help-13

[distant kite]

number that could form a fraction

[zinc crown]

Yes

[distant kite]

am I right?

ok

[zinc crown]

Integers are also rationals, by the way

[distant kite]

yup

[zinc crown]

(as, for example, 3 = 3/1)

[distant kite]

ok

[zinc crown]

Now, what does it mean for a and c to be coprime?

[distant kite]

a and c does not share any common factor

gcd(a, c) would be 1

I created this question so I think I know the wordings, lol

?

Wlmmm?

[zinc crown]

Ok, yes

So the possible zeros are in the form a' / c' , where a' and c' are factors of a and c respectively

For the rational zeros that is

[distant kite]

so a'/c' = 1?

[zinc crown]

Not necessarily

But the point here is, a' / c' can be an integer, if c' = 1

It can also be a rational and not an integer

So the answer to your question is: it depends

Sometimes yes, sometimes no

[distant kite]

umm

But it is just a possible zero

doesn't mean it's correct

[zinc crown]

It doesn't mean

Indeed

[distant kite]

So how do u proof that

need to be definitely sure

if it's possible or not

[distant kite]

still need proof

[zinc crown]

My wording here is bad as it was implying that both are possible from my argument

While both are possible, it only suggests it

But to prove it, just find two polynomials

With a,c coprimes

[distant kite]

i tried

[zinc crown]

One of them having integers has roots

[distant kite]

like 2 and 3

[zinc crown]

x^2 - 2x + 1

[distant kite]

Doesn't have integer roots

1 and 1???

Huh

wait

lol

How I didn't think that

Thanks for ur help, wlmmm

.close

[cedar kiln]

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[distant kite]

.reopen

[cedar kiln]

✅

[distant kite]

Wait a minute

[distant kite]

b belongs to positive integers

nvm i found one of the solutions

.close

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[wicked badge]

(0 + 20) + (0 - 21) = r^2 right?

so -1 = r^2, which is, impossible?

nvm

I seeee

it makes sense!

[carmine whale]

..

[wicked badge]

I forgot the squared

.close

[cedar kiln]

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[carmine whale]

U missed the squares

[wicked badge]

.reopen

[cedar kiln]

✅

[wicked badge]

@carmine whale sooo

is it squre root 401

cause it's 20 squared plus 1 squared right

which is 400 plus 1

which is 401?

.close

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[crimson sedge]

why do you keep closing your channels

[wicked badge]

ahhh

[crimson sedge]

You're wrong btw

[cedar kiln]

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[tropic goblet]

Hello need help on integerating factors

[tropic goblet]

Solution says it's this

That's how how far i gotten but im unsure where to go from there

it's cool

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[trim pine]

@cedar kiln i need help

[tropic oxide]

pinging Bot is pointless

anyway tho

!da2a

[cedar kiln]

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/

[trim pine]

lmao

then whats the point of being in here if its not gonna help ??

[tropic oxide]

you get help from PEOPLE

not bots

the bot is just there to clarify server policy and be QoL generally

and also make the channel system work

anyway

you need to post the question you need help with

[trim pine]

ohhh

and will it help me ?

[tropic oxide]

bruh

do you still not get it??

you get help from PEOPLE

NOT from bots

[trim pine]

OHHH LMAOOO

[tropic oxide]

!volunteers

[cedar kiln]

Helpers are just people volunteering their time to help you. Be polite and patient.

[trim pine]

mbbb

[tropic oxide]

seriously just post your question already it's been like 3 pages of back and forth now

[trim pine]

okay lemme send a ss

lmaooo its alr

i tried solving it my self and i got it wrong but

[tropic oxide]

jptfq

[trim pine]

lmaoo

[tropic oxide]

ok so

do you know how to graph straight lines generally

[trim pine]

i got all of them right except this one

yes

[tropic oxide]

ok can you show your last wrong attempt

[trim pine]

idk if it lets me go back but lemme try

[tropic oxide]

failing this, try again, but instead of submitting it, only screenshot and send it here

[trim pine]

no bc its like i only get two trys

so if i get it wrong im done

but like ill still get like a 93 or something

[tropic oxide]

but you CAN graph the line but just hold off on submitting it, no?

[trim pine]

ofc

[trim pine]

i think so

[tropic oxide]

then do that

let's see you do it

[trim pine]

i mean x is 2 right

[tropic oxide]

the x intercept is not 2 here no

[trim pine]

mm

[wraith dagger]

QuasiStar 超新星

[trim pine]

find the intercepts

??

ohhhh got it

[tropic oxide]

you're confusing some shit here but i don't know exactly how to articulate it

[trim pine]

lemme do this on paper and ima see if its right hold on

oop my teacher closed the assignment 😭

[cedar kiln]

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[little junco]

How do I show b? help

[cedar kiln]

@little junco Has your question been resolved?

