help-10

[mossy fjord]

And then you ultimately get f(g(x)) = x, which was the original goal.

[obtuse pebble]

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[heady turtle]

where am I wrong

[heady turtle]

The answer is -2

[red ice]

ah because you should have $-\frac{1}{2} \cdot -f'(1)$

[warm shale]

south

[red ice]

f'(1) would be ( f(1 - x) - f(1) ) / (-x)

[heady turtle]

Ooh!

Tysm!!

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[median siren]

How do I find the midline of a trapezoid if I only know that there's a circle in it and it's perimeter?

[obtuse pebble]

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[tiny epoch]

Can anyone simplify this

[tender tusk]

do u know how to factorise the numerator

[obtuse pebble]

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[grand dune]

could someone look over this?

[obtuse pebble]

@grand dune Has your question been resolved?

[tiny epoch]

That’s the problem

[crude moss]

@grand dune My german seems a bit rusty. What you have done seems to be okay. Replacing by -1,0,1 in your polynomial in the end gives the right result. Is the question saying second degree polynomial and/or find if it exitsts ?

[grand dune]

it says find the polynom of second degree

with the information givem

given

[crude moss]

the thing is that it does not exists lol

there is a unique polynomial of degree 2 going through 3 points

and the first point of the question is not the actual one lol

,w f(x) = -0.5x^2 + 2.5x + 3 with x = -1,0,1,2

[warm shale]

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[crude moss]

There is a polynomial of degree 3 surely, but not of degree 2

sorry

[grand dune]

shit i will check that

[crude moss]

,w interpolation (-1,0) , (0,3) , (1,5) , (2,1)

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[crude moss]

What you did is right, I question the rightfullness of the question

[grand dune]

maybe i need to approximate it diffrent

[crude moss]

If indeed the question says find the polynomial of degree 2 such that ... if it exists then you can give a simple argument that it does not

[crude moss]

Errrr idthink so

xD

[tiny epoch]

1

[obtuse pebble]

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[timid silo]

hoi

[timid silo]

what are these specific examples called??

I wanna learn more about them

idk if this is out of the scope of maths tbh

I think these are called ramp problems?

and the broader topic would be momentum/kinematics/energy/force....

[timid silo]

thank you!!

https://www.youtube.com/watch?v=ufgY237M5KQ

you might like this video

[supple void]

These come under Newton's law of motion I assume.

[timid silo]

this guys amazing

yup that as welll

https://tenor.com/view/adrianjacobsen-jonaspjac-feziwigen-tvsentralen-gif-13439446607587169539

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[timid silo]

any ideas folks

[vagrant pecan]

looks like a formula

[opaque dome]

You can first put them together

[rotund cedar]

Looks like binomial theorem(?) but I’m not familiar enough with it to properly explain

[timid silo]

sure i put them together

[timid silo]

so (x^3+8)^7

that leads me to nowhere the values from the exapnsion i got dont come up here

[opaque dome]

May i see what you have done so far ?

[timid silo]

yeah for sure, let me write it in a presentable form

[opaque dome]

,w binomial theorem

[warm shale]

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[timid silo]

There's what I got to

[opaque dome]

,rotate

[opaque dome]

You inverted x and a in the formula

[timid silo]

Would still result in 56

[opaque dome]

Uhm

[timid silo]

Wait

So it's 7c3 you mean?

Oh

It's the rule that's written for me in the book

I didn't know I'm supposed to invert the

[opaque dome]

,w expand (x^3+8)^7

[warm shale]

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[opaque dome]

:)

Actually you are right about the (x³ + 8)^7

,w expand (x+2)^7*(x^2 -2x + 4)^7

[warm shale]

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[opaque dome]

:D

[timid silo]

So the question is just plain wrong

😭😭

[opaque dome]

Somehow

Well

Good thing is that you still find the correct answer

The real one

[timid silo]

Yeah that's why I was super confused, I did a question like it just a second ago and it was harder but this stumped me

I'm gonna give my teacher a piece of my mind hahaha

[opaque dome]

Kekw

[timid silo]

Anyways thanks for your help ❤️❤️

[opaque dome]

You're welcome

[timid silo]

.close

[obtuse pebble]

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[timid silo]

Wait I wanna try something

[opaque dome]

What you wanna try

[timid silo]

,w expand (1/x^3)*(x+(1/x^2))^10

Let's goooo

Alright

[opaque dome]

Bot is solving Riemann 💪

[timid silo]

Apparently

😭😭

[obtuse pebble]

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[keen apex]

am i going crazy

[frigid mauve]

bruh

[frigid mauve]

the answr is there

[keen apex]

which

[frigid mauve]

34

cuz its 25 divided by 72

its gon be 0347 somethig lke that

[keen apex]

ty

[obtuse pebble]

If you are done with this channel, please mark your problem as solved by typing .close

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[vapid halo]

if your question has been answered you can close this chat with .close have a good day<3

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[bronze birch]

chat

is the answer not log(3^(3/4)) divided by k

[azure wraith]

simplify the right first

[bronze birch]

i got 3^3/4

for it

if i did it right

[azure wraith]

so

Given, ${\log a^b = b \log a}$, then ${\frac{3}{4} \log 3}$ is the expression we want in terms of k and we know ${\log 3 = k}$?

