help-49

[untold oyster]

,rotate

[sleek cloud]

Sorry I think you just need to multiply and add all of them up

For simplification

[midnight plank]

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[untold oyster]

How?

[sleek cloud]

Do you know how to multiple two products?

[midnight plank]

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[kind coyote]

Hi

[tawdry laurel]

you could find the pattern

[slow thorn]

oh hi

[kind coyote]

Hi

[tawdry laurel]

找 pattern

[slow thorn]

the wha??

[tawdry laurel]

find an expression for the number of rods in terms of n

@kind coyote 找到了吗

[kind coyote]

I only see multiples of 6

[slow thorn]

indeed

and there is a reason for that

lets say you finish the first hexagon

now you put six rods around the hexagon

[kind coyote]

6 rods?

Or hexagons

[slow thorn]

like this

and then put 18 more rods

to complete 3 sides for 6 hexagons

[kind coyote]

Ok

[slow thorn]

just see if you can find a pattern with the number of surrounding rods to be attached for each n

for 1, it is 6 as we did now

or you can even count and figure it out

[kind coyote]

Hmm

[slow thorn]

can you find for the next step?

from 2 layers to 3 layers

[kind coyote]

36

[slow thorn]

no like

only the rods surrounding the layer

like what we did for the first one

[slow thorn]

over here

[kind coyote]

12

[slow thorn]

ye

and for 3?

[kind coyote]

18

[slow thorn]

crct

so what pattern can you find?

[kind coyote]

6 12 18 24 …

6n

[slow thorn]

yep

now finally we need to attach the surrounding rods to complete the structure

[kind coyote]

Ye

[slow thorn]

if we take n = 3 as an example we can see that there are two different types of hexagons

there are six hexagons at the corners

which need 3 more rods to be completed

and six hexagons at the sides

which need 2 more rods to be completed

and now for n = 4

we still have 6 corner hexagons, but now we have 12 hexagons at the sides

can you find the formula for the number of side hexagons for a given n from this?

[kind coyote]

No

N = 5 will have 6 corner hexagons and 18 at sides\

[slow thorn]

yes

6, 12, 18... 🤔

[kind coyote]

Ye

[slow thorn]

and it is 6 at n = 3

so,
n = 3: 6 * 1
n = 4: 6 * 2
n = 5: 6 * 3

and so on

can you find a relation between n and the number we multiply with 6?

[kind coyote]

Making the formula

Omg

[slow thorn]

hint: 3 - 2 = 1

4 - 2 = 2

and so on

[kind coyote]

N-2

[slow thorn]

yes

6 * (n-2)

is the number of side hexagons

and the number of corner hexagons is always 6

so how many rods do we need to complete all corner hexagons?

[kind coyote]

3

X6

18

[slow thorn]

yes

for now we have 6n surrounding rods and 18 corner rods

so how many rods do we need to complete all the side hexagons?

[kind coyote]

2

Hmmm

If 2 hexagons on each side

24

[slow thorn]

okk

so for 6 * (n-2) hexagons

how many rods would you need

[midnight plank]

@kind coyote Has your question been resolved?

[kind coyote]

Doesn’t it vary depending on n

[slow thorn]

yes

so here

you can just multiply the expression by 2

so 6 * (n-2) * 2 = 12 * (n-2) rods to complete the side hexagons

and now we finally can say how many rods we would need to go to n from n - 1

can you figure it out?

we have 6n surrounding rods, 18 corner hexagon rods and 12 * (n-2) surrounding rods

and you need to add them all up now

[kind coyote]

48

[slow thorn]

no, like in general

do you know algebra?

[kind coyote]

Little

[slow thorn]

okk

so can you expand 12 * (n-2)?

[kind coyote]

12n -24

[slow thorn]

ok nice

so that + 6n + 18 will give?

[kind coyote]

18n-6

[slow thorn]

ok

so thats the total number of rods you need to get from layer n-1 to n

wait

no

i think its 6(n-1) instead of 6n

we need 6 rods to go to n = 2

so we just need to add 6n - 6 instead of 6n

to get 18n - 12

to go to n = 2 from n = 1 we need 18 * 2 - 12 = 24 rods

okk

[kind coyote]

Ok

[slow thorn]

now finally

we can make an expression for the number of rods for a given n

say we start at n = 1 (6 rods)

and we want to reach n = 3

so first we will go to n = 2

which needs 18 * 2 - 12 = 24 rods

and then go to n = 3

which needs 18 * 3 - 12 = 42 rods

so totally we would need 6 + 24 + 42 = 72 rods

for n = 3

so to reach a given n

we can start from n = 1

and then go step-by-step

6 + (18 * 2 - 12) + (18 * 3 - 12) + ... + (18 * n - 12)

giving us the total number of rods

[kind coyote]

Ok

Can we do this? 6 + 18n - 12 < 2000

[slow thorn]

hmm

[midnight plank]

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[kind coyote]

Next time thank u for your patience I have to go

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[grave kernel]

Not really a true math question
I know that, well, if there are some x white balls and y blue balls
The probability of taking out a white ball is
x/(x+y)
Now if I take out a ball, without observing what it is
and then a second ball is picked
the probability that the second ball is white is still x/(x+y)
but what is the logical reasoning to this?
this is very counter intuitive and also relates to the double slit experiment
but can't really understand the logic to it
if there is one

