help-10

[timid silo]

it should be sec

idk how ot explain it just 1 sec

[drifting loom]

you wanna find derivative of this

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[timid silo]

.reopen

[obtuse pebble]

✅

[timid silo]

okay sowe need

function horizontal and vertical asymptote

and function local min and max

i hope google translated it right

local min max f'(x) and asymptote

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

For horizontal asymptotes: what happens to f(x) when $x \to \pm \infty$

[warm shale]

tofu

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

Channel closed

Closed by @lean pier

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[primal river]

I was wondering if there are word problems for power series I have a project to make a situation involving power series but I cant find any exampels

[drifting loom]

what is power series?

[primal river]

a infinite series

[drifting loom]

ohh

of infinite series

lemme give you an example

@primal river

see this if it help you

[primal river]

I think I was supposed to think of something like a word problem

[drifting loom]

you can create a word problem of this

[zenith raft]

also that's not a power series and it's not totally clear what the pattern is but it doesn't seem to equal 1

[final thunder]

1/10

[obtuse pebble]

@primal river Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[royal basin]

distinct positive integers b, c, d

[hot hazel]

apologies for the disruption.

[neon eagle]

ok so you do realise the question boils down to $c^3=d^4$

[warm shale]

ed

[royal basin]

to reiterate what i said that got purged,

a needs to be a cube and a fourth power at the same time

[neon eagle]

where c and d are integers and you're trying to minimise the values of c and d

yes what ann said

[sharp pecan]

thats cool and all, but please don't just give the man the fish, teach them how to fish

[neon eagle]

jigglyproff be quiet.

@nocturne quail use fundamental theorem of algebra

every positive integer can be written as a unique factorisation of prime factors

[nocturne quail]

Yes, I think I got it. It's a = 4096

Thanks a lot all of you.

[zenith raft]

i think that's the wrong FTA

[neon eagle]

wait is it

shit

[zenith raft]

arithmetic haha

[neon eagle]

OOPS

same thing.

wait im so bad at math

why am i helping people i need help.

[zenith raft]

same

[neon eagle]

mood

[sharp pecan]

I think it would be more helpful on why the approach is the right one, instead of just giving it

[neon eagle]

dude

he might not even know FTA

also its obvious why its the correct approach

thats how math is

once a solution is presented its hard to forget

[nocturne quail]

.close

[obtuse pebble]

Channel closed

Closed by @nocturne quail

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

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

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

Channel closed

Closed due to the original message being deleted

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

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

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

[royal basin]

are you sure that this inequality is in the right direction

bc as-is it feels like a contradiction is about to surface

or maybe not a contradiction but the problem might become a little too trivial

can you show your proof?

it's strange that your proof would depend on a_2 itself being nonnegative as opposed to say the difference between a_2 and a_1

$a_n = (n-1)a_2 - a_1$?

[warm shale]

Ann

[royal basin]

er

≥ not =

that looks odd to me though

you could say $a_n - a_1 \geq (n-1)(a_2 - a_1)$

[warm shale]

Ann

[royal basin]

are you sure though

sorry, got distracted

Using this I managed to get
a(n)>=(n-1)a(2)-a(1)
i think this is nonsense, but given how you jump from the givens to this with nothing inbetween i can't really comment on where you went wrong

here's how i would've done this: let $d = a_2 - a_1$, then for all $n$ you have $a_{n+1} \geq a_n + d$, and from this it is transparent that $a_n \geq a_1 + (n-1)d$

[warm shale]

Ann

[royal basin]

from this you can conclude that if d > 0 then the sequence is unbounded, so d has to be ≤0

because $a_{n+1} - a_n \geq d$

[warm shale]

Ann

[lethal cloud]

are you sure the inequality at the top is right?

If the sequence is monotonically increasing, then we have $\lvert a_{2} - a_{1} \rvert \leq \lvert a_{n+1} - a_n \rvert$. Hence we have some $\varepsilon$ so that $\varepsilon < \lvert a_{2} - a_{1} \rvert \leq \lvert a_{n+1} - a_n \rvert$` for any n, therefore it is not cauchy, and cannot converge.

[royal basin]

let $b_n = a_{n+1} - a_n$, then you're told that $(b_n)$ is an increasing sequence

[warm shale]

Ann

Shell

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[heady tendon]

can anyone help with this

[faint prism]

Calculate every surface area and add them together.

[heady tendon]

yh ik but i keep getting it wrong

can u help me

[faint prism]

Calculate for each figure. Rectangle, 2 of 4x6, 4 of 15x4. And do it the same way with the triangle.

Take your time and calculate each side of the figures.

[heady tendon]

i still dont understand D

[faint prism]

Do you know what surface area is?

[heady tendon]

i not very goof with it D:

[faint prism]

Like it is, the area of the surface. You got a table that is 4cm long and 6cm wide, the areal of the surface is 24cm^2.

[heady tendon]

6x4 ok

[faint prism]

If we include the height, then it's just area.

[heady tendon]

ok

so the answer is 174

so the answer is 174????

umm @faint prism

[faint prism]

How you got it?

answer:||4* 6 + 4(15* 4)+(3* 4)+5* 6+3*6||

[heady tendon]

that 3798 right

@faint prism its was wrong

[warm shale]

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!

[faint prism]

324 dosen't work?

