help-0

[alpine sable]

It says those are hidden

Why do you have 1K+ pings

[scenic falcon]

uhmmm cus i dont check my messages

its from old groups i joined

[alpine sable]

Not muting the servers?

[scenic falcon]

yeah

but pls can you help i jus have one maths quesxtion

[wary stream]

Open your own channel

[scenic falcon]

hooow

brah my exam in 5 mins im at school

i just want someone to double check 😭

[wary stream]

Then open your own channel

[scenic falcon]

i cannot

i cannot even send message

can you make one for me

or send me link

can you pls send me link to unoccupied channel

omg i got it now

[lone heart]

@alpine sable Has your question been resolved?

[tacit arch]

@alpine sable you got raided. Please open a new channel with your question and progress

.close

[lone heart]

Channel closed

Closed by @tacit arch

Use .reopen if this was a mistake.

[alpine sable]

.reopen

[lone heart]

✅

[chrome flower]

'

[alpine sable]

@alpine sable heres your song for helping me

im not going to lie

its very hard lol

i cant really do it

only the first part

https://tenor.com/view/clap-gif-25211694

Marvelous

ill be using this channel still wont be closing yet

[tardy tapir]

keep going!👍

[tacit arch]

It's gonna be hard to find your question

When you open a new channel, your first message gets pinned

[alpine sable]

What is This question asking?

^ My Question Below ^

2 nd second (i.e., from time t= 1 s to t = 2 s)? What does this mean?

does it just mean 1-2s

[near apex]

Yes

[alpine sable]

weird way of asking it

[near apex]

During the second second, i.e. in the interval of 1-2 second.

[near apex]

Not really. Made total sense to me.

[alpine sable]

doesnt the first two mean the same thing

first two seconds

[near apex]

No.

See. Object travelled some distance in first second. Let's call it d.
Then, it travelled some more in the second after that. Let's call it e.

They have given d+e.

They have asked you for e.

[alpine sable]

so my time would be 1

[alpine sable]

and my total distance would be half of 19.6?

hmm

im gonna go ask someone i know in real life

if i have no luck ill ask again

.close

[lone heart]

Channel closed

Closed by @wintry garnet

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[lone heart]

Please don't occupy multiple help channels.

[low holly]

Can someone help with this on asap please

I’m on a deadline just need this last problem

<@&286206848099549185>

[lone heart]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[lone heart]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[low holly]

Ok

[tacit arch]

Closed your other channel

[low holly]

Help 🥲🥲🥲

It’s just 1 problem 😭

[tardy tapir]

use difference of squares in the numerator

[low holly]

I forgot how to do that

How do I do that

[ocean seal]

Dyssrupt

[low holly]

But there’s a coefficient?

[tardy tapir]

i meant the question

not what you did

[low holly]

Oh ok

[low holly]

Do I plug in A for x now?

[tardy tapir]

its wrong

[low holly]

Bruh

[ocean seal]

Dyssrupt

[low holly]

There’s no 2x^2

It’s -4x^2

[tardy tapir]

whats (2x)^2

[low holly]

O

I’m dumb

Wait

But it’s not negative

Yours

[tardy tapir]

bruh

[tardy tapir]

you see that - in between

[low holly]

Yah

Ok

[low holly]

4x

So you got rid of the square?

[tardy tapir]

try again

[low holly]

Oh 4x^2

Can I plug in A now

[tardy tapir]

here a is (x+a) and b is 2x

[low holly]

What?

[tardy tapir]

im literally spoon feeding you now

[low holly]

I did that

I put

(X+a) (x-a) -2x^2

[tardy tapir]

no

(x+a)^2 is x^2 + 2ax + a^2

[low holly]

Ohh

[tardy tapir]

not (x-a)(x+a)

[tardy tapir]

now use this

[low holly]

Better?

[tardy tapir]

not ^2

[low holly]

Oh yeah your right

Fixed

[tardy tapir]

now simplify the paranthesis

[low holly]

I can plug in a now rigjt just making sure lol

[tardy tapir]

no

first you need to remove the denominator

[low holly]

Multiple by the inverse of x-a?

*multiply

[tardy tapir]

one of the paranthesis of numerator will cancel the denominator

[low holly]

X-a/1?

[low holly]

Oh

[tardy tapir]

try simplifying the terms inside the paranthesis of numerator

[low holly]

I can cancel out x -a right cause -x+a is the same thing?

[tardy tapir]

first factor out a -1

[low holly]

Can I cancel the 1?

[thorn monolith]

Is everything on numerator factorized? Regardless of its or not, text at bottom is wrong

I cant tell if its $\frac{(3x+a).(x-a)}{x-a}$
OR $\frac{(3x+a) + (x-a)}{x-a} $

[ocean seal]

Cyrenux

[lone heart]

@low holly Has your question been resolved?

[low holly]

First one I think

[low holly]

The first one they’re being multiplied

[thorn monolith]

After that you turned multiplication somehow

On the numerator

Numerator should have been (3x+a).1

Remember $\frac{a}{a} = 1$ , for $a \neq 0$

[ocean seal]

Cyrenux

[low holly]

I didn’t make it 0?

[low holly]

Times 1?

