help-10

[low fossil]

So similarly for 0

[dense dagger]

Umm

Let me try

Thanks anyway

[dense dagger]

I think I can solve part a

But how about part b

So I put u=abs(xy)

[compact shadow]

f(x)<=g(x)<=h(x) then limf(x)<=limg(x)<=limh(x)

What is the limit of 1/(1+u^2) when u approaches 0?

[dense dagger]

1

We can do the direct computation

[compact shadow]

And what’s the limit of constant function 1 when u approaches 0?

[dense dagger]

Still 1?

[compact shadow]

Yeah so solved

[dense dagger]

Oh

So that's just sandwich Theorem

[compact shadow]

Oh didn’t know that name now I know

[dense dagger]

Lol I just not sure about the notation

A vector tends to a zero vector

[compact shadow]

Can’t see any vector here…

[dense dagger]

Umm (x,y) is in R^2

[compact shadow]

By definition

[dense dagger]

So I can't conpare the left hand limit and the right hand limit

[compact shadow]

That limit (in terms of x and y ) is actually equal to

Limit u approaches 0+ tan^-1(u)/u

By definition

I mean the right limit of tan^-1(u)/u when x approaches 0 coming from the rand hand side

And since tan^-1 (u)/u is continuous at 0 so right limit equals limit

[dense dagger]

Ok

And abs(xy)>0

[compact shadow]

Yeah

[dense dagger]

So we can apply the ineq in part a

Yeah

[compact shadow]

Idk, someone answered a for you before I came, I only read b

[dense dagger]

Lol it is LMVT

I get that already

I want to know how to do part b

Part a is obvious, we can simply factorize the common factor

It looks like the limit x approaches 0 (exp(x)-1)/x in real number

But this is the R^2 version

[compact shadow]

You prove by definition that the limit equals -1

[dense dagger]

Umm

So u mean the epsilon delta definition?

[compact shadow]

Yeah any ε… there exists δ … any x^2+y^2 <δ… …

Using the fact that limit (1-e^u)/u=-1

[dense dagger]

Ok

I will try

[compact shadow]

(Hopital rule for that fact)

[compact shadow]

No you can’t

Points near (1,2) even apart from (1,2) still might satisfy 2x-y=0

Like (1+t,2+2t)

[dense dagger]

Oh i see

[compact shadow]

Oh yes you can…

Those kind of points are ruled out from domain in the first place…

So nvm

[dense dagger]

Ok thank you

[compact shadow]

How do I make my messages hidden, that other people need to click on it to read them?

[wicked bolt]

you mean spoilers?

do || message here ||

or you could just type /spoiler

[compact shadow]

|| test ||

Thanks. Okay got to go, in case you need c) d) e)
|| c) Consider x=ky ||
|| d) let x=rcos(t), y=rsin(t) ||
|| e) consider x=ky^3 ||

[dense dagger]

Thank you. I will read it later

.close

[obtuse pebble]

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[warm olive]

when we write the integral why is the Jacobian 1/2v and not just 2v

nm im stupid

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[obtuse pebble]

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[scenic kestrel]

I can identify specific points and whether they are continuous or not but I’m not sure about how to determine a range. In this text book, [a,b] is a notation of a <= x <= b. And (a,b) is a < x < b

[scenic kestrel]

I think I just need a refresher on the difference between specific points and ranges

[latent otter]

If f is continuous in a<x<b, then it is continuous for every point between a and b (exclusive), and the same for [a, b] but inclusive

[obtuse pebble]

@scenic kestrel Has your question been resolved?

[scenic kestrel]

So in the example of [0,2] it would be continuous because there are no gaps?

[desert sparrow]

What is the definition of continuity you are given?

[leaden ibex]

I would assume it's the standard $\lim_{x\to a} f(x) = f(a)$

[warm shale]

Remavas

[obtuse pebble]

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[random egret]

Ok so

[random egret]

Anyone yoohooo

[hollow sun]

lmao

you didnt even ask anything

[random egret]

Oh

Shit

[random egret]

It’s making a the subject

[hollow sun]

you want to reshape it to a?

[random egret]

Uhm

Transpose for a

Idk how to explain it

[hollow sun]

you want a=... ?

[random egret]

Yes

[hollow sun]

ok where are you stuck?

[random egret]

So I’ve squared it to remove the square root

And now I have z-ax=t^2

[hollow sun]

yes you got $z-ax = t^2$

[warm shale]

Big xdddd

[random egret]

Yes

[hollow sun]

and what is your next step?

[random egret]

Not sure

Do I need to move the z over

[hollow sun]

you could add or subtract things on both side

[hollow sun]

for example yes

what do you get if you do that?

[random egret]

So now I have -ax = t^2 -z

[hollow sun]

$-ax = t^2 - z$

[warm shale]

Big xdddd

[hollow sun]

ok thats good

[random egret]

Yes

Now I’m stuck

[hollow sun]

now you can write $ax$ as $a\cdot x$

[warm shale]

Big xdddd

[hollow sun]

if you only want a what do you have to do on both sides?

[random egret]

Divide

By x?

[hollow sun]

yes

what you get?

