help-19

[mystic saffron]

and x+270

[spring nacelle]

Do you see how to do it from here?

[mystic saffron]

wait let me try

so 10/3 x + 120 = 5/6 x

wait

no way something is wrong

[spring nacelle]

Yes, you need to remember how you defined things

[mystic saffron]

yea yea

10/3 x + 120 is the second year

x+270+10/3 x + 120 = total of final

x+10/3x = total of original

total of final = total of original + 390

man

i thought i had something

[spring nacelle]

Total is unnecessary here

You only care about ratio of first year and second years

[mystic saffron]

alright thanks

.close

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[blissful quail]

im having trouble solving dy/dx x^2-4xy+y^2=4

[blissful quail]

i tried solving it and i got (2y-x)/(-2x+2y) but apparently thats not right

[carmine idol]

what did you try

[blissful quail]

i tried finding derivative of both sides

so d/dx (x^2-4xy+y^2) = d/dx 4

[carmine idol]

wait, what is the question?

it doesn't seem right that you are doing that

[blissful quail]

find dy/dx by implicit differentiation

[carmine idol]

oh

[slender jacinth]

aight bet

[carmine idol]

i thought you wanted to solve the differential equation 💀

[blissful quail]

nah

so im supposed to find y' = something

[slender jacinth]

yea

u do d/dx of both sides

[blissful quail]

i did

[slender jacinth]

lemme try

hold on

[blissful quail]

wait for y^2

[slender jacinth]

ye

[blissful quail]

i thought you just do the same as you would for x but multipy y' to it after

but im assuming thats wrong

[slender jacinth]

its 2yy' or 2y dy/dx

ur right

[blissful quail]

ye

hold up lemme post my work

[slender jacinth]

do yk the answer

[blissful quail]

ye

[slender jacinth]

i got (2y-x)/(y-2x)

[blissful quail]

ye thats right

[slender jacinth]

oh

wait lemme check where ur thing went wrong

[blissful quail]

oh wait

i forgot to divide 2yy' by 2

oops

NOO

ai word thanks

.close

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[slender jacinth]

aight

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[odd edge]

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

[elfin zodiac]

What did you try?

[stiff wind]

Geometric combination.

Show me ur work

[mystic saffron]

i counted

and got 13

i used inspection

[elfin zodiac]

Well, that is a way but your counting is wrong

Lets inspect in a different way

[stiff wind]

What is the answer?

[elfin zodiac]

Can you see the big triangle is divided by 2 horizontal lines?

[elfin zodiac]

Irrelevant right now, the important thing is the process of thinking

[elfin zodiac]

Take the smallest triangle in the top, divided by the first horizontal line

And tell me how many triangles can you count

[stiff wind]

I found the answer but need to be sure before explaining. C(4,2) x C(3,1)

[elfin zodiac]

I don’t want to explain how to get the answer, I want him to get the answer himself

[stiff wind]

Ok

[stiff wind]

A side note: Watcjing a basic video bout this topic would lead him to answer also

[mystic saffron]

ohh what's the video?

[elfin zodiac]

…

Can you do what I asked you?

There is no sense on asking something you don’t want to do

[mystic saffron]

ok so

top triangle = 4C2 = 6

[elfin zodiac]

That is the top one

Now

Imagine the first horizontal like does not exist

Only the second

How many triangles afther the second (new first line)?

[mystic saffron]

so find triangles using only the second line?

[elfin zodiac]

Yes

[mystic saffron]

7?

[elfin zodiac]

How?

[mystic saffron]

oh one sec

[elfin zodiac]

Triangle with 2 vertical lines and one horizontal (first triangle of the top)

Repeat the same qith second horizontal line

The answer cannot be different

It is just the same triangle but bigger

[mystic saffron]

oh i see

[elfin zodiac]

You repeat this for the third horizontal line (which is the main triangle bottom line)

[mystic saffron]

it should be the same right? assuming theres no first line, and second line

[elfin zodiac]

Exactly assuming no intermediate horizontal lines

[mystic saffron]

ok so we got 18

[elfin zodiac]

Yes, but you need an intermediate question

If you add the two horizontal lines

The figures inside created by these two lines

Are triangles?

[mystic saffron]

yh it will give us all sort of triangles

[elfin zodiac]

Or they have 4 sides?

I am talking now about the original pixture

[mystic saffron]

it will also give us 4 sides

but 3 sides aswell

[elfin zodiac]

The small figures created due to the horizontal lines with the vertical lines are triangles? (Without counting the first small triangles)

[mystic saffron]

oh wait

yh

[elfin zodiac]

They are all 4 sided

Except in the top

[mystic saffron]

i see

[elfin zodiac]

But we already know the top is 1+2+3

[mystic saffron]

yh

[elfin zodiac]

So we get the total number as

3(n+1)(n+2)/2

[mystic saffron]

i see

so what's n?

