help-23

[obtuse jackal]

as you're talking about elements of the multiplicative group of bijections on a given set

which leads to this general notation

that's why it's the same for matrices

etc

[obtuse jackal]

though you haven't studied abstract algebra yet

you can just write that as 1/f already

[fickle trail]

https://tenor.com/view/safety-death-gif-25739095

[obtuse jackal]

so it's a bit silly to add a notation

[fickle trail]

I am hoping this is pure math only and not a CS requirement

[obtuse jackal]

it is

[fickle trail]

but who knows, maybe I would enjoy it

[obtuse jackal]

if it starts with "abstract" it's a sign already

though it is quite nice

imo

[fickle trail]

I keep hearing it's where many go to die, and only the real ones survive

[obtuse jackal]

that's the nice part of the French system: you see these things straight out of HS as a math major and it just gives you so much mathematical maturity in a way

[fickle trail]

I'm not familiar with the French system. Today I learned for bracket notation (-Inf, 2) it's common to see ]-Inf, 2[ in the French system

closed brackets but flipped

[obtuse jackal]

yes

[fickle trail]

to mean the same thing as (-Inf, 2)

[obtuse jackal]

it's not common. It's the only one

[fickle trail]

meaning it's the only right way?

I always see () but today I just learned about ][

[obtuse jackal]

it's the only notation any French would use when doing math in French

I still prefer it when using english

[fickle trail]

with programming I am more used to seeing ().. I mean, I'm OK with both, but I don't think I have seen ][ in programming (yet)

[obtuse jackal]

because even though it's unusual, it leaves no ambiguity

intervals don't exist in programming

[fickle trail]

array would be []

function or method would be ()

object would be {}

[obtuse jackal]

depends on the programming language

[fickle trail]

true

I'm more familiar with JS

[safe radish]

@fickle trail Has your question been resolved?

[safe radish]

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[stray axle]

can someone help me understand the last line

[lean otter]

which part of it

its a bunch of inequality algebra

[stray axle]

dont really understand how they first get |x|<sqrt(epsilon ) and are just moving things around

like i get it until the second line where |x^2|<epsilon(1+x^2) but after that i dont get

[lean otter]

we chose delta=sqrt(epsilon)

and we are saying that |x|<delta

[lean otter]

1+x^2 is always positive

[stray axle]

yea that makes sense

[lean otter]

then use the fact that x^2=(x^2+1)-1

[stray axle]

ok wait

so how do we get delta=sqrt(epislon)

[lean otter]

its cleverly chosen to make it work

see the paragraph before the last line

[stray axle]

im just hella confused on it

[lean otter]

do you understand the epsilon delta definition of limits?

[stray axle]

sorta

just started learning it

[lean otter]

https://www.milefoot.com/math/calculus/limits/DeltaEpsilonProofs03.htm

for every choice of epsilon, we choose a delta (depending of epsilon) so that for all x delta close to the limit has f(x) epsilon close to the limit

so we construct a delta that works

which is the hard part of every epsilon delta proof

[lean otter]

most of this is about choosing delta

the last line is the actual proof

where we use the clever choice of delta we derived earlier

[stray axle]

😵‍💫 alright, i'll look further into this with the link and your explanation

thanks so much!!

[lean otter]

yw

[stray axle]

though quick question

i get that they are factoring out the (x^2+1) and then multiplying it to both sides, why does the |-x^2| not just become a regular x^2

why does it still have the absolute value signs after they've been "applied"

[lean otter]

we can ignore the absolute values

but we want a condition on |x-0|

[stray axle]

alright thanks

[safe radish]

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[sharp cypress]

isn't the second line wrong ?

.close

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[coral sierra]

thanks so much!!!

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[cold saffron]

I need help with identifying the type of math problem it is

That’s all

[true glacier]

just to solve for v it looks like

[cold saffron]

Yea but how would I like enter it in able to get more examples of that?

Like on khan academy

[true glacier]

probably something to do with algebra 1

[cold saffron]

Algebra 2

Well it’s basically algebra 1 but I’m taking algebra 2

And completely forgot how to do some basic functions

[true glacier]

do you need help with this question

[cold saffron]

Yea I do

I know you’ll find the LCM

Of the denominators

But after that I’m confused

[true glacier]

you can multiply both sides by the lcm

of 9, 5, 3

[cold saffron]

Ok

Thank you

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[lean otter]

if f(x) = 4x -1, f(x-3) would mean that x is replaced with (x-3)

[patent mortar]

so it'd be 4(x-3)-1

which would be 4x-12-1=4x-13?

