait why sorry

[brazen gorge]

well U({0, 1}) means it's uniform over the set {0, 1}

which means it's either 0 or 1 with equal probability, namely half

[brazen gorge]

yes true

[fossil crag]

I think this problem would be a lot easier to think about

if you just started simplifying this

And I'm assuming the B_k are mutually independent

maybe it's unneeded

You could see what happens when all B_k = B

[zinc badge]

to what? (2^(-i))*p?

[zinc badge]

what do you mean when all Bk equal B sorry 😭

[fossil crag]

It is supposed that all B_k are random variables that follow a "uniform law" on {0,1}

[zinc badge]

would writing p as (2^(n+1))-1 help or nah

[fossil crag]

What if B_1 = B_2 = ... are all equal to each other

[zinc badge]

OH okay

[fossil crag]

the exact value of the indices doesn't matter much, rather notice that each floor(p/2^i) is different

there aren't two indices in that sum that are the same

[zinc badge]

bruh 😭

[fossil crag]

which is why I was inclined to think there was an assumption missing on independence

[zinc badge]

it's probably assuming independence

i mean most problems do

[fossil crag]

as stated before, "uniform law" on {0,1} is the same as bernouilli of probability 1/2

do you know what is the law followed by a sum of independent bernouillis?

[zinc badge]

idk 😭😭

or is it geo

[fossil crag]

not geo

Imagine this bernouilli R.V of probability 1/2 is tossing a coin

each coin toss is independent

you get heads on each coin toss with probability 1/2

if you do n coin tosses

what's the distribution of the number of heads?

[zinc badge]

but isn't n infinite

[fossil crag]

right now no

[zinc badge]

oh then bin

[fossil crag]

we're working for a fixed, always finite value of n

[fossil crag]

yes

[zinc badge]

i thought n was infinite mb 😭😭

[fossil crag]

it's only THEN that we make n grow larger and larger

[fossil crag]

yeah n is never infinite in the limit, we can only approach it

[zinc badge]

ill have a look at this later and see if i can make any progress this has been helpful so far thanks

[fossil crag]

This looks a lot like a branching walk problem

maybe the result is trying to say that you can almost always find a path that maximizes the sum

So my bet's on the result being 1, but hard to prove just like that

[obtuse pebble]

@zinc badge Has your question been resolved?

[zinc badge]

no bro

[obtuse pebble]

@zinc badge Has your question been resolved?

[zinc badge]

ok sure if it makes u sybau

[obtuse pebble]

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[burnt chasm]

I'm really scared for maths

[burnt chasm]

Like I forget everything in exam hall

Can't solve simple stuff

[hybrid cosmos]

right, so you need help with something?

[burnt chasm]

Everything about math actually

[slow island]

Wha?

[burnt chasm]

I'm very bad at maths

[hybrid cosmos]

um, I don't think "all of math" would be a good idea for one help session...

is there, like, a specific concern you have? like a specific question that particularly troubles you.

[burnt chasm]

Okay can I atleast know how to make my maths strong and stop fearing it

And stop forgetting things in exam hall

[slow island]

I think practice would be the best advice I could give.

[hybrid cosmos]

okay so those two questions are different questions with different answers, so let's take it one at a time.

[royal basin]

have you tried making mock exams for yourself at home

[brazen gorge]

#study-discussion might also able to give better support, these channels are mostly for help with specific math problems

[hybrid cosmos]

1. how to get better at math.
   this one, other than practice, try to look for intuitive explanations about whatever it is you're studying. try designing experiments around what you've learnt.
   learnt about the general vertex form formula for a quadratic? test it out with different quadratics!
   learnt about the centers of a triangle? draw some triangles, test them out!
   try to find some other properties of stuff that is connected or related to what you've learnt.

also, try finding out how you learn best. do you learn best by doing? watching? hearing? reading?

[burnt chasm]

Oh okay sorry

[slow island]

yeah right

[royal basin]

like, print off a copy of a past paper, sit down at home and give yourself as much time as you'd get on an exam and then see how well you did.

this way you can tell if it's a math issue or a test-anxiety issue

[burnt chasm]

Doing and watching

[hybrid cosmos]

2. how to not fear exams.
   this I suppose comes with experience and perhaps also with knowledge. but on top of that, remember to also take care of yourself during the time leading up to an exam.

many students overprepare thinking that they can get away with poor meals and a sleep debt into an exam. that's a no-go for sure.

