help-27

[woven radish]

TheKingPin

[gaunt galleon]

that is just a tautology

[dire slate]

oh

[gaunt galleon]

back to the example: let's assume a > b is true and there is some c such that a+c <= b+c

it follows from a+c <= b+c that (b+c)-(a+c) >= 0

but this means that b - a >= 0 which is equivalent to saying a <= b

a <= b is exactly the same as "not a > b"

so we have a contradiction

in this case you can also do a direct proof, but ultimately it doesn't matter, because $A \implies B$ is logically equivalent to $(A \land \neg B) \implies \neg A$

[woven radish]

belabutter

[gaunt galleon]

if you aren't convinced still, fill out the truth tables for $A \implies B$ and $(A \land \neg B) \implies \neg A$ and you will see that they are equivalent

[woven radish]

belabutter

[dire slate]

had to write it down to see it

so what i gathered

you have the implication, you went with the proof by contradiction which assumed the left side is true and the right side false

you worked through it and somehow found a piece that showed information that contradicted the left side of the implication a > b

[gaunt galleon]

yes

[dire slate]

oh wait, you went with the a and not b approach which is the same thing

[gaunt galleon]

yes

[dire slate]

i get the approach

but its not intuitive

[gaunt galleon]

you will get used to it by seeing more examples

and proving things by contradiction yourself

[dire slate]

true, it seems like its something you have to know ahead that its going to contradict and not something you just work throuhg

[gaunt galleon]

not always

sometimes you just assume A and not B and see where it leads

[dire slate]

so where in your work did you use your idea (A and notB) implies notA?

[gaunt galleon]

so A and not B is what i assumed

you mean my example right?

[dire slate]

yes

i saw the inital a implies b and then you went with a and not b

[gaunt galleon]

right, i am trying to prove A implies B

in this case: a>b implies a+c>b+c for all c

[dire slate]

right

[gaunt galleon]

not B in this case is there is some c such that a+c <= b+c

[dire slate]

right

[gaunt galleon]

by definition of <= this would imply that (b+c)-(a+b) >= 0

in a written proof i would have the double sided arrow on the left side of each step

(b+c)-(a+b) >= implies b-a >= 0

[dire slate]

oh, i see it now

so for the contradiction proof, you start with the statement that you assume true which is (A and notB) and with this information youre trying to prove notA which you do

[gaunt galleon]

yes

[dire slate]

ok

[gaunt galleon]

one reason why proof by contradiction can be useful is that the for all quantifier gets turned to an existential quantifier when negated

so in this case i just need a single c for which a+c>b+c doesn't hold, instead of proving a+c>b+c for all possible values of c

[dire slate]

right

[gaunt galleon]

the classic example for a proof of contradiction would be proving that there are infinitely many primes

in that case too, you assume there are finitely many primes and then that leads to a contradiction.

a finite amount of primes are easier to handle than infinite primes

[dire slate]

I see why (A and notB) implies notA is a contradiction. Because we assume A and notB is true which makes A true. So if we assume its true and then find that notA is also true at the same time thats why its a contradiction

[gaunt galleon]

exactly!

[dire slate]

that makes sense, thats why its called proof by contradiction

[gaunt galleon]

yup

[dire slate]

just using A and notB would be proof by a negation

this is making sense now

[gaunt galleon]

great

[dire slate]

so, is this format for all logical statements?

[gaunt galleon]

wdym?

[dire slate]

instead of A implies B, lets say we have A and B, would the proof by contradiction be (notA or notB) implies A?

since the negation of A and B is not A or not B and we're trying to prove a contradiciton so imply not A is true and lead to A also being true?

which also leads to a contradiction

[gaunt galleon]

my understanding of proving things is that you take some assumptions A and show that B follows from those

i don't think you would prove something like "A and B"

[dire slate]

ah, so proof start with an implication

[gaunt galleon]

yes

[dire slate]

good enough with me

that answers all my questions unless you have anything to add?

[gaunt galleon]

yes

ima go to bed now, good night ❤️

[dire slate]

thank you so much for the help

get some good rest

good night

[gaunt galleon]

np

[dire slate]

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[prisma pendant]

I need help with a maff question 🙂
X produces 50l waste water every 1.2 seconds.
Y need to take a supply of 100l waste water every 0.5 seconds.

The fact that Y has a production rate of 0.5s and X has a production rate of 1.2s confuses me in this matter and I can't figure out how to pair the two together whilst having no backlog in either blocks and Y block is not being overfilled at any point by X.

Finding out the optimal number of X's compared to Y's would be great! (Like, if the most optimal numbers would be 9 X's to 2 Y's, if you understand).

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@prisma pendant Has your question been resolved?

[prisma pendant]

<@&286206848099549185>

Idk what that means, what slope?

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[tall mist]

I just came across a fault in my work. To solve for B in y=mx+b I got two answers...
y-mx=b and y/mx=b
The second one is what I originally thought it was because m and x are multiplying each other but when I do the other way of subtracting y I get y-mx=b

[tiny oyster]

its the first one

y-mx=b

if thats what you are asking

you can only divide if the mx was multiplying with the b

[tall mist]

so with mx, we are just assuming they are adding each other and not multiplying?

I dont understand

Or is it just a rule that we only divide if its multiplied towards the targeted variable?

[tiny oyster]

$ab=y, b = \frac{y}{a}$
$b-a=y, b = y + a$

[tall mist]

Like for example if it's like s+pieR^2 H and solving for H it would be s/pieR^2=H because they are multiplying each other?

[tiny oyster]

i can't explain because to be honest i don't completely understand your problem, is your issue why we subtracted mx and not divided it ?