[vapid halo]

awwww k is answering for k thats so cute

[warm shale]

k

[azure wraith]

this is a spiritual jouney to find myself

@bronze birch

[bronze birch]

yh

[bronze birch]

i think so

i just wrote it as log(3^(3/4))

tho

is the answer not log(3^(3/4)) divided by LOG3

and then log 3 is k

[azure wraith]

${\log 3^\frac{3}{4} = \frac{3}{4}\log 3}$

[warm shale]

k

[bronze birch]

cuz of change of bases

[azure wraith]

why division by log3

[bronze birch]

because log base 10 3

cuz the orginal base was 3

[azure wraith]

i think original base is 10

[bronze birch]

no for the

3^3/4

chat i think i cooked

[azure wraith]

where did base 3 come from 😭

[bronze birch]

oh mb

[bronze birch]

i meant base 3 here (first it was 3^3/4)

because then it was log3(x)=3/4

and then i converted it to that

[azure wraith]

sorry i dont follow

[bronze birch]

😦

if the original expression simplified to 3^3/4

i can write it like that right

??

[azure wraith]

i dont think 3^3/4 = log_3{3^3/4}

[azure wraith]

by writing this, u say that 3^3/4 = 3/4

which is very wrong

[bronze birch]

oh wait

so its just this?

[azure wraith]

$3^{3/4} = 3^{\log_3{3^\frac{3}{4}}}$

[warm shale]

k

[bronze birch]

oh

cooked 💀

wait no

thats wrong

wait no

thats right

nvm

uhhh

then what do i do with the 3^3/4

[azure wraith]

so the expression we want in terms of k is $\log{3^\frac{3}{4}}$

[warm shale]

k

[bronze birch]

i think so

wait no

why is it base 10

[azure wraith]

.

[bronze birch]

bruh

[bronze birch]

i may have missed that

bruh this is so ez then

its just $k^(3/4)$

[warm shale]

GangsterBeamer03

[bronze birch]

oh nah that didnt work

[azure wraith]

yes

[bronze birch]

alr ty

[azure wraith]

wait

no

${\frac{3}{4}k}$

[warm shale]

k

[azure wraith]

no power

[bronze birch]

oh yeah

w @azure wraith found k

thanks for the help

.close

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[olive totem]

I need help on number 28

Could I do ln on both sides? Not sure if i can do ln of f(x)

[dark stirrup]

Yes. That is a popular way to do it. It would require implicit differentiation, but it's perfectly fine

[olive totem]

ahh ok, ill try that then

[dark stirrup]

Another perfectly good way is to use your power and log rules. Recall that $e^x$ and $\ln x$ are inverses of each other. Meaning $f(x)=e^{\ln f(x)}$

[warm shale]

SWR

[olive totem]

so i could do ln of both sides and then do e to both sides

is this right so far?

[static marten]

this is wrong

[olive totem]

hmm ok

what could i do instead

oh i can do e to both sides and that gets me 2^sin 2x

[olive totem]

is this right

[gilded venture]

thats what you started with

youre good up until and including $ln(f(x)) = \sin(2x)*ln(2)$

[warm shale]

Adum

[gilded venture]

then you can use both sides as a power of e and use the chain rule

[olive totem]

oh ok

lemme try

this is what i got so far

but i think I made a mistake 😐

im gonna move onto the next problems

.close

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[split dew]

in what cases does an eigenspace have a dimension greater than 1?

[tardy epoch]

Very broad question

Can you give more context

[split dew]

i was kinda thinking what would be the "opposite" of this

like in what cases is the dimension not 1 cause tbh i don't really understand the dimensions too well

[tardy epoch]

Have you learned about repeated eigenvalues

[split dew]

as in something like (5 + lambda)^3 with lambda having a algebraic multiplicity of 3?

[worn yoke]

the geometric multiplicity of an eigenvalue is the dimension of its eigenspace. it obeys the inequality
1 <= geo. mult <= alg. mult

[obtuse pebble]

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[azure roost]

I don't understand the steps taken in the solution

[obtuse pebble]

@azure roost Has your question been resolved?

[obtuse pebble]

@azure roost Has your question been resolved?

[oak bolt]

hello

are you still there?

@azure roost

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[glass pilot]

why can you calculate mean and sd when it doesn't meet the criteria for central limit theorem?

