help-13

[sly swan]

do u know anything else?

such as the slant height perhaps?

[feral juniper]

sec I havent checked it so far mb

[dim gulch]

its all given bro just see the latex part

[sly swan]

latex?

isn't that what u get from rubber trees?

or was that smth else

[feral juniper]

nah it works out

again please show your calculation

[sly swan]

tobi

what does latex mean?

[feral juniper]

$A = \pi r^2 + \pi r s = \pi r (r + s)$

[wraith dagger]

tobi

[feral juniper]

this

[dim gulch]

@feral juniper

[sly swan]

bro he obviously didn't mean it like that

my question is

is the second picture related to the first

cuz it looks like two entirely different questions to me

[dim gulch]

i dont want these huge formulas

[sly swan]

in the first and the second picture

and you dont have enough information in the first picture

[feral juniper]

what is that calculation supposed to represent?

[dim gulch]

this is my working

[dim gulch]

read the ENTIRE question bro

u will get it

[sly swan]

is it TSA or CSA?

oh wait now i understand

mb

[feral juniper]

okay how did you do b) ?

[sly swan]

$\pirl - piRl$

[wraith dagger]

bagelguy3
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[feral juniper]

go to latex testing pls

[dim gulch]

#latex-testing

[sly swan]

I thought u were trolling

I had no idea it was a real channel

[dim gulch]

@feral juniper

curv surface area of cone = pi * r * l

thats how i did part 2

[feral juniper]

yes nice

so for the full surface area you are missing the area of the circle

[dim gulch]

that gives 4000pi

just read it and do the last part

[feral juniper]

I already did

do you agree with me that the surface Area for the cone can be represented with this formula

$A = \pi r^2 + \pi r l = \pi r (r + l)$

[wraith dagger]

tobi

[feral juniper]

this is what you derived

where A is the surface area
r is the radius (in the case of b) its 40cm))
and l is 60cm (in the case of b))

so A is adding up the curved surface area which is

pi r l

and the area of the circle

which is

pi r^2

[sly swan]

$A = \pi{R}l - \pi{r}l$

[wraith dagger]

bagelguy3

[sly swan]

thats the formula

for the CSA of the part

part of the cone

like in the figure

curved part

[dim gulch]

whats the answer?

[sly swan]

of?

I still cant figure out if they are asking for TSA or CSA

total surface area of curved SA?

[dim gulch]

the fkn question

ye

[sly swan]

"ye" isn't an answer to a question where you have to choose

[dim gulch]

oh

[dim gulch]

no

[sly swan]

great

alright u want me to gone, I get it

I wont bother u again

[blazing zephyr]

what happend here

[dim gulch]

please solve the question

[blazing zephyr]

!nosols for the helpers :shrug:

[cedar kiln]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[dim gulch]

bro i dont know how to do it

i maen

i know how to do it

see my method and correct it

its just the last part

im not even asking u to do more

[blazing zephyr]

where is your method

[dim gulch]

read the question, the entire question

and then see this

and tell me why its wrong

i already know the answer

the answer is 3800 pi

i dont need to copy

from shit

[blazing zephyr]

1000 is also including the small circle

[dim gulch]

so i removed that

and added the small circles area

[blazing zephyr]

if you think about it, 4000pi is the whole cone including the base, the 1000pi is all of the top cone including the base, so essentially you want to add the area of the smaller circle, but instead your subtracting the small circle then adding the small circle which cancels eachother out, leaving the top circle still not calculated

[dim gulch]

which is 20^2 pi

[dim gulch]

i am subracting the entire upper cone, and adding the small circle

[blazing zephyr]

yes but subtracting the entire upper cone also subtracts the small circle

[dim gulch]

i got it

thanks

.close

[cedar kiln]

Channel closed

Closed by @dim gulch

Use .reopen if this was a mistake.

[blazing zephyr]

[cedar kiln]

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[lusty thunder]

help

[lusty thunder]

how do i solve x+5 = 2x-1 i forgot it

[dim gulch]

walahi

[lusty thunder]

?

[proud vigil]

move the x terms and the constants to one side

[cedar kiln]

@lusty thunder Has your question been resolved?

[lusty thunder]

yep got it

[cedar kiln]

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[inland bone]

If I've got an equation log(a) = b, how do I see that increasing b will increase a?

[uneven quartz]

to what base?

[round geode]

Presumably 10 if none is shown

[uneven quartz]

then 10^b = a

by def of log

[round geode]

Do you want to see or prove ?

[inland bone]

In reality the equation looks like this so it's harder for me to see it. They claim that increasing I will increase gamma

[twilit bison]

note that x^(1/2) and log(x) are both increasing functions

[inland bone]

oh wait it causes a decrease in gamma

my bad

cause of the negative sign

[twilit bison]

oh yeah

[inland bone]

So it'll be like gamma = 10^(-RHS)

[round geode]

What does the plus minus imply by gamma?

[inland bone]

and then the bigger the I in the RHS is, the more negative the power will be, and then the gamma will be smaller

[inland bone]

long story lol

https://en.wikipedia.org/wiki/Activity_coefficient

if you want to learn about it

[round geode]

Eeeek, I live in denial of maths outside of pure maths

[inland bone]

Physical chemistry course 👍

[round geode]

I'll check it out though

[twilit bison]

it seems to refer to electrical charge

so nothing to do with sign

[inland bone]

Yeah afaik it's the average activity coefficient of a + and - ion

something like that

It's a positive number though

[round geode]

Makes sense to be, since coefficient are generally measures of ratios

[inland bone]

The equation I sent is called the Debye-Huckel limiting law

[round geode]

Seeing maths applied to describe real world phenomena is the coolest shit

[cedar kiln]

@inland bone Has your question been resolved?

