help-27

[prime narwhal]

well since its two items it would be .09x+.09y=total pst

so

.09x+.09y=13.14

[lean plover]

yup!

[prime narwhal]

so this one?

[lean plover]

yes!

[prime narwhal]

how do i do this type of question?

[lean plover]

so intuitively for me is to match up the "coefficient" of the variable and subtract the equations.

so just as an example, y's have "opposite" coefficients, namely +1, and -1.

now adding these two equations, then what do you get?

(hint: you won't have any y's left!)

[prime narwhal]

so am i looking for the intercepts?

im confused

[lean plover]

at a higher level you're looking at the intersections of these two equations, yes!

but i was looking more for you add these two equations together:
(x+x) + (y-y) = 2 - 4

[prime narwhal]

so what am i suppost to do

[lean plover]

try solving it out

[prime narwhal]

how thought

[lean plover]

can you simplify (x+x) + (y-y) = 2 - 4...?

[lean plover]

im also getting off discord soon just letting you know

[prime narwhal]

ok

[prime narwhal]

i dont know what i would do first

like do i need to get x alone?

[lean plover]

just add x's togehter, subtract y's together, etc etc

just do it out

[prime narwhal]

but i dont know what x='s

so how do i add it

[lean plover]

x+x=2x??

so there are two x's

so x + x = 2x

[prime narwhal]

(2x) = -2

?

since y-y would be nothing

so x=-1?

[analog gale]

If you subtract y IT will Land on the other Side If im right

[prime narwhal]

so what do i do

[prime narwhal]

(x2) + (y-y) = -2

what do i do next

[prime narwhal]

<@&286206848099549185>

.cloise

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[prisma frost]

why is saying u cant divide by x wrong? for c

[upper schooner]

Can you show the original part (i.e. what comes before the "continued")?

I note they say "isolated reasons" [and so, it's more that they want more of a justification than that, rather than it being "wrong"]

[upper schooner]

Yea, they want "more" of a justification, they want you to refer to the fact that stationary points/tangents don't really "work" with NR

[prisma frost]

ok ty

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[scenic surge]

for b)

My attempt:

mean: integrate from 0 to 1 for x(6x-6x^2) = 2 - 3/2 = 1/2

variance: (integrate from 0 to 1 for x^2(6x-6x^2) ) - mean^2 = (2 - 3/2) - (1/2)^2 = 1/4

but the solution in my book says

the variance is also 1/2

what am I doing wrong?

am I integrating wrong for the variance?

integrate from 0 to 1 for x^2(6x-6x^2)

= 6x^3 - 6x^4 dx = 6x^4/4 - 6x^5/5 |_0^1

= 6/4 - 6/5 =

oops

im an actual idiot

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[scenic surge]

.reopen

[devout snow]

✅

[scenic surge]

wait but even then it cant be 0.5

6/4 - 6/5 = 3/10

3/10 - (1/2)^2 = 1/20 = 0.05

omg again im an idiot I misread the solution it said 0.05

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[scenic surge]

.reopen

[devout snow]

✅

[scenic surge]

for part c) I am definitely doing something wrong but im not sure what it is

For the mean I am expected to integrate form 1 to inf for 2x/x^3 dx

which can be simplified to

2(integral from 0 to 1 for 1/x^2 dx) right

the antideriv of that should be 1/x

but when pluggin in the bounds for inf I have to take the limit

so it should be:

2((lim x -> inf 1/x) - 1/1)) right

but wont that = 2(0 - 1) = -2 ?

the answer should have been +2

what did I do wrong this time

[tall knoll]

Should it be?

[scenic surge]

omgggg -1/ x right

i forgot that im bringing down -1 not 1 smhhhh mb

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[atomic drum]

i need help with these

[faint gorge]

For the first one

try to get the form

2^(x-1) = ... first

For the second one you can apply the following log rule:

log(a.b) = log(a) + log(b)

[atomic drum]

how i do that

[faint gorge]

wdym

you can apply arithmetic operations

like addition and division

etc.

[atomic drum]

lemme try

oh that was easy

i just needed to think for a sec

but i have no idea how to do the second one

[faint gorge]

did you solve the whole?

[atomic drum]

yes

[faint gorge]

What did you get?

for x

[atomic drum]

3

[faint gorge]

very good

I almost was afraid you didnt know about logs too

anyway

well done

[atomic drum]

yes

[faint gorge]

for the 2nd one

if you see summands of logs

you can apply this rule

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

This helps to simplify the experessions

So naturally what would be the first step?

[atomic drum]

simplify prob

[faint gorge]

yea like what would your 2nd line looks like

[faint gorge]

after applying this

[atomic drum]

lemme think

yep i got no clue

[faint gorge]

cmon

start with the left side

,,\log_3(x) + \log_3(x-1)

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

it's a sum

according to the log rule

[atomic drum]

oh

[faint gorge]

we can write it as one log

and multiply the arguments

[atomic drum]

x(x-1)

[faint gorge]

yes

,,\log_3(x) + \log_3(x-1) = \log_3(x(x-1))

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

what's the right side looking like?