[heady tendon]

yes

and now i'm in recovery

doing this

thanks alot

[faint prism]

Ah, I saw my mistake. It should be ||4* 6 + 2(15* 4)+2(15* 6)+(3* 4)+5* 6+3*6||

[heady tendon]

i already failed u goota help me with recovery pls man

[faint prism]

I mean, does the question just ask for surface area of one pyramid?

[heady tendon]

like 20 questions

of this but it gets harder

[faint prism]

If it's just one, then 21.

[void swift]

no..?

[heady tendon]

no

[void swift]

hi you need help?

[faint prism]

Alright, I'm in a doubt. You should probably find other helpers.

[heady tendon]

yes pls

[void swift]

you see those shapes they gave you?

how can you break them up into

(smaller shapes)

[heady tendon]

u mean this

[void swift]

yes

what are the shapes you can see?

[heady tendon]

pyramid\

[void swift]

can you see a square in the middle?

[heady tendon]

yes and fyi cool pffp

[void swift]

and four other triangles that connects the square?

[void swift]

ty

[heady tendon]

yh

[void swift]

okay now what you need to do is, in order to calculate the entire surface area, you calculate the surface areas of the smaller shapes (square, triangle), and add them up

make sense?

[heady tendon]

not at all

[void swift]

gimme a sec

[heady tendon]

ok

[void swift]

see how the larger shape's surface area is just the combination of the smaller ones

[heady tendon]

nice drawing mine is ass

[void swift]

do you understand this or not?

[heady tendon]

yh i think

[void swift]

alright

so we just find the area of each individual shapes that we just got

tell me how do we find the surfacd area of a square?

[heady tendon]

is the answer 5`

[void swift]

nonono

tell me

[heady tendon]

51

[void swift]

yep

[heady tendon]

is it 51?

[void swift]

how did you get it?

you break them up into smaller shapes

so we calculate the small shape's area and sum it up

for the square, its 9inches squared

for one of the triangles, its 10.5 inches squared

[heady tendon]

bruh i just did 7 x 3 / 2 and times it by 4 and i did 3 x 3 and added it

[void swift]

yep its the same thing as what im doing

its just a breakdown of it

[heady tendon]

ok ty pls dont just go

i will need later pls pls pls dont go

[void swift]

i have things to do, if i can ill come back but no guarantees..

[heady tendon]

D:

pls can u come back at 3

[void swift]

how old are you btw?

[void swift]

we dont have the same timezone yea..

[heady tendon]

whats ur time

@void swift

[heady tendon]

jk jk

i'm 13

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[obtuse pebble]

@heady tendon Has your question been resolved?

[heady tendon]

<@&286206848099549185>

[timid silo]

@heady tendon wait let me see

@heady tendon @heady tendon @heady tendon @timid silo

[heady tendon]

hi

srrry

[timid silo]

@heady tendon

[heady tendon]

done with that

i need help with this

wait pls try and solve it with me pls pls

is the answer 314

is 489.84 @timid silo

umm pls can I get another <@&286206848099549185>

[timid silo]

@heady tendon come in vc?

[heady tendon]

srry its very very loud

so i cant chat

which one

[timid silo]

@heady tendon sorry?

[heady tendon]

i have to watch a 15 min vid

before i can do anything

[timid silo]

@heady tendon so can't come in vc

[heady tendon]

i think i can

[timid silo]

?

[heady tendon]

what vc?

[timid silo]

Wait a sec I am inviting you

@heady tendon

@heady tendon come in 384kbps

@heady tendon

@heady tendon fast

[heady tendon]

ook

hi

i cant chat

[timid silo]

@heady tendon why?

[heady tendon]

@timid silo

[timid silo]

@timid silo why?

[heady tendon]

like is aid real loud

hi

ok

hold on a sec still in a vid

[timid silo]

.

[heady tendon]

do u know surface area of composition figures

[timid silo]

Yes

[heady tendon]

ty

nah already done that

too easy

i need to do 7 question

@timid silo

[timid silo]

Yes

[heady tendon]

i need to do 7 questions

🥺 help

@timid silo u here

[timid silo]

@heady tendon it is the same type of sum I helped you now try on your own

[heady tendon]

54?

is the total?

@timid silo

📚

[timid silo]

@heady tendon donno

[heady tendon]

I manage to finish that one but an even harder one

<@&286206848099549185>

pls can anyone help me

[proven lotus]

stop

[drowsy cosmos]

no one will help you if you behave like that

[timid silo]

its only the green part right

2 x base x height + 5x4^2

so 2 x 5 x 4 + 5 x 16

if i am not mistaken

[light geyser]

poor @jaunty stream

Please don't spam the helpers

It does the opposite of the desired effect

[timid silo]

,, 3(12\times3)+2(3\times4)+13\times4+2(5\times12)+5\times4

[warm shale]

Vortism

[timid silo]

@heady tendon

[balmy mortar]

<@&268886789983436800> . And i got a random ghost ping here

[upbeat plinth]

.close

[obtuse pebble]

Channel closed

Closed by @upbeat plinth

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[vital frost]

hey

[obtuse pebble]

Please don't occupy multiple help channels.

[vital frost]

.close

[obtuse pebble]

Channel closed

Closed by @vital frost

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[hollow grail]

How?

