help-19

[urban swallow]

i get it now i'll come back to you for verification

thx so much

$a_c = 7.5 m/s^2$

[clever fjord]

00x0*97&!3

[urban swallow]

$a_t = 12.99 m/s^2$

[clever fjord]

00x0*97&!3

[urban swallow]

is this right or is it the other way 'round?

[undone basin]

umm

,calc 15 cos(pi/6)

[clever fjord]

Result:

12.990381056767

[undone basin]

the other way around

[urban swallow]

okay thank you very much

[odd edge]

@urban swallow Has your question been resolved?

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[sweet oracle]

can someone explain me how does he come to the conclusion that discriminant is 9 - 24?

[sweet oracle]

i dont see how this is a discriminant or why is there a discriminant in the middle of the equation, i started doing them today literally

[undone basin]

what is the full question?

[rapid reef]

The D is the discriminant of the quadratic inside the radical

[tall veldt]

It's the discriminant of x²-3x+6

[sweet oracle]

i dont see how he got 9 - 24 though

im confused what is what compared to b^2 - 4ac formula

[tall veldt]

Well what do a b c stand for in the formula

[undone basin]

what is the question tho? why D < 0

[sweet oracle]

find number of all real solutions

[sweet oracle]

numbers?

[rapid reef]

ax^2 + bx + c

The coefficients of each term

[sweet oracle]

oh i see it now i didnt realize that there was a quadratic equation under the root, i thought that x^2 - 3x + 6 was b^2 - 4ac

thanks

.close

[odd edge]

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[steady tide]

Given $x, y, z\in[2,8]$ and the following function [
P = \log_2^2(xyz)-150\sqrt[3]{2xyz}+75x+75y+2907
]
Given $\min P$ is a four digit number, calculate the digit sum of that number

[steady tide]

not sure where to start

[odd edge]

@steady tide Has your question been resolved?

[peak thunder]

as you are being asked about the min P, you can plug in the values that will make P the smallest

I suggest you consider the sign. For minimization problems, the negative values need to be higher, and positive should be smaller.

@steady tide

[mystic saffron]

probably xyz = 2^8

idk how to do this other than using the fact that its a four digit number

[steady tide]

hm

i did consider 2 2 8

but it isnt correct

[mystic saffron]

thats 2^5

[odd edge]

@steady tide Has your question been resolved?

[odd edge]

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[flint maple]

hello

[flint maple]

can you please explain how this works

im trying all trig identities but all i get is (1+cos4x)/8

[viscid crest]

lemme try

[flint maple]

let me show what i do to get that

i use the hald angle identities first on both

which gets me ((1+cos2x)(1-cos2x))4

then its (1-cos^2(2x))/4

so then i use the identity again

(1+cos(4x))/8

idk why i dont get the right answer

[torn marlin]

whats sin2x formula

[flint maple]

2sinxcosx

[torn marlin]

so you got the first step

and then cos 2x formula

[flint maple]

why not just do difference of two squares and do it that way

[torn marlin]

multiple and divide the expression by 4 you get that

[flint maple]

i dont get it

we never get a sin(x) after we do the half angle ones

[torn marlin]

which one are you talking about

im suggesting a simpler way

but do tell what you are thinking

[flint maple]

sin^2(x) = (1-cos2x)/2

those ones

half angle identities

[torn marlin]

just a sec

[viscid crest]

hey uhm i got something but i got cos4x-1/8 🥲

im sure i did a mistake in my equations imma try again

[flint maple]

ive been trying to figure out what they do but to no avail 😭

[torn marlin]

that method would take a little work

i have an alternate

do you wanna hear it?

@flint maple

[flint maple]

sure

[viscid crest]

i looked at the solutio and sighed

[torn marlin]

so you did get the first step right

did you?

[flint maple]

so

the first step is to amke it (1-cos2x)(1+cos2x) ?

[torn marlin]

no no

[flint maple]

ohh

ok wait let me try your way

[torn marlin]

sin 2x is 2sinxcosx

multiple and div the question with 4

so numerator you get (2sinxcos)^2

[flint maple]

why by four?

ohh ok i get it

[torn marlin]

so you get that

[flint maple]

you square all

[torn marlin]

yeah

[flint maple]

ok sure

then what

[torn marlin]

cos 2x formula in terms of sin

half angle

[flint maple]

?

this

hmm

[torn marlin]

no theres 2 inside bracket

i said multiple and divide

[flint maple]

why tho

[torn marlin]

dont you get 4sin^2cos^2

in numerator

( 4sin^2cos^2)/4

thats the entire expressoin

[flint maple]

what i dont get is where that comes from

because isnt that for sin(2x)

[torn marlin]

yeah we just multiply and divide

yeah we do

hold on

ill send you a picture

[flint maple]

thanks

[flint maple]

sure

[torn marlin]

you got taht?

