help-0

[sleek briar]

pattern with 3

any help?

dont exactly notice a pattern

[cunning coyote]

that looks off

i think

[sleek briar]

where are my mistakes?

[merry iris]

i think you just colored it wrong

[sleek briar]

which one

[cunning coyote]

when you color D out fully, you realize that you skipped out on a couple C's

oops wait

the first one i overdid

the coloring* was correct the first time

[sleek briar]

How is it correct if you just overlaid it?

[cunning coyote]

it isnt

im just making it more apparent

this is something like CDCCC CDCCC DCC

[sleek briar]

is this correct?

[cunning coyote]

yea that looks right

do you notice any patterns

[sleek briar]

no

thats why im kinda stuck

[cunning coyote]

well, one thing to notice is that D is always on the left side

that is because each D has 4 C's between them, which you could think of as a full rotation

of course, the direction of the spiral alternates each time

[sleek briar]

But like i think c is asking the area

[cunning coyote]

yes

i mean, if you really wanted to, you could just draw out all 10 of them and count the area

but thats obviously not very efficient

[sleek briar]

you could but also for (d) i think you need to know a pattern

[cunning coyote]

right

so lets just try to analyze this as best as we can

now.. another thing i notice is that each red line is 2 squares apart

if you just look at the vertical

[sleek briar]

do you think for (c), i should just draw the 10 out manually or actually find a pattern

[cunning coyote]

i think you should find a pattern

woah

theres a big dragonfly

on my desk

anyway

okay look at the right side

do you see how it only increments by 1 blue line?

every time we do CDCCC, the left side increases in width by 2 and the right side increases by 1

[sleek briar]

yes thats true

maybe lets look at up and down. is it just increase 1 each time right?

[cunning coyote]

looks like it

[sleek briar]

ok i think thats a great strategy

[cunning coyote]

its important to note those 2 yellow lines at the beginning, they add an extra bit of width

[sleek briar]

That doesn't change anyways

lets look at (d) now

[cunning coyote]

yes that part is constant

okay

it says it must end with CCC, is that in the definition somewhere or did they just mention this now?

am i missing any info?

[sleek briar]

i'll repost everything here for your reference

[cunning coyote]

okay

ill use the bathroom rq, brb

[sleek briar]

I think i just missed 2 pages before

[cunning coyote]

thank you

okay i think we should try to generalize our results from c

hmm

[sleek briar]

It doesnt actually necessarily say the sequence is a repetition of CDCCC

But the maze does have to end in CCC

[cunning coyote]

yes but we should generalize how C and D effects the area

affects?

whatever same thing

whenever C is instructed, the dimensions of the rectangular region increase by 1 unit on the "next" side

since C moves the robot around it

when D is instructed the dimensions increase by 1 unit on the same side

[sleek briar]

is this necessarily true or just for the sequence of CDCCC

[cunning coyote]

i believe it is true in general

C always moves to the next side

by next side i just mean a side adjacent to the previous side

so lets say the area of our rectangular region was a*b and we were on the a side

instructing C thereafter would give us an area of a*(b+1)

[sleek briar]

i think we should first test with some examples

[cunning coyote]

instructing D would give us an area of (a+1)*b

[cunning coyote]

sure

[sleek briar]

i need to go to toilet now. if you can try test some and tell me on your findings

[cunning coyote]

alright

[cunning coyote]

i think i got these backwards

it would be (a+1)*b for C and a*(b+1) for D

uhh

[sleek briar]

im back

so basically do you mean by the multiplication

[cunning coyote]

its hard to explain

this is some previous example

[sleek briar]

i sitll dont understand. because technically there are both cs and ds

[cunning coyote]

i mean to draw it the other way

mb

anyway

you see how you get that 1 unit of width after applying C?

[sleek briar]

it depends

yes C definitely has to move 1 before turning, D has to straightaway turn

[cunning coyote]

oen second

[sleek briar]

?

[cunning coyote]

okay

whenever you apply C it increases in the same direction whereas when you apply D it increases in the orthogonal direction

[sleek briar]

yes so we know blue adds one left/right and red adds one left/right

is that correct understnading

what is orthogonal direction

[cunning coyote]

orthogonal means perpendicular

so C actually does two things

it adds a unit

AND it shifts the side

D never shifts the side

[sleek briar]

doesnt D add an unit right/left

[cunning coyote]

if you keep applying D you will always increase the same side by 1

[sleek briar]

wdym

[cunning coyote]

but if you apply C you will alternate between increasing one side and increasing the other side

[cunning coyote]

see how it always increases that side

it never deviates from going in the direction its already going in

[sleek briar]

ok so what do you mean by shifting a side

[cunning coyote]

so lets say your current area is ab

and lets say applying C once gets you to (a+1)b

then applying it again will get you to (a+1)(b+1)

applying it once more will get you to (a+2)(b+1)

[sleek briar]

is a up/down or left/right

[cunning coyote]

and so on

[cunning coyote]

it swaps between the two. which is which in any particular moment is not really relevant (or at least im ignoring that for now and just thinking about it abstractly)

if you did the same thing with D you would get (a+1)b, (a+2)b, (a+3)b,... and so on

so the other dimension will never increment until you apply C again

[sleek briar]

i think you still need to choose a side so it stabilises the pattern

[cunning coyote]

does this make sense at least

[sleek briar]

yeah kind of

[cunning coyote]

okay i will try to explain more concretely now. so we always start with just the X in the middle, so our default area is 1. lets say that our current width is x and current height is y

or

idk maybe a and b was fine

w and h?