[quasi bison]

indeed

[patent mortar]

oh thanks i didn't expect it to be this easy

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[finite crypt]

#3.

my book on this:

f(x) is a rational function and -2 is given to me in the instructions

so, i perform direct substitution using the values in the chart from the book

correct?

[safe radish]

@finite crypt Has your question been resolved?

[quasi bison]

i think you don't need to trouble yourself with this evaluation

and there is nothing sacred about points specifically 0.1, 0.01 or 0.001 away from the point of approach

[quasi bison]

if you want to evaluate f at -2.1, -2.01 etc then you can do that. but there is no need for that

you have a graph

the graph can tell you right away that the function explodes upward, i.e. +infty.

[safe radish]

@finite crypt Has your question been resolved?

[finite crypt]

what evaluation?

oh, you mean the calculation

what's the purpose of the function, then?

the book's example calculated points

shouldn't I mimic that?

[quasi bison]

no

you can but you don't need to

[finite crypt]

the exercise literally just wants me to look at the graph and decide on - or + infinity?

[quasi bison]

yeah

i mean that is a fine method from my pov

coupled with the knowledge that the function really does explode at that point

[finite crypt]

yes, the official answer seems to be this:

[quasi bison]

because of a 1/0 kind of situation

[finite crypt]

https://www.calcchat.com/book/Calculus-11e/1/5/3/

good for me, I guess, as that's really easy!

[quasi bison]

that's only the answer with no indication of the method used to arrive at it

but yeah

this is not a hard exercise rly

[finite crypt]

ok

ty

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[hushed saffron]

write the eq of the plane that cuts on the Ox and Oy axes segments equal to 5 and -7 and passes through the point (1,1,2)

[hushed saffron]

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[hushed saffron]

.reopen

[safe radish]

✅

[wild basin]

hi

[hushed saffron]

hello

[wild basin]

any doubt buddy :).

[hushed saffron]

all of them

[wild basin]

wht ?

[hushed saffron]

idk, what did you write lol

[wild basin]

ok

np

:).

you are creepy!

[hushed saffron]

huh?

can you help me solve or what

[wild basin]

see this

maybe it will help u

https://www.toppr.com/ask/question/find-the-equation-of-the-plane-passing-through-the-point33/

[hushed saffron]

my intercepts are not equal

that is the problem

[wild basin]

ohh

just put x=5 and y=-7

[hushed saffron]

what does that give me

[wild basin]

it will give z

[hushed saffron]

sorry but how

[wild basin]

see this

put x=5 and y=-7 and a=1 b=1 c=2

find z

[hushed saffron]

abc is the point ?

[wild basin]

yes

[hushed saffron]

so i got z is 6

sorry for the delay

what next

how do i write it in standart form

[wild basin]

use

that

z

and

all points

1,1,2

put in this

[wild basin]

here

you will get a relation

in x and y

[hushed saffron]

whats the point it will give me a number

what

[wild basin]

thats your anser

[hushed saffron]

this is my answer 👀

[safe radish]

@hushed saffron Has your question been resolved?

[hushed saffron]

<@&286206848099549185>

[glad silo]

Can I see your question @hushed saffron

[hushed saffron]

i solved it

[glad silo]

Ok

[hushed saffron]

can you help with this ?

[glad silo]

What grade is this?

[hushed saffron]

well i have it in first sem in uni but for someone it could be hs level

[wild basin]

its a simple question

just put r vector in the nr = D

@hushed saffron are you on?

then find t from here

and put back t in r=r0+ta

[hushed saffron]

im here had to help mom

[wild basin]

did you understand

?

[hushed saffron]

are there any special rules or do i do it like the normal math?

vectors make me go boom in the head

[wild basin]

dont think they are vectors

that is the representation that ,make you go bloom

just think this as simple math

[hushed saffron]

alr sec

[wild basin]

ok

[hushed saffron]

do those letters hold any meaning or are they just placeholders

[wild basin]

well in this question its just representation

and no meaning

just use simple substitutions

if you understand the vectors

then also you dont need to panic

[hushed saffron]

so i got (vector)r= to the result that is given

is that it?