[random egret]

Wait

But what do I do with the -

[random egret]

If it’s -ax

[hollow sun]

if you want you can divide -x in one step or you can divide first by x and then by -1

[random egret]

Ok

Lemme do that

Wait

So I divided first by x and now I have -a = t^2-2/x

Now how do I get rid of the - from a

[hollow sun]

yes

[random egret]

E

[hollow sun]

think of this: $-a = -1\cdot a$

[warm shale]

Big xdddd

[random egret]

It’s a t

Lol

Oh wait

Now I’m confused

[hollow sun]

if you only want 'a' what do you have to do on both sides?

[random egret]

Uhhhh

Divide ?

[hollow sun]

yes

[random egret]

By what tho

[hollow sun]

think about what you did here..

[random egret]

Erm

I’m running dry on this part

[hollow sun]

Bro?

[random egret]

What

[hollow sun]

what did you do here to get rid of x?

[random egret]

Divide both sides by x

[hollow sun]

YES

[hollow sun]

what do you have to do here?

[random egret]

Divide both sides by…

-1?

[hollow sun]

yes

[random egret]

So I divide the t^2 -z/x BY negative one ?

[hollow sun]

yes

[random egret]

Ok

So a= t^2-z/x/-1 which breaks down to t^2-z /x

I think

[hollow sun]

the first one is correct

[random egret]

Uh

[hollow sun]

where did the -1 go in the second one?

[random egret]

Well if you divide t^2 -z/x by -1 you get t^2-z/x

[hollow sun]

what

[random egret]

Oh wait

[hollow sun]

$4 = \frac{4}{-1}$ ?

[warm shale]

Big xdddd

[random egret]

So the answer would be -t^2 -2/x ?

[hollow sun]

ok

so i got a question for you

[random egret]

Ok

[hollow sun]

this is your answer right: $\ -t^2\frac{-2}{x}$ ?

[warm shale]

Big xdddd

[random egret]

No

It’s all over x

[hollow sun]

is this your answer$\frac{-t^2-2}{x}$ ?

[warm shale]

Big xdddd

[random egret]

Yes

[hollow sun]

WRONG

[random egret]

HIW

HOW

[hollow sun]

you had: $-a=\frac{t^2-2}{x}$

[warm shale]

Big xdddd

[hollow sun]

right?

[random egret]

Wait is it a=t^2-z/-x

It’s a z not 2 sort mb

Sorry*

[hollow sun]

then you divided by -1 so you got$a=\frac{\frac{t^2-z}{x}}{-1}$

[warm shale]

Big xdddd

[random egret]

Yes

[hollow sun]

now simplify this..

[random egret]

-t^2-z/x

What did I do wrong

Oh

Wait

[unique solstice]

Distribution

[random egret]

T^2-z:-x?

[hollow sun]

you say that: $\frac{\frac{t^2-z}{x}}{-1} = \frac{-t^2-z}{x}$ ?

[warm shale]

Big xdddd

[random egret]

T^2-z/x?

Wait I typed wrong

t^2-z/-x

[hollow sun]

okay

thats correct

$\frac{t^2-z}{-x}$ can you simplfy this even more?

[random egret]

So a=t^2-z/-x

[warm shale]

Big xdddd

[random egret]

Oh

Uhm

[hollow sun]

this looks terrible tbh

but its correct

[random egret]

I don’t think so ?

[hollow sun]

i do think so

[random egret]

Oh

How

[hollow sun]

well for example you can write: $\frac{t^2-z}{-x} = \frac{-(t^2-z)}{x}$

[warm shale]

Big xdddd

[hollow sun]

now dissolve the clamp

[random egret]

Oh

[random egret]

So the answer is

Uhh

[random egret]

This?

[hollow sun]

no

there is a clamp

[random egret]

Could I just leave it like this

[hollow sun]

NO

[random egret]

Oh

[hollow sun]

dissolve the clamp

i give you a hint

like literally the answer

$-(a-b) = -a + b$

[warm shale]

Big xdddd

[random egret]

So it’s -t^2+z/-x

[hollow sun]

bro

what have you done

[random egret]

I see that the sign has changed

[hollow sun]

WHERE DID YOU GET THAT -X IN THE DENOMINATOR

[random egret]

TYPO

[hollow sun]

tell me the answer right now

[random egret]

-t^2+z/x

[hollow sun]

$\frac{-t^2+z}{x}$

[warm shale]

Big xdddd

[hollow sun]

yes

[random egret]

Lmao ty

[hollow sun]

if you want to make it look better you can write:

$\frac{z-t^2}{x}$

[warm shale]

Big xdddd

[random egret]

Ok

Ty

[hollow sun]

no problem

have a nice day or night

[random egret]

Hollup

.close

[obtuse pebble]

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[humble coral]

I need help

[humble coral]

-6x > 66

[timid silo]

how would you isolate x

[obtuse pebble]

@humble coral Has your question been resolved?

[humble coral]

I already done it sorry

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[obtuse pebble]

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[slow pivot]

can somebody quickly check if the following definition of expectation in maths is correct. Couldn't find one which defines expectation in the general sense so I made my own:

[slow pivot]

maybe use approach instead of be, but then again it would be an unprecise definition of expected value

[obtuse pebble]

@slow pivot Has your question been resolved?