[cedar kiln]

@little junco Has your question been resolved?

[unkempt tartan]

You put the functions on top of each other in a limit, and just find if it converges to a number

h^a/h^b is well, h^(a-b)

If a-b is negative, it's essentially just 1/h^C where C is a positive constant. Which converges to 0 as h gets bigger

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[sour folio]

ik its

[sour folio]

Ik its rly simple but im just dumb

6*180
————
360

then divided by 2?

but i got 3

so 1.5?

[distant wyvern]

sum of all angles in octagon?

[sour folio]

what?

8

[distant wyvern]

yeah this is it

[sour folio]

wait

its

6*180
———-
8

[distant wyvern]

6x180 = ?

[sour folio]

1080 i think

[distant wyvern]

yes

so thats the sum of all angles in an octagon

[sour folio]

idk wymm

[distant wyvern]

do you think you can find the sum of each individual angle here from that?

[distant wyvern]

like when you add all the interior angles in a polygon

you get a particular value

for a triangle its 180

for a quadrilateral its 360

[sour folio]

ik

so

1080/360

=3

[distant wyvern]

similary, for a octagon its 1080, given by the formula S = (n-2) $\cdot$ 180

[wraith dagger]

SirGareth

[distant wyvern]

why the 1080 by 360?

[sour folio]

Bc of what u said

[distant wyvern]

ahh i thought they were blank spaces

nevermind, that was wrong

[sour folio]

Divided

(6*180)/360

…

i forgor

[distant wyvern]

what would each of these angles be?

i marked some, but all the angles are equal (because regular octagon)

and they sum up to 1080

[sour folio]

135?

[distant wyvern]

yes

now, do you think you can find x?

[sour folio]

divide by 2?

…

[distant wyvern]

yes

[sour folio]

ok

[distant wyvern]

thats your answer

[sour folio]

@distant wyvern welp

[distant wyvern]

?

[sour folio]

with this to last one

[distant wyvern]

is this from a test?

[sour folio]

bro if this was how would i be typing in dc

[distant wyvern]

no, just wondering since it shows the 4 marks

[sour folio]

ye its DrFrostmaths

idk why doe

[distant wyvern]

m

[sour folio]

m?

[distant wyvern]

to work out the interior angles you can use the fact that P is a hexagon

idk how we would show its a hexagon tho

[sour folio]

MHM

Wait

its 120

ty for all the help @distant wyvern

.close

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[clever shard]

So, question:
I know what’s pmf (short for probability mass function)(individual probability of each event, e.g. pmf of rolling the numbers on a die are all 1/6)

But the word “marginal” pmf got me stuck.Does it mean something different? Or is it the exact same thing? I tried to find it in my online notes, but nothing about it was mentioned other than the normal pmf…

[tropic oxide]

marginal just means you consider X and Y each on their own

[clever shard]

Ah, I see

Non-english mother language in a nutshell I suppose :p thanks!

.close

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[blazing epoch]

Solve the task without a digital tool.

For a straight line y=kx+m
applies:

◗ k
-value is two more than m

• value
  ◗ y=3
  when x=4

Determine the equation of the line.

[blazing epoch]

so what i did was that i did it like this:
3=4x+2x
3=6x
but thats clearly wrong

<@&286206848099549185>

oh shoot i thought it was 10 mins

[livid hound]

can you write an equation that represents

k value is two more than m

[blazing epoch]

k=2m?

[livid hound]

no

that would be k is two times m

[blazing epoch]

oh u mean 2k=m

[livid hound]

no

[blazing epoch]

sorry i am not the best in english

[livid hound]

if i had 6 apples
and you have 2 more apples than me,
how many apples do you have?

[blazing epoch]

8

[livid hound]

yes

same idea applies here

[blazing epoch]

oh i thought multiplication

k+2=m?

[livid hound]

no

what is 2 more than m?

[blazing epoch]

k right

wait no

k=m+2

[livid hound]

yes

now substitute that into

y=kx+m

[blazing epoch]

hang on

3=4x+m+2

and i just solve that

right?

