help-43

[glacial tusk]

then i would include all of this that i wrote

[upper bane]

try it!

[glacial tusk]

so the answers are k=10 and k=-10

[upper bane]

check by substituting

[glacial tusk]

yaya

thank you

[upper bane]

nps!

[compact pewter]

@glacial tusk Has your question been resolved?

[compact pewter]

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[manic mist]

Just studying for this and I'm confused on something. When we do 2/3 - 5/6 we do 2 x 2 and 3 x 2 to get 4/6 - 5/6. But when we get to 1/7 - 4/3 we multiply both sides? so i just want some clarification on when we Multiply Fractions on one side and when we do it on both sides by different numbers?

[kind viper]

the idea is always that you want them over a common denominator to add or subtract them.

the common denominator has to be a multiple of both denominators involved.

in "most" cases, the simplest one is just their product (as in the case of 1/7 - 4/3, where you take 7*3=21 for the common denom and work towards that)

but in the case of 2/3 - 5/6, you can notice that 6 is actually already divisible by 3 and so you can take 6 as the common denom (and this looks like not touching the second fraction).

[manic mist]

Oh okay I think i see it

I have my study notbook rn should i write down as a mental note that as

in simple terms Multiply the denomiators by themselves To get a number then work from there?

[kind viper]

i don't like that simple-terms phrasing

but i don't know if i can offer you one

i tried to make it as simple as possible for you

[manic mist]

okay mb thank you for that 😅

[paper anchor]

Are you aware of cross-multiplication

If you are, then remember to do that, that's all

[manic mist]

so when i have 1/7 and 4/3 just cross multiply?

[paper anchor]

yes, imagine you're adding them

so, ((3 into 1)+(7 into 4)) upon 7*3

[manic mist]

ok so something llike this

one sec

Then add them

[paper anchor]

Yes, add the numerators and multiply the denominators

[bleak dock]

yes it comes from $\frac{2}{3} \cdot 1 - \frac{5}{6} \cdot 1 = \frac{2}{3} \cdot \frac{6}{6} - \frac{5}{6} \cdot \frac{3}{3}$

[manic mist]

so essentaly It does matter if it 1/7 - 4/3 add the numerators and multiply the denomiators

[bleak dock]

but 3 * 6 = 6 * 3 so you can combine the numerators, same denominator

[boreal girder]

south

[paper anchor]

no no, the sign stays, so it there's a minus then you subtract the numerators

[manic mist]

oh mn

MB*

ok sorry about this but from what it is 1-4 and 7 x 3 correct?

[bleak dock]

that's not clear

[bleak dock]

yeah then, what do you get if you do this

but then you subtract the numerators instead of adding

[manic mist]

from that i got -3/21

mb if thats wrong

[bleak dock]

not -3

[bleak dock]

like using this criss-cross method, you get 3 and 28 right

[manic mist]

Oh i see where i messed up i did 7 x 3 and not 7 x 4

[bleak dock]

ahhh yeah

so yep, you get (3 - 28)/21 = -25/21

that explains it

[manic mist]

alright i made a note for the cross multiplaction steps to not mix that up

then i see where it goes from there

[bleak dock]

cool! if you're done type .close

[manic mist]

OKay also im really sorry for this i see where the 3/28 comes from

but where the 28/21

[bleak dock]

are you forgetting that you're multiplying the numbers like this

so 1 * 3 = 3

7 * 4 = 28

the denominator is the product of the two denominators, so 7 * 3 = 21

[manic mist]

Oh crap my bad

[bleak dock]

that's ok!

[manic mist]

ok im just write some notes done and thats it

Thank you all for your time and help

😅

Made some notes on the cross multiplaction where its 1/3 to get 3 then 3 x 7 to get 21

then 7 x 4 to get 28 and 7 x 3 to get 21

.close

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[dreamy zenith]

hi im wodering if the value of x is 67 or not?

[pseudo rapids]

67?

[dreamy zenith]

yea

[upper bane]

fellow Malaysian 👀
could you show your working please?

[dreamy zenith]

eh how yk in malaysian

x+36=103
x=103-36
x=67

[pseudo rapids]

language?

[upper bane]

kan ada bahasa melayu kat situ

i'm more than familiar enough with typical Malaysian secondary school workbooks even though i don't do them as often as i should

[coral halo]

hot take: those are not that good

[upper bane]

i think it might be right based on alternate segments

[upper bane]

make that a based take. you're talking to someone who never tried any of them

[dreamy zenith]

so its correct?

but whats the purpose of it saying the congruent

[upper bane]

wdym?

[coral halo]

yeah its correct

[dreamy zenith]

the question say angle pqr and pts is congruent

[dreamy zenith]

thankss

[upper bane]

well it's not needed for part a perhaps, but that's only part a

could be needed for future parts

[dreamy zenith]

ohhh okayy thank youuu

[upper bane]

no problem

!done

[compact pewter]

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

[dreamy zenith]

.close

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[kind glen]

,rccw

[kind glen]

how do I solve this?

I wrote $A_2 =\int^{1}_{0}x-x^2$ and so on...