[obtuse pebble]

@slow pivot Has your question been resolved?

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[buoyant arch]

Is there anyone who can explain exponential growth and decay in calculus to me? I have an bacteria culture with initial population 200, population 660 after an hour, and I have to find things like the population and rate of growth at different points, as well as the equations representing that.

[obtuse pebble]

@buoyant arch Has your question been resolved?

[obtuse pebble]

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[sharp ingot]

.

[long sinew]

Why does volume of a sphere use 4/3...?

[short spruce]

you could prove it with volumes of rotation i think

[long sinew]

What does that mean?

[short spruce]

revolving the equation of a circle around the x axis to form a sphere

and taking the integral of that

using disks

[long sinew]

Welp sadly I do not know integrals

[short spruce]

well neither did archimedes

there's probably a youtube video that explains it although i don't know easy it is

[long sinew]

Hmm

What about cones?

Maybe something to do with a right triangle revolved around an axis?

I will google it

Oh, it's literally 1/3 of a cylinder

[obtuse pebble]

@long sinew Has your question been resolved?

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[vernal saffron]

Can someone explain to me how they get the u-1 and the 1/25 works?

[low fossil]

Can you write x in terms of u

This actually explains the linearity

[vernal saffron]

i kinda get it, u is 5x+1

[low fossil]

No

X be the subject

[vernal saffron]

?

[low fossil]

Alright I mean :”x=(u-1)/5” here

The denominator is 5

[vernal saffron]

oh

[low fossil]

Also the dx change to du/5

Give u another 5

[vernal saffron]

so you get 25 on the bottom

[low fossil]

Yeah

U get it

[vernal saffron]

got it, ty

.close

[obtuse pebble]

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[elder vale]

confused on how to find number of possible solutions

without nasty casework

<@&286206848099549185>

[short spruce]

• After 15 minutes, feel free to ping @Helpers.

[obtuse pebble]

@elder vale Has your question been resolved?

[obtuse pebble]

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[elder vale]

<@&286206848099549185>

[elder vale]

.reopen

.close

[obtuse pebble]

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[steep vault]

would %15+%0 be 0?

[steep vault]

i am got in an agrument

and google doesnt want to tell me

[short spruce]

i never thought i'd suggest the additive identity to anyone

[obtuse pebble]

@steep vault Has your question been resolved?

[steep vault]

lol

i said it wrong

%15 of %0 is nothing

im dumb

[obtuse pebble]

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[proven musk]

Hey if a/b = c/d and therefore ad = bc does da = cd and from that a/b = c/d?

[short spruce]

?

[proven musk]

self explainatory

[short spruce]

uhh

"does da=cd and from that a/b=c/d" makes no sense

[proven musk]

no I mean if from a/b = c/d you can cross multiply to ad = bc can you get a/b = c/d from da = cb?

a, b, c, d are integers d and b aren’t 0

[short spruce]

in your og question you didn't have a b term

in the 2nd equation

but yeah you can

[proven musk]

so the product of cross multiplying comes from

identity a/b = c/d = I

I and I^-1?

[short spruce]

not sure i understand what that means

cross multiplication is just the "shortcut" for algebraic manipulation

[astral ivy]

What is this I business

[obtuse pebble]

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[restive thorn]

how do I calculate the transformations when doing this

So the teacher plugged in -2, -1, 2, 7 into the function

how do I judge where to draw the opposite dots (flip the x axis)

[obtuse pebble]

@restive thorn Has your question been resolved?

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[wanton dagger]

This is what I did

But it’s wrong

[nocturne creek]

That line passes through (5, 6)

You want it to pass through the midpoint of (5, 6) and (1, 10)

[hearty zephyr]

Wouldn't the midpoint be (2, -2)

[nocturne creek]

It wouldn't.

[hearty zephyr]

What the guy above me said

Damn I'm dumb then

3,8

[nocturne creek]

It would be odd for the midpoint of two things that are above the x axis to be below the x axis.

[wanton dagger]

So do I do?

[hearty zephyr]

Use midpoint formula ❤️ for the two points

That's where your point should pass through

[wanton dagger]

Ok

[hearty zephyr]

Use that in the equation in place of 5,6

[wanton dagger]

Ok but why? Just curious

[hearty zephyr]

Because according to the general equation of lines, the values where X1 and Y1 are substituted are coordinates for any point that is ON the line

That lies on the line

Now you substituted (5,6) which lead you to get an equation that passes through point 5,6 with the slope 2/3

[wanton dagger]

Ok

[hearty zephyr]

However, the question demands that the line pass through the midpoint of (5,6) and another point (1,10)

[wanton dagger]

Ok I got it thanks

[hearty zephyr]

Alright ❤️

Take care

[wanton dagger]

You too

✌️

.close

[obtuse pebble]

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[raven holly]

Hi guys, I hope you are all fine and doing well, I need some help
c/d=pi
d=diameter, c=circumference, pi=3.1415926535897
r=radius
/=÷
If we took pi from both sides of equation would be
c/d-pi=0
Then if we multiply both sides of equation by d we get
c-pi×d=d×0
c-3.14×2r=0
c-6.28r=0
Then we add 6.28r to both sides of the equation we get
c=6.28r
Then we divide both sides of equation by 6.28 we get
c/6.28=r
Since c=1017.36 we get
1017.36/6.28=r
162=r
Is that right?