,,\log_3(2) + \log_3(x+1)

[woven radish]

𝔸dωn𝓲²s

[atomic drum]

2(x+1)

[faint gorge]

yes

,,\log_3(2) + \log_3(x+1) = \log_3(2(x+1))

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

So the 2nd line would look

,, \log_3(x(x-1)) = \log_3(2(x+1))

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

Now

what can you do on both sides

to get rid of log base 3

[atomic drum]

subtract

[faint gorge]

easier

two log with same base are the same with their arguments are the same

[atomic drum]

divide

[faint gorge]

have you ever seen this

[atomic drum]

can't you just like

get rid of them

like ignore them

[faint gorge]

,,b^{\log_b(x)} = x

[woven radish]

𝔸dωn𝓲²s

[atomic drum]

no

[faint gorge]

kinda you should still know the background

[faint gorge]

you can do on both sides 3^(...)

[atomic drum]

but it would be x(x-1)=2(x+1)?

[woven radish]

𝔸dωn𝓲²s

[faint gorge]

yes

that's the reason why above

[atomic drum]

chill that stuff is confusing me

[faint gorge]

it's easy

log disappears if base is the same

[faint gorge]

this rule

It is because 3^x is the inverse function of log_3(x)

and f(f^-1(x)) = x = f^-1(f(x))

it's like

arccos(cos(x)) = x

inverse function

[atomic drum]

what do i do again if i have an x^2 and a x

if i'm tryna solve for x

do i gotta square root

[faint gorge]

?

show it

[atomic drum]

x^2-x=2x+2

[faint gorge]

bring everything to one side

and use quadratic formula

[atomic drum]

ohhhh

yea

ok i got

3 plus or minus square root of 17 over 2

lemme put that into desmos rq

[faint gorge]

qe

[atomic drum]

ok it's 3 plus square root of 17 over 2

right?

[faint gorge]

ye

[atomic drum]

ight

i'm done

[faint gorge]

me too

[atomic drum]

i got my final tmr

[faint gorge]

good luck

[atomic drum]

i only studied last minute

[faint gorge]

one tip

[atomic drum]

yea

[faint gorge]

think

and relax haha

[atomic drum]

alr

i'll try

alr bro

[faint gorge]

alr bro

[atomic drum]

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[scenic surge]

I do not understand what happened after P(|X| >= 3/2)

why is it 1 - the integral

oops nvm I remember now

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[eternal tapir]

one of these times you'll post a question that you don't immediately answer yourself

[verbal matrix]

lol very relatable

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[devout snow]

@night hatch Has your question been resolved?

[devout snow]

@night hatch Has your question been resolved?

[uncut crow]

oh hey i had to make a model like this in my numerical analysis class

mine used a different runge kutta variation

and not in... python is that?

oh boy looking back at my code this is a mess

maybe. RK4 i think it's called

force like the gravitational force the bodies have on each other?

function Xgrav = Xg(M,p1,p2) %computes the X acceleration effect p2 has on p1 
G = 6.67430*10^(-11);
if p1 == p2
    Xgrav = 0;
else
    Xgrav = G*M*(p1(1) - p2(1))/((p1(1) - p2(1))^2 + (p1(2) - p2(2))^2 + (p1(3) - p2(3))^2)^(3/2);
end
end```

so i had 3 different ones

one for x, y, and z directions

that comment i wrote is a little misleading

oh actually no it's not

might as well send the other two

function Ygrav = Yg(M,p1,p2) %computes the Y acceleration effect p2 has on p1 
G = 6.67430*10^(-11);
if p1 == p2
    Ygrav = 0;
else
    Ygrav = G*M*(p1(2) - p2(2))/((p1(1) - p2(1))^2 + (p1(2) - p2(2))^2 + (p1(3) - p2(3))^2)^(3/2);
end
end

function Zgrav = Zg(M,p1,p2) %computes the Z acceleration effect p2 has on p1 
G = 6.67430*10^(-11);
if p1 == p2
    Zgrav = 0;
else
    Zgrav = G*M*(p1(3) - p2(3))/((p1(1) - p2(1))^2 + (p1(2) - p2(2))^2 + (p1(3) - p2(3))^2)^(3/2);
end
end```

p1 is supposed to be a triple with with the x, y, and z position of the body

similar with p2

yea

i mean i wrote it and i don't even understand it now so

what dumbass wrote this?

matlab

one second i need to grab food

ok back

i will try to type out the x gravity one

$$GM\left(\frac{A_x - B_x}{(A_x - B_x)^2 + (A_y - B_y)^2 + (A_z - B_z)^2}\right)^{3/2}$$ where $A_x$ is the $x$ position of body $A$, $B_x$ is the $x$ position of body $B$, etc., G is gravitional constant, $M$ is the mass of... $B$ i think

crickets

oh yes this is the acceleration

i may have forgotten what we were doing

oops missed a parentheses

[woven radish]

slayla

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[muted goblet]

https://cdn.discordapp.com/attachments/903486480075354182/1245694359228383242/Screenshot_2024-05-30_at_9.04.05_pm.png?ex=665a5787&is=66590607&hm=91172ed0d574a284806ab934f0d4dd837ef79d5dd337d41b02aca3a89a2cd659&

[muted goblet]

how to prove this ?

i did this and got stuck

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@muted goblet Has your question been resolved?