[thorn monolith]

And you can split $\frac{ba}{a} = b \frac{
a}{a}$

[low holly]

I factored out the -1

[ocean seal]

Cyrenux

[low holly]

3x+a-1 is that correct

[thorn monolith]

I didnt say you made it 0, im showing the a/a works for non zero a,since 0/0 is uncertain

[low holly]

Oh ok

[thorn monolith]

Already told you its not

[thorn monolith]

Split it like this

[low holly]

It’s so shoukd it be 3 (x+a)/1

[thorn monolith]

Then use this

You are confusing that with
$$\frac{b+a}{c} = \frac{b}{c} + \frac{a}{c}$$
\ ( also suppose $c \neq 0$ so numerator is defined)

[thorn monolith]

But yeah this is correct

[low holly]

What do I do after

[ocean seal]

Cyrenux

[thorn monolith]

Go back to your original question, what is it asking you to find?

[low holly]

The limit as x goes to A

[thorn monolith]

How did you obtain this?

[thorn monolith]

Do you remember why we didnt plug a as x before?

If you do, you should see the difference now

[low holly]

I’m gonna give up

This is impossible

[thorn monolith]

At the end

After all that work

[low holly]

What else

Am I supposed to do

I’m stuck

This is the last probelm I need

It’s due in like half an hour

[thorn monolith]

I can only help if you answer

So you understand why

[low holly]

I can’t click your original reply

[low holly]

Because we hadn’t fully factored out the function?

[thorn monolith]

Beginning of question where it wants you to take the limit

Close, our original equation gives 0/0 when you plug in a in place of x

[low holly]

Oh yeah

So it’s indeterminant or whatever

[thorn monolith]

To fix that we factored 0s out of numerator and denominator, (x-a) was the 0 in our case

[thorn monolith]

Yes

[low holly]

Oh ok

[thorn monolith]

Now by cancelling (x-a) you got rid of the 0s

You can plug a in place of x

Do you see how i knew (x-a) is a zero though?

[low holly]

Yeah because I’m the beginning if we plugged it in a -a=0

Right?

[thorn monolith]

Also you should type $$lim _{x \longrightarrow a}$$ every time you put a new equal symbol

[low holly]

Oh ok

[thorn monolith]

Oops wait

[ocean seal]

Cyrenux

[thorn monolith]

There we go @low holly

[low holly]

So is that the final answer?

[thorn monolith]

Let me see if i can show it better $\lim_{x \longrightarrow a}$

[ocean seal]

Cyrenux

[thorn monolith]

Ok this is a bit better

[low holly]

Ok

[thorn monolith]

Uhh where is the minus 1 from

[low holly]

Remmeber there was a minus 1 before

[thorn monolith]

Also you said numerator was multiplied when i asked the first time, why is there addition now

[low holly]

I went from that to the previous picture

[low holly]

Ohhh so it’s still being multiplied

Oh I got confused

I forgot to put parentheses

[thorn monolith]

You also forgot to put multiplication symbol so it looks whack

[low holly]

Yeah my bad

[thorn monolith]

And eventually you made a mistake

[low holly]

Is that good now

[thorn monolith]

If you are going to send this to someone though rewrite the whole progress,putting the limit symbol on EVERY equation

[low holly]

Yeah ima start next weeks homework early i procastinated that’s why my stuff kinda messy my bad

[thorn monolith]

Otherwise you cant cancel 0s

[low holly]

Ok

[thorn monolith]

Yeah its true now, but dont type limit anymore when you already typed a in place of x

[low holly]

Oh ok

[thorn monolith]

There are already no terms in x

So its unnecessary

[low holly]

So when I plug stuff in no need for lim

[thorn monolith]

Yeah, as there will be no more of that variable after

In our case x is absent after we plug a

Putting limit is unnecessary

[low holly]

Thank you

A lot

Sorry if I was dumb

[thorn monolith]

Its about experience not intelligence, good luck on next studies

[lone heart]

@low holly Has your question been resolved?

Channel closed

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[safe tartan]

How would you implicit xy? Wrt x

[safe tartan]

I’m really only familiar with like y^2 or something where y is independent where it becomes dy/dx y’

[wheat isle]

product rule

[safe tartan]

xy’+y

[wheat isle]

And what is y’

You still haven’t differentiated it yet

[safe tartan]

Not sure why you mean

[wheat isle]

What is the derivative of y with respect to x

[safe tartan]

dy/dx

[wheat isle]

So what’s the final answer

[safe tartan]

dy/dx= -y/x?

[wheat isle]

Well I can’t say because you haven’t told me what you’re trying to solve

You just asked implicit derivative for xy

[safe tartan]

xy-2=0

[wheat isle]

it should equal something

[safe tartan]

Find dy/dx

But using implicit

Could you write this in terms of the cancelation of dy and dx things

[wheat isle]

Yes

[safe tartan]

Like d/dx f(y) = d/dy (f(y)) x dy/dx

How would this case work

Because it isn’t really fully f(y)

[wheat isle]

Do you have a specific example I can see?