[obtuse pebble]

@raven holly Has your question been resolved?

[timid silo]

yes its right

[royal basin]

what exactly are you trying to do here?

are you trying to calculate the radius of a circle whose circumference is 1017.36 m?

if so, you are correct, but your work is extremely roundabout

[raven holly]

Thanks

[raven holly]

Yeah I know but I wanted.I wanted to see if it's working like this

[raven holly]

I know I can say c=d×pi directly

Also I thought there is a mistake in this way so now I am sure that we can solve it in both ways

[obtuse pebble]

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[wheat gyro]

for b when i split the integral and do x^2 do i integrate with 0 to 3 or 2 to 5

because of the f(x+2)

or does it only apply to that function

[barren wing]

@wheat gyro

[wheat gyro]

ohh okay so you do it individually when it’s adding inside of a function

[barren wing]

yeah

[wheat gyro]

thank you!

[barren wing]

also

the whole things not multiplied by 2

I just realized that

[wheat gyro]

only the f(x+2)?

[barren wing]

yeah

[wheat gyro]

alright yeah that makes sense

[barren wing]

https://tenor.com/view/wheelie-rollin-they-see-me-rollin-jeep-kid-gif-19615923

[wheat gyro]

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[near timber]

if a graph has a hole in it does the inverse point in an inverse graph have a hole in it too?

[near timber]

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[distant shard]

Im thinking on how to write a problem statement for a linear regression problem. Can somebody give me some idea

I know that linear regression is used to establish a relationship between an independent variable x and a dependent variable y as a function y = f(x), but I dont really have an idea on how to write a concise problem statement

[obtuse pebble]

@distant shard Has your question been resolved?

[robust sleet]

just think of an experiment

that has a lineair relation

then take one in dependant variable

change its value and see what happens to the other variable

plug it on graph

and do lineair regression

also it can be a power relation

for power relation plug the values in a loglog graph

for a power you can think of a pendulum with Force and Period

[distant shard]

I got some idea more or less

Thank you

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[rich holly]

why can't I do this?

[royal basin]

$\frac{106}{103} \overset?= 102$

[warm shale]

Ann

[rich holly]

what does that mean?

[royal basin]

"does 106/103 equal 102?"

[rich holly]

no?

ohh

yeah, it's because of the x right

[royal basin]

you cannot cancel like you tried to do.

[rich holly]

then why is this possible?

isn't x+6 just x+3 twice?

[timid silo]

No…its 2x+6

[rich holly]

oh right, it would be x² bla bla bla

[royal basin]

do not compose multiplication by two with the composition of a function with itself

[timid silo]

?
No?

[rich holly]

x+3 twice would be (x+3)(x+3) wouldn't it?

that's squared

alright

[timid silo]

No

2(x+3) = 2x+6

[rich holly]

yeah twice means 2(x+3)

[timid silo]

Yeah

[rich holly]

got confused for a bit there lol

[timid silo]

Yeah it’s ok

[obtuse pebble]

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[patent knoll]

My brain is getting fried

[high lily]

what have you tried?

[patent knoll]

what do you mean?

[high lily]

what part of
what have you tried?
don't you understand?

[patent knoll]

what does it mean when its "shorter than twice the length"?

[high lily]

do you know what it would mean for something to be
5cm shorter than 10cm?

[patent knoll]

5cm is the answer to that question?

[high lily]

and how did you arrive at that value

[patent knoll]

subtraction?

[high lily]

yes

and do you know what it means for something to be twice the length of 16cm?

[patent knoll]

32cm?

[high lily]

yes, that "something" will have a length of 32cm

and combine and apply the same ideas to
width is 4 units shorter than twice the length

[patent knoll]

alright let me give a few minutes to get my brain

OH GOD DAM IT

alright my brain is working fine once again

[obtuse pebble]

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[obtuse pebble]

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[obtuse pebble]

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[brittle swan]

hi

[brittle swan]

can someone walk me through finding the area of the loop of x³ + y³ = 3xy?

i converted it into
r³ (sin³θ + cos³θ) = r² sin θ cos θ

not sure where to go next

[dusk nexus]

use the double angle identity for sin

assuming u converted it correctly, u should be able to get $\frac{1}{2} r^2 \sin {2 \theta}$

[warm shale]

sean

[dusk nexus]

uhh

for the other side

yeah nvm.

[royal basin]

so you have $r = \frac{\sin(\theta)\cos(\theta)}{\sin^3(\theta) + \cos^3(\theta)}$

is texit dead

whatever

you have r expressed in terms of theta now

so you can integrate this region in polar coordinates

[cloud patio]

Yea it's dead

[royal basin]

int[0,pi/2] 1/2 r^2 dθ

[timid silo]

Rare lol

[brittle swan]

$9$

[timid silo]

Not working

@warm shale

Its online

$sin(x)$

[brittle swan]

[timid silo]

[brittle swan]

$1+1=3$

[timid silo]

for some reason 1+1=3 is reminding me of that stupid proof of 2+2=5

[brittle swan]

so you have $r = \frac{\sin(\theta)\cos(\theta)}{\sin^3(\theta) + \cos^3(\theta)}$

$3+3=7$

LOOAOADADDADAFSHDJKDK

[raven spire]

._.