[muted goblet]

<@&286206848099549185>

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[junior fossil]

$.$

[woven radish]

Mövlan

[junior fossil]

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[sharp root]

??

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[wraith horizon]

it's 1/6 . (M . x^-3)

[faint hearth]

Correctly apply the power rule

[deep hound]

,,M/(6x^3)

[woven radish]

paperpanda

[faint hearth]

$\frac{M}{6x^3}=\frac{M}{6}x^{-3}\newline.$ and when you differentiate above expression you have $\frac{M}{6} \cdot -3 \cdot x^{-3-1}=\frac{-3M}{6}x^{-4}=\frac{-M}{2}x^{-4}$

[woven radish]

quickdoom

[eternal tapir]

bepo

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[short barn]

Do i claim a help channel here?

[eternal tapir]

[short barn]

Ge/ef=m.
How do i define ah/ge using m?

I know i need to use thales theorm for it

[fringe dove]

ABCD is a rectangle?

[short barn]

Oh yes

Its a triangle

And angles fea=gea if that helps

[fringe dove]

Is D a random point on GE?

Is AD parallel to FE?

[short barn]

No

[short barn]

Yes just a random point

[fringe dove]

So ADE is not a right angle?

[short barn]

It is.
Imma just write everything they give here:
Abcd is a rectangle
Angles fea=gea
Ge/ef=m

Currently i know that
Fh/fe=fa/fg=ah/ge
From thales theorm

[fringe dove]

Okay got it

Do you know angle bisector theorem?

[short barn]

Nope, whats that?

[fringe dove]

(If PS is the angle bisector)

(that is, angle QPS = angle SPR)

[short barn]

Onn i know it

[fringe dove]

Your question becomes easy using this theorem

[short barn]

Ohhh

Wait i wanna try now myself

Thanks

Ah/ge=1/m?

[fringe dove]

No

[short barn]

Oh wait so i did something wrong there

[fringe dove]

Yup, do it slowly

[short barn]

Fa/fg=ah/ge Thales
Fe/ge=fa=ag Bisector
Kind of lost from here, can you help me?

[fringe dove]

Yup

So FE/GE = 1/m = FA/AG

You need to find AH/GE = FA/FG = FA/(FA + AG) = 1/(1 + AG/FA) = 1/(1 + m)

You figured out everything, just needed to use that GA + AF = GF

[short barn]

Why does fa/(fa+ag)=1(1+ag/fa)?

[fringe dove]

Dividing both numerator and denominator by FA

You can see it like this too: FA/(FA+AG) = FA/(FA + mFA) = 1/(1+m)

(Since AG = mFA)

[short barn]

Ohh ok got it

Thanks! I fully understand now

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[dusty mesa]

Hi, is this making any sense? The question is SIMPLIFY

[winter patrol]

you made a mistake

[dusty mesa]

Where

[dusty mesa]

I see noe

Thank you

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[vernal grail]

$A=\begin{pmatrix}\frac{1}{2} & 0 & 1 \ 1 & 2 & 0 \ 1 & 1 & h\end{pmatrix} \Rightarrow \begin{pmatrix}\frac{1}{2} & 0 & 1 \ 1 & 2 & 0 \ \frac{1}{2} & \frac{1}{2} & \frac{h}{2}\end{pmatrix} \Rightarrow \begin{pmatrix}\frac{1}{2} & 0 & 1 \ 1 & 2 & 0 \ 0 & -\frac{1}{2} & 1-\frac{h}{2}\end{pmatrix}\Rightarrow \begin{pmatrix}\frac{1}{2} & 0 & 1 \ \frac{1}{2} & 1 & 0 \ 0 & -\frac{1}{2} & 1-\frac{h}{2}\end{pmatrix}$

[woven radish]

alee

alee

[vernal grail]

did I reduce it well?

i can explain the steps, if needed

[exotic dagger]

looks good

[vernal grail]

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[rain carbon]

Given that the function has no asymptotes perpendicular to the axes find a

a is posetive

definition domain x>-a

i need to find a

[tiny beacon]

well x+a cant be zero since the definition domain is x>-a

means no vertical asymtotes

[rain carbon]

yeah

ik that

[tiny beacon]

alr

[rain carbon]

thats why i am stuck

there is no way for me to find a

[tiny beacon]

you need to check for horizontal asymtotes

[rain carbon]

like y=?

horizontal is with y right?

[tiny beacon]

yes

some line like y=b

[rain carbon]

ok s othere is a rule that if the biggest number for x on the top in our case is x^2 is bigger then the biggest x down then there is no horizontal asymtote and for us its like that

so there is none

@tiny beacon

so how do i find a?