[tall mist]

Yes

[tiny oyster]

its because you can't actually divide, it will cause the equation to be weird and hectic

try doing so

divide both sides by mx

[tall mist]

y=mx+b
y/mx=b

I don't understand whats so wrong about this answer since it divides the Multipication

[tiny oyster]

your thing would work if you divided by x or m, not the entirety of mx

[tall mist]

I see, I think I sorta understand

So with another example problem like if it's like s=pieR^2 H and solving for H it would be s/pieR^2=H because they are multiplying each other?

[tiny oyster]

s+pieR^2 H = ?

you can't solve for something if it isnt an equation

oaky

okay*

[tall mist]

Sorry forgot to capitalize

[tiny oyster]

you're fine

[tall mist]

to make it an equal sign

[tiny oyster]

yes thats correct

[tall mist]

So that would be fine?

[tiny oyster]

yes thats fine

[tall mist]

Okay, I think I understand now, thanks for the help

[tiny oyster]

no problem

[tall mist]

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[bold harbor]

Hi, I would love to have some help with part 2 of this qustion, thanks!

[bold harbor]

plz help

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[smoky gyro]

a sum of $1000 is invested at the beginning of each year in superannuation fund. at the end of 30 years, how much will be available if the interest rate is 12% p.a?

[smoky gyro]

i dont get what to do

[lost laurel]

,w superannuation

[woven radish]

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[lost laurel]

so compound interest ?

[smoky gyro]

its superannuatinon

idk

[lost laurel]

what's that

,w define superannuation

[woven radish]

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[smoky gyro]

isnt that common in every country?

[lost laurel]

I've never heard of it

ok, so this is basically a compound interest problem

with the principal amount being 1000

[winter patrol]

there's an annuities formula

[lost laurel]

It's a GP

[smoky gyro]

yeah i know i have to do GP stuff

but i dont know whats going on

[lost laurel]

ok, so what's the formula for sum

[smoky gyro]

for money

?

idk

[lost laurel]

for a GP

[smoky gyro]

um

a(r^n-1)/r-1

[lost laurel]

good

use that here

a=1000

[smoky gyro]

why is a 1000?

[unborn atlas]

what does the p.a mean at the end?

[smoky gyro]

per annum

[unborn atlas]

so like per year?

[lost laurel]

because that's the initial amount

[smoky gyro]

um

[lost laurel]

oh

right

1000 at the beginning of each year

hmm

you can define this as $a_n= 1.12(a_{n-1})+1000$

right?

[woven radish]

ƒ(Why am. I here)=I don't Know

[smoky gyro]

no i dont know what that is

[lost laurel]

the amount at the nth year

is 1.12 the amount in the previous year

• 1000

so the amount at the end of 30 years is $1.12(a_{29})+1000$

[unborn atlas]

1.12

[lost laurel]

my bad

[unborn atlas]

😭

[lost laurel]

like this you can derive a general formula in terms of $a_1$

[woven radish]

ƒ(Why am. I here)=I don't Know

ƒ(Why am. I here)=I don't Know

[lost laurel]

so this is $1.12(1.12(a_28)+1000$

[woven radish]

ƒ(Why am. I here)=I don't Know

[lost laurel]

yes?

[smoky gyro]

immm super lost i think im just gonna watch a video

thanks tho

[lost laurel]

sorry

[opal cloak]

have u done it using the table?

[smoky gyro]

whats that?

[opal cloak]

its what they get taught in standard

[smoky gyro]

ohhh

[lost laurel]

so this is $(1.12)^{29}*1000+1000$

[smoky gyro]

i just started the finance topic so i havent really touched financial maths since like year 10 i hate it so much 😭

[opal cloak]

+1000

[unborn atlas]

i havent done this stuff in years

[woven radish]

ƒ(Why am. I here)=I don't Know

[lost laurel]

I've never done this stuff

[opal cloak]

i skipped over it, since its not necessary for ext 2, but i can breifly explain it

[smoky gyro]

yeah lucky u lol finance and stats areeeeee sooooo dumb

[unborn atlas]

its 1000 a year so its atleast 30000 💀

[smoky gyro]

sure

[unborn atlas]

i think this follow this

https://www.mathsisfun.com/money/compound-interest.html

[lost laurel]

that's different though

here you're adding 1000 each year too

[opal cloak]

lets let A_n represent the amount of money in our account at the nth year

[unborn atlas]

idk how to account for the 1k a year

why not just add 1000

er

[opal cloak]

on our first year, A_1 we'd put in $1000 and that will gain interest

therefore A_1 = 1000(1.12)

[unborn atlas]

$$1000 * (1+0.12)^{# of years}$$

[woven radish]

Nathan
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[unborn atlas]

this gives it if the value isnt adding 1k

but i feel like it shouldnt be hard to manipulate

[opal cloak]

on our second year A_2 we add another 1000, and the money from last year and this year will gain interest

A_2 = 1000(1.12)(1.12)+1000(1.12)

[smoky gyro]

so 1.12 is from 12% interest rate?

[opal cloak]

yes

have u understood whats happening thus far?

[smoky gyro]

wait why is there (1.12)(1.12) next to the 1000 on the left

[opal cloak]

so on our first year we got 1000(1.12)

[smoky gyro]

yep

[opal cloak]

but that money is going to gain interest again

so it becomes 1000(1.12)(1.12)

[smoky gyro]

ohhh

okay i see

[opal cloak]

now if we look at the 3rd year, A_3, we get 1000(1.12)(1.12)(1.12)+1000(1.12)(1.12)+1000(1.12)

now we can simplify this

to 1000(1.12)^3 + 1000(1.12)^2 + 1000(1.12)^1

[smoky gyro]

yep

and we are looking at A_30 or somethin?