[sleek briar]

a and b, x and y are interchangeable but choose one

maybe w and h is better

[cunning coyote]

hmmm

okay lets do w and h

so when you apply D first it increases in h

[cunning coyote]

as we see here

[sleek briar]

are we working on (c) or (d) for these cases?

[cunning coyote]

im going to work on (d)

i think (c) is mostly solved already

[sleek briar]

C is already concreted. you can always actually draw it out to verify

[cunning coyote]

okay so it wants the number of mazes that cover an area of 25..

and it always ends in CCC

we can do a simplifcation then

i think

[sleek briar]

It has to end in CCC

That means that the last part is constant

[cunning coyote]

exactly

so we could actually just subtract that from our area i think

well

so it would be a 5x5 obviously

[sleek briar]

How do you know its definitely 5*5

[cunning coyote]

then undo 3 C's

[sleek briar]

oh yeah true i didnt even notice that

so we know its always rectangular right?

[cunning coyote]

yea

so okay if we undid the last 3 C's that would be something like 3x4

(or perhaps 4x3, but i dont think the difference really matters)

idk if this simplification is helpful but now we can just consider all the possible sequences

we no longer have the restriction that the last three have to be C

[cunning coyote]

okay wait maybe it does matter actually lol

um..

okay well a 4x3 is easy to get if you just did like 5 C's in a row

it starts out as a 1x1, apply C gives 2x1, apply C again gives 2x2, so on

[sleek briar]

Maybe we can try with the most Cs and decrease from there?

[cunning coyote]

wait a minute

[sleek briar]

honestly i dont even know if the order of w and h doesnt matter

[cunning coyote]

um

i think that might be the only possibility

or im crazy

wait i think im actually stupid

we had to find a square, right?

but whenever you apply D

it increments the same side

oh

but i guess you could just apply D again on the opposite side

ohhh

yeah okay so you could do

CDCDC

i think

[sleek briar]

is that the only possibility

because what if just did all Cs

and cut down on Cs from there?

[cunning coyote]

1x1 -> 2x1 -> 3x1 -> 3x2 -> 3x3 -> 4x3

[sleek briar]

honestly i need a refresher on what C and D does. can you write it in a more clear way so i am able to better understand it?

[cunning coyote]

let me think for a sec

i feel liek the number of possibilities is really small

like maybe 2 or 4

given the constraints

[sleek briar]

but can you still give a refresher on the properties of C and D, writte n in one message so i can understand it?

thanks

[cunning coyote]

okay i think its important to remember if we are currently on a horizontal or vertical line

u know what ill just write it all out

if horizontal, then C: a x b -> (a+1) x b and horizontal -> vertical
if vertical, then C: a x b -> a x (b+1) and vertical -> horizontal

if horizontal, then D: a x b -> a x (b+1) and horizontal -> horizontal
if vertical, then D: a x b -> (a+1) x b and vertical -> vertical

kind of making up notation on the fly

i would probably be more helpful if i actually knew what i was doing lol

[sleek briar]

i dont really think we need notation

[cunning coyote]

does this make sense

a is horizontal length and b is vertical length

[sleek briar]

Also in regards of width and height, it might matter as it starts like this

i think w h is the best

[cunning coyote]

it starts as horizontal

so the default is 1x1 and horizontal

[sleek briar]

depends if it starts with C or D

• after that

[cunning coyote]

yea if it starts with C it becomes 2x1 and vertical

or

uh

yea

i defined it incorrectly i think

oops

[sleek briar]

what is the correct C or D

[cunning coyote]

okay i fixed it

or

i didnt fix it

UGH

okay

[cunning coyote]

NOW it should be good. i think. hopefully.

because D adds in the perpendicular direction and C adds in the same direction

i think i forgot what the problem was about at this point

it is 5am...

my sleep schedule is so cooked

[sleek briar]

you woke at 5am or havent slept yet

bc i feel for you

[cunning coyote]

havent slept yet

[sleek briar]

if you want get someone else to carry on

but like its 7pm for me now so i've still got a lot of time to go

[cunning coyote]

i feel like we were getting somewhere

cuz the number of possibilties feels pretty low

[sleek briar]

but we need to first define c and d

and how they affect and i think you might have gotten confused based on what u said

[cunning coyote]

okay so the last state either needs to be 3x4 and horizontal or 4x3 and vertical

so that when you apply C three times

it results in 5x5

[sleek briar]

43 and 34 are different

bruh why does discord always change to italics if i use *

[cunning coyote]

dunno

wait you know what i realized i dont even need to keep the vertical / horizontal information i could just swap the variables instead

[sleek briar]

anyways i was using i think the q tells you 3x4 and 4x3 are two different cases, meaning more cases to be found

yeh but it starts with and you cannot turn it around. hence if you turn the grid. you turn the starting

[cunning coyote]

hm

okay i think im going to take advantage of symmetry here

i really want to simplify this

i might be overengineering the problem

theres probably a much simpler way to do this...

C: a x b -> (b+1) x a
D: a x b -> (a+1) x b

let me check if that makes sense

C(1x1) = 2x1
D(2x1) = 3x1
C(3x1) = 2x3
D(2x3) = 3x3
C(3x3) = 4x3

[cunning coyote]

that matches what i did here

the only difference is now the width and height are like

ambiguous

well you could probably deduce which is which based off the starting point

but the point is that we shouldnt care which is which! we only want the area

as long as the operations still give the same result then i think its fine

[cunning coyote]

this is starting to look like a transformation lol, maybe i could represent axb as just a vector

or like a multivariable function

C(x,y) = (y+1, x)
D(x,y) = (x+1, y)

ive definitely lost my mind

[cunning coyote]

okay so based off of this definition we are gonna try to find all the combinations of C and D that lead to 4x3, starting from 1x1

and i KNOW that its 4x3 because CCC(4x3) = 5x5

3x4 would lead to 6x4

[lone heart]

@sleek briar Has your question been resolved?