[flint maple]

yes

[torn marlin]

then

now

numerator

4sin^2 x cos^2 x

[flint maple]

sure

[torn marlin]

can this be written as (2sinxcosx)^2

[flint maple]

based on context i assume so

[torn marlin]

simplify it what do you get

isnt it the numerator?

[flint maple]

so this is what i have so far

ohh ok

then it becomes sin^2(2x) ?

ohhh

ok

but why not use the half angle formulas

[torn marlin]

you could try

[flint maple]

i get this way is easier but wouldnt the half angle formulas be more straight forward of an approach

[flint maple]

let me show u what i do

[torn marlin]

this is technically half angle formulas

[torn marlin]

and now cos2x = 1-2sin^2(x)

so you get sin^2 (x) from here

substitute there

youll get it

[flint maple]

let me show u

is this not correct

(ik i get the wrong answer but what did i do wrong)

[viscid crest]

i just got it n came here to write lmao

@flint maple need to ask soemthing

[flint maple]

surde

[viscid crest]

turns out my notes were incorrect

on operating on identities

can u tell me are these incorrect

[torn marlin]

in second step shouldnt denominator be 4

[viscid crest]

oh wait

uhm

[flint maple]

yes

sorry about that

[viscid crest]

i mean prefectus

[torn marlin]

yeah?

[viscid crest]

i mean both of u can explain

are u done with the question

[flint maple]

see because everything is kinda correct but idk why i get the wrong answer

[torn marlin]

just a sec

[viscid crest]

i dont wanna interrupt in the middle of

alr

[torn marlin]

type it out

ill read it

[viscid crest]

go ahead

alr

so i think i got my operating on sin (multiplication) notes wrong can u check them if i send it here

i kept getting wrong and i solved the question 3 times

[flint maple]

so it should be a (1+cos(4x))/8 at the bottom ( i wrote it wrong on the thingy, (i get the wrong answer))

[viscid crest]

these are my notes here can u check out the product rule

if ur done with the question *

[flint maple]

man

ive been at this question for too long

i think i shouldve skipped it long ago

[viscid crest]

happens the same to me but at least we're learning lol

[torn marlin]

yeah seems right

[viscid crest]

here cos(a-b) - cos(a+b )

this is what i hve been doing wrong tbh

like do i need to adjust my noes

notes

[flint maple]

i think

[torn marlin]

i mean i dont see anything wrong with it

[flint maple]

my understanding of half angle identities is wrong hence why im wrong

[viscid crest]

i got the question wrong using my notes though i kept getting cos4x -1 over 8

i was so confused lol

[torn marlin]

nah you substituted wrong again

theres a 2 in denominator remember

after you find lcm

you get the same thing

your approach is also right

[flint maple]

what

what is wrong

[flint maple]

yea but what does that have to do with it

[torn marlin]

2nd step to third step

is wrong

[flint maple]

ok

[torn marlin]

you substituted cos^2(x) wrong

youll get the same answer

(cos2x+1)/2 = cos^2 (x)

[flint maple]

yea

thats what i did

[torn marlin]

so when you substitute it

[flint maple]

this

on 4 sorry not on 8

[flint maple]

hm

ohh

ok

[torn marlin]

so you get the same thing

[flint maple]

thanks !

[torn marlin]

:)

[flint maple]

.close

[odd edge]

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[hallow lintel]

how do i do this? i alrdy got it wrong on the quiz just want to know

[hallow lintel]

https://tenor.com/view/kitty-cat-hello-chat-gif-635688626936566701

[viscid crest]

lol this is a test right?

[hallow lintel]

nah

it was homework

but the same qeustion is gonna be on my quiz

[viscid crest]

okay

!status

[odd edge]

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

[viscid crest]

show us ur work if u got any

[hallow lintel]

1

i dont know how to do this

its a sync so no one is teaching me

ive been using chat gpt to teach me

but its dumb when it comes to this

[viscid crest]

been there

what grade is this

uni?

[hallow lintel]

ya its finite math

uni

wait are u going to help

[flint maple]

ah

ok

is this

just determinant?

ad-bc=0

[hallow lintel]

its async so idk what to do

[flint maple]

ok

i think i know how to do this

but im not in uni so im not sure if its correct

[hallow lintel]

chatgpt and mathway both say no solution but its wrong 🥲

[flint maple]

ah sorry bro i have no clue

[hallow lintel]

its okay thank u for trying

[flint maple]

but

im

like 70% sure its this

(unless im wrong)

[hallow lintel]

i think thats wrong

but thanks

for trying at least

😔

[flint maple]

gl

cant u just solve it like a simultaneous equation

[hallow lintel]

thats what i thought

but its wrongngngnggngg

idk how to do all this augmented matrix methods

[flint maple]

ok

so

infinite solutions no?

if we multiply bottom for 2/3

[hallow lintel]

idk all these words bro 😭

<@&286206848099549185>

[viscid crest]

bruh i had to leave im sorry but i am a 12th grader i solved it using a way but i am sure it is 99.9% wrong

found this

https://www.cuemath.com/algebra/augmented-matrix/#

[viscid crest]

.