[livid hound]

no

trying to skip ahead and making multiple mistakes along the way

[blazing epoch]

then i dont know

[livid hound]

yes
now substitute that into
y=kx+m

replace the k in that with (m+2)

[blazing epoch]

ohhhhhhhhhhhhhhhhh

y=2m+x+2

but then i have 2 variables

ohh but i know x

so m is m=(-3)/2

[livid hound]

no

you're not substituting properly

[blazing epoch]

whats substuting

[livid hound]

*substituting

typo

[blazing epoch]

yeah but what is it

[livid hound]

replacing something with something of equivalent value

in this case the replacement of k with (m+2)

[blazing epoch]

oh

[livid hound]

$kx + m\redneq k + x + m$

[wraith dagger]

ℝam()n()v

[livid hound]

$kx$ represents the product of $k$ and $x$, upon substituting, that component of your expression should still represent the product of:
$x$ and whatever $k$ is equal to

[wraith dagger]

ℝam()n()v

[blazing epoch]

oh wait

3=5m+8

so m=-1

[livid hound]

skipped ahead again
but work is right this time

[blazing epoch]

i wrote it on paper

i dont like writing math on keyboard

[livid hound]

writing all steps allows people to interject immediately if mistakes were made so that you can stop going down a bad path

[blazing epoch]

i did write all steps

just on paper

[livid hound]

shows though process etc, making it easier to identify what issues you need to fix

[blazing epoch]

do u wanna see it

[livid hound]

no need anymore since its right

[blazing epoch]

okay

[livid hound]

just try to make an effort to show all steps in the future

[blazing epoch]

okok

but the answer should be y=4x-1?

[livid hound]

no

[blazing epoch]

oh k is supposed toi be 2 more than m

[livid hound]

you've found m to be -1,
what will k be?

[blazing epoch]

so y=x-1

[livid hound]

yes

[blazing epoch]

thank you

.close

[cedar kiln]

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[silver fable]

does f=O(g) means that g grows faster than f ??

[fair geyser]

yes

it could grow equally fast too

[silver fable]

what's the difference between O and o then ?

[fair geyser]

i'd guess it's the opposite?

oh ok small o is strict, in addition to "opposite"

[silver fable]

wait so if f=O(g) and g=O(f) then can we say that f is equivalent to g ?

[fair geyser]

i don't know this stuff well

one important thing is it's up to a constant, we would say f(n) = 2n grows faster than f(n) = n

but that's like not enough, it's factored out

[opaque pivot]

it would differ by a constant. im pretty sure it just means f' = g'

[fair geyser]

that doesn't sound right

[silver fable]

the derivative of both functions is not the same here

[fair geyser]

it's clearly not about first derivative

[opaque pivot]

yes i realised as soon as i said that

sorry

[silver fable]

I found the same question on stackexchange in case u're interested

https://math.stackexchange.com/questions/580948/if-fx-ogx-gx-ofx-then-can-we-write-fx-sim-gx-or-any-othe

[cedar kiln]

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[crystal tinsel]

I need help with a system of polynomial equations in a ring. Extending fermat's little theorem to polynomials: number n is prime iff (X + a)^n = X^n + a (mod n) when n and a are not coprime. I'm trying to figure out what happens when n and a are NOT coprime and compute a number n s.t. (X+2)^n = X^n + 2 mod n. n is orders of magnitude larger than 2.

[cedar kiln]

@crystal tinsel Has your question been resolved?

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@crystal tinsel Has your question been resolved?

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@crystal tinsel Has your question been resolved?

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[sly canopy]

how would I start writing a formula for this

[prime notch]

sin^2 + cos^2 = 1

for any angles

[cedar kiln]

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[urban ermine]

how do i do a ?

[cedar kiln]

@urban ermine Has your question been resolved?

[urban ermine]

<@&286206848099549185>

[cedar kiln]

@urban ermine Has your question been resolved?

[wary storm]

dk

[cedar kiln]

@urban ermine Has your question been resolved?

[urban ermine]

<@&286206848099549185> if you cant answer the second part of the question, could you help me with the "assuming that it starts in state 1" im not sure what to do from there

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[urban ermine]

.close

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[obtuse citrus]

hi

[obtuse citrus]

if the half life of strontium 90 is 29 years use a differential equation to find the equation for the amount of strontium-90 p(t) as a function of years

so i know that t=29 when p=0

and i know when u do 1/29 it represents a small change and that would be the equation of p(t)=t/29

i just dont know how to show that in a differiencial

oh wait

this is the half life equation

so 0=1/2^(29/((1/2)^29))

[violet flume]

👀 I'm not sure if what you wrote there makes sense

maybe it'd be easier to solve $\frac12 N_0 = N_0 \qty( \frac 12 ) ^{\frac{t}{t_{\sfrac12}}}$

[wraith dagger]

jan Niku

[violet flume]

oh wait, you already know what this will tell you

[obtuse citrus]

what i said i was wrong

[violet flume]

nah, you got it, theres nothing to do but plug into that equation

just have to do it right

[obtuse citrus]

i got 1

[violet flume]

yea, thats my bad

[violet flume]

just substitute into this equation the facts you have

no

[obtuse citrus]

idk what other facts i have besides the fact that strontium-90 half life is 29 years