[hollow grail]

how is cos/sin^2 = -1/sin

[distant moth]

Try substitution

[solar trellis]

It isn't

[hollow grail]

its question 4

i dont get it

[hollow grail]

he didnt use substitution

ok i got it now

.close

[obtuse pebble]

Channel closed

Closed by @hollow grail

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[loud tangle]

having some issues with this problem

in the center figure, ABC is a right triangle with BC = 2AB, BCD is a right isoceles triangle right in D and E is on AC s.t AE = AB.

i need help with b) prove BDE is equilateral

[loud tangle]

i got 1 60° angle down but i just cant see what to do to get the 2nd

or prove 2 sides are congruent

cube people i am hereby reaching outto any of you

<@&286206848099549185>

been 15 minutes

i counted

come onnnnn 😭

[vital bramble]

if im not mistaken abc should be a 30 60 90 triangle

[loud tangle]

yeah ik

did that already

i tried messing with the angles a whole lot and all i got was ebd is 60

anything else involves proving edc is isosceles

which idrk how to do cause all i can rly compute is bec to be 135

if anyone finds a solution ping me

ill be skipping and doing the next problems

[vital bramble]

ACB = 30, BEC = 135, (solve for third angle of triangle BEC) 180 - 135 + 30 = 15, DBE = CBE + DBC, DBE = 15 + 45, DBE = 60

isoceles right triangles are 45-45-90 triangles so DBC = 45

so thats one angle of the equilateral triangle if that helps @loud tangle

[loud tangle]

its just 1 anglei

angle

i already did that

i need to find a 2nd one

just cause 1 angle's 60 doesnt mean its equilateral

ebd is dbe

[loud tangle]

literally said here ebd is 60

[obtuse pebble]

@loud tangle Has your question been resolved?

[obtuse pebble]

@loud tangle Has your question been resolved?

Channel closed

Closed by @loud tangle

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[stark hollow]

the answer has to be 84, but i didn't understand how to get that

[proud haven]

I mean I got 36 so if anyone else can pitch in and verify before i give out any misleading info that'd be great

[stark hollow]

oh have u drawn the altitude from the vertex A? i also did that and got 36 as well, but it is only part of it

[obtuse pebble]

@stark hollow Has your question been resolved?

[stark hollow]

PLS HELPPP

[obtuse pebble]

@stark hollow Has your question been resolved?

[obtuse pebble]

@stark hollow Has your question been resolved?

[obtuse pebble]

@stark hollow Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

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

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

[real cobalt]

how did they solve this,?

[smoky slate]

It's written there

[real cobalt]

ik

but its not helpful

its ixl, ixl scks

[cedar lichen]

What do you mean it's not helpful

It says exactly what they did

[real cobalt]

it says subtract 14x, but from where?

[cedar lichen]

From both sides of the equation

[real cobalt]

wait nvm

now i get it

.close

[obtuse pebble]

Channel closed

Closed by @real cobalt

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[obtuse pebble]

@clear leaf Has your question been resolved?

[real cobalt]

is this savvas?

[clear leaf]

yes😭

[real cobalt]

ok

idk how to this

[clear leaf]

i hate savvas sm

[real cobalt]

same

type "<@&286206848099549185>"

[clear leaf]

oh alr

<@&286206848099549185>

[obtuse pebble]

@clear leaf Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[jagged edge]

how did they know it was x and y?

randomly?

[tardy epoch]

that's the definition of polar to cartesian

[jagged edge]

wait wdym

[tardy epoch]

https://math.libretexts.org/Courses/Truckee_Meadows_Community_College/TMCC%3A_Precalculus_I_and_II/Under_Construction_test2_08%3A_Further_Applications_of_Trigonometry/Under_Construction%2F%2Ftest2%2F%2F08%3A_Further_Applications_of_Trigonometry%2F%2F8.3%3A_Polar_Coordinates

[jagged edge]

Ohh

the formula

i see it now

xd

[obtuse pebble]

@jagged edge Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[static glacier]

can anyone help with the last part

answer is x+2y+3z=d

[obtuse pebble]

@static glacier Has your question been resolved?

[obtuse pebble]

@static glacier Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[obtuse pebble]

@timid silo Has your question been resolved?

Channel closed

Closed by @bleak elbow

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

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

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

[stone flint]

Hello

[stone flint]

A+b=10
Ab= 36

How to get a and b

[gilded needle]

well the first one gives you b = 10 - a, try substituting that into the second equation

[stone flint]

A×10-a=36
10a - A^2 = 36

[gilded needle]

so far so good

[stone flint]

LOL.

But nau what

[timid silo]

Quad quad

[copper spade]

Ye quadratic formula

[stone flint]

I don't get what I need

[copper spade]

Ya applied formula

Or better way ya know how to take factors and split 10a?

[stone flint]

Where

[copper spade]

Like that equation is 10a-a²=36
a²-10a+36=0

Split 10a

Ya know how to?

[stone flint]

A(a-10)

?

[copper spade]

Nah wait ig you can't split here

Use quadratic formula

[stone flint]

I keep getting wrong numbers ffs

[obtuse pebble]

@stone flint Has your question been resolved?

[copper spade]

How?

Send me your working

(Ping me if u do send me the workin)

[obtuse pebble]

@stone flint Has your question been resolved?

[stone flint]

@copper spade

[copper spade]

Slide

[stone flint]

Can u tell me precisely what are a b and c as numbers

C 36

A is 1

B is -10

Am I right or wrong about it

<@&286206848099549185>

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[timid silo]

What can I do for this

[timid silo]

$x^{2009} + 2009x + 1=0$

[warm shale]

beijing

[timid silo]

Prove it only has one solution in [-1, 0]

[sharp pecan]

in what field are you learning this?
did they teach you with modulo or with cyclic sets?