@tiny beacon i think i got it

if there is no asymptote then in x=-a there needs to be an hole which means that it needs to be 0/0 so

x^2-9 =0

x=3 x=-3

then a is 3 bc its always posetive

[tiny beacon]

but you said this

[rain carbon]

yeah

if there is an hole then x is not defined there

so x= -a is not defined but i am told its not an asymptote so ik its an hole

yk what is an hole right?

or is it called something else in english?

[tiny beacon]

x cant be -a, it will make the function undefined everywhere so you wont get a function

[rain carbon]

here

it is undefined ik

but its not an asymptote

its an hole

[tiny beacon]

ik yes

[rain carbon]

i am sorry if you dont call it an hole in english

but it means that you have a number devided by 0 that is also 0

so 0/0

[tiny beacon]

it is called a hole

[rain carbon]

x= -a is fine with the sqrt its only not ok bc you cant have something devided by 0 bc it makes it an asymptote exept if the top number is also 0

so bc ik that there is no asymptote ik that the top number has to be 0

[tiny beacon]

i still dont see how you are supposed to find asymptotes, i have not solved that many asymptote questions tbh

are you going to college or high school?

[rain carbon]

you cant find one wdym? thats what i said in the start there is no asymptote

[rain carbon]

10th grade

what grade are you?

[tiny beacon]

im an engineering college student, just finished first grade

[rain carbon]

so its 13th grade right?

[tiny beacon]

well if you look at it that way yes

anyways sorry i cant be helpful to you on this one

you can ping the Helpers role if you want to

[plush swallow]

No more information than this?

[rain carbon]

wait

a is 3

is the derivative

bc i think i am wrong

ia m closing this one and asking something else ok?

.close

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[frail sail]

i was taught that we should pull out in front of the parenthesis the coefficient at the highest power

[small urchin]

wdym

[vernal grail]

Numerator : Is ~ to 2n
Denominator : sqrt(n^2+1) Is ~ to n , so 2n+n = 3n
So 2n/3n = 2/3

[small urchin]

all clear?

[restive river]

yoooooooooo

imnew

i love maths

[devout snow]

@frail sail Has your question been resolved?

[devout snow]

Channel closed

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[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[lament schooner]

for u sub here

do i have to u sub 2x

wait no i can just integrate this

with inverse anti chain rule

[stable storm]

works too

[lament schooner]

ah okay ty

.close

[devout snow]

Channel closed

Closed by @lament schooner

Use .reopen if this was a mistake.

[eternal tapir]

wow

[lament schooner]

why are you cat pee now

i wish i could change my name 😭

[eternal tapir]

I am cat girl

[lament schooner]

meow

[eternal tapir]

what was your cat girl persona again

[lament schooner]

had i ever?

[heavy current]

[eternal tapir]

were you not there last night

ok important question for you bofa

if you were a cat girl what would you be named

[lament schooner]

i think you meant slayla-

uh

uh

[lost laurel]

you can

[lament schooner]

uh

uh

u

[lost laurel]

on discord

[lament schooner]

uh

uh

eys on discord

[lost laurel]

you mean, right?

[lament schooner]

smh

uh

cat girl name

huh

can a cat be called Syrex

[eternal tapir]

if that is who you wish to cat girl as

[lament schooner]

i would totally name my cat syrex

[lament schooner]

what if i am already

[eternal tapir]

You've misunderstood the assignment

[lament schooner]

https://tenor.com/view/rule-505-rule-505-gif-16834093529872838245

[eternal tapir]

https://tenor.com/view/homura-homura-cat-madoka-madoka-magica-madoka-dance-gif-23721015

@heavy current

[heavy current]

what's up

[eternal tapir]

if you were a catgirl

what would your catgirl name be

[heavy current]

what is this question

[eternal tapir]

if you were a catgirl what would your catgirl name be

[heavy current]

idk, I've never given this any thought before lol

never crossed my mind

[eternal tapir]

high cat

[heavy current]

flying cat

cats that fly would go higher!, right?

[lost laurel]

yes

[eternal tapir]

@lost laurel

your turn

[lost laurel]

my name?

if I were a catgirl ?

[eternal tapir]

[lament schooner]

https://tenor.com/view/pain-are-we-suffering-swag-kruglik-yahicord-gif-21495687

[heavy current]

:despair:

[viscid ivy]

We re suffering

[faint hearth]

What exactly js happening gere

[eternal tapir]

hello doom cat

[faint hearth]

Hello peeing cat

[heavy current]

quickdoom... or smidgen?

you were smidgen a while ago, right?

[faint hearth]

Why not both

[heavy current]

[faint hearth]

Smidgin , yeah

[heavy current]

average Syrex channel

[faint hearth]

Results came out?

[lost laurel]

answer key did

right

hope you did well

[faint hearth]

Dont wanna check

😊 same to you

[lament schooner]

me tmrrow

[heavy current]

when're your exams

[eternal tapir]

Do we collectively impede syrex from studying by causing distractions in her help channels

[lament schooner]

tmrrow

[faint hearth]

Looks like it

Feels like it

[heavy current]

[devout snow]

Available help channel!

Send your question here to claim the channel.

