help-0

[past sundial]

not sure

[slate jolt]

i solve for b

[past sundial]

did you divide all and shift?

[slate jolt]

i divide the equation by pi/6 * 18^3

[past sundial]

9 = bpi/618^3

96/pi=b18^3

96/pi18^3 =b?

[slate jolt]

yes but dont put multiple * symbols

discord doesnt like it

[past sundial]

oh ok

[slate jolt]

so now you use your calculator to get b

and then you can compute the volume in pumps for the unknown ball

[past sundial]

Where did the formula for v small p = b*v come from?

[past sundial]

Yeah but is it like one of those formulas that come down on ancient history or is it like a use this to find this formula?

like 4/3pir^3 is a ancient formula

so it is set in stone

[slate jolt]

well the equation Vp = b* V is just the mathematical expression of Vp and V being proportional

it litterally means that there exists a ratio b such that Vp = b*V

[slate jolt]

its not the fact that its ancient that makes it true

i know how to prove it

on my own

[past sundial]

my teacher says it invovles a lot of calculus

But I don’t know calculus

so she said that it is just is for now

[slate jolt]

things are not given true to you by some authority in math, they are proven first

but since you trust your teacher

you can use some facts she gives you

[past sundial]

I got confused once with my teacher though

[slate jolt]

as if u saw the proof

[past sundial]

like area of cone

[slate jolt]

area or volume?

[past sundial]

Volume

So if you use radius * height/2 u get triangle

And then multiply by circumference

But then there is no 1/3?

[slate jolt]

you just multiply by 2pi

since you rotate it

[past sundial]

yes

But then there is 1/3 in formula

Idk why

1/3pi r ^2 *h

that’s formula

but when I try to map it in my mind I can’t

(r(h/2)2r)pi

and then she say there is a square when I do that so it doesn’t work if you spin it?

[slate jolt]

actually thats wrong

[past sundial]

?

this is what I think

but it doesn’t match up with cone formula

Google say /3 not 2

[slate jolt]

yeah your formula is weird

but i dont know whats going on in your head to tell you what is wrong with your thinking tho

you try to apply the same logic as with a cylinder maybe

[past sundial]

So to make one triangle

H*r *1/2

[slate jolt]

the area of the triangle is right

the one of the circle too

[past sundial]

But then you need a 360 of triangles

[hardy gulch]

if you want the volume from "spinning the triangle"

you need to integrate

[slate jolt]

but why would you just multiply them both together

[slate jolt]

i dont think they're there yet

[past sundial]

So you multiply d pi which is 2*r times pi

[past sundial]

What’s that

[slate jolt]

its calculus

[past sundial]

oh
I don’t know calculus yet

[hardy gulch]

the best way to explain why there is 1/3 in the formula is this

take a cube

cut it up into 3 equal pyramids

[past sundial]

Yep

[hardy gulch]

let the length of the cube be 1

then the volume is 1

then it means that each pyramid has a volume of 1/3

this is how you cut the cube up

these 3 pyramids are identical

see https://www.cuemath.com/measurement/volume-of-pyramid/

the problem with this argument is that it doesn't extend to cones very obviously

[past sundial]

ok I have to go right now thank you though

[hardy gulch]

np

[past sundial]

!close

?close

[hardy gulch]

.close

[past sundial]

.close

[lone heart]

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[fluid flame]

$3,4,7,7,13,13,21,22,31,34,x$

Find x

[ocean seal]

Plumdan

[rare gale]

doesn't seem like enough information

[vale wigeon]

and precisely because of that, the answer is 19

malicious compliance theorem

[fluid flame]

...

That's not even in the choices my g

[vale wigeon]

ain't my fault you withheld the problem statement from us.

[fluid flame]

There's no statement

That's the literal sequence

Choices are

[vale wigeon]

there is only this data set and you are told nothing else about it at all?

[fluid flame]

Yep.

[vale wigeon]

nothing like "the mean of this data set is 69" either?

[fluid flame]

Nope

Nothing

Just

[vale wigeon]

then it's impossible.

[fluid flame]

find x

[vale wigeon]

it's a guessing game.

[fluid flame]

Bruh

[vale wigeon]

you are given zero information to go on.

[fluid flame]

B r u h

Bet thanks for the support ;-;

[prime badge]

one choice may make significantly more sense

[fluid flame]

Well

Here they are then

D)52
C)51
B)43
A)42

[brittle elbow]

It’s C, always guess C my guy 💪

[rare gale]

What if i gave you this start of a sequence, what would be the next number: 1, 2, 4, 8, 16, ... ?

[fluid flame]

32

[rare gale]

right, if you were talking about the sequence of powers of 2

[fluid flame]

I hate you

💀

Just what the fuck is that

[rare gale]

so the point is, just because you know the first terms of a sequence, is doesn't define the rest of it

[brittle elbow]

I-

[vale wigeon]

that was also my point, though i made it a little less directly

[rare gale]

that is the sequence that is defined by the number of regions formed by dividing a circle by placing however many points on its edges then connecting those with segments

so your question if that's how it was given to you is just badly formed

[fluid flame]

So its the Question's fault?