[boreal girder]

Prathmesh

[shy fable]

yuh so what's ur expression for that

for An in general

[prime loom]

see

$An = \int^{1}_{0}x-x^n dn$

[molten badger]

That doesn't seem like th right backslash

$An = \int^{1}_{0}x-x^n dn$

[boreal girder]

Alexis_Fx

[prime loom]

yeah thnx

[boreal girder]

AlmondAxis987

[prime loom]

$f(x) = x-x^n$

[boreal girder]

AlmondAxis987

[prime loom]

$F(x) = x^2/2 - x^{n+1}/(n+1) + C$

[boreal girder]

AlmondAxis987

[molten badger]

$F(x) = \frac{x^2}{2} - \frac{x^{n+1}}{n+1} + C$

[boreal girder]

Alexis_Fx

[shy fable]

for general n you should get $\frac{n-1}{2(n+1)}$

[boreal girder]

Bronze Jade

[shy fable]

for An

and if you multiply everything a lot cancels out

[prime loom]

$An = F(1)-F(0) \ An = \frac{1}{2} - \frac{1}{n+1} - 0$

[boreal girder]

AlmondAxis987

[prime loom]

$An = \frac{n-1}{2(n+1)}$

[boreal girder]

AlmondAxis987

[prime loom]

$A_2 A_3 A_4... A_{n-1} A_n = \frac{n-1}{2(n+1)} \frac{n-2}{2(n)}\frac{n-3} {2(n-1)}... \frac{1}{2(3)}$

[boreal girder]

AlmondAxis987

[prime loom]

so we see that we have n-1 terms (as we go from A2 to An)

and All (n-k) after n-1 cancel out with numerator

so were left wiht 2^(n-1)(n)(n+1) in denominator

and 2 in numerator

so it becomes

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

[boreal girder]

AlmondAxis987

[prime loom]

and the 2s cancel further to give

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

[boreal girder]

AlmondAxis987

[prime loom]

so answer is option (D)

@kind glen ?

[molten badger]

uhh, you shouldn't have solved the problem

[prime loom]

lol

fun anyways

[molten badger]

!sols

!nosols

[compact pewter]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[prime loom]

shouldve told before lol

Im sorry

https://tenor.com/view/kuronewbie-sorry-gif-11690506020320418165

[molten badger]

It's okay, I didn't want to interrupt

[prime loom]

so... what, close the channel now?

or will OP have to close

[molten badger]

ig OP will close it when he finishes

@kind glen are you here

[compact pewter]

@kind glen Has your question been resolved?

[kind glen]

yea

just a min, lemme try it myself now

[kind glen]

It clicked me when you did this, but still thanks for solving the whole question

[prime loom]

ur welcome

[kind glen]

.cloe

.close

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[kind viper]

!status

[compact pewter]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[pine osprey]

Computation, computation and computation

[strange pendant]

1

[kind viper]

do you know what the symbol $\circ$ means?

[boreal girder]

Ann

[strange pendant]

yes fun comp

[kind viper]

ok

so you know that $(f \circ g)(3,4)$ means $f(g(3,4))$ yes?

[boreal girder]

Ann

[strange pendant]

yes

[kind viper]

ok so go and calculate that then

[kind viper]

... there wasn't any need to post the question again

[strange pendant]

46

[kind viper]

great now do the other two as well

[strange pendant]

72

[strange pendant]

9 div 6?

[pine osprey]

Seems good

[kind viper]

are you trying to ask me a question?

[pine osprey]

No

[kind viper]

if you want to ask something then you have to phrase it as a complete sentence.

i do not know what "9 div 6?" means.

[strange pendant]

what is the definition of divisi

[jade wind]

T-T

[pine osprey]

Also its 6 that has to divide n not the inverse

[kind viper]

"divisi" looks like a word that's half-eaten

[strange pendant]

, I mean, what is the definition of divisibility?

[kind viper]

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

[pine osprey]

if the remainder of the division is 0

[kind viper]

remainder*

also let's try to help renato develop such useful life skills as:

• googling
• looking shit up on wikipedia

[strange pendant]

if x is div by y

then it is said y | x and x | y?

[kind viper]

only y | x

[strange pendant]

is the divisibility relation symmetric?

[prime loom]

not at all

[kind viper]

what do you think, renato

[lavish gull]

well, how about you figure it out

take 6 and 3

[kind viper]

i'm gonna throw your question right back at you: is the divisibilty relation symmetric?

[strange pendant]

i dont understand why rhey sawp places for example, y div x is written x | y and x div y is written y | x

[kind viper]

the | symbol is read as "divides", and "x divides y" means the same thing as "y is divisible by x".

[lavish gull]

y is divisible by x is the same as saying x divides y, if that helps

oh oops

[kind viper]

it is the same relation written backwards.

why this switch even exists, i unfortunately can't answer as i don't know the history of it.

[strange pendant]

is confusing

to say the very least, any tips?

[strange pendant]

any tips on remembering?

[kind viper]

no

it's just vocabulary

[strange pendant]

whats the reminder of 9/6?

[lavish gull]

are you seriously asking others to do division...?

