help-17

[faint matrix]

wtf

[hollow thunder]

dont worry about it

[faint matrix]

is there someone who could help with this question?

<@&286206848099549185>

[hollow thunder]

There might be something on khan academy or youtube

thats prob your best bet rn

[faint matrix]

is there anyone in this server?

[hollow thunder]

I would check youtube

[faint matrix]

bruhh

[hollow thunder]

sorry but if midterm is tmrw there arent many options

[drifting jackal]

Your notes would be one place to go to

You need the formulas, and they could be in your notes

[hollow thunder]

idk how it works but if the formulas aren't enough examples might be helpful

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[fallen sandal]

Is it 56

Idek

Well I guess 64 would be at 50% so 56 is 25%

Wait no

70

Cuz the last 25% have more

Than that

[vast shale]

D

[vast shale]

Q3 is 75%

q1 is 25%

Median is 50

So 25% would fall on Q3

[fallen sandal]

Yay

Ty

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[vast shale]

Oh shit

it was B

My brain wasn't working

@fallen sandal

[fallen sandal]

It's D

It was correct

[vast shale]

Oh

[fallen sandal]

(did u see my amazing deductive reasoning too 😎 )

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[barren steppe]

I'm familiar with this type of problem but typically I would be given bounds for the integral

[barren steppe]

Im unsure how to complete it without something like 0 to x

[vast shale]

it is like integral 0 to x

you just add the area manually instead of doing the integral

[barren steppe]

but how is g(0) = 1

[vast shale]

g(0) = 1 means the area under g' up till 0 was 1

[barren steppe]

but where is that on the graph

its a semicircle

so wouldn't it be pi

[dusty ice]

it's the '+c'

[barren steppe]

ohhh

okay

so that sets c to 1

[vast shale]

oh uyeah sorry im stupid

[barren steppe]

and then 0 to 3 is another pi plus one

so its pi plus 2?

is that correct

[vast shale]

prolly

[barren steppe]

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[fathom hedge]

struggling to do this one

initial thoughts was

7 positions CBA to be in

once 3 letters are confirmed

6! remaining

7 * 6! / (2! * 2!)

2! * 2! counting for the repeated A and B

but yeah, not right

this is a given hint which i basically just guessed and don't understand at all why it is what it is

[regal bane]

Are you asking specifically about the "two subword" case?

[fathom hedge]

I am

actually i think is should go back and make sure of somsething

for this question, is this the right working?

or is it straight up just

7 * 6! / (3!)

they both happen to be the same answer here but im not sure about my logic

i just wanna be sure of that first

so like

7 cases

all are basically identical just moving the positiuon of BBC

if you set the position of BBC as the first 3 letters

it would be

3 * 2 * 2 * 6 * 5 * 4 * 3 * 2 * 1 right

or if you set the position of BBC would it just be 1 * 6!

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[barren steppe]

This problem is very unlike others I've done so I'm not sure how to start

[barren steppe]

I don't know if I should do u-sub or just simple FTC

[lime gorge]

Wdym by simple FTC?

I recommend u sub btw

[barren steppe]

I guess I'm just not sure because I've never seen it set equal to a constant like that

like can I just take the derivative of both sides

idk

[lime gorge]

Nah

U sub

[barren steppe]

do I u sub the denominator

[lime gorge]

Try it and find out

[misty belfry]

ignore the right side for now

and solve the left side in terms of k

then look at the right side

[lime gorge]

Collegeboard loves these types of problems

[barren steppe]

ok

so i have

integral from 0 to k

is 1/u (1/2) (du)

[lime gorge]

Well are u gonna change back to x after integrating? If not, change the bounds

Otherwise keep going

[barren steppe]

wdym change the bounds

change them to what

[lime gorge]

U = x^2 + 4 -> U(k) = k^2 + 4

[barren steppe]

oh

so dont you mean keep the bounds

because they are in terms of k

wait

I think I have it

so it simplifies into 1/2 times ln(u)

and if you make k=0 then u = 4

wait

no

it says k >0

hmm

[lime gorge]

1/2ln(u) then sub back in the value of u in terms of x

@barren steppe

[barren steppe]

but how do I find a value of k

[lime gorge]

Dw you’ll get there

[barren steppe]

the x value has to be 0

becuase 0^2 plus 4 matches the ln4 on the other side

x=0

[lime gorge]

Wat

Just sub back in

[barren steppe]

1/2 ln(x^2 +4)

[lime gorge]

Ok

That evaluated from 0 to k = 0.5ln4

Evaluate that from 0 to k

[barren steppe]

what

I'm not understanding

[lime gorge]

Ok lemme write it down

[barren steppe]

OH

OHHHHHHH

[lime gorge]

OHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH YEAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

https://tenor.com/view/dior-pop-smoke-dance-dance-move-gif-16164289

[barren steppe]

ok

so

ln(k^2 +4) -ln(4) = ln(4)

is k = 2?