[wild basin]

u mean answer?

[hushed saffron]

yes

[wild basin]

ok

yes

thats the answer u have to find

in that ques

[hushed saffron]

like is this it

[wild basin]

yes

nice solution!

[hushed saffron]

idk what im doing

what is the meaning of this problem

im glad its over

[wild basin]

well

do u know what

is point of intersection

?

[hushed saffron]

yes

[wild basin]

it is defind as a point which satisfies the plane

so u just find the POI of line vector and plane

by substitution

[hushed saffron]

thank you

[wild basin]

now u undersatnd!

[hushed saffron]

you are the weirdest character of today but the most useful

[wild basin]

or not

[hushed saffron]

it does not entirely matter as i will not have this subject anymore

hopefully

[wild basin]

ohh

[hushed saffron]

but i do get the idea

[wild basin]

well

i can give you example

to understand it better

[hushed saffron]

i got it its fine

i didnt know how to do it

but if its that easy then im ok

[wild basin]

Do you solve this things ever in your life

5x+4y=10

17x+64y=34

[hushed saffron]

yeah

[wild basin]

these two equations

[hushed saffron]

equal the lines and find x and y

or solve system

[wild basin]

yup!

or you can find x from one equation and put into other

the same thing

[hushed saffron]

yeah

[wild basin]

is acting in this question

[hushed saffron]

i get it now but i didnt before because i did not know these are the same things

[wild basin]

well

yes

those x and y equations represent lines intersecting

and here

a line and plane is intersecting

[hushed saffron]

yup

thanks

[wild basin]

i think now u got it

[hushed saffron]

btw wanna know how to solve the one i posted before?

[wild basin]

yes plss

[hushed saffron]

,rotate

[hushed saffron]

basically since we know the length of those segments we can say where the plane intersects

and we get 2 other points

[wild basin]

ohh

you use matrices here nice

[hushed saffron]

and we can use the eq of the plane that passes through 3 noncolinear points to get the equation per se

so thats cool

[wild basin]

ok i got it

[hushed saffron]

aight see ya

thanks again

[wild basin]

see ya

[hushed saffron]

.close

[safe radish]

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[wild basin]

np

bye bye

[hushed saffron]

bb

[safe radish]

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[lean otter]

So for this question:

I divide both sides by cos(3x + 2pi/3) to begin with

I get

tan(3X + 2pi/3) = 1

Not sure where to proceed from here

[spice grove]

tan(A) = 1 for A = npi + pi/4

[lean otter]

ah

45 degrees

[spice grove]

Well, no. In this case no.

[lean otter]

Oh

I thought tan tan 45 = 1

[spice grove]

Yes that is right.

But so is tan(225)

[lean otter]

right

but

it's from 0 to 120 degrees

so tan(pi/4) if I'm not mistaken

[spice grove]

Phi is.

Not 3phi + 2pi/3

[spice grove]

This isn't in 0 to 120.

[spice grove]

Yes.

But that's for phi.

I can't explain better. Someone take over I gotta run.

[safe radish]

@lean otter Has your question been resolved?

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[lunar bluff]

How does this make sense mathematically? Top one should be at - r*at, bottom is rat/1+r

and no matter how I look at it the 2 are not equal

same for the yt term, but then the summation term still has an r multiplier for some reason?

[safe radish]

@lunar bluff Has your question been resolved?

[safe radish]

@lunar bluff Has your question been resolved?

[safe radish]

@lunar bluff Has your question been resolved?

[plucky elk]

Show the whole page

Can't tell without more context

[lunar bluff]

@plucky elk im not sure if its relevant

but this is the entire part

[plucky elk]

Yea it's probably relevant that Et is a random walk and properties given in the hint

[lunar bluff]

hmm, I thought they only rearranged something with r(1+r)

maybe im wrong then

[plucky elk]

Where's the expression for wealth

[lunar bluff]

thats the a(t+1)

budget constraint

[plucky elk]

Try grouping a_t and y_t terms then factoring

1-r/(1+r)=?

[lunar bluff]

yes, but then it should be 1-r at and yt, not r/(1+r)

thats what I got as well

[plucky elk]

What is it here

Just show your work

[lunar bluff]

the last line

I dont know how the prof got the last line from the one before it

[plucky elk]

.