[strange pendant]

no, I am asking what is the reminder how do I find it

,calc 9/6

[boreal girder]

Result:

1.5

[strange pendant]

well I dont think we should care about the reminder anyways

like

[kind viper]

remainder

[strange pendant]

9 is not divided evenly by 6

[kind viper]

a reminder is something you put in your phone calendar to remind yourself of a doctor's appointment

a remainder is something that remains

[strange pendant]

yeah exactly

[kind viper]

anyway surely you know division with remainder? like that's a school-level topic. maybe around age 10 is when you learn it in school

[strange pendant]

sort of

but I think

a | b <=> ∃k in Z, such fhat b = ak

.solved

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[cedar hawk]

could someone explain how to go about this question

[cedar hawk]

ive tried to solve it but only now that i have finished my working out i realise i was completely wrong lol

[kind viper]

initially the bowl contains 4 oranges and n-4 apples

if you want the bowl to be left with n-6 apples, what 2 pieces of fruit will you have to take from it?

[cedar hawk]

orange

i thought the opposite lol

[kind viper]

do you mean "two oranges"?

the only possible answer options are "two oranges", "two apples" and "one of each" btw.

i think either you're overthinking it or not reading me correctly

[cedar hawk]

two oranges

yeah

thats what i meant

[kind viper]

are you absolutely sure

[cedar hawk]

i though you asked which type of fruit

yeah i meant two oranges

[kind viper]

Initially, the bowl contians 4 oranges and n-4 apples.
I take 2 oranges from the bowl.
How many of each fruit remains in the bowl?

(answer as a complete sentence)

[cedar hawk]

the bowl contains 2 oranges and n-4 apples

[kind viper]

i took no apples out and also didn't put any in, and yet the amount of apples changed?

how did that happen?

@cedar hawk

[cedar hawk]

my bad

fixed

[kind viper]

right...

so taking out 2 oranges does NOT get us to

if you want the bowl to be left with n-6 apples

because after that there's still n-4 apples in there and not n-6...

[cedar hawk]

yeah

so 2 apples were taken out right?

[kind viper]

yes, glad we could come to that conclusion finally.

yes, you have to take out specifically 2 apples.

[cedar hawk]

thats what i though initially

then i continued to solve and got to some quadratic

but my answer was not 10 as shown

[kind viper]

if you could have sent your entire line of work here, we could have probably already figured out where your mistake is.

is this where i'm supposed to start looking

yes or no

@cedar hawk are you still here or what

[cedar hawk]

yes

sorry

[cedar hawk]

yeah my bad for the non sequential working out

[kind viper]

ok, then why's there an n at the beginning of the LHS?

[cedar hawk]

thats the number of fruits in the basket

[kind viper]

but why are you multiplying it with the probabilities...

like i can see (n-4)/n is P(first drawn fruit is an apple) and then (n-5)/(n-1) is P(second drawn fruit is also an apple)

but what's that n doing there afterwards

and you didn't think to show me your diagram...?

but then again the question of "how tf did that extra n come in" still stands.

like actually let's work this expression out for some specific value of n without concern as to whether it's equal to 1/3...

,calc 1000 * 996/1000 * 995/999

[boreal girder]

Result:

992.01201201201

[kind viper]

this gives you a number greater than 1, so you know it cannot possibly be a probability.

my point is this

[cedar hawk]

yeah

got it, thanks for explaining that part

[kind viper]

i was hoping you would tell me your reasoning for this n so that i could dissect it and tell you where exactly you went wrong.

but you didn't or couldn't give me that.

[cedar hawk]

its just that usually in probablity tree diagrams you strat with multiplying the number at the very beginning

[kind viper]

what number at the very beginning

[compact pewter]

@cedar hawk Has your question been resolved?

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[dire void]

bmo1 2023 q2

are u allowed to just prove the contrapositive? that given a_0 and a_1 are not consecutive then a_2023 and a_2024 are not consecutive? it's nowhere in the markscheme but it seems a lot easier to me

[quartz yoke]

yes u are allowed that

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[silent wigeon]

Am I able to cancel that 3³

With the one on the bottom

Leaving y as alone

[lavish gull]

yes

[silent wigeon]

Ok thank you!

[hushed magnet]

pls dont use x for multiplication

[silent wigeon]

Huh

Then I should use . ?

[hushed magnet]

yes

[silent wigeon]

Kk

Thanks

.close

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[lavish gull]

or just put them in parentheses

[silent wigeon]

Alr

[lavish gull]

(3^3)(y)

[silent wigeon]

I mean putting a dot would take less time than brackets

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[snow stone]

hey all, im trying to understand if this infinite series is convergent/divergent

[snow stone]

so i was trying to first find a finite Sn and see if its divergent or convergent, but couldnt find so

any hint will be wonderful

[winged lion]

have you tried partial fraction decomposition?

spoiler: ||you should get a telescoping series||

[snow stone]

where tho

[soft stone]

I like astronomy that's the hint I would give you

[winged lion]

1/(k^2-1)

[snow stone]

i mean how come they expect me to come up with that idea, i guess its the experience that tells u so

[winged lion]

because k^2-1 factors and i know telescoping series

it's not guratneed but mostly works

[soft stone]

Because why not trying since it is easily decomposable

[snow stone]

actually im new to partial fraction decomposition so..

[soft stone]

You could use a test then

Like ratio test or something

For me it is trivial that it converges bc it is essentially the armonic series with just -1 in the denominator so that essentially does not change anything and it must be convergent. That idea might be useful

Partial fraction decomposition can give you the sum but you just need to prove convergency

[winged lion]

you can also just compare it to 1/k^2

[snow stone]

yall ideas are good thx for sharing 🙂

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[winged lion]

harmonic series diverges

unless you meant something else

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[tardy lintel]

Is this series convergent or divergent? Normally I would use the limit comparison test, but that doesn't work with complex numbers.