[hallow lintel]

okay ill use that thanks tho 🙂

[viscid crest]

ur welcome i wish i could help but i am just doing pre calc im so sorry

[odd edge]

@hallow lintel Has your question been resolved?

[hallow lintel]

<@&286206848099549185>

<@&286206848099549185>

@dim lava

[dim lava]

post

[dim lava]

do yk how to put it in augmented matrix form

[hallow lintel]

its like -8 4 I 12

[dim lava]

ye

[hallow lintel]

and 6 -3 I -9 right

[dim lava]

yes

[hallow lintel]

but idk what to go from there

[dim lava]

have you learned RREF

[hallow lintel]

no i havent learnt anything this is async

[dim lava]

oh

[hallow lintel]

im getting a tutor tmr

but i have to submit this in 10

minutes

[dim lava]

if u dont know how to do RREF just solve it like a normal systems of equations for now

yk how to do that right

[hallow lintel]

yea

i got -3/2 and 0

for x1 and x2

can u check if thats right

[dim lava]

ye looks right to me

[hallow lintel]

but how do i know if it has infinite solutions

for option b

what about th is one

@dim lava sorry for pinging but its urgent 😔

is 7x1 + 3x2 = 4 and x1 + 5x2 = 8 right

[dim lava]

that u need to learn how to do RREF

[hallow lintel]

oo

[dim lava]

the way a matrix looks like will tel u

[hallow lintel]

wait so iis 7x1 + 3x2 = 4 and x1 + 5x2 = 8 right

[dim lava]

yes

[hallow lintel]

kk thanks

[dim lava]

if youre under a time constraint use a calculator cuz looks like u understand the concepts

to check

[hallow lintel]

alright got it ill try that

thank u for helping me you are a blessig

[dim lava]

lol need to learn RREEF

wait search up a calc for it

ive used one before that checks

bru one min

https://matrix.reshish.com/gauss-jordanElimination.php

use this

lol

lol chatgpt not good with numbers

but sometimes it helps with how to do it

🙂

[odd edge]

@hallow lintel Has your question been resolved?

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[jagged sable]

i literally don't get it

i rationalized it

i got 80 on top

then what

[mystic saffron]

That is 1/5 not 1/2

Rationalizing a fraction would help as long as the numbers have the power of 1/2

[jagged sable]

wdym

no like thats the answer

the pic i sent is the correct answer

idk how thats the answer

[jagged sable]

wait what

[torn marlin]

1/2+1/2 = 1

thats why rationalizing works

[jagged sable]

i'm so confused rn

[torn marlin]

product of power rule

[jagged sable]

so i can't rationalize it???

i'm not getting what ur saying

[torn marlin]

uhh

@mystic saffron

[mystic saffron]

Mhm

[torn marlin]

i dont know how to explain

could you

[mystic saffron]

Okay

[torn marlin]

sorry to interrupt

[mystic saffron]

It is alright

[mystic saffron]

By rationalizing what did you do?

[jagged sable]

multiply up and down by the denominator

[mystic saffron]

And what would you get?

[jagged sable]

80

with radical

and denominator of just 16

[mystic saffron]

That is where you are wrong

It is (16)^1/5

[jagged sable]

huh

[mystic saffron]

Not 16^1/2

[jagged sable]

i'm so confused

i never learned it that way

my teacher always told us to multiply up and down

[torn marlin]

check that out once

[torn marlin]

thats absolutely right

[jagged sable]

guys i'm so so sorry but i literally don't have time my final is literally tomorrow

[torn marlin]

but do you know why we do that

[mystic saffron]

That would be the case when the question has a power of 1/2 I.e a sqrt

[jagged sable]

i also have bio final tomorrow i'm literally stressing

[jagged sable]

cuz we can't have a radical in the denominator

so the radical in the denominator cancels out

[torn marlin]

that is when its square root

say

1/((a)^(1/2))

when you multiply and divide with root a again

in denominator

you get root a x root a

[jagged sable]

guys i actually don't get what you're saying

[torn marlin]

hollon

do you know product of power rule

[jagged sable]

why did it work here then

[jagged sable]

no

[torn marlin]

check it out

[jagged sable]

why does it work in this qs but not in the other one

[torn marlin]

maybe they didnt do it that way

you dont even need rationalizing here

both have the same root

and 750 is divisible by 2

try reading it after a while, cramming one day before will ruin it more tbh

[jagged sable]

i'm not cramming

i've been studying for days

[torn marlin]

oh then cool

why you stressed then

[jagged sable]

cuz the exam is tmrw

[torn marlin]

like that

okay

[jagged sable]

i had to study 14 sections

that's the only qs i have

i'm not even done studying bio and it's almost 10 pm i can't do this

[torn marlin]

thats product of power rule

alright?

thats what i applied in third step

hope that made sense

[torn marlin]

thats why rationalising works like that

but in that question

you have 5th root

so denominator will be (term)^((1/5)+(1/5))

@mystic saffron

[mystic saffron]

Yes

[torn marlin]

can you like take over

[mystic saffron]

You are right

[torn marlin]

i gotta sleep

[jagged sable]

i'm so confused

[torn marlin]

thanks

@mystic saffron

[mystic saffron]

Okay

[mystic saffron]

Can you wait for some time?