[violet flume]

this is all you need

you can express the amount of strontium present as a function of the initial amount and years elapsed

[obtuse citrus]

but how can i make that into a differential

like

[violet flume]

you can derive this equation I guess

the decay of the substance is proportional to the amount

[wraith dagger]

jan Niku

jan Niku

[violet flume]

@obtuse citrus is this what you were looking for

[obtuse citrus]

kinda

im just confused on ur added statement

use initial condition of N at 0

[violet flume]

right

[obtuse citrus]

so it would be dN_0/dt_29

[violet flume]

huh? no

[violet flume]

this is a seperable first order diff eq

convention is to separate the variables with their differentials

so get to $\frac{\dd N}{N(t)} = -\alpha \dd t$

[wraith dagger]

jan Niku

[violet flume]

now integrate both sides

$\int \frac{\dd N}{N(t)} = \int -\alpha \dd t$

[wraith dagger]

jan Niku

[obtuse citrus]

[violet flume]

what happens now?

[obtuse citrus]

N= -alpha^2/2

this is weird

sorry

[violet flume]

huh? lambda?

[obtuse citrus]

i mean alpha

[violet flume]

what is $\int \frac{\dd x}{x} = ?$

[wraith dagger]

jan Niku

[obtuse citrus]

OH

ln(x)

[violet flume]

okay

what is $\int \alpha \dd t$

oops

[obtuse citrus]

i cant tell if alpha is a variable

[wraith dagger]

jan Niku

[obtuse citrus]

or a placement holder

[violet flume]

alpha is a parameter

[obtuse citrus]

oh alpha^2/2

[violet flume]

its known as the rate constant

its just a number

some unknown number

[obtuse citrus]

[violet flume]

so this is wrong

[obtuse citrus]

is it just alpha t

[violet flume]

its a constant right

are we integrating with respect to alpha

[obtuse citrus]

alpha^2/2 + C?

[violet flume]

stop guessing

[obtuse citrus]

oh wait

[violet flume]

what is the variable of integration in this integral

by stop guessing i just mean like

[obtuse citrus]

idk t

[violet flume]

stop just throwing out random answers its not gonna help you if you just randomly guess answers until i tell you which one is right

[violet flume]

right

so then $\int -\alpha \dd t = - \alpha \int \dd t$

[wraith dagger]

jan Niku

[obtuse citrus]

makes sense

[violet flume]

whats $- \alpha \int \dd t = ?$

[obtuse citrus]

t

[wraith dagger]

jan Niku

[obtuse citrus]

-alpha*t + C

[violet flume]

yea!

okay, were closer

now were here

[obtuse citrus]

[violet flume]

$\log N(t) = -\alpha t + C$

[wraith dagger]

jan Niku

[violet flume]

how do we get just N(t) on the LHS

[obtuse citrus]

rase everything to e?

[violet flume]

$e^{\log N(t)} = e^{-\alpha t + C}$

[wraith dagger]

jan Niku

[violet flume]

simplify this for me 🙏

[obtuse citrus]

n(t)=e^(-alpha*t+C)

[violet flume]

simpler

we can fix up the RHS

[obtuse citrus]

rhs?

[violet flume]

right hand side

[obtuse citrus]

oh ok

[violet flume]

this is an initial value problem

[obtuse citrus]

ohhh

[violet flume]

weve been given an initial value

[obtuse citrus]

yuh

[violet flume]

that means we can solve for C

but its in a weird place

can you get this into a form thats more natural using exponent rules

no

[obtuse citrus]

wait

[violet flume]

$e^{-\alpha t + C}$

[obtuse citrus]

sorry

[wraith dagger]

jan Niku

[violet flume]

use exponent rules here

get C out of the exponent

feel free to define a new constant

maybe you use $e^{a+b} = e^a e^b$

[wraith dagger]

jan Niku

[obtuse citrus]

n(t)=e^(-alpha*t+C)

i am moving down for refrence

not answe

r

n(t)=e^(alpha*t) + e^C

[violet flume]

youve misapplied the rule

here

$N(t) = Ce^{-\alpha t}$

[wraith dagger]

jan Niku

[violet flume]

this will be the result of this step

now, use the initial condition to solve for C

$N(0) = N_0$ is given. What's C?

[wraith dagger]

jan Niku

[obtuse citrus]

1

if u plugin 0 for t then multiply alpha by zero it becomes e^0 then thats 1*C

unless u want me to rework the equation

[violet flume]

👀

you dont need to rework just do it more carefully

we want to solve for C

we have an initial condition

so $N(0) = Ce^{- \alpha \cdot 0}$

[wraith dagger]

jan Niku

[violet flume]

then $N(0) = ?$

[wraith dagger]

jan Niku

[violet flume]

you say this is $N(0) = C$?