[timid silo]

Calculus

[sharp pecan]

I see

[timid silo]

I think I’m supposed to use derivatives or something related

[sharp pecan]

with this it means that the graph is only considered from -1 to 0 right?

[timid silo]

Yes

I think youre supposed to use differential proving method

[sharp pecan]

you can use derivatives

it has something to do with minimum and maximum

[timid silo]

Yea yea

[sharp pecan]

I just have a hard time explaining that without telling the answer right away ;-;

[sharp pecan]

do you already have an approach or are you stuck at the beginning?

[timid silo]

The derivative is 2009x^2008 + 2009

[sharp pecan]

mhm yes but do you already know what to do with that information?

[timid silo]

I know how to prove its always growing but I can’t prove the solution is exactly in interval-1,0

[sharp pecan]

mhm... if you know that its always growing in some interval, how could you check if it passes some certain y value inside that interval?

[timid silo]

Do you know first and second order derivatives? Or how fo calculate maxima and minima and input interval values on the derivative?

[sharp pecan]

i think they already did that to find out its always growing in the interval

since its basically required to have no min or max inside that interval then

[sharp pecan]

hint: ||plug -1 and 0 into f(x)||

[timid silo]

Second order derivative is 4034072x^2007

Odd power

Idrk but try rolle's theorem or other ways to find.. wish i could help

[sharp pecan]

I think they already figured out there is no extreme locality in that interval though...

since thats required for steady growth

[novel knoll]

Is your function continious? Do you have a theorem that says smth about what points a continious function attains? You have shown its increasing so it has at most 1 solution

Figure out what theorem gives you it has 1 solution

[timid silo]

Bro what are u doing

[stone flint]

@sharp pecan help

[sharp pecan]

lets try to not overtake someone else's channel

[obtuse pebble]

@timid silo Has your question been resolved?

Channel closed

Closed by @viral hornet

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[quaint finch]

I feel like generalised pigeonhole principle is required

ans is 225

[sharp pecan]

is there more to the task?
if I get it right, you may have level 1 with 29 people, then the other levels have 0 and its still true

[quaint finch]

I don't understand what ur tryin to say

[sharp pecan]

I don't quite know how to interpret the task.
there are a total of n objects and 8 containers
One of the containers must have at least 29 objects

is that correct?

[quaint finch]

yes

[sharp pecan]

so, the minimum n is 29 right?

[quaint finch]

yes

[sharp pecan]

since one object has 29, the other 7 all have 0

so how 225?

[quaint finch]

the others have to be more than 29

[sharp pecan]

sorry, but where does that condition come from?

[quaint finch]

cuz it says at least 29

so 29 or more

[sharp pecan]

yes, but ONE level has to have 29 or more, not all of them

right?

[quaint finch]

yes

[sharp pecan]

so if level 1 has 29 people the condition is fullfilled.
how do you know the other levels need 29 too?

[quaint finch]

I get you

[sharp pecan]

at least 29 people in at least one level I think that really means that ONE level needs to be 29+

[quaint finch]

yh ur right

meaning that the other 7 can have 0?

[sharp pecan]

I think

something like {29 ,0 ,0 ,0 ,0 ,0 ,0 ,0} could be a distribution over the levels

since you have at least 29 people in at least one level

so min(n)=29

so the function for n(x) could be n(x)=29+x, x is a natural number

[quaint finch]

u reckon we have a max value too

[sharp pecan]

do we?

in practice yes, but there is nothing said that you cannot have infinite people in a single level

and you have 8 levels

something like {12321321312312321 ,12312321321312, ...} would be possible

[quaint finch]

so there are many possible answers but the lowest it can get to is 29

[sharp pecan]

yes

to formulate n you could say
${n=29+x|x\in \mathbb{N}}$

[warm shale]

Jigglyproff

[quaint finch]

alright, that helps, thx

[sharp pecan]

yw

[quaint finch]

do u go to uni

[sharp pecan]

yes

[quaint finch]

no wonder

[sharp pecan]

im here to prognosticate on my own math tasks XD

[quaint finch]

ha

.close

[obtuse pebble]

Channel closed

Closed by @quaint finch

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

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

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

[tribal iris]

Am currently stuck at example A the last part

How did we get that( -xe^-x )+(-e^-x) | 0 to infinity equals 1

[mighty geyser]

take limits as x -> 0 and x -> inf

[tribal iris]

What do i do for the lim at infinity

[mighty geyser]

might be easier if you write as a fraction

[tribal iris]

How

[obtuse pebble]

@tribal iris Has your question been resolved?

[mighty geyser]

(-x-1)/e^x

[tribal iris]

Ohhhhhhhhhh

Then i do Lhopital

1/e^x

So =0

So [0] - [-1] =1

[tribal iris]

Kk thanks alot

[obtuse pebble]

Channel closed

Closed by @tribal iris

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[balmy scaffold]

wait

yes

(3; -2) (4; -2) (3; -4)

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[balmy scaffold]

|x-1|+|x-2|+|x-3|+...|x-n|>=59

[obtuse pebble]

Please don't occupy multiple help channels.

[balmy scaffold]

n is the smallest possible natural number
to find n
and x is any real number

[timid silo]

Wat

[balmy scaffold]

|x-1|+|x-2|+|x-3|+...|x-n|>=59

[timid silo]

send screenshot

[balmy scaffold]

how do i calculate it?