[jagged sable]

okay

[mystic saffron]

@jagged sable

[mystic saffron]

Are you aware of this?

[jagged sable]

yea i get that

[mystic saffron]

What would 1/(a^1/5)(a^1/5) be?

[jagged sable]

wait i get it

so i multiply up and down

by just 2

since 16 is 2^4

then it becomes 2^5

[mystic saffron]

You mean 2^1/5

[jagged sable]

huh

one sec i'll show you my work

[mystic saffron]

You are on the right path but you are missing the powers

If that made any sense lol

[jagged sable]

it's kinda messy but yk

[mystic saffron]

Everything is right

But

You might want to check the numerator again

[jagged sable]

wdym

[mystic saffron]

I mean that it is wrong

[jagged sable]

which part

[mystic saffron]

It isn't t^5

[jagged sable]

oh the 5 is for the radical

[mystic saffron]

Oh

[jagged sable]

the index

[mystic saffron]

Then it's right

✅

[jagged sable]

thank you soooo much

i owe you my life

same with the other guy

[mystic saffron]

You figured it out yourself

[jagged sable]

thank you so much

[mystic saffron]

And gl

[jagged sable]

i'll need it

.close

[odd edge]

Channel closed

Closed by @jagged sable

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[odd edge]

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[outer temple]

i feel stupid for asking this but i forgot. if there are two lines that are the same, are they parallel? for example x=3 and x=3

[outer temple]

or no because they meet

i tend to think not

but then what are they

[nimble blaze]

coincident lines

not considered to be parallel

[odd edge]

@outer temple Has your question been resolved?

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[pliant oar]

Using the cylindrical shell method

[pliant oar]

<@&286206848099549185>

<@&286206848099549185>

[sullen ferry]

!status

[odd edge]

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

[sullen ferry]

https://tutorial.math.lamar.edu/classes/calci/volumewithcylinder.aspx

[pliant oar]

1

All i know is the integral would be from 0 to 2

Of 2pi*something

I don’t know what the radius and height would
Be

<@&286206848099549185>

I just need to know how to set this up

.close

[odd edge]

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

how do i solve this?

[vale horizon]

i dont even know where to start

wait actually

im thinking of possible avenues, could taking u = x^7 - 7x work for u sub?

[wanton bison]

yea

I would also pull out the 3

and multiply by 7/7

[vale horizon]

sry wdym by pull out the 3

[wanton bison]

from the numerator

3(x^6-1)

[vale horizon]

OHHHHH

[wanton bison]

and instead

multiply by 7/7

[vale horizon]

du/7

[wanton bison]

so it cancels out

[vale horizon]

bet ty gimme a sec to work this out

this feels pretty good

[wanton bison]

uhm what did you do with w

[vale horizon]

integrated 1/w to get ln(w) and plugged u back in

[wanton bison]

uhm

[vale horizon]

oh oops i meant 6^u not 6^x

[wanton bison]

,,\frac{3}{7} \int \frac{1}{6^u} : \dd u = \frac{3}{7} \int \frac{1}{w} \cdot \frac{1}{w \ln(6)} \dd w = \frac{3}{7\ln(6)} \int \frac{1}{w^2} : \dd w = -\frac{3}{7\ln(6)} \cdot \frac{1}{w} + C

So you integrated 1/w²

@vale horizon

[vale horizon]

uhhhh

sorry gimme a sec im tryna figure where i went wrong

[wanton bison]

I just applied your sub

[clever fjord]

𝔸dωn𝓲²s

[vale horizon]

uhhhhhhh

sry im kinda slow im gonna try to work this out step by step

[wanton bison]

,, ... = -\frac{3}{7\ln(6)} \cdot \frac{1}{w} + C = -\frac{3}{7\ln(6)} \cdot \frac{1}{6^u} + C = -\frac{3}{7\ln(6)} \cdot \frac{1}{6^{x^7-7x}} + C

[wanton bison]

no worries

[clever fjord]

𝔸dωn𝓲²s

[wanton bison]

I think your mistake is you did $\int \frac{1}{w} : \dd w = \ln \abs{w} + C$ mistakenly...

