help-19

[mystic saffron]

u can simplify the ln first

[deep silo]

So ln(2t) - ln(t)?

[mystic saffron]

yes

[deep silo]

Ln(2)?

So the answer is just ln(2)

Ad t goes to infinity the 1/2t and 1/t will gets smaller and smaller

[mystic saffron]

yes

[deep silo]

So the answer is just ln(2)?

[mystic saffron]

yea

[deep silo]

One question

Could we do this with substitution to?

Or would it be harder for us ?

[mystic saffron]

lemme see

[mystic saffron]

i think yes

[deep silo]

can we really take ln(2t) - ln(t) to be ln(2)?

[mystic saffron]

yea why not ?

[deep silo]

just formally

she did ln(2t/t)

but its the same thing

we get left the ln(2)

[mystic saffron]

u did same, just skipped a step

[deep silo]

what step would that be?

[mystic saffron]

this one

ln(a) - ln(b) = ln(a/b)

[deep silo]

ohh ok

[mystic saffron]

do u have other integrals or stuff ?

[deep silo]

Ur good at geometric sums?

[mystic saffron]

ye im ok

[deep silo]

I suck at it

like really bad

but the first question

is just use the definition+

?

[mystic saffron]

definition of what ?

[deep silo]

well the sum is a geometric series of r= x/4

[mystic saffron]

ye so it converges only if absolute value of r <1

so absolute of x <4

[deep silo]

and that will happen if x =4

[mystic saffron]

diverge

bc u sum 1 to infinity

[deep silo]

ye

wait

[mystic saffron]

ok i wait

[deep silo]

1- x/4 it says

[mystic saffron]

• or +

[mystic saffron]

ok right

[deep silo]

I don’t understand what I wrote

Really

[mystic saffron]

so u have a formula that tell u the sum of all the terms of a geometric sequence

it is first term* 1/(1-r)

only if the sum is infinite and converge

[mystic saffron]

ye thats it

[deep silo]

wtf does it tell me?

[mystic saffron]

the first equality

1+r+... +r^n

this is ur sum

but with r=x/4

[deep silo]

so this is 1 + x/4 + (x/4)^2...

[mystic saffron]

yes

[deep silo]

this is the geometric series

[mystic saffron]

ja

[deep silo]

swedish?

[mystic saffron]

no

[deep silo]

oh okay

haha

but what does converge means really?

[mystic saffron]

it means that ur sum tend to a finite value

[deep silo]

ohhhh

[mystic saffron]

not inf or - inf for example

[deep silo]

so a value that has a "limit"

or it has defined values

[mystic saffron]

just this

[deep silo]

okey..

[mystic saffron]

so now u know f(x)

[deep silo]

yes

and we know that if they take the values -4 or 4

the x

the denominator will be 0

[mystic saffron]

it diverge first so u cant get the denominator

[deep silo]

ok

but if this series takes the value 4

4/(4-x) = 4/0

and 4/(4+4) = 2/4

[mystic saffron]

it will diverge and so f(4) undefined

thats all

[deep silo]

0.o

how

will it be undifined

[mystic saffron]

bc the serie diverge

so u cant calculate sum of sigma 1

[deep silo]

4/4 is undefined ye

is that how you thought?

[mystic saffron]

4/4 = 1

and sigma 1 is undefined

bc it tend to inf

[deep silo]

😦 this mindset is to hard for me

Imma need to watch a video i think in order to grasp this

[mystic saffron]

what dont u understand ?

[deep silo]

first question

show that f(x) is defined on the interval (-4,4)

we know that r= x/4

how could you from this tell me that it will converge

[mystic saffron]

is is because if x=4 u do 1+1+1+..... till infinity so it will diverge with x=4

[deep silo]

ohhhhhhhhh

and -4

-1-1-1-1-1-1-1

[mystic saffron]

-1-1-...

=- inf

[deep silo]

Righht

[mystic saffron]

i see ur question

[deep silo]

but what does the other functions before tell me ?

just from quriosity

what do i do with that

before the geometric sum

[mystic saffron]

if n= inf, the r^n tend to infinity only if absolute value of r is <1

[mystic saffron]

uh it is just a part of the function, no problem on this side

[deep silo]

ok

[mystic saffron]

so u understand ?

[deep silo]

I think so ye

so it will actually be -1-1^2-1^3-1^4...

and it will just keep going

until - infinity

no wait thats wrong

[mystic saffron]

yes

it will be +..-..+..-..

if x<0

[deep silo]

oh ok

but as answer do i just say that if X= -4 it converges and if x=4 it converges aswell? xd

Does it converge on both sides?

[mystic saffron]

it diverge on both side if x=4 or x=-4

[deep silo]

ok

[mystic saffron]

not converge

[deep silo]

oh

but u said converge?

[mystic saffron]

where ?