[tardy epoch]

Mean and sd have formulas that exist independent of CLT

The CLT requires SD and mean, not the other way around

[obtuse pebble]

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[raven surge]

I need help

[obtuse pebble]

Please don't occupy multiple help channels.

[raven surge]

i need help

with a proof

can someone prove a^kx using base e

its derevative

[gloomy vector]

!original

[obtuse pebble]

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.

[raven surge]

do i have to?

it was a book question

and i dont have my book with me

[gloomy vector]

what

[raven surge]

i did a bit and got this far

(elna)^kx

[azure wraith]

Bring kx to the front

ln a^b = b ln a

[raven surge]

yeah but i need to have a^kx as well

[gloomy vector]

do you want to find the derivative of a^kx using natural logarithms?

[raven surge]

no i want to prove it

thats what the book said

exactly

using base e

log

[gloomy vector]

ok that makes more sense

[azure wraith]

Just take derivative

in form of e^(kx ln a)

Using chain rule

[raven surge]

lemme tryt

that gives me

lna * a^kx

i need the seperate constant too

nvm

ur right

i had to bring down lna * k

not just lna

[azure wraith]

Yes

[raven surge]

so is it just

lna * k * (elna)^kx

that was easier than. ithought

[azure wraith]

e^ln a something simplifies back

[raven surge]

thank you so much

[azure wraith]

To a^kx

[raven surge]

yeahh

damn

thanks

a lot

[azure wraith]

👍

[raven surge]

.close

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[indigo harness]

I’m trying to simplify this idk where I want wrong

I need to get w on its own

[obtuse pebble]

@indigo harness Has your question been resolved?

[heavy shadow]

Can you show me the original equation

[obtuse pebble]

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[kindred scaffold]

Help please

[kindred scaffold]

for solving 8t-125^-2

[upper sandal]

!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

[kindred scaffold]

2

[upper sandal]

!show

[obtuse pebble]

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

[kindred scaffold]

so i got to multiplying by t

kinda

[upper sandal]

so far so good

remember its =0 right

[kindred scaffold]

yes wait ill write that

[upper sandal]

yup

[kindred scaffold]

what do i do now?

[upper sandal]

if $a\cdot b=0$, then what are a or b?

[warm shale]

Bonk

[heavy shadow]

You kinda made a mistake on factoring the t

[kindred scaffold]

0

[kindred scaffold]

which part?

[upper sandal]

what is 0

[kindred scaffold]

a=0 or b=0

[upper sandal]

exactly

and now a=t and b=8+125/t^2, right?

[kindred scaffold]

yes

[upper sandal]

okay, continue from there

[kindred scaffold]

does t=0?

[upper sandal]

that is one conclusion yes, but we have to check if thats valid

what do we get when we plug t=0 into our equation?

$8\cdot 0+\frac{125}{0}=?$

[warm shale]

Bonk

[upper sandal]

yes, that solution is not allowed, since we divide by 0

so we ignore that case

so lets move onto the other case, which is?

[kindred scaffold]

-125/t^2?

[kindred scaffold]

lemme factorjse your onr

no wajt

expand-

[upper sandal]

almost

[upper sandal]

the original question is $8t+125/t=0$

[warm shale]

Bonk

[upper sandal]

actually nvm, i misread

@kindred scaffold how did you get to 8t+125/t?

[heavy shadow]

Took the derivative of the distance function

[kindred scaffold]

i differenciated the displacement

[heavy shadow]

I mean displacement

[upper sandal]

did you do it correctly?

[kindred scaffold]

uhh lemme redo it to check

and the last one wasss

OHH I HAD A + SIGN

[kindred scaffold]

why t^3 tho?

[heavy shadow]

I only followed your way that u factored t

But I suggest starting again from here

I suggest u multiply both sides by t²

[kindred scaffold]

Like this?

[heavy shadow]

What happened to the t^-2?

[kindred scaffold]

wait um

it becomes just t

lemme do this on notebook wait

Is this fine so far?

[heavy shadow]

Not quite, we want to multiply both sides by t² right? So how do we do it on the left hand side

[oak bolt]

$t^2(8t-125t^{-2})=t^2 \cdot 0$

[warm shale]

Axe

[kindred scaffold]

ohh

[oak bolt]

$8t-125t=0$ is wrong

[warm shale]

Axe

[kindred scaffold]

yeah I meant the last line sorry

[oak bolt]

can you distribute now?

[kindred scaffold]

like expand?

[oak bolt]

yes

[kindred scaffold]

ok

[oak bolt]

$t^2\cdot 125t^{-2}=125t^{2-2}=125t^0=125$

[warm shale]

Axe

[kindred scaffold]

uhh one sec

YEAH SORRY

question

doesnt 0xt^2 =0?

[oak bolt]

yes

[kindred scaffold]

so can i write just 0?