[elfin zodiac]

Where n=2

Number of vertical lines

And 3 is the number of horizontal lines counting the bottom from the main triangles

[mystic saffron]

i see

that makes so much sense

thanks

[elfin zodiac]

You are welcome

!done

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[mystic saffron]

.closee

.close

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[sharp solar]

hi! did i do this right? i basically just counted (left to right) the distance between D and p which is how i got 3,0

[odd edge]

@sharp solar Has your question been resolved?

[sharp solar]

<@&286206848099549185>

.close

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[tribal basin]

So like:

Mike wants to invest in a annuity that pays 100$ per month for 4 years (48 months). The annuity has an annual rate of 3.5%, compounded monthly. What is the present value of this annuity?

[odd edge]

@tribal basin Has your question been resolved?

[tribal basin]

<@&286206848099549185>

[keen latch]

is that right?

if then, why did you **(-48) insted of **(+)48

[tribal basin]

Ther formula was given already.

I just need to make a problem based on it

so I did it wrong i am assuming.

[keen latch]

yeah maybe.

if i follow your problem, i have to ** 48 instead of -48

just think about this.

"Mike wants to save for a future expense by making monthly deposits into an account that earns an annual interest rate of 3.5%, compounded monthly. He plans to make 48 equal monthly payments of $100 each. What is the present value of this annuity?"

[tribal basin]

Sounds better than mine

They are taking money within those 48 payments so its -

right?

Something like that?

[keen latch]

yeah.

[tribal basin]

It would be this one?

it is reworded better tbh idk where I messed up

48 equal monthly of 100$

[keen latch]

that sounds good.

your name is Mike?

[tribal basin]

no i just like using mike for every problem i make

very easy and generic name

[keen latch]

which grade?

[tribal basin]

12 mathematics

I have new units every week its kind of awkward ngl

thank you for your help.

.close

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[keen latch]

so this year you will graduate?

[tribal basin]

Yes

[keen latch]

good for you.

[tribal basin]

🫡

[keen latch]

us?

[tribal basin]

born and raised

[keen latch]

and now?

[tribal basin]

same place lol

[keen latch]

cool...

[tribal basin]

yeppers

[keen latch]

i wanted to make an native en speaker..

can't we become friends?

[odd edge]

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[nocturne belfry]

no

[silk quail]

uh, it's undefined because 0 is in the interval

[nocturne belfry]

you need to split the bounds

use two improper integrals

but it will diverge

[glossy lotus]

Then every time we do partial integration we need to check if between the given values diverges?

[nocturne belfry]

you cant integrate over a singularity like that

the function needs to be defined over the region of integration

if its not you use an improper integral

[uneven lodge]

You're using the Fundamental Theorem of Calculus in the 3rd step, which relies on continuity over the whole interval you're integrating over

[nocturne belfry]

hello mikkel

[uneven lodge]

Hello jan Niku

[glossy lotus]

Huh?

[glossy lotus]

Improper integrals are the ones that the bounds are infinitive?

[silk quail]

improper integrals also occur if a function is not bounded for the interval (discontinuity)

[nocturne belfry]

improper integrals have bounds that are limited

or infinity, yea, which is just a special case

in your case you need a limit towards 0

[glossy lotus]

So we need to break this integral into two

-1 to 0 and 0 to 1?

[silk quail]

yes

should be like this [\int_{-1}^1\frac{1}{x^2},{dx}=\int_{-1}^a\frac{1}{x^2},{dx}+\int_a^1\frac{1}{x^2},{dx}.]

[clever fjord]

fish

[silk quail]

and take limit of a to 0

[nocturne belfry]

you want $2\lim _{a\to 0} \int _a ^1 x^{-2} \dd x$

[clever fjord]

jan Niku

[nocturne belfry]

its even so we combine

[glossy lotus]

Oh so it is 2(-1+lim(a->0)1/a)

[uneven lodge]

$a \to 0^{+}$

[clever fjord]

Mikkel Angelo

[nocturne belfry]

sure lol

[glossy lotus]

But this is infinitive $\lim _{a\to 0^{+}} 1/a$

[clever fjord]

Khwu

[glossy lotus]

Aha so we have infinitive as the solution?

[glossy lotus]

Ah ok so it is undefined

[silk quail]

well, say it diverges

[glossy lotus]

Thank you guys for helping me!!

.close

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[blissful depot]

Hey guys. I have to solve this system in R, though I've found out that x = y = z, I'd like to prove that also, and I don't know how.

[blissful depot]

So please help me prove that x = y = z

[proper solstice]

what is 'in R'?

[odd edge]

@blissful depot Has your question been resolved?

[blissful depot]

real

numbers

anyways, I solved it with calculus

thank you anyways

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[pastel orbit]

does this proof of mine work?
#1260424130822541403 message

[odd edge]

@pastel orbit Has your question been resolved?

[sand horizon]

Ima read this

[sand horizon]

Had to remember topo but yeah, are you unsure of somewhere in the proof ?

[pastel orbit]

no, I'm just looking to verify its correctness

[sand horizon]

Ok, i dont see anything wrong while reading, not familar with this version so maybe someone who could double check would be better ig

[pastel orbit]

[sand horizon]

What level is this ?

