help-42

[lyric ice]

im lost

lol

[remote mural]

its okay

[lyric ice]

how do i do y^2

square root both sides

so then im left with

square root of -x-5

[remote mural]

square root of x+5

Bc you can't have a -y

[remote mural]

So you divide the negative out

[lyric ice]

oml

dude

[remote mural]

Real

[lyric ice]

sorry pardon my language

[remote mural]

ur good

[lyric ice]

so im tryna get this into y=mx+b form

right

[remote mural]

Uh, no, bc equation b is a square root function

[lyric ice]

okay

so

i got y = square root of x+5

[remote mural]

Yes

[lyric ice]

now

can i graph that?

[remote mural]

Yes,

[lyric ice]

how do i graph a x on this shit

[remote mural]

click on graph

on ur calc

[lyric ice]

okay

i have this bar chart

idk what its from

[remote mural]

i wish i had my calculator rn oml

[lyric ice]

ok now i clicked graph

and its empty

did u want me to click y= first?

[remote mural]

Yeah, but input equation a into y(1)=

[lyric ice]

okay

so

how tf do i get x into it

[remote mural]

do you see the button where it has some shapes and letters on it (X,T,0,n)

[lyric ice]

yeah

so i got that

[remote mural]

click on it

[lyric ice]

but hwen i click graph

it shows a bar graph

[remote mural]

oh god

[lyric ice]

Im cooked

[remote mural]

OMG

UR ON STATS

[lyric ice]

oh ya am i not supposed to be

i took ap stats

not calc

or precalc

[remote mural]

OH

[lyric ice]

this was in hs

im in college now..

[remote mural]

OH

Okay okay

bc im taking precalc rn, reset ur calc

[lyric ice]

how tf

im taking pre calc too.......

uh

anyway

idk how to reset this shit

[remote mural]

hold on

[lyric ice]

can i just set it to calc instead of stat

[remote mural]

That too

i pulled out my graphing calc for u

[lyric ice]

ur a goat

ima go fck myself bruh

my vape is dying

jesus

time for zyns

anyway

[remote mural]

click on window first

[lyric ice]

yes

[remote mural]

Set this up:

Xmin=-10

Xmax=10

Ymin=-10

Ymax=10

then after that click "2nd" and click on the "mode" button to the right of that button

[lyric ice]

right

[remote mural]

Okay ur screen is now blank right

[lyric ice]

yes

[remote mural]

click on "y="

[lyric ice]

okay

put equation in?

[remote mural]

i'm gonna help u graph equation a

Yeah

[lyric ice]

okay

[remote mural]

If it looks weird, u maybe did it wrong

[lyric ice]

no it looks right

i got a parabola

reverse

wtvr

[remote mural]

Now graph equation b

On Y(2)=

[lyric ice]

okay i got that

[remote mural]

The graphs aren't symmetrical btw

[lyric ice]

okay

so

im confused

what am i trying to solve for

how do i find the intercepts

for (a) it would be (-2,0) and (0,0)

[remote mural]

yeah

[lyric ice]

for (b) it would be (-5,0) and (0,2)

[remote mural]

Yes

[lyric ice]

right

so how do i prove that

[remote mural]

Show work?

Or u could solve the equations algebraically

[lyric ice]

algebraically

how

[remote mural]

is a way of proving it

[lyric ice]

would i do that

[remote mural]

Solve for x on equation a

[lyric ice]

i need algebraically for the actual intercepts and symmetry

[remote mural]

jesus christ

[lyric ice]

right

[remote mural]

Okay,

[lyric ice]

thats what im saying

lmao

r u in college

or hs

[remote mural]

hs junior

[lyric ice]

oh wow

ur like

much younger than me

lmaooo

[remote mural]

I am

[lyric ice]

ight well in college they like the actual math behind it

unfortunately

[remote mural]

Well i think u can do the algebraic work for the intercepts

[lyric ice]

Sooooo

This is what i have

I plugged in 2 for x for A

And for b idk im just stuck

holdon

[remote mural]

did u figure it out?

[lyric ice]

chat gpt ing it

rn

[remote mural]

chat gpt is my savior

[lyric ice]

bruh

im gonna actually shoot myself

like

this is all j useless bs

atp

gonna fkn drop out

[remote mural]

wait

[lyric ice]

fck this shit

[remote mural]

😭

[lyric ice]

dude

lik

like

this

is

such

bullshit

lmao

[remote mural]

well

what r u gonna do now

[lyric ice]

kms

lmao

idk

thug it out

[lyric ice]

how does

0 = -x^2 - 2x = -x (x+2)

then x = 0, x = -2

that does not make sense

[remote mural]

bc its two intercepts

[lyric ice]

right

yeah

im cooked

lol

[calm coral]

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[calm coral]

@steel mason Has your question been resolved?