[clever fjord]

𝔸dωn𝓲²s

[vale horizon]

ohhhhhhh my god yeah

i forgot to transfer the derivative of 6^u to dw side

[wanton bison]

but $\dd u = \frac{\dd w}{6^u \ln(6)} = \frac{\dd w}{w \ln(6)}$

[clever fjord]

𝔸dωn𝓲²s

[wanton bison]

I think now this makes more sense

@vale horizon

[vale horizon]

ok i finished the problem and got the same answer as you i think

lemme send work

.

yeah i think this works now, ur a godsend @wanton bison thank u

wait shit my constants flipped

😭

one sec

alright got it, tysm for the help

.close

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[wanton bison]

ln(6) in the denominator

not numerator

[wanton bison]

.

then you got it

[vale horizon]

yeah accidentally divided lol

tysm for the help bro

[odd edge]

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[steel nest]

I get how to do A and B, its just C

I just started logs like a day ago

[nocturne belfry]

try writing 1 as a log

lets start here: can you think of any way to write 1 in terms of an exponential?

[steel nest]

https://tenor.com/view/ohno-meme-monkey-ohno-ohno-monkey-ohno-emote-ohno-twitch-emote-gif-119989999548046247

[nocturne belfry]

lets say i have e^x

can we pick an x that makes e^x = 1?

[steel nest]

1

wait no

0

right

[nocturne belfry]

yea

okay so e^0 = 1

now take the log of each side

actually i dont think any of this is necessary rereading the problem

[steel nest]

Oh

[nocturne belfry]

lets use this

sorry my brain is slow

rule 5 gives us a way to rewrite 1

what b do you want to use?

[steel nest]

wait what

isn't it just b

I don't understand

[viscid flint]

does your log mean base e or base 10

[nocturne belfry]

hi hayley

[steel nest]

I just started yesterday and my tutor gave me some of the questions to do. I tried asking gpt for some steps and that didn't work out 😭

I dont get what e and 10 means

sorry

[nocturne belfry]

so heres the idea

theres a way to write 1 in terms of logs

thats rule 5

[steel nest]

right

[nocturne belfry]

then, rule 2 gives us a way to combine two logs into 1 if they are minus each other

so you would have something like

log( b ) = log( something )

after you apply rules 5 and then 2

[steel nest]

theres 2 b's tho

[nocturne belfry]

you can solve this just like you solved a

thats fine

do you know what that means?

$\log _b (b)$

[clever fjord]

jan Niku

[nocturne belfry]

i mean, the under-b

the little b

[steel nest]

nope

[nocturne belfry]

it just means base

its not important if youve never seen it

what you should know is that $\log _{10} ( 10 ) = 1$

[clever fjord]

jan Niku

[nocturne belfry]

and a person just learning logs may not write the little 10

they might just write $\log (10) = 1$

[clever fjord]

jan Niku

[steel nest]

oh ok

how do I apply this? because its 1-4loga, is it log(1/4loga)

[nocturne belfry]

these problems are actually arranged in kind of a clever way which is rare

because you solve problem a right

then, for problem b, you apply rule 2

and it looks like problem a

then, for problem c, you apply rule 5, and it looks like problem b

[steel nest]

https://tenor.com/view/ohno-meme-monkey-ohno-ohno-monkey-ohno-emote-ohno-twitch-emote-gif-119989999548046247

[nocturne belfry]

first you need rule 5

$\log b = 1 - 4 \log a$

[clever fjord]

jan Niku

[nocturne belfry]

how do we rewrite the 1 again?

using logs

[steel nest]

B is 0

right

[nocturne belfry]

no

[nocturne belfry]

go back here

[steel nest]

10

[nocturne belfry]

i want you to rewrite the whole equation to be clear

[nocturne belfry]

rewrite this whole equation, but with a log where 1 is

we know log(10) = 1

so rewrite the equation using this

[steel nest]

$\log b = log(10) - 4\log a$

[clever fjord]

dane249

[nocturne belfry]

alright

hmm, looks like we need one more rule

you notice that rule 2 has just log - just log

theres no 4 out front

can you spot a rule to rewrite $4 \log a$ as $\log \qty( \text{something} )$?

[clever fjord]

jan Niku

[steel nest]

uh

3

it will be loga^4

right

[nocturne belfry]

yea

okay so $\log b = \log 10 - \log \qty(a^4)$

[clever fjord]

jan Niku

[nocturne belfry]

were close

rule 2 says ...

[steel nest]

uh

log10/loga^4

[nocturne belfry]

not quite

the rule actually combines them into a single log

[steel nest]

wait what

logb(10/a^4)

[nocturne belfry]

the b's might be mixing you up

just ignore the b's in this particular image

okay $\log b = \log \qty(\frac{10}{a^4})$

what now?

[clever fjord]

jan Niku

[steel nest]

uh

do we get r id of the logs

[nocturne belfry]

yea, i think this would be good

[steel nest]

so b=(10/a^4)

[nocturne belfry]

yea

[steel nest]

whats the next step

do we multiply both sides by 10?