[plucky elk]

Of this

[plucky elk]

You probably just have some simple algebra mistake

[lunar bluff]

thats literally it

I get 1-r outside the bracket not r/1+r

[plucky elk]

Do this with the stuff inside the parentheses

[plucky elk]

Don't bring factors into the parentheses

[lunar bluff]

ohhh wait, yeah im dumb

got it

thanks

[safe radish]

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[cerulean walrus]

I skipped this question and it told me that the correct answer is 6.2m^2. I have no clue on how that answer was obtained. When attempting to solve, I did 153-12=141 and 2.3-0.5=1.8. To convert the degrees to radians I did 141 * pi/180 to equal 141pi/180. From there I use the circle sector area formula A=(1/2) * r^2 * theta and multiplied (1/2) * (141pi/180) to get 141pi/360. I did 1.8^2=3.24 and then did 3.24*(141pi/360) to get (456.84pi/360). I then solved that to get 3.97m^2. I've been wracking my brain on this for hours and idk what I'm doing wrong

[dawn jacinth]

how come you converted to radians

[cerulean walrus]

i was told to express theta in radians

i have no clue what im doing wrong

[dawn jacinth]

um okay

theta/360 * pi * r^2 is area sector

141 degrees is right

then calculate large sector to 2.3

• small sector to 0.5

and youll have your answer

@cerulean walrus

[cerulean walrus]

so i dont do 2.3-0.5?

[worthy hemlock]

Not directly

[dawn jacinth]

no

cause youd create

[worthy hemlock]

You want to find the area of the sector up when it's at 2.3m first then subtract the area at when it's at 0.5m

[dawn jacinth]

a differet shape

[cerulean walrus]

ok, that makes sense. thank you both so much

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[steel scroll]

can someone explain how my summation is wrong?

there are 6 rectangles, so k should go from 1-6

it is a right rieman sum, so the first rectangle is at x = 2.5

[real swan]

6 rectangles?

[steel scroll]

and the width of each rectangle is 1/2

*12 rectangles

goes from 2 to 8

total width is 6

nvm got it

thanks anyway!

I thought it was 6 rectangles lol

.close

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[foggy raft]

The answer is 2.4 but I keep getting a diff answer, could someone tell me what am I doing wrong?

[thin bridge]

you're not plugging values into the cos law properly

[worthy hemlock]

Also, why use cosine rule?

[thin bridge]

you don't just sub in any way you feel like

[worthy hemlock]

You can use sine rule

[thin bridge]

(but then you have to apply angle sum of a triangle)

from labelling conventions
a is the side opposite vertex/angleA
b is the side opposite vertex/angleB
c is the side opposite vertex/angleC
and how you sub should be based on that

[worthy hemlock]

You still need that for this, wouldn't you? That's why you did strikethrough, right?

[thin bridge]

i did strikethrough because its an effortless task

there's sufficient info to apply cos rule directly

[foggy raft]

If I use sine law, I’ll find the side right?

[thin bridge]

If I use sine law, I’ll find the side right?
wdym

[foggy raft]

on c

this one’s looking for the side length of it

[thin bridge]

that's not depending on what law you use
that's just the end goal

[foggy raft]

oh okay, well still I’m not getting what im plugging in wrong

[worthy hemlock]

Are you using cosine rule still or sine rule now?

What are you plugging in?

[foggy raft]

Cosine Law

[worthy hemlock]

from labelling conventions
a is the side opposite vertex/angleA
b is the side opposite vertex/angleB
c is the side opposite vertex/angleC

You want the angle across from side c

What is that angle?

[foggy raft]

hold on let me just find that real quick

[thin bridge]

note that there are multiple forms of the cosine law
you don't necessarily need to use the one with c^2 = ... and use the values matching those variables in that form

note that from the diagram,
the side of 4.3cm is opposite the 78° angle

[foggy raft]

That was the only way I was taught.

[thin bridge]

the variables can be changed, you just need to understand what they represent

note that in that form
c represents the side opposite vertex/angleC

a,b would represent the other two sides

[foggy raft]

So they want me to find the angle C

So that I can find the opposite side

?