[oak bolt]

yes 👍

[kindred scaffold]

Done ty

do I add 125 now?

[oak bolt]

yeah

[heavy shadow]

Nice

Now you found the time when the velocity is 0

[kindred scaffold]

And I substitute that into the displacement?

Since displacement is the distance

[heavy shadow]

Yes

[kindred scaffold]

This?

[obtuse pebble]

@kindred scaffold Has your question been resolved?

[upper sandal]

sry, had to make dinner

do you have the solution yet?

@kindred scaffold

[kindred scaffold]

so i got 75 at the end and I'm not sure if it's correct or not, but other than that it's all good :D

[kindred scaffold]

nws lol

[upper sandal]

can you send the full question in a photo again?

[kindred scaffold]

ok

I got t=2.5

(if that's correct)

and then substituted it

and got 75 as the distance for a final answer

(If that's also correct-)

[upper sandal]

yeah i think 2.5 is right

and then distance is 75

yup, totally correct

gj

[kindred scaffold]

THANK YOUU-

:D

.close

[obtuse pebble]

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[rapid plover]

hi im stuck on this one
i have tried using laplace transform and then use the finite value theorem to solve it

but then im stuck at that point

[upper sandal]

!show

[obtuse pebble]

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

[rapid plover]

can someone point me to the right direction? thanks

oh ok sure hold on

not sure where to go after this

[upper sandal]

dont you first need to inverse it?

[rapid plover]

u mean the last part?

[upper sandal]

you have Y(s)

but not y(t)

[rapid plover]

yea

[upper sandal]

try finding y(t) first

then taking the limit

[upper sandal]

ah, i see

[rapid plover]

but if i use this one i can kind of not inverse it?

yea thats what im trying to do

[upper sandal]

then just fill in s=0, no?

i dont see anywhere it can go wrong

[rapid plover]

then ill just know the final steady value is just a ratio of the 3 initial conditions

but i think he's asking me to write sth like
y(0) = 0
dy/dt(0) = 0
sth like that

[upper sandal]

$\lim_{t\to\infty}f(t)=\lim_{s\to 0}sY(s)=\lim_{s\to 0}\left(y(0)+\dv{y}{t}\ (0)\frac{s-1}{(s-2)(s+1)}+\dv[2]{y}{t}\ (0)\frac1{(s-2)(s+1)}\right)$

wtf...

looks shit but w/e

[rapid plover]

its ok lol

[upper sandal]

anyway, there is no divide by 0 when you fill in s, so its fine

[rapid plover]

no hold on

[upper sandal]

wha.....

[warm shale]

Bonk

[upper sandal]

thats better

[rapid plover]

$y(0) + \frac{1}{2}y(0) + -\frac{1}{2}y(0)$

$y(0) + \frac{1}{2}y`(0) + -\frac{1}{2}y``(0)$

[warm shale]

Arctic

[upper sandal]

yup

[rapid plover]

yea thats what i get in the end but

doesn't this basically mean i can put anything in those initial conditon and it will not blow up?

[upper sandal]

yeah, as long as theyre bounded i dont see an issue

[rapid plover]

oh okay

thanks m8

.close

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[steady quest]

can some one help me

[rapid osprey]

It would be helpful to convert the fractions to decimals first

[steady quest]

I tried that but I still couldn't figger it out

[olive roost]

when converting, what did you get for 1/4

[steady quest]

I got 0.25

[olive roost]

ok so number 1 should be above or below 0.25?

[steady quest]

above?

[olive roost]

the thing named "1"

like the marker

[steady quest]

oh below

[olive roost]

yeah so which options are below

[steady quest]

there are no options

[olive roost]

wdym?

Like a,b,c,d...

[steady quest]

oh

thats wat u mean

the options below are g

[olive roost]

There's one more

[steady quest]

is it I?

[olive roost]

yes

[steady quest]

so its G first?

[olive roost]

noo

marker 1 is closer to 0.25 than 0

which of the two, 0.1 and 0.2 is closer to 0.25?

[nova vale]

lol

[steady quest]

I

[olive roost]

kindergarten 😭

[olive roost]

Yeaaa that's right

[nova vale]

true lol

[steady quest]

bro really

[nova vale]

what grade are u in

this is kindergarten question

are u even allowed to be on discord

13 is the minimum age

lol he left

[waxen mirage]

💀

[slate shoal]

escaping before getting caught...

[olive roost]

Bullying a 7 year old 😭

[nova vale]

figger it out

[normal rivet]

hi

[obtuse pebble]

@steady quest Has your question been resolved?

[obtuse pebble]

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[runic grotto]

I got 0.09m rounded off 2 deci points. I'm just confused if i'm going to write the "i" showed in my calculator

[worn yoke]

what do you mean by "i"?

can you show a picture of your calculator output?

