help-13

[high kernel]

Am I supposed to keep it as 1.05s

[buoyant latch]

Look what I wrote

[high kernel]

If so it would be 1.09 sc

[buoyant latch]

What is the percentage difference between A(s) and A(1.05s)

[high kernel]

It would be 1-1.05

So 0.5

[buoyant latch]

What

A(1.05s)-A(s) ≠ 1-1.05

Where are you getting 1-1.05 from

[high kernel]

I thought s =1

[buoyant latch]

No!!!

Where does it say s=1

[high kernel]

I thought that’s the way you’d do it

[buoyant latch]

No one nowhere said s=1

You can’t just be changing stuff out of thin air

[high kernel]

You’d get 1.05A- As

[buoyant latch]

What???

A is a function

[high kernel]

When I tried doing this

[buoyant latch]

When you type A

It should be like A(something)

A is a function that takes an input and returns an output

[high kernel]

Wait then what would you like me to solve I’m confused

[buoyant latch]

What is A(s) equal to

It’s in the question

[high kernel]

Cs1.84

[buoyant latch]

Ok

Whats A(1.05s) equal to

[high kernel]

C(1.05s)^1.84

[buoyant latch]

Ok

[buoyant latch]

Compare

[buoyant latch]

And

To find the percentage difference

Not quite

[high kernel]

Can you please tell me how to do that

[buoyant latch]

Do you know how to find the percentage difference of 2 numbers

[high kernel]

Well I know how to find for 1

[buoyant latch]

Ok show me

[high kernel]

You just divide it by 100

For example 2 is 0.02%

[buoyant latch]

What

How is 2 = 0.02%

[buoyant latch]

This needs 2 numbers

Like what’s the percentage difference of 1.05 from 1

[high kernel]

Sorry I meant divide it by the total and then multiply by 100

I thought you said something else

[buoyant latch]

Ok doesn’t matter, try do this

[high kernel]

It would be 105%

[buoyant latch]

No that’ll be 2.05

[crimson sedge]

you guys know of a good physics server?

[buoyant latch]

#old-network

[crimson sedge]

thanks

[high kernel]

Why?

[buoyant latch]

And try ask in #discussion next time

[buoyant latch]

Because we’re asking percentage difference here

[crimson sedge]

it says no access?

[high kernel]

So 1.05-1/1

[buoyant latch]

It’s also the first sentence of #info

[buoyant latch]

Right

[high kernel]

0.05

[buoyant latch]

Which is what in percentage

[high kernel]

So I’d multiply that by 100 right

[buoyant latch]

Yes

[high kernel]

It’s 5

[buoyant latch]

5 what

[high kernel]

%

[buoyant latch]

Ok

Now what about this

What’s the percentage difference of 2.5 from 2

[high kernel]

25%

[buoyant latch]

Write out the steps

[high kernel]

Wait

I messed up

[buoyant latch]

(It’ll be important for this next step)

[high kernel]

12.5%

[buoyant latch]

.

[high kernel]

25% is right I did something on my calculator

Because

2.5-2

=1/2

/2

=1/4

[buoyant latch]

Or in other words

[high kernel]

X100

=25%

[buoyant latch]

(2.5-2)/2 * 100

Yeah?

[high kernel]

👍

I understand this part

[buoyant latch]

Ok this next one might be a bit more tricky

[high kernel]

Ok

[buoyant latch]

What’s the percentage difference of b from a

[high kernel]

B-a?

[buoyant latch]

Think back to this

There’s a reason I wrote this out

[high kernel]

So it would be this

(B-A)/A *100

[buoyant latch]

Ok good job

Now last step

[high kernel]

Ok

[buoyant latch]

What’s the percentage difference of A(1.05s) from A(s)

[buoyant latch]

Refer to this part for what A(s) and A(1.05s) is

[high kernel]

Question which one would be a and b from the example above

[buoyant latch]

Oh

I wrote it wrong

The baseline is A(s)

We want to compare how much the error in measurement affects the result

So we want to know how far is A(1.05s) from A(s)

[high kernel]

So (A(1.05)-A(s)/A(s))*100

Like that?

[buoyant latch]

Yep

[high kernel]

A(s) would cancel out no?

[buoyant latch]

(x-y)/y ≠ x - 1

So no

[high kernel]

Ok

Anyway now what?

[buoyant latch]

Look back through here

[high kernel]

Cs1.84

Is what I said before

Haha it just showed up when I typed

[buoyant latch]

So replace A(s) and A(1.05s) with what you said here

[high kernel]

Ok

Let me show you on paper

[buoyant latch]

Yes

[high kernel]

I don’t include A() right

Everything after the equal sign is what I include or do I do all of it

[buoyant latch]

Everything after the equal sign

The equal sign means the left side is exactly equal to the right side

Always.

[buoyant latch]

(Although there’s usually some conditions, but at this level omitted)

[high kernel]

Like this?

[buoyant latch]

Man what happened to the s

[high kernel]

It’s there?