[deep silo]

oh no

sorry

[mystic saffron]

really all come from this formula, the fact that it diverge or converge

it is becaus of the r^n

when n goes to infinity, r^n converge only if -1<r<1

and so r^n tend to 0

[deep silo]

and our r will be = -1 and =1 which lead to diverge?

[mystic saffron]

yes if x=4

[deep silo]

ok

so in x values, the function is defined if -4 < x <4?

[mystic saffron]

yes

[deep silo]

and thats becuase if x is either those values, our r will be =1 and =-1 and will tend to + inf and - inf

[mystic saffron]

exactly

[deep silo]

u are such a lifesaver

I really really thank you alot

[mystic saffron]

np

and for the next question u know how to do ?

[deep silo]

take the function = 0 ?

derivate first*

the derivative of the function = 0

and found the critical values that in the boundaries

is in*

[mystic saffron]

ye looks cool

[deep silo]

But the thing that hunts me again..

the geomtric sum

wtf do i do with that?

i just take the function before times the x/4?

[mystic saffron]

u know what the sum is equal to when -1<x<1

since the function is defined between -4 and 4

th sum just is 1/(1-x/4)

[deep silo]

when -1<x<1?

[mystic saffron]

no imeant -4 and 4

[deep silo]

oh ok

ye

So just this function times the 1/(1-x/4)

[mystic saffron]

yea

[deep silo]

ohh ok

but that is also like saying 4/(4-x)

[mystic saffron]

yes

[deep silo]

and how come really

that its the same thing

[mystic saffron]

what ?

[deep silo]

How can we rewrite 1/(1-x/4) to 4/(4-x)

we can take it times 4?

so 14 /(14 - x)

[mystic saffron]

wdym ?

[deep silo]

is tha that how we got = 4/(4-x)

we multiply with 4

[mystic saffron]

yes

[deep silo]

ohh ok

and then this would be easier ofc to work it

[mystic saffron]

si

[deep silo]

so 4* the whole equation / (4-x)

and just use qutient rule

[mystic saffron]

idk what u mean but this is good

[deep silo]

?

to get x free from x/4

we can multyply by 4

thats how we get the 4/(4-x)

or?

[mystic saffron]

if u try to explain what u did on the photo i agree

[deep silo]

ye xd

I just wrote it very fast

Okey then imm on it, ill try to do it on my own and stop harrasing you! thank you so much, I think im getting a grip of it. Ill try alone, see ya and thank you

[mystic saffron]

ok cya

[deep silo]

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[brave torrent]

Quick question :
What are these equation? I'm not sure to recognize them.

[brave torrent]

They seems to be some formlaes deriveas

[faint knot]

do you know about chain rule?

[brave torrent]

But I thought that I should write dx/dy?

[faint knot]

not correct

do you know about chain rule?

[brave torrent]

I'll need a refresher.

[faint knot]

theres a general formula for d/dx f(g(x))

d/dx f(g(x)) = f'(g(x)) g'(x)

for example if you have d/dx sin(x^2)

your result is cos(x^2) 2x

do you understand?

[brave torrent]

Yes I do.

So basically, these are the formulaes for the derivaes, but written in a chain rule?

[faint knot]

no

[brave torrent]

Well I know that the derivae of x^4 is 3x^2

4x^3*

[faint knot]

do you recognize the chain rule portions of these formulas though?

[brave torrent]

Like this: ?

[faint knot]

yep

now the purpose of these more specific formulas is in case you dont remember chain rule

they work out three examples that would be the result of chain rule

however you shouldnt memorize these

its easier to memorize chain rule which applies for these cases anyways

[brave torrent]

But say I don't have the chain rule, like...

I need to apply hospital rule.

Oops n should be x.

f(x) = 2x for example?&

[faint knot]

your formulas give you a more restricted version of chain rule, so youll be able to differentiate those

[brave torrent]

Say I have to derivate 6^log_4(x)

Would I use this formlae?

Oops, not this one

[faint knot]

formula, not formulae

[brave torrent]

This one.

Oops. Thank you.

[faint knot]

sure, as an example

[brave torrent]

I would need to use the 2nd formula, then I would have to use the first one.

[faint knot]

formula

[brave torrent]

It was a typo, my bad.

[faint knot]

yes

[brave torrent]

Gotcha. Merci!

.close

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[mystic saffron]

the solution to this equation should be x1= 9/2 and x2 = 25/8, instead it gives me an imaginary number, what am I missing?

[hybrid bone]

you don't get rid of the logaritm when carrying the coefficient as an exponent to inside of the logarithm

[mystic saffron]

interesting

.close

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[hybrid bone]

oh wait

I am in the wrong I didn't notice that you took the base 2 of both sides

[mystic saffron]

.reopen

[odd edge]

✅

[mystic saffron]

yeah i checked again and it still the same

i just skip that

[hybrid bone]

the quadratic you have gives the roots 9/2 and 25/8 as you said it should

[mystic saffron]

Fr?