[cunning coyote]

the only combos i can think of are CCCCC, CDCDC, DCDCD, DCDCC, CCDCD, DDCDD

you definitely cant have more than 2 D's in a row, otherwise itll go too far

i think 6 is the max

which feels right somehow, though idk how i would prove it

alright, i think its time i slept. hope i helped a little

goodnight

[lone heart]

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[sharp stag]

I took ln on both sides and made rhs fraction into 3 things multiplied

seem like the right idea?

[vale wigeon]

kinda but your wording looks off. can you show your work thus far?

[sharp stag]

sure

ln(f(x)) = ln[(x^8) * (x-9)^7 * (x^2 + 5)^-8]

[charred jewel]

are you sure its multiply?

[sharp stag]

plan was to then use this

[sharp stag]

uh, idk I think so

I just moved the denominator up

[zinc bolt]

Please use latex if you know how to

[zinc bolt]

$\ln(f(x)) = \ln(x^8 \cdot(x - 9)^7 \cdot (x^2 + 5)^{-8})$

[ocean seal]

@zinc bolt

[charred jewel]

i think you got the log rules confused

[sharp stag]

maybe, idk what they all are though

[zinc bolt]

Everything they did so far looks correct tho?

They haven't even used log rules yet

[charred jewel]

oh nvm im blind

i thought you did log(ab) = log(a)log(b)

[zinc bolt]

lol

[sharp stag]

$\frac{f'(x)}{f(x)} = \frac{d}{dx} \left[ \ln\left(x^8 \cdot (x - 9)^7 \cdot (x^2 + 5)^{-8} \right) \right]$

[ocean seal]

Paul04

[zinc bolt]

Yes, that seems about right

[sharp stag]

is this the right thing to apply? if so how should I apply it?

is it just 1/[inside stuff] * deriv of inside stuff?

so (1/inside stuff) * a long product rule

where this is inside stuff

[zinc bolt]

Um...$\log(ab) = \log(a) + \log(b)$ furthermore $\log(a^b) = b\log(a)$

[ocean seal]

@zinc bolt

[zinc bolt]

Use these on the RHS

[vale wigeon]

yeah like

[sharp stag]

ngl idk how I would

[vale wigeon]

the idea is that you break up the log as much as you can

before diff

[vale wigeon]

log(abc) = log(a)+log(b)+log(c)

[sharp stag]

is ln(ab) = ln(a) + ln(b) and ln(a^b) = bln(a) true?

okay then ig ik how to do it

just needa remember all these formulas somehow

okay cool, thanks everyone

[zinc bolt]

!done

[lone heart]

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

[sharp stag]

.clolse

.close

[lone heart]

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[vale hollow]

Could someone help with this question please

[vale wigeon]

do you have a ruler and protractor on hand?

[vale hollow]

yess

[vale wigeon]

ok

draw a ray starting at point S at a bearing of 160°

[vale hollow]

ok

I've done that

[vale wigeon]

show what you got

[vale hollow]

I'm not sure I've done it correct

[vale wigeon]

well that's why i am asking you to show it

so that i can look at it and tell you if it is correct

[vale wigeon]

that bearing definitely isn't 160°

you should line up the 0 of your protractor's clockwise scale with North

[vale hollow]

okay let me try that

[plush wing]

you cant calculate second angle like that

[vale wigeon]

the line from S looks good to me

[vale hollow]

.close

[lone heart]

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[slender nova]

i have tried to this question by organizing it into a matrix that is symmetric and in a sudoku like manner

but don't know how to proceed

[lone heart]

@slender nova Has your question been resolved?

[slender nova]

<@&286206848099549185>

[ocean seal]

@zinc bolt

[lone heart]

@slender nova Has your question been resolved?

[slender nova]

@zinc bolt i didn't get how this proves it

[pallid scarab]

if you manage to show that f(r,r) = 1 for some r, then you'll have succeeded (why?)
So now, here are two hints to set you on the right path:
Hint 1: ||Prove that 1 has exactly n inverse images by f.||
Hint 2: ||Prove that |{(r,s) : f(r,s) = 1, r != s}| is even. (!= means "not equal to")||

[zinc bolt]

It doesnt' complete the proof but you can work from there

[slender nova]

oh

wait i got that

so

proving those two statements is enough for the full proof right?

cause it proves that there is a f(r,r) = k for all k belonging to S

@pallid scarab

correct?

[pallid scarab]

yeah

because you can just replace 1 by k

[slender nova]

alr thanks

@pallid scarab i have one more doubt can i just ask it here?

.close

[lone heart]

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[slender nova]

maximise and minimise cosx(sinx + root(sin^2x +3))

[slender nova]

i tried differentiating but i couldn't simplify the differential

[shell agate]

$cosx(sinx+\sqrt{sin^2{x}+3})$

Is this your question?

[slender nova]

yes

[ocean seal]

devthemasked

[slender nova]

thanks

[lone heart]

@slender nova Has your question been resolved?

[random cypress]

try consider
$f(x) = cosx(sinx+\sqrt{sin^2{x}+3})$
$$ f(x) = cosx(sinx+\sqrt{sin^2{x}+3}) (\frac{sinx-\sqrt{sin^2{x}+3}}{sinx-\sqrt{sin^2{x}+3}})$$
$$ f(x) = \frac{-3cosx}{sinx-\sqrt{sin^2{x}+3}} $$
instead may help you ,since denominator is always negative..