[opal cloak]

so after 30 years, A_30, we will get 1000(1.12)^30 + 1000(1.12)^29 + 1000(1.12)^28 + ... + 1000(1.12)^2 + 1000(1.12^1

[opal cloak]

yes

[smoky gyro]

ohh

[unborn atlas]

vwala

[opal cloak]

now, if we write this in reverse its actually easier

[smoky gyro]

so r is 1000 here?

[opal cloak]

1000(1.12)^1 + 1000(1.12)^2 + ... + 1000(1.12)^28 + 1000(1.12)^29 + 1000(1.12)^30

[opal cloak]

r is the ratio between each term

[smoky gyro]

yah this looks more familiar now

a is 1000(1.12)?

[opal cloak]

yes

and r is 1.12

[smoky gyro]

ohhhhhhh

okay

thanks that made so much more sense lol

[opal cloak]

all good

[smoky gyro]

what are u doing in ext 2 btw?

[opal cloak]

atm its mechanics

which is just a calculus version of physics

[smoky gyro]

o yeah

my mates are doing further integration rn

and their last assessment is just a presentation on the binomial distribution stuff

[smoky gyro]

what subjects are u doin?

[opal cloak]

i do English Advanced, Math ext 1, Math ext 2, Physics and Software Design and Development

hbu?

[smoky gyro]

im doing Eng ADV, Maths ADV, Physics, Legal, Ancient

[unborn atlas]

whats ext

[opal cloak]

extension

[unborn atlas]

is that some highschool program?

[opal cloak]

solid subjects, I despise anything to do with history tho

[opal cloak]

yeah its our high school subjects ig

[smoky gyro]

haha yeah i was debating on whether doing ancient or modern but i decided to go with ancient cuz it seemed for interesting

[unborn atlas]

@opal cloak could you lend me a hand in https://discord.com/channels/268882317391429632/409596215697866773 if you know how when you get the chance

[smoky gyro]

but dammmmnnn ext 1 and 2, atar scaling gods be putting in work

[opal cloak]

problem is im lazy

[smoky gyro]

arent we all tbh

what are you up to in physics rn?

[opal cloak]

nature of light

its so boring

[smoky gyro]

are u in the middle of it?

[unborn atlas]

sounds super cool to me

[opal cloak]

not sure tbh, but I have an exam on Tuesday

and its all theory

[smoky gyro]

ya theres a fair bit

have u done photoelectric effect?

[opal cloak]

yep

[smoky gyro]

ive just finished special relativity which is at the end of mod 7

SR is cool

i think youll like it

[opal cloak]

hopefully

[smoky gyro]

trials next term tho lol

spoooky

[opal cloak]

yep

[smoky gyro]

what week is trials for you?

[opal cloak]

5 & 6 but english is at the end of 4

[smoky gyro]

ah nice

mine are week 3

[opal cloak]

oh rip

[smoky gyro]

when would the hsc be i know its like in 3 months and a bit

[opal cloak]

ah i forgot

i know its in october

[smoky gyro]

is that in term 3?

[opal cloak]

nah thats 4

[smoky gyro]

oh ok good

[opal cloak]

what course do u hope to get into?

[smoky gyro]

wait do u wanna take this to dms

[opal cloak]

sure

[smoky gyro]

aight

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[quaint pawn]

could anyone explain please

[lost laurel]

is this a test?

[opal cloak]

in a triangle, if u pick any 2 sides, their combined length must be larger than the 3rd

[quaint pawn]

no, it's APEX

so can u tell me why 2 wouldn't work

[opal cloak]

?

[quaint pawn]

the answer, 2, is not correct right

[opal cloak]

oh sorry, yeah that would be correct

[lost laurel]

and the difference of two sides must be less than the third

[topaz axle]

you add two smallest sides

2+7 > 9 is false

[quaint pawn]

ahh that makes sense

appreciate it

adding the smaller sides helps

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[granite arch]

is turned to

but i dont really get how this is done thx

[opal cloak]

looks like they squared everything

nah that woudlnt work

[granite arch]

i dont know if theyre really equivalent but it's the answer format

[lost laurel]

!original

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[lost laurel]

please

[granite arch]

ok

@lost laurel

[lost laurel]

yeah, I'm reading it

thanks

[granite arch]

thx

[arctic field]

$\alpha$ is a root of $x^4 + 3x^3 + 2x + 6 = 0$

\medskip
implies

\medskip
$\alpha^{-2}$ is a root of $(x^{-1/2})^4 + 3(x^{-1/2})^3 + 2x^{-1/2} + 6 = 0$

hm

this is probably badly written

[granite arch]

ye

i got to here

[arctic field]

so you just do algebra

,, -2y^{3/2} = 1 + 3y + 6y^2

[arctic field]

square

,, 4y^3 = (1 + 3y + 6y^2)^2 = 36y^4 + 9y^2 + 1 + 6y + 36y^3 + 12y^2

[granite arch]

Oh so just square both sides?

[arctic field]

yes

[granite arch]

thanksss

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[nimble turret]

in ap i have seen question where sum of first m and first n terms are same but they arent equal . can anyone explain how does it work

[nimble turret]

like in sn=sm but an=!am

since ap increases or decreases in constant values which are same
for example 2,5,8,11,14,17...........
lets say n =3
2+5+8=15
now how can there exist another term where the total value is 15 too?

