help-10

[viral tree]

M is Molarity right which is not the same as Mole

[paper surge]

molarity is like how many moles per a unit

[viral tree]

Molarity is in a solution

[paper surge]

concentration directly is a moles over dm(1000ml)

[polar fossil]

correct, mol just means 6.02 x 10²³, it's just a number

[viral tree]

ok than how do I get to 1.1M?

[paper surge]

so 0.05=moles/volume

[polar fossil]

it's just fractions, you multiply top and bottom by 10³

[viral tree]

because if I multiply 0.0011 by avogadro I dont get that answer

[paper surge]

and moles is mass over mr

so 0.05=(mass/mr)/volume

[polar fossil]

correct, you don't need to do that

[viral tree]

but if I do that I get 1.1/1000?

[paper surge]

this

[polar fossil]

$\frac{0.0011 mol}{10^{-3}L} \times \f{10³}{10³} = \frac{1.1 mol}{L} = 1.1 M$

[warm shale]

هايلي

[viral tree]

that is what I dont understand

[polar fossil]

do you understand what I wrote

[lapis quartz]

you have written in terms of ml

[viral tree]

why is 1 liter = 0.001

[paper surge]

it not 1000 liter

[polar fossil]

I don't understand the question

[paper surge]

you times 1000 from L to get ml

[viral tree]

im thinking that I need to write liters as 1000 mL

[polar fossil]

that would work yes

this kind of unit conversion problem is going to come up a lot in science

[viral tree]

I'm a bit confused but I feel like im getting there

[polar fossil]

so whatever you can do to get comfortable with it

[paper surge]

concentration<whatever that is

[viral tree]

I still dont quite understand

[polar fossil]

M just means mol/L

[viral tree]

then why is the number different

[polar fossil]

wdym

[viral tree]

why its not 0.0011 M = 0.0011 mol/L

instead its = 1.1 M

[polar fossil]

the numbers $\frac{0.0011}{10^{-3}}$ and $1.1$ are the same number

[warm shale]

هايلي

[lapis quartz]

i think i understand your problem,the above statements used ml

[viral tree]

this I need to understand

[lapis quartz]

but molarity is defined for litres

thats why they used 10^-3

[lapis quartz]

that must have been a misprint should have been mL

i think so anyway

[viral tree]

im trying to wrap my brain around it im new to these terms

[polar fossil]

nope, not a misprint

what step of the problem are you having issues with

[viral tree]

the very last one

from 0.0011 mol = 1.1 M

[polar fossil]

the very last one where we go from 0.0011 mol / 10^-3 L to 1.1 M?

[polar fossil]

no

0.0011 mol per 10^-3 L to M

[viral tree]

yea I dont know why its written 10^-3

[polar fossil]

10^-3 L = 1 mL

[viral tree]

I know that 10^-3 = 0.001

[lapis quartz]

ye mb i just got the quetion you are right i am sorry for confusing

[polar fossil]

let's back up one step

[viral tree]

so thats 0.0011/0.001

[polar fossil]

you have 0.0011 mol NaOH / mL, do you agree?

[viral tree]

yes

[polar fossil]

okay, now that's a fraction

we don't want those units, we need L not mL

but we can convert it

by multiplying by 1000 mL / L

[viral tree]

why it has to be over L?

[polar fossil]

so that it's multiplying by 1

[viral tree]

so I get 1 L at the end

in the denominator

[polar fossil]

great that's what we want

and what's in the numerator?

[viral tree]

0.0011

[polar fossil]

no?

what happened to the 1000

[viral tree]

• 1000

[polar fossil]

also what units

[viral tree]

ml

times 1000

[polar fossil]

hang on let me write this out

[viral tree]

I need algebra

[polar fossil]

when multiplying fractions, we multiply the tops and the bottoms

the mL cancels out

what is the result?

[viral tree]

but shouldnt it be 0.0011 mol NaDH/100 mL?

[polar fossil]

no? where are you getting 100 mL from

[viral tree]

like

I'm gonna write my process

so this is 0.011 molar if we are looking at a solution that is 100ml

[polar fossil]

,calc 0.044 / 40

[warm shale]

Result:

0.0011

[polar fossil]

what is the unit on that last number?

because I think it's mol / mL

[viral tree]

so that is what I'm thining that its 0.0011 over 100 ml solution

that was gram

[polar fossil]

where are you getting 100mL from

[viral tree]

is it not true that I have 0.0011 molar in 100 ml solution?

[polar fossil]

where are you getting 100mL from?? it's nowhere in the question

[viral tree]

ok thats what I thought, that since w/w is a sum of gram / 100 grams

[polar fossil]

it's not

it never has been

it's always been g/g

[viral tree]

and weight per volume? is gram/ 100 ml? right

[polar fossil]

not usually

the point is that units are everything

you must keep track of them

you start with g NaOH / g solution

and you need to end with mol NaOH / L solution

[viral tree]

ok so saying that 0.0011/mL is the same as saying 0.0011mL correct?

[polar fossil]

no?

one seems to have mL in the denominator and the other one has it in the numerator

[viral tree]

idk what to do...

maybe its not that important for now , and whats important is that I got to 0.0011 M

[polar fossil]

write down a fraction: 0.05 g NaOH / 1 g solution

then, to the right of it, we're going to multiply it by 1, so that we don't change its value, but we're multiplying by a special form of 1

we multiply by: 0.88 g solution / 1 mL solution

so go ahead and write that

and then do you see that "g solution" cancels out?