[ocean seal]

Sea

[lone heart]

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[lone heart]

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[ripe hamlet]

Is what I have selected correct? Series 1,2,9,7 are convergant but absolutely so I didn't check them

[ripe hamlet]

<@&286206848099549185> Please help

[lone heart]

@ripe hamlet Has your question been resolved?

[lone heart]

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[pseudo seal]

I am trying to find the closest set of whole numbers needed to multiply to reach the closest to 8760, starting with 12.
Only using numbers between 10 and 50.

For example: 12 * 24 * 30 = 8640. which is off by 120.
Or similarly, 12 * 40 * 18 = 8640 as well.

[modern sedge]

what do you mean by closest set of whole numbers?

17 * 43 * 12 = 8772

off by 12

[pseudo seal]

like 12, 24, 30 is a set of numbers. They are also whole numbers.

I can easily find numbers to multiple to get exactly 8760, but then they are not whole numbers. They are decimals.

Example, 36.5 * 12 * 20. But it has a dcimal

[modern sedge]

27*27*12 = 8748

also off by 12

[pseudo seal]

❤️

I'm new here, can I give thanks before I close? how does that work?

[modern sedge]

yeah, ofc, no problem

oh and btw here is the way I got those - 8760 is 12 * 730, so i just checked the multiples of 12 closest to it

so like 12 * 729 and 12 * 731

[pseudo seal]

ohhhh I see!

[modern sedge]

as you mentioned, 8760 itself doesnt work

[pseudo seal]

Right

within 12 is very nice.

[pseudo seal]

thanks

.close

[lone heart]

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[prisma night]

how did they got the third last line

[prisma night]

.close

[lone heart]

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[jolly notch]

i have a quick concept clearing qs.

so first off just making the function for gf came out to be $\frac{2}{\left(x-2\right)^2+2:}$

now for the range first off we need the domain for gf correct? we know that for gf to exist, the range of f must lie within the domain of g, which is true here. next up what i thought we had to do was find the value of f(x) at 1 which is 4 and judging by the fact that it's to the left of the turning point in a concave up, the values will only become greater.

now here comes my question. will we input these values i.e. from 4 to infinity in g(x) or in gf.

[ocean seal]

Sam | TBND

[jolly notch]

to find the range

from my understanding, we'd input the domain of gf into gf to get the range, or otherwise we'd find the range of f, use it as a domain for g and get the range for g which becomes the range for overall gf?

so how would we find the domain for gf

ping me when/if you reply <3

<@&286206848099549185>

[oak swallow]

you have lready found out an expression for gf right?

[jolly notch]

yep

$\frac{2}{\left(x-2\right)^2+2}$

[oak swallow]

okay so clearly x belongs to R for the function f

[ocean seal]

Sam | TBND

[oak swallow]

and for the function g, x > 1

[jolly notch]

x < 1 for f(x)

as per the question's requirement

[oak swallow]

Oh yes sorry I just read the question fully

my bad mate

[jolly notch]

np

[oak swallow]

okay I have returned

can you find the turning point of f(x)? the minima?

[jolly notch]

i did that, it's 2,3 and thats why I know that for x < 1 the graph is increasing

for f

smth like this

[oak swallow]

I guess than you can use x > 4 as the domain for gf

if the minimum value that f can provide is 4

[jolly notch]

would we not put it in g?

x > 4 becomes the domain for g, not gf

[oak swallow]

yeah that is what I meant

like you calculate the value using the function f

and then put it in g

I think I get what you are saying

[jolly notch]

my question then becomes how do I find the domain for gf

by using THIS range that i'll get?

[oak swallow]

basically teh largest possible value gf can achieve is

$$\frac{2}{4 - 1} = \frac{2}{3}$$

[ocean seal]

Klein Bottle

[oak swallow]

and then any subsequent value it will have is greater than 4

so we can find

$$\frac{2}{\infty -1} = 0$$

[ocean seal]

Klein Bottle

[jolly notch]

that i got, gf lies between 0 and 2/3

because i understand that for the composite function the range of gf becomes the range of g when you take the range of f as a domain for it (holy word salad) because thats literally just what gf means

however, let's say the question was to find the domain of gf, then that would be?

equating $\frac{2}{\left(x-2\right)^2+2}$ to 2/3 and finding $x$?

[oak swallow]

the domain would be negative infinity to 1

[ocean seal]

Sam | TBND

[oak swallow]

did you get why?

[jolly notch]

im computing it gimme a sec

neg inf gives 0

and

1 gives 2/3

[oak swallow]

yup! and ofc negative infinity is not really a number so we put a nice little > sign in front of it

and same thing with 1. f(x) < 1, we put the < synbol

[jolly notch]

so the domain of gf is just x < 1

[oak swallow]

precisely

[jolly notch]

so the domain of gf is just the domain of f?

given that the range of f lies within the domain of g

[oak swallow]

two reasons

--> any value less than 1 will satisfy the domain of g(x) once the output of f(x) is taken as the input of g(x)

--> any value less than 1 will obviously satisfy f(x)

[oak swallow]

yup you got it

[jolly notch]

would you recommend just directly working with the domain of the inner function as the domain of the composite given that it's defined for all values, or would you rather I do the proper like find the range of f and then use it for g

just to confirm, the former always holds true?

[oak swallow]

the former always holds true. Yes

[jolly notch]

brilliant so i think i'll do the latter to show working and just confirm with the former

[oak swallow]

yupadoodly do

[jolly notch]

.close

thanks

[lone heart]

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[oak swallow]

one thing to note though

[jolly notch]

.reopen

[oak swallow]

sometimes fucntions can have breaks

[jolly notch]

oop

[lone heart]

✅

[jolly notch]

yeah

i get that

[oak swallow]

sometimes functions aren't nice and continuous

they are nasty

[jolly notch]

asymptotes

yesyes

[oak swallow]

so you have to be careful

yup

anyways I think you understood quite well

cheers good day!