[safe tartan]

I think that’s what implicit differentiation is defined as

For example

d/dx (y^2) = dy/dx times d(y^2)/dy

Since the fractions cancel (not really but it works)

[wheat isle]

All implicit is, is differentiating y like you would x but slapping on a dy/dx onto it

And then solving for dy/dx

[safe tartan]

So would we treat xy as product of two functions of x

So xy’+yx’

And then multiply by dy/dx ?

[wheat isle]

Well y is it’s own thing because y can be anything

You only multiply a y term by dy/dx if you are differentiating it with respect to x

Otherwise you leave it alone

[safe tartan]

So for xy-2=0 and I’m tryna find dy/dx

So d/dx (xy) = 0 and then I don’t understand really how product rule can be used here

[wheat isle]

xy = 2

x*y = 2

by differentiating both sides with respect to x you get

[x*y]’ = 0

There’s no other way you can differentiate this without the product rule since it’s a product of functions

[safe tartan]

And then what does lhs becomes

[wheat isle]

dy/dx = uv’ + u’v

x*dy/dx + y = 0

do you see how to get the derivative from here?

[safe tartan]

Yes rearrange

So you can just use product rule regardless of if there are multiple different letters

[wheat isle]

You use product rule when a function is multiplied together

Tell me what rule you would use to differentiate this

(x^2 + 5x - 7)(sqrt(6)-7x)

[safe tartan]

Well yeah product rule but I’m confused as there are multiple variables

But i guess since we assume y is a function of x then it works

[wheat isle]

What, for xy?

[safe tartan]

Yes

[wheat isle]

Well we aren’t really assuming we are treating it as a function of the other

[safe tartan]

I see

.close

[lone heart]

Channel closed

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[novel raft]

i need help finding the equation of a circle on a coordinate plane that passes through 2 given coordinates on the plane. the two coordinates are basically all im working off

[novel raft]

for my maths task im doing a psmt and the task is to model the outline of a lake on a coordinate plane and i need to show how i got each shape, those being parabolas, rectangular hyperbolas, circles and straight lines.

i need to show algebreically how i calculated each formula of these shapes

its all on desmos

i have already calculated the perfect shape using trial and error, but i havent been able to show mathematically how i came up with the formula. i showed how i came up with restrictions using substitution

[nimble fern]

is there a picture for more info for this problem

you did explained a lot, whcih is nice
but without the original question, it's hard to see what's happening

[lone heart]

@novel raft Has your question been resolved?

[novel raft]

i have found a solution using 3 simultaneous equations

[ripe zodiac]

hi ahston lmao

[novel raft]

hi do ik you

[ripe zodiac]

yes im bence

[novel raft]

HAHAHAHA

[ripe zodiac]

i joined this server as well yesterday bc i needed help w my psmt

[novel raft]

are you here bc u need help too

LMFAO

small world

[ripe zodiac]

yea

i clicked the name bc it was so random and then i saw the lake and i thought it looked familar

[novel raft]

💀

[vapid shuttle]

omg bence and ashton?

whats up

[novel raft]

hi austin hows your lake coming

[vapid shuttle]

it is coming along

[novel raft]

nice

[vapid shuttle]

honestly I'm stuck on the same part as you

this is funny

I will just watch your channel then along with bence

[ripe zodiac]

do u go to our school as well??

[ripe zodiac]

im have to go do my psmt now bc im behind cya

[vapid shuttle]

yeah I'm also not doing great on the psmt lol

cya

[ripe zodiac]

ok

[lone heart]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[echo socket]

Could you show your work?

[solid fog]

need help with this again.

yeah sure hold on

lemme take a picture

hm for crossing out the h im sure i did it wrong.. do i recipricate it?

[echo socket]

Yeah that's not how terms should be cancelled

Also you probably meant to multiply by 1/h rather than divide by 1/h

[solid fog]

its been a while since ive done math.. lol

[echo socket]

So the last line should be $\frac{-\cancel h}{3h + 27}\cdot\frac1{\cancel h}$

[ocean seal]

A Lonely Bean

[echo socket]

Leaving you with $-\frac1{3(h + 9)}$

[ocean seal]

A Lonely Bean

[solid fog]

ah yes

yes

thank you so much

[lone heart]

@solid fog Has your question been resolved?

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[high sapphire]

How to find Dearrangement of 6 objects when 2 are identical?

The question originally was

Find the possible distinguishable arrangements of BHARAT, such that no letter pertains to it's original position(Consider the two A identical)

[wanton pebble]

i have seen this question 3 times here...

divide by 2

[lone heart]

@high sapphire Has your question been resolved?

[lone heart]

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[tacit pine]

I am trying to understand how to solve this question. (Studying for math exam), and the sum is supposed to be 48, but idk how to go about it..

[rocky grove]

Hmm

You evaluate from the inside then you go outside

[tacit pine]

I first solve the (2m-1) then when I have that number I solve for k?

[rocky grove]

Essentially yeah

[tacit pine]

(21-1) + (22-1) * (3 + 4 + 5)?