[hollow oyster]

for what i understood that picking without observing is the same as not picking could be wrong

[grave kernel]

yeah i mean i had the same thought too
like its as if youre picking it up for the first time

but i didnt know if it was really correct or not

.close

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[twilit field]

how does this look

The set ${v}$ is a linealry independent set, we know that any linearly independent set can be extended to a basis set, thus any non-zero vector in a finite dim vector space is aprt of a basis

[runic hamlet]

well ok you can extend it to a spanning set

what does that have to do with a basis

[twilit field]

I can extend it to a basis too

I've proven that earlier

[runic hamlet]

then why are you writing spanning set instead of basis

[grand pond]

A dense set

[twilit field]

here will a single counterexample in $\R^4$ suffice

[grand pond]

A dense set

[prime hornet]

yes

[twilit field]

Let $v_1,v_2$ be basis of $U$. We also first extend this to a basis of $V$; ${v_1,v_2,v_3,v_4}$ consider another basis of $V$, $v_1+v_3, v_2+v_4, v_2+v_3, v_4$. Clearly none of these vector are a basis of $U$, nevertheless they span $V$. Thus the statement is false

[grand pond]

A dense set

[runic hamlet]

why are none of these vectors a basis of U?

[twilit field]

none of them lie in $U$

[grand pond]

A dense set

[runic hamlet]

why not?

[twilit field]

$U$ is spanned by $v_1,v_2$ Let $\alpha_1v_1+ \alpha_2v_2 =v_1+ v_3$, that would imply that $v_3$ lies in the span fo $v_1, v_2$ ,which is clearly not possible

[grand pond]

A dense set

[twilit field]

similarly $v_2+v_4$ can't either

[grand pond]

A dense set

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[twilit field]

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[chrome ibex]

Question:Determine a base for this vector spaces a)

Can someone give me a hint or something like that ?

i thought of giving "random" values to x y and z such that i get 3 different vectors , since its R^3 i have at maximum 3 linear independent vectors , but since i am under some restrictions i can have less Linear independent vectors right ?

and then i will check for linear independence

but i think this is not the right process

[cedar coral]

giving random values is indeed pretty much never the right process

on the first one, you have the vector space defined by three equations in R3.
Your first step is to determine if the equations are redundant or not. Gauss method should help there.
For b and c, you have the linear space generated by 3 vectors. Since first and 2nd vectors are not proportional in either, they are independant. Check if the 3rd is dependant. If it is, the base is the 2 vectors. If it isnt, the base is the 3 vectors.

[chrome ibex]

tyyy

I was a little bit confused about b) and c) cause i was like "If a vector space is generated by 3 vectors doesent that mean , that they form a basis" , but now that i said that outloud and with your text,i think that i got why , it means that at maximum it is generated by 3 linear independent vectors , but if some are linear dependent of each other that reduces the amount of vectors in the base right ?

My english is not the best i am sorry for that

[cedar coral]

no, they form a generating system

a generating system lets you get all the vectors of the space. A base is a generating system with the minimum amount of vectors, aka, all of them are linearly independant

to get a base from a generating system, you remove all the linearly dependant vectors

[chrome ibex]

yeah exacly!!! maybe i didnt express my self quite well in the previous text , but thats the ideia i got

and btw in the first one i found out by the Gauss Method that the 2nd line and 3nd line are redudant equations

so that means that i can write the vector (-2,5,-3) as a linear combination of (2,-1,1) and (-2,3,-2) right ?

[cedar coral]

it might be a language issue, but only one of these is redundant. Assuming your next line is correct, the 3rd would be the redundant one

because you can write the third as a LC of the other two

your vector would be (-2, 5, -3) tho, btw

[chrome ibex]

i am doing this on paper , sorry for my handwritting here "online"

thank you for helping btw

[chrome ibex]

yeah sorry for that

(-2,5,-3) = (2,-1,1) + 2(-2,3,-2)

now it is correct

i think

.close

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[plucky atlas]

which derivation is correct? the second one is the derivation my teacher used in her solution. she ignored the the digit in without x "x" in the product rule but in the third derivation she uses the digit without "x". so she made a mistake either in the first or third derivation.

[surreal moon]

Images are preferred over pdf

[plucky atlas]

ok one minute

[surreal moon]

$f(x)=1+2x^2 e^{-0.1x}$

[plucky atlas]

no 1+2x²

[grand pond]

SWR

[surreal moon]

The first f'(x) calculation is wrong

The second f' is correct

And I see no third

[plucky atlas]

$f‘‘‘(t) = (−0,002x² + 0,12x − 1,2)$

[grand pond]

ir0nfly
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[plucky atlas]

why is there -1.2 ?

[keen saddle]

can you translate to eng?

[keen saddle]

this

[plucky atlas]

it's an exponential function for bacteria growth in x minutes.
assignement a: how many bacteria are there after 10 minutes.
assignement b: when will there be most bacteria.
assignement c: when will be slowest growth

that's my progress so far. but my third derivation looks different than hers

[keen saddle]

can you use a darker pencil or pen pls?

i cant read anything

[plucky atlas]

i tried editing for better visibility. is that betterm

[keen saddle]

ohk gimme a min

f(x) = c + 2x^2 . e ^()

right?

e^(0.1) ?