[pastel orbit]

what do you mean?

[sand horizon]

Like grade or idk

When topo is taught ?

[pastel orbit]

it's not tied to any specific grade I suppose

different places will teach it at different times

[eternal glade]

Here in germany, this basic topology is taught in "Analysis 2" in second semester. 🙂

[sand horizon]

Yeah i start in end hs so i was asking

For basic

[sand horizon]

And what could you say about his proof ?

[eternal glade]

I havent really read it, just went over it and scanned if there are concepts in it i havent yet heard of

[odd edge]

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[heady plover]

any idea of this

[zinc glacier]

a u-sub of u=lnx seems like the way to go

[heady plover]

how to deal with this part

[odd edge]

@heady plover Has your question been resolved?

[heady plover]

<@&286206848099549185>

[clever fjord]

wtr.on

[heady plover]

ok then if u=ln x, it will be $\frac{\ln(\frac{e}{x^2})}{u}$

[clever fjord]

nino

[heady plover]

$\int_{1/e}^{e} (u)^{\frac{\ln(\frac{e}{x^2})}{u}\right)} \left(1 - \ln(u)\right)du$

[clever fjord]

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

[heady plover]

$\int_{1/e}^{e} (u)^{\frac{(1-2\lnx)}{u}\right)} \left(1 - \ln(u)\right)du$

[clever fjord]

wtr.on

[heady plover]

ooh

how did you come up with that

[clever fjord]

wtr.on

wtr.on

[heady plover]

$du=\frac{\frac{-2}{x}\ln x-\frac{1-2\ln x}{x}}{(\ln x)^2}dx=\frac{1}{\ln^2 x}dx$

[clever fjord]

nino

[heady plover]

$\int_{1/e}^{e} \frac{(\ln x)^u \left(1 - \ln(\ln x)\right)}{x} (\ln^2 x)dx$

[clever fjord]

nino

[heady plover]

wait how to continue

[simple ether]

no

[simple ether]

look at it

[heady plover]

ye i've known that

[simple ether]

dx/x is d(lnx)

[heady plover]

ye but i stuck, you can take a try, i dont think taking u as lnx will work

[simple ether]

Ohhhh

i know where is the problem

I cant solve it now

but i believe the first step is d(lnx)

[heady plover]

i dont think so

ok but still ... how to continue

[clever fjord]

wtr.on

wtr.on

[heady plover]

$\int_{1/e}^{e} \frac{(\ln x)^{\frac{1-2\ln(x)}{\ln(x)}} \left(1 - \ln(\ln x)\right)}{x}(-x\ln^2(x))dx$

[clever fjord]

nino

wtr.on

[heady plover]

lnx=1/(u+2)

[heady plover]

ok then how can we continue

[clever fjord]

wtr.on

[heady plover]

darn....

how did you come up with that...

any tricks?

so once you see the power part being like a form a(x) then treat it as u?

You mean the problems which treating the power part as u?

darn...where can i find problems like that....

I never met such problems before

I guess yes

But igtg rn, if i didn't make it successfully,I'll ask again, Anyway thank you so much

.close

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[odd edge]

@signal oar Has your question been resolved?

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[celest echo]

help

[ember oak]

One channel please

.close

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[small zinc]

I must decide values for x and y

Ive come to: log(2y)=y/5

But I dont know how to solve further

[mystic saffron]

How did you arrive here?

[small zinc]

eq 1. can be simplified to log(xy)=1, xy=10 gives that x = 10/y

Putting that into the next equation, 10/y * log (2y) = 2 => log(2y) = 2y/10 or just log(2y) = y/5

[ember oak]

I would just guess random integers for y.

[mystic saffron]

Might help to get y in terms of x and plug into the logarithm? Haven't solved through though

[ember oak]

Maybe try to make the log expression a power of 10

[small zinc]

yeah thats what they say in the solution

but isnt there a way to solve this without guessing?

[ember oak]

No

You would need the Lambert W function

I believe it's also known as the product log

[sinful grove]

If youre able to manipulate such that you can apply the lambert W function i would say its totally legit

[small zinc]

ah alright

fair enough

[ember oak]

But it's not very helpful for you, as it is simply defined as the solution

[sinful grove]

its really no different than applying the square root to something

[small zinc]

alr thanks for the help

.close

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[small oxide]

I'm doing some conversion practice and I need help. I know that the answer for the numberator in the second box is 28.3168 L/ft^3 but there is no 28.3... in the blue answer bank. So what do I do? Am I converting incorrectly? What do I need to do to get the correct answer as 99 L/s?

[mystic saffron]

There is an extra conversion you have to do

You can go from Liters to cm to ft

In order to cancel the ft correctly

[small oxide]

Do you mean in the second box to go from L/cm x cm/ft ?

[mystic saffron]

Yes

[small oxide]

Do you mean cm or cm^3

Okay, I fixed the second box, now the third box is incorrect

The numerator needs to be 30.48 cm I think

[odd edge]

@small oxide Has your question been resolved?

[small oxide]

Not yet

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@small oxide Has your question been resolved?