[winged lion]

@bleak flower

[foggy mortar]

it can still work if you take the absolute values nvm

[bleak flower]

thats a good one ty

[foggy mortar]

you can use this to prove something converges but not that it diverges

[bleak flower]

i would write the function and test if it continues

@tardy lintel

[cold mural]

would it be valid to multiply by complex conjugate then splitting into real and img sums

[tardy lintel]

Probably

[tardy lintel]

What do you mean by this?

[foggy mortar]

that will work

a series converges if and only if both its real and imaginary parts converge

[tidal raptor]

I means multiply and divide by (n^2-4i)

[tardy lintel]

And I just want to check my working out for this limit:
|Re(z)|^2 / z
= |x|^2 / z
= x^2 / z
<= [(x)^2 +y^2 ]/ z
= z_bar * z / z
= z_bar
= 0
Therefore using squeeze theorem, the limit is 0

[winged lion]

You can't do inqualities with complex numbers

You would have to write |z|

[tardy lintel]

Fair enough. I dont' know how to do it then

[winged lion]

z_bar • z = |z|²

[frosty imp]

Why don't you use the ratio test?

[tardy lintel]

That looks too complecated for that series. I didin't even want to try

But I guess I can try

[tardy lintel]

Are you suggesting I do basically the exact same thing, but initially multiply by z_bar/z_bar?

Wait that changes nothing

I'm not 100% sure what you are implying

[winged lion]

I am saying in your proof x²+y²=|z|²

So you get |z|²/|z|

[tardy lintel]

Yeah but how did you get the denomenator to be |z| instead of just z?

[winged lion]

because that's how squeeze theorem works with complex functions

you take the absolute value of your function

[tardy lintel]

But then you are not solving the same limit?

[shy current]

It feels easier to just multiply and divide by the conjugate, that is n^2 - 4i and then split into imaginary and real sums, then show that the imaginary sum diverges

[tardy lintel]

Yeah I figured that one out thanks, but now I am stuck on this limit

[winged lion]

If |f| <= |g| and g goes to 0 then f goes to 0, because we know that for all epsilon there exists delta such that
|g| < eps, so by that there exists a delta such that |f| < eps, for all eps.

Are you convinced now?

[tardy lintel]

Yeah that's fair enough thanks

.close

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[abstract bronze]

Can anyone give me a nudge in the right direction as to how I can do the proof question in the challenge section? Part a specifically

[molten badger]

,rccw

[abstract bronze]

I dont really understand where to start, so I just subbed in values for a and c as 1 and 1

[molten badger]

Have you learnt about discriminant?

[abstract bronze]

Yes

I know that for that proof it needs to be b^2 - 4ac > 0

[soft stone]

Generalized harmonic series with exponent greater than 1, sorry. That's what I meant

[upbeat gorge]

this is a good start :) we need to "choose a value of b," so can you solve this inequality for b?

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@abstract bronze Has your question been resolved?

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[tardy lintel]

I am stuck on the first line of this proof. Why is that series convergent?

[dusky nymph]

because |z| <= R1 < R

and R is the radius of convergence

[tardy lintel]

This is how my notes define it. Note it only has to be convergent at |z| = r (not <= )
So I get why it would be convergent for some R1 < R, but I don't get why it is for z element of B_R1(0)

[dusky nymph]

well R1 is less than R, so it's less than the sup of that set

which means there exists some z with |z| = R1 and the series converges

you need to show that in fact for any z with |z| = R1 the series converges

[tardy lintel]

Here is the proof in my textbook

here is the proof in a past test

[tardy lintel]

I feel like at the beginning where the textbook uses |z1|, the past test should be using that R1, but it just uses z
Idk the proof in the textbook makes way more sense to me

[abstract bronze]

Ohhhhhhh thank u

I get it now

This is why I like maths yk

[tardy lintel]

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[short jewel]

is it converging, and if it is what does it converge to cuz denominator contains an imaginary unit

[soft stone]

already closed you have opened a question

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[soft stone]

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[forest token]

i was watching this yt video and this guy was teaching kings rule property and he said the shortcut is the answer comes to be (upper limit - lower limit)/2

[forest token]

now when can i use this ?

nvm

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[compact pewter]

Please don't occupy multiple help channels.

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[kindred isle]

can someone explain this to me.... I do have some work but i dont think its working out lol

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[kindred isle]

oops

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[rapid pike]

Divide a triangle with angles of 15°, 105°, and 60° into three equilateral triangles.

[rapid pike]

🤔🤔🤔

[dry canopy]

That's going to be slightly difficult 🙂

[random mica]

but any way we divide it theres gonna be an angle <=15 right?

[molten badger]

yeah

I'm wonder how're we gonna do that

[dry canopy]

There's certainly "cheaty" ways

[molten badger]

We can't divide 15 degree angle smaller but we also can't add

[dry canopy]

E.g. cut out pieces and rearrange them.

[molten badger]

lol

oh okay I understand the context

[random mica]

ok i got the solution if we are allowed to rearrange

wait nvm

[dry canopy]

But yeah, that's wouldn't be just "divide", it would be "divide and reassemble".