[plucky atlas]

e^-0,1x

[keen saddle]

ohk

[plucky atlas]

$f(x)=1+2x^2 e^{-0.1x}$

[grand pond]

ir0nfly

[keen saddle]

yeah 1.2 is incorrect

$f'''(x) = -0.002x^2 + 0.12x + -0.8$ is correct

[grand pond]

Wumpus Man

[plucky atlas]

ok perfect thank you very much

[midnight plank]

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[umbral scroll]

does anyone know the electrical potential energy in electronvolts of an electron subjected to a potential of one volt?

[olive matrix]

surely 1

[umbral scroll]

ok

so we agree the E = q x U right ?

[lethal path]

it's E = qV

[umbral scroll]

yes sorry

V the potential

but you also agree that q of an electron is -1.6 * 10^-19 right

[grim vector]

Oui

[vague remnant]

J

[umbral scroll]

pourquoi E n'est pas égal à -1 du coup ?

why is E then not equal to -1

if the charge of the electron is negative

$E= q_{e-} * V$

[grand pond]

Poposagrado

[lethal path]

cause $W = -\Delta U$, work done is negative of change in potential energy

[grand pond]

south's secret twin brother

[umbral scroll]

sorry ?

the potential is positive

as it is of 1 on the given point

[keen saddle]

work done by a conservative force is negative of dU/dr

[keen saddle]

so this is correct

[olive matrix]

it's the same reason that when you lift a weight you add potential energy to it

[umbral scroll]

but how do i correct the formula

i dont understand the link with the work

[weak parrot]

potential is kQq/r i think

work done by pulling an charge q from infinitely far away to the point in the centre of the field

[umbral scroll]

i dont understand

[weak parrot]

its negative because the forces opposes the motion, otherwise the charge would accelerate

[umbral scroll]

U=1
q(e-) = -1.6 * 10^-19
E = (U * 1.6 * -10^-19)/ 1.6*10^-19
so E = -1

[keen saddle]

idk myself i was just saying that W = -delta U is correct

[umbral scroll]

maybe it is something with the formula ?

E = | q * U| ?

does it still work for a proton for example ?

i guess it would

[weak parrot]

e is the negative derivative of potential against distance r

kQq/r^2

[umbral scroll]

ok but the formula

E = q* U

@grim vector tu peux m'aider ou pas

stp 😢

<@&286206848099549185> anyone ?

[midnight plank]

@umbral scroll Has your question been resolved?

[keen saddle]

this is the relation between electric potential

and electrical potential energu

wait no

bro can you tell me what E and U are

i think they are the same thing

ateast have the same units

[umbral scroll]

sory

E is the potential electric energy

and U is the potential (in Volts)

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[umbral scroll]

.reopen

[midnight plank]

✅

[umbral scroll]

<@&286206848099549185> ?

[midnight plank]

@umbral scroll Has your question been resolved?

[midnight plank]

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[last slate]

Prove that the perpendicular bisectors of the sides of a triangle are concurrent.

[nova pike]

what have you tried?

[last slate]

nothing

!status

[midnight plank]

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

[last slate]

1

nothing that worked

[nova pike]

hint -> || prove that the perpendicular bisectors meet at the circumcenter of the triangle ||

[last slate]

we have to, then, prove that a triangle can be inscribed in a circle

any triangle

[keen saddle]

I think that is an axiom

[last slate]

Idts

[keen saddle]

No need to prove

[last slate]

suppose I didn't know that circumcentre existed

also we're not allowed coordinate geometry

[nova pike]

use perpendicular bisector theorem

[rain wasp]

you can assume that there exists a circle passing through 3 points

a unique one at that

what you want to do is to let some points, call it O, be the intersection of two perpendicular bisectors

your goal is to prove the third perp bis also goes through O

it helps to draw a picture

[nova pike]

yes, I also suggest that this link is helpful to prove the concurrency of the perpendicular bisectors of a triangle (has diagrams too): https://jwilson.coe.uga.edu/EMAT6680Fa07/O'Kelley/Assignment 4/Perpendicular Bisectors of a Triangle.html

[lethal path]

the circumcenter isn't always inside the triangle

that's the tricky part

[midnight plank]

@last slate Has your question been resolved?

[lyric iris]

Find the area of shaded region

[last slate]

open yo own channel

#❓how-to-get-help

[keen saddle]

@last slate is your doubt still there?

[last slate]

i can't do it

<@&286206848099549185>

[severe basin]

hi

[last slate]

hi

[severe basin]

so

what have you tried

[lone wind]

This ur question?

[severe basin]

wait what question is this

[lone wind]

Look at the bottom triangle ..now that we have proved the congruency for all three triangles

We get all three sides are equal

Hence equilateral triangle

Interior angles are hence 60

U got it till now ?

[last slate]

how is it an equilateral triangle

I get it for an equilateral triangle

[lone wind]

Look at the congruent triangles

[last slate]

Is this any random triangle?

[lone wind]

The question it is said perpendicular bisector of sides right ?