[jolly notch]

you too

.close

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[jolly notch]

have a good one

[lone heart]

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[graceful jungle]

Only using A-Level circular motion - I can't use F = Gmm/r^2

[graceful jungle]

I'm not really sure how you would do this without knowing the orbital period of the ISS

[potent mirage]

u are supposed to use only circular motion for this?

[graceful jungle]

yea

[potent mirage]

huh the inward force would be mv^2/R or mw^2 R but we dont know the velocity or w

[graceful jungle]

yep

[potent mirage]

i dont know how u calculate the gravitational force without using gravitation??

also why would they give u the mass of the earth if they didnt want u to use gravitation

[graceful jungle]

for later parts i think

nvm i asked my teacher he said we're allowed to use F = Gmm/r^2

.close

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[potent mirage]

bro

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[keen pewter]

I get most of this DMT1 question but confused by the graph built for G_m+1

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@keen pewter Has your question been resolved?

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[fair crypt]

god damn i cant figure out these convex stuff is destroying me

[fair crypt]

Given a convex $f:(a,b) \to \bR$ , and $a < s < t < u < b$ , i want to show $\frac{f(t) - f(s)}{t - s} \leq \frac{f(u) - f(s)}{u - s} \leq \frac{f(u) - f(t)}{u - t}$

[ocean seal]

Prelude to archbishop

[fair crypt]

i don't know if there's supposed to be an idea that i'm just not getting

rn all im doing is brute forcing random inequalities hoping to get anything similar to the result

[forest marsh]

what does it means about the derivatives values if f is convex ?

[fair crypt]

i can't use derivatives

It's continuity chapter

[forest marsh]

i don't see anything from continuity that you can use here

[fair crypt]

im just gonna have to switch to sequences ig

doesnt seem like i can do anything in the realm of epsilon-delta

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[calm nebula]

The math test scores of students in a school follow a Gaussian distribution. If the average score is 70 and 15% of students get less than 60, what percentage of students get more than 85? Report the data with a tolerance of 0.1
.

this is a probability question what I don't understand is how chatGPT uses the solution in the photo and where it gets that -1.28. You say you get it from tables but I can only use geogebra for the exam any help?

[lone heart]

@calm nebula Has your question been resolved?

[calm nebula]

please

it's a probability problem

[tacit arch]

!nogpt

[lone heart]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[tiny fable]

idk what the tolerance of 0.1 has to do with anything, but if we ignore that, then we know that

P(X < 60) = 0.15, not 0.10

chat gpt is very bad at math

[calm nebula]

I used it because my prof. created an "AI tutor" and recommended us to use it

[tacit arch]

awful

[calm nebula]

sorry the tolerance of 0.1 is the proximity of error

like if the result is 0.3 and i write 0.5 is wrong

while if I write 0.4 it is correct

[calm nebula]

i know but i can't understand how to solve this problem. And that's the only solution that i have 😭

[tacit arch]

don't waste helpers' time reviewing nonsense

just post your question and your understanding

[calm nebula]

i wrote the question

i gave the full problem

[tacit arch]

you gave ai crap that just wastes everyone's time

[calm nebula]

and my question is how to solve it

[calm nebula]

so I rephrase the question how do I calculate the standard deviation with geogebra?

[twin nimbus]

You probably cannot. @calm nebula

[calm nebula]

😭

[twin nimbus]

You would either need a way to computer the inverse normal cdf, which is available from online calculators or WA

[calm nebula]

The math test scores of students in a school follow a Gaussian distribution. If the average score is 70 and 15% of students get less than 60, what percentage of students get more than 85? i really can't understand how to solve this problem

[twin nimbus]

Or you'd need a computer algebra system that supports the function

Or a table of values

[calm nebula]

yeah but i can't use them

[twin nimbus]

This is from a class?

[calm nebula]

yea

[twin nimbus]

Did you receive a table of normal cdf values?

[calm nebula]

nope

[twin nimbus]

Ok, your professor would have told you how to calculate this, how did he tell you?

"use geogebra?"

[calm nebula]

yes

it's blowing my mind

[twin nimbus]

@calm nebula https://geogebra.github.io/docs/manual/en/commands/InverseNormal/

I was wrong, it seems to be a built in, but not one you can access with your phone without explicitly typing it in

[calm nebula]

i'm gonna try it tomorrow. Thank you so much

[lone heart]

@calm nebula Has your question been resolved?

[lone heart]

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[ivory ivy]

I believe f(x) is differentiable everywhere and has a relative minimum at (0,0), yet f'(0) does not equal 0. Is there something I am missing here?

[ivory ivy]

Given a function f(x) that is differentiable on (a, b) that has a relative minimum at (c,f(c)), I thought that f'(c) must equal 0, but it seems I might've found a contradiction, can someone explain?

[foggy vector]

wdym f'(0) isnt 0

[ivory ivy]

its undefined

based on desmos

[placid zinc]

Green definitely isn't the correct derivative, very high when it should be low

[somber crest]

it sounds like you're seeing that f'(x) is nearly vertical at x=0 and getting worried because of it

oh, well, nevermind I guess

[ivory ivy]

yeah soo

[patent vale]

maku no sensu

[ivory ivy]

are u sure its not correct

it checks out tbh

[placid zinc]

Or, that might be the scale tripping me out

[grave matrix]

It's zoomed in a lot, yes

[ivory ivy]

regular scale

[patent vale]

so strange

[median oar]

What’s the black function

[somber crest]

f

[foggy vector]

yeah i have to agree it looks fine

[ivory ivy]

look at the original

[median oar]

f is not differentiable at 0

[ivory ivy]

ahh

how do u know

[median oar]

f doesn’t meet the assumptions here

[ivory ivy]

it looks smooth enough

[median oar]

Absolute value

[ivory ivy]

so i made that assumption

[placid zinc]

Very small cusp, perhaps

[median oar]

That’s not a differentiable function at 0

[ivory ivy]

okay

one more question

[median oar]

Just the abs value function’s issue

[grave matrix]

Oh yh right

[ivory ivy]

Given a function f(x) that is differentiable on (a, b) that has a relative minimum at (c,f(c)), must f''(c) exist?