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[last lodge]

yo guys

[last lodge]

i have a chem question

can I ask it here?

[ember oak]

Can't guarantee an answer

[last lodge]

why is Hydrogen ion not considered basic radical

[odd edge]

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[edgy junco]

why does lim x approaches 0 of (e^x-1)/x equal 1?

[sand horizon]

Its a rate change so its the derivative of e^x in 0

Ima share the image

Let f(x) = e^x

[edgy junco]

is there some other proof of this?

[sand horizon]

L'hôpital

[edgy junco]

?

[sand horizon]

Like you can apply l'hôpital rule to the limit

[edgy junco]

I mean some proof of this that doesn't depend on derivatives

[weary pelican]

that's worse than using the derivative limit

[sand horizon]

Yes

But without derivative i dont see actually

[edgy junco]

since a version of this limit appears as you try to differentiate e^x

[edgy junco]

?

[weary pelican]

l'hôpital is literally taking derivative of both numerator and denominator

[sand horizon]

No i meant a proof without derivative

[weary pelican]

what's your definition of e^x

[sand horizon]

Mb

[edgy junco]

in the classic def of the derivative of e^x, you get lim as dx approaches zero of e^x times (e^dx-1)dx after a bit of fiddling, I remember my teacher giving me a nice explanation of why the (e^dx - 1)/dx part equalled zero but I forgot.

[weary pelican]

?

That doesn't answer my question

what does "e^x" mean to you

how was this introduced to you

[edgy junco]

i guess, lim as x approaches infinity of (1 + 1/x)^x

[weary pelican]

not quite

that's e

[edgy junco]

i don't know any other definitions

[weary pelican]

ok then so

you know log?

[edgy junco]

yeah

[weary pelican]

how was log defined

[edgy junco]

the inverse of a exponential

[weary pelican]

ok this is really weird

because how did you even define e^x then

did they teach you what it means to take a real number to the power of a real number

[edgy junco]

just e to the x power?

[weary pelican]

uh huh, what does it mean when x is any real number

[edgy junco]

I don't really know

[weary pelican]

then your definition is bad

so

I agree about the definition of e, whatever it may be

it's a number

but if we can't explain what it means to take e to the power of a real number

then we can't define it that way

so instead

we can define the exponential function first

and it's not gonna depend on "e"

have you seen infinite sums before?

[edgy junco]

yeah

[weary pelican]

ok

so let's fix some real number $x$

[clever fjord]

rafilou2003

[weary pelican]

and let's consider the infinite sum $\sum_{n=0}^\infty \frac{x^n}{n!}$

[clever fjord]

rafilou2003

[weary pelican]

does this converge?

[edgy junco]

yeah

[weary pelican]

yeah, you can take ratio test for example

so this, converges to some value

that we will call $\exp(x)$

[clever fjord]

rafilou2003

[weary pelican]

we have defined a function that takes real numbers as input, and outputs real numbers

now

we're gonna prove two things:

1. this function's derivative is itself

2. exp(1) = e

then, we can give sense to the expression "e^x"

as it will refer to exp(x)

and it's gonna correlate with the value we expect it to be when x is an integer

(we know how to define what a^b is when b is an integer)

so

let's try point 1)

[weary pelican]

how do we differentiate an infinite sum?

[edgy junco]

that should be infinte sum of nx^(n-1)/n! , if I didn't goof up

[weary pelican]

yep, exactly

just a small problem when n=0

that we would better solve

the derivative of x^0/0! = 1

is 0, since it's a constant

so we can remove the "n=0" part

and the other terms are good

$\frac{d}{dx}\sum_{n=0}^\infty \frac{x^n}{n!} = \sum_{n=1}^\infty \frac{nx^{n-1}}{n!}$

[clever fjord]

rafilou2003

[weary pelican]

but notice

that n! = n * (n-1)(n-2)... = n * (n-1)!

so we're left with $\sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!}$

[clever fjord]

rafilou2003

[weary pelican]

and if you push all of the terms to the left

[edgy junco]

oh damn, I see it now

[weary pelican]

we can rewrite that into

$\sum_{k=0}^\infty \frac{x^k}{k!}$

[clever fjord]

rafilou2003

[weary pelican]

boom, $\exp(x)$ is back

[clever fjord]

rafilou2003

[weary pelican]

we have found that $\exp' = \exp$

[clever fjord]

rafilou2003

[weary pelican]

point 1) done

[edgy junco]

that's an approach I've never seen before. I guess it's becuase they teach series pretty late in the curriculum

[weary pelican]

yeah, the problem with trying to use exponential without REALLY defining it is that it leads to shitty definitions like the previous one

where you can maybe intuitively see what e^1, e^3, e^integer is... even e^(rational)

but e^pi ?

so with exp' = exp

by definition of the derivative

[edgy junco]

I always just though they took the limit as x approaches some irrational number when trying the wonky stuff

[weary pelican]

$\frac{\exp(x)-\exp(0)}{x-0}$ converges when $x\to 0$ to $\exp'(0) = \exp(0) = 1$

[clever fjord]

rafilou2003

[weary pelican]

is it well defined?

like if I take two different sequences of rationals that approaches an irrational, do they have the same limit?