[topaz beacon]

consider 2,1,0,-1

S1 = 2, S4 = 2

[nimble turret]

wait what

lemme think

ohh

so it can happen only when we go down?

wait nvm i got it

[nimble turret]

thankyou

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[topaz beacon]

it can happen when we go up

consider -2,-1,0,1

S1=-2, S4=-2

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[topaz beacon]

it does have to cross 0 though

[topaz beacon]

oh great

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[nimble turret]

lol

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[dry iron]

Can someone help me, how do I do these? i tried part a by adding 1/10 and 3/10 but its not right

[umbral raft]

What is the correct answer?

[naive shuttle]

Is it 3/10?

[dry iron]

0.3 for a, 0.7 for b, 1 for c

[dry iron]

for part a its 0.3 so i think yeah, but how did you get 3/10?

[opal ferry]

I’m slowly learning even in elementary

[naive shuttle]

So umm it’s cumulative

So a) asks for P(X<=3) but since its cumulative if we sub x= 3 in F(x) it already includes everything for x=1 to x=3

For b) u rewrite P(X>4) as 1-P(X<=4)

But there’s no x=4 since ,x is just 1,3,6,10

So we go down to the next smallest x that is available in F(x) which is X=3

So we get 1-P(X<=4) = 1-P(X<=3) = 1-0.3 = 0.7

c)

Umm same logic as a) F(x) already included all possibilities up to X=10

So it’s just 10/10 = 1

[dry iron]

thank you

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[manic sigil]

Hey, I was hoping someone could help me understand whether I should use confidence interval for a population mean, using a normal distribution or a Student's T distribution

[manic sigil]

I've got this table in excel. I calculated values for "Absorbance" (Y axis) based on "Protein concentration" (X axis)

Now, the green columns are the values 4 groups got from doing the experiment.

The red column is the average of each 4 values.

Each row is a different concentration, from 0 to 0.7

The blue column is the standard deviation that I found using excel's standard deviation function based on a sample

The confidence interval has an alpha of 0.05, so 95% confidence interval

My issue is that I don't know if I should be calculating the confidence interval using Normal distribution or using Student's T distribution (excel asks me for one of the two)

[devout snow]

@manic sigil Has your question been resolved?

[devout snow]

@manic sigil Has your question been resolved?

[devout snow]

@manic sigil Has your question been resolved?

[rich crow]

I would use the T distribution. In general what you use depends on the sample size. Since you only have 4 values in each group the T-distribution is better.

[manic sigil]

Great, thanks 🌟

I'll update my friends as well, since they picked one of them randomly and stuck with it lol

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[quiet mason]

How do I continue

[quiet mason]

Or restart

Probably restart tbh

[echo lagoon]

i'd say try trig sub before feynman

[quiet mason]

Ok ill try

No certain results

I put it into wolfram

Was absolutely wild

Ill try no trig sub

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[foggy walrus]

hi all,

im having huge trouble with linear algebra 2. specifically, quaternions.
in our exam every year there is a seemingly simple problem involving an equation with quaternions. but, i always fail at this, and i don't know what i'm doing wrong.

here's an example problem with necessary information and where i get stuck (for example). i'm repeating this subject for the third time this year and figuring this out would be a huge help.

[fierce remnant]

So what's the question?

[foggy walrus]

what am i doing wrong?

im getting J2+J1 when the solution for z2 should be just j2

[fierce remnant]

What's z_2?

[foggy walrus]

in the equation, a variable we're supposed to find the solution to

we're given 3 sets (as seen above) and asked to find z1, z2, z3

[fierce remnant]

oh ok

are they simultaneous equations?

or are you just trying to solve 1), 2) and 3) separately?

[foggy walrus]

simultaneous

z1 in the first is the same as the 2nd and third

[fierce remnant]

You should write words in your working out

so you can keep track of the equations you're using

(and it'll be easier to mark)

[foggy walrus]

right

give me a second

[fierce remnant]

I can see you've rearranged (3)

as your first line of working

[fierce remnant]

what are you doing in the 2nd line?

did you multiply the line above by J_1?

[fierce remnant]

this is what I mean by ^

it's painful to read

[foggy walrus]

i mean it does say that i multiply by j1 in the second line. yes

[fierce remnant]

so you're multiplying (3) by J_1?

[foggy walrus]

yes

so i can get j1z1 which appears in 1)

[fierce remnant]

why are you writing → (1) at the end of the line?

[foggy walrus]

because of the equation that's marked in blue and enumerated as 1)??

[fierce remnant]

that doesn't help at all

use words

literally write down the narration in your head as you work through the question

[foggy walrus]

this is common notation where i'm from. perhaps it's academic difference

[fierce remnant]

that's how you demonstrate to an examiner what you're doing

yea the academic difference is between doing it properly and leaving it up to the examiner to figure out your hieroglyphics

[foggy walrus]

i can show you the notes we used in class and our profs do it like this too lol

anyway, i'll do as you ask

[fierce remnant]

I'm following your work until the last line

I can only imagine the error is there

You aren't allowed to just make matrices commute

[woven radish]

ekafeman

[fierce remnant]

@foggy walrus

[foggy walrus]

A^_1 B?

[woven radish]

ekafeman

[foggy walrus]

because i'm multiplying from the left on the left side so i should do the same on the right?

[fierce remnant]

yea

and because

in general

'multiplying on the left' and 'multiplying on the right' are not the same operation

[foggy walrus]

that seems to give the right solution. trouble is prof's notes on this aren't terribly consistent

[fierce remnant]

BC isn't necessarily equal to CB for matrices

it's not the prof

[fierce remnant]

look at your final line where you write z_2 = ...

you messed it up there

[foggy walrus]

👍

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[raven swift]

help?