My calculator its bad maybe

[hybrid bone]

try wolframalpha

[hybrid bone]

huh...

am I blind or isn't that the same thing

[mystic saffron]

Yep

Ig we miss the analysis

Idk

[hybrid bone]

it makes sense when you think about the parabola being turned upside down it changes from not intersecting the x axis to do intersecting it

but it never accoured to me that you could get no real solutions if you decide to move terms to the one side of the equation as to the other

[mystic saffron]

It was 225

[random summit]

ooh i see the problem you did + 255 and he did + 225 that why you two get different answers

[mystic saffron]

Not 250

Ye

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[hybrid bone]

I feel stupid

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[mystic saffron]

What about me

[mystic saffron]

.close

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[quasi sparrow]

,calc 2^7 / (3^4 * 12)

[clever fjord]

Result:

0.13168724279835

[quasi sparrow]

,calc 128/972

[clever fjord]

Result:

0.13168724279835

[quasi sparrow]

,calc 32/248

[clever fjord]

Result:

0.12903225806452

[quasi sparrow]

there you go

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[cursive meadow]

My answer in question 25 is 3 times the actual value. Anyone see where I went wrong? Did I misapply chain rule?

[cursive meadow]

Should be 7/root5 not 21/root5

[quasi sparrow]

mistake here

you sorta differentiated twice

g'(x) = 3 is correct

but you combined that with $\frac{d}{dx}[f(g(x))]$

[clever fjord]

riemann

[quasi sparrow]

you either do one or the other, not both

[cursive meadow]

Wdym

[quasi sparrow]

which part is confusing

[cursive meadow]

The fact that I double differentiated

OHH

[quasi sparrow]

you differentiated both f(g(x)) and g(x) and combined them

[cursive meadow]

because the new function within the composed would no longer be (3x-2)??

[quasi sparrow]

maybe you just don't understand chain rule

[cursive meadow]

Nah I think I get it now

[quasi sparrow]

$f'(g(x))$ is the derivative of $f(x)$ evaluated at $g(x)$

[clever fjord]

riemann

[quasi sparrow]

but you differentiated the whole $f(g(x))$ and multiplied it with $g'(x)$

[clever fjord]

riemann

[cursive meadow]

Yeah bc the inner function is no longer (3x-2)

Tysm for showing me I was staring at it for way too long

[quasi sparrow]

[cursive meadow]

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[boreal crag]

Do you know the slope formula

[willow plover]

yes

[boreal crag]

Apply it

[willow plover]

then ?

[boreal crag]

Then solve for r

[willow plover]

ok

r-5 = ?

hello ?

[strange jetty]

Which question are you doing?

Or are you asking for general concept to solve these questions?

[willow plover]

42

[strange jetty]

Have you solved the other questions?

[willow plover]

no

[strange jetty]

Okay

[willow plover]

I dont know how to solve them

[strange jetty]

Well all the questions are the same concept

The formula for slope is dY/dX

[willow plover]

what id dy/dx

is*

[strange jetty]

Which pretty much means you do y1-y2 divided by x1-x2

[willow plover]

oh

yeah i know this

[steep mantle]

then plug in the values

[strange jetty]

so for question 42

(r, 3) is one point

(5, 9) is another point

each point is written as (x , y)

[willow plover]

yeah

[strange jetty]

so, you use the equation

[willow plover]

9-3

and

5-r

?

[strange jetty]

Yes

[willow plover]

9-3 is 6 but what ir 5-r

this is what i dont get

[steep mantle]

you dont know what 5-r is

[willow plover]

yeah i dont

[steep mantle]

but you do know that $\frac{6}{5-r}=2$

[clever fjord]

Skill_Issue

[willow plover]

ohhhh

okay now i do understand

thank you

[steep mantle]

im sure you can solve it from here, nunber 44 is the same

[willow plover]

3-2 over 6- r = 1 over 2

for question 44

right ?

[strange jetty]

Ya

[willow plover]

wow

[strange jetty]

You pretty much just spam the formula

(y1-y2)/(x1-x2) = m

it's as braindead as that

then solve for r

all these questions are the same

[willow plover]

cool

thanks !

[strange jetty]

Good luck

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[red socket]

can anybody help with some quantum mechanics, i’m trying to normalize a wave fnx that depends on the first and second stationary states

[red socket]

after using the kroneker delta to factor out the integrals i am left with modulus squared of c1 plus mod squared c2

is this then multiplied by A^2 and set equal to 1?

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@red socket Has your question been resolved?

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[short lantern]

implicit differentiation

i keep getting $$\frac{6x^2-5}{10y}$$

[clever fjord]

Devil Wears Prada

[short lantern]

but the answer key says otherwise

how do i do this correctly

[nocturne belfry]

what is $\dv x \qty( 5y^2 )$?

[clever fjord]

jan Niku

[short lantern]

10y(dy/dx)

[nocturne belfry]

and 2x^3?

[short lantern]

6x^2]

[untold blade]

power rule

[nocturne belfry]

how about 5y

[short lantern]

5

[nocturne belfry]

no

[short lantern]

5(dy/dx)

?