[steel mason]

No

[calm coral]

@steel mason Has your question been resolved?

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[blazing coyote]

$\int \left(\frac{\left(1-x\right)}{1+x}\right)^{\frac{1}{3}}\frac{dx}{x}$

[potent lotus]

Why am. I here

[blazing coyote]

any hints?

The only thing that comes to mind is multiplying by the conjugate

but I doubt that will help

,w $\int \left(\frac{\left(1-x\right)}{1+x}\right)^{\frac{1}{3}}\frac{dx}{x}$

[potent lotus]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by physicsrocks

[blazing coyote]

,w \int \left(\frac{\left(1-x\right)}{1+x}\right)^{\frac{1}{3}}\frac{dx}{x}

[potent lotus]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by physicsrocks

[blazing coyote]

wolfram's output involves the hypergeometric function

[marsh agate]

then it is safe to assume you can't solve it

[quaint sphinx]

it sometimes does that even when it's solvable

[blazing coyote]

oh, just saw the soln

have to take the entire surd as u

[quaint sphinx]

i was doing that

(there is not much else you can do here)

[blazing coyote]

yeah, didn't even try that because of how nasty differentitating that would be, oh well. Thanks a lot!

.close

[calm coral]

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[quaint sphinx]

differentiating a rational function is not nasty btw

but yw

[blazing coyote]

yw?

[blazing coyote]

I mean lengthy

[quaint sphinx]

u said thanks so i reply yw (means you're welcome)

[calm coral]

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[maiden crater]

hi guys i need help for 61

[maiden crater]

i dont knowwhere to start

but i know the answer is -sqr root of 2 / 2

i haveno idea where they got the 2

[spiral patio]

Do you know the unit circle

[maiden crater]

yes i do

[fleet verge]

In which quadrant is 225º?

[maiden crater]

the third

[fleet verge]

So, what is the sign of cos

[maiden crater]

negative

[fleet verge]

ok so, which angle is referencing 225?

forget about the sign, just tell me which one from the first quadrant

[maiden crater]

first quadrant?

ima show you a drawing of what i did

[fleet verge]

ok

[maiden crater]

OH WAIT I REMEMBER SOMETHINGNOW

my teacher said that reference angles alwayshave to be by the x axis

[maiden crater]

this is what i did to get the reference angle

[fleet verge]

if u use the x axis

it would be 225 - 180

which is also 45º

[maiden crater]

oh

[fleet verge]

so now you know cos(225º) = -cos(45º)

next step is trivial

[maiden crater]

whats the next step

i find the hypotenuse right

[fleet verge]

you have to know whats cos(45º)

it's the angle of reference the exercise is asking you to use

[maiden crater]

how do i do this

wait i'llshow ya what i did

or actually nevermind

i havent started yet

[maiden crater]

ok so how do i do this

.close

[calm coral]

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[calm coral]

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[sweet briar]

the height is given by h = (b/2)/tan(36) from step 2

multiply both sides by h and divide both sides by tan(36)

yeah

that's just the first b/2 being multiplied by the other b/2

no, that's just the tan(36) being written in terms of sin and cos

1/4 is from the b^2 / 4, they just moved it to the front to look nice
the 5 is because the entire shape is made of 5 of those triangles

so total area is 5 times area of one triangle

you see the 1/4 from b^2 /4 here?

they just moved it to the front because it looks better

literally nothing has changed in the math from that

$A_{\Delta} = \frac{1}{2}\cdot\frac{b^2}{4}\cdot\frac{\cos 36}{\sin 36} = \frac{1}{2}\cdot b^2 \cdot \frac{1}{4}\cdot\frac{\cos 36}{\sin 36}$

[potent lotus]

nalin

[sweet briar]

sure

yep

[crude pecan]

I seee

My bad man.

[sweet briar]

all good

[crude pecan]

I just had to take some minutes.

Tysm.

[sweet briar]

np

[crude pecan]

That's all!

Have a nice day 👍

.close

[calm coral]

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[gaunt forge]

Need help with the equation of the loci of complex numbers

[crimson cliff]

$z=x+iy \implies re^{\theta}$

[calm coral]

@gaunt forge Has your question been resolved?

[potent lotus]

Dubs

[crimson cliff]

$z^2= r^2e^{2\theta} \implies$ arg$(z^2)= 2\theta$

[potent lotus]

Dubs

[crimson cliff]

$2\theta = -\frac{\pi}{4} \implies \theta=-\frac{\pi}{8}$

[potent lotus]

Dubs

[crimson cliff]

This is a ray.