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[ornate sinew]

Find the maximum

[ornate sinew]

Of a finction given with this equation

(x^2+y^2)^3-3(x^2+y^2)+1=0

So I started doing this

What do I do wrong?

I can't find any point that works

I have a historia to use a Darboux theorem to estimate the y(x0)

.close

[cedar kiln]

Channel closed

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[cedar kiln]

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@dusty lava Has your question been resolved?

[woven bluff]

have you used the compound angle identity before?

It states that $\sin{(x+y)} = \sin{x}\cos{y} + \sin{y}\cos{x}$

[wraith dagger]

rep

[queen stirrup]

i dont think this can be simplified further

[woven bluff]

o

did you convert the 30 degrees ones into their actual values

with the one on the right

also you can convert a sin x + b cos x to Rcos(x+a)

so maybe harmonic form can be useful

but maybe that would just put it back to where it was before idk

yeah its definitely hard

[wicked grotto]

Use this and you obtain an expression in terms of sinx and cosx

Then use half angle to obtain a further expression

[cedar kiln]

@dusty lava Has your question been resolved?

[cedar kiln]

Channel closed

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[tall ibex]

Can somebody explain differentiate (x+y) wrt x will it be 1 or 2

[tall ibex]

I'm getting 2 while it should be 1

[frigid dust]

if you differentiate wrt x you assume y is constant

,rccw

[frigid dust]

$\dv{(x+y)}{x}=\dv{x}{x}+\dv{y}{x}=1+0=1$

[tall ibex]

That i kind of get it but how would i know where should i take y as a constant and where not i mean in which ques

[wraith dagger]

Bonk

[frigid dust]

i dont understand what you mean

[tall ibex]

So should i always take y as constant when it differentiating wrt x

[frigid dust]

yes, always

and similarly when diff wrt y

[tall ibex]

Ok that is what i wanted to know

[frosty orchid]

i wouldnt say always

[frigid dust]

(assuming y isnt a function of x)

[frosty orchid]

yes that

[tall ibex]

It isn't

[frigid dust]

if y is a function of x then almost always not constant wrt x

[tall ibex]

Like can u look in my ques it isn't right fn of x

[frigid dust]

you are basically doing chain rule

[tall ibex]

Yeah but i didn't do it for some time so i'm forgetting very basic things

[frigid dust]

where

[tall ibex]

.

Here

[frigid dust]

$f(x,y)=e^x+e^y$?

[wraith dagger]

Bonk

[frigid dust]

this one?

[tall ibex]

e^x + e^y = e^x+y differentiate

[frigid dust]

what?

[frigid dust]

which line

first of all, that first line is wrong

[tall ibex]

1st one in black

[frigid dust]

$e^x+e^y\neq e^{x+y}$

[wraith dagger]

Bonk

[tall ibex]

That is a ques

[frigid dust]

like, for which x and y its true?

[tall ibex]

No it was something else i had to prove differentiate

[frigid dust]

can you just sned the original question?

!original

[cedar kiln]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[tall ibex]

Okay a sec

This is the original

[tall ibex]

Are u still here?

[frigid dust]

this is quite a different question that from what you asked

[tall ibex]

How so?

[frigid dust]

!xy

[cedar kiln]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[tall ibex]

I had to differentiate all of it differentiate e^x + differentiate e^y = differentiate e^x+y

[frigid dust]

with respect to what?

x or y

btw, in this question i think y(x)

[tall ibex]

So should i not take y as a constant

[frigid dust]

no, its dependent on x

otherwise dy/dx=0

and thats not an option

[tall ibex]

Can u try to solve it

[frigid dust]

!nosols

[cedar kiln]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[frigid dust]

continue from here

what do you get

[tall ibex]

This is how guy did it on yt whom i was learning

[frigid dust]

yes but try doing that yourself

[tall ibex]

I have

[frigid dust]

$\dv{}{x}\left(e^x+e^{y(x)}=e^{x+y(x)}\right)$

[wraith dagger]

Bonk

[frigid dust]

,rccw

fk

[frigid dust]

almost right

the RHS is wrong

[tall ibex]

Yeah i got confused with differentiation of x+y

[frigid dust]

$\dv{(e^{x+y(x)})}{x}=e^{x+y(x)}\dv{(x+y(x))}{x}=e^{x+y(x)}\left(\dv{x}{x}+\dv{y(x)}{x}\right)=e^{x+y}\left(1+\dv{y}{x}\right)$

[wraith dagger]

Bonk

[frigid dust]

make sense?

[tall ibex]

A sec lemme try it out by myself once more

[tall ibex]

Hey thanks a lot for help

But just one last

[frigid dust]

then i think you can solve from here

[tall ibex]

So should i just bluntly take d(x+y)/dx equal to 1

[frigid dust]

then its just algebra

[frigid dust]

?

i just showed you its not 1

[tall ibex]

I'll try it once more all of it got too mixed up

How do i close it?

[frigid dust]

.close

[tall ibex]

.close

[cedar kiln]

Channel closed

Closed by @tall ibex

Use .reopen if this was a mistake.