[vale wigeon]

yes

[rare gale]

basically lol

[vale wigeon]

it's the question's fault

[fluid flame]

Bet thanks then

.close

[lone heart]

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[tight locust]

cursed

[tight locust]

.close

[lone heart]

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[lone heart]

@past dagger Has your question been resolved?

[lone heart]

@past dagger Has your question been resolved?

[vale wigeon]

@past dagger have you made any progress so far on this

[lone heart]

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[lone heart]

@mint bronze Has your question been resolved?

[lone heart]

@mint bronze Has your question been resolved?

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[lethal stump]

i was thinking about maybe doing left hand/ right hand limits but idk if that works because the domain is only positive reals

bruh

desmos saying lim is undefined

how about using a sub

u = 1/x

then evaluating the limit as u -> infinity

which is equivalent to evaluating the lim as x -> 0

[bitter vault]

Is that the whole question? F is not defined at 0, but has domain [0,inf).

[lethal stump]

F is not defined at x = 0 because F(0) = 0/0?

[hasty bramble]

oh

[bitter vault]

You've solved (a) wrongly. We can see that F(0) = 0, and F(x) = x/(1-e^-x) for x>0. So since lim_{x to 0+} = 1 =/= F(0) = 0, F is not continuous at 0

yes

[lone heart]

@hasty bramble Has your question been resolved?

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[manic harness]

A car is traveling in a straight road at a speed of 10km/hr. While passing by a man on the roadside, the driver throws an object at 10√2 km/hr towards him. At which angle he had to throw the object parallelly with ground with respect to the car speed direction so that the object reaches the man's hand? (neglecting "g")

[manic harness]

Cool question, but what do I do here

[olive loom]

if the driver throws the object straight at him, the object wont reach the man - bcz the object will have a speed in the direction of the car as well

to make the object go straight to the man - the driver will have to throw the object a little in the backwards direction

u will have to throw it at an angle such that the objects forward and backwards speeds cancel out

[manic harness]

Is it angle with the horizontal or angle with the vertical?

[olive loom]

angle with the cars speed

direction of the car's speed

like this
the car's speed direction is red
the person is P
and the object is thrown in the green direction

[manic harness]

Okay makes sense

Thanks

.close

[lone heart]

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[fast cargo]

anyone know how to do this?

[worn fox]

#help-5 you guys have the exact same question

[rigid smelt]

is this a test because you seem really urgent with these questions being answered

[rare gale]

It seems your teacher misspelled differentiable

NOT differential function

[lone heart]

@fast cargo Has your question been resolved?

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[golden frost]

Hey guys, need your help

A number with 7 digits, with a composition of 2 digits 1, 2 digits 2 and numbers 3, 4, 5 each with 1 digit. How many ways to randomize the number without the same number next to each other

[crisp pine]

Let's say they were all different numbers

Do you have any ideas about how many ways you could order them in that case

[golden frost]

hmm

without the same number next to each?

5040 ways?

[crisp pine]

Well we are thinking about a different example. Where you just have the numbers 1,2,3,4,5,6,7

Yes

How did u calculate that

[golden frost]

7!

[crisp pine]

Exactly

[golden frost]

factorial

[crisp pine]

Let's make it slightly harder, and say that we have one repeated digit

So 1,1,2,3,4,5,6

If they were all the different, it would still be 7!

[golden frost]

so this is still 7!?

[crisp pine]

Not quite

Because 1123456

And 1123456

Would be counted twice

For example

So swapping the 1s doesn't change the number

Let's say we just have two 1s, how many different ways can they be placed

If the order matters

[golden frost]

hmm

[crisp pine]

Or more exactly, how many ways can you order two things

[golden frost]

im confused

i dont know

[crisp pine]

Sorry

[golden frost]

noo, dont sorry

[crisp pine]

So its 2! I.e 2 right?

[main ridge]

How can I determine if an index lies on
0 ,1, 2, or 3
with a single equation

[golden frost]

Ohhhh, yeahh

[crisp pine]

Could you open a new help channel

[main ridge]

sur esorry

[crisp pine]

No worries

[crisp pine]

So if we divided our 7! By 2!

Then that accounts for each of the times where 1s have been swapped around

Does that make sense?

[golden frost]

yess

[crisp pine]

However

We have repeated 2s and 1s

Do you have any ideas what we should do in that case

[golden frost]

7! / 2! 2!

?

[crisp pine]

Yes :)

And that's the solution

https://brilliant.org/wiki/permutations-with-repetition/

Here is a good article about this topic

I hope that was helpful

[golden frost]

So, there are 1260 ways?

But in these 1260 ways, are the 1s and 2s could be next to each other?

like 1134522

[crisp pine]

Does that matter?