[kind tiger]

have you done completing the square before

for solving quadratics

[raven swift]

have but forgot about them

[kind tiger]

ok this is how it works

if you have a quadratic x^2 + bx + c you complete the square by thinking about (x + b/2)^2 because this expands as x^2 + bx + b^2/4

do you see how that lines up with the x^2 + bx in the original quadratic

[raven swift]

but c is already on the right hand side

[kind tiger]

i havent written any equations yet, what right hand side do you mean?

[raven swift]

u have to subtract c, right?

[kind tiger]

we'r egonna add c

we have (x + b/2)^2 = x^2 + bx + b^2/4

so x^2 + bx = (x + b/2)^2 - b^2/4

and finally x^2 + bx + c = (x + b/2)^2 - b^2/4 + c

[raven swift]

im not understanding this

we have two variables

[kind tiger]

yes but now im just going over how to complete the square

forget about the problem at hand for the moment

[raven swift]

oh alr

ok i think i get it

u subtract/add the c value

and divide b by 2

and square it

and add that on both sides?

and then u also have ur factor

[kind tiger]

yh

now we need to deal with a slightly hard case when we have ax^2 + bx + c

but we can just factor out the a to get a(x^2 + b/a x + c/a)
and then the bit inside the brackets is in the same form that we just did, so we can go on from there

[kind tiger]

do you think you could write out this in the complete the square forrm

[raven swift]

im not understanding it through the typing

[kind tiger]

$ax^2 + bx + c = a\left[x^2 + \frac{b}{a} x + \frac{c}{a} \right]$

[woven radish]

Acman

[kind tiger]

does that help?

[raven swift]

we never learned thisd

type of form

[kind tiger]

wdym

i just factored out a

ill finish out the process and maybe thatll help

[raven swift]

is this conics?

[kind tiger]

no

this is just algebra, quadratic equations

[raven swift]

nvrm i got it

its 4

i see what u were saying

[kind tiger]

\begin{align*}
ax^2 + bx + c &= a\left[x^2 + \frac{b}{a} x + \frac{c}{a} \right] \
&= a\left[x^2 + 2 \frac{b}{2a} x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a}\right] \
&= a\left[\left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a}\right] \
&= a\left[\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{c}{a}\right] \
&= a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c
\end{align*}

[woven radish]

Acman

[kind tiger]

yep

[kind tiger]

though i would take some time to go over completing the square, particular this ^ so you understand how it works
both a useful tool common to come up in tests

[raven swift]

thank you tho

[raven swift]

wait i think i do see it

we factor out aa 4

and then get it in two quad forms

and we find c value

[kind tiger]

yeah basically, nice

[raven swift]

alr take it easy

[kind tiger]

and you

[raven swift]

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[chilly lodge]

Can someone explain why this is not the inverse of f? It looks to be the inverse right? And what is the "preimage?"

[devout snow]

@chilly lodge Has your question been resolved?

[chilly lodge]

<@&286206848099549185>

[dawn iris]

note that the inverse has to be a function, if you give it an input, it only gives you one output

so like if you take f(x) = x^2, this doesn't have an inverse because f^-1(1) can either be -1 or 1

the preimage is a more general idea; you just take the set of points that get mapped to the range

so the preimage of 1 under f is {-1,1}

its like a more general definition of an inverse

[west pewter]

Invertibility refers to the function having an inverse by composition

You shouldn't confuse that with the multiplicative inverse

We say that for a function $f:A \rightarrow B$, the preimage of $C \subseteq B$ is the set $f^{-1}(C) \coloneq {x \in A | f(x) \in C}$

[woven radish]

Uta

[west pewter]

Do you need the function to be invertible to use this? Definitely not

[chilly lodge]

yes it is not needed

[kind tiger]

similarly a function $f : A \to B$ has \emph{image} $f\left(A\right) = \left{f\left(a\right) , | , a \in A\right}$

[woven radish]

Acman

[west pewter]

Yep

[chilly lodge]

So the image is basically just the function

[kind tiger]

might help you see why its called a 'preimage' here

[chilly lodge]

so is the preimage like having the function being applied twice?

[kind tiger]

wdym

[chilly lodge]

if the image of $f(a)$ is $f\left(A\right) = \left{f\left(a\right) | a \in A\right}$

wait

so the image is the range of the function

[kind tiger]

yes

its a bit of an abuse of notation

[chilly lodge]

okay so I get the image

but what does that make the preimage?

[kind tiger]

you can also have the image of a subset $X \subset A$, $f\left(X) = \left{f\left(x\right) , | , x \in X\right}$

[woven radish]

Acman
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[chilly lodge]

so that's for a restricted domain

[kind tiger]

yes

then the preimage is the same concept in the other direction

[chilly lodge]

so going from range to domain?

[kind tiger]

yes

[chilly lodge]

so the range of the domain

is the preimage of $f(x)$?

[woven radish]

Decoway

[kind tiger]

image is "given these points, what outputs can f give"
preimage is "given these outputs of f, what are the possible inputs"

[chilly lodge]

oh okay I get it now

thanks a lot!

[kind tiger]

nws !

[chilly lodge]

so for this image

z is the set of ranges given

and $f^{-1}({z})$ is the list of domains for those ranges?

[woven radish]

Decoway

[kind tiger]

yes, though i wouldnt call these 'domains' and 'ranges'

a function $f : A \to B$ has \emph{domain} $A$, \emph{codomain} $B$, and \emph{image} $f\left(A\right)$

[woven radish]

Acman

[chilly lodge]

👍

yes that makes sense

[kind tiger]

id try to avoid "range" entirely tbh

[chilly lodge]

so "domain" and "image?"

[kind tiger]

yes

[chilly lodge]

okay thanks!