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

[heavy current]

.woah

.what happened

[lament schooner]

oh

[lament schooner]

it just clossed what

[heavy current]

[eternal tapir]

she doesn't know

[heavy current]

I don't know either

[lament schooner]

wait it was a closed channel this whole time

LOL

[heavy current]

oh

LOL

[eternal tapir]

I was referring to how you can preppend a . to avoid opening a available channel

[heavy current]

I forgot about this

[eternal tapir]

that's why I said wow

[lament schooner]

btw chartbit weirdly silent today

[eternal tapir]

you scared her

[lament schooner]

https://tenor.com/view/cat-look-cat-look-at-camera-silly-cat-in-a-cage-gif-889392959852579879

[heavy current]

they've been lurking a lot

apparently they've decided today is the day to be quiet

[faint hearth]

It be like that sometimes

[lament schooner]

i have to

tattoo +C

on me

cuz i keep forgetting

[fervent hornet]

To never forget about it?

[lament schooner]

eys

yred

yes

[eternal tapir]

meowvergence

[fervent hornet]

The pain of forgetting+c

[lament schooner]

last semester i did midterm in degree mode

https://tenor.com/view/monkey-gif-18890916

[fervent hornet]

I prefer catvergence

[lament schooner]

https://tenor.com/view/arlecchino-arlecchino-leak-arlecchino-gameplay-arlecchino-ult-arlecchino-q-gif-6632082696836024604

is it meow or mew

[fervent hornet]

What r u referring abt?

[eternal tapir]

it's mow

[fervent hornet]

That's new

[upper schooner]

Hey, I'm still here just doing some other bits

[lament schooner]

sooo why is your name chartbit

[upper schooner]

Cause I'm a chatbot

[fervent hornet]

This is a certified chatbot classic

Source: trust me bro

Lol

[eternal tapir]

[fervent hornet]

So how it's like being a chatbot

[upper schooner]

Not bad, just have to give canned responses to questions and select random emotes

[fervent hornet]

On what model is ur responses based on

[upper schooner]

That's a secret

[fervent hornet]

I see

Which topic are you interested in?

[upper schooner]

Maths-wise, or?

[if so, mostly analysis/algebra ]

[fervent hornet]

Other

[upper schooner]

I don't know at all, honestly

[fervent hornet]

Lol

Hola

[devout snow]

@lament schooner Has your question been resolved?

[devout snow]

Channel closed

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[devout snow]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[indigo radish]

the first one was easy as you can see I did it but I’m stuck on the other

[viscid ivy]

Ang PSQ = Ang PRQ = 22°

[indigo radish]

wait but why

[viscid ivy]

Angles subtended on the same chord of a circle , or a chord of equal length are equal

QP is the common chord

[indigo radish]

hold up I’m sorry that I’m not very good at this but what’s a chord

[indigo radish]

ohhh

alright that all makes sense now I think

thank you for your help 🙏

[viscid ivy]

A chord of a circle is a straight line segment whose endpoints both lie on a circular arc.

[indigo radish]

yeahhh that makes a lot of sense

😎

[viscid ivy]

There is a theorem

And a proof for this

You can google it

[indigo radish]

I will when I finish these questions that are far too long lol

thanks

should I close this now?

[viscid ivy]

Welcome !

[viscid ivy]

Yeah ofcourse if your doubt solved

[indigo radish]

yeah it is defo thanks to you

.close

[devout snow]

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[devout snow]

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[marsh osprey]

how do i find the range of k for maximun solutions?

[marsh osprey]

i know there are 5 maximum solutions

but fonding range is harder

i tried making a graoh but couldn't

you also cannot use a calculator for this and is supposed to be done within an hour among many other questions so is there a quick way to solve it

[devout snow]

@marsh osprey Has your question been resolved?

[devout snow]

@marsh osprey Has your question been resolved?

[devout snow]

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[lusty sapphire]

If $2a+8b=13x$, would you agreed that you can divide both sides by the same value?

[woven radish]

SWR

[lusty sapphire]

Ask yourself this first

[devout snow]

@restive river Has your question been resolved?

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[prime narwhal]

I don't understand what im totally doing/where to start

am i just looking for the intercept of each line or what

[stable wraith]

yes

[prime narwhal]

so i just look at the picture given to me and i check for what intercepts

[stable wraith]

well you could try to set up a system of equations, but you could solve this by the information given

[prime narwhal]

how though

i dont understand what im doing

[stable wraith]

well for a system of equations to have a solution the lines given by the equations have to intercept

[prime narwhal]

but there is two lines

what are the answers

like it has (number, number)

but there is two things

[stable wraith]

the lines are given from the equations in the question, and for the system of equations to have a solution they will need to intercept

so if the lines dont intercept there are no solutions

[prime narwhal]

so right now i have y=mx-3 and y=mx+3

with -1 slope for m

[prime narwhal]

i dont know what im looking for still

like what is my answer suppost to be?>

two random numbers that i pick?

i dont know what the heck if my answer is suppost to be

[prime narwhal]

<@&286206848099549185>

[queen vine]

Basically

[prime narwhal]

okay so 1 and 2

done?

like

[queen vine]