[golden frost]

yeah

A number with 7 digits, with a composition of 2 digits 1, 2 digits 2 and numbers 3, 4, 5 each with 1 digit. How many ways to randomize the number without the same number next to each other

[crisp pine]

Ohhh yes it does

[golden frost]

without the same number next to each other

[crisp pine]

Hold on a sec then

[golden frost]

okay

[crisp pine]

That makes it a bit trickier

This might take me a while, I'll get back to you if I think of anything

[golden frost]

yeahh sure

take your time

[crisp pine]

Yeah I'm not sure

My permutations knowledge only goes so far lol sorry

[golden frost]

Ohh okay, that's fine

thankyouu

A number with 7 digits, with a composition of 2 digits 1, 2 digits 2 and numbers 3, 4, 5 each with 1 digit. How many ways to randomize the number without the same number next to each other

<@&286206848099549185>

[olive loom]

i think the way to do it is - first calculate the number of ways such that theres always a same number next to each other
and then subtract it from the total number of ways possible to randomize

[golden frost]

5040 ways (same number next to each other)

[golden frost]

hwo to calculate total number of ways possible

[olive loom]

no i dont think thats right

[golden frost]

Ohh

1260

[olive loom]

thats the total number of ways

u have 7 items to rearrange and in which 1 is 2 times and 2 is 2 times and rest are once

so the total number of ways u can rearrange that would be 7!/2!2!

[olive loom]

and thats 1260

[golden frost]

yeahh

and then

is that the total number of ways possible to randomize?

[olive loom]

yeh

now u take the cases of when at least one of the numbers (either 1 or 2) are together

so let say 1 is always together

[golden frost]

okay

[olive loom]

its like we tied those two 1s together

it now becomes a single object

so now u will have total 6 objects to randomaise with the number 2 being twice

so total number of ways to do that would be?

[golden frost]

the two 1s become a single object so there now there are 6 digits? with two 2s

?

[olive loom]

yeh

[olive loom]

how many ways to arrange this?

[golden frost]

7!/2!

[olive loom]

6*

[golden frost]

ohh yeah

6!/2!

[olive loom]

right

now similarly now assume we tied the two 2s together

now we have again 6 objects with two 1s

how many ways to arrange this?

[golden frost]

so the two 1s together and two 2s together?

5 objects

[olive loom]

no the 1s are free

they are not tied

[golden frost]

Ohh

the same I think, 6!/2!

?

[olive loom]

yep

now finally take the case when both of them are tied

[olive loom]

this one

how many ways to arrange this?

[golden frost]

5!

[olive loom]

yep

so we have 6!/2!+6!/2!+5! ways to arrange those 7 digits
such that theres always atleast one same number together

[golden frost]

yes

[olive loom]

so whats the total number of ways such that none will be together?

[golden frost]

hmm

[golden frost]

is it similar like these?

[olive loom]

no its like saying
if there is x probability of me passing an exam
then 1-x will be the probability of me failing

[golden frost]

okay

So x = 6!/2!+6!/2!+5! ?

[olive loom]

so basically u have subtract this with total number of ways to arrange the 7 digitts

[olive loom]

this

[golden frost]

OHHHH

so 1260 - (6!/2!+6!/2!+5!)?

[olive loom]

put brackets

yep thats right

[golden frost]

Ohhhh i see

so the answer is 420?

[olive loom]

it should be

[olive loom]

the calculation is right

[golden frost]

Okayy

but

the calculation is the same as 7!/2! 3!

any idea or just a coincidance?

[olive loom]

hmmm let me think

[olive loom]

is this in given answers or something?

[golden frost]

no

its my calculation earlier

[olive loom]

oh how did u get it?

[golden frost]

7 digits

2! is bcs 1212 and 2121

and 3! is bcs in every 1212 they can move a side like 1 _ 212

[golden frost]

is it true LOL

idk I was just trying everything

[olive loom]

hmmmm doesnt seem like it

[golden frost]

HAHAHAHA

[olive loom]

do u have the answer key or solution for that problem?

[golden frost]

no i dont have

[olive loom]

i see

then i dont know if 7!/2!3! is just a coincidence or not

but 420 should be right unless we missed something

[golden frost]

Okayyy

THANKYOUU SM

GBU @olive loom

[olive loom]

u can do .close to close the channel

[golden frost]

ohh yeah

.close

[lone heart]

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[gritty slate]

Could anyone help with https://www.reddit.com/r/askmath/comments/uoouy4/understanding_als_alternating_least_square/ ?

[lone heart]

@gritty slate Has your question been resolved?

[lone heart]

@gritty slate Has your question been resolved?

[tacit arch]

just copy paste your question here

[cold tangle]

can anyone help with a question

[lone heart]

@gritty slate Has your question been resolved?

[wicked relic]

#❓how-to-get-help just ask the question. Don't ask if anyone can help, and don't ask in the occupied rooms ask in the available ones

[lone heart]

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[wide cradle]

hi

[red timber]

go on

[wide cradle]

u do it

pls

mommy

[red timber]

can anyone help us solve this

[minor peak]

is this gcse

[wide cradle]

yes

[minor peak]

😹

[red timber]

whats gcse

[wide cradle]

can u help

[minor peak]

yeah sure but could u post a straight picture pls

[merry depot]

,rotate

[red timber]

ohhh

smart

[minor peak]

True @merry depot

[red timber]

aight someone help pls

[minor peak]

yeah so how many 1p coins are there?