[edgy junco]

i got no idea

[weary pelican]

so see I feel like this "definition" is even harder to properly define than the one I'm showing you

anyways

[weary pelican]

with this we finally have $\lim_{x\to 0}\frac{\exp(x)-1}{x} = 1$

[clever fjord]

rafilou2003

[weary pelican]

now we only have to show that exp(x) is the "same thing" as e^x

so let's start with showing point 2), that exp(1) = e

this stackexchange discussion shows it better and quicker than I would manually https://math.stackexchange.com/questions/4025559/how-do-i-prove-exp1-e-if-exp-is-defined-in-a-series-expansion

[edgy junco]

alright, thanks man

[weary pelican]

and final thing

after proving that exp(1) = e

for more reassuring it would be nice to know if we also have exp(n) = e^n for any integer

this is where the "exponential property" comes into play:

$\forall x,y\in \bR, \exp(x+y) = \exp(x)\exp(y)$

[clever fjord]

rafilou2003

[weary pelican]

(just like you would expect with exponents)

to prove this:

fix some value of y

and we're gonna show that the functions $f(x) = exp(x+y)$ and $g(x) = exp(x)exp(y)$ are equal

[clever fjord]

rafilou2003

rafilou2003

[weary pelican]

oh it is the way oops

now, what we would like to do is to study the function $\frac{f(x)}{g(x)}$

[clever fjord]

rafilou2003

[weary pelican]

problem: if g(x) = 0 somewhere that's not possible

so what we need to do is to show that exp function doesn't equal 0 anywhere

hope you're still following

we're almost out of the rabbit hole I promise 😅

[edgy junco]

I'm following what you're saying, but I'm not really following why proving that exp(x+y) = exp(x)exp(y) would help with proving that exp(n)=e^n. I haven't really figured out what they're saying in the stack overflow chats so that might be why

[weary pelican]

haha it's kinda by induction

once we have exp(x+y) = exp(x)exp(y)

and since we know exp(0) = 1, exp(1) = e

then exp(2) = exp(1+1) = exp(1)exp(1) = e^2

exp(3) = exp(2+1) = e^2 * e = e^3

do you see it?

[edgy junco]

oh, I see

to prove that exp(n) = e^n for any number we just gotta prove exp(x+y) = exp(x)exp(y) becuase it's already established that exp(0)=1 and exp(1)= e

[weary pelican]

yep

so

[weary pelican]

going back to this, let's prove exp can't be 0 anywhere to make this function have sense

welp, it's by looking at $h(x) = exp(-x)exp(x)$

[clever fjord]

rafilou2003

[weary pelican]

we can compute its derivative using chain rule and product rule

$h'(x) = -exp'(-x)exp(x) + exp(-x)exp'(x) = -exp(-x)exp(x) + exp(-x)exp(x) = 0$

[clever fjord]

rafilou2003

[weary pelican]

so h is constant

and h(0) = exp(0)exp(0) = 1

so

for all x

h(x) = exp(-x)exp(x) = 1

we've proved exp(x) can't be 0

and 1/exp(x) = exp(-x)

[weary pelican]

so this function can be defined

anyways you get the gist, we can differentiate f(x)/g(x)

we'll get the derivative is 0

so f(x)/g(x) is constant

and f(0)/g(0) is, you guessed it, equal to 1

so f(x) = g(x)

meaning $\exp(x+y) = \exp(x)\exp(y)$

[clever fjord]

rafilou2003

[weary pelican]

so we're finally done

[edgy junco]

I see

[edgy junco]

I'm just slightly confused about these functions. f(x) = exp(x+y), for these functions the "x+y" in the exp() has nothing to do with the
"x" in f(x) right?

anyways thank you

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[weary pelican]

it does, x is the variable

[weary pelican]

y is a constant

it's like f(x) = exp(x+1)

just replace 1 by "y"

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[cedar magnet]

How does this derivation go?

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@cedar magnet Has your question been resolved?

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@cedar magnet Has your question been resolved?

[cedar magnet]

<@&286206848099549185>

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[low locust]

its unclear what exactly you are asking. the text says how the argument works

what exactly do you want to hear

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[fluid stratus]

hii how do i do 5.(iii)?

[fluid stratus]

i have this so far but im not sure if i put the correct term in the bracket

<@&286206848099549185>

[serene folio]

maybe try binomial theorem ( newton's binom)

use this for your second braket

[fluid stratus]

i did this so far but im not sure if i need the x fully gone or if theres supposed to be one x term

[serene folio]

x has to be gone

"independent of x"

[fluid stratus]

oh

ugh

wait what term can give me that?

[serene folio]

?

you will see

[fluid stratus]

no like i mean in the second bracket if i use the 6th term, theres no x

but if i open the brackets then thrs gnna be the 1/x^3 left

no matter what term i use i cant get the x gone😭😭

[serene folio]

could you transform me the second braket into a sum using this formula?