[buoyant latch]

The first one

It’s not A(1.05)

It’s A(1.05s)

[high kernel]

My fault I was looking at what I wrote above

Is it good now

[buoyant latch]

Yes

Now, actually do notice you can factor the c and the s^(1.05) out and cancel them

You could’ve done this from the beginning

But I wanted you to write the steps properly and know where everything comes from

[high kernel]

I see thank you for that

I actually have learned a lot from you

[buoyant latch]

👍

[high kernel]

When I evaluate my answer I got 9.39

[buoyant latch]

,calc 1.05^1.84

[wraith dagger]

Result:

1.0939269211317

[high kernel]

Wait I have a question

[buoyant latch]

Yea?

[high kernel]

Which C are you talking about

And S^1.05

[high kernel]

Would this be the answer?

[buoyant latch]

$\frac{c(1.05s)^{1.84}-cs^{1.84}}{cs^{1.84}}\cdot 100\
\frac{cs^{1.84}(1.05^{1.84}-1)}{cs^{1.84}}\cdot 100$

[wraith dagger]

Frosst

[high kernel]

What would the answer be?

[buoyant latch]

This is right

[high kernel]

I see so it would be 9.39%

[buoyant latch]

Yeah

[high kernel]

That’s the final answer

[buoyant latch]

Its 9.39% of the correct measurement

[high kernel]

Thank you so much

[buoyant latch]

So let’s say it measures the leaf area to be 100m²

[high kernel]

I’m so sorry it took so long I’m still very new to this

[buoyant latch]

Then the real measurement would be any number that if you look at +- 9.39% will have your measured amount

[high kernel]

Thanks 🙂

[high kernel]

So 9.39 is the final answer

[buoyant latch]

Yea

[cedar kiln]

@high kernel Has your question been resolved?

[cedar kiln]

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[stone charm]

am i doing this right

[cedar kiln]

@stone charm Has your question been resolved?

[cedar kiln]

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[quasi sentinel]

Wtf does this 2nd point mean

[quasi sentinel]

@buoyant latch

<@&286206848099549185>

[buoyant latch]

!15m

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[buoyant latch]

We are just volunteers that help around please don’t directly ping us out of the blue

[cedar kiln]

@quasi sentinel Has your question been resolved?

[slow thicket]

the more than 400m squared one?

[cedar kiln]

@quasi sentinel Has your question been resolved?

[cedar kiln]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[hollow kraken]

why this one has no meaning ?

[slow thicket]

cause the pi affects the -2 not just the 2

[hollow kraken]

can i not put an exponent on a negative number ?

[wraith dagger]

Stephen

[upbeat lotus]

$(-1×2)^{π}$

[slow thicket]

counterexample:

[wraith dagger]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[upbeat lotus]

Right?

[slow thicket]

(-2)^2

[hollow kraken]

yeah

[slow thicket]

thats 4

[hollow kraken]

yes

[slow thicket]

but -(2^2) is -4

you cant just take the -1 out

[upbeat lotus]

Just split it

[hollow kraken]

so ?

[slow thicket]

so (-2)^2 is not the same as -(2^2)

therefore you cant take the -1 out

(-1)^2 = 1

[upbeat lotus]

$(-1)^{π}×2^π$

[wraith dagger]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[slow thicket]

how do you get that × symbol

[upbeat lotus]

$e^{iπ}=-1$

[wraith dagger]

𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛

[upbeat lotus]

Multiplied by

[slow thicket]

what

?

i know what it means lol

how do you get the symbol typed

[upbeat lotus]

Answer would be in complex number right

[cursive harbor]

\times

[upbeat lotus]

×

It is in my keyboard

[slow thicket]

oh but that didnt show in the code

oh what

didnt know that existed

[slow thicket]

anyways @hollow krakendo you get it now

[upbeat lotus]

Use this identity

[slow thicket]

we're not actually solving for it

just picking what it can be represented for

[slow thicket]

just basic exponent rules

the pi is there to trick them

[upbeat lotus]

Oh I already solved it without knowing 😂

[hollow kraken]

oh okay

thanks

.close

[cedar kiln]

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[crimson sedge]

Let $H$ be a Hilbert space and $A \in \map \BB H$. Show that

\vs{3 mm}
a) $\overline{\textxs{image}A} = \parens{\ker A^*}^\perp$

\vs{3 mm}
b) $\ker A = \parens{\textxs{Image}A^*}^\perp$

\vs{5 mm}
\textbf{My attempt:}

\vs{3 mm}
a) Take any $x \in \textxs{image} A$. Then there is a $y \in H$ such that $x =Ay$. For any $z \in \ker A^$, we have [
\angles{x,z} = \angles{A_y , z} = \angles{y, A^ z} = \angles{y, 0} = 0
]
Hence, $x \in \parens{\ker A}^\perp$ . This proves that $\textxs{Image}A \subset \parens{\ker A^*}^{\perp}$

\vs{3 mm}
But how do I exactly proceed from here? This isn't sufficient to prove a) I believe

[crimson sedge]

I did this like a month ago but I completely forgot how

Any hints

[gilded elm]

time to go to #point-set-topology

[crimson sedge]

This is more analysis tbh

[gilded elm]

sure

#advanced-analysis

[crimson sedge]

Smh PIGEON cultivate your analysis to help me with my analysis

[gilded elm]

my analysis is too low level

you can call ryc

[crimson sedge]

Yeah sure I have his number of course

Let me ring him up

[gilded elm]

he is asleep

go to adv analysis bruh

you gotta stop opening the overly advanced questions here

[crimson sedge]

I have to land my fish in both seas derpz..