[calm coral]

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[dull wagon]

show your work

[rough ferry]

I found the zero from 3x-2 which is 2/3, so (x-2/3) is the factor, then I plugged it into the equation

and -37/27 was the rseult

[dull wagon]

,w x^3-3x^2-2x+1 when x=2/3

[potent lotus]

Results provided by WolframAlpha

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

Requested by ramonov

[dull wagon]

looks ok

[rough ferry]

so my remainder is correct?

[dull wagon]

yes, i said it looked ok and your result matched the bot's calculations

[clear delta]

$\polylongdiv{x^3-3x^2-2x+1}{3x-2}$

[potent lotus]

hayley!

[rough ferry]

wow, latex can do that!??!

cool, thanks

[rough ferry]

👍

.close

[calm coral]

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[calm coral]

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[latent magnet]

How can I determine the number of solutions for a linear equation without plotting them, especially during an exam with limited time?

[glad plover]

so you can notice that

8x + 12y = 4(2x + 3y)

so when this happens, the lines will be parallel

so there will either be no solutions or infinite solutions

(infinite only if they are literally on top of each other)

[latent magnet]

yes true

[twilit lagoon]

ax + by = c
a'x + b'y = c'

if a/a' ≠ b/b' -> unique solution

if a/a' = b/b' = c/c' -> infinite solutions

if a/a' = b/b' ≠ c/c' -> no solution

[latent magnet]

ohh thank you sir
it answers my questions well

[stray tundra]

Have you studied matrices yet?

[latent magnet]

yes

[twilit lagoon]

this is kind of technique to determine the solution quickly

[latent magnet]

is there a way through matrix also?

[stray tundra]

then you can also work from writing the system as a matrix, and working with the ranks

you order the equations in the form ax+by+cz+...=k

and then work with the rank of the coefficients matrix, and the augmented matrix

it's more efficient for systems with 3+ variables

[latent magnet]

yes yes but that's will take quite of time in case of 2 variables

I will keep note of this and learn if needed till now I haven't seen any questions which require that 😅

thank you, sir. I appreciate your help 😄

.close

[calm coral]

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[calm coral]

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[remote mural]

Why is limit of f(x) as x approaches 1, not defined?

[pure kayak]

should be

[remote mural]

I mean both the LHL and RHL would be 6

The answer key says that it does not exist

[pure kayak]

may i see

[remote mural]

Answer key?

[pure kayak]

3(x+1) is continuous, so there shouldnt be an issue

yeah

[dull wagon]

the author of the book is high

[quick karma]

which book is it

[remote mural]

It’s a problem from ncert but this book is by rd Sharma

[quick karma]

oh lol aap 11 th?

[pure kayak]

they seem to think the left limit of 1 is coming from left of 0

crazy stuff

[lyric ravine]

Writing a book is tough you know...

Even one for lower levels

[dull wagon]

books go through reviews, subsequent editions etc

[lyric ravine]

I guess the reviewers weren't looking that carefully at the book then 😆

[remote mural]

Not giving an incorrect solution to a problem they themselves selected out of all the questions isn’t

Can I ask another one please?

[dull wagon]

typos could happen.
but this seems like a misunderstanding/major lapse in judgment especially based on the question asking about x→0 as well where considering both pieces was appropriate

[lyric ravine]

!status

[calm coral]

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

[remote mural]

One minute let me rewrite it

[lyric ravine]

Cos(π/2-h)=sin(h)

Use this fact

[remote mural]

Okay wait

Let me see what I can do with that

[lyric ravine]

This step is faulty

Im not sure what you did here

[remote mural]

I just simplified the denominators

[lyric ravine]

What you did is not allowed, -cos(π/2+h)≠cos(π/2-h)

However it does give you the right answer

In this case

You should have used the fact that cos(π/2+h)=-sin(h) to simplify

[remote mural]

I don’t understand

[lyric ravine]

How did you go from the LHS to the RHS?

[remote mural]

That is just what the question gave me

[remote mural]

Wdym?

[lyric ravine]

How did you simplify $\lim_{h \rightarrow 0} \frac{kcos(\pi/2+h)}{-2h}$ into $\lim_{h \rightarrow 0} \frac{kcos(\pi/2-h)}{2h}$

[remote mural]

I just plugged x= pi/2 +- h in the function for both my left hand and right hand limits. Then I simplified the denominator

Oh I just equated my Left hand limit and right hand limit since we know for a fact that the limit is 3

[lyric ravine]

Oh you are doing it that way, in that case it is good to use different variables for the limits

Use some other letter for one of the limits

You are doing it correctly . use the fact that cos(π/2+h)=-sin(h) and cos(π/2-h)=sin(h)

[remote mural]

Ohh okay I understand what you mean to say now

But both my limits become the same thing

[lyric ravine]

Yes

And they should,

[remote mural]

How can get info out of that?