[burnt chasm]

Understood

[hybrid cosmos]

but without more specific concerns that's all I can say, unfortunately.

[burnt chasm]

You can close this chat I'll continue in the other channel

[neon vector]

Same for every single thing in existence. Do more of it

[hybrid cosmos]

you're the OP. I cannot close it.

but all the best!

[burnt chasm]

How to close

[hybrid cosmos]

type .close.

[burnt chasm]

.close.

Uh

.close

[obtuse pebble]

Channel closed

Closed by @burnt chasm

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

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[warm shale]

Renato

[obtuse pebble]

@brave burrow Has your question been resolved?

[obtuse pebble]

@brave burrow Has your question been resolved?

[twilit pelican]

can you do the first question

[brave burrow]

i think its

11! / 8!

11 x 10 x 9 x 8 x 7 x6 x 5 x 4 x 3

,calc 11109876543

[warm shale]

Result:

1.99584e+7

[brave burrow]

11 C 9?

[twilit pelican]

hmm idts

or well, unless you mean 11 C 9 as 11!/2!

[brave burrow]

11 x 10 x 9 x 8 x 7 x6 x 5 x 4 x 3

this is what I mean

[twilit pelican]

oh ye

we chillin

onto the next

[brave burrow]

good

im not that bad at counting it seems

[brave burrow]

as long as 0 is even we good for reflexive

[twilit pelican]

yep it is

symmetric?

[brave burrow]

xRy => yRx

[twilit pelican]

yup

does that hold here?

[brave burrow]

the equality part is good but idk if f(5)-g(5) is even

like

idk if f(5)-g(5) even implies g(5)-f(5) is even

[twilit pelican]

well lets think abt it

[brave burrow]

say for example 1 - 3 vs 3 - 1

[twilit pelican]

-2 vs 2

so long as you define an even integer to be 2k for some k in Z i think we’re chillin, no?

[brave burrow]

where are you going with this

[twilit pelican]

because 2 = 2*1

and -2 = 2*-1

so they’re both even

[brave burrow]

yes

but

[twilit pelican]

if f(5) - g(5) = 2k

then g(5) - f(5) = -2k = 2(-k)

[brave burrow]

according to who

[twilit pelican]

we said f R g right?

[brave burrow]

ya

[twilit pelican]

so f(5) - g(5) is even

[brave burrow]

ya

[twilit pelican]

ergo we can say this

[brave burrow]

ya

[twilit pelican]

right

then multiply both sides by -1

[brave burrow]

ahh I see

[twilit pelican]

you get g(5) - f(5) is also even

[brave burrow]

i see

very nice

what about transitivity

x R y , y R z => x R z

[twilit pelican]

try that one out

you might have to do some equation manipulation like adding and stuff

[brave burrow]

the equality doesnt bother me

[twilit pelican]

can you tell me what our assumptions look like about some f g and h?

like f(5) - g(5) and g(5) - h(5)

[brave burrow]

f(3) - g(3) = f(4) - g(4)
g(3) - h(3) = g(4) - h(4)

you sum them both

f(3) - h(3) = f(4) - h(4)

[brave burrow]

well

this is where it gets nasty

assumptions

1. f(5)-g(5) even
2. g(5)-h(5) even
   ---‐-------------------

f(5)-g(5) = 2k
g(5)-h(5) = 2m
you sum them
f(5)-h(5) = 2(k+m)
let k + m = q
f(5)-h(5) = 2q
@twilit pelican

[twilit pelican]

looks great!

[brave burrow]

ok equivalence relation proven

any ideas on c) ?

[twilit pelican]

also im a bit unfamiliar on c)

what does your problem mean by r_9?

[brave burrow]

r9(n) = n (mod 9)

is the remainder function

[twilit pelican]

oh ok thats what i thought

[brave burrow]

any pointers

.close

[obtuse pebble]

Channel closed

Closed by @brave burrow

Use .reopen if this was a mistake.