[molten badger]

I somehow got two triangles and not 3 lol

[random mica]

same lol

[random mica]

i mean its impossible without reassembling right

[dry canopy]

Yes

[rapid pike]

im back

lol i think the problem meant 2 triangles

[random mica]

are we allowed to reassemble pieces

[rapid pike]

yes ofcoure3

[random mica]

try to create an equilateral triangle from the given 60 and see what you can do from there

[rapid pike]

hmm ok

i thnk i got ti

a lot of construction

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[young flame]

Fk

[young flame]

What is this shit

[chilly basalt]

a....a.. square root

[young flame]

How am I supposed to mental math this

[kind viper]

sqrt(10000) = ?

[chilly basalt]

you mean removing the sqrt?

[young flame]

100*sqrt10

[pseudo shale]

Let a = 10000

10000+ 200 + 1

[young flame]

It is 10^2

[kind viper]

https://tenor.com/view/are-you-sure-john-cena-ru-sure-about-dat-gif-14258954

[pseudo shale]

a^2 + 2a + 1

Which is equivalent to a+1

If you take sqrt

[young flame]

Fk it is 101^2

[pseudo shale]

Modulus I get it

[young flame]

I feel stress when doing this quiz

My mindset has changed

I used to feel relax when doing it

[pseudo shale]

this is easy

[young flame]

What my mind try to tell me

[pseudo shale]

You were the guy with the Diophantine equation right

?

[young flame]

I think so

[pseudo shale]

11a-14c=15

What is factor of a

[young flame]

See -14c will never result in something that has 5 as the last digit

[pseudo shale]

Idk

I don’t remember the question properly tho

It was 1 week ago

[young flame]

I actually solved that one with my teacher

I think I got it now

[pseudo shale]

Damn

[young flame]

I feel tired easily

I cannot do academics for too long

Properly 4 hours a day is ideal

5 is a bit too much

[pseudo shale]

U can use .close

Tho

@young flame

[young flame]

.close

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[quartz yoke]

Am I on the right track to finding the answer to no. 87

[quartz yoke]

The question is to the left underlined in pink with the direction up top

[upper bane]

good start

[quartz yoke]

And I can’t simplify the radical 21 anymore right?

[kind viper]

what's this x doing here

yes sqrt(21) cannot be simplified

[upper bane]

oh wait yeah i thought that was a remnant of something else

[quartz yoke]

That’s just a show of what I need to find i forgot the equal sign

[kind viper]

also make your 5 not look like S

other than that yes you're on the right track

[quartz yoke]

Okay 😔 thank you 🫰

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[inner sonnet]

is this question even posible?

[dusky quartz]

yes, if there are 2 equations (which aren't just the same one but multiplied by a factor), you can find the exact value of any 2 variables

picture it as 2 lines intersecting

[inner sonnet]

my answer is (2401/195,9/5)

i didnt simplify btw

is it correct?

wait

i think it is wrong

[dusky quartz]

let me check

it is wrong

[kind viper]

show your work tho

there's a lot of places where you could mess up your arithmetic here

[dusky quartz]

for 3 variables, you can picture this as a 3d space. where each equation becomes a straight sheet of paper. with 3 sheets of paper, you can find any exact point

note: they can't be the same equation twice

[inner sonnet]

btw

how to know when to use fraction as answer when to use decimal

[ripe ether]

I dont think this is necessary. its just the same plane twice

[inner sonnet]

because the answer i got is (2401/405,-2158/135)

btw it just take too much time

will it be better if i just write decimal if the fraction is too big?

[ripe ether]

i just always use fraction. and its fine to always use fraction. cuz writing down decimal might take more work.

[inner sonnet]

but both will also accepted right?

unless it ask for specific one

[ripe ether]

make sure the fraction is simplified ofc

[inner sonnet]

ty guy

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[vagrant kelp]

is my answer correct the table below

[vagrant kelp]

<@&286206848099549185>

[ionic python]

what

i don’t understand the question

[mighty lance]

real

if you're just asking for whether you filled the table below based on graphs, yes it seems correct

[vagrant kelp]

yes im asking that

thanks

I have another one

how about this one

<@&286206848099549185> im asking if i filled the table below correct hehe

[mighty lance]

yes

it is

[vagrant kelp]

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[frosty reef]

I'm not getting anywhere close to finding a multiple of 5

in the equation 3*2^2n + 2*3^2n

[kind viper]

thats not an equation

thats an expression

stop using the word "equation" to refer to things without an equals sign in them

anyway: consider that 2^(2n) = 4^n and 3^(2n) = 9^n

[frosty reef]

oh ok

[frosty reef]

yeah?

[kind viper]

also to be clear i am going to push the modular-arithmetic angle

[frosty reef]

but then i end up at

4^n + 2(13^n)

iirc

that's what i ended up at

[kind viper]

how did 13^n happen?

actually how did any of that happen

this looks sus

[frosty reef]

because it's 2^2n + 2(2^2n + 3^2n)

= 4^n + 2(4^n + 9^n)

[kind viper]

4^n + 9^n is not equal to 13^n...