[nocturne belfry]

yea

[short lantern]

oh

wait how do u simplyify this then

[nocturne belfry]

collect up all of the dy/dx

[short lantern]

(dy/dx) + (dy/dx)

what is that

2(dy/dx)?

[nocturne belfry]

$10y \dv{y}{x} = 6x^2 - 5\dv{y}{x}$

[short lantern]

yes

[clever fjord]

jan Niku

[nocturne belfry]

so collect all of your derivatives

[night dragon]

then put dy/dx on the same side

[short lantern]

like wdym

[nocturne belfry]

maybe you do something like

[short lantern]

yes they are

but what happens when its

[nocturne belfry]

$10 y \dv{y}{x} + 5 \dv{y}{x} = 6x^2$

[clever fjord]

jan Niku

[short lantern]

(dy/dx) + (dy/dx)

yes

[night dragon]

factor then

[short lantern]

but when its

added

[nocturne belfry]

ye

factor

[short lantern]

what happens

what

factor what

[nocturne belfry]

you factor out the derivative

its just a function

[night dragon]

factor dy/dx then solve

[nocturne belfry]

$\dv{y}{x} \qty( 10y + 5) = 6x^2$

[clever fjord]

jan Niku

[night dragon]

yup

[short lantern]

oh

[night dragon]

from there its easy to get dy/dx by itself

[short lantern]

thanks

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[short lantern]

.reopen

[odd edge]

✅

[short lantern]

i got $$8x = 6y\frac{dy}{dx}^2 + 4(\frac{dy}{dx})$$

[clever fjord]

Devil Wears Prada

[short lantern]

what to do now

[nocturne belfry]

you need to isolate

wait up

this is not right

$\dv x y^3$

[clever fjord]

jan Niku

[nocturne belfry]

what is it @short lantern

use the chain rule

[short lantern]

hi

um

oh

is the answer $$\frac{dy}{dx}=\frac{8x}{6y^2+4}$$

[clever fjord]

Devil Wears Prada

[short lantern]

i did chain rule wrong i noticed

but this is right

.close

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[short lantern]

.reopen

[odd edge]

✅

[short lantern]

i got this one good

but

how do i use the chain rule on -3xy

which is the outside function and which is the inside function

[thick shell]

implicit differentiation

[short lantern]

yes

how to use chain rule on -3xy

which is outside and which is inside function

[thick shell]

you use product rule

[short lantern]

that is just wrong

wait

[thick shell]

-3xy is not a composite function

[short lantern]

no

right

thanks

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[short lantern]

.reopen

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✅

[short lantern]

i got $$\frac{dy}{dx}=-\frac{4xy^2}{2x^22y}$$

[clever fjord]

Devil Wears Prada

[short lantern]

did i do it right

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[radiant garnet]

Why does i*i = -1

[radiant garnet]

If i is root -1

When you multiply square roots its just like normal multiplication

So wouldn’t it just be

Root 1

= 1

[amber schooner]

i * i = i^2 = (sqrt(-1))^2 = -1

[full marsh]

Because squares cancel out roots

And it's a square bcs ur multiplying by itself

[radiant garnet]

So is multiplying the inside of the square root wrong

[full marsh]

Inside?

[radiant garnet]

Like the way I remember

[amber schooner]

he means

[radiant garnet]

When we multiply two square roots together

it’s like root 2 * root 2

root 4

[full marsh]

Woah

Okay

What is root 2

Round it

[amber schooner]

$\sqrt{-1} \cdot \sqrt{-1} = \sqrt{-1 \cdot -1} = \sqrt{1} = 1$

[clever fjord]

knief

[radiant garnet]

Okay good point that wouldn’t equal root 4

Weird so why do I see that method so much

Fuck

[full marsh]

It's fine

[radiant garnet]

Like even just searching online how to multiply root by root

That’s how they seem to do it

[amber schooner]

wrong

[full marsh]

No I meant

Root 2 * root 2 won't be root 4

[amber schooner]

$\sqrt{2} \cdot \sqrt{2} = \sqrt{4} = 2$

[clever fjord]

knief

[amber schooner]

it is

[full marsh]

Hold on chat

[radiant garnet]

Wait but

[amber schooner]

2 is positive

[radiant garnet]

If you do root 2 individually

And then multiply the simplified root 2a

2s

You wouldn’t be getting the answer of root 4

They aren’t similar right

So how is it possible

[amber schooner]

huh

[radiant garnet]

Wait

[amber schooner]

yes you would sir

[radiant garnet]

Hmm

[full marsh]

He's right

[amber schooner]

who?