Your looking for when the two lines cross each other

[prime narwhal]

oh so no solution?

since they never will

becuase htey are parrel and not on top of each other

[queen vine]

Yeah that’s what I’m assuming because it says select one

[prime narwhal]

same answer for this one right?>

[stable wraith]

not infinite

there is no solution

[queen vine]

No solution

Yeah ur right

[prime narwhal]

so this is the same thing but i need to draw the graph on my own

then check

so 1, -4

[queen vine]

If thts where the two lines meet, yeah

[prime narwhal]

also i just used a graphing calc for that, but if i didnt have one would i just need to plug in a point to figure out where it goes

then sue the y intercept as the other point

[queen vine]

Yeah if you need to draw the graph

If you don’t need to draw the graph

Then

[prime narwhal]

graphing calc?

[queen vine]

Set both equations equal to each other

[prime narwhal]

wdym

[queen vine]

I’ll show you with ur question

One sec

@prime narwhal

[devout snow]

@prime narwhal Has your question been resolved?

[prime narwhal]

how did you get this

[queen vine]

So before can you see how I rearranged the equations

They are both equal to y

And if they are both equal to y

Then they must be equal to each other

[devout snow]

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[devout snow]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[safe fractal]

What do you need help with?

[elder harness]

i dont understand the question

[viral hornet]

The first image put costs on y-axis and number of people on x-axis. For every x you get y increase by the listed costs except the room cost, so should be linear.

[dense jay]

whats the initial cost
then whats the additional cost per person

[elder harness]

is the initla cost 100£

wait

115&

[viral hornet]

f(x)=ax+b. the hotel room cost is b.

[elder harness]

wait wut

[viral hornet]

b is where the line cuts the y-axis

since the hotel room cost is static and always there no matter what even with no people y=a*0+b would be b: the hotel room cost.

of course there will be minimum 20 people, so that should be the inital cost.

but y= a*0+b = hotel room cost, line still cuts y-axis at b

although it does state "LIKELY" between 20-80, but I guess we'll assume "likely" to be the case

Or perhaps they mean to say that if below 20 people there will be no prom and no hotel room cost because then we cancel it. In that case initial cost wil be 0.

That is a bit unclear.

[devout snow]

@elder harness Has your question been resolved?

[devout snow]

Channel closed

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[devout snow]

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[vast quiver]

Question 6

[vast quiver]

I get what we do but

When you want to integrate square roots you’ll do that step with u^2=…

And then you say u>0

But then we do the roots for the denominator

And then we use the root u=-1

But that isn’t greater than 0?

[covert root]

Is it possible to type the question and working out. Writing is hard to make out

[vast quiver]

Question 6

sub 4-x²=u² with u >= 0

and then you'll end up getting the integral of -integral(1/(1-u²))

but if you do the roots for 1-u² then youll get -1 and 1

but if u should be greater than 0 why do we use the -1

[covert root]

So the first thing I cannot understand is why are you letting u >=0?

[vast quiver]

because u is also equal to sqrt(4-x²) if youd quadrate then u should be >= 0

[covert root]

yes that principal sqrt

So we have |u|

[vast quiver]

but we dont want the negative sqrt

thats why >=

[covert root]

You can just write |u|. No need to add non-negativity constraints

[vast quiver]

ok but then u should still be positive

and if we use u = -1 then it isnt

[covert root]

How do you get this?

[vast quiver]

,rotate

[woven radish]

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[vast quiver]

,rotate

[covert root]

I think you should let u = 4-x^2

instead

But your approach matches the integral calculator I am using

[royal valley]

@vast quiver use method of substitution but use, x = 2sinA

it would simplify ur complexity

[vast quiver]

no i want to use my approach

and its right using both roots

but i dont get why the second root works

[covert root]

what does integral have to do with the roots

[vast quiver]

how would u solve the integral on the very right

[covert root]

Using ln

function

d/dx ln(x) = 1/x

d/dx ln(f(x)) = f'(x)/f(x)

[sharp harness]

does that other variable looking like capital Lamda constant?

[covert root]

This result comes from d/dx ln(x) = 1/x and applying chain rule

Oh wait there is square

[sharp harness]

if yes, then expand just expand

[vast quiver]

with partial fractions youll need the roots

and if u = -1 then im lost

[sharp harness]

idk but i couldn't understand what you did due to intermixing of many variables (like lamda, u, mew, etc.)

[vast quiver]

theres only u

[sharp harness]

really?

wait

[vast quiver]

my handwriting sucks

[sharp harness]

this is u, not mew?

[vast quiver]

no

its u

[sharp harness]

what is this?