!?

Ann gave you to integrate something I suppose?

[tardy epoch]

this help channel has turned into @warm shale shitpost

[raven spire]

hm

[tardy epoch]

$\int \mathrm{Ansh} d\mu(help) = + \infty$

[brittle swan]

why 0 to pi/2?

[royal basin]

that's the range of angles that your loop spans

[brittle swan]

how did you solve for them

[royal basin]

sin(θ)cos(θ)/(sin^3(θ)+cos^3(θ)) = 0

[brittle swan]

won’t π, 3π/2 etc work?

:[

[tardy epoch]

calculate both integrals and see what you get

[royal basin]

consider that pi to 3pi/2 covers the third quadrant

[raven spire]

I believe that's the best approach to chromium questions

make him work more

[brittle swan]

so π, 3π/2 etc aren’t even in its domain?

[raven spire]

can you please write a parametric point on the curve in terms of θ?

[brittle swan]

parametric point?

polars are heavily related with parametrics?

[raven spire]

?

can you elaborate how and what did you convert into that?

[brittle swan]

x = r cos, y = r sin

[raven spire]

also, tf you mean polars? the curve you have is drawn on a cartesian plane, not the argand plane

[raven spire]

And what the heck are r and θ, if not parameters?

[brittle swan]

there lol

[raven spire]

sure, give me the polar coordinates of any point on the curve :o

[restive swift]

Its desmos

Amazing

[brittle swan]

((sin θ cos θ)/(sin³θ + cos³θ), θ)

[raven spire]

plug in π

(x, y) = (0, 0)

which is clearly in your domain

[raven spire]

Answer this though

[obtuse pebble]

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[vagrant dawn]

Can someone tell me wtf I'm doing wrong here? It's telling me to add up the solutions and I get pi. But the correct answer is 4pi, how?

[vagrant dawn]

<@&286206848099549185>

[tardy epoch]

it's just your last line

you didn't divide by 2 in all terms when solving for x

should be x = pi/2 + pi/2 * k like x = 3 pi / 2 + pi k. then check different k values

[vagrant dawn]

man wtf how did I not notice that

thank you

.close

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[long basin]

Ok what abt this?

Am I doing it right so far?

[potent hearth]

I think so

[long basin]

what do Ido next?

Just to clarify, ln3 is the same as log3

correct?

[leaden ibex]

$\ln x = \log_e x$

[potent hearth]

I don’t think it is

[leaden ibex]

right, bot is down

[obtuse pebble]

@long basin Has your question been resolved?

[tardy epoch]

nah rema is right

[obtuse pebble]

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[obtuse pebble]

@idle reef Has your question been resolved?

[obtuse pebble]

@idle reef Has your question been resolved?

[halcyon tulip]

Well you can plug slightly higher values than 2 like 2.001

From x²-4, It should tend to zero as we get closer to 2

[idle reef]

I know but I couldn't do the rest
is the answer 0??

[tardy epoch]

use squeeze theorem

[idle reef]

I know the theorem but ı can't see how can ı use
if u can show me how to start iıcan bring the rest

[obtuse pebble]

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[idle reef]

.reopen

[obtuse pebble]

✅

[tardy epoch]

|cos(y)| <= 1 for all y

[obtuse pebble]

@idle reef Has your question been resolved?

[obtuse pebble]

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[lilac grove]

$n^0=1$

[warm shale]

JUGisMUG

[lilac grove]

How is this possible?

Like why isn't that 0

It should be 1

[brave bramble]

If increasing the power on n is multiplying the number by n

Then decreasing the power on n is dividing the number by n

So, for example, to get from 2³ to 2⁴, you multiply by 2

[flat anvil]

^ n^0 = n^(1-1) = n^1/n^1 = 1

[brave bramble]

But to get from 3² to 3¹, you divide by 3

How do you get from 6¹ to 6⁰?

[lilac grove]

Damn

U divide by 6

6/6=1

$n^0 = n^(1-1) = n^1/n^1 = 1$

[warm shale]

JUGisMUG

[brave bramble]

Can also think of 1 as "The product of no things"

[lilac grove]

Damn

$n^{x-1}=\frac{n^x}{n}$

[flat anvil]

$n^{x-1} = \frac{n^x}{n}$

[warm shale]

JUGisMUG

Shuri 4 honorable (Yottachad)

[lilac grove]

I see

So that statement is correct?

That is correct but it makes me feel weird that I can cancel out the n's and get x, which says 4=3 in this case if n=2, x=3

[royal solar]

seems like it

[royal solar]

think it as n^(1-1)

= n/n

=1

[lilac grove]

Ok

[royal solar]

as long as n is not 0

[lilac grove]

But

[lilac grove]

How this

[royal solar]

where have you cancelled out ns?