[lime gorge]

plug in and see if it works

[lime gorge]

into this

[barren steppe]

it does

[lime gorge]

u sure?

[barren steppe]

it does not.

[lime gorge]

simplify this

[barren steppe]

i got

k^2 +4 = 8

am i wrong

[lime gorge]

show me how u got that

[barren steppe]

ln (k^2 + 4) = 2ln4

and then

wait

I feel lame

algebra moment

I'm unsure how to remove both ln's

[lime gorge]

well what did u get after that last step u showed me

[barren steppe]

I don't know

I can't remove the 2

[lime gorge]

$a\ln b = \ln b^a$

[twin meteor]

Stephen

[barren steppe]

k^2 + 4 = 16

the answer is rad 12

[lime gorge]

let the collegeboard judge your response

[barren steppe]

thank you good sir

have a wonderful day, alligator man

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[nimble gorge]

Help on 1

I don't know how to determine the numbers for angle a and b

Can I just count the distance???

[vast shale]

is this all you're given can i see the original question

[nimble gorge]

Yes

[vast shale]

wdym count, using tan^-1 meaning you would know tan

[vast shale]

huh what are the sides?

[nimble gorge]

When I mean count I mean count the distances between each angle

Tan -1 is a calculator function to calculate tan properly

[vast shale]

what did you erase here, i see 6

[nimble gorge]

That's the number I got counting in-between a and b

[vast shale]

if you have tan x you can use tan^-1 to get x

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[nimble gorge]

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[fading valve]

helloooo

[fading valve]

trying to simplify the top

answer given did not reflect mine

there is a dtheta playing with me

[sudden compass]

wait

how does v=T imply that dv=dTdTheta

are you saying v(theta)=T?

cus that would make sense in ur context

[fading valve]

hang on ill fix the notation

[sudden compass]

the problem seems tobe with udv. specifically, dv

I have a feeling its missing something

[fading valve]

im just trying to get line one at the top to be true

[sudden compass]

is it necessary? and is this part of a bigger problem?

[fading valve]

yes, modelling a suspension bridge's tension and wires

[vocal sleet]

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.

[fading valve]

thats not my issue though dude

[sudden compass]

who knows u might be a making a mistake elsewhere

its also possible u were entirely correct everywhere

we dont really know

if you are sure ur correct till this point then its fine

[fading valve]

i'm following someone do it online and that first line is where i get confused

[sudden compass]

Ah I see. have u checked if they rather use what u obtained in their solution after the first line? or do they keep using that first line only?

[fading valve]

they could've made a mistake but idk

[sudden compass]

cus its possible they forgot to write a dtheta down in the first line, then proceeded to use it to solve the rest of the problem

[fading valve]

ill have a look good point

this is what they've done

i redid the calculations and i've still got the same issue

the only things im unsure of is

cos(dtheta) is approximately 1 and sin(dtheta) = dtheta

but i've head of these before and they make sense

if you scroll down to the bottom of this page on maths stackexchange u can find this guy gets the same result too

https://math.stackexchange.com/questions/4259959/deriving-the-shape-of-the-suspension-bridge

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@fading valve Has your question been resolved?

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@fading valve Has your question been resolved?

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[pine dragon]

how do u figure out what bandwidthis most appropriate for data when looking at SAMSE and LSCV

[vast shale]

I don't think this can be easily answered

it depends on your data and you have to be careful with over smoothing and over fitting the data

for SAMSE

I'd say start by trying out a range of different bandwidths

then try to find the bandwidth that minimises the SAMSE criterion

also worth to note that SAMSE is a non parametric way of estimating probability density functions

LSCV is data driven more so

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[dry yacht]

Take your focus on the denominator.

[dapper iris]

Factorise, then u=e^x and partial fractions?

[dry yacht]

Take it step by step. Factorize it first.

[paper depot]

!occupied

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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).

[dapper iris]

Got it 👍

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[hushed cosmos]

anyone know why theyve multiplied (16-10.6) by 4 here to get the period?

[young vortex]

this is the shape of a sinusoidal graph, right ?
notice that you can divide it into four quadrants that are of exactly equal length:
the start to the peak;
the peak to the middle (where it intersects at y = 0);
the middle to the trough;
and the trough to the end

in the graph, the period starts at 10.6 and reaches the peak at 16

[undone pond]

sinusoidal?

[young vortex]

wavy

periodic

it just refers to the same shape in your question

[undone pond]

Can it be referred as: in the shape of a sin wave

[young vortex]

anyways this is normally what periodic graphs look like, and because the distance between the start and the peak is exactly one fourth of the total period, you can calculate it by taking (x at peak - x at start) * 4 = (16 - 10.6) * 4

[young vortex]

yeah

[hushed cosmos]

yah i got it

thanks

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[quick fjord]

what are you having trouble with?

they want to reduce it into the simplest form

[safe shore]

make it as the simpelest form

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[vast shale]

By what number can you devide both 21 and 30 to make them smaller?