[lyric ravine]

Evaluate the limit

[remote mural]

I am getting 0/0 form

[lyric ravine]

Yes

Im assuming you are aware of the
limit sin(x)/x x->0 = 1

You have that here

[remote mural]

So k=6

[lyric ravine]

No, the limit is equal to k/2

There is a 2 in the denominator

Yes

[remote mural]

I understand now

Thank you so much:)))

I really appreciate it:)

Have a great day ahead

.close

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[final leaf]

i NEED help

[blazing coyote]

!da2a

[calm coral]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

@final leaf Has your question been resolved?

[calm coral]

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[orchid abyss]

How do i determine the parametric equation of this ellipsoide? Do I use spherical coordinates?

[calm coral]

Please don't occupy multiple help channels.

[civic dirge]

Yes, but adjusted

Shifted and stretched

[orchid abyss]

Do i substitue these values in for x y and z?

[civic dirge]

So for example,
y = 1/√2 rsinφsinθ + 1

Need the +1, and the 1/√2, in order to better cancel with the stuff in the equation

[orchid abyss]

aah okay thanks. i also need to identify the values that the parameters can have, what does that mean?

[calm coral]

@orchid abyss Has your question been resolved?

[calm coral]

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[dry isle]

yo guys ive been trying this for past 30 minutes and i am going absolutely batshit crazy someone please help me

[dry isle]

i took log x=a

so 3^a-2^a=2^a+1-3^a-1

[gloomy badge]

Wait, wouldn't that be 3^a - 2^a on the left side?

[dry isle]

uh yea edited that

so then 4(3^a)=3(2^a)

[gloomy badge]

Also 2^(a+1) - 3^(a-1) just to follow order of operations

[dry isle]

and im stuck here

[dry isle]

yea bro but i need help

[gloomy badge]

OK, what is m?

[dry isle]

wdym

oooooooooooooof

[gloomy badge]

You had m in an equation, you said 4(3^m) = 3(2^m) -- what is m?

[dry isle]

whyd i type m

that is a

sorry

[dry isle]

am i correct till that step tho

[gloomy badge]

No worries, still not seeing how you got there, though

[dry isle]

wait a second

[full tree]

split up the power

2^(m+1) = 2(2^m)

[gloomy badge]

^

You can combine terms easier if the 2^(thing) and 3^(thing) are on different sides

Then you can factor

[dry isle]

what is was doing was correct

it becomes 4(3^m) = 2^m * 9

i forgor 3 as denominatoor

sorry bro

[gloomy badge]

OK, you might be right. Can you show me HOW you got there?

[dry isle]

ok
put 3s and 2s on one side
3^m + 3^m-1 = 2^m + 2^m+1
= 3^m + 3^m/3=3(2^m) [take 2^m commons so 2^m(1+2)]
also take 3 lcm on lhs
3(3^m)+3^m/3=3(2^m)
again take 3^m common
so 4(3^m)=3*3(2^m)[cross multiply by 3]

[hollow aurora]

Write the equation as $2^2 \times 3^a = 2^a \times 9$ and try to figure it out

[potent lotus]

Kakashi_Hatake

[dry isle]

ye i got the answer

m=2

logx=2

x=100

[gloomy badge]

Wait: 3^m + 3^m/3 = 3(2^m)

OK, good deal

[hollow aurora]

Ye

[gloomy badge]

I'm pretty sure 3^m + 3^m/3 = 3(2^m) isn't true (for m=1, this would be 3 + 1 = 3(2), which isn't true)

But if you've solved the problem, that's cool

[dry isle]

wait what

[dry isle]

but i got m as 2

[gloomy badge]

Right, but I thought you were simplifying the expression. Did I miss something?