[lilac grove]

I can

It is possible isn't it

[royal solar]

logn(

??

you can take log

[lilac grove]

Wait nvm

[royal solar]

but i need you to show me what you’re cancelling from

[lilac grove]

I can cancel x from both sides

n^-1=n/n

[royal solar]

which two sides?

[lilac grove]

Here

[royal solar]

they mean the same thing tho?

it’s just a different form of writing it

it’s like asking you to multiply and divide 6 by the same random number to get 6 again

if that makes sense

[lilac grove]

Bro yes but

[royal solar]

you’re dividing both sides by n^x

[leaden ibex]

or adding and subtracting 1

[lilac grove]

On the RHS

[leaden ibex]

the most useful trick in calculus

[lilac grove]

Can I

Cancel out the n

From the fraction

[royal solar]

sorta, if you’re doing Riemann integration

[leaden ibex]

Ey even when solving integrals

[royal solar]

but you don’t see that very often either

[leaden ibex]

Adding and subtracting a constant can be really helpful

to simplify it

[royal solar]

highschool integration? that’s what i said too

[leaden ibex]

I thought you were referring specifically to solving integrals using the Riemann sum

[royal solar]

no? unless you divide both sides by everything

[leaden ibex]

Now having re-read it I understand

[royal solar]

and get 1=1

[royal solar]

ahh, cool. nws

[royal solar]

so, what is your real question

what do you wish to achieve ?

[lilac grove]

$n^{x-1}=\frac{n^x}{n}$ so here, if I cancel out the n's on both sides (RHS, LHS), what will this become?

[warm shale]

JUGisMUG

[lilac grove]

How will the dxponents be dealt with

[royal solar]

you can’t cancel ns

you know the indices rules right?

[lilac grove]

If there was another n^x-1

Then I could?

[royal solar]

substitute n = 2 and compute the values

see what you get

[royal solar]

it should present on both sides

then yes

but you’ll get 1= 1

not particularly useful

[lilac grove]

2^x-1=2^x/2

ex: x-1=x/1
ex: x-1=x
ex: x-x=1
ex: 0=1

[royal solar]

your math is wrong

[leaden ibex]

x-1=x
no

[royal solar]

you CANNOT cancel exponents like that my friend

[lilac grove]

??

[leaden ibex]

and you getting 0=1 is an indication that you can't

[lilac grove]

Bases are the same

I can

[royal solar]

no

[lilac grove]

Just solve directly in the exponejys

[leaden ibex]

$2^{x-1} = \frac{2^x}{2}$

[royal solar]

you can’t

[warm shale]

Remavas

[leaden ibex]

this is what you have

[royal solar]

yep

and your bases are not the same

you have 2 and 2/logx(2)

is it log or root

it’s root right ?

[lilac grove]

Beo idk log

[royal solar]

yeah so 2/2^(1/x)

[royal solar]

having the same bases and cancelling is the equivalent of taking log to the base of whatever your base is

[royal solar]

that’s the base on your right hand side

having the same base means

having

b^f(x) = b^g(x)

hence f(x) = g(x)

[leaden ibex]

$log_2{2^{x-1}} = log_2{\frac{2^x}{2}} \implies x-1 = x-1$

[warm shale]

Remavas

[leaden ibex]

Unless my 1 am math is wrong

[royal solar]

yep

[leaden ibex]

Ultimately useless

[royal solar]

your math is on point my friend

[royal solar]

yeah i mean. it’s an identity

difference between identity and eqn is identities are valid for all values in the domain

[leaden ibex]

Well yeah, but that's how the base would actually cancel

in the real world 😄

[royal solar]

yeppp

[royal solar]

^^

[leaden ibex]

I haven't been here from the start, so. Is this all about just showing that $n^0 = 1$?

[warm shale]

Remavas

[leaden ibex]

Looking at the pinned message

[royal solar]

that’s what i thought

[royal solar]

i showed him this

as long as n is not 0

this is always true

0^0 has no defined value and you can find what it is when it approaches 0 at your specific curve

[leaden ibex]

I did hear something along those lines, though I haven't had the chance to perform complex analysis myself

Since I am pretty sure in the real numbers it just approaches 1

lim x^x as x to 0

[royal solar]

if y = 0^x, as x approaches 0, it’s still 0
if y = x^0, as x approaches 0, it’s 1

[royal solar]

not always the case you see

that’s why it’s undefined

[leaden ibex]

Ah

[royal solar]

yeee

depends on your curve

but in majority of the cases, you’re right in saying it’s more likely to tend towards being 1

but it doesn’t change the fact it’s undefined in reality.

[leaden ibex]

interesting interesting. At some point after my uni calculus course I should get a book on real analysis. Doubt I'll ever get that studying physics.

[royal solar]

oooh, a physics student.

i wanted to try theoretical physics because i like abstract concepts and stuff

but rn i’m doing maths

[leaden ibex]

I haven't decided quite yet, but probably theoretical too

or possibly computational

[royal solar]

fair plays,

it’s just a me thing tho. i did fairly alright in physics practical stuff. but i much prefer thinking about stuff and using just some paper and pen or a whiteboard to express my ideas

[leaden ibex]

same here, same here. Since I am definitely going astrophysics anyways, I won't be doing anything practical anyways

It will be observational at most

Which I guess is practical in a sense?