[frigid dust]

$\dv{(x+y(x)}{x}=\dv{(x)}{x}+\dv{(y(x))}{x}=1+\dv{y}{x}$

[wraith dagger]

Bonk

[frigid dust]

and $\dv{y}{x}\neq 0$

[wraith dagger]

Bonk

[tall ibex]

Isn't this is how u do it?

Or am i really really stupid that i forgot the very basics

.close

[cedar kiln]

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[raw gull]

Hello, I need help with this. How do I solve these?

[raw gull]

Oh and a=4

[upper ruin]

The same way you solve them when there are 2 eqns in 2 unknowns

[raw gull]

And how do you solve those?

[upper ruin]

Well, there are several methods, I don't know which ones you've been taught

[raw gull]

kramer method

[rain pollen]

for the first question add all simaltaneous equations

which cancels all z

[upper ruin]

In my country they are called reduction, substitution , comparison, Cramer
Or Gauss reduction of you're studying linear algebra

[rain pollen]

gauss best

[upper ruin]

Yep, but at highschool I've never seen it being used

[rain pollen]

can anyone help me?

[raw gull]

How do I do that?

[rain pollen]

mine is help-34

[rain pollen]

i need answer and explanation

[upper ruin]

Literally you add all three equations, side by side

[cedar kiln]

Please do not advertise your help channel or thread in other parts of the server. There are many people who need help, so advertising can quickly turn into spam.

[upper ruin]

So on the RHS you'll get -13-4+5 which equals -12

[rain pollen]

that doesnt work

[upper ruin]

Why not? It was you who suggested this btw 😅

[raw gull]

I don't think that's how it's supposed to be

If I'm not wrong

[rain pollen]

yeah now i see why because we would only get 1 equation for 2 terms , rest would be for 3 terms

[upper ruin]

Well yeah but from the addition of the three you get x in terms of y or viceversa, since z cancels out

That's the main goal of adding equation

[rain pollen]

bro we need atleast 2 equations to solve 2 term simaltaenous equation

wait i gotta go get a book and pencil

i think i have an idea

[cedar kiln]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[upper ruin]

Yeah the last two together

[raw gull]

I need the answer to see how to solve them tho

[rain pollen]

i think i understand it now lowkey

ok so first step

use the first equation and get a value for z in terms of x and y

[cedar kiln]

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.

[rain pollen]

bro if someone is outmatched then?

[raw gull]

So basically I solve this one

[rain pollen]

yeah

[raw gull]

How am I supposed to learn how to do them if someone doesn't show/explain how to do them tho

[rain pollen]

i got z= (13-4y+x)/2

what u got?

[raw gull]

But in my notes I solve it in a different way

I don't know how to explain it in English

[rain pollen]

just send a photo

[upper ruin]

There are a lot of ways to solve it

[raw gull]

Basically I start by making it into

|1 4 2|
|3 5 1|
|4 7 1|

[upper ruin]

It should be clear that it's better to find x, instead of y or z. Because x has a coefficient of 1

[raw gull]

The entire thing

[rain pollen]

best way for anyone to understand is to align the new information with previous information'

[rain pollen]

its probably gauss method

[upper ruin]

Or Cramer, since there's a determinant

[raw gull]

Yea we learned only Gauss and Cramer

Those are the only two I've written about

[upper ruin]

Well, then tell it! Otherwise we're explaining things you've not seen in class 🙈

[raw gull]

Checked the Gauss one is the one they told me to focus on

It's easier apparently

[rain pollen]

i will try to teah gauss

so basically the first step is to write it in form of
1st equation| x y z | (answer of the equation)
2nd equatio| x y z |(answer of the 2nd)
3rd equatio| x y z|(answer of the 3rd)

i dont know carmer , they dont teach me carmer or gauss in school

learned it for fun

[rain pollen]

tell me what u get

|() , () , () | ()
|() , () , () | ()
|() , () , () | ()

[rain pollen]

copy and edit this'

[raw gull]

1st I get

X= 4y+2z-13

Is this the correct way?

[rain pollen]

nah just write the number not the variable

like 1 , -4 , -2 | -13

[raw gull]

|1 -4 -2| -13
|3 -5 1| -4
|4 -7 1| 5

So like this?

[rain pollen]

yeah

[raw gull]

Ohhhh I see

| x -4y -2z = -13
| 3x -5y +z = -4
| 4x -7y + z = 5

|1 -4 -2| -13
|3 -5 1| -4
|4 -7 1| 5

(I'm just writing it so I can write it later easier)

After that we do?

[rain pollen]

now 2nd step is to get 1st value(in 1st equation) = 1 , other values in 1st equation should be equal to 0(except answer of equation) and 2nd value in 2nd equation = 1 , (others should be equal to 0, except answer of equation) , same with 3rd equation

we do this by subtracting 2 equations

we can multiply (the equations)

[raw gull]

Soo it needs to be:

|1 0 0| -13
|0 1 0| -4
|0 0 1| 5

[rain pollen]

yeah but i bet the equation answers would be changed

when u are subtracting equation the answer of it also gets subtracted

or multiplying

i suggest give variable to each equation line

think of it like vector

[raw gull]

Soo I go like this?

Nvm

[rain pollen]

yeah

[raw gull]

Wait it's correct m

[rain pollen]

it is , this litterly was the easiast part

[raw gull]

Don't I just get this?

Wait no

I'm stupid

[rain pollen]

how

[raw gull]

This is what I'm getting on the 1st line

[rain pollen]

u divided 1st line by 4?

[raw gull]

Yea?

Or am I confusing it?