[twilit pelican]

oop

did ya get it?

i just had to brb

[brave burrow]

i kinda surrendered

shit is too hard

[twilit pelican]

nah we got it trust

reopen it

[brave burrow]

. reopen

.reopen

[obtuse pebble]

✅ Original question: #help-10 message

[twilit pelican]

okay so

we're trying to find functions g that are equivalent to our choice of f

i.e., f(3) - g(3) = f(4) - g(4), and f(5) - g(5) is even

here's the first hint, rearrange the first condition so that it looks like f(3) - f(4) = g(3) - g(4)

for brevity, ill call this number x

so f(3) - f(4) = g(3) - g(4) = x

how many choices do we have for x?

@brave burrow

[brave burrow]

wdym

[twilit pelican]

like for this number

x

note that x is f(3) - f(4)

[brave burrow]

its injective

both f and g are injective

[twilit pelican]

right

i understand that

but like

okay if i asked you whats f(2)

how many choices do i have for f(2)

like its output

(whats the size of our codomain)

[brave burrow]

9

[weak needle]

that's size of range

[twilit pelican]

are you sure?

how many possibilities are there in the codomain

[brave burrow]

the codomain is size 11 but since A is made of injective functions the size of the image is equal to the size of the domain

[twilit pelican]

injective functions doesnt necessarily mean that you're mapping 1 to 1, 2 to 2, etc etc

yes all 9 elements are being mapped to 9 distinct outputs

[brave burrow]

yes

[twilit pelican]

but if i just ask you what can f(2) be, what can you tell me

it can be anything between 1 through 11 right?

[brave burrow]

sure

[twilit pelican]

okay wait, do you see why it can be anything 1 through 11, or no

[brave burrow]

yes

[twilit pelican]

okay sick

so g(3) - g(4) is just a number in the codomain right

aka, x has 11 choices

[zenith raft]

[brave burrow]

sure

[twilit pelican]

g(3) = g(4) + x is completely determined by what we pick for x, so g(3) has just the 1 choice

so right now we have 11

now g(5) has to be even

how many evens do we have

[brave burrow]

2 4 6 8 10

[twilit pelican]

hopefully you know that $g(5) \equiv f(5) \bmod 2$

[warm shale]

blanketism

[brave burrow]

how

[twilit pelican]

because f(5) - g(5) is even?

[brave burrow]

why

[twilit pelican]

sorry i mistyped that, i meant f(5) - g(5) has to be even

[brave burrow]

how

[twilit pelican]

it is a little tricker, but can you utilize $f(n) = r_9(n) + 1$ to deduce some facts about the parities of $g(4)$ and $g(3)$?

[brave burrow]

[warm shale]

blanketism

[twilit pelican]

since f and g are equivalent according to the relation R

[weak needle]

r9(1)=1, r9(2)=2... r9(8)=8 and r9(9)=0, right?
So we know f...
-# am I wrong?

[brave burrow]

nice

[twilit pelican]

yeah i just realized halfway through we do know f, but im just counting the number of g related to it

at first i thought about having to worry about the parity of the initial f(3) and f(4) but its already explicit

[weak needle]

now it's ezpz

[twilit pelican]

a little mishap it happens

anyways

so we know f(4) = 5 and f(3) = 4, so f(4) - f(3) = 1

[weak needle]

Renato do it now I m here to cheer

[brave burrow]

.

[brave burrow]

is it f(4)-f(3) or f(3)-f(4)

[twilit pelican]

oh buns, its f(4) - f(3)

f(5) = 6, so g(5) must also be even

[brave burrow]

even + even = even

even + odd = odd

[twilit pelican]

yep

how many choices do we have for g(5)?

[brave burrow]

5

[twilit pelican]

not quite

[brave burrow]

2 4 6 8 10

[twilit pelican]

you're right that there are 5 evens, yes

but do you remember how g(4) - g(3) = f(4) - f(3) = 1?

[brave burrow]

g is injective

[twilit pelican]

indeed

what can you tell me about g(4) and g(3) if their difference is 1

are they both odd, both even, consecutive, etc

[brave burrow]

consecutive

[twilit pelican]

good

so one must be even and one must be odd, right?

in any case, we are using up an even integer

[brave burrow]

yes or the other way around

[twilit pelican]

right, so instead of having 5 evens

we only have 4

[brave burrow]

4

[twilit pelican]

oh good

and now for the rest, we just need to assign them injectively

so we have 6 inputs remaining with 8 possible outputs

should be pretty simple, 8!/2! like in part a

[brave burrow]

what ?