[wide cradle]

aww we love fat people btw

[minor peak]

Lmao

[red timber]

@wide cradle how many 1p coins

[wide cradle]

@red timber u answer

[red timber]

okay okay

there r as many 5p coins

as 1p coins

[minor peak]

yeah it tells u that in the q, but what number is that

(it says 3/8 of the coins are 1p)

[red timber]

3

[wide cradle]

240

[red timber]

hold up what now

[wide cradle]

240 1p coins

[red timber]

who told u that

[minor peak]

correct @wide cradle

[red timber]

LMAOOOO

[wide cradle]

3 divide by 640

[red timber]

OH WAIT

[minor peak]

vlz u have hope for passing ur gcses

[red timber]

I UNDERSTAND NOW

[wide cradle]

times by 3

[minor peak]

astrid idk about u

[red timber]

I DONT HAVE GCSE

[wide cradle]

aw fat and proud can u help me

and her

[red timber]

I DONT EVEN UNDERSTAND MATH IN ENGLISH

im just guessin

[minor peak]

wait why are you solving this then @red timber

[wide cradle]

did u pass ur gcses

[red timber]

bc vlz wanted me to

[minor peak]

yes i did

[wide cradle]

are u like 20 now

[minor peak]

fucks sake u guessed my age in one try

[red timber]

@wide cradle why did u get me to answer like u knew it easily😭

[wide cradle]

LOL

i didnt know at first

[minor peak]

so 240 1p coins

how many 2p coins?

[wide cradle]

160

[minor peak]

yes

[red timber]

kenzie u dont need me u just did this to set me up 😪

[wide cradle]

NO

@minor peak can u explain this

[minor peak]

yeah we have 240 1p coins, then how many 5p coins are there

[wide cradle]

can u explain how to do this

[red timber]

EW I HATE MATH LIKE THAT

[wide cradle]

LOL ME TOO

[lone heart]

@wide cradle Has your question been resolved?

[lone heart]

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[buoyant sorrel]

I haven’t even started, i’ve not gotten a clue what this means

[alpine sable]

just google what subset is... this is very simple question

[buoyant sorrel]

I have googled it

can you give me the answer at least

[buoyant sorrel]

ya i know, worth a try i guess 😟

.close

[lone heart]

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[misty bobcat]

100 competitors will take part in the race. The competitors are numbered from 1 to 100. The competitors start gradually. Competitors with numbers 1 to 25 are from Club A, competitors with numbers 26 to 50 are from Club B, competitors with numbers 51 to 75 are from Club C and competitors with numbers 76 to 100 are from Club D. The athlete from Club A starts first, followed by competitor from club B, then competitor from club C, competitor from club D. A again, according to clubs ABCDABCDABCD.... What are the all possibilities of the starting order ?

[tight locust]

25!^4 is my guess

[wide cradle]

i did not get help

😡

[misty bobcat]

i think the same

[tight locust]

Lets verify this works

Let's say we only have 2 people from each club

Such that our pattern is:
ABCDABCD

And each letter can either be a 1 or a 2

[misty bobcat]

i think it works

[tight locust]

Yeah that should work

Because we have 4 different sequences:
A1, A2, A3... A25
same for the other clubs

[misty bobcat]

thanks for help

[tight locust]

Go to an open channel

[misty bobcat]

.close

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[devout latch]

Hi, I have a problem with this equation:

z^4 - 2z^3 + 3z^2 - 2z + 2 = 0

z is a complex number that is changeble in (a + ib)

I have to find the solutions

[lone heart]

@devout latch Has your question been resolved?

[devout latch]

<@&286206848099549185>

[tacit arch]

,calc 1 - 2 + 3 - 2 + 2

[ocean seal]

Result:

2

[tacit arch]

dammit

[devout latch]

I tried using Ruffini's theorema

I tried with polynomial zeros 1, -1, 2, -2, 3 and -3

but nothing could solve this demoniac equation

[tacit arch]

,w factor z^4 - 2z^3 + 3z^2 - 2z + 2

[ocean seal]

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[tacit arch]

bitch

how did i not see that

from 2 being a root

[devout latch]

Is 2 valid? Oh I wrong calculus

[tacit arch]

oops

i meant the function equals to 2 when z = 1

i.e. all the coefficients sum to 2

(z^2 + 1) can be factored out so that gives you 2 roots

[tacit arch]

convince yourself with polynomial division

or just plug in the roots by "guessing and checking"

[devout latch]

okk I'm doing in this way

[lone heart]

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[ancient oxide]

the question is if me writing it as 4•10^5•1.08 is year one and then just writing 1.08 as t and for each next year adding a "t" resulting in it looking like so for 5 years for example 4•10^5•5t is that a correct way or am i wrong?

[ancient oxide]

on second thought im thinking about doing it like t=4•10^5•1.08 resulting in 1.08 written as x and for 5years resulting in it looking like 5t=4•10^5•1.08^5 or something like that

[slate jolt]

it looks as though you are trying to come up with a foprmula, eventhough its useful to be able to do so, if u are not confident just state what you know and do the calculations afterwards

here we know that if P(t) is the population at time t

P(0) = 4.10^3

and for all t

P(t+1) = 1.08 *P(t)

so if u know what a geometric sequence is, it is easy to carry on

[slate jolt]

if not try to apply this formula to get P(1), P(2), P(3) ..... to see if u can spot a pattern

[ancient oxide]

so its basically as looking on it as sequence question where i can write it as a1=4•10^5•1.08 and a2=4•10^5•1.08•1.08 which is like a2=a1•1.08 and so on?

if u say yes than you just helped me simplify that question 😄

[slate jolt]

not sure what that means but it seems that you are onto something

[ancient oxide]

@slate jolt thanks for the help 🙂

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[midnight owl]

can someone explain to me how to solve this infinite limit algebraically? my textbook doesnt make it clear

[worn fox]

What techniques do you know

[midnight owl]

i have only done regular limits

i just dont know how to do infinite limits

im assuming im not supposed to evaluated the function at a

[worn fox]

I don't know how rigorous you need to be for your course

What do you think the limit is?