[fluid stratus]

sure but im not sure what the n Z k=0 looking thing is

[serene folio]

it's a sum

sum after k=0 to n

1+2+3+4+5+6+...+n = n Z k =1 (k)

[fluid stratus]

girl i have no idea what your talking about

[serene folio]

umm

what grade are you in?

[fluid stratus]

10th i think?

sec 4

[serene folio]

i dont know if in us or what state are you from the binomial theorem or sums are in 10th class

hmm

[fluid stratus]

im frm sg

and we learnt them last year😭

[fluid stratus]

but how do i get a term independant of x frm here?

[serene folio]

We look for the terms for which his power 𝑥 is zero. That is, we determine the terms for which the exponent of x is zero.

[fluid stratus]

but if i multiply that in thrs still the 1/x^3 left

[serene folio]

that's your second bracket using binomial theorem

[serene folio]

we will make it disappear

then replace k = 6 in first bracket and k =5 in second

calculate and you will find that the independent term for x is 252m^6 - 125m^5

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@fluid stratus Has your question been resolved?

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[viral steppe]

help😭 how to solve for c

[mystic saffron]

ac+bc = 30

a+b = 5

a= 5-b

5c - bc + bc = 30

5c = 30

c= 6

ayyy im smart

[indigo crater]

!noans

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The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[mystic saffron]

oh mb

[indigo crater]

Anyways

[subtle sleet]

It is so hard

[viral steppe]

oh thanks

[viral steppe]

😭

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[peak maple]

I got up until

the last step

I don't get why

the negative sign

becomes positive at the end though

[dawn tiger]

they switched the bounds

[pastel orbit]

^

[peak maple]

oh...

dang

thanks

ok

cool

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[tame oasis]

Sooo i have summer school and this teacher is saying 2+2 is not 4 its some weird equation can sum1 help me solve this

[tame oasis]

Plsss

😔

[worn sandal]

erm

[fair token]

Yeah, don't do that.

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[somber adder]

Idek where to start

<@&286206848099549185>

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[somber adder]

oh sorry

[viscid tendon]

R = 9

[somber adder]

Wrong

answer key says wrong

[viscid tendon]

What is the answer?

[somber adder]

4

I know how to do it

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[median galleon]

How does one take on this?

[sonic topaz]

b_7 = 96 or b_7 = 960, right?

i think just solve for the first 9 terms in terms of b_7??

[median galleon]

b6 = 96 and b8 = 960

[sonic topaz]

yeah but we know that b6 | b7 | b8

and 960/96 = 2 5

so it can't be 96 2 or 96 5

so it has to be either 96 or 960

[median galleon]

Hmmm

[sonic topaz]

we also know
b8 = b7(b6/b5 + 1)
-> 960 = 96b7 / b5 + b7
-> (96b7)/(960 - b7) = b5

so b7 better not be 960 then

?

[median galleon]

Hmmmm

[sonic topaz]

yeah this actually just locks b7=96

wait

no

what's the mistake?

because b7=96 gives noninteger for b5

oh i'm too dumb ain't i

[median galleon]

So b7 = 960

[sonic topaz]

this all is wrong

it can actually be 2 96 or 5 96

[median galleon]

Yah i didn't even get how one get that

Ohhhh

[sonic topaz]

sorry too fried

but either way we have
(96 b7) / (960 - b7) = b5
b5 = 24(if b7 is 192) or 96(if b7 is 480)

[median galleon]

Okay Okay

[sonic topaz]

b7 = b6(b5/b4) -> b4 = b5 b6 / b7 = 24 96 / 192 or 96 96 / 480 = 12 or 19.2

but 19.2 is not integer

so b4 = 12

[median galleon]

Hmm

[sonic topaz]

b5 = 24
b7 = 192

[median galleon]

Ohhh

Damn

[sonic topaz]

b6 = b5(b4/b3 + 1) -> 96 = 24(12/b3 + 1) -> 3 - 1 = 12/b3 -> b3 = 6
b5 = b4(b3/b2) -> 24 = 12(6/b2) -> b2 = 3
b4 = b3(b2/b1 + 1) -> 12 = 6(3/b1 + 1) -> b1 = 3

b9 = 960 (192/96 + 1) = 2880

giving average of 464

[median galleon]

Wait qs

How does one get b7?

[sonic topaz]

there are 4 options it can be because of the divisibility rule

then just take it as unknown for now and rearange the definition for b_8 to solve for b_5 in terms of b7

you will see that if b7 = 960, then there is no solution for b5, and if it is 96 then the solution is noninteger

so now you know that b7 is either 192 or 480

and then you can do it on case by case basis

again

solve for b4

and you will see that b4 is only ever integer if b7 was 192 to begin with

[median galleon]

Oh, I see it came from factors of 960

I was confused where did you get it from

[sonic topaz]

because b7 = b6 n and b8 = b7 m

so really i was thinking b8/b6 = 960/96 = 10 with factors of 1, 2, 5, 10
so b7 has to be 96 x one of those factors

[median galleon]

OH I SEE

So what are the values for b1 to b9?

b1 = 3, b2 = 3, b3 = 6, b4 = 12, b5 = 24, b6 = 96, b7, = 192, b8 = 960, b9 = 2880

Holy

The problem though

There is a floor A

It's equal to 464 or its meant to put us off?