Some of the smart people here never check the advanced channels

[gilded elm]

thats untrue

lmao how r u doing adv analysis but asking me how to factor a quadratic

[crimson sedge]

[vagrant elbow]

the smart people only check the advanced channels

[crimson sedge]

Life of a mathematician

[gilded elm]

you are smart if and only if you check advanced

[vagrant elbow]

only and if if

[gilded elm]

lix did you see the

super sullyable message neon sent in the server

[crimson sedge]

What is it

Also what is "the server" There are a lot of them now

[gilded elm]

#1082281192033890354 message

"i like computations more than proofs"

its not even the most sullied message there atm

[crimson sedge]

I think the one below it is more... Concerning

[gilded elm]

owner pinged everyone and got 7 sullies lmao

[vagrant elbow]

derp

[gilded elm]

i think that one was an intentional joke

my crankery also got sullied

[crimson sedge]

OMG CAN I DO IT TOO

I wanna do it

I willl do it

[vagrant elbow]

:/

[crimson sedge]

Wait who is the owner

[gilded elm]

not me

[paper trout]

yo

who can help me

[gilded elm]

nobody

[paper trout]

with a question really quick

[crimson sedge]

me but I charge 35 euros per question

Sorry

[paper trout]

35?

[gilded elm]

if its a 1 line question maybe

[paper trout]

JESUS

[gilded elm]

nope

[crimson sedge]

[gilded elm]

go make your own channel

[crimson sedge]

Okay fr go to #❓how-to-get-help

[paper trout]

bro

i have my exam in 20 minutes

[gilded elm]

it doesnt cost money to get help here

[paper trout]

wheer

[gilded elm]

but be patient

oh

oof

lix will answer if you pay 69420 euros

[vagrant elbow]

procrastination destroys lives

[gilded elm]

yeah sorry if ur exam is in 20 mins ur not gonna be able to learn this in 20mins

[paper trout]

i got 3.6

[crimson sedge]

I'm a high end mathematician Derpz! Time is money!

[paper trout]

idk if its right

[gilded elm]

who knows

now

dont invade

make your own help channel

[paper trout]

alr mb

[gilded elm]

good luck

with ur exam

[paper trout]

yea thx

i am prepared 90%

the other 10% is that question

[crimson sedge]

This channel: discussion 4

Let's get it boys

[crimson sedge]

The cow in ur profile is so funny to me idk why @gilded elm

Why does it have such a derpy face

[gilded elm]

lol

its the

sphere = cow shit

[crimson sedge]

Hey derpz

Topology is basically like

When the cup

Turns into a donut right

[gilded elm]

donut = coffee cup yes

i wish it was like that

i am so sick of closed and open sets

[crimson sedge]

Do clopen sets then my boy

Have the best of both worlds!

[vagrant elbow]

if donut equals coffee cup, and human = donut, then human = coffee cup?

[gilded elm]

theyre even more irritating

is a human a donut tho

oh actually yes

[vagrant elbow]

there is one continuous hole all the way through

[gilded elm]

holes cant be continuous

only functions can

[vagrant elbow]

mb

[gilded elm]

a manifold is a second countable hausdorff space that is locally euclidean

[crimson sedge]

Manifold is such a funny name

[sacred grail]

[crimson sedge]

I always imagine some lump of clay that u just start shaping stuff with when I hear it

[gilded elm]

omg lix

#algebraic-geometry message

most confused user

they really went into alg geom channel for this

[crimson sedge]

LOL

They really need to like

[paper trout]

i solved it

[crimson sedge]

Call abstract algebra something else

[sacred grail]

because its geometry, and uses algebra

[crimson sedge]

Or something

[sacred grail]

[paper trout]

it is 6.2

[crimson sedge]

U r so smart snow

[gilded elm]

commutative too

[crimson sedge]

Well solve it while you are at it then derpz

Duh

Are u working hard or hardly working??

Fuck I'm smelling someone cooking some nice chicken

MAan I'm hungry

Let's go to subway or something at 8 or something my g @sacred grail

I will roll up to ur place in 15

[sacred grail]

[gilded elm]

im in a meeting

[wraith dagger]

User cancelled member selection.

Mar 9, 2023, 8:16 AM

[gilded elm]

,ti 118663865264242688

[wraith dagger]

This user hasn't set their timezone! Ask them to set it using ,ti --set.

[gilded elm]

,ti --at Asia/Sydney

[wraith dagger]

No matching timezones were found!

[gilded elm]

,ti --at Australia/Sydney

[wraith dagger]

The time in Australia/Sydney is 07:16 PM (AEDT) on Thu, 09/03/2023.
DerpZ is 3 hours behind, at 04:16 PM (AWST) on Thu, 09/03/2023.