[thin bridge]

that'll work

though there are other ways to apply the law

you don't need to have the variables in the diagram/question match the law you're given

e.g. the same techniques could be applied if your triangle had p,q,r,P,Q,R
isntad

[foggy raft]

True

[thin bridge]

it is possible to apply the cos law directly

note that from the diagram,
the side of 4.3cm is opposite the 78° angle
which gets you
4.3^2 = c^2 + 4.1^2 - 2 * 4.1 * c * cos(78°)

which you can then solve for c

alternatively note that
the side of 4.1cm is opposite the 69° angle
and you can use that pair instead

or if it makes you more comfortable first determine angle C
which gets you something that matches the variables in the cos law you know

[foggy raft]

Ohhhh

[foggy raft]

Well I did find the angle , I found 33.

I Found it

Thank you!

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[ancient viper]

how does this even work...

a is a constant btw

tried using the basic form a(ar+a+1)+b(ar+1)=1

it ends up with a system of 3 unknowns

[ember sphinx]

is r a constant?

[ancient viper]

considering the context of this question I don't think I should be finding for 3

well this is the full question, I'm just having trouble with the partial fraction part

r varies

[ember sphinx]

Ahh ok

I was gonna say partial fractions but idk how that works with 2 variables

[ancient viper]

just confused on how to work out the form shown on the right

I really shouldn't have problems with this at this level

quite disappointing ngl

[ember sphinx]

I don’t know either

My guess is to expand the bottom see if u can simplify it more

[ancient viper]

the +1 will be a problem

maybe I'll try to solve simultaneous and try to eliminate a somehow

alright I did get a solution by solving simultaneously

but it is very unusual for a question of this calibre to need so much working

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[crude star]

how is this wrong

$\Delta x = \frac{9}{n}$ so $x_i = \frac{9i}{n}$ right?

[flat frigate]

Worth

[crude star]

and then I just plug that in f(x)?

and then * delta x

what did I do wrong? Im lost

[plucky elk]

Your x0 should equal 1

And xn=10

[crude star]

wait what

ur talking about $\Delta x = \frac{10 - 1}{n}$ ?

[flat frigate]

Worth

[crude star]

idgi

oh wait

oh shit

so $x_i = 1 + \frac{9i}{n}$ ?

[flat frigate]

Worth

[crude star]

I am a proper idiot man

ahhhhhhhhhhhh

ty

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[merry dragon]

I got 6L for this question but it turned out wrong. Can anyone help?

[empty gyro]

show what you tried

[merry dragon]

oh whoops wrong question

oh nvm nvm i got it. thank you for your time.

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[dusk stratus]

how do i do this, i forgot 💀

[spiral bane]

do what

[dusk stratus]

facotr

factor*

[silver venture]

find a number that multiplied gives you AC and added together gives you B iirc

[dusk stratus]

thats if a is 1

i dont remeber how to do it if a is more than 1

[silver venture]

I dont think so

pretty sure it works for all real and/or rational coefficients

[empty gyro]

You do (ax+b)(cx+d)

[dusk stratus]

can you show me the steps please

[empty gyro]

Expand it and get (ac)x²+(ad+bc)x+(bd)

Find an a and c such that ac=2

Assuming they're integers, you get either a=1, c=2 or a=2, c=1

Find b and d such that bd=6

It can be either
b=1, d=6
b=2, d=3
b=3, d=2
b=6, d=1

Finally, to determine which of these combinations is correct, play around with different configurations such that ad+bc=7

[dusk stratus]

i think i got it

thank you

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[charred flax]

what does it mean for two angles ot be supplementary?

[upper dust]

When they add up to 180

[silver venture]

correct

now, how can you "expand" a ratio?

for example, 4/1 = 20/5

[upper dust]

You multiply the top and the bottom by the same number?

[silver venture]

yes, now is there anything you can multiply by that would be helpful?

you're trying to find an exact value using an equation

[azure vessel]

Suppose two quantities are in a:b ratio. You can first quantify each unit quantity by x. Then the first quantity can be quantized by ax and the second by bx

[silver venture]

basically that^

[azure vessel]

For two angles x,y in a 4:1 ratio, you can say that x=4k and y=k for some unknown k. Notice that x/y=4/1 which is the same as saying 4:1

[dull sequoia]

you can pretty much this think of this like this
2 angles, a and b
a+b = 180 (supplementary)
a:b = 4:1, or a/b = 4/1

now you just have 2 equations and 2 unknowns

[azure vessel]

Then you just find what the value of k should be from the extra condition x+y=180

[dull sequoia]

solve simultaneously

[upper dust]

oh ok

[silver venture]