[frosty reef]

actually yeah wtf am i doing

nvm

[kind viper]

your expression equals 3 * 4^n + 2 * 9^n

your blue hint says 4^n ≡ (-1)^n (mod 5)

do you see how to extend this idea

[frosty reef]

ih

oh

can i turn 4^n into something like

5m + (-1)^n then

[kind viper]

you are overthinking it

[dusky quartz]

4^n can be expressed as (5-1)^n

[frosty reef]

oh

i still dont fully understand how that's going to help

[dusky quartz]

try expanding out (5-1)^n (mod 5)

[kind viper]

unnecessary

[simple maple]

there's no need to that considering you were explicitly given the result of it

[kind viper]

the problem sheet already gives 4^n ≡ (-1)^n (mod 5)

[simple maple]

which is this

[kind viper]

we can circumnavigate the world to rearrive at that

but in the interest of not wasting 80+ days

let's not do that

[frosty reef]

wait so

[frosty reef]

how do i use this though

[frosty reef]

if this is "overcomplicating"

[kind viper]

what is 9^n congruent to mod 5

[frosty reef]

oh

it's (-1)^n + 5

so it's the same modulus

[kind viper]

why the +5

[frosty reef]

well not +5

sorry

[kind viper]

like you're technically correct but how did the +5 happen

[frosty reef]

im not good with modulus functions

i barely use them

[frosty reef]

what i meant is

4 and 9 are 5 apart

[kind viper]

4^n ≡ (-1)^n (mod 5) because 4 ≡ -1 (mod 5)

[frosty reef]

so the modulus would be the same?

[kind viper]

and by the same token, since 9 ≡ -1 (mod 5), so too do you have 9^n ≡ (-1)^n (mod 5)

thus your original expression

3 * 4^n + 2 * 9^n ≡ 3 * (-1)^n + 2 * (-1)^n ≡ ? (mod 5)

[frosty reef]

okay

[frosty reef]

ah

that's 5*(-1)^n

[kind viper]

yes

[frosty reef]

so the modulus would be 0?

[kind viper]

"the modulus" is the wrong word.

you may say the remainder mod 5 is 0.

[frosty reef]

oh okay

mb

.solved

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[worldly ermine]

when i solve it i get the answer 1 but chatgpt says it should be -1

why?

[ionic python]

$u=\sqrt{t}$?

[boreal girder]

Allen

[real hazel]

show ur work

[ionic python]

u sub i guess

[quiet sentinel]

Integration by parts?

[quiet sentinel]

!nogpt

[compact pewter]

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

[ionic python]

i think a u sub is ok

lemme try

[quiet sentinel]

U sub still leaves two bits tho

[potent berry]

!nogpt

[compact pewter]

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

[worldly ermine]

sry i ment i am saying it should be -1 and gpt is telling -1

my work isnt pretty thb

[quiet sentinel]

$$u=\frac{1}{\sqrt{t}}$$
$$dv=e^{-2\sqrt{t}}dt$$

[real hazel]

just tell us what you did

[boreal girder]

ImOakley

[real hazel]

ok so everything up to the last integral seems right

now lets think about it

[ionic python]

i calculated 1

[quiet sentinel]

You used e^-v

[real hazel]

Whats the antiderivative of e^(-u)? @worldly ermine

[quiet sentinel]

That doesnt work

[ionic python]

just u sub $u=2\sqrt{t}$

[boreal girder]

Allen

[ionic python]

i need some time to write out my steps

[worldly ermine]

wait i need to google i dont do maths in englisch sry

[real hazel]

what language

[worldly ermine]

okok

[worldly ermine]

german

[real hazel]

tell me answer to this

$\int e^{-u} ,du$

[real hazel]

@worldly ermine

[worldly ermine]

-e^-u+c

[real hazel]

correct

[worldly ermine]

oh

[real hazel]

but thats not what you did

[ionic python]

$$\int_{0}^{\infty}\frac{1}{\sqrt{t}}e^{-2\sqrt{t}} dt$$
$$u=2\sqrt{t}$$
$$du=2\cdot\frac{1}{2\sqrt{t}} dt=\frac{1}{\sqrt{t}} dt$$
$$dt=\sqrt{t}du$$
$$t=0\Rightarrow u=0, t=\infty\Rightarrow u=\infty$$
$$=\int_{0}^{\infty}e^{-u}du$$
$$=-e^{-u}\Big|_{0}^{\infty}$$
$$=-e^{-\infty}-(-e^0)$$
$$\boxed{=1}$$

[real hazel]

@worldly ermine u see now?

[worldly ermine]

i forgot to integrate after the substitution

i get it now

thx

.close

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[ionic python]

ok I’ll continue writing anyways

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[boreal girder]

Allen

$$\int_{0}^{\infty}\frac{1}{\sqrt{t}}e^{-2\sqrt{t}} dt$$
$$u=2\sqrt{t}$$
$$du=2\cdot\frac{1}{2\sqrt{t}} dt=\frac{1}{\sqrt{t}} dt$$
$$dt=\sqrt{t}du$$
$$t=0\Rightarrow u=0, t=\infty\Rightarrow u=\infty$$
$$=\int_{0}^{\infty}e^{-u}du$$
$$=-e^{-u}\Big|_{0}^{\infty}$$
$$=-e^{-\infty}-(-e^0)$$
$$\boxed{=1}$$

[worldly ermine]

u the goat

[worldly ermine]

.close

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[ionic python]

uh oh

[ionic python]

ChatGPT stupid

[kind viper]

!nogpt

[compact pewter]

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[fair thunder]

number 6
what exactly is the initial condition like, im so confused

[kind viper]

that looks weird

what's that expression doing there

[fair thunder]

yeah

[kind viper]

it doesn't look like an initial condition for sure

[fair thunder]

idk

prolly a misprint maybe

anyway, just wanted to confirm that im not dumb

thank you

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[worldly ermine]

is this correct?