[rain pollen]

u confusing it

[rain pollen]

even if u did the first value would be 1/4

[rain pollen]

and the goal is to get only one ''1'' value on LHS others should be 0, for the 1st equation

[raw gull]

How do I do that?

[rain pollen]

u can divide it by some number or subtract 2 lines(can also multiply like 2(line c)-line a)

[supple pollen]

Cramer method or elementary row operation

[rain pollen]

its different for each simaltaenous equation

[raw gull]

Like this?

[rain pollen]

u dont have to divide every thing u can also divide one line

[rain pollen]

anyways the point is to get a single value(even work if its not 1) on LHS but only in a single row, what i am trying to say is the should only be 1 value in 1 row(not zero) LHS

[raw gull]

Like this?

[raw gull]

This seems helpful

[rain pollen]

do what ever to get sigle value(rest should be 0 on lhs)

[raw gull]

Soo something like this?

[frigid dust]

exactly right!

[rain pollen]

yo can u carry from here?

[frigid dust]

sure np

[raw gull]

So now I do this?

[frigid dust]

thats not quite right

-3*-4-5=7

[raw gull]

Is what I get if I do what I do?

[frigid dust]

you subtracted wqrong

[raw gull]

Is this better?

[frigid dust]

almost

3rd column is wrong

[raw gull]

Is this better?

[frigid dust]

yes!

[raw gull]

The minuses are so confusing

😭

[frigid dust]

yea they are

[raw gull]

Wait do I have a time limit on using the Chanel? I need to go to sleep but still need help

[frigid dust]

after a while itll ask if your problem has been solved

and if you dont respond it times out and closes it

but you can always just opena enw one

[raw gull]

Okay I will try to solve this part and tomorrow will start from there

Is this correct?

[frigid dust]

yeah its the same as the last one, which was correcct

[raw gull]

Okay thank you

I will now close the channel and continue tomorrow

Thanks for the help so far

.close

[cedar kiln]

Channel closed

Closed by @raw gull

Use .reopen if this was a mistake.

[frigid dust]

goodluck!

[cedar kiln]

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[rain pollen]

help

[proud vigil]

well, we have a lot of information given here.

let me draw it out

[rain pollen]

the annoying thing is to write it in the for off cospbq

[proud vigil]

oh boy

im thinking cosine law bash?

[rain pollen]

it is cosine law bro but the form

[proud vigil]

yeah

[rain pollen]

wait lemme go beyond ultra plus\

[proud vigil]

ok

[rain pollen]

yeah i got cooked again

i did it

what do i do

i am the best

help new one

[cedar kiln]

@rain pollen Has your question been resolved?

Channel closed

Closed by @rain pollen

Use .reopen if this was a mistake.

[cedar kiln]

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[odd breach]

can someone explain how to do this question i don't understand the solution

[dull oxide]

Which part do you not understand?

[odd breach]

why can't x not equal x^-1

i can think of groups where some of the elements are it's own inverses

<@&286206848099549185>

[dire geode]

Then multiplying both sides by x would imply n = 2 and contradict n is odd

[cedar kiln]

@odd breach Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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

[mental holly]

My reasoning (I eventually got stuck trying to find the root of the discriminate)

since m is complex to

[frigid dust]

$m^2+4mi-4=(m+2i)^2$

[wraith dagger]

Bonk

[frigid dust]

i think you know how to carry on from here

[mental holly]

Wait so can I just cancel out the power?

[frigid dust]

wdym?

[mental holly]

Without solving the system

[mental holly]

I don't think this is correct

?

[frigid dust]

you know the determinant is m^2+4mi-4 right?

well, m^2+4mi-4=(m+2i)^2

thus sqrt(determinant)=m+2i

[mental holly]

but you can't just square root a complex?

[frigid dust]

why not?

if you can square it, you can take the root

[mental holly]

wait hold on

According to my book it isn't

[frigid dust]

what does it say in your book

[mental holly]

please wait

Let me search

[frigid dust]

i mean, you technically get a second solution

[mental holly]

Well in theory, this should be equal to 0?

Unless I messed up mid way

[frigid dust]

-8-4mi=0?

is that what you mean?

[mental holly]

no

[frigid dust]

im not sure what youre trying to say

[mental holly]

you said sqrt(determinant)=m+2i

I found first solution

Then replaced it in said equation and didn't get 0

(Unless I made a mistake)

[frigid dust]

what do you have for z1?

[frigid dust]

$z_1=\frac{-(3m-2i)+(m+2i)}{2}=\frac{-2m+4i}{2}=-m+2i$

[wraith dagger]

Bonk

[frigid dust]

should be the first root

[mental holly]

oh so I did second root

In theory they both should be correct?

[frigid dust]

$z_2=\frac{-(3m-2i)-(m+2i)}{2}=\frac{-4m}{2}=-2m$

[wraith dagger]

Bonk

[frigid dust]

is the second root

so you made a mistake in your calculations

[mental holly]

wait no

[frigid dust]

did you do $z_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}$?