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@lament path Has your question been resolved?

[vast shale]

@lament path answer these questions before clicking ❌

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@lament path Has your question been resolved?

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[summer ore]

Hi, can anyone guide me on this?

This is what I did

<@&286206848099549185>

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[analog stag]

i’m having trouble with this question and finding the answer to it

[outer warren]

draw a triangle where cot(A) = 4/9

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[vocal sleet]

Please don't occupy multiple help channels.

[fleet salmon]

,rotate 270

[fleet salmon]

i feel like this is wrong

[near tartan]

(x⁴/4)³ + (8)³

[fleet salmon]

is the other side

right?

[near tartan]

Left side yeah, that's the original question

Now you can apply that formula to it

[fleet salmon]

no lke

the right

[near tartan]

Oh, nah yours doesn't look correct

[fleet salmon]

omg

😭😭

im confused

on how

[near tartan]

Lemme check

[fleet salmon]

to do this

[near tartan]

Well now you know a is x⁴/4 and b is 8

Just have to plug it in the formula

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[covert flume]

hello, im trying to understand this statement, what do these symbols from the sequence mean? :(

[paper depot]

what two symbols

[covert flume]

here at the top, what do the parenthesis (n / i) and the weird z do?

[heavy yoke]

by "weird z" do you mean Σ?

[stiff arch]

the (n i) is n choose i

[paper depot]

the "weird z" is a capital greek letter Sigma and it means summation

and yeah those are binomial coefficients.

[covert flume]

yes!

thank you guys, I can finally look them up, I wasn't googling right

🥹

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[warm atlas]

Hello, I’m having a problem solving the circled problems. I’m wondering how to get to 1/9 in the table shown..

[paper depot]

P(X=0) is the probability that José pays nothing

under what circumstances will José pay nothing? @warm atlas

[warm atlas]

When E occurs

So 1/3

[paper depot]

no, it's when E occurs on both days.

i.e. if he misses the bus both times.

that's when he pays nothing.

[warm atlas]

Ah, kk got it

[paper depot]

he pays 3 euro if he misses the bus one day and catches it the other

and 6 if he catches it both days

[warm atlas]

Kk, I didn’t realise it was about the days. Tyvm

I’ll try it out first then

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[shut hearth]

how to integrate \sqrt{x^{2}-1}

[shut hearth]

@twin meteor

,help

[rugged vortex]

let x = sin(u)

[twin meteor]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[shut hearth]

ohh ok ill try that

smthing like this?

[stiff arch]

you can do this, but it might be easier to apply
$$\sin^2(x)+\cos^2(x)=1$$

[twin meteor]

Cycadellic

[shut hearth]

ok. thanks for the help

.close

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[waxen hawk]

Why A E options are incorrect

[waxen hawk]

IMO the computations are all legit

[raven owl]

It is only true for positive numbers

[waxen hawk]

it basically just put two individual sqrt shits into one sqrt

[waxen hawk]

why

[raven owl]

Try -1 and -1

Nvm give me a sec

Yeah that works

It would imply -1=1

[waxen hawk]

wdym

[raven owl]

What is sqrt(-1*-1)

Obviously its 1

What is sqrt(-1)sqrt(-1)

i*i

-1

But -1≠1

[waxen hawk]

.close

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[open coral]

is Re(1/z) equal to 1/x
if z = x + iy

[outer warren]

how are you getting that?

[open coral]

1/x + iy

so Re(1/z) is

[outer warren]

1/x + iy
how are yo getting that

[open coral]

z is a complex number that is (x + iy)

so 1/z

is 1/(x+iy)

but idk what the real part of that is meant to be

[outer warren]

realise the denominator

(using conjugates)

[open coral]

ohhhhhhhhh yes

completely forgot about that

ok so now I got (x-iy) / (x^2+y^2) < 1/2
(thats the whole question)

wait i only need the real parts of the left side

holup

x/(x^2+y^2) < 1/2

x < (x^2 + y^2)/2

and then y^2 > x(2+x)

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[unborn quail]

Why am I not getting 0005T

[unborn quail]

,rotate

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[sleek umbra]

Find all functions f such that:
f(x)f(y) = f(x+y) + xy
for all real x and y.

[cobalt ocean]

seems difficult

!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

@sleek umbra Has your question been resolved?