I think the simplest way is just 4x+1x=180

since ratio between 4x and 1x is 4

[upper dust]

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[turbid raptor]

A man has to travel 50 km in 4 hours. He does it by walking the first 7 km at x km/h, cycling the next 7 km at 4x km/h and motoring the remainder at (6x + 3) km/h. Find x

[misty bay]

first of all, see if you can figure out how long the motoring distance is

then see if you can write an equation involving x

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[fervent dome]

I have tried to expand the thing needed to be proved

But I failed

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[lean otter]

hey i have a few questions

[lean otter]

i really need help understanding these

these too

also this one

fuck this im doomed

<@&286206848099549185>

please anyone

<@&286206848099549185>

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[final sigil]

why is 5x 180

[thin bridge]

linear pair

total angle on a line

[final sigil]

linear pair?

[thin bridge]

look it up as well as
angle sum on a line
supplementary angles
those should all give you what's relevant here

[final sigil]

thank you

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[narrow vault]

for b), do I use the orthogonal projection formula on each of the vectors u_1 and u_2? and what is the next step? Will I have 2 orthogonal projections, x_1_hat and x_2_hat? because in the formula for orthogonal projection : x_hat=[(x*u)/u^2]*u, I cant multiply the vector by the matrix W right?

[nocturne quiver]

How can I solve this problem

[frank glen]

!help

[safe radish]

Please read #❓how-to-get-help

[frank glen]

This channel is already occupied

Thank you

[narrow vault]

this is my channel right? it says stefan

[stray socket]

It's yours

[narrow vault]

for b), do I use the orthogonal projection formula on each of the vectors u_1 and u_2? and what is the next step? Will I have 2 orthogonal projections, x_1_hat and x_2_hat? because in the formula for orthogonal projection : x_hat=[(xu)/u^2]u, I cant multiply the vector by the matrix W right?

like this?

basically i know how to calculate the proijection from one vector to another. But how do you do it with a matrix?

🙏

at least tell me if it is maybe the sum of all the projections on each vector..

[safe radish]

@narrow vault Has your question been resolved?

[safe radish]

@narrow vault Has your question been resolved?

[wild cape]

yes

you probably should check your class notes to confirm anyway

[narrow vault]

what if I tell you the practice exams dont have solutions and my friends dont know the answers..

[wild cape]

if you think a question involves a certain concept but forgot exactly how it goes then check your notes

[narrow vault]

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[lean otter]

12.012 9.172

lol can i get help pls

[zinc token]

your question is cut off

please post it again

[stray socket]

Btw

You need to show the full question

[lean otter]

that full question

[stray socket]

Are you multiplying the two number on the numerator

[lean otter]

sry

[lean otter]

two

[stray socket]

Percentage is just $\frac{A}{B} \cdot 100%$

[flat frigate]

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

[earnest wagon]

If you're finding the increment percentage, that would be (increased number - original number)/original number * 100

[lean otter]

thzx

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[fresh lion]

sorry my papers dirty lol. anyways, does anyone know what i did wrong?

[safe radish]

@fresh lion Has your question been resolved?

[flat frigate]

Mehdi_Moulati

[shut inlet]

since it's an arithmetic series

[flat frigate]

Mehdi_Moulati

[shut inlet]

you can then solve the quadratic equation

[safe radish]

@fresh lion Has your question been resolved?

[fresh lion]

ngl i aint ever seen that formula before, is there another way i can do it? similar to what i was doing? im assuming there is because i was never taught your method

[shut inlet]

Ehy did you multiply-2 by 2

[fresh lion]

lmao 5 mins after i sent that i realized that

its supposed to divide everything in the brackets by 2 right

[shut inlet]

Yeh

n/2 × (-2+6n-2)

[flat frigate]

Mehdi_Moulati

[fresh lion]

ahh i got it, awesome

thank you very much @shut inlet

idk how to close it lol

!close

[shut inlet]

.close

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[fresh lion]

ah

[shut inlet]

anytime bro

[shut inlet]

this is the rule in general

when you have an arithmetic series
and you want to calculate the sum of series

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[true estuary]

hi

[true estuary]

how did i go from top to bottom?

<@&286206848099549185>

[pine moon]

Aye

...

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[true estuary]

@junior smelt sry to ping mate any luck with this?

[shut inlet]

I think chartbit is asleep

[true estuary]

ah nws then

he deserves it

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[gaunt inlet]

what does a general term mean of an arithmetic sequence?