You got the right answer

[dry isle]

yess why

wait so is it wrong

[gloomy badge]

The answer is right. Saying that 3^m + 3^m/3 = 3(2^m) is always true is incorrect

[dry isle]

hmmm

[dry isle]

how so

[gloomy badge]

Well, if you plugin m=1, you have 3^1 + 3^1/3 = 3(2^1) or 3 + 1 = 3*2

[dry isle]

ye but isnt finding m the entire point tho?
thats like saying if you put x as 2 in 15x=10x+5 then itll be incorrect

i might be wrong tho

[gloomy badge]

OK, you said 3^m + 3^m-1 = 2^m + 2^m+1 = 3^m + 3^m/3=3(2^m) [take 2^m commons so 2^m(1+2)] so I thought you were making a general simplication

[dry isle]

oohhhhh

[gloomy badge]

Oh wait... you were simplifying both sides at once, my ba

My mistake, sorry. You're right

I thought you were working the left side first, but you were working both sides

[dry isle]

oh ok

ye thats why

thanks bro

[gloomy badge]

Sure thing

[dry isle]

.close

[calm coral]

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[indigo hinge]

anyone know how 5.89 was obtained?

[leaden thunder]

did you find all 3 of those numbers

R_n, N_A, V

[calm coral]

@indigo hinge Has your question been resolved?

[indigo hinge]

Well I know V

but not sure what 1/Na is

[leaden thunder]

that's a chemistry thing, not a math thing

look in your book or notes

[calm coral]

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[remote mural]

lol am i brainfarting right now ? Does 200r/r**2 simplify down to 200/r ?

[dull wagon]

cancel common factor of r

assuming your use of ** is notation for exponentiation

[remote mural]

Yep

So it does simplfiy down ?

[dull wagon]

yes

[remote mural]

Sweet ! Thanks alot.

[calm coral]

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[swift marsh]

need help with part b

praying for a physicist in this discord

[leaden thunder]

#old-network for physics server and or chemistry

[swift marsh]

i know

[leaden thunder]

then why is this channel still open

[swift marsh]

its an eigenvector eignavalue problem

is it possible to solve for c1 or c2

[leaden thunder]

where's your math question then

[swift marsh]

coz this is just ridicvulous

im hoping for a physics person to see if there was an error but i think its just long

[leaden thunder]

there are more physics people in physics server

[glad plover]

so the first one is like (k - S)c_1 - tc_2 = 0 where S is the square root and k is 0.5(e_1 - e_2)

then the second one is -tc_1 + (-k - S)c_2 = 0

so i don't see why you can't determine whether or not this can be solved pretty easily

[swift marsh]

no

one has epsilon_1 - ....

[glad plover]

i know

[swift marsh]

other is epsilon_2 - ...

[glad plover]

i know

[swift marsh]

Yea

[glad plover]

e_1 - 0.5(e_1 + e_2) = 0.5(e_1 - e_2)

e_2 - 0.5(e_1 + e_2) = -0.5(e_1 - e_2)

so k and -k

right?

[swift marsh]

Yes i see

ahh

ok

yes

I will try with this

[swift marsh]

e1 - 0.5e1 = 0.5e1

[glad plover]

?

[swift marsh]

oh

nvm

[calm coral]

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[swift marsh]

.close

[calm coral]

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[spark adder]

can someone please help me write this as latex

[hasty fiber]

What have you tried?

[spark adder]

I tried to ask ChatGPT to do it but it expresses it as a fraction

I mean I can do (5n+2)^4n+3 but I wanna know how do I get this one exactly

[hasty fiber]

So you don't know any latex of your own?

[spark adder]

no i dont really use latex

this is probably one of the few times I have to use it

[hasty fiber]

Eh whatever alright

$\lim_{n \to \infty} \sqrt[4n+3]{5n+2}$

[potent lotus]

Steakanator

[spark adder]

thanks

.close

[calm coral]

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[dusky verge]

Hii is this math correctly

[dusky verge]

number 1

ping me when you respond

<@&286206848099549185>

[calm coral]

@dusky verge Has your question been resolved?

[dusky verge]

huh

i dont think thats for me lol

<@&286206848099549185>

[calm coral]

@dusky verge Has your question been resolved?

[dusky verge]

<@&286206848099549185>

[dusky verge]

<@&286206848099549185>

[calm coral]

@dusky verge Has your question been resolved?

[calm coral]

@dusky verge Has your question been resolved?

[calm coral]

@dusky verge Has your question been resolved?

[trail isle]

.

[calm coral]

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[echo granite]

pls help

[echo granite]

I'm struggling with this remainder question 3c

This is what I've came up with so far

This is my answer to part a if it helpz

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[calm coral]

@echo granite Has your question been resolved?

[echo granite]

<@&286206848099549185>

bro

[echo granite]

<@&286206848099549185>

<@&286206848099549185>

[winter elbow]

You’ve been waiting for an hour and half…

[echo granite]

ikr

[winter elbow]

gl gl

[echo granite]

ty

<@&286206848099549185>

[calm coral]

@echo granite Has your question been resolved?