[nocturne belfry]

you just need to isolate a

[steel nest]

do we multiply both sides by a^4

[nocturne belfry]

yea you could start there

then divide across the b

[steel nest]

so a^4 = 10/b

[nocturne belfry]

yea, almost done

[odd edge]

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[worn mauve]

did i find the derivative correctly

[zinc glacier]

you cannot cancel things out like that

[worn mauve]

what should i do at the second line then

[boreal crag]

The numerator is wrong

what's u'

[worn mauve]

forgot the bottom squared

[boreal crag]

no, that's the denominator, which is fine

[wanton bison]

you can use partial fraction decomposition if you wanna simplify it

[worn mauve]

idk what that is

[boreal crag]

Anyways fix your numerator and it should be easier to simplify

[worn mauve]

u' would be 0 right

[viscid flint]

lo dee hi minus hi dee low, over the square of what's below

yes

[worn mauve]

so that cancels out my (x^2-4x)

leaving me with -(2x-4) on the numerator correct?

[viscid flint]

yes

[worn mauve]

ahh thank you

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[worn mauve]

.reopen

[odd edge]

✅

[worn mauve]

@viscid flint is this more correct

did i find the derivative correctly

.close

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[viscid flint]

isn't that what i just said? yes

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[robust pollen]

Can anyone tell me how to do the following question.

[robust pollen]

$\frac{d}{dx} x!$

[clever fjord]

DaWhiteXD

[robust pollen]

and how's that possible

Les say we know the gamma function

[dim igloo]

yikes what is that

[robust pollen]

but how to derive gamma function?

[robust pollen]

Formula of factorial

or we simply write as

$\frac{d}{dx} \int_{0}^{\infty} t^{x-1}e^{-t} ,dt$

[brittle beacon]

\infty

[robust pollen]

thx

[brittle beacon]

(the space as well at the end where the $ is too, remove it)

[clever fjord]

DaWhiteXD

[robust pollen]

done

thx for helping me to make that thing, chartbit

So what's the first step

to derive this thing

[brittle beacon]

Also bear in mind that when it comes to positive integers, how you'd write n! in terms of the Gamma function, I'd say

[robust pollen]

ok

x(x-1)(x-2)...

Do we actually do intergral by parts?

[brittle beacon]

Erm, more that the suggestion was how you'd express x! "properly" in terms of the Gamma function

But in terms of actually differentiating it, erm, I'm a bit fried and don't think I can help

[robust pollen]

no clue, I thought gamma function is the 'most proper' way to express x!

[robust pollen]

it's ok

[merry finch]

i think you need to do a bunch of ibp's

[robust pollen]

I think if we do ibp, it comes to be x(x-1)(x-2)...

[merry finch]

i cant really remember tbh

[robust pollen]

oh...

[merry finch]

it looks like you do some fuckery that moves the derivative inside the integral sign

you're supposed to get

[robust pollen]

umm

just adding ln(t) inside

I'm thinking of partial derivative

that you may actually put the d/dx inside the integration

and ye

I think the problem solved, lol

we can use parital differentiation to get e^-t \parital d/dx {e^(x-1)ln(t)} and it should be equal to somehow e^-t t^x-1 ln(t)

[robust pollen]

thx man

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[dusk mica]

I'm given the function 3000sin(0.08(t - 9.875)) - 4500 and need to to find the first four solutions to four significant digits when f(t)=-5900 and this is what I have so far, but I'm pretty stuck after this

[light agate]

whats the inverse of sin

[dusk mica]

sin^-1

and i tried that but moving it to the other side / arcsining it doesnt work when i try to solve it

like i just really dont know how to find the 4 answers besides that each answer is when the sine function does a full cycle and repeats

[light agate]

wdym it doesnt work

are you not allowed to use a calculator

[dusk mica]

like

i have isolated t before but then when i try to check my answers in desmos they never end up right

idk how to explain it besides that sorry

[light agate]

hmm i see hold on

the issue might be radian vs degree

have you double checked that youre being consistent with that

[dusk mica]

yeah im always in radians

when i solve i remember getting a 0.4 decimal somewhere

for what t is

hold on ill try and solve for it again because that part is consistent

-0.4855 rounded to 4 sig figs

[light agate]

right thats what im getting too

i mean you can just track it step by step on desmos and see where it differs

[dusk mica]

alright thats giving me some confidence over here thanks

alright ill do that

im getting 15.9438 as my first answer which isn't right @light agate

[light agate]

where did the minus sign go

[dusk mica]

i added sorry its faint

the little 11 looking thing next to the first isolated t is a drag down

[light agate]

right but

[light agate]

here you said -0.4855

[dusk mica]

omg ur right

thanks

[light agate]

npnp

[dusk mica]

yeah im getting 3.8063 and its still not right

@light agate

[light agate]

not right according to whom

[dusk mica]

desmos

LOL

he said its susposed to be whenever a cycle repeats which would be at -4500 as the y value right?