[sleek umbra]

I don't know how to begin

[hard atlas]

early on its good to just start trying some values for x and y and see what you get

for example x=0 or x=1 is usually a good start

[cobalt ocean]

i have a nontrivial example of a function that works

psyche nvm i misread

do any of these even exist

[sleek umbra]

Idk but it just wants me to find all functions

[vast shale]

we can find f(0) by putting x=0 but that's about all i have rn lol

another observation, f(-1)f(1) = f(0)-1 = 0 so f(1) = 0 or f(-1) = 0

ah i got it

[vast shale]

in each case here plugging y = -1 or y = 1 yields one function. you can check both the functions do indeed work

and of course the trivial f(x) = 0
edit: never mind

[cobalt ocean]

no, f(x) = 0 doesn't work

you would have 0 = xy for all x, y

[vast shale]

oh oop

the other 2 work

[cobalt ocean]

yes

[vast shale]

i think that should be it

[cobalt ocean]

ok yeah

let f(1) = a, then you can get f(0.5), f(0.25), f(0.125), ...

and therefore f(1.5), f(1.25), f(1.125), ...

i think

[vast shale]

we also have f(-1)

which must be 0 if a is not 0

[cobalt ocean]

yes

[vast shale]

^^ can just split into cases f(1) = 0 and f(-1) = 0 and in each case a general formula for the function can be found easily

[cobalt ocean]

it's not completely trivial to me

like with f(1) = a you can get it for all natural numbers easily

0.5, 0.25, 0.125 a little more effort

and then the other rationals seem like a bit of a pain maaaybe

[vast shale]

i mean f(1) = 0: put y = 1
f(-1) = 0: put y = -1

[sleek umbra]

I have been trying to put in x = 0 etc but I have found nothing so far

[cobalt ocean]

and then the irrationals are like completely unhooked

[vast shale]

does this not work? this accounts for all cases right since f(1)f(-1)=0

[cobalt ocean]

it does

wait

wdym

so what is the function

like let's take the first case

[vast shale]

1-x and 1+x

[cobalt ocean]

let f(1) = a, let f(-1) = 0

[sleek umbra]

I dont think we can assume anything

[vast shale]

we haven't assumed anything

[cobalt ocean]

then we have f(2) = f(1)f(1) - 1 = a^2 -1

f(3) = f(1)f(2) - 2 = a^3 - a - 2

etc.

?

ah but we do have f(x)f(-x) = 1 - x^2 which restricts it a lot

[vast shale]

my solution

[cobalt ocean]

oh i'm an idiot

ok ok ok ok

[vast shale]

@sleek umbra

[sleek umbra]

Yeah i understand now

Thank you guys so much

Is that all though?

[vast shale]

happy to help

we haven't made any assumptions and we narrowed it down to those 2

just check those 2, and that should be all

[sleek umbra]

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[stoic knoll]

Hey, i got this task:
If m^2 + n^2 can be divided by 3, proof that m and n can also be divided by 3. Here is my start but i dont know how i can proof that it can be divided by 3

[stoic knoll]

m, n ∈ N

[cobalt ocean]

not the best start

[stoic knoll]

i also tried binomical formula, i dont know if thats the correct english word for it

i had m^2 + 2mn + n^2 = 3k + 2mn

but then i end up with root(3k - 2mn) which doesnt help me further

[cobalt ocean]

generally with this stuff you don't want to involve the square root straight away

[desert hornet]

you will need to use modular arithmetic for this

[cobalt ocean]

ok basically here's one way to approach it

every number n is either of the form 3k, 3k+1 or 3k+1

so like

3 is 3 * 1
4 is 3 * 1 + 1
5 is 3 * 1 + 2
6 is 3 * 2

etc.

[stoic knoll]

Yea

[cobalt ocean]

so you can consider like

let m = 3a + 1, n = 3b + 2 for example

and then square it

so like

m^2 + n^2 = (3a + 1)^2 + (3b + 2)^2

so it's like, when can this divide by 3

or like if m = 3a + 2, n = 3b

m^2 + n^2 = (3a + 2)^2 + (3b)^2

when can this ever divide by 3

[stoic knoll]

uhmm

[cobalt ocean]

try expanding those things out

[stoic knoll]

before having a deeper look in this, we used a different scheme for the previous task (if 100m + n can be divided by 7, then m+4n can also be divided by 7), thats where i got the start from. Ill post a picture. Can you explain why exactly this start is bad for this new task then ?

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[scarlet bough]

How do I work out the uncertainty?

[scarlet bough]

The resolution of a vernier caliper is 0.01cm so u do 1/2 resolution to work out uncertainty which is 0.005cm

Ooohhhh thanks

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[outer warren]

!da2a

[vocal sleet]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

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[tidal dock]

in geometry or statistics

median is the value right in the middle of a dataset

just... look at the middle of your dataset

take the mean of the middle 2 values

yes

if you have 1 2 3 4 5 6 then your median is (3+4)/2

=3.5

write it down like a list

132, 132 (two girls), 134, 136, 136, 136 etc.

rounding up i think

yes

but like, a complete one

data table is also a good word for it

look at the middle 2 values

yes

yeah median is 138

are you sure he did it on the same table?

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[cinder cobalt]

i just have no idea what is the difference between getting absolute maxima of open interval, i can get only in closed interval

[cinder cobalt]

like in closed interval we find critical points, then we put critical point, left endpoint, right endpoint in the normal function and get the abs max and abs min

but what we do in intervals like this

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@cinder cobalt Has your question been resolved?

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@cinder cobalt Has your question been resolved?