[upbeat lotus]

Asia Sydney 💀

[gilded elm]

idk some timezones are weird

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

Woops forgot to close this

[cedar kiln]

Channel closed

Closed by @stray spruce

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[cedar kiln]

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[long mauve]

Hwlp

[long mauve]

I'm stuck after I convert 1/u + 1/v = 1/f into linear eqn

what should I do next

[cedar kiln]

@long mauve Has your question been resolved?

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
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[wary blade]

When we say a set is closed under addition, doesn't it have to be an infinite set?

[wary blade]

take the largest element and the 2nd largest element and add them, that value must be larger

so its infinitely expanding right?

[crimson delta]

no

do you know modulo?

[wary blade]

yes

[trail rivet]

if you mean 'is there a finite set of integers closed under addition?'

then {0}

[wary blade]

yes

yes

ok makes sense

[trail rivet]

otherwise no

but it depends how you define addition and what the elements of the set are

[wary blade]

unless we consider

remainders?

modulo stuff

[trail rivet]

that comes under how you define addition

[wary blade]

alright

nice

[crimson delta]

in theory you could define whatever operation you want and just call it addition

[wary blade]

interesting

[cedar kiln]

@wary blade Has your question been resolved?

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

[crimson sedge]

What is going on in this question?
solving one way it give different answer and other way its giving completely different answer.

I have to find derivative of the function

[crimson sedge]

which is the correct method

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@uncut zodiac Has your question been resolved?

[cedar kiln]

Channel closed

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[cedar kiln]

Available help channel!

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Remember:
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[prisma phoenix]

me

[prisma phoenix]

okay i don't have a mobile so can i type the answers

[distant cedar]

@prisma phoenix What is your question?

[prisma phoenix]

wait a sec

i was studying chem so

heres the question

the angles of a triangle are (x-40) , (x-20), and (x / 2 - 10). find the value of x

i am in 6th grade so don't judge me

[crimson sedge]

the sum of the angles in a triangle is 180°

[distant cedar]

This is not a chemistry question

[distant cedar]

Yep use this

[prisma phoenix]

okay

[crimson sedge]

solve the equation (x-40) + (x-20) + (x/2 -10) = 180°

[prisma phoenix]

3x + 70 = 180?

3x = 180 - 70?

3x = 110?

[crimson sedge]

hmm nop

[prisma phoenix]

x = 110/3?

[distant cedar]

There is a x/2 for the 3rd term

[crimson sedge]

yep

[prisma phoenix]

yes

[distant cedar]

(x/2) = 0.5x

[prisma phoenix]

okay

[crimson sedge]

so it would be 5x/2 - 70° = 180°

[prisma phoenix]

how 5x?

there are only 3 variables

[distant cedar]

It's not 5x , it's 5x/2 or 2.5x

[prisma phoenix]

okay

[distant cedar]

You have two x's and one x/2

[prisma phoenix]

so 2.5x - 70 = 180?

2.5x = 180 + 70

[distant cedar]

Yes

[prisma phoenix]

so it is

2.5x = 250

x = 10?

[crimson sedge]

5x/2=250

multiply both sides by 2

5x=500

[distant cedar]

This is correct as well , but what you get from here isn't

[prisma phoenix]

okay, so what should i do next?

[crimson sedge]

divide both sides by 5

[prisma phoenix]

wait its 100

[crimson sedge]

yea

[prisma phoenix]

cuz 2.5 * 10 is 25

[distant cedar]

Yes

[prisma phoenix]

2.5 * 100 is 250

[distant cedar]

That's correct

[crimson sedge]

and the answer is 100°

[prisma phoenix]

First angle = 100 - 40 = 60
second angle = 100 - 20 = 80

third angle = 100- 2 = 50- 10 = 40

[crimson sedge]

yes

[prisma phoenix]

60 + 80 + 40 = 180

[crimson sedge]

and the sum will be 180

yep

[prisma phoenix]

thx a lot

[crimson sedge]

np!

[prisma phoenix]

my exams are coming up next week and i had no way to connect to my math teacher so thx a lot

.close

[cedar kiln]

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[worthy ridge]

how do I do this question? Is this even an additive or multiplicative group?

[cedar kiln]

@worthy ridge Has your question been resolved?

[cedar kiln]

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[river ermine]

How many seven digit odd numbers can be made using the digits 0, 1, 2, 3, 4, 5, 6, 7 and 8 if no digit may be used more than once and zero cant be the first digit?

[river ermine]

8 * 8 * 7 * 6 * 5 * 4 * 3

= 161280

then i tried dividing by 8 then multiplying by 4 because theres 4 odd numbers that can be at the end of the 7 digit number

161280/4 = 40320

[tropic oxide]

kind of convoluted, that

[river ermine]

wait nvm

[tropic oxide]

you could have simply written 8 * 4 * 7 * 6 * 5 * 4 * 3

ah, wait.

[hushed talon]

40,320

[tropic oxide]

hold up.