[wraith dagger]

jan Niku

[violet flume]

then given $N(0) = N_0$, we could say $C = N_0$

[wraith dagger]

jan Niku

[obtuse citrus]

no

N_0=ce^-(alpha*0)

ln(n_0)=(alpha*0)*ln(c*e)

[violet flume]

you're overcomplicating this, sorry

Here's how I'd think about this step

[obtuse citrus]

[violet flume]

$N(t) = e^{- \alpha t + C} = e^{-\alpha t} e^C$

[wraith dagger]

jan Niku

[violet flume]

this is by exponent rules.

[obtuse citrus]

oh

[violet flume]

now we can use the initial condition

$N(0) = e^{-\alpha \cdot 0}e^C = 1 \cdot e^C = e^C$

[wraith dagger]

jan Niku

[obtuse citrus]

oh

im dumb

[violet flume]

given $N(0) = N_0 = e^C$, we can rewrite our equation $$N(t) = N_0 e^{-\alpha t}$$

[wraith dagger]

jan Niku

[violet flume]

I have to go so maybe ill leave the last piece to you?

You have one more piece of information

we can solve for the rate constant

we can do this because youre given the half life

that is, $N(29) = \frac 12 N_0$, right?

[wraith dagger]

jan Niku

[obtuse citrus]

[violet flume]

after t=29 years, you have only half as much left

[violet flume]

this is an additional piece of information, use it to solve for $\alpha$ in $$N(t) = N_0 e^{-\alpha t}$$ and you've rederived the equation you wanted from scratch

[wraith dagger]

jan Niku

[violet flume]

i gotta go to the gym

ill keep an eye on the chat tho, feel free to post might just be a minute til i ca nreply

[obtuse citrus]

thank u

i was moving my laundry

thank u i am gonna rederive this from scratch thank u for the steps

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[winter swan]

Hi would appreciate some help on this Riemann Sum problem
Specifically 1c.

my expression for 1b evaluated to N* lnN - N

[cedar kiln]

@winter swan Has your question been resolved?

[winter swan]

<@&286206848099549185>

[cedar kiln]

@winter swan Has your question been resolved?

[obtuse citrus]

ok

so

draw a graph of ln(x)

so if ur going lN is 1 to N or 1 to Rn infinity why is L_N Smaller than R_n

[winter swan]

Oh I did 1a and 1b already

[obtuse citrus]

nice

[winter swan]

I’m just confused on the 1c part like how they want me to do the

Inequality thing

I have no idea how they get to that expression

[opaque pivot]

ln(N!) = lnN + ln(N-1) + ln(N-2) +... + ln2 + ln1

which is the sum of areas of rectangles of width 1 under y = lnx from 1 to N

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[bright karma]

just finding the derivative. can someone check

[rain drift]

yep looks good

[cedar kiln]

@bright karma Has your question been resolved?

[bright karma]

btw do u know the difference between d/dx and df/dx

nvm

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[verbal pawn]

T

[verbal owl]

T

[verbal pawn]

the graph is loading

[verbal owl]

its ok lol

[verbal pawn]

,rotate

[verbal pawn]

that what I have so far

[verbal owl]

yes and what would be ur question here ?

[verbal pawn]

is it so far correct and then can u explain what concav is again

concav up and concav down

[verbal owl]

do the arrows on the graph mean that it goes away to infinity ?

just makign sure

[verbal pawn]

ya infinity

minima here would be global I forgot to write that

[verbal owl]

concave down is like ur coming on or off a high

and concave up is like ur coming on or off a low

!

[verbal pawn]

so

the concav down

starts where and ends where

[verbal owl]

why

do you think the graph is concav down

[verbal pawn]

because it is going down on the left of the parabola and up on the right

so it’s concav down and up? lol idk

[verbal owl]

concave down is ur going back down to the right

concave up is ur going back up to the right

[verbal pawn]

ok

[verbal owl]

like imagine its "arms"

[verbal pawn]

ok

[verbal owl]

if they are dangling to the top

its up

if they are dangling down its down

[verbal pawn]

I’m going to need a visual of you doing that I can’t really understand

can u take a picture of ur arms up and down

[verbal owl]

ahhah

[verbal pawn]

so this would be concav up

[verbal owl]

yes !

hes coming back from a rough time

[elfin otter]

Sad = down
Smiling = up
||if this helps||

[verbal owl]

and this... hum idk what inc and dec mean neither maxima and minima

[verbal pawn]

increasing decreasing

global, local maxima minima

[verbal owl]

oh i see

[verbal pawn]

I don’t think there is any global maxima right

[verbal owl]

when is it losing money and when is it gaining money

when are u stonking pretty much

[verbal pawn]

yea so for decreasing

I said infinity to 1

[verbal owl]

always read from left to right

[verbal pawn]

• mean

Omg

decreasing

[verbal owl]

yes

at first right

then what happens ?