[timid silo]

I dont understand you writing send pic

[balmy scaffold]

no its in another language

[timid silo]

Send idc

[timber island]

no its not 1

[balmy scaffold]

waht 1

@timid silo

?

[timid silo]

Idk 1

[mighty geyser]

consider |x-1|+|x-n|, what's the minimum value of this

[balmy scaffold]

nonono sorry its

smallest possible natural value

[mighty geyser]

in terms of n, if x can vary

[balmy scaffold]

x is any real number

pls help

[loud tangle]

so what do you plan to acomplish with this?

like find an n that follows this rule

[balmy scaffold]

ye

[loud tangle]

well it has to at least be bigger than the middle term of the sum

so its at the very least positive

no wait

its mod anyway

[ebon remnant]

what do you need to find from this

n?

[balmy scaffold]

i think its 16 but i cant prove it without a graph

[balmy scaffold]

yes yes

[ebon remnant]

is there any specific things for n?

[loud tangle]

can you set x to say 0

since youre finding the smallest value

[balmy scaffold]

its natural number

[balmy scaffold]

its any real number

[loud tangle]

yeah but you wanna find the min value of n

[balmy scaffold]

umm so

[loud tangle]

which would require the min value of x

and in mod its 0

[ebon remnant]

this is always correct if you said n is natural number cause everything on abs is always +, so n is real number?

nvm wait

wait

[loud tangle]

not really

[ebon remnant]

<59

wait

[balmy scaffold]

x can be any real number i have to find the n (smallest) that is possible to prove thin inequality

[loud tangle]

so set x to 0

then work out the sum

[balmy scaffold]

im sorry for lettin yall comfused its more or equal to 59

[loud tangle]

do u wanna prove the inequality for all x

for all x and with the smallest n

?

[balmy scaffold]

the Arithmetic progression of "|x-1|+|x-2|+|x-3|+..." goes until it hits |x-n|

[loud tangle]

for all x?

like for every value x can have you wanna find n?

or just the smallest n

[balmy scaffold]

x is any real number

[loud tangle]

arbitrary from x

so for all values of x

[balmy scaffold]

yes

[loud tangle]

alr then

the graph idea was good tbh

thats what i thought of first

[balmy scaffold]

but i was set to like

prove it without graph

[loud tangle]

honestly it try induction

set x to like the minimum value then see if u notice anything that is the same for x is 0 and 1

then by induction works for all x

[balmy scaffold]

i think it is induction

x is like

one number right

[loud tangle]

yes

[balmy scaffold]

yea i think so

[ebon remnant]

hmm

[ebon remnant]

can't we say its = 59

cuz we need to find smalled n natural number to get it right

[balmy scaffold]

yea

[ebon remnant]

sry nvm

the number of the sum can;t be 59

like 60 something

[loud tangle]

wouldnt do anything

if were doing induction you kinda have it be in inequality form

[balmy scaffold]

but if its =59 and if n has to be natural number it wont be possible

[loud tangle]

if x is 0 it would just be 1 + 2 + 3 + ... + n >= 59 => ((n)(n+1))/2 >= 59
=> n(n+1)>=118

if x = 1 => 1 + 2 + 3 + ... + n-1 >= 59
=> n(n-1) >= 59 => n²-n >= 118
.
.
.
meaning for all x the series reduces to 1 + 2 + ... n-x>= 59 => (n-x)(n-x+1)>=118
maybe that general formula does something

x is real right?

meaning it can be irrational

[balmy scaffold]

yea

[balmy scaffold]

umm how did u get ((n)(n+1))/2>=59

[loud tangle]

gauss' formula for sums

for sums of n terms and ration 1

[balmy scaffold]

ohhhh

i get it now

[ebon remnant]

i don't understand about n(n-1)

idk about that

[loud tangle]

(n-1)(n-1+1) = n(n-1)

[ebon remnant]

@balmy scaffold can u give me the question pic?

[balmy scaffold]

umm sure but

its in another language

[ebon remnant]

um can u copy paste it?

can u translate it with google translate?

or u may copy the question

[balmy scaffold]

sure

For any real number x, find the smallest natural number n such that |x-1|+|x-2|+|x-3|+...|x-n|>=59 is satisfied.

@ebon remnant

[ebon remnant]

thanks

[loud tangle]

n is natural right?

meaning it cant be negative

[ebon remnant]

ye

[balmy scaffold]

ye

[ebon remnant]

start from 1

[loud tangle]

working on that final equation and assuming its 118 cause its minunum

[balmy scaffold]

man my brain goin real smooth ngl

118?

[loud tangle]

i got $n = \frac{2x \pm \sqrt{437}}{2}$

[warm shale]

Kel.plush

[loud tangle]

by me i mean symbolav

symbolab

[ebon remnant]

the n can't be decimal

[balmy scaffold]

how did u got the 437?

[ebon remnant]

its -b +- thingy

[balmy scaffold]

oh

[loud tangle]

i dont really think there is a minimum value that satisfies all values of x

cause x is dependent on n

so if we have to define n automatically x gets restrained

hold on i fucked up

i realized i did

[balmy scaffold]

where

[loud tangle]

ok so here i assumed for the values where n > x it would be negative and cancel with a following term

but its in mod

so its always +

[balmy scaffold]

ye

[loud tangle]

ok so u can split it in 2 sums

1 for x>n and one for the opposite

and if x=n its 0 anyway

[balmy scaffold]

and?