4 choices for the last digit (1, 3, 5, 7), then 7 choices for the first digit! since we can't have 0 and we already used up one for the last

[hushed talon]

4 × 8 × 7 × 6 × 5 × 4 × 3 = 80640

80640/2 = 40320

For the first digit, we have four choices (1, 3, 5, or 7)

For the second digit, we have eight choices

[river ermine]

they cant be repeated tho

[hushed talon]

For the third digit, we have seven choices (all digits except the one already chosen for the second digit)

[river ermine]

so wouldnt the options decrease

[hushed talon]

no no for second, all digits are available since the first digit has already been chosen

[river ermine]

ohh yeah mb

[hushed talon]

For the fourth digit, we have six choices (all digits except the two already chosen)

For the fifth digit, we have five choices.

[river ermine]

but the final digit has to be odd to make it an odd number

[hushed talon]

For the sixth digit, we have four choices.

[river ermine]

however when multiplying the order doesnt matter right?

[hushed talon]

For the seventh digit, we have three choices (only the odd digits 1, 3, and 5 can be used to make the number odd).

nope

[river ermine]

cant 7 be used

[hushed talon]

it can

[river ermine]

the answer is 70560, i just want to be able to get to that point

zero cant be used in the first digit so 8 choices

.

<@&286206848099549185>

can someone push me in the right direction?

8x8x7x6x5x4x3

=161280

[cedar kiln]

@river ermine Has your question been resolved?

[stray river]

what?

$8^2 x 7 x 6 x 5 x 4 x3$

bruh

[stray river]

?

ok

[river ermine]

flame me all you want, as long as i understand how to do it by the end

[stray river]

144

[stray river]

haha

not going to do that

[river ermine]

whats 144?

[stray river]

combinatorics questions are hard sometimes

let me explain

[river ermine]

please do

[stray river]

wait

this is 7 digits

ok no

it's not 144

[river ermine]

yeah thought so

[stray river]

lol

[river ermine]

this is like introduction level

[stray river]

lemme explain how to do it tho

[river ermine]

alright

[stray river]

you have to fill the last place first

because that is what makes a number odd

[river ermine]

4 choices right?

[stray river]

why four?

you have 7 digits

lol

wait

yes

damn it

was answering another question

got confused

[river ermine]

all good

[stray river]

yes 4 choices

[river ermine]

0 cant be at the beginning so 8 choices for first digit

and with the second digit 0 can be used so its 8 choices again right?

[stray river]

yes

[stray river]

7

right?

[river ermine]

0, 1, 2, 3, 4, 5, 6, 7, 8 are the given numbers, 9 in total

7 digit number made from those, trying to find the odd ones

[stray river]

ok I see

and first digit cant be zero

so 8

alr

so

8,9,9,9,9,9,4

right?

[river ermine]

no repetition

mb

shouldve told you that

[stray river]

hm

nah nah

I shouldve read the question

let me think bout this

give me a moment

do you have any ideas

[river ermine]

8x8x7x6x5x4x4 shouldnt work because from the four chosen in the 6th digit there should only be 3 left

so i tried 8x8x7x6x5x4x3 and got 161280 as the total amount of 7 digit numbers possible

[stray river]

hmm

[river ermine]

i tried to get the amount of odd numbers from 161280 by first dividing by the amount of different starting numbers

that being 8

[stray river]

hmm

[river ermine]

then multiplying by 4 because thats the amount of odd numbers there are

but that didnt work

[stray river]

but 8x8x7x6x5x4x3 should work

[river ermine]

but thats the total amount of 7 digit numbers that can be made right?

[stray river]

ooh

odds

yeah

[stray river]

yeah yeah

do you know the last odd number in the set?

[river ermine]

i also tried (161280/9)x4 which equals 71680, this is the closest ive gotten to the actual answer

[stray river]

no

[sturdy inlet]

Let's fix last spot Firstly

4 options

[river ermine]

yup

[sturdy inlet]

Now we have 6 spots left

[river ermine]

0 cant be first digit so 8 options

[sturdy inlet]

First spot has 7 options , second has 7 , third has 6 , fourth has 5 , fifth has 4 sixth has 3

[stray river]

what is the last odd number @river ermine

?

[river ermine]

why 7

[river ermine]

wdym by last odd number? is it not 7

[stray river]

bro

[sturdy inlet]

We already fixed one at last, so we are left with 7 numbers

[stray river]

161280

in 161280

[river ermine]

there are 9 numbers in total

[stray river]

last odd num

[sturdy inlet]

7765434 = 70560

[river ermine]

youre going to have to bear with me here

idk what you mean by last odd number

what, is it just 161279

[stray river]

no sorry

one sec

[river ermine]

could you explain why its 7 for both the first and second digits?

[stray river]

80640 odd numbers?

[river ermine]

i want to be able to do this in the future without bothering strangers on discord

the answer in the textbook is 70560 so spector is correct

but i want to know why it isnt 8 for the 1st and 2nd digits

wait

[stray river]

ic

[river ermine]

there must be 4 at the end

then leaving 8

but 0 cant be at the beginning

meaning that its 7

[stray river]

0, 1, 2, 3, 4, 5, 6, 7 ,8

8 options available right?

excluding 0 ofc

I just did

[sturdy inlet]

We have 9 total ways of filling a spot ,but the last number must be odd so we pick a nunber out of the four possible odd nunbers , now we are left with 8 nunbers but out of the eight one is zero so have 7 numbers for the first spot now we pick another nunber for the first spot and we are left with 7 again as we can use zero now and so on

[river ermine]

yeah for the first digit

[stray river]

yes

[river ermine]

i see,

[stray river]

but why is it not 8 for the first spot

?