[median oar]

Nope

[ivory ivy]

i didn't think os

but any counter examples?

[median oar]

You never said f has 2 derivatives

[ivory ivy]

sure but I just couldn't imagine a function that wouldn't have f''(c) existing

[median oar]

Yeah take the weisstrass function and integrate it once

[grave matrix]

You mean f needs to be twice-differentiable?

[ivory ivy]

if u integrate it

is it differentiable

and does it have a relative minimum

[median oar]

The weistrass function is nowhere differentiable so if you integrate it once it is differentiable once

[ivory ivy]

yeah but read

what i said

[median oar]

It does have a relative minimum because it is continuous

[ivory ivy]

Given a function f(x) that is differentiable on (a, b) that has a relative minimum at (c,f(c)), must f''(c) exist?

[median oar]

On a bounded interval the function has a bounded image

[ivory ivy]

would an integrated weierstrass function

meet that criteria

[median oar]

I did read what yoy said and the function does satisfy your conditions

[ivory ivy]

okay

[median oar]

And it does not have a second derivative anywhere

https://en.m.wikipedia.org/wiki/Weierstrass_function

[ivory ivy]

i've seen the function before

but never considered its antiderivative

but that is good to know

thank you

i can tell you know quite a bit of math

what branches are you studying

and where do u go

[median oar]

Well yeah a nowhere differentiable function will be exactly once differentiable if you integrate it

[ivory ivy]

just curious

[ivory ivy]

wait exactly once?

[median oar]

Fwiw I saw this trick in a YouTube video

[ivory ivy]

my f(x) had to be differentiable on an interval (a,b)

not just exactly once

[median oar]

Well once you differentiate it once it will become the weierstrass function and be nowhere differentiable

[ivory ivy]

oh

i see what u mean

sorry

yeah

thats a perfet

counterexample

[median oar]

Im interested in probability atm but I’m still undergraduate

[ivory ivy]

what college

[median oar]

Macquarie university

It’s shit don’t go here xd

[ivory ivy]

ah

nothing crazy

well thanks anyways

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[surreal crystal]

https://en.wikipedia.org/wiki/Archimedean_spiral

[keen jungle]

doesnt really answer my question

[surreal crystal]

it has the parametric equations

plug in the parametric equation, plot the point at a number

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[keen jungle]

Alr

[surreal crystal]

just plot (x(t), y(t))

[keen jungle]

it says v and w are for velocity, so do i ignore those?

[surreal crystal]

you can choose any number for them

just make t the variable

[keen jungle]

yeah i'm confused... can you give me an example for r=100(5theta)

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[surreal crystal]

lol

.close

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[vale hornet]

Determine how many 3 of a kind hands contain the card Queen of Spades in a standard deck

[lone heart]

@vale hornet Has your question been resolved?

[lone heart]

@vale hornet Has your question been resolved?

[modest robin]

You are talking about this with the standard 13 * 4 =52 deck and 5 cards hands?

[shell cave]

probably

[vale hornet]

yes

[modest robin]

I have to go but I'm thinking something like this

but this is wrong because the two remaining card and the 1 remaining card can't make the hand more than a three of a kind, for instance you could get a 4 of akind if you don't restrict further

the separation of cases looks plausible tough

[honest nacelle]

yeah you'll want to split this into 2 cases
more specifically, one case will be 3 Q (one queen of spades) and 2 (different kind non-Q) cards
other case will be Q of spades, 3 identical kind non-Q cards, and one card that isnt of the same kind as any of the preceding cards

ill do the first case for you: first you choose the queen of spades (one way)
then you choose 2 queens from the remaining 3 (3 choose 2 ways)
then you choose a kind for the remaining 2 cards out of the 12 kinds left since queen is taken (12 choose 2 ways)
then you choose a suit for one of those cards (4 ways)
then you choose a suit for the other card (4 ways)
so case 1 can unfold in 1 * (3 choose 2) * (12 choose 2) * 4 * 4 ways
try to figure out case 2 yourself

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[rigid dirge]

Doing integrals specifically solving by Parts and I need to find the anti derivative of e^(x^2) I have no idea have to find the anti derivative of this. I’m thinking it might be possible to use substitution before solving by parts. The question asks find the integral of x^3 times e^(x^2) dx

[rigid dirge]

Ok so I was able to solve it using substitution first. But I still don’t understand how I would find the anti derivative of e^(x^2)

[wintry wadi]

you can't, that integral is non elementary

if they wanted you to do IBP first, you can do xe^(x^2) as one of them

[rigid dirge]

Sorry I don’t know what that means

[alpine sable]

Heyp

Heyo

[rigid dirge]

Does it mean that the anti derivative includes imaginary numbers

[alpine sable]

I think I can help

[rigid dirge]

Ok

[alpine sable]

It's a bit tricky

But u need to first apply substitution then parts

[rigid dirge]

Yeah I did that already. I solved it easily like that lol

[wintry wadi]

it means that the antiderivative isn't a combination of normal function

[rigid dirge]

I’m just curious about the anti derivative stuff now

[alpine sable]

Then?