OKAY I GET IT

THANK YOU VERY MUCH

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[waxen helm]

Relative to a fixed origin O, the point A has a position vector -2i + 3j. The point B has a position vector 3i + pj, where p is a constant. The point C has a position vector qi + 7j, where q is a constant. Given that |AB| = 5 * sqrt(2), (a) find the possible values of p. Given that the angle between AC and the unit vector i is pi/3 radians, (b) find the exact value of q.

I got part a pretty easily but using the dot formula produces two possible answers for q, ignoring how these answers aren't nice numbers, the question said that there was one exact solutions. the specification normally specifies if there are multiple possible solutions in the question

[waxen helm]

it could just be the answer i originally got though

Some more context

[cold sage]

it does say find the possible "values" of p

oh wait youre talking about b

[waxen helm]

ik i was talking about q

[waxen helm]

yh

[cold sage]

alr, one minute

[waxen helm]

cool

[cold sage]

so
(q+2)=1/2 [sqrt(4^2+(q+2)^2]
(q+2)^2=1/4 [ 4^2+(q+2)^2]=4+1/4(q+2)^2
then 3/4 (q+2)^2 = 4 (q+2)^2=16/3 q=+-4/sqrt(3)-2
since it says theres one im guessing one of the cases doesnt actually fulfill the dot products equality

you can check that

[waxen helm]

oh okay cool thank you

/close

thats not teh command

[cold sage]

.

[waxen helm]

.close

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[subtle sleet]

what?

[waxen helm]

i closed the channel

.avaliable

idk

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[wide drift]

I mean I got the right answer but I kinda just guessed

I have no idea how to do this question

Like theres nothing inside the integral\

[atomic epoch]

i can help

i think

[wide drift]

Yes please haha im a bit confused

[atomic epoch]

it’s basically taking the integral of 1

[golden marten]

^

Think about DX as DX • 1

[wide drift]

Sry if this is a dumb question but wheres the 1

[atomic epoch]

the antiderivative which you got correct is x +c since it’s undefined ( no lower or upper )

[atomic epoch]

here

[wide drift]

Hmm ok

I think I get it

[atomic epoch]

it’s not dumb but think of it as multiplying it by one like gizmic said

[wide drift]

Feels weird tho haha

[atomic epoch]

in what way, i can try to explain further if i can

[wide drift]

Nah its nothing, I think I get it 😄

[atomic epoch]

you sure ?

[wide drift]

Yeye haha

[atomic epoch]

if you have any more questions feel free to dm me

[wide drift]

I will 🤗

Thank you youre very kind!

❤️

[atomic epoch]

okayy good luck

[wide drift]

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[autumn vine]

By 'circle O', is it meaning that the centre of the circle is labelled O?

[rough rivet]

ye

it is ussully written as {O;R}

[rough rivet]

"given {O;R}" means given a circle with a center of O and a radius of R

but circle O can still be understandable depending on context

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@autumn vine Has your question been resolved?

[nimble blaze]

lone circles are usually named by their centres

if there are multiple circles present with the same centre, then more info would be included to distinguish between them

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[vagrant python]

National Oric Mathematics Competition 2024
Problem 1)The diagram below displays a permutation of the letters "A, B, D, E, F, J, O,.....". This permutation permits repeated letters, and no letter or order shall be removed. What is the letter at the 2024²⁰²⁴th position? (10 Marks)

[vagrant python]

What's this question

Crazy question

[summer cradle]

it looks like there is missing info

also what is this competition?

[vagrant python]

private competition. non public

the question is just following the pattern, and 2024²⁰²⁴ i believe can be simplified

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[red egret]

in (0,0)

[peak lake]

use the formal limit definitioin of differentiation

[weary pelican]

we're in 2D, only partial derivatives have limit definitions

but you can certainly try to compute the gradient this way and see if it makes sense

[red egret]

I've already did the partial derivatives

1 for x 2 for y

but I don't remember the formula after that

[peak lake]

there is a theorem that implies if you have cts SO partials around a point, then it is diff'able at the point

[weary pelican]

?

just recall the formula for 1D differentiation

$\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$

[clever fjord]

rafilou2003

[weary pelican]

and for multiple dimensions partial derivatives

just place the "+h" on the direction of the derivative

$\frac{\partial f}{\partial y}(x,y) = \lim_{h\to 0}\frac{f(x,y+h)-f(x,y)}{h}$

[clever fjord]

rafilou2003

[red egret]

no I mean differentiable in (0,0)

forgot to write

[weary pelican]

yes

so (x,y) = (0,0)

[red egret]

but isn't that just the derivative for x

y*

sorry

[weary pelican]

?

isn't that EXACTLY what partial derivatives are?