[rocky grove]

Yeah

4 * 12

[tacit pine]

I see

[rocky grove]

Don't forget parenthesis

[tepid drum]

you can do this by just substituting m=1 and add to that at m=2 and after that o the same process with k fr the second sum and you are done

or you can split the sum inside and use some facts then evaluate

[tacit pine]

Thnx

[rocky grove]

[tepid drum]

so for the first way you get $\sum_{k=3}^{5}{(2(1)-1)k+(2(2)-1)k}=\sum_{k=3}^{5}{k+3k}=\sum_{k=3}^{5}{4k}=4(3)+4(4)+4(5)=48$

[ocean seal]

calculus is fun

[tacit pine]

jeez

[tepid drum]

using second method you get $\sum_{k=3}^{5}{\sum_{m=1}^{2}{2m}}-\sum_{k=3}^{5}{\sum_{m=1}^{2}{1}}$

[tacit pine]

.close

[lone heart]

Channel closed

Closed by @tacit pine

Use .reopen if this was a mistake.

[ocean seal]

calculus is fun

[tepid drum]

now use facts like $\sum_{n=1}^{k}{c}=ck$

[lone heart]

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[ocean seal]

calculus is fun

[ruby sinew]

How do you add non perfect squareroots that cant be simplified but of the same variables? Such as (sqrt23 + sqrt23)

[tiny panther]

How to solve Y over 4 + Y over 2 which equals to 9

[ruby sinew]

I got a solution for ya

[tiny panther]

okay

?

[ruby sinew]

Wait give me a sec

[tiny panther]

okay

[ruby sinew]

Times both sides by 4 I think

[zealous lichen]

$\sqrt{x}+\sqrt{x}=2\sqrt{x}$

[ocean seal]

WhereWolf

[ruby sinew]

Thanks

[zealous lichen]

no problem

[ruby sinew]

I found your solution I think

[tiny panther]

Ok

[ruby sinew]

Times both sides by 5

Then you will end up with y + 4y = 36

Since y + 4y are like Variables you can add them into 5y

5y = 36

[tiny panther]

Ohhh

Alr

[ruby sinew]

Divide both sides by 5 then there you have it

[tiny panther]

Okay

Thank you

[ruby sinew]

y + 4y are no differemt it just have a 4 on it

.close

[lone heart]

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[unborn dock]

i need a partner to do the feynmen's technique

[unborn dock]

anyone willing to hear me explain a projectile

[lone heart]

@unborn dock Has your question been resolved?

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[safe tartan]

Show that the normal to the circle $x^2+y^2=a^2$ at any point $(x1, y1)$ on it passes through the origin

[ocean seal]

annyeong

[safe tartan]

so i have found dy/dx = -x/y

and using point gradient formula,

[wind cloak]

it might be better to use parametric coordinates here actually

but sure

[safe tartan]

um

[wind cloak]

so what's the slope of the normal

[safe tartan]

x/y

[wind cloak]

[safe tartan]

so my problem is

umm is that wrong

[wind cloak]

yes

[safe tartan]

fuck

whats the normal again

wait y/x ?

[wind cloak]

It's perpendicular to the tangent

yes

[safe tartan]

ahhhhhhhhhhhhhhhhhhhhhh

ok ill just redo the other questions

[wind cloak]

lol

F

[safe tartan]

but for this one, does showing that when you sub x=0 and y=0 into equation, the equation remains true

is that enough to prove that

[wind cloak]

So you have th slope and a point, can you find the equation of a line

[safe tartan]

yep

so y-y1=y1(x-x1)/x1

[wind cloak]

The slope is y1/x1

[safe tartan]

oh yeah since u sub it in

[wind cloak]

yep

So now just show 0,0 satisfies

[safe tartan]

ah

so simply showing that this equation passes through 0,0 is enough

hmm yeah that makes sense

tyty

.close

[lone heart]

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[celest spire]

Guys help needed with finding the argument of (tan1-i)^2

[celest spire]

Since it will lie in 4th quadrant

And for the 4th quadrant the Principle value of argument = 2pi - theta -2pi

The theta came out to be -2

(- 2 radians )to be precise

Plz ping me

<@&286206848099549185>

Where are you helpers ,😭

[lone heart]

@celest spire Has your question been resolved?

[alpine sable]

<@&286206848099549185>

[lone heart]

@celest spire Has your question been resolved?

[lone heart]

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[terse turtle]

I need help with 12-15 and 17

[zealous lichen]

!show

[lone heart]

Show your work, and if possible, explain where you are stuck.

[terse turtle]

I literally have no clue what to do

the first two I got from google

[zealous lichen]

a useful thing to do is to set answer to x

[terse turtle]

ok

[somber owl]

12 and 13 needs complex

[terse turtle]

huh

[zealous lichen]

$\log_3{-10}=x$

[somber owl]

Complex number

[ocean seal]

WhereWolf

[zealous lichen]

simplify the expression

[terse turtle]

3^x=-10

[zealous lichen]

find the x value if you can

[terse turtle]

how

[somber owl]

Have u learn about complex number?

[zealous lichen]

you don't know how to deal with exponent?

[terse turtle]

Yeah

I don’t know how to find it

[shy dove]

Guys, they might not have gotten to complex numbers, and the teacher just wants them to say that it is not a real number / no answer

[somber owl]

Oh

Ok

[terse turtle]

no they don’t want that

i would get a zero

[somber owl]

Wtf

[shy dove]

Then, have you seen complex numbers?