[radiant garnet]

Root 2 is like

[full marsh]

U

[amber schooner]

i know

[full marsh]

Because squares cancel out roots

[amber schooner]

brother you can’t round it

💀

if you round it it is no longer root 2

[radiant garnet]

so its an irrational number

[amber schooner]

yes

[full marsh]

For the sake of prooving

[radiant garnet]

okay so lets say

root 4 * root 4

[full marsh]

If you had that point u would say dx is not 0

[radiant garnet]

= root 16

2 * 2 = 4

[amber schooner]

that’s root 16 which is 4

[radiant garnet]

root 16 = 4

[amber schooner]

more generally;

[radiant garnet]

so the multiplying inside the root works

[amber schooner]

$\sqrt{a} \cdot \sqrt{b} = \sqrt{ab} \quad a, b \geq 0$

[radiant garnet]

hmm

[clever fjord]

knief

[radiant garnet]

so why does root - 1 and root -1

not equal root 1

-1 * -1

[amber schooner]

because by convention we only consider root 1 to be 1

we take the positive root

and neglect the negative root

however

[radiant garnet]

oh root 1 technically has like 2 answers right

but we only consider the 1 answer which is 1?

1 and -1

fuck im still a bit confused

like in this example of -i * i

its like -1 root -1 * 1 root -1

-1 root 1

but apparnelty i * i does not equal that positive

[amber schooner]

what

you want to find -i * i?

[radiant garnet]

yes

but first i want to understand why

i * i = -1

and not 1

cause obviously if i know that i guess itll be easy to find -i * i

[amber schooner]

$(\sqrt{-1})^2 = -1$

[clever fjord]

knief

[radiant garnet]

i understand that using squares cancels out the square root

[amber schooner]

^

[radiant garnet]

but why cant we do the multiply inside by the inside strategy again?

[amber schooner]

i explained already

only if the two numbers are nonnegative

[radiant garnet]

so the answer of root ab just has to be positive

not allowed to be negative

if its negative what would we do? just reverse the signs or

[amber schooner]

again, the rule youre referring to only works when the two numbers are nonnegative

[amber schooner]

^

as in, the two roots are real

[radiant garnet]

so we can only do the inside mutliplication thing when the numbers are positive

real numbers

what is th erule called when tis negative numbers

[amber schooner]

what is the rule called?

wdym

[radiant garnet]

i guess im just trying to understand why its not allowed to do that method when the numbers are negative

[amber schooner]

because by convention when we take the sqrt of a number we only consider the positive root

we say sqrt(4) = 2

not -2

even though

(-2)^2 = 4

[radiant garnet]

okay

lemem write it down one sec

[amber schooner]

if you multiplied the inside with two negative numbers you’d get a positive

which would lead us to this scenario

[amber schooner]

^

however the two sqrts on their own aren’t defined for real numbers

we can’t take the sqrt(-1) and get a real outcome

[radiant garnet]

so it has to be = i

because there is no real otucome

fuck idk why im struggling so much with this

like say i have root -2 * root -2

if i tried to use my root a * root b method it would be root ab which would be 4, but root -2 is not a real number

so that isnt a possible solution

i have to incorporate the i

so like ig it also technically looks like

[amber schooner]

yea or some multiple times i

[radiant garnet]

i root 2

i root 2 * i root 2

[amber schooner]

mhm

[amber schooner]

again it only works for positive inputs

you fw grilled cheese sandwiches?

[radiant garnet]

yes

aloot

[amber schooner]

fire

[radiant garnet]

maybe all the grilled cheese lowered my intelligence

[amber schooner]

damn

[radiant garnet]

so if root -2 is technically equal to

i root 2

[amber schooner]

yes

it is

[radiant garnet]

i guess im back at the issue i had before where im trying to figure out how to represent root - 1 times root - 1

because those 2 are imaginary

even if i multiply inside and pretend i get 1, that isnt a possible solution

so i have to say

i 1

i = root 1

root -1*

[amber schooner]

$(-1^{\frac{1}{2}})^2$

[radiant garnet]

root -1 * 1

[clever fjord]

knief

[radiant garnet]

= -1

i mean shit i guess i found a solution 😭

this is so weird soundiung

but say i got

i root 2 * i root 2

i = root -1

[amber schooner]

yea

[radiant garnet]

so (root-1)(root2)*(root-1)(root2)

those two negative roots multipled togehter with my method would be 1 but thats not possible theres an imaginary number in there because root -1 isnt possible to begin with

so i could call those two multipled together i root 1

[amber schooner]

stop saying your method please it’s just breaking rules of algebra

[radiant garnet]

ik my shitty method

that is wrong

[amber schooner]

it isn’t a method to begin with

[radiant garnet]

but its the way my brain is stuck on

[amber schooner]

it’s just false

[amber schooner]

?

[radiant garnet]

i * i = i^2

root -1 ^ 2

so just -1

[amber schooner]

yup

[radiant garnet]

its also 2i

technically speaking ig

i = root - 1, root - 1 is impossible there is no real number that squares to -1

[amber schooner]

wyat

[radiant garnet]

is it not technically

[amber schooner]

x^2 ≠ 2x

[radiant garnet]

oh yknow what ive been thinking about it likes it a fucking

real number

when its not

thays why its so hard for me to wrap around it

[amber schooner]

how is i times i 2i?