[last slate]

yes

[lone wind]

Okay so what I did is I assumed a random triangle ABC ...and then took 2 triangles within triangle ABC to prove congruency

Konsa class hai yeh ?

[last slate]

but this would imply AB = AC which is not true for every triangle

[livid summit]

man

[zealous schooner]

the altitudes of a triangle don't bisect the sides..

and the medians are not perpendicular to the sides

you just proved it for an equilateral triangle

[lone wind]

I would suggest go check the question

[lone wind]

.

[keen saddle]

bro you assumed perpendicular bisectors pass from the vertex

which is not the case for any random triangle

[zealous schooner]

any way you look at it you're assuming something that's not general to all triangles

[last slate]

yes

[keen saddle]

bhai tune hi question galat samjha h no offence

[livid summit]

just intersect 2 of the bisectors

[lone wind]

Kaise bhai

[livid summit]

and then prove an altitude from that point to the third side is also an bisector

of that side

[lone wind]

How can a side be bisected and perpendicular..if it is not from the opposite vertex

[zealous schooner]

yeah this is the traditional proof i've seen for this

[zealous schooner]

why should it?

the only time that happens is either for isosceles triangles on the non-equal side, or in an equilateral triangle

[zealous schooner]

otherwise the perpendicular bisector of a side necessarily cannot pass through the opposite vertex

[keen saddle]

dekh bro, MN perp bis h but B se pass nhi kar rha

[lone wind]

Oooh

[keen saddle]

sirph equi triangle mai vertex se pass karta h

[last slate]

you can conclude you've done something wrong when you get AB = AC

[keen saddle]

(aur iso mai ek perp bis pass karta h baaki dono nhi)

[zealous schooner]

@last slate you can also prove this by proving that a point lies on the perpendicular bisector of a line segment if and only if the point is equidistant from both end points

be careful though, you need to prove both sides of the theorem

[last slate]

wait

I had done the same way earlier

I got it

[zealous schooner]

Consider F to be the point of intersection of any two perpendicular bisectors, say the bisectors of AB and BC.
(These two must intersect as AB and BC may not be parallel)

Then by the theorem above, F is equidistant to A and B, and F is equidistant to B and C. This means F is equidistant to A and C, which means F lies on the perpendicular bisector of AC, hence proving all three are concurrent at F.

[last slate]

yup

👍

thanks guys

.close

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[wanton vault]

lads

i cant even begin solving this cursed limit

some context, b part

my work till now

i feel like there is alot of wrong here

[midnight plank]

@wanton vault Has your question been resolved?

[wanton vault]

i feel like no one gonna answer this one so

.close

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[midnight plank]

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[nova coyote]

how do i find out what t is equal to when A and B meet? i used trial and error to get my answer although im sure there is a much more efficient way

[olive matrix]

what's A's x-value as a function of time?

what's B's x-value as a function of time?

when will they be at the same x value?

[nova coyote]

oh right

i see thanks

yeah i got t = 4

.close

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[tranquil lily]

I'm doing this so wildly wrong

[tranquil lily]

Yea

How am I actually meant to do it?

I have no intuition for this

[midnight plank]

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@tranquil lily Has your question been resolved?

[tranquil lily]

@help

<@&286206848099549185>

I keep trying to use cartesian / cooridnate geometry and I just get F is 0,0

Which it isn't but idk where I've gone wrong

[tranquil lily]

..

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[echo heart]

Hello can somebody help me with a difficult math equation?

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[echo heart]

so ive been trying to calcutate the farm value with this

[echo heart]

and these are the numbers ive gotten Farmvalue = 30000 * (1.1) x (10000000) * (230000000000/300000000/13800000000000) * (1+6627000000/100) * (130-4)^0.25 * 13 * (300000000 (930000000000/230000000000) + 0.2 * 1 + 5670000000 * 0.6 + 0.25 * 2046000

ive tried doing them in multiple parts but i'm not sure if that's right, and even AI couldn't solve it

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[frozen igloo]

how would you apply the divergence theorem here? because S isn't closed

[sharp coral]

we can say
(integral of side surfaces) + (integral of caps) = (volume integral)

[surreal moon]

Close it maybe?

[frozen igloo]

TRUE

[frozen igloo]

ah ok

so evaluate all but integral of side surfaces

tysm :)

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[fair plume]

I don’t know how to solve question 14

[woven patrol]

Translate?

[fair plume]

We can construct a sequence of whole numbers by adding to each number, the double of the sum of the numbers composing it. (Look at the example). So how many numbers from the start that are less than 2025, let us get 2025 using this equation?

[fair plume]

Yeah was just doing that

[west iron]

well the sum of 4 digits can't be more than 36

when the number is <2025 it can't be more than 28

so we only have to check from 2025-2*28 to 2025

that's still annoying, so the next step is to find another way to rule out numbers

Which is look at the value on each side mod 9

[fair plume]

ok

um idk if its important

but if you look at the last digit of each origingal number, and compare it to the last digit of the new number, you can possibly conclude that you take the last digit, multiply it by two, then add it by 2 from its orginal form.