[runic grotto]

an imaginary number I suppose

[worn yoke]

if you are getting an imaginary number then that means something went wrong

[wary badger]

did you use -2.6 x 10^-7 C

[runic grotto]

yeah

[wary badger]

you’re not supposed to

you’re only concerned with the magnitude of the charges

[runic grotto]

oh, what am I supposed to put in the equation

[wary badger]

use the magnitude instead so instead of -2.6 x 10^-7 use 2.6 x 10^-7

[raven spire]

$R^2 = \left| \frac{kQ_1 Q_2}{F} \right|$

[runic grotto]

there you go, I got it

[warm shale]

Arya

[runic grotto]

thank you guys!

[wary badger]

you’re welcome

[runic grotto]

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[copper falcon]

Hey can I have some help. Ive been running this find the limit problem and I just cant seem to get the right answer. The original function s(t)=5sin(t)
You are supposed to use the equation (s(t1)-s(t0))/t(1)-t(0) and I'm using it but I cant get an answer that the computer accepts. Anyone know if I'm doing somthing stupid?

[lean rune]

what do you have so far?

[ancient drift]

Do you mean
$\frac{s(t_1)-s(t_0)}{t_1-t_0}$?

[warm shale]

denzio321

[copper falcon]

I got to -3.18310 then starte//

yes I do

So I got to the -3.18310 number then did the +.1 and got .24979 but then the change in the denominator and the numerator cancel

so it should stay the same

but it wont accept it

[ancient drift]

For which time interval?

[copper falcon]

pi/2 , pi

[ancient drift]

what is the value of sin(pi/2)

[copper falcon]

1

so s(t0)=5

s(t1)=0

so you would end up with 5/pi/2

or 10/pi

[ancient drift]

Yeps

[copper falcon]

which ends up at that -3.18310 number

but after that it says im wrong

[ancient drift]

Why would it end up negative?

[copper falcon]

because its - the first function

[ancient drift]

Oh yea mb

[copper falcon]

s(t0) = -5

ye

[ancient drift]

I mean can you answer in terms of pi

[copper falcon]

no has to be a number to 5 decimal places

[ancient drift]

Try to 5 significant figures instead?

[copper falcon]

I did

no dicew

[ancient drift]

Can we see the whole question

[ancient drift]

Well the average velocity doesn't stay constant

It approaches 0

[copper falcon]

yeha

idk

I ran it through but Im always off by just a little

I really dont know what to do

[obtuse pebble]

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[obtuse pebble]

@cold wagon Has your question been resolved?

[cold wagon]

<@&286206848099549185>

.close

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[untold terrace]

Like get to know how to use previous answer to evaluate d(ii)

[untold terrace]

I am trying to reshape the integral

[gilded jungle]

Firstly, what is the answer to d(i)?

[untold terrace]

This

[untold terrace]

But wait I think I have some mistakes in it

[untold terrace]

Yeah the 2nd term denominator should be 12 instead of 3

[obtuse pebble]

@untold terrace Has your question been resolved?

[untold terrace]

<@&286206848099549185>

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[wanton dune]

Why does this hold even tho those sequences have different graphs?
[\sum_{n=1}^{\infty} \frac{1}{n} \ln \left(\mathrm{e}^{\frac{n+1}{n}}-\mathrm{e}\right)=\sum_{n=1}^{\infty} \frac{1}{n}\frac{\ln(e^\frac{n+1}{n})}{\ln e}=\sum_{n=1}^{\infty} \frac{1}{n}\frac{\frac{n+1}{n}}{1}=\sum_{n=1}^{\infty} \frac{n+1}{n^2}]

[warm shale]

Slowaq

[gilded needle]

this is not true:

log(a-b) is not the same as log(a)/log(b)

[wanton dune]

then why did people here say that it is?

[gilded needle]

you're thinking of $e^{a-b} = e^a/e^b$

[warm shale]

Bungo

[gilded needle]

who said that?

[wanton dune]

look into that message i fowarded

[dusk widget]

I think moni overlooked the mistake

[wanton dune]

its interesting that 3 different people overlooked it

[gilded needle]

yea they misspoke

[wanton dune]

but anyway thanks for clarification

[gilded needle]

sure, yw

[wanton dune]

.close

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Please don't occupy multiple help channels.

[gilded needle]

srsly?

this is 3 help channels now

.close

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[upper sandal]

$x(x+1)=20(x+1)130=2600(x+1)$???

[warm shale]

Bonk

[dapper bloom]

I think you dropped the + on accident

"130 more" as in add 130 rather than multiply

Other than that it seems right

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[foggy hearth]

Isnt this just the slope of the line?