[radiant garnet]

if i pretend its x then that makes alot more sensew

its not

[amber schooner]

i + i = 2i that’s true

go watch a series on imaginary numbers

hold

https://youtu.be/T647CGsuOVU?si=X3fyBle5n3QIUS64

great series

maybe it’ll demystify

[radiant garnet]

ok

thank you for the help

.close

[odd edge]

Channel closed

Closed by @radiant garnet

Use .reopen if this was a mistake.

[odd edge]

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[midnight spindle]

Although this is physics, I don’t understand this at all, and I’m desperate to understand this

[midnight spindle]

Help please 🙏🏻

[deep mason]

ok

the points represent position

like camera frames of a movie

at equal time intervals

so if you try to visualize this

you can maybe imagine what the velocity should be

or also

you can just use math

[midnight spindle]

I don’t really understand that

[deep mason]

ok

maybe math then

or logic

in points C and D, it is on the way upward, yeah?

so it's slowing down

so is speed greater at C or D

[odd edge]

@midnight spindle Has your question been resolved?

[midnight spindle]

I also have another one like this

[deep mason]

yeah that sounds similar

acceleration is always the same for projectiles in the air

it's always downward with the same magnitude, g = 9.8 m/s^2

so answers are

B up, down

D zero, down

E down, down

[midnight spindle]

If you don’t mind me asking, would it be possible to join one of the vc’s so I could get a better understanding? If not that’s completely fine.

Through text is pretty confusing

[deep mason]

i'm sorry i can't tonight

[midnight spindle]

Ah Alr np

[deep mason]

another day possibly

[midnight spindle]

Tmmr?

[deep mason]

maybe 🪦

feel free to add me. idk when i'll be able to, right now is a weird time

[midnight spindle]

Yeah yeah np, tysm though

I’m pretty desperate to get this stuff since my teacher practically doesn’t teach

[deep mason]

that happenss

i can still answer in text for now anyway

the whole picture in your head should be of a ball that starts being thrown upward, moves up, and then falls back down

btw ping me if i'm not here

[odd edge]

@midnight spindle Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

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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.

[glossy plover]

I would like to know how it is possible to go from the 1st line to the 2nd line. It's easy go from the 2nd to the 1st. But I can't figure out a way to go from 1st to the 2nd.

[glossy plover]

I need a simplification of this math.

[uneven osprey]

by using partial fraction decomposition

[glossy plover]

I've tried it but can't get to the 2nd line

[steady tide]

its kinda a shortcut here

[thin kelp]

jesus, ur profile scared me until i saw it was facing the left

[thick shell]

factorize denominator

[glossy plover]

Lol. It's from an anime

[cursive field]

you know it's not just a nazi symbol? even the hindus use it as a religious mark

[steady tide]

they were referring to the manji

[glossy plover]

Can anyone do the math and post a photo? I would really appreciate it

[steady tide]

!noans

[odd edge]

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

[thin kelp]

im fully aware, thats why i pointed out that it was facing left

[steady tide]

anyway, can you show your work regarding your fraction decomposition?

[glossy plover]

Been trying it this way

But can't seem to go anywhere

I just can't find a way to get to the 2nd line

[steady tide]

eh, no

you might want to take a look at this

[odd edge]

@glossy plover Has your question been resolved?

[odd edge]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[odd edge]

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.

[obtuse light]

sorry for low quality

[weary pelican]

you know gronwall's inequality?

[obtuse light]

eh no?

[obtuse light]

jesus that thing can be written into a whole document

but i think i can turn the original inequation into that

[weary pelican]

I mean it helps to show f(x) >= 1

[obtuse light]

huh?

f(x)=1?

It's hard to consume new thing

I don't think this is a normal question for average calculus 1-2 student

[weary pelican]

that's good

[weary pelican]

I don't think that's it

[obtuse light]

idk

I tried dumb stuff bc f(x) >=1

[weary pelican]

yeah the problem is that not every f(x) >= 1 function works

if you focus on f(x)-1 = g(x)

there are tons of functions such that g' + ag >= 0

[obtuse light]

True

[weary pelican]

but if you account for g(0) = g(1) = 0

mmh

[obtuse light]

I mean i'm trying to solve ridiculous question for my prep at college olympiad

so learning new thing is useful

[weary pelican]

stuff like e^(1/(1-4(x-1/2)^2))

I don't know if this works

[obtuse light]

I don't think that will work well

[weary pelican]

maybe not but lemme check

g' = -8(x-1/2)/[(1-4(x-1/2)^2)]^2 * g

so g'+ag = (a-8(x-1/2)/[(1-4(x-1/2)^2)]^2)*g

wait

[obtuse light]

I think the g(x) may be like this

g(0) =< g(x) =<g(1)

we can prove this quicker

[weary pelican]

oh wait I have an idea

[weary pelican]

I'm really not sure

that would imply g(x) = 0 everywhere

and I believe there are solutions

[obtuse light]