[river ermine]

from the 9, 1 must be at the end

leaving 8 options

but you must exclude 0

so its 7, it think

thats what i understand

[stray river]

but that doesn't explain why 7 for first and second right?

hmm

wait

it does

[river ermine]

i thought i understood what was going on but i think im losing it

1 must be at the end, 0 cant be at the beginning meaning 7 options for 1st digit

2 have been chosen so 7 remaing including 0 for the 2nd digit

nvm

i got it

thanks a lot, ill try attempting the second part of the question. if i fail miserably ill come back ig

[stray river]

nice

thanks @sturdy inlet

[sturdy inlet]

You shouldn't sentence it like that , there is an ODD number at the end not necessarily one , that is why we multiply by four at the end for four different cases of four different odd nunbers

[river ermine]

yeah i meant one of the numbers have been chosen

[sturdy inlet]

Oh , my bad

[river ermine]

nah you were right tho, the way i wrote it was technically incorrect

[sturdy inlet]

If you are having troubles with pnc look up some solved examples first and then ponder on some smilar questions . If your answer seems to be correct but it doesn't match make sure you understand where you went wrong .

[river ermine]

sure thing, thanks a lot

ok im still stupid, part b asks how many of the numbers from the previous question are less that 4,000,000

i dont think my teacher has ever done questions like these with us in class

[stray river]

4 million?

we got 70,560?

[river ermine]

thats the amount of odd numbers

[stray river]

yeah?

[river ermine]

4 million is the actual number in the 7 digit format

4000000

[stray river]

how many total numbers?

[river ermine]

its out of the odd numbers that we had

70560

[stray river]

we can get 4 million different numbers from 7 digit format

so isn't it just 3 million 999,999

only 4 million is equal

[river ermine]

70560 is the amount of odd numbers that could be made with 0, 1, 2, 3, 4, 5, 6, 7 and 8

[stray river]

how many total numbers can be made?

[river ermine]

no its just out of 70560

[stray river]

I don't understand the question at all

all 70,560 numbers are less than 4 million

[river ermine]

b) how many of these numbers from part a (70560) are less than 4 million

[stray river]

oooh

ok

[river ermine]

nah cuz 6543217 is greater

[stray river]

I get it

yeah yeah

nvm I'm just dumb

[river ermine]

better off than me

i swear my teacher never went through questions like these in class

[stray river]

we should bring spector

@sturdy inlet

[sturdy inlet]

Yes?

[stray river]

...

help?

[sturdy inlet]

Well for less than four million thr first digit can only be 1 , 2 ,3

[stray river]

ic

4 million

is 2^8 x 5^6

[river ermine]

3 options for first digit

[stray river]

can you explain why?

[river ermine]

4>3,2,1

[stray river]

yeah

but how do you explain that

in terms

[sturdy inlet]

The digits should still be 7 right , or are lower digit numbers allowed

[river ermine]

7 digits

[river ermine]

same amount of digits so starting digits must be less than 4

[stray river]

yeah but I meant in mathematical terms

but nah

that's ok

I guess you just say

4>3,2,1

[sturdy inlet]

X< 4* 10^6
x÷10^6 < 4
If we let x be 5million the x = 5 * 10^6
Now x÷10^6 = 5*10^6÷ 10^6 =5
So x÷10^6 greater then four hence 4 or higher integers are not allowed since the question asks for less than four million we exclude four , if it said less than equal we include it .

[stray river]

?

wait no

so just (3)(6)(5)(4)(3)(2)(1) right?

[river ermine]

i think its 4 at the end still

cuz theyre still odd

[stray river]

but why?

oh right I forgot

so just 3 x 6 x 5 x 4 x 3 x 2 x 4

?

no

it's wrong

because

[river ermine]

3x7x6x5x4x3x4

?

[stray river]

yeah

[sturdy inlet]

Just use the same framework as before

[stray river]

30,240

[river ermine]

the textbook says the answer is 25200

[stray river]

oof

[river ermine]

idk what else could be done

[stray river]

lets think

try cutting up 25200

ok so can a number be:

3321570

[river ermine]

for less than 4 million or something else?

[stray river]

4million

less than 4 million

[river ermine]

no repetition and it has to stay odd im pretty sure

god im so fried

[stray river]

so why would it be (3)(7)

we can't repeat the first 3 digits

those are gone right?

those can't be one of the values

[sturdy inlet]

We choose one at a time not all three

Try and find what went wrong , if you can't ping me , but try for a while first

[river ermine]

3x7x6x5x4x3x2 because 1 and 3 are less than 4 but cant be used in the last digit

but that gives 15120

[sturdy inlet]

You are the wrong way about it

You don't use 1 and 3 at first place but 1 or 2 or 3 , if you even use 1 in first place then you would still have a odd number to pair it up with.