[strange ermine]

Yes

[wheat pasture]

(don't forget the dx's for the integrals )

[compact pewter]

@worldly ermine Has your question been resolved?

[worldly ermine]

ok thx guys

[compact pewter]

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[modern spindle]

Hello I need help with some of these questions I feel like they will be on my test and indont know how to like do them, also yes they are ch!tg!t but I asked him to create question

[modern spindle]

For the first one

Ik u can just plug in the values and solve for b and a

But like the second one I'm a little lost

The 3rd one I would just do the same as the first

Same with 4th

But I don't get the last one

[faint tendon]

well, for the second one, after 9 days 0.6 * 0.6 * 0.6 (since it decays by 0.6 every 3 days) of the original is remaining

[modern spindle]

wait what

whats *

[faint tendon]

multiply

sorry

[modern spindle]

So after 6 days its 100% of the original remaining?

I don't get it

[faint tendon]

so look

[modern spindle]

Wouldn't it be plus

[faint tendon]

let's say the original is x

[modern spindle]

yes

[faint tendon]

after 3 days, we have 0.6x

[modern spindle]

oh yeah

[faint tendon]

but after 3 more days, we have 0.6 * 0.6x

[modern spindle]

Ahh

[faint tendon]

and after 3 more, 0.6 * 0.6 * 0.6x

[modern spindle]

And then at 9 days its

Yeah

Ahhh I see

[faint tendon]

you got it from here?

[modern spindle]

and then

I would basically

plug in these values?

[modern spindle]

wait it says that x is days tho

[faint tendon]

sorry, this is a different x

i was being confusing

it's 0.6 * 0.6 * 0.6 * b

and you want b

[modern spindle]

so we let b be the original amount

wait would it be y=b times 0.6 ^x

and then we can plug in at 3 days

wait but where would we plug in 0.6

To find b

[faint tendon]

i think it would be y = b * 0.6^(x / 3)

[modern spindle]

yeah it is that like x/3

But indont understand it

And also how do we plug in the value for y

[faint tendon]

wdym?

like, you don't understand how it works?

[modern spindle]

wait never mind

I use the second question

I don't understand why its x/3

why can't it just be x as that denotes the day

[faint tendon]

ok let's try applying it where it's just x

for x = 3:
y = b * 0.6^3

,calc 0.6^3

[boreal girder]

Result:

0.216

[faint tendon]

but you see, after 3 days it's not supposed to be 0.216b, it's supposed to be 0.6b

of course, you could also do it by using a different value instead of 0.6

[modern spindle]

ahh I see

ok can we do the last one

I get this one now

[faint tendon]

sure

[modern spindle]

^

[faint tendon]

so y = a * b^t right

[modern spindle]

ye

[faint tendon]

it's currently 35,000 so a = 35,000

[modern spindle]

oh wait so in an exponential is the value of a is original amount all the time

[faint tendon]

now, if it doubles every 4.2 years we can do an approach similar to the last question

[modern spindle]

ahh that would make sense cause ur timing it

[faint tendon]

yes
at least in the form used there

[modern spindle]

👍

[faint tendon]

y = a * 2^(years / 4.2)

[modern spindle]

So its y= 35000 b^t

wait uh

Oh yeah

why 2..

[faint tendon]

it doubles every 4.2 years

doubles = 2

[modern spindle]

oh true

oh so it would be y=35000 * 2 ^years/4.2

and then we plug in 12years to find the b

[faint tendon]

yes

[modern spindle]

Ok thanks so much

I have a test

I'm like 30 minutes its about parabola and non linear

I hope I get good mark

🙏

[faint tendon]

gl!

remember to .close if you're done!

wait you're not cheating right

[modern spindle]

no imnnot

haha

I just wanted to touch on that question because inwasny confident

.cloae

.close

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[full plinth]

hi

[full plinth]

can someone please help me w calculus ab

question 10 here

it doesnt come w a graph or anything i dont rlly get it

cuz if i factor it i get x+3 but it says what is wrong

[molten coral]

theres nothing 'wrong'

idk why they phrased it that way

[pseudo rapids]

well i can plug in x = 2 on the right side, but on the left side it's undefined...

yeah the question is wrong

[full plinth]

its for limits

[molten coral]

the left hand side has a removeable discontinuity

[molten coral]

oh, interesting

i get what they mean

sure so they're not equivalent

you see why, yea?

[pseudo rapids]

they still phrased it wrong

[molten coral]

yea its phrased weird but i see what they mean now in context

@full plinth you get what they mean in the key?

[full plinth]

ohhh

i think i get it now

smth else

[molten coral]

yea

[full plinth]

for questions like these (any one of them) i cant plug x in until i completely factor it?

[molten coral]

usually you don't get so lucky

the easiest way to solve a limit is if the function is just defined at the limiting value

so if it is, you can

they don't usually give people problems like this, other than to demonstrate that it's possible

[full plinth]

oh ok thanks got it

[molten coral]

[full plinth]

.close

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[pliant canopy]

chat is this good so far i feel like i messed up

[compact pewter]

@pliant canopy Has your question been resolved?