Yeah i did ask a really hard question

[weary pelican]

does gronwall give us something if we consider g(x) = f(1-x) instead

[obtuse light]

nah i don't think that will work

[weary pelican]

I'm gonna try tho

[obtuse light]

How about g(x) = (e^2003x/2004)(f(x)-1)

[weary pelican]

$2003g'(x) = -2003f'(1-x) \leq 2004f(1-x) - 2004 = 2004g(x) - 2004$

[clever fjord]

rafilou is not not born in 2003

[weary pelican]

we already know f(x) >= 1

won't bring us more

ok so 2003g'(x) - 2004g(x) <= -2004

consider h(x) = g(x)-1

2003h'(x) - 2004h(x) <= 0

AH there we go

the proof we needed that f(x) <= 1

[obtuse light]

f(x) is actually equal to 1

and we need atleast 2 paper to proof it

God damn

[weary pelican]

we need to use gronwall on both sides (x = 0 and x = 1)

[obtuse light]

yeah

I really don't think it is a calculus 1-2 student's question

but gronwall is actually good for the inequality

[patent matrix]

You need help?

Nah ur good

See ya

[obtuse light]

yeah

.close

[odd edge]

Channel closed

Closed by @obtuse light

Use .reopen if this was a mistake.

[odd edge]

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.

[kindred nest]

is this a thing?

[verbal python]

Yes

[inland rover]

yes

[kindred nest]

what's the concept behind it

I'm trying to find ways to remember it

though simple

[inland rover]

the properties of powers

[verbal python]

For positive integer exponents this is simple since

But for real exponents you need to prove some properties

[cursive field]

$\left(\frac 42\right)^2 = \left(4\cdot 2^{-1}\right)^2 = 4^2\cdot 2^{-2} = \frac{4^2}{2^2}$

[clever fjord]

ren

[mystic saffron]

I mean in a way

actually your explanation is good too

[kindred nest]

alright I see it now

I'll keep this in mind

ty everyone

[mystic saffron]

we all know the actual answer is cause of the illuminati 🔺 👁️

but no one would believe me if I said that

[cursive field]

in general, $(ab)^x = \underbrace{\left(a\cdot a\cdots a\right)}{x\text{ times}}\right) \underbrace{\left(b\cdot b\cdots b)\right}{x\text{ times}} = a^x b^x$

[inland rover]

this reminds me of rudin i dont like it.....

[cursive field]

idk who rudin is

oops-

uh

[inland rover]

[cursive field]

sh

[inland rover]

$\text{\mcr{red}{cooked}}=ren$

[clever fjord]

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

[mystic saffron]

@cursive field your latex be so fancy

[cursive field]

what NOW texit

[cursive field]

no not really

[mystic saffron]

okay :(

if you say so

[cursive field]

in general, $(ab)^x = \underbrace{\left(a\cdot a\cdots a\right)}{x\text{ times}} \underbrace{\left(b\cdot b\cdots b)\right}{x\text{ times}} = a^x b^x$

[clever fjord]