[echo granite]

<@&286206848099549185>

[calm coral]

@echo granite Has your question been resolved?

[calm coral]

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[lone flame]

Hello, can someone explain how we could solve the following cases in R:
• |x + 1| + |x − 3| = 6.
• |x + 1| + |x − 3| ≤ 6.

and Prove that for any x, y ∈ R:
• ||x| − |y|| ≤ |x − y|.
• |x| + |y|| ≤ |x + y| + |x − y|.
• 1 + |xy − 1| ≤ (1 + |x − 1|)(1 + |y − 1|).

[royal basalt]

Graphs. Use graph

[lone flame]

I'm not allowed to use graphs unfortunately

[royal basalt]

Ohhh

[lone flame]

we have to show it using mathematical logic

[royal basalt]

Are you atleast allowed to break the mod in different cases ?

[lone flame]

for the first one i just solved it like a normal equation in two cases:
(x+1+x+3) = 6 and -(x+1+x+3) = 6
but i'm not sure that is correct

[royal basalt]

There'll be four cases

[lone flame]

yes

oh is that like when we change the signs for each of the mods?

but in that case how could my answer differ from the second case involving the inequality/

[fleet verge]

|x + 1| + |x − 3| = 6 has only 2 solutions.

Sorry you said cases.

Show what you did for the first equation.

$x - 3 = 6 - |x + 1|$ \quad \text{or} \quad $x - 3 = |x + 1| - 6$

[potent lotus]

Samuel

[fleet verge]

from here i think u can continue easily

[lone flame]

i'm getting 4 and -2

and then for the second one i'm getting [-2,4]

is that correct?

[fleet verge]

yes

but you can always check your results by yourlsef

plugging your solutions into the original equation

[lone flame]

oh perfect thank you!

could you also give me some guidance on the last part i mentioend above:
Prove that for any x, y ∈ R:
• ||x| − |y|| ≤ |x − y|.
• |x| + |y|| ≤ |x + y| + |x − y|.
• 1 + |xy − 1| ≤ (1 + |x − 1|)(1 + |y − 1|).

<@&286206848099549185>

[soft tinsel]

similarly to the previous part, consider the 4 cases when x is positive or negative and y is positive and negative

[lone flame]

can u explain further?

[calm coral]

@lone flame Has your question been resolved?

[calm coral]

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[limpid flame]

$\text{is it possible to prove there is some } (a, b)\text{ such that }\\sqrt{(x_1-a)^2+(y_1-b)^2}\neq \sqrt{(x_2-a)^2+(y_2-b)^2}\neq\cdots\\text{ for all }x_n,y_n\in \mathbb{N}$

[potent lotus]

Captn

[glad plover]

square both sides, expand, simplify, attempt to solve

[limpid flame]

i tried but you cant do much other than cancel out the a^2 and b^2 terms

let me clarify that the (x_n, y_n) pairs have to be unique, not every x and y

so (x1, y1) could be (1, 3) and (x4, y4) could be (1, 6)

so x_1 = x_4 in that case

but since theres no consistency in which are equal that doesnt seem helpful

[glad plover]

you can do something i think

intuitively we should be looking for non-integer a, b

because like if we have integer a, b then it obviously just isn't true

so try and sorta group together all the integers in the thing you get

and then you can just sorta ignore them

you can sorta quotient R by Z

[limpid flame]

so I can ignore the integers if I group them since were just checking for all integers anyway? seems scuffed but ig

[glad plover]

essentially we want all the integers on one side and all the expressions with non-integers in on another

and then we want that second side to never be an integer somehow

[limpid flame]

did you notice the dot dot dot its an infinite chain of inequalities

[glad plover]

yes

but like

if it holds for all distinct pairs

then that's enough

[limpid flame]

ohhhh i see

so I end up with $x_1^2-x_2^2+y_1^2-y_1^2 \neq -2ax_1-2by_1-2ax_2-2ay_2\\text{so I can just say let a, b be irrational and since x_n y_n are rational they never equal therefore since one works they all work blah blah proven}$

[potent lotus]

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

[limpid flame]

wtv just read it im gonna eat 😭

[glad plover]

yeah

[limpid flame]

huzzah

.close

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[lime trench]

last line

how did he multiply 2 above equations

help!!!!!!!

[karmic sonnet]

do you mean why did he multiply or how did he multiply?