[wraith dagger]

Bonk

[mental holly]

my equation is this

There is a minus

[frigid dust]

ah, i see

$z_1=\frac{(3m-2i)+(m+2i)}{2}=\frac{4m}{2}=2m$

[wraith dagger]

Bonk

[frigid dust]

is the first root

$z_2=\frac{(3m-2i)-(m+2i)}{2}=\frac{2m-4i}{2}=m-2i$

[wraith dagger]

Bonk

[frigid dust]

is the second root

[mental holly]

when I replace that second root

I didn't get 0

[frigid dust]

$(m-2i)^2-(3m-2i)(m-2i)+2m^2-4mi=m^2-4mi-4-3m^2+2m+6mi+4+2m^2-4mi$

[wraith dagger]

Bonk

[frigid dust]

im missing an i somewhere

but it should be correct

[mental holly]

how did you get a +4

-2*-2 is 4

[frigid dust]

$-(-2i)(-2i)=-(4i^2)=-(-4)=4$

[wraith dagger]

Bonk

[mental holly]

oh there is a minus

didn't see it

[frigid dust]

happens 😄

[mental holly]

yeah it does equal to 0 lol

both actually

this was book's methode

Of finding a complex determinant

root*

Anyways thank you so much for giving me some time/aid

🙏

[frigid dust]

ofc, np

the hard part was seeing that its a square

[mental holly]

yeah I didn't see the -4

.close

[cedar kiln]

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[frigid dust]

if youre not used to complex numbers its easy to write off

since with real numbers thats nto possible

[cedar kiln]

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[blazing dew]

Hi there. I'm relatively new to calculus, I'm trying to learn how to do problems involving exponential growth and decay. There's a general way to show how the change of the variable (mass, population, etc) over time is proportional to the derivative of that (the rate of change of that variable), and derive the equation. But there's a step which involves integrating both sides that I'm not wrapping my head around.

Q: Why are we allowed to integrate each side with respect to different things? It seems like we end up needing to integrate the left side with respect to P, and the right side with respect to t (in my setup). Why is that allowed?

[ancient lodge]

you're looking for this (the variable names are different, but the idea is the same)

Source: https://math.stackexchange.com/questions/27425/what-am-i-doing-when-i-separate-the-variables-of-a-differential-equation

[blazing dew]

YES! Ok so my question becomes, why does integration by substitution work. So i guess i have to follow the link here

oh which is U sub

which is next on my list of things to learn lol

If the justification is using u sub, then I think I know what direction to head in now

so ill close it, but thanks a lot Pigeon 🙏

.close

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[tranquil reef]

Okay, I have a problem. I’m trying to figure out, the growth percent to make 62.5 into 9 million over 100 multiplications of 62.5 times X where X equals the growth percent

[tranquil reef]

Basically exponential growth but i already have the time, final value and initial value, but not the growth function

[proven nova]

So like 62.5x^100 = 9000000?

[tranquil reef]

Yeah

Assuming 62.5x is 62.5 times X and that’s just not shown for simplifying reasons

[proven nova]

yes

so the growth percent would just be (growth factor -1) * 100

the growth factor here being x

[tranquil reef]

Yeah

But we don’t know what the growth factor is

[proven nova]

well we have the equation 62.5x^100 = 9000000

so solve for x in that equation

[tranquil reef]

Maybe..

square root on both sides?

[proven nova]

we want to first get rid of the 62.5 that's being multiplied to the x^100 before moving on to getting rid of that ^100 by square rooting on both sides

you have the right idea, just a little too soon in the solving process

please lmk if that makes sense

[tranquil reef]

Ok

So

I’m not sure how we remove 62.5

I feel like subtracting wouldn’t work

[proven nova]

62.5 is being multiplied

[tranquil reef]

Yeah

[proven nova]

so do the opposite of multiplication

[tranquil reef]

So we divide 9 million by 62.5

[proven nova]

yes

[tranquil reef]

That would be

144,000

[proven nova]

yep

[tranquil reef]

How do you square root and remove ^100

[proven nova]

you do the one hundreth root on both sides

like y'know how roots sometimes have numbers

i'm explaining this really badly

give me a sec

$\sqrt[100]{144000}$

[wraith dagger]

Planet

[tranquil reef]

ok

thats

1.126117286

[proven nova]

yeah

thats the growth factor

how do we find the growth percent

[tranquil reef]

times by 100

i think?

[proven nova]

yes

wait no thats wildy wrong

sorry

it's the (growth factor - 1) * 100

[tranquil reef]

so

(1.126117286 - 1)*100

[proven nova]

yeah

but you can round the growth factor

to like the thousandths

depends on if the exercise specifies

or not

but generally you don't need to put all the decimals the calculator gives you

[tranquil reef]

so

if i input this into an exponential growth equation

it should equal 9 million

[proven nova]

yes

[tranquil reef]

how do i do that

on desmos scientific calculator

do i need a graphing one

[proven nova]

look up desmos graphing calculator

but if you want to input it into a scientific calculator you can

you can just do

62.5(1.126117286)^100

and it should give 9 mil

[tranquil reef]

thanks so much

[proven nova]

no problem

you can close the channel with .close

[cedar kiln]

@tranquil reef Has your question been resolved?

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[silver ibex]

Just need some quick clarification, does the 0.1912 mean 0.1912 percent or 19.12 percent? The video example said 0.1912 percent, but should it not be 19.12% (as I wrote down)?

[flint plinth]

,rotate

[flint plinth]

0.1912 means 19.12%, you are correct

[silver ibex]

Alr cool, thanks for clearing it up for me 👍

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[sly fern]

Does the standard form of quadratic tell you the y intercept the factored form tell you the x intercepts and the vertex form tells you the vertex?

[livid hound]

yes

[sly fern]

And in the factored form it says x-r and x-r , is the r itself the x intercepts or the -r?