ren

in general, $(ab)^x = \underbrace{\left(a\cdot a\cdots a\right)}_{x\text{ times}} \underbrace{\left(b\cdot b\cdots b)\right}_{x\text{ times}} = a^x b^x$
```Compilation error:```! Missing delimiter (. inserted).
<to be read again> 
                   }
l.1421 ...nderbrace{\left(b\cdot b\cdots b)\right}
                                                  _{x\text{ times}} = a^x b^x$
I was expecting to see something like `(' or `\{' or
`\}' here. If you typed, e.g., `{' instead of `\{', you
should probably delete the `{' by typing `1' now, so that
braces don't get unbalanced. Otherwise just proceed.
Acceptable delimiters are characters whose \delcode is
nonnegative, or you can use `\delimiter <delimiter code>'.```

[cursive field]

WZKJXEHC DUHXXUIEC SC'

[mystic saffron]

In general, $(ab)^x = \underbrace{\left(a\cdot a\cdots a\right)}{x\text{ times}} \underbrace{\left(b\cdot b\cdots b\right)}{x\text{ times}} = a^x b^x$

[clever fjord]

Edmund Cloudsley

[cursive field]

@kindred nest are you done? if so, do .close

bruh.

[mystic saffron]

I love how this became #latex-testing

[kindred nest]

I let you guys have fun

this is so cooked

.clo9se

.close

[odd edge]

Channel closed

Closed by @kindred nest

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[odd edge]

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.

[warm seal]

REQUESTING HELP

[mystic saffron]

which part and what have you done uptill now?

@warm seal

[warm seal]

none

[mystic saffron]

well which specific part are you having a problem in?

[warm seal]

all of it

idk where to start

[mystic saffron]

okay let

us

tackle this part by part

starting a

we know that in total, the people spend $$11.2 \text{ tril.}$ in a year on themselves

[clever fjord]

Edmund Cloudsley

[mystic saffron]

and we have to find the per capita consumption

what does the word per capita mean in your understanding?

[odd edge]

@warm seal Has your question been resolved?

[warm seal]

i have no idea

[odd edge]

@warm seal Has your question been resolved?

[mystic saffron]

per capita means per person

[warm seal]

alright so i’m going to using division?

[odd edge]

@warm seal Has your question been resolved?

[warm seal]

<@&286206848099549185> you guys are slackin

it’s been 2 hours

[mystic saffron]

I mean that’s no way to talk to the helpers

We all are just volunteering here

We do have real life commitments and responsibilities

Not sure if I want to help you anymore :/

[floral stratus]

Someone replied with an in depth explanation and you chose to move to a new channel 😭

[warm seal]

bro they responded while i was asleep and it closed tf u want me to do

[mystic saffron]

I feel like he/she is just trolling

Mate then probably be nicer to helpers

[warm seal]

i’m not being mean

[mystic saffron]

Hmm okay ⬆️

[warm seal]

ur a lil snowflake if it comes off that way

[mystic saffron]

Also mate

You cannot expect us to help you with an entire worksheet

U must try to do some work yourself as well

[warm seal]

i can’t do that if i dont know where to start

[mystic saffron]

U might wanna revisit your course content then

[floral stratus]

Oh okay

[mystic saffron]

<@&268886789983436800>

[floral stratus]

It definitely did not come off that way right now!

[warm seal]

i mean yeah it should dumbass

glad u understand

[pulsar elbow]

don’t be rude to other users please

[mystic saffron]

Doesn’t sound like you really do want any help lol

[magic idol]

You know that in a particular year, all 275 million people in a country combined spent around 11.7*10^(12) dollars on personal consumption.
'Per capita' means 'per person'.
If 275 million people spent 11.7 trillion dollars, how much did one person spend to the nearest dollar?

[magic idol]

Note that I am continuing off your previous conversations.
Next time if this issue occurs, you can look at who mentions you in the 'inbox' or search 'mentions:' then your user or your last messages 'from:'

[odd edge]

@warm seal Has your question been resolved?

[odd edge]

@warm seal Has your question been resolved?

[odd edge]

@warm seal Has your question been resolved?

[daring flint]

so this is the question?

[warm seal]

yes

[mystic saffron]

yessir

late notif sorry

[warm seal]

send this yesterday morning btw

[mystic saffron]

Guys

Is there no question for prep 3

Bruh

[odd edge]

@warm seal Has your question been resolved?

[warm seal]

this server is ass don’t even bother askin for help

[mystic saffron]

Why are you still here then🤨

[pale monolith]

Hello

I'm new here but I need help

[mystic saffron]

#❓how-to-get-help

[verbal python]

This generation doesn’t know how to be patient

[warm seal]

i asked this 2 days ago

[verbal python]

Helpers are volunteers and you acting rude doesn’t help

[mystic saffron]

Mate if you were nicer to people and more precise with your questions, you would have gotten help much sooner

In my experience, this server has the best response time from volunteers I have ever seen.

I have been on larger servers and not one was as helpful as this one...

[sly hatch]

Not to be a dick but if the the units give 99% of the work to do and the % is essentially just division with the values you are given helpers wont be doing your homework for u

[mystic saffron]

https://tenor.com/view/travis-fran-healy-you-said-a-bad-word-bad-word-swearing-gif-27378431

[quasi sparrow]

Gee I wonder why nobody's helping you

[mystic saffron]

I think no one would disagree this channel's dead now, and hence no one would object to me closing it

@warm seal In case you still wish to obtain help from this server, please open a new channel when you know you have the time to work through it with a helper, and please be nice to them. If you don't get a reply within a few hours, there's many factors that may explain this, one of which is bad luck (though, in this channel specifically, the main factors were you being rude and the channel getting clogged up). What it does not mean is that the helpers are slacking: Helpers are volunteers, they're not paid for this, and yet I can assure you that almost all of the people coming here for help who didn't solve the problem on their own right after posting it get help and end up with a much better understanding of the problem

PS: If you have a response to this, my DMs are open. Please do not send a message here, as this channel is not meant for this type of discussion

.close

[odd edge]

Channel closed

Closed by @wintry storm

Use .reopen if this was a mistake.

[odd edge]

✅

[fringe solstice]

this channel is occupied, please post your question in one of these to claim a channel of your own

[odd edge]

@warm seal Has your question been resolved?

[remote sinew]

Help!!!!!!!!!!!!!

[mental spear]

Pls help me solve this

guys

4/612/9+(3/8)(-12/9)

[grand dove]

Use trigonometry on triangle CDE to get CE
Now you have BC
Use trigonometry on triangle ABC to get AC
Now subtract DC to get AD

[odd edge]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[rotund hawk]

Do you know how to calculate per capita?

[latent crescent]

How was the Tmua

[rotund hawk]

@warm seal Do you still want help? Cause if not I’ll just close this