Choose two points and apply y2-y1/x2-x1

Yes the height of the point

Yeah any two points will work

Yes minus the two y'es then divide by the two x'es minus each other

[obtuse pebble]

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[tardy mortar]

We can do a u-sub

To find such a value

Let u = sqrt(x)

Yes, to find what value of alpha is needed

Assuming you know the p-values

Mhm

So lets do a u sub and see if we can change anything

If we let u = sqrt(x)

Then du = dx/2sqrt(x)

So we can get rid of that sqrt(x) on the bottom

Then we get $\frac12 \int_0^{\infty} \frac{\dd u}{1 + u^{2\alpha}}$

[warm shale]

hiidostuff

[tardy mortar]

Nice

Sorry one second

Gotta go do smth for a moment

[obtuse pebble]

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[rare owl]

does someone know why this solution of a differential equation isnt valid for R

it looks valid to me

already found it

[worn yoke]

what was the differential equation?

[pseudo swift]

cube root(x) isn't differentiable at x=0

so if the thing inside the cube root ends up being 0 somewhere

you end up with a point where the "solution" isn't differentiable

& they found one such point

@rare owl

[rare owl]

indeed

thats it

i forgot how to close this

[pseudo swift]

.close

[obtuse pebble]

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[rare owl]

thx

.close

[pseudo swift]

can't redo it lol

[rare owl]

i thought i reopend it by texting something haha

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[steel jay]

K = C + 273 F (F - 32) x 5/9
C = K - 273 C = (F-32) x 5/9

[steel jay]

Use above formulas above to convert the following:

1. 123 Celsius to kelvin
2. 79.6 Fahrenheit to kelvin
3. 183 Celsius to Fahrenheit

[tardy epoch]

typo in upper right eqn

[steel jay]

Oh yeah

It’s

F = (C x 9/5) + 32

[obtuse pebble]

@steel jay Has your question been resolved?

[brisk sequoia]

HI somone understand integral has parametre??

[obtuse pebble]

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[wide grove]

Help

[wide grove]

I got some hw i need help with I sont understand

9th grade geometry

[wide grove]

Im confused on #5 rn

Im unsure how to find the scale factor

[sage dagger]

can u take a more clearer pic

hard to see

[worn yoke]

,rccw

[wide grove]

is this alright

[raven spire]

,rccw

[raven spire]

@wide grove You got scale factor of 5/7 right? Therefore, perimeters of ABCDE and PQRST are in ratio 5/7 as well

[wide grove]

uh idk

[raven spire]

So, ABCDE = 65 => PQRST = 91

[wide grove]

I never got the scale factor

[raven spire]

Oh, scale factor means - as per the question, ABCDE is similar to PQRST and so the ratios of corresponding sides would be a constant number, which is defined the scale factor

Now, as AB and PQ are given to you, you can find the scale factor by doing AB/PQ

[wide grove]

Yea

15/21

OH

[raven spire]

[wide grove]

wait I disnt see that it could be

Sinplified

whoops

[raven spire]

You got it ^-^

[wide grove]

so then

[raven spire]

Now perimeter ratio will also be = the scale factor

[wide grove]

To get 91

its 65 x 7/5?

[raven spire]

✅

[wide grove]

oh

ok thank you very much

If you dont mind could you also help me with some other stuff

[raven spire]

!da2a

[obtuse pebble]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[wide grove]

ok ty

How do i tell if the triangles are similar in 1 and 2

Using the angle angle concept

[raven spire]

For Q1, check that angles A, B of ABC are equal to angles Z, Y of triangle XYZ

[gilded needle]

for Q1 you have enough info to find the missing angles

[wide grove]

So they are similar

B and Y are equal

Wait no

Sorry, im kinda confused still

[raven spire]

which part?

[wide grove]

I didnt auote get what u meant by the L part

Where 2 angles make that and that thing

[raven spire]

Did you get Q1?

[wide grove]

For the most part

The only rhing is why do I compare ab to zy

Instead of xy

or does it not matter bc they equal?

[raven spire]

Yes

[wide grove]

So the triangles are similar?

[raven spire]

because it's not always necessary that ABC be similar to XYZ, ABC ~ ZXY or ZYX is also a possibility

Here, as Z = A, Y = B, we have ABC ~ ZYX

[wide grove]

ooooooh

I see

Ok i have another wuestion about 4 and 5

When im soing the proportion or wtv

For example like 8/6=x+7/x-1

Am I supposed to like add them together for one of the triangles because the picture of the triangles is like a 2 in 1

[obtuse pebble]

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[ocean plover]

I'm trying to find a family of equal partitions for the perms of EFGH (so each partition is 12+12, 8+8+8, 6+6+6+6, 4+4+4+4+4+4...). These perms should be reasonable natural and independent from each other. How does one go about this?