[light agate]

what value is desmos giving

[dusk mica]

its giving 9.875 as the first point at -4500

which is already in the equation

so idk how it could be less unless im thinking about this wrong which could 100% be likely

@light agate

mb keep forgetting to mention when i reply

[light agate]

no worries

[light agate]

oh hold on youre plugging in the original equation

[dusk mica]

wdym

[light agate]

idk desmos seems to agree with you here

[dusk mica]

omg what

[light agate]

unless i messed up really bad i think youre right

[dusk mica]

omg ur right

yeah

okay the next problem i have which is likely way easier, i completely forgot how to find the next cycle

@light agate

[light agate]

wdym next cycle

[dusk mica]

like im susposed to find the first 4 answers

and i forget to find the next 3

[light agate]

oh uh

at the arcsin step you just add or subtract 2pi

[dusk mica]

i know its adding 2pi somewhere

ya

[light agate]

depending on what you mean by the first four answers

[dusk mica]

so like

[dusk mica]

at one step i'd do (0.08(t-9.875)) = sin^-1(-7/15) + 2pi

@light agate

[light agate]

yeah

[dusk mica]

then again with (0.08(t-9.875)) = sin^-1(-7/15) + (2)2pi

[light agate]

yep. you can verify for yourself

[dusk mica]

alright thanks! that's all

[light agate]

np

close the channel if youre done

[dusk mica]

oh ok uh

is there a button

nvm

.close

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[rocky ocean]

Inequalities of 3 variables

I'm a bit stuck on finding an elegant solution to this problem. I've tried fiddling around with Cauchy-Schwarz, rearrangement, AM-GM but I keep running into walls of way too long expressions for anything elegant.

Any ideas?

[odd edge]

@rocky ocean Has your question been resolved?

[odd edge]

@rocky ocean Has your question been resolved?

[rocky ocean]

<@&286206848099549185>

[odd edge]

@rocky ocean Has your question been resolved?

[mystic saffron]

AM Gm inequality should work

Lemme try

@rocky ocean

[rocky ocean]

Go ahead :D

[mystic saffron]

Have you tried something with HM ?

[rocky ocean]

That works. I think you could also apply AM-GM just twice in a row too

[mystic saffron]

I tried this

Oh waitttt

Now we do for s-x , s-y , s- z .

Uh

@rocky ocean

[rocky ocean]

Cool!

[mystic saffron]

Last line should be 5s/3 my bad there

[rocky ocean]

That’s a cool approach. I also found a nice solution using Chebyshev's inequality and combining that with AM-GM-HM

[mystic saffron]

Can you please show me ?

[rocky ocean]

Sure thing!

[mystic saffron]

Damn too good, well I thought of the LHS to be cyclic terms( symmetrical) could not think more of it

[rocky ocean]

Thanks for the help @mystic saffron !

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[summer cradle]

could you make two different ticket price setups that would be consistent with the equations?

[signal oar]

Yeah

[summer cradle]

then yea, who knows

[light agate]

assuming your row reduction is right, any setup where
x1 + x3 = 17
x2 = 6
is valid

[signal oar]

Oh wait, we really can atmost say they can be correct because it could happen that in one case, children had to pay more than adults and then the other way around

[light agate]

yes

[signal oar]

Thanks

[light agate]

i mean it really comes down to what you mean by "the prices are correct"

"the prices are necessarily correct for all x1 x2 and x3" isnt right, however there are still infinite x1 and x3 that satisfy this

[signal oar]

Yep, thanks

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[tawdry wagon]

Hi everyone, I was hoping someone could check this proof. I have to prove that differentiability implies continuity. But I honestly think I can prove that differentiability implies lipschitz cts which I already know implies cts. Idk if this is correct, there’s not too much on it online. So my proof goes: we have $|\frac{f(x) - f(y)}{x - a} - f’(a) |< \epsilon$ then by the triangle inequality we have $| |\frac{f(x) - f(y)}{x - a}| - |f’(a)| | < \epsilon $ then it follows that $ |\frac{f(x) - f(y)}{x - a}| < |f’(x)| + \epsilon$. And thus, $|f(x) - f(y)| < (|f’(a)|+\epsilon)|x - a|

[tawdry wagon]

Then just choose epsilon = 1 or smth

[clever fjord]

theaveragejoe6029
Compile Error! Click the reaction for more information.
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[tawdry wagon]

Idk if this works, but it’s definitely easier to remember this than the proof for cts for my exam ig

[lusty nova]

Differentiability implies continuity is a straight forward proof

[tawdry wagon]

I know I can remember both, but is this still solid?

Plus imo this is easier, but that’s besides the point

[lusty nova]

Yes

[tawdry wagon]

Fr, I thought something might be wrong with it only working for a specific delta or smth

[weary pelican]

this works, but if you want something easier :

(f(x)-f(a))/(x-a) converges to f'(a)

so by product of limits, f(x) - f(a) converges to f'(a)*(a-a) = 0

f(x) converges to f(a)

[tawdry wagon]

Yeah, I know that one, but my course is pretty anal about rigour 😦

I don’t wanna be marked down for not using epsilon delta defintions

[weary pelican]

wdym? You can't use the limit def of continuity?