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[mild carbon]

can someone please explain these rules to me?

I don't really understand them, it doesn't make sense

[pale perch]

theyve just chosen an interesting way to write it, but the red underlines are the common factors

they then pull those values out

in the same way as the one in the box

[mild carbon]

Oh wait im stupid

i was interpereting the rules wrong

thanks anyways : )

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[stiff light]

My task is to simplify but theres too much going on

[desert hornet]

go term by term

shouldn't be too hard

[stiff light]

What do you mean

[desert hornet]

well

the first term is $8x^5y^3\cdot 2.5$

[stiff light]

Theres like almost nothing i can do

[twin meteor]

kheerii

[desert hornet]

can u think of how to simplify this?

[desert hornet]

sure there is

[stiff light]

No

[desert hornet]

can't you just multiply the numbers?

$8x^5y^3\cdot 2.5 = (8\cdot 2.5)x^5y^3$

[twin meteor]

kheerii

[stiff light]

Can i one has variables thats why i cant no?

[desert hornet]

it doesn't matter if there are variables, the basic rules of arithmetic and exponents still apply

[stiff light]

So i can just make a 20 out of the 8

[desert hornet]

out of the entire first term, yes

now try to do something about this term

expand the bracket first of all

[stiff light]

Huh

[desert hornet]

how can you expand $(-3y)^3$?

[twin meteor]

kheerii

[stiff light]

This ones gonna make a -6x³y⁵

[desert hornet]

not quite

notice that the 3 is also under the power

$2x^2\cdot -3y^3 \ne 2x^2\cdot(-3y)^3$

[stiff light]

What

[twin meteor]

kheerii

[desert hornet]

what you said would be correct if it were written as the first expression

[stiff light]

-1x²y³

I dont know really

[desert hornet]

how did you arrive at this?

[stiff light]

I dont know

[stiff light]

I dont understand what ur trying to say

[desert hornet]

you multiplied 2 and -3 to get the -6 yes?

[stiff light]

Yes

[desert hornet]

and then multiplied x^2 and y^3

what i'm trying to say is that $-3y^3$ and $(-3y)^3$ are NOT the same.

[twin meteor]

kheerii

[desert hornet]

can u see the difference?

[stiff light]

OH

im blind

Explains alot so

First the -3y times 2x³y²

Right?

Thatll be

[desert hornet]

no, first you have to cube the (-3y)

[stiff light]

y times y times y

Y³

Still makes me get the same thing

-6x²y⁵

[desert hornet]

again, you need to do the same thing with the -3

[stiff light]

Oh mb

[desert hornet]

$(-3y)^3=(-3y)\cdot(-3y)\cdot(-3y)$

[twin meteor]

kheerii

[stiff light]

-27y³

[desert hornet]

yes'

now multiply that with the 2x^2

[stiff light]

My brain is on afk mode

-44x³y⁵

@desert hornet

[desert hornet]

no

[stiff light]

What

[desert hornet]

27x2=54, not 44

[stiff light]

Sorry my brain wont work

54x³y⁵

[desert hornet]

yes

now expand the final term

use the fact that $a(b+c)=ab+ac$

[twin meteor]

kheerii

[stiff light]

At this point im wasting my time

Our teacher gave us such an annoying term dude

[vocal sleet]

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[junior pollen]

I think my old tab got closed but in summary would someone please help me and expain to me how I was firgure out the asnwers to these questions. They are physcis question s but not very ahrd as I amm still 13

[vocal sleet]

Please don't occupy multiple help channels.

[junior pollen]

but can somone please help if u have the tme

[pale perch]

we cant help since you havent told us what the questions are

[junior pollen]

number 2 was teachers working out but I make no sense of it

and for the last question I am only sure about a and b becuase I don;t know for b wethear I should say 100 or say 99 and estimmate and a should I just say nothing

no pressure

sorry forgot to upload thougfht I did

[vast shale]

i may be able to help

for the seconf pic

area is 0.005 msq

atm pressure is 0.7

you have to conver it into pascals

so 0.7x100000 is 70,000 pa

now just add the values into p=f/a

so its

70000 = f/0.005

then

70000x0.005 = f

350 = f

[junior pollen]

then force is how much it weight so 350 kilos?

[vast shale]

so force is 350N

[vast shale]

nono

[junior pollen]

damn I thought if it was 350 n it woudl wegh 350 kilos

[vast shale]

si unit of force is newton

[vast shale]

nope

[junior pollen]

yes and weight is meauresed in n?

and mass is measured in k

kilos

or completey wrong

[vast shale]

weight in physics is a vector quantity

[junior pollen]

? is the answer to the question 350 n

[vast shale]

oh wait

yes

350 N

not 350 kg

because they arent asking mass

[junior pollen]

thank u

[vast shale]

lemme try to solve the other question too

why does it look like a

[junior pollen]

thanks

[vast shale]

wait is the whole barometer 76cm?