[gaunt inlet]

what does it do

or how do you use it

[ember sphinx]

So like

The 4th term would = t_4 = -8(4) +11

[gaunt inlet]

Ohhhhh

That makes sense thank you

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[fringe viper]

Is it possible to find all the subgroups of an arbitrary finite group?

[merry swift]

I mean yeah, just write out every possible subset and check if each is a group.

[fringe viper]

sorry there was an error in the question. meant to ask arbitrary group, not necessarily finite

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[fringe viper]

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[timid void]

Suppose a set of test scores is approximately symmetric (bell-shaped), with a mean of 80 and a range of 40. Approximately, what would the minimum and maximum test scores be (in points)?

[timid void]

I don't know how to find the minimum of this.

[hollow siren]

So where is the median approximately on a symmetric curve?

[timid void]

i'd assume in the middle?

[hollow siren]

Yes

the range represents the span of the minimum to the maximum value

So if the range is 40, and 80 is in the middle of that span, what do u think the min and max is

[timid void]

the max is 100. i still have no idea about the min

60?

[hollow siren]

Yup

[timid void]

thank you!

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[stray socket]

Ok so uh, for legal reasons I cannot post an image of what I'm tryna do but I'm with the boys and we decided to graph a dick on Desmos. We have it be defined as f(x)

If I let h be some number, and we want f(x) to horizontally travel across a function, t(x), would that just be f(x + t(h))?

[stray socket]

I

I regret every life decision I've made

[dusk iron]

Graph a what... LMAOOO

[stray socket]

Look buddy

I ain't got nothing much anymore

[dusk iron]

Like in general

?

Dude what's wrong

[dusk iron]

Then be my friend lol

[stray socket]

Anyways, my friends and I wanna know how to make the funny dick travel along a curve

Oh wait

Wouldn't it just be f(x-h) + t(h)?

[junior smelt]

Why did you not define it as d(x)

[stray socket]

NO

[frank glen]

[dusk iron]

[stray socket]

Anyways will this work for all curves? (Assuming t(x) is continuous for all x)

And does t(x) have to be smooth

Because I could just make it spasmodic in motion if possible

I hate everything

[frank glen]

[stray socket]

Will it work though

Regardless of what t(x) is

It will "travel" along that curve

[frank glen]

I guess you can have a look at how SVG curves work and maybe apply the concept there

[stray socket]

Wdym SVG

Oh I see

Idk how to put that into desmos

[frank glen]

[stray socket]

Well that indirectly helped me prove whatever I'm tryna do

Ty

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[solar cradle]

Help

[safe radish]

@solar cradle Has your question been resolved?

[silver venture]

Look closely at the expression on the right with the arrow pointed to it

You can cnel something out

Cancel

[solar cradle]

Can I cancel and x and a y??

[silver venture]

Yep

[solar cradle]

It’s still doesn’t look like an answer

[silver venture]

Are you just simplifying

In that case you can also simplify the sqrt125

And the xs

[solar cradle]

Ohhh

5x square root of 5x?

[silver venture]

Yes

[solar cradle]

Ty

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[little sorrel]

how do find general term

[quasi bison]

it is not entirely clear in what form they want you to do that, but it sounds like you are expected to apply the binomial theorem one way or another.

[little sorrel]

this the answer

ah

ok

i see

mb

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[winged hull]

Let $f:\mathbb{R}→\mathbb{R}$ be a function verifying

$$f(x^2+y)+f(1)=2f(x)y+f(y-1)+f(x^2)$$ $\forall x\in\mathbb{R}

Question : If $f(0)=1$, then what is $f(3)$?
(No information for $y$)

[flat frigate]

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

[quasi bison]

so we are explicitly not told that f(x^2+y)+f(1) = 2f(x)y + f(y-1) + f(x^2) is true for all real x and y?

i.e. the statement of the problem is an open sentence?

@winged hull can you share a screenshot or picture of the problem in its original french

[spiral crescent]

Is the answer 10?

[winged hull]

This is an exercise

Not a problem

But that seems like a problem

I this we can let : x=0

[spiral crescent]

I did x=1,y=1, then y=2,x=0, then y=2,x=1

[winged hull]

Yes this ye equivalent

Thanks I will conclud with that .