[verbal pawn]

1 to inifty increasing

[verbal owl]

yep !

if u notice also

the closer u get to the middle the less intense the curve is

meaning that the speed at which its decreasing or increasing is smaller in the middle

and more intense at the poles

[verbal pawn]

yeahh

[verbal owl]

imagine putting bigger and bigger number to multiply eachother

1x1 = 1
5x5 = 25
100 x 100 = 10 000

[verbal pawn]

mhmmm

okay

[verbal owl]

makes sens ?

[verbal pawn]

Yes

[verbal owl]

any other questions ?

[verbal pawn]

ok

there’s no global maxima right

or local maxima

[verbal owl]

well did it ever stop stonking and peaked in his life

or is he limitless

global maxima is like the cooler version of local maxima

[verbal pawn]

he’s a loser

[verbal owl]

local maxima means that like at one point it stopped gaining

so that would mean that he got a sad face

if its the highest sad face then its the global maxima of all

makes sens ?

[verbal pawn]

yes

[verbal owl]

did he have any sad moments

[verbal pawn]

yes

at 1,0

[verbal owl]

that parabola is winning though no ?

[verbal pawn]

so he’s winner?

[verbal owl]

well to the right yes

he made a smiley lol

[verbal pawn]

a long one

I see

[verbal owl]

yes ahhah

but basically like he has a gloval and local minima

because at first he was losing but he started winning big u know

if he starts winning and then he peaks its a maxima

but if u lose and you have a glow up then it was ur worst time (minima )

[verbal pawn]

ya okay

[verbal owl]

those would be local

[verbal pawn]

hmm

ok so

u are saying his maxima would be infinity

the bottom, his minmima is (1,0)

the top, his maxima is infinity

[verbal owl]

globally he kinda doesnt have any because infinity is not a number

[verbal pawn]

OK

so he doesn’t have any

[verbal owl]

but the idea is there

[verbal pawn]

right

[verbal owl]

hes limitless

c:

[verbal pawn]

ur an idiot

thank u

[verbal owl]

well we need idiots like me sometimes

ill take it as a compliment

np

[verbal pawn]

ur explanations are good

u use good visuals

[verbal owl]

tbh try to look at it like a game it helps

[verbal pawn]

I see

[verbal owl]

for ur own future

youll see it becomes quite fun

(not for most but for me ig )

if not fun, less painful

gl in ur endeavors thanks for you time

[verbal pawn]

ty

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[floral marten]

i've been doing this question for a while, and I think i know the answer I don't think im formatting it right

[floral marten]

im pretty sure the 5th equation is being added from the 1st equation

so it would look like R5->R5-R1

ive been typing it forever in the format and it doesnt work

[peak minnow]

it says to write it with 1 space between every character

maybe try R 5 - > R 5 - R 1?

ur answer is right tho

[floral marten]

well it would go something like e 5 - e 1

but let me see

[upper abyss]

Looks like "E 5 - E 1"

[floral marten]

nope doesnt work

[upper abyss]

Terrible question all around

[floral marten]

it is

the website is really finnicky with answers overall

ill try the other way around

• 1 e 1 - e 5

[upper abyss]

That would be wrong, but who knows

[peak minnow]

that cant be right , -8-5 = -13 not 13

[upper abyss]

Could be anything at this point lol

[floral marten]

im sorry i meant + instead of -

but that didnt work

[upper abyss]

Maybe they want
E 5 + - 1 E 1

Sorry, -1 is allowed with no spaces

E 5 + -1 E 1

[floral marten]

no, omg this is annoying

[rare vault]

most practical online math software

[floral marten]

exactly, never have anyone use webwork

ill go through one more time and if it doesnt work. Ill tell my professor, ive finished every question except this.

and also close this too

alright thats it im done sorry guys, thanks for trying

ill let my prof know about this

.close

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[cedar kiln]

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[olive heart]

4x^2<13x-3

[olive heart]

i dont get why the interval notation is (1/4,3) and not (-inf,1/4] U [3,inf)

[crimson sedge]

Can you show your work?

[olive heart]

(-4x+1)(x-3)>0 is what I used to find my x's

i mean theres nothing really to show

but ok

[crimson sedge]

This is correct

[olive heart]

no ik but the previous answer i had was wrong

[crimson sedge]

There's no equality with 0 in this inequality

[olive heart]

after looking at the correct answer i changed it

wdym

[crimson sedge]

So you can't put close bracket with 1/4 and 3

[crimson sedge]

And second of all you are looking for >0 means +ve region so no inf

[crimson sedge]

Answer is (1/4 , 3) right?