[twilit pelican]

8 x 7 x 6 x 5 x 4 x 3?

we've only used up 3 of the integers from 1 through 11

[brave burrow]

what the fuck is going on

we need to do a little recap

[twilit pelican]

oka

so you know how we were talking about how g(4) - g(3) = 1 right?

and how there were 10 choices for the pair g(4) and g(3)

aka it could be 11 -10, 10 - 9, 9 - 8, etc

we have used up 2 of the integers from 1 through 11

we got 9 left

then we talked about how g(5) had to be even since f(5) = 6 is even

[brave burrow]

11-10
10-9
9-8
8-7
7-6
6-5
5-4
4-3
3-2
2-1

[twilit pelican]

right

there's 10 of those

and we also mentioned that g(5) had 4 choices right?

since f(5) was even

[brave burrow]

where did one of the even went

[twilit pelican]

in determining g(4) and g(3)

remember how we said they were consecutive?

so one of them has to be even

[brave burrow]

oh right

[twilit pelican]

yea

so we assign one of the remaining evens to g(5)

how many numbers do we have left in the codomain?

[brave burrow]

we used 2 integers from the 11 of the domain for the pair g4g3

[twilit pelican]

yup

[brave burrow]

we got 9 integers left

[twilit pelican]

wat

o

okay coolio

then we use another one for g(5)

[brave burrow]

yes

[brave burrow]

8

[twilit pelican]

okay

and how many inputs do we have left

[brave burrow]

6

8 x 7 x 6 x 5 x 4 x 3

[twilit pelican]

x 6 x 5... so on and so on

yup

[brave burrow]

crazy hard

[twilit pelican]

okay, now just take that and multiply it by 10 and 4 and we have our solution

yeah, thanks to @shubh i forgot that we had an explicit f(n) lmao

@weak needle

[brave burrow]

was it hard or easy

[twilit pelican]

hmm it wasn't easy at first when i had to think about an arbitrary f(n)

but my bum ass forgot we were given one

[brave burrow]

I appreciate the help though i gotta say counting is lowk hard af

[twilit pelican]

nw

glad i could assist

[brave burrow]

I appreciate it aswell shrivam @weak needle

.close

[obtuse pebble]

Channel closed

Closed by @brave burrow

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

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[tacit light]

.close

[raw wharf]

I am not understanding themethod of characteristic in PDE please help

[winter cargo]

buh..

<@&268886789983436800>

[neon vector]

<@&268886789983436800>

Oh damn you

🙁

[winter cargo]

🏆

[frozen wing]

Any certain ques?

[winter cargo]

what in the

<@&268886789983436800>

thankouy

[quiet wadi]

oh cool it's logan paul this time

[raw wharf]

Yea so

I have likethis

I got myself up to w(a,0)=1/3e^(3(0))+k

The solution states

I am notreally sure how they jumped this

[frozen wing]

A min, i have not touched pde in some time

[raw wharf]

I think I am getting 'lost inthe suace' but I amnot sure

[frozen wing]

Wdym

[raw wharf]

When I asked a senior they said I set k as something then substitute in

but I am not sure how to logically go about defining K

Cause the implication is K = w(a,0)-1/3

But I am not sure how you go about stating this

[frozen wing]

Ffs i forgot the same thing

[raw wharf]

;-;

[frozen wing]

As far as i remember the final constant is as a function of the variable you integerated as

In this case its a function in terms of s

[raw wharf]

Oh

I can rearrange the expression itself to be expressed in terms of k

Then what would I do with w(a,0)

[frozen wing]

You have on ques that w(x,0) is e^-x2

[raw wharf]

Not w(a,0) tho

[obtuse pebble]

@raw wharf Has your question been resolved?

[obtuse pebble]

@raw wharf Has your question been resolved?

[obtuse pebble]

@raw wharf Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

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• Show your work, and if possible, explain where you are stuck.
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[static cedar]

How do I do 7c?

[woeful edge]

try proving vectors BS and TA are equal, and same for AN and MB

[static cedar]

After I do that, what do I do after?