[loud tangle]

and that may partlemy be why it resulted that goofy ahh result

oh you literally just put a - in front of the first bit

[balmy scaffold]

fisrt bit?

[loud tangle]

hold on

ok so the first half of the sum is solely dependant on x

[balmy scaffold]

yes?

[loud tangle]

lemme write it out in latex

if x is 0 it would just be $1 + 2 + 3 + ... + n \ge 59 => \frac{n(n+1)}{2} \ge 59
=> (n)(n+1)\ge 118 =>n²+n \ge 118\\vdot\
\vdot\
\vdot\$
if x is 2, it would mean $|2-1|+|2
-2|+|2-3|+...+|2-n| = 1 + 0 + 1 + 2 + 3 + ... + |3-n| = 1 + \frac{|3-n|(|3-n|+1)}{2}\
\vdot\
\vdot\
\vdot\$
for a general scenario, it could be interpreted as 2 sums, one of x-1 terms and one of n-x+1 terms (the +1 being the case where x = 0)

therefore, a general formula would be:
$|x-1|+|x-2|+...+|x-n| = \frac{x(x-1)}{2} + \frac{|x-n|(|x-n|+2)}{2}$

im still editing hold on

[balmy scaffold]

oke

[warm shale]

Kel.plush

[balmy scaffold]

still editing?

[loud tangle]

if x is 0 it would just be $1 + 2 + 3 + ... + n \ge 59 => \frac{0(0-1)}{2}+\frac{n(n+1)}{2} \ge 59
=> (n)(n+1)\ge 118 =>n²+n \ge 118\\vdot\
\vdot\
\vdot\$
if x is 2, it would mean $|2-1|+|2
-2|+|2-3|+...+|2-n| = 1 + 0 + 1 + 2 + 3 + ... + |3-n| = \frac{2(2-1)}{2} + \frac{|3-n|(|3-n|+1)}{2}\
\vdot\
\vdot\
\vdot\$
for a general scenario, it could be interpreted as 2 sums, one of x-1 terms and one of n-x+1 terms (the +1 being the case where x = 0)

therefore, a general formula would be:
$|x-1|+|x-2|+...+|x-n| = \frac{x(x-1)}{2} + \frac{|x-n|(|x-n|+2)}{2}=\frac{x(x-1)+|x-n|(|x-n|+1)}{2}$

i think thats it

i think

this is very confusing tbh

[balmy scaffold]

it is

[loud tangle]

wait lemme write the first sum in a clearer way in the first 2 examples

[warm shale]

Kel.plush

[loud tangle]

there

phew

done

@balmy scaffold

[balmy scaffold]

umm

[loud tangle]

i might have fucked up somewhere

[balmy scaffold]

so the general formula right

[loud tangle]

yeah i think thats it

thats the sum rewritten for all x and n

[balmy scaffold]

$\frac{x(x-1)+|x-n|(|x-n|+1)}{2}$

[warm shale]

water bottle

[loud tangle]

yes

[balmy scaffold]

is it equal to 0?

or 59

[loud tangle]

its equal to the sum

[balmy scaffold]

on

[loud tangle]

if were taking the minimum its 59

[balmy scaffold]

ohh

as you were saying

n is directly Proportional to x

?

[ebon remnant]

its from other server

i don't understand the whole thing (I'm on grade 10)

why theres sigma

[balmy scaffold]

same tbh

[ebon remnant]

yo

[loud tangle]

is that the same exercise?

[balmy scaffold]

but still getting the grasp of it

[loud tangle]

cause idk where the floor came from

[ebon remnant]

i asked the same question

Anonymous

I'd probably approach the problem by finding the value of x for which the sum is minimal, which is x=(n+1)/2
then I'd split the summation into two parts, one for k less than (n+1)/2 and one for the rest
then I'd make the second summation negative and remove the absolute values
we have summation formulas that would apply to this scenario
and then you should be able to solve for n

wait that's actually almost right
but the minimal value of n for which this is true is 15
but it should be 16

[balmy scaffold]

what server did u ask tho

[ebon remnant]

its a mathematics server too

[balmy scaffold]

can u like inv me

[ebon remnant]

sure

[ebon remnant]

Anonymous

I said it was floor((n+1)/2) to simplify the algebra, but I think it was supposed to be floor((n-1)/2) or floor((n+1)/2)-1

[balmy scaffold]

hm

also this problem is from 10th olympiad

but i think thats it

thank you for your time

.close

[obtuse pebble]

Channel closed

Closed by @balmy scaffold

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[wintry sundial]

Why am I getting p = .5

[wintry sundial]

P is supposed to be .125

idk what i did wrong

in the 2nd question

[obtuse pebble]

@wintry sundial Has your question been resolved?