[dusk goblet]

x intercept —> y = 0 hence at least one of the factors is zero, if one of the factors is x-r then x-r = 0 hence x = r

is an x intercept

[sly fern]

Ok an lets say the x intercepts is negative hence its x+r does that mean its still just r?

[dusk goblet]

no of course not

x+r = 0

x = -r

[sly fern]

Ah ok thx

[cedar kiln]

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[coral salmon]

i need help with graphing x & y intercepts and angles

[coral salmon]

so how can i graph x on a line plot

[rustic pier]

like how

[fallen heath]

!original

[cedar kiln]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[coral salmon]

with like y intercepting without like uhm graphing backwards or upside down.

[rustic pier]

are we talking abt a particular question or like as a general

[coral salmon]

as in general

[rustic pier]

so just graphing a function

[coral salmon]

yeah. i'm a bit slow and don't understand what my teacher been teaching me.

[rustic pier]

..

firstly take the terms dependent on x to one side of the equation

it should be like y=f(x)

and there should be no x on the left hand side

[coral salmon]

thank you sm my teacher talks a bit fast and is very confusing

[rustic pier]

i havent even done anything yet

[coral salmon]

theres more. its going to be long isn't it.

[rustic pier]

not that long

[coral salmon]

alright

[rustic pier]

after this

take any value you like for x

usually we take something in between -5 and 5

but it depends on the function

[coral salmon]

mhm.

[rustic pier]

the functions are usually given in particular types

like a linear

or a quadratic

and so on

each function tupe has a characteristic graph

*type

dude reply

[coral salmon]

yeah sorry

sorry

[rustic pier]

.

you need to know abt some of these types

to be more efficient

but its optional

[coral salmon]

okay

[rustic pier]

firstly linear

it will be simplified to y=mx+c

this will give a straight line

the original may be given as Ax+By+C=0

[coral salmon]

alright

[rustic pier]

here, m is known as the slope and c as the y-intercept

slope is nothing but the (change in y)/(change in x)

and the y-intercept is the value of y when x is zero

[coral salmon]

okay

[rustic pier]

next is quadratic

it will be simplified to y=ax^2+bx+c

it will give a parabola

a u-shaped curve which may open upward or downward

depending on the value of a

[coral salmon]

mhm

[rustic pier]

the maximum/minimum of the parabola is called as its vertex

and occurs at x=-b/2a

[rustic pier]

if a>0

then it will be an upward parabola

and will have a minima at x=-b/2a

and if a<0 it will be a downward parabola with a maxima at x=-b/2a

[coral salmon]

okay

[rustic pier]

.

you need only linear

or do you want other types too

[coral salmon]

no thank you i haven't gotten to that yet actually.

[rustic pier]

ok

what are you studying

[coral salmon]

now we just moved onto supplementary, vertical and adjacent stuff.

[rustic pier]

what grade

[coral salmon]

7th advanced math

[rustic pier]

ok

straight liines/linear equations

is enough

[rustic pier]

do you understand till here

[coral salmon]

mhm

[rustic pier]

now place any value you like for x

and record the output

i.e. y

[coral salmon]

okay

[rustic pier]

let the points be (x1,y1),(x2,y2)..

for different values of x

[coral salmon]

okay

[rustic pier]

and now plot the points on a graph

you will get a straight line

[coral salmon]

mhm

[rustic pier]

the points where it intersects the x and y axes are known as intercepts

if x=0 then y-intercept

and if y=0 then x-intercept

[coral salmon]

i did 15 & 5 and gotten a straight line. thank you

[rustic pier]

ok

youre welcome

[coral salmon]

.close

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[obsidian crest]

i dont understand whats wrong on the increasing

[livid hound]

you want open parentheses not square brackets for the first one
and you used y coordinates instead of x coords for the second

[obsidian crest]

why did the video i watch tell me to use brackets TT

[livid hound]

what video

[obsidian crest]

https://www.youtube.com/watch?v=t8s5Iso2a4E

[livid hound]

no idea

[obsidian crest]

on this is concave meaning the same as the earlier problem?

[livid hound]

no

concavity relates to how fast **the slope of **the function is changing

[obsidian crest]

what did i do wrong

[livid hound]

explain your reasoning

[obsidian crest]

idk i just followed the video my teacher gave

[livid hound]

the whole red curve is concave up, the whole blue curve is concave down

use those for reference

[obsidian crest]

where am i supposed to stop the points?

[livid hound]

try determining when the slope starts/stops decreasing
look up some vids on concavity
a ruler may help

have it tangent to the curve, follow along,
see when you change from rotating the ruler between clockwise and anti-clockwise

[obsidian crest]

Oh I don't really understand what that means

Is it at x=-3?

[livid hound]

yes

[obsidian crest]

it still says its wrong?

[livid hound]

(-inf,-3) would be the interval for concave down

i don't know where the others are coming from

[naive talon]

Do you know what "concave" means?

[obsidian crest]

i havent touched those

[obsidian crest]

yes idk why it doesnt use convex

its mega confusing

concave up and down is just dumb TT

[naive talon]

Well, things can be difficult sometimes.

[obsidian crest]

why cant it just use convex and concave TT

[naive talon]

I've a feeling mathematicians don't want to use seperate terms for it.

Or something, I dunno.

[obsidian crest]

up and down to describe a word that already exists is just weird

is it -3,inf for concave?

[livid hound]

(-inf,-3) would be the interval for concave down

[naive talon]

(-3, ∞) is concave up.

[obsidian crest]

yeah sorry thats what i was asking

[naive talon]

I think.