[woeful edge]

after proving BS = TA and MB = AN (vectors ofc)

sum up both equations

[static cedar]

I got these

[woeful edge]

ok so u proved BS = 1/5a and so is TA

so BS = TA

[static cedar]

Yeah

[woeful edge]

u did the same for AN and MB

so MB = AN

this means that, in vectors:
BS + MB = TA + AN

all i did was sum up the two equations

[static cedar]

Ohh okay

[woeful edge]

now what vector does BS + MB equal to

[static cedar]

1/5a+4/5b

[woeful edge]

no no

[static cedar]

Oh wait

[woeful edge]

use chasles' relation

[static cedar]

what’s that😭

[woeful edge]

AM + MB = AB

in vectors

for any point A, B and M

[static cedar]

Ohh

[static cedar]

MS?

[woeful edge]

yess

[woeful edge]

and for the other side of this equation?

TA + AN?

[static cedar]

TN?

[woeful edge]

yes

[static cedar]

Ah okay

[woeful edge]

so in short u have in vectors:
MS = TN now

since those four points are non aligned, what type of quadrilateral is MSNT

[static cedar]

A parallelogram?

I’m not sure

[woeful edge]

in general if u have a vector equality AB = DC and the four points are not aligned, then ABCD is a parallelogram

[woeful edge]

so yes its a parallelogram

[static cedar]

Oh okay

[woeful edge]

since its a parallelogram, then how are the diagonals MN and ST?

[static cedar]

Equal?

[woeful edge]

noo, not in a parallelogram

[static cedar]

Oh😭

[woeful edge]

where do the diagonals of a parallelogram meet at

[static cedar]

Centre?

So S

D*

[woeful edge]

yesss

in parallelograms, the diagonals are caracterized by one thing only

which is that they bisect each other (they meet at their midpoints)

[static cedar]

Oh yeah

Idk how I missed that it’s in the question mb

[woeful edge]

yeah u answered the question

[static cedar]

So how do y prove it? I saw in the book it says u show 2 ways to get to the midpoint

But their example was much simpler

[woeful edge]

wdym how do u prove it

what we did was proof lol

[static cedar]

Wait so is the question answered??

[woeful edge]

we js answered it

yeah lol

[static cedar]

I’m confused tho lowkey how is showing MS=TN showing they bisect?

[woeful edge]

because

like i said

MS = TN (in vectors) with the addition that the four points are not aligned

gives u the ability to say that MSNT is a parallelogram

draw any two equal vectors on whatever paper u have youll notice u get a parallelogram

[static cedar]

Uh how do I draw it

Either way Cus they’re a parallelogram, how does it show they bisect?

[woeful edge]

no no im js saying it js to show u that this is a true statement in geometry

[static cedar]

Like Cus bisect means meet right

[woeful edge]

bisect means meet at the midpoints

MN and ST bisect one other means MN meets ST at the midpoint of ST, and same for ST, it meets MN at the midpoint of MN

they meet each other at their own midpoints at the same time

[static cedar]

It legit meets them at the start of the points no

[woeful edge]

wdym

[static cedar]

Idk if m making sense but like

The MS=TN

[woeful edge]

what abt it

[static cedar]

Lemme start over

So MS=Tn

Shows it’s a parallelogram

But then how does it meet at the midpoints ??

[woeful edge]

its

a parallelogram thing

the diagonals of a parallelogram always meet at their midpoints

[static cedar]

And the diagonals are MN and Tn?

[woeful edge]

yes

[static cedar]

I hate geometry honestly

okay that makes sense now

[woeful edge]

whats not making sense for u rn lol

[woeful edge]

o

[static cedar]

I was tryna

Visualise it

Like u said draw a parallelogram

And it was not making sense to me at all

[woeful edge]

nooo

[static cedar]

😭

[woeful edge]

i was js saying like

if u draw any two vectors that are equal

youll always form a parallelogram

i meant it like try it on ur own as an experiment to see for urself

[static cedar]

I was seeing it like this

But then I was like so they don’t bisect

The mark scheme does it way differently oh my

[woeful edge]

they do bisect

if u draw out ST and MN they will intersect at their midpoints

if not its cause ur drawing is kinda rough lol

[static cedar]

This is not close😭

[woeful edge]

u dont have to have the same exact method to find it

[static cedar]

The alternative method isn’t like that either

It was like this

[woeful edge]

i meannn

did i use smt confusing

[woeful edge]

that you didnt learn?

[static cedar]

I do understand it I’m js scared if I use it in the test tomorrow I might be cooked

[woeful edge]

no u wont

did u learn all this before?