[rocky blaze]

I need help

with my math problem

[pliant canopy]

Yeah me too buddy

[rocky blaze]

can someone help

im a frshman and im struggling

[pliant canopy]

Do that in another chat

[upbeat gorge]

!occupied

[compact pewter]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[shut garnet]

oh that's a factoid. Neat

[pliant canopy]

Im not gonna get help am i😔

[shut garnet]

sorry I was looking at it and my game started. Give me a moment

Yeah that looks fine so far, you still have to undo your substitution to get your final answer though. But good use of trig sub

@pliant canopy (ping if you were expecting one. Sorry if not 😅 )

[pliant canopy]

Oh oki thanks

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[limpid python]

Looking for statistics teacher immediately,even students with good command are welcome

[upper bane]

you can try asking in #discussion or #study-discussion for that, but a help channel is not the right place to ask these i believe

[short lantern]

it is indeed not the place, consider resources like khan academy too

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[spring maple]

ok i NEED help with problems b and c (i have somewhat of a clue of what im doing but for the most part i need an explanation)

[spring maple]

btw the second image is what me and my friends got from trying to understand this insanity

[kind viper]

the graph shows the rate of people arriving at the ride yeah?

[spring maple]

yes

i believe so?

[kind viper]

yes

and they get seated on the ride at a constant 800 ppl/h

[spring maple]

yes

[kind viper]

so the rate at which the line grows is equal to r(t)-800

line as in queue

[spring maple]

uhm

i dont understand 😔

im so sry dude

[kind viper]

don't call me "dude" please.

[spring maple]

oh sry ms

[kind viper]

thank you.

[spring maple]

i didnt see 🫢

[kind viper]

the line has newly arrived ppl going in at the back

and going out (and into the ride) at the front

the amount of ppl in the line changes only from these two factors

so the rate at which that happens is equal to the arrival rate minus the ride-seating rate

capisce?

[spring maple]

capisce

i think i understand

[kind viper]

ok

[spring maple]

thx

[kind viper]

so between t=2 and t=3 hours, are we getting more arrivals than we are seating, or less?

[spring maple]

we're getting more arrivals, correct?

[kind viper]

indeed we are

as you can hopefully see from the graph

[spring maple]

mhm

[kind viper]

now how do you think we can find when the line for the ride is longest?

[spring maple]

i just dont know tbh

[kind viper]

ok let's try to call on some physical intuition

imagine you throw a ball in the air, vertically

what can you say about the ball at the instant when it's at the highest point in its trajectory?

[spring maple]

thats when its got the most potential energy?

correct?

[kind viper]

correct but not what im going for

what is its velocity at that instant

[spring maple]

ohhh

[coarse plume]

0

[spring maple]

uhm

[spring maple]

is that it?

[coarse plume]

Ya because when it’s far up in the air it stops for a second then comes back down

Yk what I mean

[spring maple]

mhm

[kind viper]

would've appreciated if you had let OP come to this himself

but anyway yes the instantaneous velocity at the peak of the trajectory is 0

you may know that in general, a function's derivative at a local min or max point is 0

so what we want, then, is to find when the new arrival rate is equal to 800 ppl/h

[spring maple]

wouldn't that be at hour 3 or am i misunderstood?

[kind viper]

yeah that would be specifically at t=3.

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[urban flicker]

is there any good ways to prove the recursive sequence x_n = 0.5(x_n-1+1/x_n-1) is bounded?

[urban flicker]

or it has an upper bound

it has a lower bound 1, i've calculated by using basic inequality

[winged lion]

What is x_0?

[urban flicker]

1

[prime loom]

then sequence is constant?

[winged lion]

hmm?

[urban flicker]

ummm

sry

[winged lion]

Oh now I see the +

yea it's constant apparently

you might prove that

by induction

[urban flicker]

maybe 2 instead of 1

2/ f(n-1)

[urban flicker]

x_n = 0.5(x_n-1+2/x_n-1)

[winged lion]

,w x=0.5(x+2/x)

[boreal girder]

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[prime loom]

we can prove it is bounded for any t by monotonicity of the recursion function

let f(n) = 1/2(n+1/n)

then $x_n = f(x_{n-1})$

[boreal girder]

AlmondAxis987

[prime loom]

now if $x_0 > 0$, then $f(x_0)=x_1>0$

[boreal girder]

AlmondAxis987

[prime loom]

inductively, all $x_n > 0$

[boreal girder]

AlmondAxis987

[prime loom]

lower bound is found by AM-GM to be 1

as u did

[urban flicker]

how about upper bound

[prime loom]

now for $n>=1, f(n) = \frac12(n+\frac1{n}) <= \frac{n+n}2 = n$

[devout wraith]

cool

[boreal girder]

AlmondAxis987

[prime loom]

because $n>=1 \implies \frac1{n} <= n$

[boreal girder]

AlmondAxis987

[prime loom]

but $x_n \geq 1$for any n

[boreal girder]

AlmondAxis987

[urban flicker]

https://tenor.com/view/niubi-quentin-supercool-cool-chinese-abc-gif-26451591

[devout wraith]

damm

[prime loom]

what is it wrong?

https://tenor.com/view/sweats-gif-25346666

[prime loom]

lol pls I cant tell if ur saying Im wrong or im right 😭

[winged lion]

Did you replace x_n with n, and also wasn't it 2/x_(n-1)?

[prime loom]

.

not 2?