For example

1008 gives 1026 in the example

(8+2) + (8+2) = 26

and also

1002 -> 1008

which is (2*2)+(2+2) = 8

<@&286206848099549185>

anyone?/

<@&286206848099549185>

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<@&286206848099549185>

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[twilit field]

To solve this I suspect I have to find the values of $x$ for which $a^x-x$ has a solution.
\
To do so, we apply the intermediate value theorem, for $a>1$, it follows that for a sufficiently small value $a^x-x$ is negative, and for a suficiently large value , it's positive. It Thus follows that it has a solution

[grand pond]

A dense set

[twilit field]

wait

ooh

why is this wrtong

I think because -x>0 fo x<0?

[merry pewter]

why is a^x - x negative for small x

[twilit field]

I thought IVT at first

but then I realised a^x>0 and -x>0

But for $a<1$ this argument works for $x>0$

[grand pond]

A dense set

[twilit field]

and there is trivially a solution for $a=1$

[grand pond]

A dense set

[twilit field]

so in Q18, $A(\theta)$ is the area of the triangle - the area of the arc?

[grand pond]

A dense set

[twilit field]

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[vivid fjord]

how to solve y < 3x - 4
5x - 2y > or equal 7

[rain wasp]

one sec

[feral sedge]

so true

[carmine sigil]

Oh man

I need to use this

[lethal path]

LMAOOOOOOO

[vivid fjord]

how do without graphign

[lethal path]

it's actually relevant though!??

[rain wasp]

i get to use this once a month or so ehehe

[vivid fjord]

how do you do it without grapghing

[lethal path]

well y < 3x - 4 and y <= 5x/2 - 7/2

I guess try to imagine the graph lol

why would you not want to sketch it

[rain wasp]

i would say the graph would help you understand the solutions better

[vivid fjord]

ty

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[vivid fjord]

y^2 −4y−15=0

what are thefactor pairs

cant get them

[sudden yacht]

Use quadratic formula

[lethal path]

yeah it's not factorable

[vivid fjord]

my question says find (y-2)^2 given that x-3y=5 and 2x-2y=y^2-5

[lethal path]

,w x-3y=5 and 2x-2y=y^2-5

[grand pond]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by sooouth.

[lethal path]

indeed your quadratic equation is correct

there should be a smarter way though

y^2 - 4y + 4

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[strange ravine]

ahh im so bad with sequences

[strange ravine]

dont even know where to start 💀

[lethal path]

yeah so try using induction

the base case is $2 > 1$

[grand pond]

south's secret twin brother

[strange ravine]

So an>an+1?

[lethal path]

you want to prove that

[strange ravine]

I-

Huh

Oh wait yeah

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[zealous dirge]

what's the pattern here on the denominators? are the next few terms like these for the denominators as follows 2^9×5, 2^10×7, 2^12×7 et cetera?

[carmine sigil]

It looks like if numerator is odd = 2k + 1, the denominator = 2^(3k) * (2k-3)

If the numerator is even = 2k, the denominator is 2^(3k - 2) * (2k - 1)

@zealous dirge

[zealous dirge]

i see~ thank youu<3

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[cunning dragon]

<@&286206848099549185> pls help

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[thorny remnant]

Okay

[cunning dragon]

thnks

[cunning dragon]

but yess u know i hadnt give proofs but i had proofs

[cunning dragon]

m i correct?

[loud geyser]

what is even going on in this entire solution?

[cunning dragon]

assigning a finite value

[cunning dragon]

also by using ramanujans summation

[cunning dragon]

in this thing u have circled using series simplified

[loud geyser]

wait, you learn zeta function regularization in class 8?

[cunning dragon]

nope

@loud geyser wait pls take theta equals 45 degrees

i didnyt mentioned mistakenly

[cunning dragon]

i get

pls take theta equals 45 degrees

[cunning dragon]

actually trigonometry wasnt even studied

[cunning dragon]

i studied maths myself

[loud geyser]

the series always diverges

doesnt sum to a finite value

[cunning dragon]

u know ramanujans summation i used that

here

[cunning dragon]

i didnt use riemann zeta regularisation here

btw

[loud geyser]

does the question specifically tell you to solve with these methods? why are we overcomplicating the solution; this series diverges

[cunning dragon]

ok

i got it

[cunning dragon]

u know this q\a come to my mind

and i solve using these methods which i know

[cunning dragon]

btw u r correct it diverges as bcz of its

coefficient is increases

@loud geyser but i m trying to give it a finite value

u know where this series can be seen and using its solution ? @loud geyser

@thorny remnant did u checked bro

[limber patio]

Monkey

[cunning dragon]

who??

[limber patio]

Monkey

[cunning dragon]

i didnt understand

[limber patio]

Why tho?

[cunning dragon]

i solve this by using this technique ik

[limber patio]

U can do it without it

And it makes it too long

[cunning dragon]

yuppp riemann zeta regularisation??

[cunning dragon]

when i solve it i didnt think its very much long

[cunning dragon]

i dont know it as i dont studied it

[limber patio]

Why do you have zeta regulisation in 8 grade bro?

[cunning dragon]

bro we dont even studied trigonometric functions in ou school and i m in cbse boards trigonometry taught us in class 10th and i know this bcz of i had passion and wanted to know more about maths

[limber patio]

Ur in tenth or 8th.