[terse turtle]

wdym by complex number, give an example

[shy dove]

2 + 3i

[somber eagle]

If you don’t know about complex numbers, then the teacher wouldnt expect you to solve that question

[terse turtle]

Been a few months

[shy dove]

i = sqrt(-1), rings a bell?

[terse turtle]

nope

oh wait yes

[shy dove]

then you can't be expected to solve that problem

oh?

[terse turtle]

I know that I=-1

teacher hasn’t gone over it in this semester tho

[zealous lichen]

judging from other

questions

I'd say the teacher expect the answer to be no solution

[terse turtle]

to all 6 of these?

[zealous lichen]

no

just 12 and 13

[somber eagle]

14 and 15 can be done normally

[terse turtle]

you can have negative exponents tho

[zealous lichen]

yes

but not negative logarithm

[terse turtle]

My teacher has said otherwise…

[somber eagle]

Yeah but no real number exponent can turn 3 into -10

[terse turtle]

true

[zealous lichen]

maybe your teacher meant negative base for log?

[terse turtle]

i guess

Idk

she could have just made a mistake

Ok i put no solution for 12 and 13

What about the other 2

[somber eagle]

14: e^x = e^3

Solve for x

[terse turtle]

x=3?

[somber eagle]

Yeah

[terse turtle]

so I write 3 as the answer?

[somber eagle]

Since ln just means log based e

Yeah

[terse turtle]

what about the natural log tho

[zealous lichen]

btw you should memorize $log(a^b)=blog(a)$

[terse turtle]

ok

[somber eagle]

Natural log is log based e

[ocean seal]

WhereWolf

[terse turtle]

ohhh

THATS rights

[terse turtle]

i honestly just forgot

ok so then 15

log 150

So the base is 10?

10^x=150?

[zealous lichen]

how though lol

[terse turtle]

nvm then

mb

[zealous lichen]

no

[terse turtle]

I mixed up logx and this

[high sapphire]

That is not the answer

[zealous lichen]

I mean there's no exact solution

[terse turtle]

I know

Fractions exist

[high sapphire]

Sorry for interfering..
My ques was timed out

[terse turtle]

nw

[zealous lichen]

6!/2

[terse turtle]

whats !

[zealous lichen]

It's factorial

[terse turtle]

huh

[zealous lichen]

I was answering to Dragon King

but they got they're own channel now

[terse turtle]

oh

[somber eagle]

Are you allowed a calculator?

[terse turtle]

yep

[somber eagle]

Maybe they want decimal form?

Idk

[terse turtle]

That can be used

[zealous lichen]

wait calculator is allowed???

[somber eagle]

Because the only exact form of the answer is log(150) but that is useless

[terse turtle]

yeah

no one has taught me how to do this on a calculator tho

[zealous lichen]

what's the point of the exercise then practice using the calculator??

[terse turtle]

to learn how to do the math

i mean you still have to do work when you use a calculator-

You there? @zealous lichen

[somber owl]

Wtf?

[zealous lichen]

yo

[zealous lichen]

what's the point of entering everything to the calculator

[somber eagle]

Btw I know this isn’t the current problem, but I don’t think your answer to 16 is correct

[terse turtle]

ok then what is it

i used a compound interest calculator

[somber eagle]

Well if you do 5000 * 1.06^14.2 it gives you well over 10000

[terse turtle]

oh?

[lone heart]

@terse turtle Has your question been resolved?

[lone heart]

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[alpine sable]

The rungs on a ladder are 25 cm apart. The length of the rungs decreases uniformly from 45 cm at the bottom to 25 cm at the top. What is the length of wood needed for the rungs if the top and bottom rungs are 2.5 m apart?

[zealous lichen]

I think this can be reduced to a arithmetic sequence

you can start by finding out how many rungs are there

[alpine sable]

yea im stuck there

[zealous lichen]

the whole ladder is 250 cm long

[alpine sable]

why is total number of rungs (n) equal to total distance between top and bottom rungs/distance between two successive rungs? and why is one added to that?

[zealous lichen]

let these be trees or something

| | |

here is 3 trees

and there are 2 spaces between them

[alpine sable]

yes

[zealous lichen]

so the amount of trees = space+1

[alpine sable]

ohh

[zealous lichen]

or you can think like this

when you have 1 tree there's no spaces

|

when you add a tree there will be an extra space

| |

| | |

[alpine sable]

got it

but why is it total distance/succesive distance of rungs tho

my basics are rusty 😭

[zealous lichen]

the vertical length of a space is 25 cm

and they add up to 250 cm

so there will be 10 spaces

[alpine sable]

OH

[rough smelt]

damn

[alpine sable]

tysm omg got it

[alpine sable]

holy shiy??

[rough smelt]

I need help with maths

[alpine sable]

.close

[rough smelt]

but learning it

[lone heart]

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[zealous lichen]

#❓how-to-get-help

[lone heart]

Available help channel!

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[crimson tundra]

How is is z 0.8?

[tardy stag]

bc there's 0.2 to the right of it ig

and 0.5 to the left (it's symmetric) so 0.5 + 0.3 = 0.8

[lone heart]

@crimson tundra Has your question been resolved?

[lone heart]

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[heavy copper]

#help-0

[heavy copper]

This is a PYQ . Not able to solve this!!!