[static cedar]

there so strict if it’s not in the mark scheme

[static cedar]

I wasn’t in for the geometric stuff

So I was revising today

And it did show me the 2 pathway stuff

As I mentioned earlier but I didn’t know how I’d approach it

[woeful edge]

so like

the way of reasoning i used

isnt into ur liking

[static cedar]

Do u think it’d still b valid

[woeful edge]

mathematically

[static cedar]

as long as it’s valid and I understand it

[woeful edge]

realistically

its correct

[static cedar]

Which I did

[woeful edge]

ask any other helper

[static cedar]

I’ll ask my teacher if I can do it before the test 🙏

[woeful edge]

the question is

were u taught that two equal vectors give a parallelogram?

[static cedar]

I haven’t heard of it but then again I’ve missed like 2/3 days

So he probably has mentioned it

Do u do edexcel?

[woeful edge]

i mean u can prove it

[woeful edge]

no

[static cedar]

ah

[woeful edge]

if (in vectors): AB = DC then the lines (AB) and (CD) are parallel and (in distances) AB = CD

cause two equal vectors have same direction and distance

[static cedar]

I’ll remember that🙏

Have u tried the 2 pathway method thing

[woeful edge]

the ones in the correction of the exercice?

[static cedar]

Yeah

[woeful edge]

no i didnt see

lemme see rq

[static cedar]

Okay

[woeful edge]

this is sm work

isnt it js better to do it the way i requested

[static cedar]

Way better ngl cus I don’t understand the symbols

N what they do

Even in the simple example they show

[woeful edge]

the "or" confused me a lot

if u want me to explain this method

sure

@static cedar

otherwise u can close the channel

[static cedar]

Could u explain it please, idm if u don’t want to

[woeful edge]

okay

so for this question he uses the two previous ones

were u able to prove in the two previous questions that:

MN = a + 3/5b and ST = -3/5a + b?

[static cedar]

Yeah

[woeful edge]

now look

you will start from vector OD

and use the chasles relation

in a reverse way

[static cedar]

Also I’m pretty sure ST = 3/5a - b no?

[static cedar]

Wait so like OD = x + x

[woeful edge]

oh mb yea

[woeful edge]

to be precise OD = Ox + xD

u choose any point

in the spot of x

[static cedar]

Could you do OT+TD

[woeful edge]

yes

[static cedar]

So it’d be 4/5a + x(3/5a-b)

?

[woeful edge]

yeah

[woeful edge]

Now choose another point to put in the spot of x

[static cedar]

Could you do M?

[woeful edge]

yes

[static cedar]

oo okay so would it be 1/5b + y(a+3/5b)

[woeful edge]

yess

reason why we chose M and T to put in the spot of x (even though we can put any other point like S or A or B or C...)

its cause choosing M and T ties it up to the previous two questions

[static cedar]

If u chose any other spot would it be harder to answer

[woeful edge]

yep

way harder

probably impossible

anyway

[woeful edge]

u got this

[woeful edge]

and this

theyre both equal to the same vector OD

[static cedar]

So do you make them equal to each other

[woeful edge]

so those two are equal to each other

[woeful edge]

yess

it becomes

4/5a + x(3/5a - b) = 1/5b + y(a + 3/5b)

[static cedar]

ooo okay I got that

But like don’t u have 2 unknowns

[woeful edge]

mhm

now simple algebra stuff

i will gather up the values multiplied by a alone

and the values multplied by b alone

we have:
4/5a + x(3/5a - b) = 1/5b + y(a + 3/5b)

meaning

4/5 a + 3/5ax - bx = 1/5b + ay + 3/5by

(all i did was foil the expression between parentheses)

[static cedar]

So like expand the brackets

[woeful edge]

after:
(4/5 + 3/5x)a - bx = ay + (1/5 + 3/5y)b

i did like i said: gather up the values multiplied by a alone and the ones multiplied by b alone

[static cedar]

How come you have the by with the x

[woeful edge]

oh its bx mb

[static cedar]

Ahh okay

[woeful edge]

fixed

so is it all clear so far

[static cedar]

Yess

[woeful edge]

(4/5 + 3/5x)a - bx = ay + (1/5 + 3/5y)b

now from this you will try to deduce a system of equations

the part multiplied by a on the left side of this equation is equal to the part multiplied by a on the right

same thing for b

[static cedar]

So (4/5+3/5x) =y ?

[woeful edge]

yes (js add a parenthesis after 3/5x)