[wintry sundial]

<@&286206848099549185>

[storm tapir]

wait but what did you do to calculate p

[wintry sundial]

I just expanded

[storm tapir]

should be $\frac{1}{2}(y+4) = (x-2)^2$

[warm shale]

asakura_akio

[wintry sundial]

and got this

[storm tapir]

and then $4p = \frac{1}{2}$

wait

om

[warm shale]

asakura_akio

[storm tapir]

did you forget the 4 part

[wintry sundial]

no

(y+4)=2(x-2)²

4p=2

p=.5

what did i do wrong

how did 4p=1/2

[storm tapir]

yeah the 2 is supposed to go to the other side

you should have nothing except (x-2)^2 on the right side

you're in this form because you have x^2 not y^2 so 4p is on the y-k side

[wintry sundial]

Is (y+4)/2=(x-2)² same as 1/2(y+4)=(x-2)²

[storm tapir]

yes

[wintry sundial]

Im really sorry but can you explain why it is the same

[storm tapir]

yeah, you just divided by 2 on both sides of the equation

1/2 times something is the same as dividing that something by 2

it's just a different way to write it

its like saying 1/2 times 2 is the same as 2/2

or 1/2 times 5 is the same as 5/2

you just have y+4 here instead of a number

[wintry sundial]

omg yeah i wrote it down and got it

thank you so much

[storm tapir]

np

[wintry sundial]

.close

[obtuse pebble]

Channel closed

Closed by @wintry sundial

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[vestal niche]

Where does the 1/11 and 11 infront of the e come from

[fierce lagoon]

Application of reversing the chain rule

Basically the direct result of a u-sub

If you let u = 11x

then du = 11 dx
=> dx = 1/11 du

But they instead wanted to keep dx so they rather split 1 = 11 • 1/11

[vestal niche]

Wait what?

I have a formula sheet with integration of e and derivative of e, but neither show or explain why that is done

Same thing with like sin1/7 x, idk why it’s -7cos1/7x

[fierce lagoon]

$$\begin{align*}
\int e^{11x}dx &= \frac{1}{11}\int e^u du \
&= \frac{1}{11}e^u \
&= \frac{1}{11}e^{11x}
\end{align*}$$

[warm shale]

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

[fierce lagoon]

They basically did this but in one step

[vestal niche]

Okay but the very first step, why is there a 1/11 infront of the integral

[fierce lagoon]

If you're familiar with the chain rule, deriving e^(11x) gets you 11e^(11x). They manipulated so that you're only integrating 11e^(11x) -> e^(11x) + C, but you need to balance out that 11 with a 1/11

[vestal niche]

I thought i was familiar with chain rule but I guess not

[fierce lagoon]

$$u = 11x$$
$$du = 11 dx$$
$$\implies dx = \frac{1}{11} du$$

[warm shale]

Umbraleviathan

[vestal niche]

Okay but it’s an e not an x, does that not change anythjnf?

Anything

[fierce lagoon]

Doesn't change anything

[vestal niche]

Okay

[fierce lagoon]

If you do this

$$e^{11x}dx \rightarrow \frac{1}{11}e^udu$$

[warm shale]

Umbraleviathan

[vestal niche]

So e^ what ever just stays the same but because we have a dx we need to have a value for the dx

[fierce lagoon]

Wdym a value for dx

[vestal niche]

Like you’re taking the derivative of 11x to get du/dx = 11, then multiplying 11 by dx to get du = 11dx, then dividing du by 11 to get du/11 = dx

[fierce lagoon]

Yeah

[vestal niche]

So the value of dx is du/11 which is the same as the 1/11 x du

[fierce lagoon]

You could say that yeah

dx is just equivalent to du/11 if you let u = 11x

[vestal niche]

Right so when we have the formula of integral e^u du = e^u + c

We actually need to re arrange what ever were taking the integral of to have a du there as well

[fierce lagoon]

well you can't just forget about the 1/11

You're integrating 1/11 • e^u du

That becomes 1/11 e^u + C

You had let u = 11x so just place it back in

[vestal niche]

Right but what I meant was like we’re integrating e^11x dx

There initially is no 1/11 there

But that dx has a value of du/11, so we take the 1/11 out infront, but idk why we keep the dx, unless it’s because we’re also putting an 11 infront of the e

[fierce lagoon]

Well what they're doing is similar. You would agree that $e^{11x} = 1 \cdot e^{11x}$?

[warm shale]

Umbraleviathan

[fierce lagoon]

Because what they basically said is "let's do a u-sub but fuck showing any of that"

[fierce lagoon]

@vestal niche

[vestal niche]

Rofl yes

[fierce lagoon]

Their logic is similar to u-sub. The only reason why I'm showing you the u-sub is to show their logic behind doing that (basically expanding a step)

[vestal niche]

I agree 11/11 is 1, it’s just dumb to make it complicated for no reason but I guess there is a reason.

[fierce lagoon]

And 11 • 1/11 = 1

[vestal niche]

Okay

[fierce lagoon]

So split that 1

And then what they did was this

[vestal niche]

So they split the 1 into 11 x 1/11 to do a u sub randomly

[fierce lagoon]

They understand that if you were to derive e^(11x), you would get 11e^(11x) right

[vestal niche]

Right cause it’s e^u x u prime

[fierce lagoon]

But the integrand is e^(11x), not 11e^(11x)

[vestal niche]

Which would be 11e^11x

[fierce lagoon]

So they slapped a 1/11 to match the integrand

[vestal niche]

Right right right okay

[fierce lagoon]

But the same intuition is used for u-sub

[vestal niche]

Same as integral of 6x^2 is 2x^3

[fierce lagoon]

Yeah

[vestal niche]

Okay so even with e

It’s going to be the same integration rules as anything else like x

You need to cancel out the constant if there’s one there

Okay

That makes sense

So the formula I have for e^ du, I need to plug a du value in before

[fierce lagoon]

If need be yeah

[vestal niche]

Kk

Thank you

Appreciate you dumbing it down for me lol

[fierce lagoon]

Lol np

[vestal niche]

Have a nice Christmas man haha

.close

[obtuse pebble]

Channel closed

Closed by @vestal niche

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

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

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

[obtuse pebble]

Please don't occupy multiple help channels.