[wary stream]

One, you opened a channel for 10 minutes, you don't need to close and open a new one because no one answered you

[heavy copper]

sorry

[lone heart]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[heavy copper]

1. I don't know where to begin

[dry shard]

if |z|=1 then the conjugat of z is 1/z

[heavy copper]

next

My answer is not in the option

[dry shard]

then tell us the work you have done

and how you got there

[lone heart]

Show your work, and if possible, explain where you are stuck.

[oak perch]

Use 1=(z_1+z_2+z_3)((z_1) bar+(z_2) bar+ (z_3) bar)

[heavy copper]

oh then next

done

.close

[lone heart]

Channel closed

Closed by @heavy copper

Use .reopen if this was a mistake.

[oak perch]

Next time close it after I answer it

@heavy copper

[lone heart]

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[barren portal]

,rotate

[barren portal]

,rotate

[barren portal]

Here, I have left inverse and left identity

trying to show uniquess of left identity

[barren portal]

but I reached a point where i can only cancel if right inverse is there

I understand having left inverse and left identity imply that it has right inverse and right identity

I was wondering if it’s possible to show uniquess of left identity just using left cancellation property

,rotate

[barren portal]

Trying to see, if i can only use this to prove uniquess of left identity

[oak perch]

Left inverse and left identity can give you the whole group axioms

Then the left identity becomes identity therefore unique

[barren portal]

Suppose in some other algebraic structure, we only have left inverse and left identity

how do we prove the uniqueness of left identity?

ie if e*a=e’*a, then e=e’

[barren portal]

I agree with this, how do we proceed to do further to show e=e’

[oak perch]

I am forming a proof

[barren portal]

essentially asking, can we just multiply in left side to show

[barren portal]

sure

[oak perch]

Wait a min

Only when left identity is unique we can define Left inverse.
So you need to use only the existence of left identities to prove the uniqueness of left identity?

Which is not true or did you mean

Existence of left inverse of any left identity

Any left identity , there exists left inverse in terms of it?

[barren portal]

What do we need to show uniquess of left inverse?

[oak perch]

I mean your condition, existence of left inverse

Did you mean

Existence of left inverse for any left identity

[barren portal]

yes, when I take the inverse of any element left identity will come

suppose e is left identity, b is left inverse, then b*a= e

[oak perch]

for left identity e, any a there exists b such that ba=e
For another left identity e’, any a there exists c such that ca=e’

Left inverse exists for each left identity, is this what you meant?

[barren portal]

Yes, if it’s not unique, then left inverses will be different

but we have to show they are same, ig

[oak perch]

No I am saying the condition, the beginning of your question

“We have such a set G, such that left inverse and left identity exist”

When you say this

“Left inverse exist”. Did you mean left inverse exist for each left identity?

[barren portal]

yes

[oak perch]

Okay

[barren portal]

if e1 is an identity, b1 is inverse for a simarly, if e2 is inverse b2 is inverse of a

[oak perch]

Okay

[barren portal]

if identities are separate, then sperate inverse exists but operating will work accordingly

[oak perch]

And do you require left inverse be unique: a left identity e, any x, there exists a unique y such that yx=e?

Or exists a y

[barren portal]

Let me share a set of alternate axioms for group I’m working with

my goal is essentially to use them

please wait

please check if those are correct to start with.

[oak perch]

Sure

[oak perch]

You see iii has to come after iv

[barren portal]

mb

It is

[oak perch]

So we know that the e in iii is one chosen e

[barren portal]

It’s just a left inverse

[oak perch]

Still I don’t think it’s rigorous at all

What is e in iii

[barren portal]

the reason why i wrote it is

[oak perch]

So your question is not clear, you know what I mean? Your iii is not well-defined

[barren portal]

I came accross that, if something is closed and associative, has left inverse and left inverse it implies that it’s left inverse and right identity is same

[oak perch]

Just look at your iii

[barren portal]

I mean, every element has a left inverse

[oak perch]

what is that e? A chosen left identity? Or for any e we have a corresponding iii?

[barren portal]

e is a left identity

[oak perch]

This sentence is not well-defined?

A chose left identity?

So iii is a statement depending on e

[barren portal]

e is the left identity and every element has an inverse b such that b*a= e (left identity)

[oak perch]

You can view it as a statement iii(e): any a, there exists b such that ba=e

[barren portal]

is this good?

[oak perch]

So is one iii(e) true or iii(e) is true for any e satisfying iv

So the former

You have iii(e) for one e satisfying iv

[barren portal]

Do you mean is the identity is unique for different inverses?

[oak perch]

No I say it again:

iii this axiom has an e in it, it depends on e

So instead of iii, I better use iii(e) to denote it

Axiom iii(e) : any a in G, there exists b in G such that ba=e

Now

The question is

Do you give me iii(e) for any e satisfying iv?

Or you give me one iii(e), for one e satisfying iv?

[barren portal]

just to make sure by iv) you mean there exist e in G (e*a=a)?

[oak perch]

iv: there exist element e such that any a in G, ea=a

So iv: {e: any a in G, ea=a} is not empty

This is a well-defined statement

iii is not

iii(e) is

So question is your iii is iii(e) for some e? Or it’s Λiii(e) for all e defined in iv?