[polar fossil]

okay, I had asked you to include solution in there but that's fine

so what's the result of that multiplication? with units

[polar fossil]

with UNITS

[polar fossil]

good

okay, now next we take into account the molar mass of NaOH

which lets us convert between g NaOH and mol NaOH

this will be another multiplication

[viral tree]

but first I need to know why its over 1g solution and the other over 1mL solution from the start

[polar fossil]

what

[viral tree]

basically I have the fraction on the left given and I need to multiply by the fraction on the right to convert it

[polar fossil]

yes

[viral tree]

so I have from mass to volume and from grams to density

[polar fossil]

err yes, density lets you convert between mass (measured in grams) and density (measured in litres or millilitres)

in this case, we know that the density of the solution is 0.88, which implicitly is g/mL although they probably should have specified that

so that is why we can multiply by 0.88 g solution / 1 mL solution

[viral tree]

I didnt know density is g/mL that would help

[polar fossil]

water is 1 g/mL so implicitly if it's just a number it means relative to water... but yes they should have clarified

so continuing, how do we use the fact that the molar mass of NaOH is 40 g/mol?

[viral tree]

divide by 40 mw

molar weight

[polar fossil]

well, sure, but let's use our "multiplying by 1" framework

what do you want to multiply by?

we want to cancel that "g NaOH" unit

[polar fossil]

with units please

[polar fossil]

MW isn't really a ynitt

unit

again we want to cancel that "g NaOH" which is in the numerator

[polar fossil]

okay, but you're not multiplying by 1 anymore

the molar mass of NaOH is 40 g / mol

in other words, you can think about 40g NaOH = 1 mol NaOH

[polar fossil]

you want to multiply by 1 mol NaOH / 40g NaOH

do you understand why we're allowed to do that?

[viral tree]

let me repeat this on a sheet of paper to see if it clicks in my head

[polar fossil]

remember how you used to multiply by like 5/5 to find a common denominator for stuff, this is the same thing

[polar fossil]

okay great, and remember that the mL on the bottom left is mL solution

wait no not great

oh you just misspelled mol

on the top right

[polar fossil]

but yeah now simplify those numbers and you should get 0.0011 mol NaOH / mL solution

[viral tree]

so thats how I get the answer

[polar fossil]

almost done, now we need to convert those mL into L. again this will be a multiplication. we know that 1000 mL = 1 L. what would you like to multiply by?

[viral tree]

and its actually a fraction

and because I want it in liters I need to multiply it by 1000

[polar fossil]

yes, you multiply by 1000 mL / L

so that the mL cancels out

[polar fossil]

okay, please write down units and don't be lazy about it

[polar fossil]

1000 mL / L

[viral tree]

wow this is so confusing

[polar fossil]

at each step you multiply by something that is 1

[polar fossil]

i.e. that is the same on top and bottom

[viral tree]

so mol/ml needs to be converted to mol/L when ever I get this kind of question

[polar fossil]

if you want to answer in terms of M, then yes you need mol/L

because M is defined to be mol/L

[viral tree]

so basically I put L in the denominator and I need it to equal to 1

and 1000mL = 1 L

[polar fossil]

yes

[viral tree]

I think I'm gonna get this

thanks a lot

I might need to come here a few more times with more questions but I'll get it

[polar fossil]

most important advice is to keep writing the units do not drop them

[viral tree]

yea it definitely helps

[polar fossil]

until you're confident you know that you're doing

[viral tree]

it gives confidence too

[polar fossil]

yes, and it lets you go back and check

[viral tree]

like I know what I'm doing , like I'm emphasizing it

thats what it feels like

I just started learning chemistry for the first time and physics and bilogy for the first time this week

this will be crucial to understand fully

.close

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[strange fiber]

can someone help me

[strange fiber]

PLS

anyone that can do algebra

[timid silo]

#1 or #2?

[strange fiber]

both 😭

[timid silo]

well for #2 i can help

[strange fiber]

yes please

[timid silo]

first step is to identify the given variables

do you know how to find a, b, and c?

[strange fiber]

yes

[timid silo]

ok then

first step write those out

a = ..., b = ..., etc

[timid silo]

ok yea next step is plug it in

dont forget the /2a part

[strange fiber]

like this right?

[timid silo]

the full formula is x = -b ±√(b^2 - 4ac) / 2a

[strange fiber]

oh wait

OOPS

[timid silo]

so just do what ur doing but over 2*3

[strange fiber]

like thisssss

[timid silo]

yea

dont forget ±

instead of +

[strange fiber]

TY

[timid silo]

ofc ofc

[strange fiber]

also whats +_

?

[timid silo]

± means plus or minus

so for example 3±3 means 6 or 0

[strange fiber]

ohh

[timid silo]

the reason for that is if x^2 = 4, x can be +2 or -2

which is why the formula has ±√

[strange fiber]

oh okay, tysm!!

[obtuse pebble]

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[timid silo]

Help

[timid silo]

I’m supposed to get 0

I did this 100 times

Jk just 3 times

[inland dagger]

can you include full calculation

on the second page

[timid silo]

I did

how do you rotate

[inland dagger]

i dont understand how are u trying to calculate the deterinant

on the second column right

[timid silo]

Oh

-25 plus 4 is -21 not -19

.solved

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[exotic dawn]

Anyone know why the bound is from x=0 and not x=y when swapping the order of integration for this double integration?

[polar fossil]

seems right to me

hard to tell your a and your x apart

[obtuse pebble]

@exotic dawn Has your question been resolved?