@winged jacinth Has your question been resolved?

Channel closed

Closed by @winged jacinth

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[timid silo]

How would I rationalize the numerator

[timid silo]

Rationalise the Numerator? There is not much else you can or should do to that tbh

Ok

Maybe provide context on what you're trying to do

thats all

but that the answer

I got it wrong

You can multiply with (sqrt(r)-sqrt(7))/(sqrt(r)-sqrt(7)) if you want to make the thing be rational ib the numerator then

Oh hm

how about this one

Yeah idk should be right, maybe expand the thing

[fierce lagoon]

Multiply by the conjugate

[timid silo]

conjugate is sqrt (x^2+3) +x

@fierce lagoon

[fierce lagoon]

That ain't the conjugate

Where did the 2 come from

Also the conjugate is a fraction

[timid silo]

thats what I meant

[fierce lagoon]

Multiply by $\frac{\sqrt{x^2 + 3} + x}{\sqrt{x^2 + 3} + x}$

[warm shale]

Umbraleviathan

[obtuse pebble]

@timid silo Has your question been resolved?

Channel closed

Closed by @dawn lotus

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[hybrid tartan]

yo

[hybrid tartan]

Can anyone correct me if im wrong

(i) == 0.00604
(ii) == 0.00035
(iii) == 3

[hybrid tartan]

These are "past" exam papers that im doing, not homework.

[timid silo]

.6 == 0.6%?

[hybrid tartan]

60%

Are you referring too the chance of a server being busy?

[open raven]

how did you reach (ii)?

(i) and (iii) seem to be correct

but on the (ii) is giving me another answer

[hybrid tartan]

are you using binomial distribution?

for part (ii) i just done .6^3 * .4^7

[open raven]

no, my logic was "less that 4 busy" = 0 busy + 1 busy + 2 busy + 3 busy

[hybrid tartan]

Aight give me a min, ill try it again

[open raven]

and im assuming order doesn't matter, right?

[hybrid tartan]

no

would it be 0.01229

For (ii)

wait

[open raven]

i dont think so

[brazen jungle]

bro?

[hybrid tartan]

.05476

[open raven]

it didn't give me that value, what was your way to thinking, i may be forgetting something

[open raven]

oh yeah, okay, i forgot the binomials

[hybrid tartan]

but instead of <= i done <

[open raven]

you are write if you did from x=0 to x=3

[hybrid tartan]

yes

[open raven]

its been a while since i have done this 😅

[hybrid tartan]

Well i think im right

[open raven]

yeah, i agree!

[hybrid tartan]

.close

[obtuse pebble]

Channel closed

Closed by @hybrid tartan

Use .reopen if this was a mistake.

[solemn crypt]

.reopen

🧍🏿‍♂️

[hybrid tartan]

.reopen

[obtuse pebble]

✅

[hybrid tartan]

.close

[obtuse pebble]

Channel closed

Closed by @hybrid tartan

Use .reopen if this was a mistake.

[solemn crypt]

☹️

🦌

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

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

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

[hybrid tartan]

yo

[hybrid tartan]

Can some correct me if im wrong

[broken dove]

Yeah

[hybrid tartan]

(i) == 1/5
(ii) == 1/25
(iii) == 1/625
(iv) == 24/625
(v) == 11/25
(vi) == 1/25

[broken dove]

give me a sec

[rigid lintel]

ii is wrong

1/5

iii correct

iv is wrong

v is correct

[hybrid tartan]

Ok i know why i got q2 wrong, but what about iv

idk what i done wrong

[rigid lintel]

vi is correct

well if theres no balls of the same colour

then theres 5 possible colours for the first ball

4 possible for the second

3 2 1

5 * 4 * 3 * 2 * 1

= 120

1/120

[broken dove]

isnt iv like 0

or am i trippin

[rigid lintel]

you have 5 spots for 5 different colours

so they can all hjave different colours

so not 0

[broken dove]

bro i got
1

1000000

is my math off

[rigid lintel]

how did tyou get that

wait actually

oh you cant type that lmao

,calc 1 * 4/5 * 3/5 * 2/5 * 1/5

[warm shale]

Result:

0.0384

[rigid lintel]

,calc 1/120

[warm shale]

Result:

0.0083333333333333

[rigid lintel]

ohhhhhhhhhhhhhhhhhhhhhhhhhhhh

,calc 24/625

[warm shale]

Result:

0.0384

[rigid lintel]

you were right

lol

iv correct

my bad

[hybrid tartan]

I got it by 5/5 * 4/5 * 3/5 * 2/5 * 1/5

[rigid lintel]

yeah

[rigid lintel]

soi this

[hybrid tartan]

I also got another question, how would I be able to approach this

These questions are also not homework, they are past exam papers which im doing as revision

Can anyone give me a hint?

[solemn crypt]

12

[hybrid tartan]

.close

[obtuse pebble]

Channel closed

Closed by @hybrid tartan

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

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

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

[timid silo]

hi my teacher sent us this for a task and i’m confused by the task like what am i suppose to do and how if you could give hints

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?