[kind tiger]

then $f^{-1}\left(\left{z\right}\right)$ is a set of values \textit{from} the domain

[woven radish]

Acman

[chilly lodge]

oh okay so if I'm understanding

z is the image of some restricted domain

and then $f^{-1}({z})$ will give that domain

[woven radish]

Decoway

[kind tiger]

yes this is absolutely perfect

or {z} but yes

[chilly lodge]

okay thank you so much!

you explain well :)

[kind tiger]

thanks :D

[chilly lodge]

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[woven radish]

Decoway

[devout snow]

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[zinc pewter]

For the love of god, please help me

[zinc pewter]

Here is my work:

I have been stuck on this for an hour

Ok, according to some guy on Chegg, my y is correct.

LMFAO I never integrated with respect to x

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[unkempt dawn]

Answer for A) is i according to the answer key

But why is it not also possibly iii)?

What am I missing

Both are negative leading coefficients, both are cubic

[scarlet sequoia]

you can simply say that for i) we have x = 0 as a root

and for iii) we don't

since i) passes through the origin it fits

[unkempt dawn]

One moment

Where do you get x as a zero

For i

[scarlet sequoia]

plug x = 0 and check

[unkempt dawn]

K hold up

[scarlet sequoia]

if every single term has x in it then it's obvious that x = 0 is a root

[unkempt dawn]

Right because anything time 0 is 0

[scarlet sequoia]

yep

exactly

[unkempt dawn]

Ok I get that's how you find the y intercept

Oh is it because 0+3 is 3?

[scarlet sequoia]

there are plenty of ways, in general, to show that i) fits in this case and iii) not, not only the y-intercept, but this one is pretty simple and fast

[scarlet sequoia]

for iii)? yeah

[unkempt dawn]

Oh I see because if thst were true then y would equal +3 which it doesn't

How do I determine where the line begins I'm reference to quadrants

Specifically with cubic polynomial functions?

[scarlet sequoia]

well, it such a picture (with limited ranges of the axes) it depends, but generally it's enough to look at a leading coefficient

[scarlet sequoia]

if its negative then it starts from the left (more precisely in 2nd quadrant) since when x tends to -inf it's inf

and if its positive then it starts from the left (III quadrant)

[devout snow]

@unkempt dawn Has your question been resolved?

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[grave stratus]

i have no idea what the answer is fuck

there's no example for show me how to work it out either

can someone translate what they want from me

and give me a step by step on how to answer

[scarlet sequoia]

I suppose they're talking about the definition

[grave stratus]

okay

which means

[scarlet sequoia]

which means f(x) = b^x

[grave stratus]

okay

whichhh meanssss?

i just don't know what part is the exponential function

obviously it's b, 0 or 1, right?

or maybe x?

but x is the variable not the exponent

[scarlet sequoia]

variable in the exponent -> exponential function

[grave stratus]

oh i'm right

okay

let's try it

the options are is/is not

[scarlet sequoia]

is x^4 of the form b^x?

[grave stratus]

i don't think the first one is an exp function but i think the 2nd one is

wait

or maybe reverse

can u explain why the first one isn't

[scarlet sequoia]

because it doesn't have x in the exponent

[grave stratus]

that's what i thought sweet

okay so now the options are increasing decreasing constant zero

i'm reading the ebook to see what it says and if there's an easy way to figure it out

[scarlet sequoia]

yeah it is

increasing for b > 1 and decreasing for 0 < b < 1

you can use this picture to determine it

[grave stratus]

that's helpful

this one is increasing too, right?

[scarlet sequoia]

,calc 2/7

[woven radish]

Result:

0.28571428571429

[scarlet sequoia]

< 1

---> ...

[grave stratus]

ohhhhhhhh

so it's decfreasing

i just did a bunch of interval notions but this one doesn't have any numbers

how do i tell which part is which

oh wait

is f(x) the domain and b^x the range?

or does it actually want numbers

[scarlet sequoia]

possible set of numbers

domain is possible set of numbers of x and range for y

[grave stratus]

it wants numbers so how do u do that

the ebook has this which looks like fx=b^x so i just have to find the domain and range by looking at this

sooooooooooooooooooooooooooooooooo

so the domain would be [-5, 5] ???????

or (-infin, infin)?

bc it looks like it goes on in both directions

and the range would also be (infin, infin)

oh u hate me now @scarlet sequoia

okay so i did the infin bc it's obvi this bitch is infin asf

<@&286206848099549185> what is an asymptote

i see this but

how do i write the equivalent

[modern rivet]

Should be y=0?

Not too sure

[grave stratus]

i want you to be sure

i need you to be sure

i can't retake this

[copper sonnet]

y = 0 is correct

[grave stratus]

❤️

fucking bet

[modern rivet]

LETS GOOOOO

[grave stratus]

apparently i can retake it yay

[modern rivet]

Tf this is a different question-

[grave stratus]

yes

it is a different question

good eye

i clicked submit and 2 of my answers were wrong

[modern rivet]

I’m blind 😞

[grave stratus]

dw me too

[modern rivet]

Oh mb

Yay!

[grave stratus]

OHHH

Fx=bx^ whatever is the answer

it wanted the whole thing

[modern rivet]

An asymptote btw is a “line” that your curve in a graph approaches

Well tries to

[modern rivet]

OHHHh

[grave stratus]

okay this actually makes sense i think

h(2), i would simply take 1/4 and power it to 2 to get ....... 1/16?

h(-1) would just be 4 because you'd flip the fraction and change the changes ig?

then h(-2) would be 16 for the same reasons?

oh my GOD I WAS RIGHT

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[cunning roost]

why does the graph look like that?

doesnt that mean theres a hole at x =2?