[rigid dirge]

Oooh. I see

[rigid dirge]

But thank you anyway=)

And thank you Sepdron

[wintry wadi]

it explains it more precisely here
https://en.m.wikipedia.org/wiki/Elementary_function

[rigid dirge]

.close

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[alpine sable]

Can someone calculate this for me and explain what im doing wrong please, I keep getting the wrong answer

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

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[rugged pond]

@ionic rampart what have you tried so far?

[ionic rampart]

lemme send a photo

[rugged pond]

Kk

[ionic rampart]

I didn't do any calculations yet

But this is my idea

Of how it looks like

[rugged pond]

What does A.P. mean?

[ionic rampart]

arithmetic progression

like theres a common difference were tryna find here

[rugged pond]

Okay, so that should mean that b = a + d for some constant d

And c = a + 2d

Right?

[ionic rampart]

hmmm

i'm not sure if i'd do it that way

we need to find the common differences for a, c and b

so

since c is the hypotenuse c=x+d

because we need to find x also

[rugged pond]

What is x?

X is not in your diagram

[ionic rampart]

fine we can do theta

if that make sit easier

[rugged pond]

Why have we jumped to theta

[ionic rampart]

were trying to find the sines

[rugged pond]

What is theta?

[ionic rampart]

did you not read the question

[rugged pond]

Yes I read the question

And there is no theta in it

[ionic rampart]

the sines are theta

[rugged pond]

I thought theta would represent an angle

[ionic rampart]

yeah we're trying to find the sine angles

so we can represent them by theta

that's what i'm trying to say here

[rugged pond]

Okay, which angle will you choose as theta?

A B or C?

[ionic rampart]

c

[rugged pond]

Ok, very good

[ionic rampart]

actually no

wait

[rugged pond]

C is the 90 degree angle

[ionic rampart]

yeah

[rugged pond]

If you want to use sin or cos, you can't use the 90 degree angle

So you must choose either A or B

(note the capitalisation please)

[ionic rampart]

b is theta then

i knew it couldnt be c

[rugged pond]

Very good

So I assume you would like to take sin(theta)?

[ionic rampart]

do you know what a middle side of the triangle is

*the

alright lets start over

[rugged pond]

Middle side?

[ionic rampart]

nvm you can close this

chat

I think i'm getting it

the variable x is the middle side

[rugged pond]

I'm not really getting your thought process

What is this middle side

[ionic rampart]

the side in between the smallest and largest sides

it's not the hypotenuse

it's in between the smallest side and the hypotenuse

thats what x ius

alright you can close the channel

ggs

[rugged pond]

Oh, the adjacent side to theta

That's what you meant

[ionic rampart]

here's the answer from the texbook

i basically gave up

also

can you close the channel

[tacit arch]

!done

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[undone herald]

not trying to cheat, just having trouble interpreting what this question even means. I think it's asking to find the fourier coefficients for x(t), do, something to them with the frequency response, and rewrite the coefficients?

[undone herald]

I would provide working but i can't even really figure out how to start

[surreal crystal]

H = Y/X

since its lti

transform x,y,
mulitply HX,
simplify,
rewrite X side with e^{j2\pi f_0 kt}

then you have them written in terms of each other

[surreal crystal]

dont forget to transform F(x)=X, F(y)=Y

[lone heart]

@undone herald Has your question been resolved?

[undone herald]

like it's the division of the two functions?

[surreal crystal]

yes

its the advantage of frequency transforms

they always have transfer functions in certain cases

[undone herald]

okay, i'll look into that further, thanks

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[hybrid ravine]

how to do c

[lyric gale]

I think you just need to find the derivative of both.

[lone heart]

@hybrid ravine Has your question been resolved?

[hybrid ravine]

i dont get what finding the derivative has to do with max distance

[junior vigil]

optimization

set up a function that is the distance between Henry and Thomas, then find the maximum of this function

(calculus!)

[hybrid ravine]

OH

[tight shuttle]

Have you learned about optimization?

[hybrid ravine]

Ok got it thanks

[hybrid ravine]

Yea i was overcomplicating

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[wet magnet]

for iv

i get that z=0

but what then?

[vale wigeon]

find the equation of line l

by putting z=0 into that of plane Pi from part iii

[wet magnet]

wait put z=0 in the plane equation?

[exotic canopy]

yes, the one you found in (iii)

[wet magnet]

alright but i also have y term, 3x - 4y= 2

we dont know y or?

[vale wigeon]

that is the equation of line l

[exotic canopy]

well you are looking for a whole line not a point

[vale wigeon]

now you can pretend you are working in the xy plane (bc you are)

and look at this line and maybe sketch it

find its intercepts with both axes

[wet magnet]

i understand it becomes a line but dont we only have two sides for the triangle?

oh wait

its x and y axis

[exotic canopy]

yes

[wet magnet]

but x and y are unkwones?

unknown

we find intersection of line and axis??

cuz then i get 2/3 for x

and since the its on the axis the base will be 2/3 and the height is -1/2????

[exotic canopy]

yes!

[wet magnet]

THANMK YOU SM.

thanks you @exotic canopy and @vale wigeon , awesome help

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[strange schooner]

Does there exist a non commutative ring with identity such that its only ring automorphism is the identity map?

[strange schooner]

(in case an example is overly hard to construct, and in case it is possible, a proof the latter hypothesis does not imply commutativity would also suffice, thanks in advance)

[mortal trellis]

https://math.stackexchange.com/questions/133031/is-there-a-non-commutative-ring-with-a-trivial-automorphism-group

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[carmine topaz]

when finding average distance of polar curve why do i have to do 1/bounds1 -bounds2 int r(theta) dtheta but not 1/bounds1-bounds2 * arc length formula?