[exotic dawn]

oh fair enough my bad but the video i got it from and chat gpt both say it should be from x= 0 not x=y

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[exotic dawn]

.reopen

[obtuse pebble]

✅

[strong aspen]

?

[exotic dawn]

sorry i still dont have an answer

chat gpt and the video i got it from both say im wrong and i just wanna know why

[strong aspen]

What’s the question

[exotic dawn]

.

[exotic dawn]

this one

[strong aspen]

I’m not familiar with the topic sry

[exotic dawn]

np

[exotic dawn]

<@&286206848099549185>

[sudden sage]

the limits for the outer integral need to be constant

for dxdy, region = 0≤y≤a, y≤x≤a but for dydx the region needs to be described as 0≤y≤x, 0≤x≤a

[obtuse pebble]

@exotic dawn Has your question been resolved?

[exotic dawn]

thank you that makes sense

appreciated

[obtuse pebble]

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[runic arrow]

In how many ways can 12 identical coins be divided among 4 people?

[runic arrow]

using counting principles

[topaz thunder]

can someone explain this to me please i dont get it that much

[brave bramble]

@runic arrow
Read the introduction on this page
https://brilliant.org/wiki/integer-equations-star-and-bars/

[silk plinth]

!status

[obtuse pebble]

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

[runic arrow]

see im confused bc i did 4C12

[obtuse pebble]

@runic arrow Has your question been resolved?

[obtuse pebble]

@runic arrow Has your question been resolved?

[runic arrow]

.close

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[timid silo]

Need help figuring this out

@brave bramble

@sudden sage

not sure

filling it

[warm matrix]

just memorise it: $V =\frac{1}{3}\pi r^2h$

[warm shale]

Obotron

[timid silo]

do you know the answer?

i have no clue

[warm shale]

faiyrose

[timid silo]

mind telling me the answer?

[warm matrix]

$48\pi$ i will not round for you xD

[warm shale]

Obotron

[timid silo]

please tell me the answer, i dont wanna learn anything

what?

[obtuse pebble]

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.

[timid silo]

ah

ok

pi 9 times 4 to the power of 2

[warm shale]

faiyrose

[timid silo]

1/3 times 3.14 times 144

not that i see

48

[warm shale]

faiyrose

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

Is the correct answer 27?

150.72=150.72

54

Awesome, this was actually fun

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@timid silo Has your question been resolved?

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[muted grove]

can anyone suggest what test i can use to see if this is convergent or not

[strong pike]

The simplest one @muted grove

Actually wait, divergence wouldn’t work.

[muted grove]

the simplest one (term one) was actually the right one

thanks

[strong pike]

How so? The limit is 0

That series converges to about 1.8. Looks like the root test.

[muted grove]

somewhere inbetween it becomes 1/e

then overall it goes to 1

[night lodge]

can you send the soln i am wondering how to do it

[muted grove]

I would but I myself dont understand the solution

once i figure it out ima send

[night lodge]

ok send it when u do figure it

[strong pike]

That’s not right. Your limit reduces to e^(-inf). You might be able to recognize it intuitively but if not, raise the limit to e^ln(f) and then lhopitals over the exponent

It’s tedious but you’ll end up with zero

You should be able to recognize it after the first application of l’h actually.

[muted grove]

@strong pike @night lodge here is it

You were right, this is supposed to be root test

But still I do not understand how is it divergent if the result is 1/e

[obtuse pebble]

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[obtuse pebble]

@timid silo Has your question been resolved?

[fickle turret]

Hint: One out of three consecutive numbers is divisible by 3

[timid silo]

ik that still not able to solve that

p odd
p-1 and p+1 even
then how p^2 -1 is divisble by 3

i taken p-1/4 as p1 adn p+1/2 as p2

[fickle turret]

Since clearly p ≠ 3 we have 3|p-1 or 3|p+1. Deviding by 2 or 4 does not change divisibility by 3, so one of the terms on the right is a prime that is divisible by 3. So one of the terms on the right in fact needs to be 3

[timid silo]

then 8 p1 p2 = p^2 -1
then multiplied p both sides
8 p p1 p2 = p (p-1)(p+1)

then what to do ?

| is it modulus or just line which separates numerator and denominator

?

[obtuse pebble]

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[fickle turret]

Try to understand my last message. Don't think what you are trying leads to something

.reopen

[obtuse pebble]

✅

[timid silo]

and p cant be 3 means p-1)/4 or p+1)/2 one of them is 3

[fickle turret]

Yes

[timid silo]

gotyou

thank you smm 🙏

[fickle turret]

You're welcome

[timid silo]

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[hollow flint]

hi i want to know the proof of the exponential laws . i mean like the rule / law{ a^m * b^m= (ab)^m } is true when the a and b both not negative i guess but why is that . i want to see the proof behind that .

[fathom flicker]

what is a^m

a times by a, m times

what is b^m

b times by b, m times

so what is (ab)^m

ab times by ab, m times

ababab..ab

just rearange it

aaaa..abbb...b

both m times

which is a^m b^m

[hollow flint]

yeah but this is when the rule is m must be and integer what about a rational number

or m as a complex number

[fathom flicker]

Q* and C* are abelian over *

[hollow flint]

yes non negative thats what i know. but the textbook only shows the proof of m as being an integer

[hollow flint]

idk .what u mean?