[midnight owl]

evaluating at a would be undefined

[worn fox]

Correct

But what happens to it as we go closer and closer to 1 from the positive side?

[midnight owl]

i would need to graph it

here hold on

[worn fox]

You don't need to graph it

[midnight owl]

My professor did a question like this

I have no idea what he did

[worn fox]

Ah okay fine thats how you're doing it

So in our question, what is the numerator going to?

[midnight owl]

i dont know how to do that algebraically

this is all he did on the board

no work

so im confused on how he did that for the numerator and denominator

[worn fox]

We just have 2 on top

That just goes to 2 as x goes to 1+ right?

[midnight owl]

oh yeah then its 2

[worn fox]

For the bottom

Is 1-x positive or negative if we take x>1?

[midnight owl]

negative

[worn fox]

Correct

So as we get closer and closer to 1 from the right we are negative, that's good to know

[midnight owl]

but where did the x > 1 come from

[worn fox]

Because of $x \to 1^+$

[ocean seal]

iCaird

[midnight owl]

oh

okay

[worn fox]

The + means you're approaching 1 from the right side of it

[midnight owl]

that makes sense

[worn fox]

So we know the bottom is always negative

Now as we make x closer and closer to 1 what does 1-x get closer and closer to?

[midnight owl]

-infinity?

[worn fox]

Not quite, that's gonna be the final answer

Think again, x is getting really close to 1 so 1-x is getting really close to?

[midnight owl]

1?

or -1

[worn fox]

What's 1-1.1?

[midnight owl]

-0.1

so its 0?

[worn fox]

Yes

[midnight owl]

oh ok

[worn fox]

BUT

$0^-$

[ocean seal]

iCaird

[worn fox]

Because we decided it was always negative

[midnight owl]

oh

okay

that makes sense

so numerator approaches 2 and denominator approaches 0-

[worn fox]

Yep so the whole thing approaches?

[midnight owl]

you said -inf but i dont know how to get there from here

[worn fox]

Well you know $\frac {2}{0}$ "=" $\infty$ right

[midnight owl]

what do the quotes mean

[worn fox]

Ugh

[ocean seal]

iCaird

[worn fox]

The quotes are just because infinity isn't really a number so its not equality in the usual sense

[midnight owl]

isnt 2/0 undefined

[worn fox]

It is, but 2/x goes to infinity as x goes to 0

That's what my "=" is saying

[midnight owl]

what?

[worn fox]

Don't worry

In our question, the top is 2 and the bottom is getting really really close to zero, but always negative

So the whole thing is going to $-\infty$

[ocean seal]

iCaird

[midnight owl]

but where did the infinity come from

[worn fox]

Because as the bottom goes to zero, we're dividing by something really really really small and negative

[midnight owl]

oh

[worn fox]

So the expression becomes really really negative

And in the limit, goes to -infinity

[midnight owl]

so the answer to these questions are always inf or -inf?

[worn fox]

Not always, depends on the expression

[midnight owl]

well what would be a case where it isnt

[worn fox]

Well like $\lim_{x\to 5} x^2$

[ocean seal]

iCaird

[worn fox]

You know that's 25

[midnight owl]

yeah thats not an infinite limit

no i see what youre saying

[worn fox]

Well if you have an infinite limit then of course the only answers are +/- infinity

[midnight owl]

yeah i kinda forgot

alright thanks

[worn fox]

No worries

[midnight owl]

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[neat sierra]

Can anyone please check my work?
Thanks!

[slate jolt]

its all good except a detail in reflexivity

you say for all nonzero a gcd(a,a)=/=1

but for a =1 its false

just use a counterexample its easier

just say gcd(2,2)=2 =/=1

[neat sierra]

I see, that makes sense

thanks

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[plain cipher]

How to do this

[winged rock]

You need to find the intervals where the second derivative is positive

[plain cipher]

The 1st derivative is just (x^3) + (x^2)+(1) and 2nd derivative is (3x^2)+(2x)

is that how start it

[worn fox]

Yes

[winged rock]

Now find the intervals where the 2nd derivative is positive

[plain cipher]

Is that where you set f’’ = 0 and draw out a number line then test numbers around the solution to see if positive or negative

[winged rock]

You could do that

You could graph it with desmos

Ask wolfram

But you can solve f'' = 0 like you said and then either test around it or evaluate the derivatives at the roots

[winged rock]

Good choice 🤣

[plain cipher]

does this mean it's only positive at -2/3 and 0?

[winged rock]

AT?

Looks like its positive in a lot of places

[plain cipher]

positive at intervals (-2/3, infinity) and (0, infinity)?

[winged rock]

(-infinity, -2/3)

[plain cipher]

that's it?