[lime trench]

how and why both

actually

[karmic sonnet]

first he wrote what (1+x)^n equals to

using binomial theorem ofc

[lime trench]

yes

[lime trench]

yes

[karmic sonnet]

then he differentiated the same

[lime trench]

yeah i understand that

[karmic sonnet]

then he wrote (1+x)^n again but this time , he wrote it backwards

[lime trench]

yeah

he did

[karmic sonnet]

now the terms he marked with the black pen , do you see how they all start becoming what he called "S"?

after multiplication

[lime trench]

hmm

yes

but how can you multiply eqns like that

give me an example

[karmic sonnet]

we can't simply multiply random equations until they give us some desired result

you're a JEE student as it seems , remember sequence and series?

[lime trench]

uhmmm

[karmic sonnet]

in Sequence and Series , the miscellaneous series

when the differences were in AP

we used to write the same equation again

but with a gap of 1

[lime trench]

yeah

yeah

[karmic sonnet]

there , we subtracted them to get a result

[lime trench]

yeah

i remember

[karmic sonnet]

likewise

we can add , subtract , multiply or divide

2 equations for a desired result

[lime trench]

oh

[leaden fox]

Im guessing all the undesired terms cancel out

[lime trench]

even divide

tf

bro give an example

[leaden fox]

Ur a jee guy right?

[lime trench]

yes

[leaden fox]

Come dms

[lime trench]

aya

[karmic sonnet]

you might've seen that in trignometery I guess

as I remember standard questions

[lime trench]

alr thanks mate

[karmic sonnet]

you're in 11th grade right?

[lime trench]

yes giving exams

will be in 12th soon

[karmic sonnet]

I see , all the best for JEE , next year

[lime trench]

thanks sm]

[jolly sentinel]

why are all the fellow jeetards here

[karmic sonnet]

as an ex JEE , the reason is unknown

[jolly sentinel]

woah sir, no wonder with organic chem related name on a math sv

[karmic sonnet]

all my knowledge , now I must spread it

I didn't give JEE , mine was this year

[jolly sentinel]

haha, i just got dealt with jee, boards ruined the damn flow

[karmic sonnet]

I qualified for the NDA before I could give JEE

I was like nah man , I ain't studying anymore

enough

[jolly sentinel]

army seems better yeah

[karmic sonnet]

that's what I'm medically fit for too lol

because of my eyes

I'm unfit for navy and AF

thanks to JEE lmfao

[jolly sentinel]

dw

[karmic sonnet]

@lime trench close this channel

[lime trench]

.close

[calm coral]

Channel closed

Closed by @lime trench

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[calm coral]

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[snow pulsar]

here is task c

i need help on how to do d

[leaden thunder]

!noclopen

[calm coral]

Please don't repeatedly close and claim a new channel with the exact same question. This erases all previous progress made towards your problem and is confusing for helpers, making it more difficult to help you. Please be patient, even if your channel has not received much attention.

[snow pulsar]

i need help with these tasks

[calm coral]

@snow pulsar Has your question been resolved?

[snow pulsar]

i need help with these tasks

<@&286206848099549185>

[calm coral]

@snow pulsar Has your question been resolved?

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[plain moss]

Doing what wrong? Derivative

Omg

-- can someone chck work for correction :

[calm coral]

@plain moss Has your question been resolved?

[plain moss]

i do believe in myself and think all the above are correct. lol 😂

<@&286206848099549185>

[remote mural]

yo

[plain moss]

yo

back at it again haha

[remote mural]

haha

sry i had to go pray last night

anyways, lets see

[plain moss]

nono all good i solved :))

[remote mural]

so u think u did the right thing, do u want sm1 t verify it or to help with next steps

[remote mural]

thats good

[plain moss]

yes esp the most recent one

the even numbered one my book doesnt put the solved answers soo stuck there for checking

[remote mural]

wait so

man this dont look familiar but ill try my best

[plain moss]

all good

[remote mural]

ah

ok so ur trying to find h'(x)

right?

[plain moss]

yup

[remote mural]

😭 ok lets see

[remote mural]

walk me thru what u did

[plain moss]

ai bet

question

could take the 3 out since its a constant and applied the constant rule here

skipped step but basically, in the numerator i did quotient rule

[remote mural]

whats u do in the numerator there

hmm

[plain moss]

simplifying/foil method

multiply the 3 back in numerator with each

[remote mural]

wait

chain rule or

quotient rule

in the numerator?

[plain moss]

wait

wrong

[remote mural]

thats

quotient rule

[plain moss]

did i say chain??