?

[cunning dragon]

thats why i dont much abhout it as i am studying myself

[cunning dragon]

8th

[limber patio]

The fuv

U have cbse board?

[cunning dragon]

yupp

[limber patio]

??

How

Aren't they in 10 and 12mm

??

[cunning dragon]

they teach maths very slowly they covers topic which is best for average maths student but it bores me

[limber patio]

Hm

[cunning dragon]

i dont get it

[flint trellis]

woho self study i see

[cunning dragon]

ya

[limber patio]

How do you have cbse board? Aren't they in 10 and 12 classes

[flint trellis]

aiming for any competetive exam ?

[cunning dragon]

they had but i m in 8th grade and cant go to other grades

[cunning dragon]

i had just passion for maths

[limber patio]

Ok...

Anyway can't help you im in medical

💀

Holduo

Are you doing this from a book?

[cunning dragon]

i just want synopsis of elementary results in pure and applied mathematics but cant get as i get less marks

[limber patio]

Or maybe a reference sheet

[limber patio]

Ahhh

[cunning dragon]

r u indian

[limber patio]

Yuh

[cunning dragon]

naah

[cunning dragon]

i also

[flint trellis]

haha thought so

[cunning dragon]

nope even i have sometimes borrow books from my frnds who have his siblings in iitjee prep

[cunning dragon]

u r from pcm or pcb

[flint trellis]

pcm

jee prep

[cunning dragon]

pls can u check it

[flint trellis]

which one?

your doubt?

alr wait

[cunning dragon]

i had borrow trigonometry book of cengage and studying it

[flint trellis]

woahh thats gud

[cunning dragon]

all in that pdf just check that

[flint trellis]

1+3+5+7... infinity?

do summation n^2

[cunning dragon]

and i come to know that i had to get some patience for studying inverse trigonometry

[cunning dragon]

using that series which i had mentioned

[flint trellis]

hmm

ramanujan?

[cunning dragon]

take theta = 45 degrees

[cunning dragon]

yupp ramanujans summation i used here

[flint trellis]

woah bro that is not used till pre university

im not sure about it

idk about university

[cunning dragon]

ohhk.

[flint trellis]

tho

yeah

next?

which one?

[cunning dragon]

but u will get iit for sure

[cunning dragon]

u get in pdf

[flint trellis]

yea my goal is nit suratkal tho

[cunning dragon]

i had mentioned the concept of 0th dimension in last page and wrorks on pi has been given also

[cunning dragon]

jee mains u r prep

[flint trellis]

btw if you take 45 degree

wont you get AGP?

then solve that for infinite sum

[cunning dragon]

i dont know that

[flint trellis]

arithmetic geometric progression

[cunning dragon]

i told u that i m studying myself

[cunning dragon]

i have listened

[flint trellis]

you have to divide sum by 2 and subtract by initial sum while shifting the first term in sum/2 giving infinte sum Geometric progression use that to solve

[cunning dragon]

i dont know where to start and end as i have any book for studying

[flint trellis]

ahh

i suggest go chapterwise

like jee

syllabus

if 8th is boring

[cunning dragon]

how can i go i havent any book to prep and

[flint trellis]

oh use online pdf

ncert, cengage and stuff

i barely use books

[cunning dragon]

i get less marks in all subjects and thats why this is happeing i cant see

[cunning dragon]

i dont know but when studying in pdf i dont get what it says

[flint trellis]

watch online lectures

regarding that particular chapter

[cunning dragon]

but i m studying synopsis book through pdf

[flint trellis]

ohh

just maths?

physics chem?

[cunning dragon]

yupp i like that always

[flint trellis]

ohh

[cunning dragon]

i can barely see that bcz my mom only says to watch and study my own class

[flint trellis]

ah too bad

[cunning dragon]

even i m studying from vmc online as a online student but they are teaching sst and english today

and other i cant

[flint trellis]

ahh

well

[cunning dragon]

pls check works on pi also

[flint trellis]

highscool boards arent much useful btw

like my 10th percentage barely mattered

just fees concession

[flint trellis]

sure

[cunning dragon]

yupp but who can tell my family

[flint trellis]

haha real

[cunning dragon]

you know that moderation in 10th and 12th boards

[flint trellis]

pi=3.14 wasnt it?

[cunning dragon]

yess

[flint trellis]

whats pi=1.08?

is that product with the other summation term?

[cunning dragon]

where u see

[flint trellis]

in ur pdf

[cunning dragon]

this is multiplied with the other series i mean this is multiplicand

and giving 3.14

[flint trellis]

yea yea thought so

but how do yk that tho?

i doubt 10% in my class students might know that

[cunning dragon]

that series notaion i had used?

[flint trellis]

yea?

you didnt derive it right?

[cunning dragon]

when i see ramanujans magnificient value for pi i saw that there

[flint trellis]

or just used the result from google?

ah

i mean nvm im being dumb rn

lemme see

[cunning dragon]

see

[cunning dragon]

i didnt get what r u saying

[cunning dragon]

ooh

[flint trellis]

idk

about those results

used in infinite

contiuned fraction

also most of it

lol

[cunning dragon]

ohh

np

@flint trellis can u just tell me aboout differential and integral calculus and other types of calculus like multivariable calculus for me

i will grateful to u

[flint trellis]

wait I'll tell that in direct msg

[cunning dragon]

kk

[flint trellis]

umm add me

[cunning dragon]

accept

[cunning dragon]

have u checked bro

[cunning dragon]

.close

.close

.close

.close

[near raven]

bot is broken

[cunning dragon]

ooh what should i do?