[ocean plover]

It's okay if they aren't totally independent, for example "E first or last" isn't independent from "F first or last", but this is fine

Those two particular partitions still intersect in a "nice" ratio (2:1)

[obtuse pebble]

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@ocean plover Has your question been resolved?

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[proud plover]

I understand nearly all the process but i dont understand how my prof chooses 4j? How he comes with that substitution solution?

[half plinth]

To kill off the sin

[proud plover]

I mean it seems its used so that I get only when the sin is 0 but wouldnt it have different behaviour then the other serie?

Like only calculating a part of the serie but avoiding the other possibilities

[half plinth]

Well the entire point of using a substitution is to simplify the expression as much as possible

And as long as you back sub then everything is alright

[proud plover]

So you say that even if i avoid considering how the serie would be behaving while sin isnt 0 the serie behaves in the same way?

[primal wadi]

I would say its less of a substitution, but rather picked a subsequence of elements whose sum is "nice"

instead of evaluating the sum of everything, he just sumed every 4 element, and that sum goes to infinite

he brushed off the other 3 element since they are >=0

[proud plover]

Yeah the confusion for me was if the serie behaves the same or not when I avoid some options

Like maybe that serie diverges but if I consider the other sin options it still diverges? How i know that

[primal wadi]

yeah to do that you need to guess that the series is divergent to positive infinity

[proud plover]

Mmm ok

[primal wadi]

well, since sin(i pi/2) behaves periodically

it is natural to split it into 4 parts

[proud plover]

Ok got it

[primal wadi]

the four parts are the sum of 1/4j
the sum of 1/(4j+2)
the sum of 2/(4j+1)
the sum of 0/(4j+3)

and you note that oh, three of them is divergent and the others is just 0

so surely you just get divergence

[proud plover]

Mm ok, yeah

Thanks

[primal wadi]

no worries

[proud plover]

.close

[obtuse pebble]

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[wintry lance]

.claim

[tardy pulsar]

someone explain the remark

[tardy pulsar]

thats not how it works

[polar fossil]

wdym explain

[tardy pulsar]

I dont get the remark

hayley

who shakes the world

[polar fossil]

ok.... i don't know your mathematical background

[upper sandal]

$\int_{-\pi}^{\pi} \cos(nx)\cos(mx)dx=0$ if $n\neq m$

[warm shale]

Bonk

[tardy pulsar]

oh

so any n and m

oh wow

thank you very much Bonk

.close

[obtuse pebble]

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[hidden compass]

Perfect timing if you wish:
https://youtu.be/6HpPcpYOtNM?feature=shared

just published few hours ago

I think he explains it in a clear way

[obtuse pebble]

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[misty magnet]

Need help understanding proof by induction once and for all

[misty magnet]

I will send the worksheet my teacher gave me, alongside her answers.

And I want to see if I can work out what she did

In order to redo the proof myself

I will also send my own attempt at the proof

Now here’s the worksheet and here’s the solutions by my teacher

I am attempting the first inductive proof on my own now so I can send my attempt

I’m not sure what I keep doing wrong

[restive gorge]

it'd be better if you didn't send it as pdf

[winter panther]

Send it, we can’t tell whats wrong until we’ll have seen your attempts

[misty magnet]

Here is my attempt so far

I’m stuck at the moment

I’m going to factor out the k+1 and (-1)^k

And see if this helps

Here’s where I’ve gotten to now

I’m not sure about what to do about the exponents to just be left with (-1)^k+1

[olive roost]

I think by i you meant n

[misty magnet]

We start off with n. I changed it to k to make sure its arbitrary

Because n is in the base case I don’t reuse it

[winter panther]

Always write sigma

[olive roost]

ngl I'm not quite sure what you're doing but what I meant to say was that in your initial question there was 2 variables namely i,n

And then you said "let n=1", then proceeded to replace i too by 1

[winter panther]

when you are working with summations always write sigma as you didnt write it you misled the proof after trivial case(n=1)
Then your assuming that summation has formula and this formula holds for fixed n
Your next step should be checking whether this formula holds for n+1 addition terms using this formula(every term besides n+1th should disappear) and summing it with remaining term to see whether its equal to the right side

[misty magnet]

Okay so I should write the sigma

Got it

I still want to know what to do about the exponents though

[olive roost]

Ah there's a summation that's why I got so confused

😭

[misty magnet]

That’s the one piece that is keeping me from figuring out the proof

I’ll write the summations in, in a moment.

[obtuse pebble]

@misty magnet Has your question been resolved?

[olive roost]

can you post another image of your updated solution since you seem to not have done it correctly yet

[obtuse pebble]

@misty magnet Has your question been resolved?

[misty magnet]

I haven’t finished it

I’m asking for what I’ve done wrong in the steps

[olive roost]

Well I'm not too sure of the question since you haven't written it out correctly but from what I can see, you're supposed to prove the equality holds true to n+1, AFTER making the assumption that it is true for n

I'll give you a very brief run-down of what induction is:
-start by proving it is true for some value n.
-make the assumption that it holds true for ALL n.
-try and prove it to be true for n+1.