[exotic marsh]

it’s still commutative

like imagine ab^(5/2)

ab^(5/2) = ab^2*sqrt(ab)

= ab*ab*sqrt(ab)

[hollow flint]

ok but i want to see the proof like why it is true

[exotic marsh]

= a*a*sqrt(a)*b*b*sqrt(b)

[hollow flint]

why it work for all complex numbers

[exotic marsh]

this isn’t the claim, ab^5/2 = ab^(2+1/2) = ab^2*ab^1/2 = ab*ab*sqrt(ab) is leadup to proof

[hollow flint]

yes but not as m is being only integer . but also like m as a reak number

[exotic marsh]

= a^(5/2)*b^(5/2)

[hollow flint]

ok but what u ment by this

[exotic marsh]

wdym

[hollow flint]

idk man i am confused too

[exotic marsh]

hold on I will layout proof (?)

[hollow flint]

we can prove the law for only m as being integer by using the proofing by induction methode

but what about number not being integers

[exotic marsh]

u can prove the law for real number by direct proof

u just have to manipulate the expression

[hollow flint]

can u show it?

[exotic marsh]

i just did?

[hollow flint]

wdym

[exotic marsh]

ok hold on I will layout proof

[hollow flint]

bruh that was just the use of the law

[exotic marsh]

no

[hollow flint]

do u guys read the proof in textbook

[exotic marsh]

pls use \begin{align*} plssssss

[hollow flint]

what?

this is i guess ok but do this count as proof

idk like is there is any proof

[warm shale]

faiyrose

[hollow flint]

cant we use like the proof by induction or any other way

[exotic marsh]

but direct proof is proof

[hollow flint]

so give the direct proof

[exotic marsh]

this is direct proof

[hollow flint]

what?

[exotic marsh]

direct proof is just manipulating expressions until claim is satisfied

[hollow flint]

this is an example

[warm shale]

faiyrose

[hollow flint]

like using the rule in many diffrent examples ultill we satisfied?

wait just a minute

let me show u my book

[exotic marsh]

$a^{\frac {c}{d}} \cdot b^{\frac {c}{d}} = (a^\frac{1}{d})^c \cdot (b^\frac{1}{d})^c \newline$
$= (a^\frac{1}{d} \cdot a^\frac{1}{d} \cdot a^\frac{1}{d} \dots \cdot a^\frac{1}{d}) \cdot (b^\frac{1}{d} \cdot b^\frac{1}{d} \cdot b^\frac{1}{d} \dots \cdot b^\frac{1}{d}) \newline$
$= a^\frac{1}{d} b^\frac{1}{d} \cdot a^\frac{1}{d} b^\frac{1}{d} \cdot a^\frac{1}{d} b^\frac{1}{d} \cdot a^\frac{1}{d} b^\frac{1}{d} \dots \cdot a^\frac{1}{d} b^\frac{1}{d} \newline$
$= (a^\frac{1}{d} b^\frac{1}{d})^c = (a \cdot b)^\frac{1}{d})^c = (ab)^\frac{1}{d})^c = (ab)^{\frac {c}{d}}$

[warm shale]

blahaquil

[exotic marsh]

imagine the dots are repeating the multiplication c times

[hollow flint]

okk

[hollow flint]

u used the

this formula here

[exotic marsh]

yes

[hollow flint]

dont mind bro but how u defing he c and d

like do they can be any number?

and also a question here bro why cant be they both negetive i mean a and b

I GTG i am coming in 10 minutes

[exotic marsh]

$a^{x} \cdot b^{x} = a^{\lfloor x \rfloor + { x }} + b^{\lfloor x \rfloor + { x }} \newline$
$= a^{\lfloor x \rfloor} \cdot a^{{ x }} \cdot b^{\lfloor x \rfloor} \cdot b^{{ x }} \newline$
$= a^{\lfloor x \rfloor} \cdot b^{\lfloor x \rfloor} \cdot a^{{ x }} \cdot b^{{ x }} \newline$
$= a^{\lfloor x \rfloor} b^{\lfloor x \rfloor} \cdot a^{{ x }} b^{{ x }} \newline$
$= (ab)^{\lfloor x \rfloor} \cdot (ab)^{{x}} \newline$
$= (ab)^{\lfloor x \rfloor + {x}} \newline$
$= (ab)^{x}\newline$
therefore $a^{x} \cdot b^{x} = (ab)^{x} \newline$ where $\lfloor x \rfloor$ is the floor function of $x, \newline$ ${x}$ is the fractional part of $x$

yeah

[warm shale]

blahaquil

[exotic marsh]

sorry im editing it so much, i just want it to be as clear as possible

[warm shale]

faiyrose

[hollow flint]

But the a and b both can't be negative right?

[exotic marsh]

they can

[hollow flint]

Hello bro

Bro wait a second

[exotic marsh]

m

!! CORRECTION !!

a^x and/or b^x have to be real for this to hold

[exotic marsh]

ya this is the case i just clarified

[hollow flint]

Ok

Bro I am in just 10 th grade

Can I tell me any book where I can find the proof

[exotic marsh]

also if you want to be a horrible person you can say that sqrt(-x) is ±i*sqrt(x) and both take different signs in this case which is objectively wrong but convenient

[hollow flint]

Lol

[exotic marsh]

ok bro i found a better proof

[hollow flint]

Give !!?!

[exotic marsh]

do you know that $a^x=e^{x \cdot \ln{a}}$

[warm shale]

blahaquil

[hollow flint]

Yes

[exotic marsh]

ok

$ab^x=e^{x \cdot \ln{ab}}$ yes?

[warm shale]

blahaquil

[hollow flint]

It should be ab whole over? Right?