[winged rock]

And your other interval

[plain cipher]

i need to review piecewise functions to remember all of this analyzing functions crap 💀

[winged rock]

I was just correcting your other one

[plain cipher]

ah ok

thanks guys

.close

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[errant cedar]

How should I work with sum notation when it's being squared?

[errant cedar]

it's a choose function in front of x^s

I'm trying to find what the coefficient of x^n is

[placid zinc]

This is a polynomial, and squares with the binomial theorem

But uh, this'll get pretty confusing quick

[slate jolt]

try gaussian product

[errant cedar]

yeah it's related to binomial, I'll show the whole thing

[placid zinc]

Oh wait this ain't a binomial haha

[errant cedar]

i've written this in sum notation, which is this

and i'm tryna show that this result is true

and from looking at the rhs i think i need to consider the coefficient of x^n?

[placid zinc]

I don't follow

[errant cedar]

the sum notation is from the binomial expansion of both sides

[placid zinc]

Oh

That's clever

[errant cedar]

yeah the 1 to the power of whatever is just 1, took me a while

[lone heart]

@errant cedar Has your question been resolved?

[lone heart]

@errant cedar Has your question been resolved?

[lone heart]

@errant cedar Has your question been resolved?

[velvet pelican]

@errant cedar maybe consider the multiplied things as different sums

for the coefficient of x^n, we must have the t+s=n

for example the s=0 term must be multiplied by the t=n term

[errant cedar]

Not sure if I understand, do you mean consider them as two sums multiplied by each other? Where does the t+s=n come from?

[velvet pelican]

to get x^n

we must have x^s term multiplied with x^t term

[velvet pelican]

^

if you want a more direct hint, ping me

[errant cedar]

Ohh I think I understand, it’s not as easy as subbing in n for s and t

If s+t=n, then the coefficient of x^n must be the sum of (nCs)(nCt), then nCt = nCn-t = nCs so it’s done

[velvet pelican]

yea

[errant cedar]

Thank u!!

[velvet pelican]

welcome

[errant cedar]

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[spring jackal]

9x=6x^2

[spring jackal]

if

you can solve for x using -6x^2+9x=0

then factorise to -3(2x^2-3x)=0

but why cant you do x(-6x+9)

[spring jackal]

this gives 3/2

[lament glen]

I don't see why you can't

[spring jackal]

this gives -9/6 or -3/2

but the answer is 3/2

[gray isle]

who's saying you can't factorise to
x(-6x+9)?

[lament glen]

this gives -9/-6 = 9/6

x has a coefficient of -6

[gray isle]

show how you're -9/6 from that

[spring jackal]

ok i see i didnt see the - for -6 guess i wasnt paying attention

thanks

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[clear swallow]

Can someone help me with this basic integration

[clear swallow]

,rotate

[clear swallow]

How do I find Q

[tacit arch]

have you heard of fundamental theorem of calculus?

[clear swallow]

I think I've heard of it but
I'm really weak at integration

How do I integrate sin(3pi)

[tacit arch]

https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(Apex)/05%3A_Integration/5.04%3A_The_Fundamental_Theorem_of_Calculus
and
https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(Apex)/05%3A_Integration/5.04%3A_The_Fundamental_Theorem_of_Calculus

oh you need finding antiderivatives?

[clear swallow]

Can u wait a min, i ll show u my work

[tacit arch]

https://math.libretexts.org/Bookshelves/Calculus/Map%3A_Calculus__Early_Transcendentals_(Stewart)/04%3A_Applications_of_Differentiation/4.09%3A_Antiderivatives

[clear swallow]

I don't know if I'm correct till here, but i got stuck here

[echo wadi]

After you put the limits, you are missing some brackets

[clear swallow]

30 - π/2 (cos3π-1)

[clear swallow]

Is that correct?

[echo wadi]

Yes, now put value for cos3π

[clear swallow]

Cos3π is -1 ?

[echo wadi]

Yea

[clear swallow]

Ok then i got 30 + π

[clear swallow]

Then what do I do?

There is so error from my part.
The final answer should be 30+1 = 31

Can you tell me what I did wrong

[echo wadi]

Integration of $\sin(\pi x)$ should be $-\frac 1 \pi \cos(\pi x)$

[ocean seal]

jaydamani

[clear swallow]

How did the 1/π come?

[echo wadi]

Try solving integration of sin(πx) using substitution of u=πx

[clear swallow]

-cos (u) right

-cos(πx)

[echo wadi]

You also have to put du = πdx or dx = du/π

Are you familiar with solving integration by method of substitution?

[clear swallow]

Nope... i didn't get this part

[tacit arch]

Yea I recommend reading this

All the derivative rules have corresponding antiderivative rules

[clear swallow]

Ok

[tacit arch]

So refresh those as well

The one you need specifically is chain rule

[clear swallow]

.close

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[misty bobcat]

100 competitors will take part in the race. The competitors are numbered from 1 to 100. The competitors start gradually. What are the all possibilities of the starting order, if competitor number i can not start right behind the competitor (i + 1) for any i∈{1,2,. . . , 99} ?

[lone heart]

@misty bobcat Has your question been resolved?

[misty bobcat]

.close

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[rich basin]

i'm confused what to do ext?

[lone heart]

@rich basin Has your question been resolved?