[junior pollen]

ahhhhh

no but from what I get they were both started at 0.0

[vast shale]

aight i got the answers

a) the pressure at A is zero.

b) the pressure at C is also 100kPa

c) Point E is under the greatest pressure.

https://www.miniphysics.com/barometer.html

you can also refer to this article for more explanation.

[junior pollen]

ah for the third one I only ened help with a and b becuase for a I thought what u though adn for b I didn;t know wethear 100 or slightly less

I understand that bit about lower down I said

[vast shale]

c will also have pressure of 100kpa

[junior pollen]

but its sightrly higher

[vast shale]

no its same

mercury is just very liquidy

[junior pollen]

curved line one is higher

also do u measure from top or bottom of miniscus

[vast shale]

what

wdym

alright

for b

you have to write c is 101325 Pa

because if you calculate

p = dgh

p is the pressure

d is density

g is gravity

and h is height

and density of mercury is 13.6 x 10^3 kg/m^3

and gravity is 9.8 m/s^2

and pressure is 101325 Pa

then height will be 76 cm

that is the height of mercury in the barometer

so yeah

c will be 101325 Pa or 101.325 kPa

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[peak gust]

.

$Exercise 20 we consider in Z the following two parts:
A = {x∈ Z such that; (4x²-4x+10/ 2x-1)∈ Z} and B={x∈ Z such that; (x+10/x-5) }
1)a) show that (∀x∈ Z-{5}); (x+10/x-5)=1+(15/x-5)
1)b) show that (∀x∈ Z); (4x²-4x+10/ 2x-1) =2x-1+( 9/ 2x-1)
2) determine: A;
B; A-B; B-A and
AΔB in extension$

[twin meteor]

Swedish enjoyer

[peak gust]

$Exercise 20 we consider in Z the following two parts:

A = {x∈ Z such that; (4x²-4x+10/ 2x-1)∈ Z} and B={x∈ Z such that; (x+10/x-5) }

1)a) show that (∀x∈ Z-{5}); (x+10/x-5)=1+(15/x-5)

1)b) show that (∀x∈ Z) ; (4x²-4x+10/ 2x-1) =2x-1+( 9/ 2x-1)

2. determine: A;
   B; A-B ; B-A and
   AΔB in extension$

[twin meteor]

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

[peak gust]

$Exercise 20 we consider in Z the following two parts:
A = {x∈ Z such that; (4x²-4x+10{fraq} 2x-1)∈ Z} and B={x∈ Z such that; (x+10{fraq} x-5) }
1)a) show that (∀x∈ Z-{5}); (x+10{fraq} x-5)=1+(15{fraq} x-5)
1)b) show that (∀x∈ Z); (4x²-4x+10{fraq} 2x-1) =2x-1+( 9{fraq} 2x-1)
2) determine: A;
B; A-B; B-A and
AΔB in extension$

[twin meteor]

Swedish enjoyer

[peak gust]

.close

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[fair vault]

I'm trying to solve the inequality 2/x-5<10, but my question is why does the direction of the inequality change for the second case?

[fair vault]

I understand the divide/multiply rule by word, I'm just wondering why 2<10(x-5) of Case 1 doesn't flip when 2>10(x-5) of Case 2 does

[split wind]

consider x-5 is positive or negative when x<5

[fair vault]

Can you please elaborate?

[split wind]

yes, when x<5, will x-5 be a positive number or a negative number?

if you don't know, just plug in some cases

[fair vault]

Would it be negative because in the graph it is below the x axis?

[split wind]

e.g. when x=4

that's a good way to spot out

[fair vault]

Ok that makess so much sense

I think that's all I need thank you

.close

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[old niche]

Prove that the function f : N^2 → N+ defined by f (a, b) = 2^a(2b + 1) for all (a, b) ∈ N2 is a
bijection, where N+ = {n ∈ N | n ⩾ 1}; you should not use any facts about prime numbers in
your proof.

[old niche]

im ngl, i have been staring at this for a long time

i think i will need some direction

[river minnow]

That doesn't look quite bijective

[lyric relic]

its surjective because every natural can be written like an odd numeber time an even number

[lyric relic]

it does!

[river minnow]

For which (a, b) is f = 1?

Ah, it's 2^a

[old niche]

sorry

Thats a mistake on my end

[lyric relic]

yeah

for injective part i would use the foundamental therem of aritmetic

and finda that if f(a,b)=f(c,d) then a=c

[old niche]

can we go through it

[lyric relic]

$2^a(2b+1)=2^c(2d+1)$ now $(2b+1)$ can't have 2 factors in (same for $2d+1$) hence by foundamental therem of aritmetic a=c

[twin meteor]

everg

[lyric relic]

so we get 2b+1=2d+1 iff 2b=2d iff b=d

done

[old niche]

sorry i am just trying to digest this

[old niche]

why cant it have 2 factors

[magic wasp]

I think he meant factors of 2

aka it's odd

[old niche]

why is that important

[magic wasp]

Because then 2^c does not divide 2b+1

[old niche]

also why is that necessary

[magic wasp]

Do you know the fundamental theorem of arithmetic?