[exotic marsh]

yes

[hollow flint]

Ok

[exotic marsh]

$e^{x \cdot \ln{ab}} = e^{x(\ln(a)+\ln(b))}$ yes?

[warm shale]

blahaquil

[hollow flint]

Yes

[exotic marsh]

$e^{x(\ln(a)+\ln(b))}=e^{x\cdot \ln(a)}\cdot e^{x\cdot \ln(b)}$ yes?

[warm shale]

blahaquil

[hollow flint]

Bruh u r defining exponential rules with logarithm when the log come from exponents . It's just a reverse proof. XD .

Loop hole

[hollow flint]

Yes

Go one

On*

Bro ?

[exotic marsh]

by the first law i stated, $e^{x\cdot \ln(a)}\cdot e^{x\cdot \ln(b)} = a^x \cdot b^x$

[warm shale]

blahaquil

[hollow flint]

Ok man

[exotic marsh]

therefore $(ab)^x=a^x \cdot b^x$

[warm shale]

blahaquil

[hollow flint]

Nic3eeee

But this formulas have limitations right

Like it can't be complex numbers

Right?

[exotic marsh]

yes

well

both cannot be complex

one can be

[hollow flint]

Ok

Thanks bro

Man*

[exotic marsh]

ok

[hollow flint]

. close

.close

[obtuse pebble]

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[gleaming fractal]

I need some help with this

[gleaming fractal]

Smth to do with equilibrium

[hexed gull]

what's W

[gleaming fractal]

I don’t know 😭

[hexed gull]

well if you don't know how shall we know :D

gravitational force potentially?

[gleaming fractal]

Idk this was on a past paper so this is all they got

[gleaming fractal]

Yea maybe

[hexed gull]

well we can't work with maybe

[gleaming fractal]

We haven’t been taught what W is

I can find the solution page and see if it makes more sense to u guys

[timid silo]

Is it a physics paper or maths?

[gleaming fractal]

Maths

Vectors

[uneven flame]

Is W work??

[gleaming fractal]

I have no clue

We haven’t been taught what w is

Or what it is

[uneven flame]

Nice! We're all lost together!

[gleaming fractal]

I think the solution was inverse j

[uneven flame]

Considering that the question seems to call for a proof
I'm now really confused

[timid silo]

Bruh

[gleaming fractal]

Yea I’m finding the worked solution now

[timid silo]

I got it

W is mg that is the weight

[gleaming fractal]

Mg?

[timid silo]

Yes

[gleaming fractal]

What’s mg

[timid silo]

Mass times gravity

[gleaming fractal]

O ok

So w is straight down from the ball?

[timid silo]

Write the components of H

[timid silo]

Yes

[gleaming fractal]

Ik it’s to do with sin and cos

[timid silo]

And the components of Weight roo

Too

[gleaming fractal]

Which is?

I’m so lost on this question sorry

[timid silo]

I will send a photo

@gleaming fractal

[gleaming fractal]

Thanks and then how do u get the answer

Hcos30=Wsin30?

[timid silo]

You equate the forces which are in the opposite direction

Inorder for the ball to be in equilibrium

[timid silo]

Aye

[gleaming fractal]

Which forces? The one going down and left?

[timid silo]

The ones going down w.r.t the plane are in the se direction which won't get cancelled

So you should consider the ones going right and left

[gleaming fractal]

What would be my next step from here

Sorry I’m not really understanding

Add i and j forces?

[timid silo]

Write the components of W and H

[gleaming fractal]

What’s that

[timid silo]

This

[gleaming fractal]

So like this but for w as well?

[timid silo]

@uneven flame

[uneven flame]

Ah

[timid silo]

Here

[uneven flame]

Found you

What are we doing?

[timid silo]

W is weight

[uneven flame]

....
Mass?

[timid silo]

^

Mass times gravity

[uneven flame]

Ah

Gimme a sec to write things

Got it

[gleaming fractal]

Are my angles wrong

How do u get z = Wsin30 from this??

[uneven flame]

Lowkey, no idea what you lot were doing

[uneven flame]

This works though

I can explain it step by step if you wabt

*want

[gleaming fractal]

this is similar to the worked solutions

[gleaming fractal]

here

[uneven flame]

Yeah, that's the same thing I did, they just put H on the other side, same direction though, so its the same vector

[gleaming fractal]

whats Hx

[uneven flame]

x component of H

Sorry, i component of H

[gleaming fractal]

O ok

[obtuse pebble]

@gleaming fractal Has your question been resolved?

[obtuse pebble]

@gleaming fractal Has your question been resolved?

[gleaming fractal]

.close

[obtuse pebble]

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[old herald]

I got 36pi

36 pi ×100%

Answer is -1%

I am so losy

[obtuse pebble]

@old herald Has your question been resolved?

[old herald]

No

[obtuse pebble]

@old herald Has your question been resolved?

[old herald]

No

[obtuse pebble]

@old herald Has your question been resolved?

[old herald]

Erm. I don't understand

[orchid trench]

Sorry my mistake

So dh/dt = dh/dV * dV/dt = 3/πh^2 x -3

I don't get 36pi @old herald

[old herald]

Wait let me redo my calculation on a cleaner paper

@orchid trench

My answer is wrong since question answers is -1%

[orchid trench]

You forgot to take the reciprocal of dV/dh

Since you're looking for dh/dV

[old herald]

Oh mb

Let me redo

[orchid trench]

The ans I got is -1.32%

[old herald]

I took dv/dh x dh/V = dh/h

I ended up with 3dh/h

Is 3cm^3/s useless?