[rich basin]

<@&286206848099549185>

[alpine sable]

You should start the problem with simplar cases.

[rich basin]

what simpler cases?

this is how you do recurrence?

[alpine sable]

Like try finding indefinite integral of $x \sqrt{1-x}$, $x^1\sqrt{1-x}$ , $x^2 \sqrt{1-x}$ and so on... And notice the pattern.

[ocean seal]

Contour

[alpine sable]

,w integrate x sqrt(1-x)

[ocean seal]

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[alpine sable]

,w integrate x sqrt(1-x) from 0 to 1

[ocean seal]

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[alpine sable]

,w integrate x² sqrt(1-x) from 0 to 1

[ocean seal]

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[alpine sable]

,w integrate x³ sqrt(1-x) from 0 to 1

[ocean seal]

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[alpine sable]

,w integrate x³ sqrt(1-x) from 0 to 1

[ocean seal]

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[alpine sable]

Hmm?

[rich basin]

Don't see how that will help?

There is no integration calculator on my calculator

[alpine sable]

So can you integrate x^n sqrt(1 - x) ?

[rich basin]

no

There are restrictions, our calculator only has basic functions such as add and multiply and all the simple stuff

This is proving a formula

[rich basin]

I'm trying to do the first part

[rich basin]

this is my working

[alpine sable]

But you can integrate x √(1-x), x² √(1-x), x³ √(1-x) easily.

[rich basin]

just integration by parts

but this is recurrence, and you want to prove the formula, so you can't just simply make n = to something

you need to integrate it with n

[alpine sable]

If you integrate like this, you will have numerator and denominator like:
4, 16, 32, 256, ...
15, 105, 315, 3465, ...

Now can you see any pattern?

[rich basin]

which is what i'm trying to do here

how would I even get numbers to work with

and this is proving a formula, how can finding some pattern help me integrate that

[alpine sable]

Brr

[alpine sable]

These values are dependent on n in Un.

The function

[rich basin]

takes too much time

there is only like 45 minutes and you have like 30 integration questions like that

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[rocky condor]

only 30 questions?

[lone heart]

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[rich basin]

.reopen

[lone heart]

✅

[rich basin]

integration + mechanics

[velvet pelican]

I'd differentiate x^n and integrate sqrt(1-x)

Try this @rich basin

[rich basin]

yeah, I saw the answers and they did exactly that

but shouldn't I arrive at a similar answer if I were to use the same method as mind?

If not, isn't this a matter of guessing?

[velvet pelican]

Not guessing tho

The show that requires you to get U(n) in terms of U(n-1)

Which means the power of x^n should decrease

Implying you differentiate it

I'm not quite sure how to do the way you did

[rich basin]

that make sense, but wouldn't the n still decrease if you were to differentiate the other one which I did

[little drum]

o_o

bruh

what the hell's goin on here

[bitter vault]

it is an exercise on reduction formula. you use integration by parts to reach the first formula, and then apply it repeatedly for the second

[little drum]

that I know but the heck y'all doing here smh

[little drum]

@rich basin

[ocean seal]

Arya

Arya

Arya

[little drum]

that'll give you the first part. Hopefully you get the second one too

[lone heart]

@rich basin Has your question been resolved?

[lone heart]

@rich basin Has your question been resolved?

[supple tundra]

what's the issue

[lone heart]

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[alpine sable]

How would one approach these types of inequalities?

[lone heart]

Please don't occupy multiple help channels.

[tight locust]

don't

[alpine sable]

I can't seem to find any worksheets with inequalities on the same level as this

I don't?

[tight locust]

lots of casework

[alpine sable]

I'm supposed to find all of the real number solutions to the equation

[tight locust]

well let's see

can you factor the quadratic

[alpine sable]

Other than rewriting the numerator to (x+3)^2

[tight locust]

yeah

[alpine sable]

Yes

So x is not defined for 1, or -3

[tight locust]

so just multiply both sides by that quantity

since it's a perfect square it's always nonnegative

[alpine sable]

Ahhh

[tight locust]

maintains the direction

[alpine sable]

I was thinking about multiplying with (x-1) but one can't since it may contain neative values

perfect squares dont!

You sir/ma'm are a genius!

[tight locust]

you can do that, but you have to split into cases

[alpine sable]

Where it might be negative and when it might not?

[tight locust]

yeah

in general try to avoid multiplication and division

[alpine sable]

I will

[tight locust]

for the general sort of problem that looks like this:
a/b < c/d
turn that into
a/b - c/d < 0
(ad - bc)/bd < 0
then you can multiply though by bd and just have one set of cases to deal with

[alpine sable]

Since addition and subtraction does not mess upp the inequalit

y

Ahh

Thank you, EndTimes!

[lone heart]

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[nimble field]

Need help in geometry sequences xd
But first is the S and Sn correct ? And we don’t know anything else about the q and n and a1

[nimble field]

Need to get Bn using a1,n,q

[lone heart]

@nimble field Has your question been resolved?

[nimble field]

Rip

.close

[lone heart]

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[brittle estuary]

The queue at the ticket office where the tickets are sold
for $5, consists of 2n people. What is the probability that
that none of the buyers will have to wait for change, if
before selling a ticket to the first buyer in the queue, the cashier had only 2m dollars, and the receipt of payment for each
is it possible to pay for each ticket with either $5 or $10?