[old niche]

not really

[magic wasp]

Hm. I don't know if that counts as a "fact about prime numbers" but the theorem states that every integer (> 1) has a unique prime factorization

So for example if n = 2^a * 3^b * 5^c, then a, b, and c have a unique value and there are no other primes that divide n

Now you have an equation like PQ = RS where P = 2^a, Q = 2b+1, R = 2^c and S = 2d+1

You know that Q and S cannot have a 2 in their prime factorizations

So you can split: P=R and Q=S

[old niche]

I see

do you mind if we talk about the surjective case?

i might want to think about the injective case a bit more

but i want to understand how you use induction for the surjection

[magic wasp]

You need to use induction?

[old niche]

yeah this is what the question says

[magic wasp]

I don't have the answer but we can try

2^a(2b + 1) = 1 when a = b = 0

If you can form n from 2^a(2b + 1), can you form n+1?

[old niche]

i guess that is the proof by induction right lol

[magic wasp]

Actually it should be easier to form n+2

You can just do two inductions, one for the odd numbers, one for the even numbers

[vocal sleet]

@old niche Has your question been resolved?

[old niche]

can i do the proof and show it here

[cedar burrow]

Of course

Even one was tough

I got it now though

@old niche tell me if you need a hint

[old niche]

yeah i am struggling

[cedar burrow]

With the even ones?

[old niche]

yes

[cedar burrow]

On the number line, what space is between every 2^a(2b+1) if a=1?

[magic wasp]

I might be missing something here, it looks pretty easy to me

[old niche]

im like just bad at number theorey

or baby number theory idk

[cedar burrow]

For the even numbers?

[magic wasp]

It's the same for odd and even numbers I don't know what you mean

You just start from 1 or 2 but the induction is the same

[old niche]

maybe i should just not do this question yet

[cedar burrow]

Increasing a by one increases the number by 2^a(2b+1), not two

@magic wasp are you sure this is easy? The solution I came up with is rather complex

[magic wasp]

Yeah you're right I'm probably missing something

[old niche]

i am so confused

[cedar burrow]

From a=1 we can get every other even number

Cause 2(2b+1)=4b+2

So increasing b by one adds 4 to the number

That leaves us with half the even numbers missing

How do we get the other even numbers?

And I just noticed that I’m not sure my proof is complete cause I did it the other way around

So let me just type out what I’ve got

@old niche if you want to figure it out yourself don’t read what I type now, although I can’t say if it’ll even prove anything

Take any number with a>=2

This is next to a number with n=1

The relationship is

2^a(2c+1)+2 = 2(2b+1)

2( 2*(2^(a-2)(2c+1)) +1) = 2(2b+1)

b = 2^(a-2)(2c+1)

This is so confusing

Oh hell nah

I think this is another inductive proof

Bro there needs to be a simpler way to do this

@old niche are you sure we must do this with induction?

[old niche]

Yes

[cedar burrow]

@magic wasp

Help

Pls

Oh I think I might have it

Nvm it wouldn’t be inductive

[magic wasp]

Trying to write it up...

[cedar burrow]

You got the solution?

[magic wasp]

Pretty sure

[cedar burrow]

Damn nice

[cedar burrow]

And it’s inductive?

[magic wasp]

Kinda

Ok so

$[\exists k >=0: n = 2^{k+1}(2b+1)] \implies n+2^k = 2^k (2(b+1)+1)$

[twin meteor]

Nel

[magic wasp]

Do we agree with this

[cedar burrow]

Uh

I’m not sure

[magic wasp]

$2^{k+1}(2b+1) + 2^k = 4b2^k + 2*2^k + 2^k = 2^k(4b + 2 + 1) = 2^k (2(b+1)+1)$

[twin meteor]

Nel

[cedar burrow]

What does that have to do with this tho

Oh I see it

Yeah okay

👍

[magic wasp]

You know what I'm not sure this works

[cedar burrow]

I’m not surprised 😂

This is way more difficult than it seems at first glance

[magic wasp]

I'm an idiot I did it in reverse

[cedar burrow]

Hahahaha

That happened to me too

[magic wasp]

$[\exists k >=0: n = 2^k(2b+1)] \implies n+2^{k+1} = 2^k (2(b+1)+1)$

[twin meteor]

Nel

[magic wasp]

$n+2^{k+1} = 2^k(2b+1) + 2^{k+1} = 2^k(2b+1+2) = 2^k(2(b+1)+1)$

[twin meteor]

Nel

[magic wasp]

Yeah?

[cedar burrow]

Yes

[magic wasp]

Do we also agree that all n of the form 2^a can be made by f(a,0)

[cedar burrow]

Yeah

[magic wasp]

Well there you go, an infinite number of base cases and a general induction step

I guess I should show that this covers all integers

[cedar burrow]

Yeah

That would be good

[magic wasp]

Left as an exercise to the reader

[cedar burrow]

And it’s actually the only genuinely difficult part 😅

[magic wasp]

Is it though?