You see the incompleteness of logic going on here?

[barren portal]

I don’t want to confuse, does this question arises here also?

[oak perch]

Yes

This author is not rigorous, I have to say that

Axiom 3 stands for one axiom 3(e) being true for one e satisfying axiom 4? Or axiom 3 stands for axioms 3(e) being true for any e satisfying axiom 4? The author didn’t make it clear at all

I suggest you to go to your prof using this textbook , for clearance, which case it stands for

[barren portal]

Can i link the pdf

[oak perch]

As I said

[barren portal]

It’s proving certain things

[oak perch]

The pdf itself didn’t make it rigorous

Have to ask the author or people using this textbook, which case they think it is

[barren portal]

They are trying to say any finite semi group with left and right cancellation is a group

[oak perch]

I know , you already told me

Which is not making sense

Whatever, how about we prove this when it’s the former case ? 3 stands for one 3(e) being true. If not work we prove when it’s the latter case, 3 stands for 3(e) being true for any e

[barren portal]

sure

[barren portal]

we have for any left identity corresponding left inverse

[oak perch]

You didn’t understand me

Anyway I am proving the former case

[barren portal]

Sorry, i was trying to

[oak perch]

You can read this again. You didn’t understand this

Done: I have proven it’s a group when 3 stands for 3(e) for one left identity e:

Now any a in G, there exists b such that ba=e. Now there exists c such that cb=e.
Therefore (ab)=e(ab)=(cb)(ab)=c(ba)b=ceb=cb=e.
Meaning when ba=e, we also have ab=e

Then ae=a(ba)=(ab)a=a
Now for any other left identity e’, plug in a=e’, e’e=e’
Since e’ is a left identity we also have e’e=e, therefore e=e’
And by the discussion above left inverse is also a right inverse
QED

Since it’s a group when 3 stands for 3(e) for one e, of course it’s a group for stronger condition 3 stands for 3(e) for any e. I guess that’s the reason the author didn’t make it clear. Turns out it works for either case. Never the less he should have

made it rigorous

[barren portal]

Thank you

[oak perch]

Np

[barren portal]

I’m reading multiple times to get it correctly

[oak perch]

Okay

[barren portal]

Appreciate your effort, have a great day, sorry for taking this long

[oak perch]

No it takes long because of the author, not you

But no problem

[barren portal]

Have a great day 🙂

[oak perch]

You too

[barren portal]

.close

[lone heart]

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[random slate]

why does this work?

[zealous lichen]

if x and y are solution of ax+by=c

try to simplify a(x+b)+b(y-a)

[random slate]

oh

right

i get it

thanks

.close

[lone heart]

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[zealous lichen]

no problem

[lone heart]

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[runic hull]

notice that they also found theta and phi. They substituted |z| and theta back into the first equation for z (in polar coordiantes) at the top

[ocean seal]

guinearW

[runic hull]

$$z=\sqrt{18}e^{(i\frac{-\pi}{4})}$$

[ocean seal]

LayneTheAndroid

[runic hull]

Do you know how to convert from polar form back to the form a+bi?

that's basically all you need to do from that point

It's just

$$e^{ix}=cos(x)+isin(x)$$

[ocean seal]

LayneTheAndroid

[runic hull]

it's okay. it'll become second nature with enough practice

yes! there's no reason it wouldn't work for w as well. the formula is still valid

[lone heart]

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[balmy path]

Hi , i need some help

[sharp thorn]

is that sin 200

and cos 290

[balmy path]

Yeah...

[sharp thorn]

can we write sin 200 as sin pi plus 20?

[balmy path]

Umm... yeah
Next step?

[sharp thorn]

so its just sin 20 right?

[balmy path]

Right

[sharp thorn]

now can we write cos 290 as cos 3pi/2 plus 20 right/

?

[balmy path]

Yeah we can

[sharp thorn]

which is sin 20 again/>

?

[balmy path]

Correct

[sharp thorn]

so your numerator is $2(-\sin(20)) + \sin(20)$ ?

[ocean seal]

Bettim

[balmy path]

Yes

[sharp thorn]

which is just -sin20

right?

[balmy path]

Yeah

[sharp thorn]

do the same for the denominator

[balmy path]

Gimme a sec please...

I reached to $\sin(-20) + \2cos(-20)$

[ocean seal]

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

[balmy path]

Uhhh ...

Let me write... T_T

[balmy path]

Is that correct?

Sorry for pinging

[sharp thorn]

sin 160 is positive

so it is

-sin 20 / 3 sin 20

right?

[balmy path]

Oh i forgot, right

So the answer is -1/3 ?

[sharp thorn]

yep seems right

[balmy path]

Thank you sir

.close

[lone heart]

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[clever lodge]

i need help in this matrix related question
this question is from my college graph theory course and matrix multiplication section

[clever lodge]

this is my approach for alpha and beta

3 question i have no idea how it should be solved

[lone heart]

@clever lodge Has your question been resolved?

[clever lodge]

<@&286206848099549185>

[clever lodge]

.close

[lone heart]

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[mellow frigate]

is this right

[lone heart]

Please don't occupy multiple help channels.