[near raven]

dont panic

[cunning dragon]

kk

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[bright grail]

how to rewrite this equation $y(t) = 1.11 + 0.441 \ln(t)$ into a form like $y(t) = A(1 - e^{-at})$

[grand pond]

ahtrader

[west iron]

You don't

It can't be rewritten that way

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[last slate]

I integrated (ax+b)² by substitution and got (ax+b)³/3a
Which becomes [(ax)³ +3b²(ax) +3(ax)²b +b³]/3a

Then I integrated (ax+b)² by expansion and got a²x³/3 + b²x + abx²
Then I multiplied a/a with it to have a similar form like the substitution method then I have
[(ax)²+ 3b²(ax) + 3(ax)²b]/3a

These two are not equivalent b³ is missing from the expansion method can someone help me

[sage helm]

b^3 is just a constant

this is why the + C is so important

[last slate]

Aah

Thank you

[last slate]

Hey can I add b³ in it from myself to factor it into (ax+b)³

[sage helm]

yes

[last slate]

Thank you 🙏

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[hexed elbow]

hi

[hexed elbow]

how do we decide the limits

[lethal path]

find where they intersect

so that means a cos theta = 0

[hexed elbow]

ya

theta =pi/2

[lethal path]

or theta = pi/2

yeah and then you have to see which function is closer to the origin

[hexed elbow]

but like idek what the graph looks like

[lethal path]

so you can figure out which function is on 'top'

[hexed elbow]

😭

[lethal path]

but then sec theta is 1/(cos theta) and cos theta is positive in that region

so sec theta is also positive

ah I think they are assuming a > 0 then

https://www.desmos.com/calculator/r6unoapze5

wait

[hexed elbow]

ah so thats why a(sec theta+ cos theta) is upper limit?

[lethal path]

yeah when a > 0 at least, this is bigger than a sec theta

[hexed elbow]

okk

[lethal path]

ah it's cause -2 sec x = 2 cos(x + pi)

anyway it turns out that this is true for all a not just a > 0

[hexed elbow]

how do i get an idea of what the graph looks like

bcz the solution says its 2xintegral

but if i didnt know what it looked like i wouldnt know that its 2x

[lethal path]

it's hard to know the graph unless you plot points and join them up

[hexed elbow]

ah

[lethal path]

no just integrate normally

$\frac{1}{2} [a^2 (\sec^2 t + \cos^2 t) - a^2 \sec^2 t ]$

[grand pond]

south's secret twin brother

[hexed elbow]

then id just end up doing this

this integral

without 2*

how do i know to do the 2*

[lethal path]

oh wait right sorry yeah I see what you mean now

yeah so the asymptotes bit, basically you have to realise that both curves are even functions w.r.t theta

so yeah theta is not only pi/2 but it can be -pi/2

[hexed elbow]

o ok

[lethal path]

so we are integrating over an even interval, -pi/2 to pi/2 in the outer integral

and so halve it and multiply by 2

[hexed elbow]

ahhh okok

i get it

man

thanks

[lethal path]

no worries!

[hexed elbow]

:D

.close

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[twilit field]

So we first try to find the nullity of this linear map

$BA=A \implies B^kA^k =A^k \implies O=A^k$

now what

[grand pond]

A dense set(Ping when reply)

[twilit field]

We pre-multiply both sides by $B^{k-1}$, such that $B^k$ is $O$, so $O= AB^{k-1}$

[grand pond]

A dense set(Ping when reply)

[sage helm]

is that first implication right

[twilit field]

no

[twilit field]

which is why I switched to this

[strong lava]

It's B^(k-1) A because it's noncommutative

[twilit field]

Hmm

oops

no what

*now what

[strong lava]

You went down this path of taking powers

I'm not sure it's gonna lead you anywhere

[twilit field]

hmm

okay

[strong lava]

Or at least, not far enough

[strong lava]

Cause this = 0 is useful, it tells you im(A) subset Ker(B^(k-1)), but that's not {0} so you can hardly deduce A = 0

[strong lava]

You lost information about A because B is not invertible

[twilit field]

yeah

[strong lava]

Instead, consider the equation in A
M = BA-A = (B-I)A

Is there a solution for all M?

[twilit field]

Yes

[strong lava]

Why?

[twilit field]

hmm

nvm

I'm not sure

[strong lava]

Yeah I just rewrote the question, just in an elementwise, equationistic version

[hard umbra]

this is where you take a leap of faith and blithely use the geometric series formula

[strong lava]

That's not the simplest way

[twilit field]

$BA=A$, we know that $B=I$ satisfies this equation

[grand pond]

A dense set(Ping when reply)

[strong lava]

Consider L(A) = (B-I)A
We want it to be surjective, i.e. to be invertible
What would that entail?