[brittle estuary]

P= 2^2n all options?

I don't know how to solve this(

<@&286206848099549185>

[zenith carbon]

“Only 2 million dollars”?

Or is m an variable

[brittle estuary]

Variable

[lone heart]

@brittle estuary Has your question been resolved?

[brittle estuary]

The queue at the ticket office where the tickets are sold
for 5 coin, consists of 2n people. What is the probability that
that none of the buyers will have to wait for change, if
before selling a ticket to the first buyer from the queue, the cashier had only 2m coins, and r
ticket is equally possible as 5 kopecks, as well as 10 kopecks?

There needs apply Markov

Chain

<@&286206848099549185>

X=2m , Xo, X1, X2...... Markov chain

[lone heart]

@brittle estuary Has your question been resolved?

[lone heart]

@brittle estuary Has your question been resolved?

[lone heart]

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[rocky trail]

idk what to do

[slate jolt]

you know that the integral of a function gives you the area between its curved and the x axis right?

[rocky trail]

no i do not

this was just introduced to me yesterday

[slate jolt]

ok but now you know

so here the integral is hard to compute

but luckily we can use geometry to eveluate it

since the area under the curve is composed of half disks

the question is essentially: what is the blue area?

[rocky trail]

i thought so

but whats the square roots above each one?

[slate jolt]

its the formula of the function that makes the half circles

but u dont need to use it

here

you know the area of a cricle right?

[rocky trail]

pi r^2

[slate jolt]

yeah

so whats the are inside the first half circle

[rocky trail]

ahh i see

just pi right?

[slate jolt]

its a half circle

dont forget

[rocky trail]

half pi

[slate jolt]

yeah

[rocky trail]

cause its just
pi 1^2

/2

[slate jolt]

yeah

so you should be able to get the whole blue area now

[rocky trail]

yea

whats the point of the square roots above it

[slate jolt]

well its to make the link between the function and the are beneath it i think

like integral of the first root

from x=0 to x=2

is equal to pi/2

[rocky trail]

ahh

[slate jolt]

since you are learning calculus, the point of the exercise is to make you feel what an integral is i think

[rocky trail]

yea im just finishing calc 1

[slate jolt]

so what you got as the answer?

[rocky trail]

7pi

[lone heart]

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[karmic rapids]

if $f \in C(I, \mathbb{R})$ and $f(I) \subseteq I$ is $f'(I) \subseteq I$?

[marsh rapids]

any intuition ?

[ocean seal]

illuminator3

[karmic rapids]

yes

so

$I := [-1, 1] \subset \mathbb{R}$

[ocean seal]

illuminator3

[karmic rapids]

deriving $f$ would just give a smaller value (positive or negative), because if $f(x)=ax^n$ then it would just be $f'(x)=nax^{n-1}$ which is obviously smaller

and

[ocean seal]

illuminator3

[marsh rapids]

you don't have a guarantee that f is polynomial

[karmic rapids]

oh

right

then I have no idea

[marsh rapids]

do you think there's a link between the area in which the values of f lie and that of its derivative ?

[marsh rapids]

something like this

[karmic rapids]

yea I'm thinking right now

[marsh rapids]

hint: ||f can be offset by any constant and its derivative would be unchanged||

[karmic rapids]

no idea about this

[karmic rapids]

I have absolutely no idea lmao

I meant like

sin(x^2) has range [-1,1]

but (sin(x^2))' has range R

but then I rememered

that it's f'([-1, 1]) not f'(R)

[marsh rapids]

but I can be more than [0, 1], it could be [0, 100]

[karmic rapids]

in my case it's only [-1, 1]

[marsh rapids]

you're only proving it for I = [-1, 1] ?

[karmic rapids]

yes

[marsh rapids]

my example can still be made to work. See if you can find a function whose derivative would reach infinity but whose values would stay finite

[karmic rapids]

wdym

would reach infinity but whose values would stay finite

[marsh rapids]

is C differentiable or does the derivative have to be continuous as well ?

that C(I, R) notation

[marsh rapids]

this would mean f(I) bounded but f'(I) unbounded, from which you can get, maybe at the cost of some tweaking, f(I) in I and f'(I) not in I

[karmic rapids]

latter

[karmic rapids]

yea

but I reckon it must be true

otherwise they wouldn't want me to prove something that requires it

[marsh rapids]

with some tweaking you can get sqrt(x) to be a counter example on [-1, 1] I believe

[karmic rapids]

hm

the derivative of sqrt is not continuous

[marsh rapids]

at one point, if that point is just barely out of I, it's irrelevant. That's what I mean by tweaking, among other things

[karmic rapids]

yeah but then they wouldn't want me to prove this

basically

I have to prove an easier form of brouwer's fixed point theorem

[marsh rapids]

can you show the question ?

[karmic rapids]

I thought about proving it via banach's fixed point theorem

by showing that the lipischitz constant is < 1

but for that f'(I) in I

[karmic rapids]

it's in german

[marsh rapids]

normally C(I, R) does not mean differentiable though

just continuous

otherwise it's C^1

[karmic rapids]

ah

[marsh rapids]

at least in French notations

[karmic rapids]

yea