@orchid trench

Can u demonstrate on how u got -1.32%

I am kinda lost on that part

[orchid trench]

dh/dt = -pi/4 cm/s
Did you get that?

[old herald]

for me its,
dh/dt = -1/3 (1/3pi h^2)

= -1/9 pi h^2

[orchid trench]

Reciprocal of πh²/3 is 3/πh²

[old herald]

Oh ye

I got -pi h^2

[orchid trench]

^

[old herald]

Wait so its -9 / (pi h^2)

@orchid trench

I found the answer

I am so dumb

The answer I was suppose to find was dh/h

Not dh/dt

I am so slow

Ty for ur help

If mods are here , help me to close this channel ! Ty

[orchid trench]

You're supposed to find dh/dt first, then take that as a fraction of h

[old herald]

I did dv/dh × dh/V

=dh/h

[obtuse pebble]

@old herald Has your question been resolved?

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[dark stirrup]

What is the question

@neat bramble what is the question?

[obtuse pebble]

@neat bramble Has your question been resolved?

[unreal musk]

That’s exactly the same thing what you posted before

Are you asking for a step in that to be explained?

[obtuse pebble]

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[elder reef]

<@&286206848099549185>

[green epoch]

Just solve the inequality, first take an assumption that x>2 and multiply (x-2) both sides

In second assumption take x<2 and change the sign of inequality when you multiply by (x-2) both sides

[elder reef]

after finding value of a and b whatd o i do

also can u tell me how to solve it im getting a wrong ans

[obtuse pebble]

@elder reef Has your question been resolved?

[obtuse pebble]

@elder reef Has your question been resolved?

[timid silo]

have you heard wavy curve inequality ?

$$\frac{\sqrt{(3x-2)}}{x-2}\geq 1$$
$$\implies \frac{3x-2}{(x-2)^2}\geq 1^2$$
$$\implies \frac{3x-2}{(x-2)^2}-1 \geq 0$$
$$\implies \frac{3x-2-x^2-4+4x}{(x-2)^2} \geq 0 $$
$$\implies \frac{-(x^2 -7x +6)}{(x-2)^2}\geq 0$$
$$\implies \frac{(x^2 -7x +6)}{(x-2)^2}\leq 0$$
$$\implies \frac{x^2-6x-x+6}{(x-2)^2} \leq 0$$
$$\implies \frac{(x-6)(x-1)}{(x-2)^2}\leq 0$$

[warm shale]

praee2k

[timid silo]

now draw inequality curve to get the answer

[obtuse pebble]

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[timid silo]

Hello

[timid silo]

someone speak spanish?

[honest forum]

what do you have to do with this expression?

simplify?

@timid silo

[timid silo]

yes

[timid silo]

can you help me?

pls

[honest forum]

first write out the prime factorizations for all of the numbers

[timid silo]

75: 3x5x5 48: 2x2x2x3 -27: -3x3x3

[honest forum]

for the square roots take out primes in groups of 2

for the cube root take out primes in groups of 3

[timid silo]

75: 3x5^2?

Is it so?

[honest forum]

yes

[timid silo]

√3x5^2 + √2^2 x 2^2 x 3 + 3√3^3 + √2^2x 2^2x3

correct?

[honest forum]

nice

[timid silo]

C:

[timid silo]

What do I do now?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

[cloud sparrow]

i assume you want to simplify?

$(2\sqrt5)^2 = 20 = \sqrt{20}^2$

[timid silo]

yes, just solve the equation

[warm shale]

SirGareth

[timid silo]

and -3√20?

i think -3 + 2x5

[cloud sparrow]

im not sure what the q wants, it is difficult to write as a monomial

you can take the sqrt 20 common maybe

[cloud sparrow]

what does the spanish(?) mean exactly?

[timid silo]

performs the operation

[cloud sparrow]

maybe they want you to estimate sqrt 20 idk

[timid silo]

I think it can be solved with prime factors

[timid silo]

what is sqrt20? XD

[silk plinth]

!status

[obtuse pebble]

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

[timid silo]

2

[silk plinth]

Lemme see what you’ve tried

[timid silo]

I don't know if this is right

[silk plinth]

Nope

Do you know how to break it down into perfect square

[timid silo]

2^2x5

the root is cancelled and only the 2 remains, right?

[silk plinth]

Yes

[timid silo]

ouuu I saw my mistake

wait

[silk plinth]

Okay

Lemme see when you do the correction

[timid silo]

so 20-6?

[silk plinth]

Right. 👍 good job

[timid silo]

Thanks bro

[timid silo]

bro can you helpme with this?

[silk plinth]

Square both sides

Have you tried anything actually?

[timid silo]

no idk how

[timid silo]

how? ahahah

[silk plinth]

It's an equation so whatever you do at the LHS you do the same to the RHS. So you square the LHS and square the RHS too

( (\sqrt{x+5})^2 = (2\sqrt{x-1})^2 )

[warm shale]

Gaza

[timid silo]

:0

I didn't know how to do that one

very thanks

[timid silo]

let me try this one

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[wind wedge]

i was able to do the previous questions on this topic with ease, but since this one has e^x it confused me

[wind wedge]

i know that the general antiderivative is just gonna be e^x + C

and i plug 5 into x and set it equal to 9

then solve for C

but it rejects any answers i give it

[cobalt shuttle]

what answer did you give it?