[vast quiver]

1

[sharp harness]

ah i see

[royal valley]

u r doing mistake here

[sharp harness]

1/(1-u^2) = 1/2 ( 1/(1-u) - 1/(1+u) )du

[sharp harness]

for last step, i didn't see other steps

[vast quiver]

yes correct but if u > 0 then why do we use 1+u

[sharp harness]

use modulus in log

it wouldn't matter else, you can even try in cases

[sharp harness]

but your solution might be wrong, so check it again

[sharp harness]

^

[royal valley]

Ur partial decomposition should be below
1 / (1 - u^2) = A / (1- u) + B / (1- u^2)

not A/(1-u) + B / (1+u)

[sharp harness]

here is the correct decompostion

[vast quiver]

yes i wouldve done that too

but why do we do it like this in class then

and it gives the right answer

[sharp harness]

i don't think you need to do partial fractions for this

[covert root]

the decomposition is correct (1-u^2) = (1-u)(1+u)

[royal valley]

ur ans is right in sheet

[vast quiver]

why do we use u = -1

[royal valley]

i got ur doubt
both are different things,
definitely here u > 0
but when u r considering u =-1 that is a trick to solve the value of A and B to solve partial decomposition

both are not related

[sharp harness]

function is not defined at that value

[vast quiver]

but that u = -1 is derived from the roots of the function 1-u²

[sharp harness]

u = +- 1 => u^2 = 1 => x^2 = 3 , therefore the lower term in question is not defined

[sharp harness]

^

[vast quiver]

how are they not related if we keep the same u

[royal valley]

how and why u are doing roots here in integration?
making denominator 0, by u=+-1 is making whole integration as infinite,
i guess here ur intution is not right

[vast quiver]

How else do u want to partial decompose without using the roots

[devout snow]

@vast quiver Has your question been resolved?

[devout snow]

@vast quiver Has your question been resolved?

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[wicked valley]

A boat can travel at a speed 20km/h (constant in size) in
calm water. The boat is traveling upstream. The current velocity of the river is 5km/h

The desired course of travel makes an angle of 60° with the perpendicular on the river bank.
At what angle θ, measured relative to the bank, must the boat navigate to follow the desired cruising route. I dont know how to start on this

[tight peak]

the idea is that you consider two components for the velocity and then also add the river current velocity and use the 60 degree thingy to relate the components

@tawdry torrent

what

[tawdry torrent]

nothing

[tight peak]

your reaction is misleading

[tawdry torrent]

the person isnt answering

[tight peak]

but your reaction is still misleading in case he reads my text

okay good now

[tawdry torrent]

i was bored

[tight peak]

[tawdry torrent]

[tight peak]

a help channel is the wrong place to avoid boredom if you aren't helping

please keep your messages relevant

[wicked valley]

i still don't get it

[tawdry torrent]

alr

[tight peak]

can you resolve the velocity into components

parallel to river flow and perpendicular

[wicked valley]

yes

[tight peak]

once you're done with that, you want to add the river velocity to the parallel component

now you have the velocity that is the resultant of the boat and the river current

(river flow doesn't affect the perpendicular component)

now the idea is because of that opposing river flow, for some theta the resultant velocity makes an angle of 60 degrees with the perpendicular

which would mean, you can represent the ratio of perpendicular and parallel components of this resultant as cotangent of 60 degrees

think about everything ive said carefully and lmk if you didn't get it, ill try to find a video that explains relative motion in general

[wicked valley]

yeah i still dont get it totally

[devout snow]

@wicked valley Has your question been resolved?

[devout snow]

@wicked valley Has your question been resolved?

[devout snow]

@wicked valley Has your question been resolved?

[devout snow]

@wicked valley Has your question been resolved?

[wicked valley]

.close

[devout snow]

Channel closed

Closed by @wicked valley

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[clear gale]

yo

is this even possibl

I tried a lot of things

the best formula that I have is

but it doesn't satisfy f(5) DNE

[devout snow]

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[clear gale]

yo

[acoustic leaf]

you can multiply it by something that is 1 everywhere except x = 5, where it is DNE

[clear gale]

wdym?

[acoustic leaf]

well consider that any number divided by itself is 1

except for 0, because 0/0 is not defined

[clear gale]

hold on if I unfactor it a little so it's

(4x+35)(x+5) / (x+5) (x-5)

would f(5) then be undefined

since it's (4x+35)(x+5) / (x+5) (0)

[acoustic leaf]

it would be undefined, but if it's 0 on bottom but not on top then it will shoot off to infinity

[clear gale]

so it's undefined at x=5 and x=-5?

and defined at any other number?

[acoustic leaf]

yes, but we also need to satisfy the "limit at 5 exists" requirement

[clear gale]

undefined at x = -5

the limit at 5 exists requirement seems to work on this

[acoustic leaf]

so if the limit exists but the function does not, that's called a "removable discontinuity" or "hole"

[clear gale]

how come the limit doesn't appear to have a "hole" when the function is completely simplified

[acoustic leaf]

can you think of a function that is 1 everywhere except x = 5, but undefined at x = 5?

[clear gale]

1/(x-5)

?

[acoustic leaf]

that is undefined at x = 5, but i am looking for a function that is equal to 1 at every other point

[clear gale]

why?

[acoustic leaf]

because that function would have a finite limit (1) at x = 5

[clear gale]

x^2/(x-5) ?

[acoustic leaf]

no, that is not equal to 1 everywhere

what kind of function "simplifies to 1" given (x-5) is in the denominator?