[olive heart]

no

i mean yes

but idk why

[crimson sedge]

You are looking for values of x where this expression is positive

And the region that you get it by wavy curve method is (1/4 ,3)

[elfin panther]

my teacher sent me this 3 min vid explaining what this topic means u guys understand it?

[tiny robin]

Can someone help with a question?
Y = 6.50x + 25

[crimson sedge]

Open your own channel, don't interept here Lol

[elfin panther]

wdym

[crimson sedge]

!help

[cedar kiln]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[olive heart]

hakken

[crimson sedge]

Hm?

[crimson sedge]

@olive heart

[olive heart]

ohhh

do u graph it first?

or is that too much

[crimson sedge]

I find roots, I plot them on number line, I check signs and mark it on number line

[crimson sedge]

Wdym?

[olive heart]

like im still confused

[crimson sedge]

Where?

[olive heart]

so when u graph it your x int are 1/4 and 3

eveything in between that makes it positive right?

[crimson sedge]

What does int means?

[olive heart]

intercept

[crimson sedge]

Yes

[olive heart]

ok

[crimson sedge]

Here, yes

[olive heart]

what if it was negative

[crimson sedge]

In special cases domain May exists

[olive heart]

then it would be the other answer right?

[crimson sedge]

If there's no positive sign (required sign) on graph, then there's no solution to it

[olive heart]

like if the sign between the 0's was negative it would be the other answer

wtf

[crimson sedge]

No, we needed the values to make expression +ve, if there were none, then there would be no solutions

[elfin panther]

bye

[crimson sedge]

@celest seal

[olive heart]

whats +ve

[crimson sedge]

Positive

[olive heart]

okay okay

look

what if the sign was negative

the sign in the middle

then it would be the other answer right?

[crimson sedge]

Then there would no positive signs, right or wrong?

[olive heart]

yea

the outside of the 0s would be negative

i mean positive

ohhhh

so it would be no solution

[crimson sedge]

And we need to find positive values of inequality

[olive heart]

i see

[crimson sedge]

But there's no positive area on graph

So no real solutions

[olive heart]

do u really have to take each individual value between 1/4 and 3?

to check if its positive

[crimson sedge]

The graph automatically takes it, but when I have to make sure, I check it on integers

Like 0

Or 1,2

Sometimes fractions too

[olive heart]

oof

[crimson sedge]

Just to cross check

[olive heart]

oh alr

i see

thx

[crimson sedge]

No worries

[olive heart]

appreciate the help

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[lone mesa]

.close

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[winter cairn]

Help me tho

[cedar kiln]

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[olive heart]

What is the absolute maximum

[olive heart]

Is it infinite

[dire thorn]

I would say it doesnt exist

[olive heart]

Oh alr

Thx

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[pliant juniper]

The length of one side of a rectangle is increasing at 4 in/min. The area is increasing at 10 in^2/min. What is the rate of change of the width of the rectangle when the length is 14 in and the area is 140 in^2?

[pliant juniper]

The way I tried to set this problem up is to use the relation equation of A = L * W, then I got the derivative of both sides which I think is dA/dT = L * dW/dT + W * dL/dT

then I got the value of W by setting 140 = 14 * W and solving which gave me W = 10.

Then I plugged in all values to get 10 = 14 * dW/dT + 10 * 4

Then I solved for dW/dT and got -30/14.

Is this correct?

related rates are my biggest weakness of what we've learned so far, and im not sure if my answer is right because no calculator online gives consistently good answers for related rates

[cedar kiln]

@pliant juniper Has your question been resolved?

[cedar kiln]

@pliant juniper Has your question been resolved?

[obtuse citrus]

@violet flume thank u sir or maam or it very useful i solved it

(non related sorry

[pliant juniper]

all good

[obtuse citrus]

u can simplify a bit more

-15/7

but yeh

thats right

[pliant juniper]

thanks

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[amber pewter]

Any experts on non linear PDEs and the KdV / Linear Schrodinger equation want to have a crack at this:

[amber pewter]

The notes I'm trying to follow relate to Example 1 from this webpage:
https://wikiwaves.org/Properties_of_the_Linear_Schrodinger_Equation

While I can follow for $u( x,0) = u_0 \delta( x)$ The way the problem expression is formed is throwing me,

[wraith dagger]

SubsonicSpraak

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@amber pewter Has your question been resolved?

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[delicate patio]

how should i approach this?

[cedar kiln]

@delicate patio Has your question been resolved?