[zealous lichen]

yes that's correct

[lone heart]

@mellow frigate Has your question been resolved?

[lone heart]

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[rapid vector]

Hello there. Uncertainty Question. Finding percentage uncertainty.

[rapid vector]

$(2\pm0.7)(7\pm0.2)$

[ocean seal]

HqppyFeet

[rapid vector]

So this is what I did:

using $\frac{\Delta x}{x}=\frac{\Delta a}{a}+\frac{\Delta b}{b}$

[ocean seal]

HqppyFeet

[rapid vector]

$\frac{\Delta x}{x}=\frac{0.7}{2}+\frac{0.2}{7}=\frac{5.3}{14} \times 100% \approx 37.8%$

[vale wigeon]

,calc 0.7/2 + 0.2/7

[ocean seal]

Result:

0.37857142857143

[vale wigeon]

yeah ok checks out

37.8% yes

[rapid vector]

welp thats very handy haha

[vale wigeon]

you forgot to backslash the second percentage sign

that's why tex is throwing an error

[primal rain]

pls i have 1 minute till its due

last question

[ocean seal]

HqppyFeet

[vale wigeon]

1. open your own channel
2. we don't give out answers
3. failure to plan on your part is not an emergency on our part

@rapid vector yeah so whats troubling you afterward

[rapid vector]

Yeah I just needed confirmation, because uncertaintycalculator.com uses a different method and got 35%

seems like i cant trust that website no more haha

[vale wigeon]

i mean ok if we do it like... naïvely

[vale wigeon]

eh?

[rapid vector]

i know xd

[vale wigeon]

that's the first time i'm hearing of that type of error propagation

idt thats what you are expected to do here lol

[rapid vector]

i dont think we are expected to use *that *formula

and idk if it's correct too xd

Anyhow, i got my confirmatoin, thank you <3<3

.close

[lone heart]

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[young forge]

Hello!

[young forge]

So uh, I'm stuck in this math problem.

I'm new to this sort of math.

My friend said it was Binomial Expansion.

[zealous lichen]

yes

[young forge]

(He's the one quizzing me)

I don't want the answer, I just want to know how to solve it!

[zealous lichen]

write down the kth term of a n degree binomial

[young forge]

Alright, hold on.

[lone heart]

@young forge Has your question been resolved?

[lone heart]

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[jagged timber]

is this a thing

[tardy stag]

not quite

[jagged timber]

so lnx has to be isolated

[west girder]

Try putting the -2 inside the natural log

[tardy stag]

use laws of exponents and logs to get it into the form $e^\ln\rsq$

[ocean seal]

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

[jagged timber]

like this?

[tardy stag]

yeah

[jagged timber]

okay ty

.close

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[hot bluff]

Need help with some beginner Linear Algebra.

[hot bluff]

if A is a 4x3, isn't it in R^3, not R^4?

how can Ax = b be consistent and have a unique solution with this setup? am I being dumb and 4x3 is 4 column x 3 row?

besides that, i know that if the question makes sense, then A must contain a pivot in every row when in RREF

[zealous lichen]

A is in $\mathbb{R}^{4\times3}$ i think

[ocean seal]

WhereWolf

[hot bluff]

wat

[oak perch]

Meaning three columns of A are linear independent and b is in the space generated by those three columns

[hot bluff]

haven't covered linear independence yet, this is a section prior to that.

[oak perch]

…

Nvm then

[hot bluff]

hmmm i think there's a difference between matrix equations and vector equations that im not noticing

.close

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[hot trail]

am i slow

for part 1:

how does it want me to calculate significant figures

.close

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[surreal aspen]

Can someone explain to me step by step, how did he reach that (repFromZero 3 (\x -> x+1)) is 3?

[tacit arch]

#computing-software is better for programming language specific questions

[surreal aspen]

okay

.close

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[hot bluff]

!status

[lone heart]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

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[heady pollen]

what a weird question

ah nvm i cant read

try to use all the info you are given

you are given L(1) and L'(1)

so get L' and input 1

it says so in the task

L(x)=1 and dL/dx(x)=3 at x=1

chain rule

just use the chain rule, let me know when you have a solution for dL/dx

reminder:
f(x)=g(h(x))
then:
f'(x)=g'(h(x))*h'(x)

[winter light]

Yes, this equation is correct

Now, you also are given L(1) = 1

[heady pollen]

then you can solve both for a or b and set them equal to each other, that will give you the other variable

[winter light]

Why 1/(a+bx)=1? They tell you that x=1, so you should write 1/(a+b) = 1

And then you make a system of 2 equations in 2 unknowns

Yes, but you already use that information

Now, you have to use the fact that L = 1 WHEN x = 1

Which are these ones?

Do you mean 1/(a+b) = 1 and -b/(a+b)^2=3 ?

Yep, you have 2 equations in 2 unknowns a & b...

$$\frac{1}{a+b} = 1 \implies a+b = 1$$

[ocean seal]

Alberto Z.

[winter light]

Do you agree?

[winter light]

Of course with a ≠ -b because of the denominator

b is -3, yep

So then a = 4

You're welcome

[charred summit]

@alpine sable don't forget to close the channel using
.close

[winter light]

Yes of course