[devout snow]

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[quiet mason]

I tried this integral a few hours ago and I really cant get enough of it, does anyone see a different approach in this?

[quiet mason]

I did try trig sub before Feynman, didnt work aswell

[timber pebble]

hows weierstrauss work

[quiet mason]

Sorry I dont... fully understand

[timber pebble]

not well

well maybe you can do it indefinite but i have doubts now

[quiet mason]

Indefinitive is very casual

Just integrate cosx

But its the infinity not giving a certain result

[stable storm]

i think ibp after feynman would work just fine

choosing $u=\sin(\alpha x), dv=\frac{x}{x^2+1}$

you can connect the integral part you get back to I(alpha)

[quiet mason]

Ooo

[woven radish]

y0shi

[stable storm]

not cosine mb should be sine

[quiet mason]

Ill give it a try

Dont mind the trig sub

[stable storm]

that integral doesn’t seem possible either

hmm

[quiet mason]

Not even only the integral, I dont think this converges at all lol

[stable storm]

yeah

[quiet mason]

I did use wolfram a few hours ago

I wont spoil it but the asnwer was... interesting

I didnt see the method ofc

[dense lynx]

this integral is one that can be approached using contour integration

[quiet mason]

Do i close or

[dense lynx]

though i assume there’s multiple ways to do it

[quiet mason]

Oh like the integral sign with the circle in it?

[stable storm]

yeah

[quiet mason]

Welp rip

Thanks for trying to help

[stable storm]

you would have to deal with a pole but

you can just use residue theorem

i’m sure feynman is possible

it’s just it might be more complicated

[quiet mason]

Grrr power serieses

[dense lynx]

what power series?

[quiet mason]

.close

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[near cedar]

can someone help me find the distance between H and point M

[near cedar]

Two electric poles of the same height h are erected on both sides of the sidewalk facing a wide avenue 80m. From a point M on the road surface between two electric poles, one can see the tops of two electric poles with elevation angles of 60degree and 30degree respectively. Calculate the height of the poles. (rounded to the second decimal place)

question btw

<@&286206848099549185>

someone pls save me... <@&286206848099549185>

[void snow]

which are your difficulties? You don't know where to start or..?

[outer merlin]

!status

[devout snow]

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

[near cedar]

yea i dont

[void snow]

how proficient are you with trigonometry?

[near cedar]

maybe average

or below

[void snow]

ok so you do know what is $\tan{()}$

[woven radish]

Adversing

[near cedar]

yea i know

[ocean seal]

There are 3 triangle there

You can use the angles from the other 2 triangles to help fill out the third.

[outer merlin]

Why would you name them differently? they are the same height

[ocean seal]

My mistake

Ignore that

[outer merlin]

Unnecessary

Two equations using tan and its solved

[void snow]

ok, so, for this problem you have to define first the data the problem gave you

I'd define the poles height as $h$, the distance from point $M$ to the pole with an elevation (with an angle of 60°) as $d$ and the distance from the same point ($M$) to the pole with an elevation (with an angle of 30°) as $80-d$

You can use the tangent of these angles to obtain what you're looking for

$$
\tan{(60°)} = \sqrt{3} = \frac{h}{d} \to h = d\sqrt{3}
$$

then you do the same for the other pole (elevation of 30°)

$$
\tan{(30°)} = \frac{1}{\sqrt{3}} = \frac{h}{80-d} \to h = \frac{80-d}{\sqrt{3}}
$$

since you obtain $h$ in both equations you can do an equality like this:

$$ d\sqrt{3} = \frac{80-d}{\sqrt{3}} $$

and then you solve for $d$ and you obtain the result

[woven radish]

Adversing

[void snow]

(to obtain the result you have to use again $h = d\sqrt{3}$)

[woven radish]

Adversing

[void snow]

idk why the angle is parsed by LaTeX as $r$ wtf

[woven radish]

Adversing

[ocean seal]

Wouldn’t it be easier at a pre university level to just fill out sine law twice and get an exact value?

[void snow]

the guy said that he knew how to use the tan function, hence there's no sense in doing what you're trying to suggest

if he didn't know about the tan function I would've done it with the sine law

[void snow]

@near cedar

[near cedar]

im trying to comprehend that english is not my first language

[ocean seal]

Ok

[void snow]

oh ok, no worries

[devout snow]

@near cedar Has your question been resolved?

[near cedar]

aint tan(30) = sqrt{3}\3 or my calculator just wrong

[void snow]

it's the same thing

[near cedar]

oh ok

[void snow]

$\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

[woven radish]

Adversing

[devout snow]

@near cedar Has your question been resolved?

[devout snow]

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[dry iron]

Can someone help me with this, I’m not sure how to continue it to solve for the unknown variables because I tried but it’s not going well

[opal cloak]

yo

remember that the sum of p(x) must be equal to 1

[devout snow]

@dry iron Has your question been resolved?

[devout snow]

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[devout snow]

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[fickle warren]

I don't understand it conceptually

[fickle warren]

idk

Wait

LHS of U

Since there is one row equal to 0, that means we can write one of the vectors in terms of the other, which means all three vectors are not linearly independant.

idk

For the second part, we have to analyse RHS of U I believe

[devout snow]

@fickle warren Has your question been resolved?

[devout snow]

@fickle warren Has your question been resolved?

[devout snow]

@fickle warren Has your question been resolved?

[devout snow]

@fickle warren Has your question been resolved?

[hybrid snow]

I have a question

[devout snow]

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[oblique silo]

find closed form of:

[elfin hill]

!status

[devout snow]

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

[oblique silo]

1

i mean i know i should use newtons binomial

[elfin hill]

Does this formula looks like anything you know?