,w solve y'(x) = e^x, y(5) = 9

[warm shale]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by ossihlikesblue

[wind wedge]

ok i tried something like that but the -e^5 i had -e^x....

if its just

ok

i am not ok. ok thank you

i understand

.close

[obtuse pebble]

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[safe stirrup]

What's the use of chararestic polynomial equations of a matrix?

[safe stirrup]

Like... Why do we have this? WHat's the use of this?

[tepid yoke]

To find the eigenvalues of a matrix

[safe stirrup]

And what would this be?

We doesN,t even cover these values in our course.

Unless it's for this :

[tepid yoke]

Eigenvectors are vectors that are only scaled by a transformation, and not rotated. Eigenvalues are how much those vectors are scaled by

[safe stirrup]

Would that be these things called vecteur propre?

I'm sorry for the french/english thing.

A vecteur propre of A is a vector not null where Ax= kX for a specific k

[tepid yoke]

Yep that’s the definition of an eigenvalue

And eigenvectors are the vectors “x” that make that equation you wrote true

[safe stirrup]

Okay so I now know the translation of the term.

But if we calculate P(k)

From what I,m reading it will give me a random number.

[tepid yoke]

You want to solve for k when you set P(k)=0

Then the values of k you find are your eigenvalues

[safe stirrup]

Ahhhh

So we don't plug the k in this equation.

We decide on a random value P(k)

then we find the k that works.

Say I dertmine that P(k) = 1

I solve the equation to find some ks.

[tepid yoke]

You have to make it equal zero

[safe stirrup]

Ooh.

I know this is a dumb question because I have my notes in front of me,but I,m not making the links.

Why do I have to make it equal zero.

[tepid yoke]

det(A-kl)=0 is derived from the equation Ax=kx

[safe stirrup]

Right I see that.

[tepid yoke]

det(A-kl) is equivalent to P(k)

[safe stirrup]

Ooh I think I see it.

So just to make sure I understand.

At first we have Ax = kx

With some shenanigans, we have
det(A - kI) = 0

And we have decided that P(k) = 0

And thus

P(k) = det(A - kI)

Is my reasoning alright?

[tepid yoke]

Kind of

[safe stirrup]

Good enough for me hehe.

So

If I have a random matrix of 2x2

If I need to find the euganvalues.

I can simply plug this equations and solve it (by using the quadratic formula).

Instead of using a system of linear equations.

[tepid yoke]

Yes simply set it equal to 0 and solve the quadratic

[safe stirrup]

Wow it gets ugly real fast.

I mean

The third degree (3x3) is huge.

I would need to calculate tons of stuffs.

Or am I missing something?

[tepid yoke]

Yeah can be depending on the matrix

That’s the general case

[safe stirrup]

yeah if the matrix is triangular or diagonal,. it is fast.

Since the values are on the diagonal.

But let's say it is not.

What would we do.

[tepid yoke]

Let me just show you how we get from Ax=kx to det(A-kI)=0

Ax=kx

Ax-kx=0

Ax-kIx=0

(A-kI)x=0

One important thing is that the zero vector cannot ever be an eigenvector.

So for that equation to be true, A-kI must have a solution

So we find when det(A-kI)=0

[safe stirrup]

Right.

What I would do is go with the linear equations system

instead of these bigs equations ?

[obtuse pebble]

@safe stirrup Has your question been resolved?

[safe stirrup]

Still studying this.

Oooh I think I understand.

it allows to find the eugianvalues directly instead of going by trial error!

I think I finally got it.

Thank you

.close

[obtuse pebble]

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[obtuse pebble]

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[safe stirrup]

I'm very clueless about this.

[safe stirrup]

Any idea on what even this is?

[pseudo swift]

qu'est-ce que tu piges ? qu'est-ce que tu piges pas ici ?

[safe stirrup]

Je pige juste c'est quoi le polynôme caractéristique.

C'est un polynôme qui nous prmet de trouver les différentes valeurs propres d'une matrice.

[pseudo swift]

si t'as une matrice avec polynôme caractéristique P(X) = (X-3)^4 * (X-5) * (X-6)

3 est racine de ce polynôme, donc 3 est valeur propre de ta matrice

l'idée ici c'est de caractériser un peu plus tes valeurs propres

[safe stirrup]

Que va-ton ajouter comme caractéristique?

[pseudo swift]

dans l'exemple, t'as (X-3) qui apparaît 4 fois dans ton polynôme caractérique [(X-3)^4]

[safe stirrup]

Oui.

Donc, Q(x) c'est simplement le restant de l'équation?

[pseudo swift]

oui

[safe stirrup]

On voulait juste isoler la portion de parenthèse qui a le plus grand exposant.

[pseudo swift]

donc l'ordre de la val propre 3 c'est 4

[pseudo swift]

oui la condition Q(alpha) != 0, c'est pour dire que t'as pris tous les (X-alpha), t'en as pas laissé dans Q(X)

[safe stirrup]

Ahh ok. Mais les alphas pourraient changer dans ces affaires là.

COmme dans votre exemple

3, 5 et 6 seraient les alphas.

Mais là on s'intéresse au trois

[pseudo swift]

oui tu peux regarder la multiplicité de n'importe quelle valeur propre

[safe stirrup]

Oh attends.

Donc c'est pas UNE seule chose.

Mais chacune des parenthèses aura cette caractéristique.

[pseudo swift]

t'as (X-3)^4 donc la multiplicité de la valeur 3 c'est 4

[safe stirrup]

Donc chacun des valeurs.

[pseudo swift]

pour la valeur 5 c'est 1

etc...