[remote mural]

yes

[plain moss]

omg yeahhh

quotient rule

[remote mural]

OK

[plain moss]

mb

[remote mural]

THAT MAKES

MORE SENSE

ok

got it

[plain moss]

lool idk where chain rule came from lol

still gotta learn that

[remote mural]

yh i thought my brain was melting

like welp, lack of sleep is getting to me already

ok now i understand

[plain moss]

i sleep over time 😭 i cant help myself since its reading week

[remote mural]

ok so ur here now

[plain moss]

yes

thats the final answer

[remote mural]

yes, that makes complete sense to me

i think thats right

i wouldnt have

multiplied it out tho

[plain moss]

the 3?

[remote mural]

yh

js put the numerator here

in parenthesis, and put the 3 up in the numerator

js me personally tbh

[plain moss]

i think it helps with making it less wordy

oh i see

[remote mural]

wait

[plain moss]

yeah true works asw some person did that yday so

[remote mural]

whered the 6f'(x) come frm

[plain moss]

that was on of my concerns one sec drawing

[remote mural]

oh i see

u took the parenthesis out

[remote mural]

idk if u can do that tbh

oh

hmm

[remote mural]

ig it wrks

yh that seems right then!

[plain moss]

bet,

ty

[remote mural]

yh np, good job

[plain moss]

#help-42 message

^ if anyone else wants to check the others

118 & 120, respectively

[remote mural]

oh i didnt see those

sorry

i can check if u want

[plain moss]

up to u, if u got time

[remote mural]

so 118, 120

what about 121

[plain moss]

yup

odd so i have the answer and i got it correct

[remote mural]

ok lets see

[plain moss]

this is 121 lol if u wanna check steps

[remote mural]

alright

[remote mural]

lets start with 121 then

[plain moss]

aii

ill upload new pic

no pencil in between

[remote mural]

o ok ty

imma work it out on paper and then send a pic

gimme a minute cuz im going dyslexic

[plain moss]

all good

happens to me asw

[remote mural]

what is that last equation?

[plain moss]

in 121?

[remote mural]

y-b=m(x-a)

[plain moss]

yes

[remote mural]

ion think ive ever seen that before but it seems correct

[plain moss]

point slope form

[remote mural]

oh i never learned that

😭

[plain moss]

is there a better one?

[remote mural]

i skipped a lot of basic math

[plain moss]

idk i forgot most of things so learned it through chem tutor

[remote mural]

ohh fair

[plain moss]

i forgot a lot of basic math 😂

[remote mural]

s

o

basically

[plain moss]

yes

[remote mural]

up till there it seems correct

and then over there, idrk pint slpe or anything but

[plain moss]

checks out yea

[remote mural]

i put it in desmos and it seems right?

[plain moss]

at x=1

[remote mural]

yep

(1,-1)

[plain moss]

yup

in textbook its also correct

so its fine

[remote mural]

alright thats good

so next is 118?

[plain moss]

118

[remote mural]

imma use the rr rq, brb

[plain moss]

yea

[remote mural]

ok

mb im back

[plain moss]

i think seems correct

oh i can check on desmos

[remote mural]

ive always js used y=mx+b

yh u can lol

but it seems correct to me to js by skimming thru it

[plain moss]

so b is y1

y=4x+1

oh thats also the same

[remote mural]

lol

yh, ive heard of point slope and standard and wtv else

ive always js used y=mx+b

[plain moss]

and it alwauys worked yea?

[remote mural]

checks out tho using desmos

and the math seems to mathing even if u dnt use desmos

[plain moss]

essentially the same thing

[remote mural]

for me, yh

[plain moss]

so yeah

[plain moss]

i am understand the methodology behind whys calculus is what it is

seems to be understanding/working 🫡

[remote mural]

yeah

i remember when i took calculus, i didnt understand anything till i got tot he final exam

[plain moss]

damn lol

hmm

[remote mural]

💀

ok lets see

[plain moss]

but this aint a tangent line

its a secant

[remote mural]

yep

[plain moss]

oh wait

it should be this

[remote mural]

i see

what u did wrong

[plain moss]

is it the division?

[remote mural]

(-2)^2

[plain moss]

yes

[remote mural]

is 4

no?

[plain moss]

-2/4 = -0.5

[remote mural]

yes

u put it as -2/2

is -1

try finding the equation using -1/2

[plain moss]

yeah lol the second u said found the mistake i knew it was gonan be the calculation part

[remote mural]

lmao yh

good thing is u understand the concept

[plain moss]

lol just using y=mx+b so much easier

waittt

[remote mural]

lol

does it work in desmos

[plain moss]

this is demos now

[remote mural]

LMAO

ok

pause

hm

what if u didnt use y=mx+b

do u still get 1

[plain moss]

yes

[plain moss]

^

both methods

[remote mural]

hm

lets see

[plain moss]

wait