[obsidian crest]

is there another thing i need to put it says its wrong

[livid hound]

show what yu're putting

[naive talon]

Try (-5, ∞).

[livid hound]

you used a .

instead of ,

[obsidian crest]

bruh

.close

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[lusty lotus]

how do i know if the upper bound for x is 1 and lower bound is 2y^2+2z^2-5 and not otherwise

[gritty viper]

because the set 1 > 2y^2 + 2z^2 - 5 is bounded

the set 1 < 2y^2 + 2z^2 - 5 is not

so only one of these are you actually integrating over a region with finite volume

[hushed folio]

If you take the xz and xy traces you'll find that this function is a paraboloid

And you know that because of the 5 that it has shifted 5 negatively on the x axis

So it starts at -5 on the x and opens from there towards positive infinity

[cedar kiln]

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[deft estuary]

Can someone help with these two questions

[cedar kiln]

@deft estuary Has your question been resolved?

[deft estuary]

<@&286206848099549185>

[cedar kiln]

@deft estuary Has your question been resolved?

[deft estuary]

<@&286206848099549185>

[olive gust]

hu

hi

look

first grid prob 3x3 grid: can the cat and dog have the same list of products??
5x5 & 11x11 Grids: similar question for larger grids

[deft estuary]

ok

[olive gust]

polynomial expansion (i) even/odd coefficients when is Cn(i, j, k) even or odd?

(ii) sum of coefficient whats the sum of the coefficients where i, j, k > 0 ??

[deft estuary]

hmm could you explain more @olive gust

like give a brief walkthrough

[olive gust]

Sure lookGrid 3x3 grid ur asked if its possible to arrange the numbers 1-9 in such a way that the products of the numbers in the rows for the cat match the products in the columns for the dog

grid 3x3*

the key idea is that the arrange of numbers in rows and columns cant be made to match in product

so the answer is a no

now 5x5 and 11x11 grids

this extends the same idea to larger grids

the row and column products still cant match for these larger grids so the answer is no too

If u dont understand anything react with an X to the message

[deft estuary]

Can u explain this?

and this part of the q
Can you find any conditions on n that guarantee that it is possible or any conditions
that guarantee that it is impossible for the monkey to write the numbers 1, 2, . . . , n2
in an
n × n grid so that the cat and the dog obtain the same lists of numbers?

[olive gust]

ye look small grids like n = 1 or n = 2 ite possible to arrange the numbs so that the row cat’s list match the column products dog s list
n for grids n ≥ 3it's impossib to arrange the numbers in such a way that the row products match the column products

due to the interdependencies between rows and columns so n = 1 or n = 2 works but for n ≥ 3 it doesnt

[deft estuary]

could you explain whats meant by interdependencies

tysm kareem

[deft estuary]

so I'm not getting exactly how to prove why when n>=3 its impossible

[olive gust]

yw!!

[olive gust]

wait lemme tell u

[olive gust]

it refer to the fact that the arrangement of numbers in the rows nd columns of the grid

that isnt independent the choice of numbers in one row affects the numbers in the columns and vice versa

[olive gust]

for n ≥ 3 its impossible cause the number of interdependencies between rows and columns increases

the arrangement of numbers in one row affects the columns nd as the grid size grows these interactions create too many conflicting constraints

[deft estuary]

got it! tysm

can u help me out with the q

after that as well?

[olive gust]

yw ! + sure!

we alr know that for n ≥ 3 its impossible for the row and column products to match due to the interdependencies

and for n = 1 and n = 2 its possible as the constraints are fewer and we can arrange the numbers to satisfy both the row and column product comditions

n = 1 or n = 2 is possible because the number of constraints is small and we can arrange the numbers so the row and column products

but for n ≥ 3 its impossible cause the number of interdependencies grows

and the constraints on rows and columns cant be satisfied simultaneously conclusion :possible when n = 1 or n = 2 impossible when n ≥ 3

if u dont understand it tell me

[deft estuary]

yep got this question 5!!! thanks kareem ❤️

can you help w question 6 as well?

[olive gust]

yw!!!:)

[olive gust]

ofc!

lemme check it one sec

when u expand (x + y + z)^n the terms involve multinomial coefficients and these coefficients depend on the values of n and the exponents i j k

the goal is to determine when these coefficients are even or odd

whichs influenced by the factorials in the multinomial formula and how the exponents interact with the number n

now secondly the sum of coefficients where i j k > 0 represents the number of ways to distribute

the total exponent n among three variables with the condition that each variable get a positive integer value

gets*

the total expinent n among three variables wit the condition that each variable gets a positive integer value

exponent

u get it?

[deft estuary]

one sec

[olive gust]

Alr

[deft estuary]

What;s the expression in 6(ii)?

also@olive gust are you fine if I dm you a couple more questions in around 30 more mins?

i'm solving a worksheet rn and I may have more questions/doubts

[lusty lotus]

ohh right thank you

[lusty lotus]

yea i see it now, thank you

[cedar kiln]

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[deft estuary]

.close

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[meager quiver]

can someone help me with understanding how the coeff of x^n in (1-x)^-r is n+r-1Cr-1

[dire geode]

Do you know binomial theorem

[meager quiver]

yep

dunno how to visualize for negative index

[dire geode]

(-x)^n = (-1 * x)^n

Do you know exponent rules?