Yes, write the formula for newton's binomial

[oblique silo]

yeah but what now

[elfin hill]

Derivate it with respect to r

[oblique silo]

derivitate this?

[elfin hill]

no, derivate newton's binomial

[oblique silo]

sum kr^(k-1) (n choose k) ?

[elfin hill]

yes

derivate it once more

[oblique silo]

oh

right

git it

but what if instead of k(k-1) it was a polinomial that you cant get from deriving?

[wise storm]

you can multiply by r before differentiating the second time

[elfin hill]

well in that case i think you can work things out using the formula for k times n choose k

[oblique silo]

what formula?

[elfin hill]

k times n choose k = n times n-1 choose k-1

[oblique silo]

how can i use that

let's say it was

(k^3 - 2k^2 + k + 1)(r^k)

[elfin hill]

Well you use it twice and get (n-1)(n-2) sum r^k times n-2 choose k-2

[elfin hill]

Well you can still use derivation

k^3 r^k = r times derivative(r times derivative ((r times derivative(r^k)))

So it's a bit long but it gets you something in the end

[devout snow]

@oblique silo Has your question been resolved?

[oblique silo]

but this doesnt give the same polinomial

[wise storm]

the main trick cjg is showing you is that you can get any power of k

once you have that, you can just multiply by constants and add these types of series together to make anything

idk if this will make it more confusing or not but if you have a polynomial $p(x)$ then

$$p\left(r \dv{r}\right) (1+r)^n = \sum_{k=0}^n p(k) r^k \binom{n}{k}$$

[woven radish]

Mero

[wise storm]

the weird part is the polynomial p( r d/dr) on the left

[wise storm]

but it basically just means doing this when p(x)=x^3 for instance

[oblique silo]

i dont really understand how i can get to any polinomial

[wise storm]

ok let's do it in steps, does this make sense:$$r \dv{r}(1+r)^n = \sum_{k=0}^n k r^k \binom{n}{k}$$

[woven radish]

Mero

[oblique silo]

not really

the derivative of r times the binomial?

why?

[wise storm]

the derivative turns r^k into k*r^{k-1} inside the sum

then cause r^k became r^{k-1} we multiply by r to make it back into r^k again

[oblique silo]

what does this part mean

[wise storm]

take the derivative wrt r, and then multiply by r

let me back up one step

$$\dv{r}(1+r)^n = \sum_{k=0}^n k r^{k-1} \binom{n}{k}$$

[woven radish]

Mero

[wise storm]

this makes sense?

[oblique silo]

the derivative of what?

is wrt a typo or an acronym? lol

[wise storm]

wrt means "with respect to" haha

[oblique silo]

ohh

[oblique silo]

this makes sense then yeah

lol

[wise storm]

yeah good question

ok cool then next level up is just to repeat that process

take the derivative again and then multiply by r

and you get k^2 inside

$$r \dv{r}\left(r \dv{r}(1+r)^n \right)= \sum_{k=0}^n k^2 r^k \binom{n}{k}$$

[oblique silo]

sure yeah

[woven radish]

Mero

[wise storm]

although we could just agree to write that as
$$\left(r \dv{r}\right)^2 (1+r)^n = \sum_{k=0}^n k^2 r^k \binom{n}{k}$$

[woven radish]

Mero

[oblique silo]

suqaring d/dr means do the derivative twice?

oh yeah

makes sense

yeah

[wise storm]

yeah, we gotta be careful about the order though

[oblique silo]

yeah

[wise storm]

yup

do you see how it gets up to this bit, or is this still a little murky:

$$p\left(r \dv{r}\right) (1+r)^n = \sum_{k=0}^n p(k) r^k \binom{n}{k}$$

[woven radish]

Mero

[oblique silo]

but wait

[oblique silo]

you get (k^2-k) inside

[wise storm]

ah no, that's what I mean about this

[oblique silo]

i thought you meant the order of d/dr

[wise storm]

it doesn't mean take the second derivative wrt r and then multiply by r^2

[oblique silo]

huh what then

[wise storm]

it means take the derivative, then multiply by r - and then do this again

for comparison compute both of these:
$$r \dv{r}\left(r \dv{r}r^k\right)$$
$$r^2 \dv[2]{r}r^k$$

[woven radish]

Mero

[wise storm]

the first one is what ( r d/dr)^2 means, the second is what you are doing

[oblique silo]

ohhh

rightt

got it

thanks

[wise storm]

yup you're welcome

[oblique silo]

.close

[devout snow]

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[devout snow]

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[marsh coral]

Consider the parabola y = 3x^2 + 6x - 17:
a) Write the equation for the axis of symmetry
b) Find the coordinates of the vertex of the graph of the parabola
c) Identify the vertex as a maximum or minimum
d) Graph the parabola

I am not very good at parabolas so I am kinda lost. But I did try using a perfect square.

[surreal night]

what did you get trying to write it as completing the square?

[marsh coral]

I don’t think it was right because it wasn’t even making sense when I did it so I erased it

It was 100% incorrect so I scrapped it

So I’m starting from square one

[surreal night]

alright

how can you complete the square here

[marsh coral]

Um

[surreal night]

(you want something like (x+a)^2)

[marsh coral]

I tried to get at that

[surreal night]

do you know how to complete the square?

[marsh coral]

I have an idea

Yeah scratch

That

Ik what it is but idk how to do it

I watched a video but I was still lost

[surreal night]

okay, if we expand (x+a)^2 we get x^2+2ax+a^2

we only care about the x terms, so x^2+2ax