[tight pier]

r is literally the distance, how come arc length

[carmine topaz]

ok lol

wait but isnt arc length a type of distance?

am i tripping?

[tight pier]

not really its just length of the curve

by distance from origin to curve is radius

[carmine topaz]

its like length of segment vs distance taken by shooting certain theta right?

[tight pier]

wdym by the latter

[carmine topaz]

like your reflecting the angle to the surface

[vale wigeon]

average distance?

!xy

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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.

[carmine topaz]

im asking about a general method and why the other doesnt work

$\frac{1}{\theta_2 - \theta_1} \int_{\theta_1}^{\theta_2} r(\theta), d\theta \ne \frac{1}{\theta_2 - \theta_1} \int_{\theta_1}^{\theta_2} \sqrt{1 + (r'(\theta))^2}, d\theta$

[ocean seal]

soo

[carmine topaz]

why doesnt the two result in same?

[tight pier]

what would even mean "average arc length"

just doesnt make sense

[brave verge]

The second equation doesn’t seem right

its for cartesian coords but is being applied in polar

so that formula wouldnt work

@carmine topaz

[carmine topaz]

yea lol

it looks wrong

i get it

thanks yall 🙂

[brave verge]

.close

wait i cant do that

[carmine topaz]

almost forgot this

.close

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[eager blaze]

Hi, what am I suppose to do around (0,0)?

Is it like one variable calculus, or is there something else?

[proven grove]

around 0, value is 0?

[shadow fjord]

You should have a limit definition for the partial derivative. You can use that at (0,0).

[eager blaze]

ye

[eager blaze]

I mean the derivetive at (0,0) would be 0

[shadow fjord]

Why do you say that?

[eager blaze]

like do $f_x(0, 0) = 0$, $f_y(0, 0) = 0$ and If not say that it's not differentiable at (0,0)?

[eager blaze]

the value is 0 at (0,0)

so wouldnt it's derivative be also 0?

[ocean seal]

Shachar

[shadow fjord]

The value of the function being 0 doesn't mean the derivative has to be 0.

You could think of a single variable counter example

This site has the limit definition. Try applying that. https://math.libretexts.org/Bookshelves/Calculus/Calculus_3e_(Apex)/12%3A_Functions_of_Several_Variables/12.03%3A_Partial_Derivatives

[eager blaze]

so like $\frac{f(x_0,y)-f(x_0, y_0)}{y-y_0}$ as y approach y0 and do two sides limits to check?

[ocean seal]

Shachar

[shadow fjord]

That looks right for f_y. Then the other way for f_x.

[eager blaze]

ok I got it

thanks

.close

[shadow fjord]

np

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[daring flame]

I need help understanding the definition of an ARCH(m) model

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@daring flame Has your question been resolved?

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@daring flame Has your question been resolved?

[alpine sable]

What's that?

[grave matrix]

My guess is that it's an autoregressive conditional heteroskedasticity

And now I hate myself for knowing that "heteroskedasticity" is an actual "econometrics" word, and not slang from my own generation

[alpine sable]

Oh wow it's totally beyond me good luck guys🫡

[grave matrix]

Not that I know what any of this is

[alpine sable]

That's sad

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[bronze scarab]

Ok chat I'm gonna need this answered quick cause I'm in a shootout with the cops, if one bag of meth equals $273,000, how many bags of meth do I need to cook at minimum to make 10 million?

[bronze scarab]

Worth mentioning I currently have 10 bags and I'm playing a game I'm not actually in a shootout with the cops

[modest bough]

10,000,000 / 273,000 rounded up to the nearest whole number

[bronze scarab]

So that's 37 bags I think

That's a lot of meth

[modest bough]

true

[bronze scarab]

Before anyone asks the game I'm playing is Payday 2

Anyways that's it cheers

.close

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[neat spire]

i’ve been tuck on this question for oh so long… HOW DO I GET 72 ?!?? first one is my teacher, second one is me, help please !!

[next stratus]

do you know the angle of a straight line?

[neat spire]

180 ? 😭

is taht what your asking

i’m so confused

[next stratus]

so the angles must add up to 180' since they lie on the straight line

[neat spire]

wait so

180-108

oh my goodness

[next stratus]

=2x

[neat spire]

i see, thank you so much 😣😣

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[mossy locust]

hello

I need to determine if the sum of the series from n=2 to infinity of the equation 1/n * ln(n) converges

[buoyant saddle]

$\sum_{n = 2}^{\infty} \frac{\ln n}{n}$ or $\sum_{n = 2}^{\infty} \frac{1}{n \ln n}$

[ocean seal]

knief

[mossy locust]

the second one

[buoyant saddle]

what have you tried sir

[mossy locust]

so basically the answer key is saying that it converges and the cornballs in my class are saying that too

But I tried doing an integral test and it seems to me that it still diverges

[buoyant saddle]

$\int_2^{\infty} \frac{1}{x \ln x} \dd{x}$

[ocean seal]

knief

[buoyant saddle]

how did you proceed

[mossy locust]

I did u sub with u= ln(n) and du=1/n

[buoyant saddle]

nice

and what was your antiderivative

[mossy locust]

so then its the integral of 1/u switched the bounds to ln2 and infinity still

then my antiderivative was

ln | u |

plug in for u

[buoyant saddle]

killing it

[mossy locust]

ln | ln (n) |

[buoyant saddle]

go on

[mossy locust]

okayy then we gotta evaluate from ln(2) to infinity plugging in a b for infinity to do the improper integral

so

basically