[delicate patio]

<@&286206848099549185>

[delicate patio]

<@&286206848099549185>

[neat dune]

I haven't worked out the details at all but I feel like We'll Ordering Principle will help

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[silent kelp]

i have no idea how to handle abs value

[lusty grotto]

how would u find zeros of something like (x-5)/(x-3)

[silent kelp]

x-3 = 0

x=3...?

[lusty grotto]

if x = 3 u get a zero in the denominator

[silent kelp]

yeah

x-5=0

more like x-5 = 0(x-3)
so x =5

yeah

[lusty grotto]

u cannot have zeros in the denominator

[silent kelp]

for a) it's 2 and -2

[lusty grotto]

division by zero is not defined

[silent kelp]

how to show work for a) though

[silent kelp]

yeah

well

not 2, because denom cant be 0

[lusty grotto]

because ur function has a division by x-2, 2 is probably not in its domain

[silent kelp]

yeah

[lusty grotto]

So zeroes of $f$ are the values $x$ such that $\frac{|x|-2}{x-2}=0$, and $x\ne2$

[wraith dagger]

SilverSoldier

[lusty grotto]

and u can simplify from there

[silent kelp]

okay

now for b)

[lusty grotto]

what is a piecewise function

[silent kelp]

a funciton with multiple functions..

or equations?

[lusty grotto]

something like the previous question u had

with a curly bracket

[silent kelp]

yea

[lusty grotto]

how is the function given to u defined differently based on different values of x?

[silent kelp]

is this right

idk what im doing

[silent kelp]

what do you mean

[lusty grotto]

can u write |x| as a piecewise function?

[silent kelp]

yeah

[lusty grotto]

how

[silent kelp]

x,
-x,

idk the output

[lusty grotto]

what do u mean output

[silent kelp]

oh well the right value

[lusty grotto]

and u need some stuff in front of those commas

when is |x| = x
and when is |x| = -x

[silent kelp]

so ok

|x|, x
|x|, -x

is the piecewise

[lusty grotto]

no..

[silent kelp]

ok

i dont know

[lusty grotto]

by "when" i meant for what kind of values of x

if x = 9,
is |x| = x or is |x| = -x ?

[silent kelp]

first one

[lusty grotto]

yeah

[silent kelp]

its more like

[lusty grotto]

so when x is a number like 9 |x| = x

for what other numbers like 9 is |x| = x?

[silent kelp]

|+-x| , x

every positive

[lusty grotto]

what is this

[silent kelp]

any value in the absolute value will be positive

|+-x| = +x

[lusty grotto]

yeah

[silent kelp]

or
|+x|, x
|-x|, x

[lusty grotto]

this is not good notation and it doesnt exactly convey ur idea, although u seem to be having the right idea

[lusty grotto]

same here

[silent kelp]

okay

[lusty grotto]

so u said earlier that when x = 9, |x| = x right?

[silent kelp]

yes

[lusty grotto]

when x = 9, to evaluate the value of |x|, u can simply evaluate the expression 'x'

and that just gives 9 itself

when x = -5, is |x| = x or is |x| = -x ?

[silent kelp]

its 5

its

[lusty grotto]

yeah true |-5| = 5, but how do u get that result

by doing what to -5 ?

[silent kelp]

making it negative

[lusty grotto]

yeah

[silent kelp]

yeah

[lusty grotto]

so when x = -5, |x| = -x right?

[silent kelp]

|x|, x
|x|, -x

yeah

[lusty grotto]

this is not correct again

[silent kelp]

ok

[lusty grotto]

anyway

[silent kelp]

okay

[lusty grotto]

so when x is what kind of number do u have to evaluate '-x' to get the value of |x| ?

[silent kelp]

negative?

[lusty grotto]

yes

for |x| = -x, when x < 0

then when is |x| = x ?

[silent kelp]

positive

more than 0

[lusty grotto]

right

how about when x = 0?

it doesnt matter right

[silent kelp]

its 0

yeah

[lusty grotto]

yaeh

so u can write

[silent kelp]

|x| = -x, x<=0
|x| = x, x>=0

[lusty grotto]

correct, and u can actually put an equal sign to any one of those u like

x<=0

or x>=0

[silent kelp]

oh yeah

[lusty grotto]

because when x = 0, to get |x|, it doesnt matter whether u evaluate x or -x

[silent kelp]

yea

[lusty grotto]

right

[silent kelp]

ok well how to do b)

[lusty grotto]

now

[silent kelp]

derrivative

[lusty grotto]

ur function f includes |x|

[silent kelp]

yes

[lusty grotto]

so its defined a certain way when x>=0, and a different way when x<0

can u write it as a piecewise function?