All integers are either not a multiple of 2, or a multiple of 2 but not of 4, or a multiple of 4 but not of 8, ...

[cedar burrow]

Prime factors

That’s not allowed

[magic wasp]

Hm it's only 2

We just want to show that for all n, n = 2b+1 or n = 2(2b+1) + 4c or n = 2^2(2b+1) + 2^3c and so on

[lyric relic]

how is going ?

[cedar burrow]

Bad

[lyric relic]

😦

[cedar burrow]

Induction sucks

[old niche]

i am still stuck

[cedar burrow]

So are we

We have something

But we’re unsure whether our reasoning is allowed

Cause it kinda references prime factors

[magic wasp]

Actually this is more like n = 1+2k or n = 2+4k or n = 2^c+2^{c+1}k

[cedar burrow]

Can you prove that that covers all integers?

[magic wasp]

n = 2^c + 2k2^c

n = (1+2k)2^c

I'm a genius in circular reasoning

[lyric relic]

oo wait

we can use that every natural number could be written uniquely with binary notation

[magic wasp]

That's equivalent to 2^a(2b+1)

[lyric relic]

then we just have to prove that $2^{c+1}k+2^c$ can be every binary number

[twin meteor]

everg

[cedar burrow]

That’s a very nice idea

[old niche]

What is binary notation

[lyric relic]

every binary number is of the form e.g. $10010110000$

[twin meteor]

everg

[old niche]

uh

what

ik what that is bc im a cs major but

why would u need to use that

[cedar burrow]

Neat multiplication rules

2*1 ⇒ 10

[lyric relic]

we can se it as $[100101 ]_2*2^4+2^3$

[magic wasp]

Still equivalent to 2^a(2b+1)

[twin meteor]

everg

[lyric relic]

indeed this is of that form

so c counts how many zeros there are before the first 1 ... and k is the number after those zeros

[cedar burrow]

We will have to use induction though. Which I think will complicate things

[lyric relic]

the most natural way is by using fundamental arithmetic theorem...but if you don t want to do this prime based argument you can also do this (but i think behind the proof of "every number can be written in binary" there is some prime based argument )

this last argument is without induction

[cedar burrow]

We must use induction though

It’s part of the task

[cedar burrow]

.

[lyric relic]

(p.s. this channel has been open open for about 3 hours lol)

[cedar burrow]

Dang

At first I kind of liked induction

But these tasks are getting out of hand

For one induction exercise today I needed 6 hours or so

And for this one there’s like 5 different solutions we’ve found so far

And none of them count cause they don’t use induction

Or make use of prime factorization

[cedar burrow]

@magic wasp my feeling was right though. Still proud of that 🗿

[lyric relic]

ohh..wait i get it

only surjective with induction

hint: even number+1=odd umber

and odd number +1=even number

[magic wasp]

Damn I didn't think of that

[cedar burrow]

Huhhh

I don’t get it

[lyric relic]

so if we can write every number less than n in the form 2^a(2b+1)

now what about n+1

$n+1=2^a(2b+1)+1=2^0(2^a(2b+1)+1)$

[twin meteor]

everg

[lyric relic]

in the case that n=2^a(2b+1) with a>0

if a=0

$n+1=2^a(2b+1)+1=(2b+1)+1=2b+2=2(b+1)$

[twin meteor]

everg

[lyric relic]

but now b+1<=n (because n=2b+1)

so b+1 can be writes as 2^k(2h+1)

hence

[magic wasp]

No?

[lyric relic]

$n+1=2^a(2b+1)+1=(2b+1)+1=2b+2=2(b+1)=2^{k+1}(2h+1)$

[twin meteor]

everg

[lyric relic]

by induction

[cedar burrow]

But like

[old niche]

oh

[cedar burrow]

That’s all for the odd numbers right?

[old niche]

0 i s a naturay number in my course btw

[lyric relic]

this is for even number (a>0)

[lyric relic]

and the last argumnt for odd number

[cedar burrow]

Oh I think I get it

[lyric relic]

p.s. i must need a>0 because i want to factor a 2 in this thing

[cedar burrow]

But wait

Doesn’t that only prove the odd numbers?

[lyric relic]

did you ask me that this proves only when n+1 is odd ?

[cedar burrow]

Yeah

I mean it must if a>0

[lyric relic]

this is for a=0

i.e. n is odd i.e. n+1 is even

[magic wasp]

Huh ok the b+1 < n part is clever

[cedar burrow]

Yeah that’s good

[lyric relic]

yeah ... i was blocked for a sec...then i think "ok lets finally use induction"

[lyric relic]

thx

[cedar burrow]

Thanks for solving it for us

I would have never gotten that

[lyric relic]

thanks you for check it 😉

[cedar burrow]

Learned something new I suppose