[dapper canyon]

check if the partial derivatives are continous at (0,0)

[weary pelican]

if they're not, this doesn't tell you if the function is differentiable at (0,0) or not

so since you've found the partial derivatives

you have the gradient

$\grad f (0,0) = \begin{pmatrix}1\2\end{pmatrix}$

[clever fjord]

rafilou2003

[weary pelican]

that means if $f$ is differentiable at $(0,0)$, then $f(x,y) = x+2y + o(x,y)$

[clever fjord]

rafilou2003

[weary pelican]

in fact, you need to prove that $\frac{f(x,y)-x-2y}{|(x,y)|} \to 0$

[clever fjord]

rafilou2003

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[red egret]

thanks!

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[green cradle]

I know this is a maths server but am desperate

[green cradle]

I could not even understand the solutions either

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[green cradle]

@dusk warren

.close

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[dense maple]

can someone help me

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[dense maple]

sigh

/close

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[elfin bear]

how can L < 1?

[elfin bear]

such that:

that would be $a_{n+1}/a_n \rightarrow lim(a_{n+1})/0$

[clever fjord]

Ayanokoji (ALWAYS PING ME)

[uneven osprey]

!original

[odd edge]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[elfin bear]

thats my question

can u explain this

[uneven osprey]

how am I supposed to know what a_n id

is*

[elfin bear]

...a sequence

[sterile blaze]

its false

suppose a_n=(-1)^n

L would be -1 which is less than 1

[elfin bear]

to be clear this is the if

[elfin bear]

and this is another if

[sterile blaze]

but a_n doesnt converge 0

[elfin bear]

to the result you see a_n -> 0

[sterile blaze]

however, if L is in (-1,1) then yea its true

[elfin bear]

in english these are the conditions for the "fraction" or "quotient" convergence test for sequences

[sterile blaze]

because of L is in (-1,1) then |L|<1 a_n will appraoch 0 slowly for each term in the sequence

[elfin bear]

but your example doesnt satisfy the conditions, L = 1

oh wait

i see

[sterile blaze]

yes

anyway but it doesnt work if L is less than or equal to -1

[sterile blaze]

im guessing they meant to write |L|

[elfin bear]

i found this

i highly doubt it

[sterile blaze]

yea thats the same

[sterile blaze]

doubt what

[elfin bear]

oh i omitted a detail they said, that a_n > 0 for all n

[sterile blaze]

yea then sure

[elfin bear]

it's kind of useless to put this condition when the theorem should be for absolute values

[sterile blaze]

L can not be negative

[elfin bear]

true

thank you for the help mate

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[sterile blaze]

no problem ahoy

[elfin bear]

.reopen

[odd edge]

✅

[elfin bear]

umm i just realized i still dont understand why a_{n+1} > a_n

i mean lim a_i = lim a_j as far as i know

[sterile blaze]

hm?

[elfin bear]

since both go i,j -> inf

[sterile blaze]

where did this come from

[elfin bear]

like in here

[sterile blaze]

yes, what about it?

[elfin bear]

they say the fraction can be anything other than 1

and i dont understand how

[sterile blaze]

no, if it is 1 it is inconclusive

it can be 1

[elfin bear]

true

but how can there be r < 1?

[sterile blaze]

1, 1/2, 1/4, 1/8, ...

r=1/2

[elfin bear]

whats the sequence?

[sterile blaze]

2/2^n

[elfin bear]

i see, i get it now

ty again

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[sterile blaze]

np

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[lapis bobcat]

I want to look up videos / content explaining this question but I don't know what kind of keywords I should be searching to find this.

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[rocky smelt]

I'm stuck with this double integral, especially with the domain after the swap of coordinates. I need to set integral bounders for u variable. Moreover I've already obtained them for v variable but I'm not sure it is the correct way

[rocky smelt]

From solutions it should be $u\in[0,2]$

[clever fjord]

x_Shadow_x

[rocky smelt]

For v variable $u+2 \leq v \leq 6-u$ as I obtained, but I'm not sure my procedure is right since I take the second function of $v \geq 2+u$ and i put it in the place of $v-2u \leq v$ as I show with the arrow

[clever fjord]

x_Shadow_x

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[rocky smelt]

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[tacit hound]

pls

[tacit hound]

help

[nocturne belfry]

What's the exercise?

[tacit hound]

wdym

[nocturne belfry]

The image just shows an angle

[tacit hound]

ok so

its asking this too

[nocturne belfry]

gotcha

[tacit hound]

yes thank you

ive been stuck for so longg

[nocturne belfry]

any thoughts on how to start?

[tacit hound]

no clue

i was hoping u can explain

[nocturne belfry]

hmm can you make an estimate

eyeball it

The idea is that you know $\angle DBA = 133 ^\circ$

hmm whats the command

[tacit hound]

im lost

wait

[clever fjord]

jan Niku

[nocturne belfry]

and we know part of DBA right?

so, DBA is made up of two angles

$\angle DBA = \angle ABC + \angle CBD$

[clever fjord]

jan Niku

[nocturne belfry]

Your task is to identify which of these 3 is known or unknown