[clear gale]

(x-5)/(x-5)

[acoustic leaf]

yes

so that is a function that "simplifies to 1" everywhere, except x = 5, where it is undefined

[clear gale]

right

does that imply that there's a hole?

[acoustic leaf]

that function has a hole at x = 5, yes

[jaunty mantle]

This also shows that (x-5)/(x-5) is not the same function as 1

So you shouldn’t “cancel” stuff like that when writing down a function

[clear gale]

so f(5) IS DNE for this?

[acoustic leaf]

well what would happen if we multiply our "1 except at the hole" function, with the original function you found?

[clear gale]

I suppose

since the original function is 0

(((4 x+35) (x-5))/((x-5) (x+5)))

t(5) = 0

[acoustic leaf]

what numbers do you get on numerator and denominator when you plug in x = 5?

[clear gale]

(55)(0) / (0) (10)

(0)/(0)

[acoustic leaf]

so does t(5) exist?

[clear gale]

I know that it's undefined , it doesn't exist?

[acoustic leaf]

yes, that's the same essentially

[clear gale]

but x -> 5 does exist

aight I think im good then

[acoustic leaf]

yes, since it's equal to g(x) everywhere except 5

[clear gale]

Thank yall for helping me out

!!!!!

[devout snow]

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[idle coral]

Hi, I need help to prove the validity of the following propositions please. They are integer divisibility exercises.

[idle coral]

I don't know how to make them, I would like to see how to make even one of them to try to make the rest.

[devout snow]

@idle coral Has your question been resolved?

[radiant anvil]

3|(2+4) but
3|2 and 3|4 is false
thus false

[devout snow]

@idle coral Has your question been resolved?

[idle coral]

How can you tell in these exercises what is true or false?

[radiant anvil]

u try it out

if u can find a counter example its false

if u can find a proof its true

[ionic ridge]

What are your thoughts about (b)? @idle coral

[idle coral]

hmm

I'm having trouble finding a counter example

[ionic ridge]

So, imagine that you are only working in IN instead of Z

Would (b) be true or false?

[idle coral]

true

I tried several values and it remains false -> something, which is always true.

I don't know how the formal demonstration would be

[idle coral]

I assume that with the principle of complete induction

[ionic ridge]

That's true

Because I'd n | m, then n <= m (in the naturals)

So in the naturals n | m and m | n implies that n <= m and m <= n so n = m

I'm the integers n | m implies n <= m is not true

Can you think of any example?

Where n | m but n > m?

@idle coral

[devout snow]

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[devout snow]

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[devout snow]

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[polar chasm]

.hahaggsf

[primal gyro]

When I do this problem I end up withhh

Can I just replace the cosines with 1s

since they alternate and we're taking the absolute value of that anyways

[restive river]

idts cuz it's /9

[primal gyro]

sorry whta does idts stand for?

[restive river]

i dont think so

[primal gyro]

ooh okay

[restive river]

how'd you conclude that the limit is 0?

[primal gyro]

I dont remember my exact reasoning but I didnt use the ratio test at all

ohh wait i thought it was 0 because n! is the bigger denominator

[restive river]

wdym

n! is in the numerator (?)

[primal gyro]

without using the ratio test n! by itself is the denominator

[restive river]

o

[primal gyro]

so i figured since thats probably bigger than cos of anything

[restive river]

oh

okay

good intuition

[primal gyro]

tyty

im assuming thats not what the problem wanted us to do thouuugh xD

[restive river]

lol ya try to understand how this ratio test would be used

[restive river]

this is good so far i feel like ur close just cancel some stuff maybe

[primal gyro]

n! wont cancel out riight?

[restive river]

it does

[primal gyro]

ooooh

[restive river]

lol

i dont think the cosines cancel though

[primal gyro]

wait so n!/(n+1)! cancels out tooooo (1/1!)?

[restive river]

nah

u know how factorial works right

its like

5! = 5x4x3x2x1

[primal gyro]

yeee

[restive river]

if you line things up it makes more sense wait

this is 7!/3!

[primal gyro]

OOOOOOO

sooo you're left wiiith

n+1

[restive river]

yep

[primal gyro]

ooowah mah gawhh~

[restive river]

lmao

[primal gyro]

and then im left with the cosines from satan right?

[restive river]

yeep

[primal gyro]

when a movable object encounters an unstoppable force

[restive river]

do you know what you would want to do ideally

[primal gyro]

no not sure

[restive river]

lmfao

real shit

[primal gyro]

get me out 😭

[restive river]

lmaoo

anyway your intuition was good so u should be looking to find the result for the limit to be 0

you have a limit -> infinity

so to get the answer to be 0 you want the denominator to be infinity

[primal gyro]

n is bigger than cos(n) right?

although idk how that applies to n < 1

[restive river]

depends

if n is negative

but here we have n -> infinity

[primal gyro]

ooooh i see

[restive river]

so yeah

n is bigger

[primal gyro]

n on denominator therefore everything 0

[restive river]

yes but you have to be careful as well

[restive river]

what did you mean by this

[primal gyro]

wait no nvm

i wasnt thinking about infinity when i said that

[restive river]

ah