[royal solar]

oh damnn. that’s pretty cool

[leaden ibex]

🤷

[royal solar]

it’s still cool

[leaden ibex]

Of course.

The only downside is a slightly less rigorous approach to mathematics

but oh well

[royal solar]

not necessarily, math is everywhere you wish to apply it

[leaden ibex]

Also weird how a probability theory professor ended up teaching calculus to 1st year physics students 🤔

[royal solar]

i am literally taking a year off and helping out medical researches witb maths

[royal solar]

probability tends to be just a phase. it gets heavily monotonous after a point

[leaden ibex]

hm maybe. He does apparently teach probability theory courses in the math faculty

But not for us. Someone else will teach probability theory to us...from the math faculty

Seems weird

But that's how it is

[royal solar]

ahh

which uni?

[leaden ibex]

Vilnius

[royal solar]

it sorta rings a bell, which country?

i live in england btw

[leaden ibex]

Lithuania.

[leaden ibex]

ah, nice.

[royal solar]

ooohhh, nice

[leaden ibex]

Just opposite ends of Europe then

[royal solar]

is that like the top uni in your place?

[royal solar]

if you don’t count russia, indeed

[leaden ibex]

It is. Apparently also the only one which teaches "pure" physics

[royal solar]

wow, that’s pretty impressive

[leaden ibex]

or one of the few

[royal solar]

i got scouted by oxford then i had a disruption during my enterance test and i got rejected because of that so i took a year off and i have offers from university of manchester, bath and warwick

bit of a toughie

[leaden ibex]

oh damn

[royal solar]

yeah, ik. quite a shame really

[leaden ibex]

well still

[royal solar]

have you heard of oxford btw?

[leaden ibex]

lightyears ahead

[royal solar]

it’s one of the top unis here. oxford and cambridge

[leaden ibex]

at least as far as ratings are concerned

[royal solar]

which one?

[leaden ibex]

Well I usually use the QS one

[royal solar]

oh, for uni rankings?

[leaden ibex]

mhm

So Vilnius would be at place 400

[royal solar]

that’s literally rigged tho. i’ll tell you. every top university in the use is based off university of cambridge

but cambridge places 4th

or something like that

[leaden ibex]

I mean I personally don't really care for the rating as long as I can get a good higher ed.

[royal solar]

like, look at the address

it says cambridge

why do you think that’s the case

cambridge university in england has been here since 1209

[royal solar]

to be fair

you’re right

but just saying, every top uni in the us except stanford university is actually inspired by the university of cambridge

[leaden ibex]

Also the tuition is probably 10x higher than cambridge

knowing the us

(maybe not, who knows)

[royal solar]

for residents of the uk, it’s subsidised

[leaden ibex]

A reason I chose Vilnius is the fact that the government here pays all of your tuition if your results are good.

Which is pretty neat

[royal solar]

otherwise you’re looking at 28-29k a year for just tuition fee. after that you have other expenses like your rent and stuff

[leaden ibex]

But it's cheap anyways

like physics costs

hm

[royal solar]

wow, you’re a lucky person and an intelligent person no doubt

[leaden ibex]

3k a semester

[leaden ibex]

You would think so

[royal solar]

pounds? euros??

[leaden ibex]

but they finance hundreds of spots

[leaden ibex]

euro

[leaden ibex]

in physics alone

So it's not that hard to get in

[royal solar]

oh, yeah no. i still gotta pay 9k

[royal solar]

you realise that there are people here who simply didn’t make the cut for physics right?

it’s pretty competitive

and considering it’s the best uni in your country

i’m sure you had some talent

[leaden ibex]

I mean true.

But still.

The lowest standardized grade from 10 that is accepted is around 6 I believe

Since no one takes physics here

Sadly

not really competitive

unlike Medicine

[royal solar]

i couldn’t make to oxford matey, i literally had to take a gap to get back in the game.

[royal solar]

med is cuthroat everywhere i swear

the pay here is pretty garbage for med

and you do like 5-6 years of undergrad then a post grad before you even prescribe someone with cough drops

[leaden ibex]

yeah med is absolute cancer

[royal solar]

like yes, as a med student, you get to do stuff, but you won’t have your own patient till you finish everything

[royal solar]

truee

[leaden ibex]

but people do it for their own drive (which always fascinates me and welcome them) or the money

more likely latter

[royal solar]

yeah, let’s be honest

they do it for the pay more than anything

but the amount of work you do to get that pay is unjustified in my opinion

[leaden ibex]

At least here in Lithuania the strat is: 1) Get a free med degree, government-funded 2) go abroad where they pay you more than most Lithuanians could imagine

^if your grades are good

[royal solar]

america *\ cough *

[leaden ibex]

cough

The UK is also popular

iirc

weirdly enough

[royal solar]

boy, the pay here is absolutely nothing i swear

[leaden ibex]