[meager quiver]

(1-x)^-r

[dire geode]

Derivation in the probability mass function section
https://en.m.wikipedia.org/wiki/Negative_binomial_distribution

[meager quiver]

😭 I was trying to understand with stars and bars

[cedar kiln]

@meager quiver Has your question been resolved?

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Closed by @meager quiver

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[static parcel]

Can someone give me the solution to this integral? My solution is Zero B), but in the solutions they are saying its C) They can be wrong, i already got some solution wrong in this document..

[frigid dust]

!show

[cedar kiln]

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

[frigid dust]

this looks perfect for a u-sub

[dreamy void]

In fact they suggested a t-sub!

[static parcel]

well i did 2 subs

[idle gulch]

They've already mentioned subbing lnx = t

[static parcel]

i got zero, but the solutions is saying its b)

can someone confirm wich one is ?

[frigid dust]

yup, mb

[frigid dust]

mb

[idle gulch]

Mb

[dreamy void]

,w Integrate[sqrt(sin(pi * ln(x)/2))/(x * sec(pi * ln(x)/2)), {x,1,e}]

[wraith dagger]

Results provided by WolframAlpha

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Requested by adoniszgks

[static parcel]

😦

damm

i dont understand where i got it wrong...

[frigid dust]

did you use the suggested substitution?

im seeing a theta

not the suggested sub

[static parcel]

yes

on the right

the theta is just to now write all the stuff inside sin

like 1/2 * Pi * ln(x)

[dreamy void]

Did you change the bounds when you substituted, doesn't seem like you did

[static parcel]

the bounds? the e and 1 ?

[dreamy void]

yes

seems you also did a double sub

theta = pi/2 * t I guess

[static parcel]

yes that i dd

[static parcel]

no, that is just to not write all the stuff

[wraith dagger]

anti-algebraist 𝔸dωn𝓲²s

[static parcel]

why did you change the bounds ?

not undersatnd

[dreamy void]

it is necessary when you perform a substitution

unless you resubstitute back but most students forget that part anyway

so it's good habit to do so

[static parcel]

ohhhh

so when x = e and after the sub (ln(x) = t) x = e^t
the upper bound is ln(e) (normal upper bound) = 1 ?

i mean, in the line before the last, you can seee i kinda did that...
when (3/2) is to the power...

[wraith dagger]

Arya

[fallen heath]

Can you make sense of this antoniocantomoniz

[static parcel]

yes yes

im so stupid

i got it right

in the last thing in the line before last

i got something times zero

:/

[dreamy void]

oh lol yea

[static parcel]

so changes to ( 4 / 3Pi ) - 0

damm

lol thanks anyway guys

how do i close this?

[dreamy void]

type .solved

[static parcel]

.solved

[cedar kiln]

Channel closed

Closed by @static parcel

Use .reopen if this was a mistake.

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• 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>.
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• Be polite and have a nice day!

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[shut otter]

The 1 question). G= … determine the function G, knowing that g(0) = 0

[mental trail]

!status

[cedar kiln]

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

[shut otter]

1 i dont know where to begin

[mental trail]

ok

Je switch en français si t'es plus à l'aise avec

Théorème fondamental de l'analyse?

[shut otter]

Ah merci

[mental trail]

Est-ce que tu as vu ce th

[shut otter]

Les équation différentielle d’ordre 1 je crois

[shut otter]

Oui jai deja vu mais je nai rien compris des formules à utiliser

[static parcel]

maybe to the int then use the 0 as + C ?

i would

[mental trail]

bon bah

ta fonction g' elle est bien continue

si on applique le TFA:

$\int_a^x g'(t)dt = g(x) - g(a)$

[wraith dagger]

rafilou is not not born in 2003

[mental trail]

Bon bah maintenant suffit de choisir une valeur de a qui nous arrange...

[shut otter]

Donc 0

[static parcel]

i got, g(t) = ln ( ( (t^2) + 3 ) / 3 )

just do the integral, then substitute t=0 in that

and you get C (the const)

then get the result of that integral and add the value you just got from C

[mental trail]

yes, it's about integrating g' between 0 and t (or x)

[shut otter]

But i dont undersatand when he need C, the question is determine the functiun G

[mental trail]

En fait il a procédé d'une autre manière

il a trouvé une primitive de g'

(en intégrant)

[shut otter]

Mon problème, c'est que je ne comprends pas comment on est passé de la fonction G'T à celle que tu m'as donnée.

[mental trail]

tu comprends ça au moins?

c'est le TFA

[shut otter]

Ba je comprend pas dou il provient

[mental trail]

Donc tu n'as pas vu le TFA en cours?

ou tu ne comprends pas c'est quoi le TFA

[shut otter]

Je nai pas vu le TFA

A moins quil y est une autre appellation

[mental trail]

théorème fondamental de l'analyse

https://fr.wikipedia.org/wiki/Théorème_fondamental_de_l'analyse#Premier_théorème

[shut otter]

Non

Je nai pas vu ca

[mental trail]

et pourtant tu as vu les équations diff d'ordre 1

bizarre

(et même d'ordre 2 juste en dessous)

bon

du coup comment on fait dans ton cours

pour calculer les solutions de y' + a(x)y = b(x)

[shut otter]

Ou alors je nai aps compris mon cours mais je ne vois pas ce théorème en tout cas pas le nom

[mental trail]

Tu n'as jamais vu un théorème qui fait le lien entre dérivée et intégrale ?

Aucun ?

[shut otter]

Voici la partie de mon cours qui traite sur lordre 1

[mental trail]

ok, mais je parle pas d'équa diff