[safe stirrup]

Donc Dans votre exemple

on aurait 3 valeurs propres

3, 5 et 6

[pseudo swift]

oui

[safe stirrup]

3 a une multiplicité de 4

5 a une multiciplité de 1

De même pour 6.

[pseudo swift]

yes

[safe stirrup]

Maintenant la quesiton qui tue... Why? / On faoit quoi avec?

[pseudo swift]

ça va arriver après tkt

c'est le setup pour pouvoir ensuite parler des matrices qui sont ou pas diagonalisables

[safe stirrup]

Ah oui c'est la prochaine chose La diagonalisation

[pseudo swift]

la multiplicité des valeurs propres est super importante dans ce contexte

[safe stirrup]

Deux min je regarde les prochaines présentations.

[pseudo swift]

mais c'est un assez gros thème la diagonalisation, faut prendre un peu ton temps

[safe stirrup]

Ah ^ca semble être lié à la dimension de l'espace propre de la matrice.

[pseudo swift]

oui

[safe stirrup]

C'est en effet lourd de ce que je vois.

Je regarde quand même. Minute.

Si je comprends bien, B = P^-1 A P
C'es tpour produire une matrice diagonale quelconque

[pseudo swift]

le contexte c'est on te donne A, et tu veux trouver une base (representée par la matrice P) dans laquelle A a une forme diagonale (ici appelée B)

et pour spoiler, B contiendra des valeurs propres, et P sera faite de vecteurs propres

[safe stirrup]

Ah ok spour ça qu'on fait autant d'effort pour ça.

Et je suppose qu'e faire ceci permet de se sauver des maux de t^tes dans des calculs éventuels plus savants.

[pseudo swift]

oui

ça te permet de calculer des puissances de matrices en un éclair par exemple

[safe stirrup]

J'avais vu qu'on avait à faire cela.

Mais je n'ai pas vu dans nos notes où exactement.

[pseudo swift]

ça pop dans des systèmes d'équations récurrentes

dans des équations différentielles ça sert aussi

c'est pas juste pour faire beau (c'est assez beau), ça a qques applications sympa aussi

[safe stirrup]

J'en doute pas. J'ai just epas accès à tout ça.

On dirait que ma présentation ne parle pas de monter la puissance d'une amtrice alors que les exercices en font.

Ya clairement quelque chose que je ne comprends pas enocre ahah.

Tant pis, je révise.

[pseudo swift]

c'est probablement que les profs vont en parler en séance d'exercice

c'est pas dur à faire quand t'as diagonalisé déjà

[safe stirrup]

Alright, je vais essayer d'avancer.

Un gros merci encore!

J'aurai surement d'autres questions tantôt ahah mais pour l'instant ça va aller.

.close

[obtuse pebble]

Channel closed

Closed by @safe stirrup

Use .reopen if this was a mistake.

[obtuse pebble]

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[safe stirrup]

Hello it's still me with eigenspaces of a Matrix. I have tried something.

[storm spade]

yo

[safe stirrup]

Here is hwat I expect to have

So far I have this.

[storm spade]

im alright

nothing special

you?

[unreal musk]

!occupied

[forest sinew]

#❓how-to-get-help

[obtuse pebble]

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).

[pastel wren]

if you want to ask a question open a new channel please

[safe stirrup]

I have no idea how to get to (-3, 0 1)

[modest ingot]

SIRRRR

get your channel please lol

[pastel wren]

youre taking ages man

go to another channel!!!!

[modest ingot]

#❓how-to-get-help

!!!!!

[pastel wren]

<@&268886789983436800>

someone asks for a ban

[modest ingot]

good

[forest sinew]

sorry about your channel niamek

[safe stirrup]

Not a big deal hehe.

[forest sinew]

it should be over soon

[modest ingot]

just start a new one i guess if u want

[safe stirrup]

I can.

.close

[obtuse pebble]

Channel closed

Closed by @safe stirrup

Use .reopen if this was a mistake.

[hot hazel]

,purge --from 905977117047652392

oh ok

we were dealing with it

[modest ingot]

[warm shale]

Couldn't find a member matching 905977117047652392!

Apr 20, 2024, 10:31 PM

Couldn't find the requested user, cancelling purge.

Apr 20, 2024, 10:31 PM

[safe stirrup]

,reopen

.reopen

[obtuse pebble]

✅

[safe stirrup]

Reopening then.

[storm spade]

wassup

[safe stirrup]

Looking to obtain(-3, 0 1)

Eigenspace of a Matrix

[forest sinew]

you mean by hand?

[safe stirrup]

Yes.

[forest sinew]

you were looking at the polynomial i guess

[safe stirrup]

We have the eigenvalue 2 already.

[forest sinew]

oh assuming the eigenvalue

chartbit u can take over u are better than me

[safe stirrup]

Basically, I need the eigenspace of this.

[tawny copper]

Not related to the question but did that 0 / 0 dude just got banned 💀

[unreal musk]

Anyways, notice e.g. that from that kernel you have, say, for the eigenvector (x, y, z) you'd have, say, y = 2x + 6z

From that, you can find the (1, 2, 0) eigenvector, and the other one you'd find is (0, 6, 1)

[safe stirrup]

But they are different from the calculator. is that normal?

[unreal musk]

Linear combinations, namely (0, 6, 1) - 3(1, 2, 0), gets you that (-3, 0, 1)

[safe stirrup]

Oh okay, so my approach is okay so far. so I assume I can reduce the matrice to this: (Hold on)

[unreal musk]

[and yep, as long as they're "equivalent" (so e.g. have the same span and are linearly independent) then you're fine]