I mean, what is the monthly pay, more or less

[royal solar]

my parents are both senior consultants and they deserve loads more

you get 40-50 of your money taxed

[leaden ibex]

good old taxes

same here

Income tax and social security insurance are the biggest expenses

[royal solar]

yeah no, you can search up the nhs website for pays, i’ll send it one sec

[leaden ibex]

ah, 6 figures on the high-end for consultants

[royal solar]

yeahh

and 45-50%tax

so you get only 5

[leaden ibex]

As a doctor here you'll make maybe 2k before taxes a month

1k after taxes

give or take

[royal solar]

my mum works overtime for keeping it in 6 figures. my dad runs a few of them so it’s no biggie for him

[royal solar]

boy, that’s harsh unless your living price is considerably lower

[leaden ibex]

I mean it is

[royal solar]

@lilac grove kid, where did you go?

[royal solar]

then, do you think the wages are sorta justified?

[leaden ibex]

In 2020, purchasing power parity for United Kingdom was 0.7 LCU per international dollars

In 2020, purchasing power parity for Lithuania was 0.5 LCU per international dollars.

So our living prices would be 1.4 times cheaper

-ish

roughly

[royal solar]

lcu?

oh

[leaden ibex]

I'm not an economist so idk what LCU is

Living Cost Units?

🤷

What is purchasing power parity?

Purchasing power parity conversion factor is the number of units of a country's currency required to buy the same amounts of goods and services in the domestic market as U.S. dollar would buy in the United States. This conversion factor is for GDP.

[royal solar]

it’s fine matey dw

[royal solar]

0.7 is heavily driven under considering the hospitals are practically free here. it costs money in the us

[leaden ibex]

I mean, free as in deducted from taxes

🙂

But US medical prices are really overboard

should be regulated to a reasonable level

[leaden ibex]

(if they stick with the private model, which imo can work)

[royal solar]

yessir

0.55 in the us, and 2.027 in the uk

[leaden ibex]

ah, petrol prices

I remember when they were 1.5x cheaper

seems like a few years ago

because it was

I disagree

1.55-ish

[royal solar]

price ?

i literally live in north east- merseyside

so 28.2 for me

electricity costs: 38 cents to 10 cents

cost of living in the uk is much higher than what you get in the us

by a lot

but it’s heavily skewed down because tax payed healthcare is seen as a free commodity

[leaden ibex]

europe is just expensive

The east is cheaper, but I am pretty sure the US still has lower living costs

maybe not real estate wise

[royal solar]

yeah no, it defo does

[leaden ibex]

Let's not keep this channel occupied any longer 😉
Nice talk tho

.close

[obtuse pebble]

Channel closed

Closed by @leaden ibex

Use .reopen if this was a mistake.

[obtuse pebble]

Available help channel!

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[brave bramble]

I don't think I've ever seen someone put an integral sign after the function like that

Oh wait that's a bar

I get it now

We use | not ∫ for that

You're correct so far, but it looks to be using part 2, not part 1

[silk condor]

Have to tried chain rule for F(cos x)?

[obtuse pebble]

@nocturne fox Has your question been resolved?

[nocturne fox]

The book says you get the absolute value of sinex times sine x

That’s where im stuck

[silk condor]

1-(cos x)^2 =(sin x)^2

[timid silo]

your answer is equivalent, as mango said, try using 1-(cos x)^2 =(sin x)^2 to get the form you are looking for

[nocturne fox]

is this like a trig thing?

[silk condor]

Yep, $\sin^2 x + \cos^2 x = 1$

[warm shale]

Mango

[nocturne fox]

.close

[obtuse pebble]

Channel closed

Closed by @nocturne fox

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[meager elm]

How do I graph these equations? Demand: P = 300-24Qd and Supply: P = 120+8Qs. First thing I did was S = D. Therefore, 120+8Qs = 300-24Qd. However, I can’t add Qs and Qd because they’re not like terms. I’m stuck on this part.

[meager elm]

Or should I first let P = 0 for each demand and supply equation? And then, what do I do next?

<@&286206848099549185>

[forest sinew]

theyre gonna be mad at you for pinging helpers so soon

[meager elm]

@forest sinew I’m sorry, I didn’t know

Or should I do this method: P = 300-24Qd… where P + 24Qd = 300. Idk what to do next. Same goes for P = 120+8Qs… where P - 120 = 8Qs. Also stuck here

[idle thunder]

no helper has pings on because of how many people ping anyways

[forest sinew]

ah word

[leaden ibex]

I've learnt to deal with hundreds of pings

the secret to staying sane is DnD

[forest sinew]

i have no idea how to do this problemb

i did look though

[meager elm]

😭

[forest sinew]

wait

you have numbers for qs and qd?

oh shit my ride is here

🙏 good luck

[meager elm]

Not sure if this is similar to y = mx+b formula

Thank you🙏🏻

[leaden ibex]

$P = 300-24Q_d$ for demand, $P = 120 + 8Q_s$ for supply

[warm shale]

Remavas

[leaden ibex]

So wait, what is your goal?

[meager elm]

I have to graph those two equations, but in order to do that, I have to figure out price, Qd, and Qs I’m assuming

If that makes sense

[leaden ibex]

hm