[river ermine]

but they cant be used in the last digit

leaving 5 and 7

[sturdy inlet]

They can

If you have 3 as first digit then 1 can be at last place and vice versa

[stray river]

yes because we can have 1 in first place and have 3 in last

[stray river]

or that yes

[river ermine]

yeah you two are right

mb

so is it still 3 at the beginning

[stray river]

yeah

[river ermine]

i literally dont know why its not 3x7x6x5x4x3x4

[stray river]

yeah

that's what I extrapolated

but we get 30,240

[river ermine]

but i still highly doubt that the textbook is incorrect, so theres got to be something we aren't seeing

problem is i have no idea what it is

[stray river]

which textbook is this?

ncert?

or like RD sharma

[river ermine]

AJ Sadler Mathematics Specialist units 1 and 2

an australian one

[stray river]

I saw a question like this in RD sharma

I'm certain

[river ermine]

im not familiar with rd sharma's textbooks

[stray river]

which chapter?

[river ermine]

chapter 2

[stray river]

unit as well

[river ermine]

wait

unit 2

chapter 2d

question 8b

[stray river]

which one @river ermine

[river ermine]

whered you find that from?

the question is in unit 2

[stray river]

uh

what now

what happened to spector

bro you there

[river ermine]

rip

[sturdy inlet]

Yeah

[river ermine]

@sturdy inlet do you know how we can get to 25200?

[sturdy inlet]

Yeah

[river ermine]

could you please explain the steps?

[sturdy inlet]

So you were right here and i kind of misdirected you. What i meant to say still holds true but i should have elaborated

[stray river]

3 x 7 x 5 x 5 x 4 x 4 x 3

?

is 25200

[river ermine]

but what are the reasons for using those numbers

[stray river]

Idk

I just say a pattern

and I took it

[sturdy inlet]

I am typing out the reason so just a minute

[stray river]

yeah

[river ermine]

sure thing, sorry

[stray river]

3, 7, 6, 5, 5, 4, 2

is also 25200

[sturdy inlet]

There will be two cases

1. 1 or 3 appear at last place so we can have two numbers at the first place instead of the usual 3 . For example if we take 1 at last place we can have 2 and 3 at first place .

2765432

2. Now we already considered 1/3 at last place so we have remaining 2 odd nunber and their cases are normal
   3765432

Add them and you get the required answer

[river ermine]

so as you said earlier, i was sort of on the right track (kind of, not really)

anyways, i understand why it is done like that. thank you guys so much for sticking around for almost 3 hours. I really appreciate it

[sturdy inlet]

For which class is this ?

[stray river]

I kind of enjoyed it

[river ermine]

highest grade 11 maths class

[stray river]

I learned a lot too

[stray river]

damn

I'm still in middleschool

[river ermine]

or whatever is the equivalent from where you are from

[stray river]

grade 11 math is awesome

[sturdy inlet]

You should try some jee pyq(s) they are much harder than this . And less forgiving , if you go on the wrong method of solving you will end up with wrong answer and you won't even realize it because your arrived answer will be there in the options..

[stray river]

jee questions are next level

no doubt

[sturdy inlet]

But i would suggest you get a better understanding of basic principle of counting.

[river ermine]

ill be sure to try them out

[sturdy inlet]

The main thing is you can get detailed solutions for it online .

In this questions it is obvious when to multiply and when to add . But a lot of the times it isn't

[river ermine]

i see

once again, thank you guys for everything

[stray river]

byjus or vedantu most of the time

[river ermine]

ill close the channel now

[stray river]

gn

[river ermine]

thanks, bye

.close

[cedar kiln]

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[halcyon orbit]

if $f(x\dot y) = f(x)+f(y)$ then $f(x)=\log_{a}x$

[wraith dagger]

Arian

[halcyon orbit]

How to prove this?

(x>0)

[red pumice]

its not true actually

[halcyon orbit]

Why? @red pumice

[red pumice]

suppose f(x) = 0 for all x

then f(xy) = 0 = 0 + 0 = f(x) + f(y)

for any x, y

[halcyon orbit]

I meant prove this for x greater than 0 @red pumice

[red pumice]

this is true for any x and y

so it certainly holds for x > 0

[halcyon orbit]

You're right that's my careless

Another thing is that

f(x) isn't equal to 0 @red pumice

[red pumice]

oh yes that changes things

any other restrictions?

[halcyon orbit]

The function is continuous

@red pumice

[cedar kiln]

@halcyon orbit Has your question been resolved?

[cedar kiln]

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[fair geyser]

what's x doing here

[cedar kiln]

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[cerulean bridge]

compared to the answer im not getting it right

whe ni m gonna calc dr i get it wrong

oh no wait i think i found soemthing

[crimson sedge]

$\iiint_{K}\frac{z}{1+x^{2}+y^{2}}\textup{d}x\textup{d}y\textup{d}z$

Just LaTeXing to make it easy for others

[cerulean bridge]

thx!

[wraith dagger]

3317

[royal loom]

What's your goal?

What's the region K?

[cerulean bridge]

thsi is the region

[royal loom]

have you thought about making a change of variables?

[cerulean bridge]

yes x =rcosa and y = rsina

and i keep z?z

z=z

[royal loom]

If you want to do it in cylindrical coordinates sure, it might be easier in spherical though

[cerulean bridge]

how do i know which i s easier?