It is bassically a domino effect

If we have a value n that is true, then by proving n+1 to be true, we can input n as our value tested, thus allowing a domino effect [if you tested for n=1, then it is true for n+1 (aka 1+1) which is 2, and true for n+1 (aka 2+1)... And so on]

[obtuse pebble]

@misty magnet Has your question been resolved?

[misty magnet]

The concept I understand but I’m more not understanding what’s up in this specific problem

[olive roost]

well what's the question

I still don't have it

😭

[misty magnet]

Here I’ll screenshot the original

And then put it side by side

[olive roost]

That's why I asked you to write down your updated sol

[misty magnet]

So it’s not a pdf

The original problem

Where I got stuck at

[olive roost]

well first off like the other person said, you need to write down summation

Also you don't do "n=k" or whatever you're doing

We assume n is true and prove that n+1 is true

[misty magnet]

I’ll fix the summation but I’m mostly concerned with the actual math work

[olive roost]

starting from the line "then," is all wrong

before that you should write "assume it is true for all n"

then say "suppose n=k+1"

[obtuse pebble]

@misty magnet Has your question been resolved?

[misty magnet]

I can change that but I’m still mostly asking about the algebraic manipulation

[obtuse pebble]

@misty magnet Has your question been resolved?

[obtuse pebble]

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[ocean portal]

What am I doing wrong?

[wary badger]

integral of u is u?

[ocean portal]

OHHHH

u^2/2

right? @wary badger

[wary badger]

good man

[ocean portal]

THANK YOU

.close

[obtuse pebble]

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[obtuse pebble]

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[pale valley]

i have multiple questions. i’m really struggling. it’s on logarithms. i just don’t understand/i can’t identify log=log and log=number problems. the actual lesson was completely different to what the homework has been assigned.

[pale valley]

i can kind of understand but i just need someone to walk me through it.

[gilded jungle]

Sheesh, that's a lot of logarithms.

[pale valley]

yeah 😕

thank god our teacher lets us do only evens or only odds

[gilded jungle]

Do you have to solve for the variable-

[pale valley]

no, let me send something rq

it’s either this or this

i think the majority of the questions are log=number though which is very frustrating

[ember frost]

https://www.youtube.com/watch?v=fnhFneOz6n8

[gilded jungle]

,rccw

[gilded jungle]

Can you send images of all questions?

[pale valley]

yes, the two images below this are all the questions

*my homework questions

ive tried almost every app i could think of and even itutoring which is a whole subscription and nothing is explaining to me how to do this correctly.

[gilded jungle]

I see...

[pale valley]

the two images i sent saying “this and this” was the actual lesson. it was talking about how there are two types of questions one being log=log questions and log=numbers questions

log=log is simple for me, i understand that one, but the thing is my entire homework is full of log=number. its so confusing the way it’s organized

the first image i sent was the way to solve log=log questions, second one being how to solve log=number

i’ll just sent the examples we did

(which don’t match the hw at all)

a lot of them have a number in front of the log and then some random like 8x after the log base of something

[gilded jungle]

This is... interesting.

[pale valley]

am i not explaining it right?

[gilded jungle]

So these are the questions you're having trouble with?

[pale valley]

yeah 😕

[gilded jungle]

Alright, I'll walk you through 'em, OK?

[pale valley]

oh i see only two of the questions i sent are log=log so i could do those

thank you so much :-((

[pale valley]

okey

[gilded jungle]

Alright, question number 12.
7log₄(v - 9) = 0 implies log₄(v - 9) = 0.

[pale valley]

||why does the 7 go away?||

[gilded jungle]

Divide both sides by 7(?)

[pale valley]

okey

i apologize, i’m not sure if you were waiting on me or if log4(v-9)=0 was the answer.

[gilded jungle]

Sorry, I'm not that great at logarithms either-

[pale valley]

awe it’s okay no worries

i’m not sure why it’s so hard , everyone keeps telling me to put it in a calculator but i can’t put these types of equations in just like that

thank you for still taking time to help me out

[rare hollow]

do u remember ur log laws?

[pale valley]

like product, power and quotient?

[rare hollow]

yes

the first step for these questions is oftentimes how u can apply ur log laws

[pale valley]

i was assuming power but i really can’t tell

[gilded jungle]

OK:
log₄(v - 9) = 0
v - 9 = 4⁰ ∵ logₓ(y) = n ⇔ y = xⁿ
v - 9 = 1 ∵ x⁰ = 1
I think? Sorry, I never learned logarithms at school.

[pale valley]

like if it’s multiplying by log or not

oh!

thank you for giving it a shot