[strong lava]

But B is fixed

[twilit field]

That We're mapping froma. higher dim to a lower dim

or that Im(T)={0}

oops

[strong lava]

input dim = output dim, which is why I used surj <=> inv

[twilit field]

wait

surjectivity implies inverses?

[strong lava]

I recommend reviewing your linalg basics if you doubt that

[twilit field]

We were never taught that

[hard umbra]

you know rank nullity

and you know injectivity <=> null space = 0

[twilit field]

Well, the issue is I don't know null space is 0

[hard umbra]

rank nullity implies that

[twilit field]

that's assuming it's surjective

[hard umbra]

which it must be

[strong lava]

Yeah we're making you prove under the assumption that input dim = output dim

[twilit field]

ah, assuming that it's easy

got it

thanks

gtg sleep

thanks

again

.close

[midnight plank]

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[midnight plank]

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[fallen aurora]

What should be the solution to this problem?
The limit doesn't exist or -pi/4?

[icy adder]

is that the greatest integer function?

[fallen aurora]

it seems like it is vidveshana abhichara😭
no idea what to accept

[fallen aurora]

yeah

or Floor ig

[icy adder]

then the answer should be limit does not exist

because when you approach from the left hand side of limit you get arctan(-1) which is -π/4

and from the right hand side you get π/2

[fallen aurora]

it is not pi/2

[icy adder]

umm why noy

am i missing something

[fallen aurora]

1/0 is not defined

[icy adder]

approaches infinity

[fallen aurora]

but it is in floor function
so it will be exact 0

[icy adder]

approaches infinity so arctan approaches π/2

if you say so then sure... but the limit still won't exist

[dreamy lichen]

the sided limits differ anyway

[fallen aurora]

erm

can you explain a bit please

[dreamy lichen]

The limit from LHS will be -pi/4, no?

[fallen aurora]

yeah

[icy adder]

it is simple

when you approach the limit from the left you get a different answer than from the right

[dreamy lichen]

and the limit from RHS doesnt exist

[dreamy lichen]

so the overall limit doesnt exist

[icy adder]

yes sir

[dreamy lichen]

,w lim x to 0 of arctan(1/floor(x))

[grand pond]

Results provided by WolframAlpha

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Requested by methisalwaysright

[dreamy lichen]

Yeah

[fallen aurora]

erm okay ig
my teacher told that it will be -pi/4
but wolfram alpha disagrees that's why I asked

[dreamy lichen]

I also disagree

maybe your teacher meant one-sided limit from left

[fallen aurora]

But I still can't wrap my head around that my teacher could be incorrect in maths😭

[fallen aurora]

but he was talking about the both

[dreamy lichen]

then he was wrong

even if there was no floor

[fallen aurora]

okay then
thank you so much

have a good day or night or whatever

.close

[dreamy lichen]

you too

[midnight plank]

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[shy knoll]

Just need help understanding the process to solve the questions

[sharp coral]

given that the figure is a parallelogram, what relationships can you draw between the angles?

[shy knoll]

opposite angles are congruent

[sharp coral]

yes. can you express that as equations?

[shy knoll]

for DEFC maybe /_D=/_F and/_E=/_C

[sharp coral]

yes, that works

is there anything else we know the angles must satisfy?

[shy knoll]

Im not sure do you mean like the DC = 180 degrees

[sharp coral]

yes, so how could we express that as an equation?

[shy knoll]

/_D + /_C = 180degrees?

[sharp coral]

yes

given that, can you solve for angle D?

[shy knoll]

Yea 110

[midnight plank]

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[full obsidian]

Please, help me

Two blocks of masses m1 = 200 g and m2 = 600 g lie on a smooth table, connected by a light inextensible thread. A horizontal force equal to 4 N is applied to the second block, directed horizontally to the right. a) With what acceleration will the bars move? b) What will be the tension in the thread? c) What will be the tension force of the thread if you apply a force to the first block and direct it to the left?

[full obsidian]

.close

[midnight plank]

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[daring lark]

I need to find the equations of the asymptotes of the graph of f(x). I wrote cosh() and sinh(x) in terms of the e^x version, but i don't know how to continue. I know I have to take the limit of x to infinity of the function to find a horizontal asymptote, but i also don't really know how to compute that

[midnight plank]

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@daring lark Has your question been resolved?

[peak thunder]

Hello 👋🏽 , could you show your work?

[midnight plank]

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[wraith bay]

Only using introduction/elimination rules?

[stone shoal]

yes

[midnight plank]

@stone shoal Has your question been resolved?

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@stone shoal Has your question been resolved?

[covert granite]

try demorgans on the right side first

and see that it is a contraposition

[midnight plank]

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[timid tangle]

Hello

[timid tangle]

Not sure how to start

[nova yoke]

probably you have some idea for (a) and (b)

[timid tangle]

Sooo apparently

Someone told me that

Domain is R^3

And co domain is R^2

How do I do c)

[lavish gazelle]

The problem has gave you the definition of T, just insert the numbers in.

[timid tangle]

In the right matrix?

How do I do d)

[nova yoke]

write it out a system of 2 linear equations, and solve the system

[timid tangle]

RREF?

[nova yoke]

sure

[timid tangle]

And the augmented side will be -3, 21?