[royal loom]

just depends, from looking at your bounds it looks like making x^2+y^2+z^2 into rho^2 seemed like a decent idea, but cylindrical might actually be best for this

because of the integrand itself and the other bound

So what part are you having trouble with?

[cerulean bridge]

I have started the problem further up in the chat . I cant it right when i calc dr

[royal loom]

what do you mean by "calc dr"

[cerulean bridge]

i split the integrals, first i calculate dz and then dr, we know that drho is 2pi

[royal loom]

Can I stop you for one second

[cerulean bridge]

and i get stuck after calculating the frist integral

yee

[royal loom]

I think it would be best to do this in spherical coordinates instead of cylindrical. Although it should workout both ways. But, what I think you are saying about "calculating dz, dr, drho" is where you are going wrong. When you switch coordinate systems, say from rectangular dxdydz to spherical, dphi, drho, dtheta, the jacobian you compute (and that you should know) takes dxdydz -> rho^2sin(phi)drhodphidtheta

[cerulean bridge]

yee that i know but isnt dxdydz = r*drdrhodz?

[royal loom]

from rectangular to cylindrical you get, r dr dtheta dz

[hollow osprey]

Not rho, probsbly theta

[royal loom]

if that is what you are asking

[cerulean bridge]

hmm ok maybe my defintion of z is wrong

i have r < z < sqrroot(1-r^2)

[hollow osprey]

it’s just (x,y,z) = (rcos(theta), rsin(theta, z)

That’s looks right

[cerulean bridge]

when i calculate the circled equation i get the wrong answer 😦

it should work but somewhere is wrong🤔

[royal loom]

Subbing spherical coordinates gives a nicer integral, I'll send a pic in a second

[cerulean bridge]

👍

[royal loom]

You should verify that yourself though

[cerulean bridge]

will try with spherical coordinates now!

how do you get this?

[royal loom]

from z >= sqrt(x^2+y^2)

and making the subs

you end up with

cos(phi) >= sin(phi)

which is true over that interval

but sorry for possibly misleading you, I've been trying to workout the spherical one and it doesn't seem to be easily budging, so maybe cylindrical is the way to go

have to do a trig sub immediately to integrate rho

it just has nicer bounds

[cerulean bridge]

hmmm ok i will see if i get same bounderies

theta is just 0 < theta <2pi ?

[royal loom]

yes

I no longer recommend doing it via spherical btw

[cerulean bridge]

i got the same with the other angle

[royal loom]

it turns out nasty this way

[cerulean bridge]

😳

[royal loom]

have to trig sub, and then power reduce sec^3 of your new introduced variable, and then evaluating is a mess

its all bad

cylindrical must be the way

[cerulean bridge]

yee i tried cylindrical but i do something wrong

wait

it shouldnt be:

z can not be negative and that is included here

so 0 instead of -3pi/5

/4

[cedar kiln]

@cerulean bridge Has your question been resolved?

[royal loom]

that's not a z, thats phi. The angle off the positive z-axis in spherical coordinates. It is different than z in cylindrical coordinates so you can't compare them like that

[cerulean bridge]

but with that angel z is negative and z always got to be postive

[vernal shell]

I agree, the term 1+x^2+y^2 complicates using spherical 🤔

[cerulean bridge]

z = sqrrott(x^2+y^2)

if phi is negative then z is negative which it can not be

but i still cant get iit right the integrals are just way to messy

[vernal shell]

Why are the limits from 0 to 1 in the last integral? I think you can solve this integral by "simplifying" that fraction

[cerulean bridge]

what do you mean🤔

[vernal shell]

$\frac{r-2r^3}{1+r^2}=-2r + \frac{3r}{1+r^2}$

[wraith dagger]

ELeonardo

[vernal shell]

(I first multiplied by 2 on both numerator and denominator, I mean take the 2 out of the fraction)

[vernal shell]

Then 3r can be written in terms of the derivative of 1+r^2, so you can try u=1+r^2

[vernal shell]

check the upper limit first

[cerulean bridge]

omg that is so smart

will try it now and see if i get it right😎 👍

but do u agree with the limits 0 to 1?

for the integral

[vernal shell]

No 😅, I'm sure the upper limit is not 1

don't confuse the radius of spherical coordinates, with the radius of cylindrical coordinates

[cerulean bridge]

hmmm but i cant figure out what the upper should be if not 1🤔

i restrict z to: r< z < sqrroot(1-r^2)

and then i thought it made sense that r is 0< r <1

[vernal shell]

The lower limit is r, and the upper limit is sqrt(1 - r^2)

the solid is bounded by the intersection of this limits

find that intersection

you will see that it's not at r=1

[cerulean bridge]

ahh if r = 1 then: 1 < z < 0

[vernal shell]

you can also draw a sketch

[cerulean bridge]

hmm i cant really imagine how that would look in 3d

[vernal shell]

it's not necessary rn, you can just use the "r dimension"

I mean graph z as a function of r, just a 2d plane

these surfaces depend on the radius, which makes them symmetrical around z, so you can just consider z as a function of r to find the limits 🤔

[cerulean bridge]

is r: 0 to