[tawdry wagon]

Yeah

[weary pelican]

that's kinda bs if it's the case

do ask your teacher

[tawdry wagon]

Yeah I have, they said that it’s a real analysis course so they want us using epsilon and delta shit from first principles

[tawdry wagon]

To be fair tho, they said it’s still possible to get marks using the limit definition, but not possible to get full marks

So it’s a good back up in case I forget the other proofs

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[harsh grotto]

Good morning, I have a question. I am working on a score calculation. Among the 1-5 star ratings, we currently have an average of 3.5 with 137 people, but I want to calculate how many 5-star ratings we need to move that score to 3.6. I can do this by adding 5 stars one by one and checking if the average increases. But is there a way to calculate this so that I can know all at once how many 5-star ratings we would need to raise the average score?

I looked into it a bit, but I'm not sure if this formula is correct:

(137 X 3.5 + x X 5) / (137 + x) = 3.6

But how would I calculate that in Excel, so that the ending of the formula becomes =X that way i can put it in excel so it calculates the awnser.

(used translation from chat gpt since i am not that great in english)

[merry finch]

so think about it like this

you currently have 137 * 3.5 = 479.5 stars in total

you want to get to total stars of a, with b number of people, where a/b = 3.6

let's call the extra people b - 137 = x

then we get the formula you've written up

now in order to solve for x you just need to rearrange it

$\frac{479.5 + 5x}{137 + x} = 3.6$

[clever fjord]

Frosst

[merry finch]

how would you rearrange this for x?

[harsh grotto]

not sure if this is correct but divide 479.5 by 137 and devide 5x by x?

[merry finch]

no, you cant do that

[harsh grotto]

sorry i dont know 😦

divide by 3.6 maybe?

[merry finch]

have you had any formal education in maths?

[harsh grotto]

i had like 8 years ago 😛

[merry finch]

like manipulation algebraic expressions (such as these ones)

[harsh grotto]

havent touched it in a long time

[merry finch]

so the problem we have right now is this division

we've got x on the top and on the bottom and it's giving us trouble

what we want to do as a first step here is multiply both sides by 137+x

looks like this

$\frac{479.5 + 5x}{137 + x} \times \textcolor{green}{(137+x)}= 3.6\times \textcolor{green}{(137+x)}$

[clever fjord]

Frosst

[merry finch]

does this make sense to you?

[low locust]

\cdot

[harsh grotto]

so that makes it 479.5 + 5x = 3.6 x (137 + x) right

[merry finch]

brackets!

but yes!

[harsh grotto]

yea xd

[merry finch]

now do you remember how to distribute the right side here

[harsh grotto]

so then you have 479.5

[low locust]

dont use x for multiplication when you have x as a variable

[harsh grotto]

i think so yea

so now you get 479.5 +5x = 493.20 + 3.6x ?

[merry finch]

yep

now we can collect the x on 1 side, and the numbers on the other side

[harsh grotto]

so 1.4 = 13.7 ?

[merry finch]

woah

hold up what did you do

[harsh grotto]

ah i put 5x -3.6 x

[merry finch]

what happened to the x

[harsh grotto]

and 493.20 - 479.5

[merry finch]

yep

[merry finch]

so then we have 1.4x = 13.7

dont lose the x

[harsh grotto]

so then divide both side by 1.4

[merry finch]

yep

[harsh grotto]

so you get 1x = 9.78

i think i have it 🙂

[merry finch]

there you go!

[harsh grotto]

thx a lot i was sturggeling to get that 1 x

its been a while but i got it thx to your help

[merry finch]

you've got it now!

🙂

[harsh grotto]

i should be able to formulate a formjula for excel with this thx 🙂

[merry finch]

👍

[odd edge]

@harsh grotto Has your question been resolved?

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[raw moat]

how did they do this?

[raw moat]

nvm

.close

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[slate onyx]

problem from discrete math:

In how many ways can we line up 5 red, 7 blue and 3 green balls if each red ball must be in front of all blue balls?

[summer cradle]

think about how you could place red and blue balls first

OR think about how to insert the green balls into a line of 5 red balls and 7 blue balls (in that order)

[slate onyx]

there is only one way to place red balls and blue balls

so then i must place green balls and i have 13 spaces to place

so i do $\binom{13}{1}^3$

[clever fjord]

OHHELLNAH

[slate onyx]

so thats actually the full solution

is that ok?

[summer cradle]

i do not think that’s right

oh

misremembered the numbers soz

yea that looks good

[slate onyx]

nicee ty

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[noble swallow]

@helpe

[noble swallow]

<@&286206848099549185>

[short topaz]

why u pinging them?