[quasi bison]

exercise, problem, no difference whatsoever

i still want to see what it was in french originally

[winged hull]

Yes $\forall x,y\in\mathbb{R}$

[flat frigate]

Pierre de Fermat

[quasi bison]

okay. so why lie?

"no information about y" made it sound like you really were given an open sentence

[winged hull]

This is on an math olympiad exercise

From french

[plucky elk]

ann asked you this 12 minutes ago. are you unable to do this?

[plucky elk]

,rotate

[winged hull]

I took some time...

[plucky elk]

is anything cut off?

[winged hull]

No

This a picture on my fone

Phone*

[plucky elk]

is this word unrelated to your problem?

[winged hull]

This is the let f:R->R be a function In french

"vérifiant"

[plucky elk]

can you just take a picture of the entire problem with every word

[placid kelp]

i'd ask you to take another picture because it does look like there's stuff missing, even if there might not be. 😅

[quasi bison]

such bad cropping

[winged hull]

No , thank i will conclud with that

[quasi bison]

i want to believe this wasn't the result of malice

[plucky elk]

.close if you're done

[winged hull]

Yes thank uu for this .

.close

[safe radish]

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[rapid lantern]

Hello

How do you find capital C?

[marsh raptor]

idt you could do anything other than use a calculator

[junior smelt]

Are you given an interval for where C must be? Is it e.g. in the setting of a triangle?

[rapid lantern]

Yes i think it is a triangle

[rapid lantern]

What do you have to press on the calculator?

[marsh raptor]

are you familiar with inverse of trigo functions?

[rapid lantern]

Oh inverse

[quasi bison]

you "think"?

you're the one who has the original problem statement

[rapid lantern]

I have used the non inverse function on the calculator

[marsh raptor]

is a triangle with that cosine even possible?

[quasi bison]

it is impossible for you to "think" that the problem involves a triangle and not KNOW for certain whether or not it does

[rapid lantern]

Or aka just regular cosine button

[quasi bison]

sure, an obtuse one

@rapid lantern do you have the original problem exactly as stated on hand?

yes or no

[rapid lantern]

Yes

[quasi bison]

ok then show it to us

[marsh raptor]

but trigo only applies to right angle triangles?

[rapid lantern]

Well it doesn’t change my question

[quasi bison]

bit narrow-minded

let's see

,calc (5^2 + 8^2 - 12^2)/(2 * 5 * 8)

[flat frigate]

Result:

-0.6875

[quasi bison]

ok yeah

[rapid lantern]

So yeah

What do you have to do

To find inverse cosine

[quasi bison]

so you tried calculating cos(-0.6875) in your calculator and, predictably, got some bullshit. is that so?

[rapid lantern]

No without caclulator

I wanted to do on the paper

[quasi bison]

impossible.

[rapid lantern]

But i got stuck

[quasi bison]

you can only write C = arccos(-0.6875)

[rapid lantern]

How was the calculator even made then?

Humans made the calculator so should know how to

[quasi bison]

computers calculate approximations to arccos(x) for "non-nice" values

very good approximations, mind you.

but that can only be achieved through sheer computational power that you, with your pen and paper, do not have.

[rapid lantern]

Ooh

[quasi bison]

What do you have to do
To find inverse cosine
show us your calculator and we can tell you where the arccos, or inverse cosine, or whatever button is on it.

[rapid lantern]

Oh thank you one sec

[marsh raptor]

oh right sine rule and cos rule exist

[rapid lantern]

Oh i think i figured it out

Isn’t it cos-1 button?

On the calculator

[junior smelt]

You may need to press e.g. a shift button to get that option

[rapid lantern]

Yes i pressed “2nd”

[junior smelt]

Cool cool, and did that work for you?

(As in do you get the answer you’re expecting)

[rapid lantern]

Umm

I got err:stntax 1:quit 2:goto after doing inverse cosine of -0.6875

[junior smelt]

Alright can you send a picture of the calculator please? And ideally what you’re putting as the input before you get that error

[marsh raptor]

shouldn't there be a bracket at the end

[rapid lantern]

It still gives me an error

Yeah still

[junior smelt]

Hmmm that’s really weird, it shouldn’t give an error for that

,w arccos(-0.6875)

[flat frigate]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by chartbit#3171

[junior smelt]

Ohhhhh no I see what you probably did!

[junior smelt]

Is that a subtraction symbol? As in did you do the big minus?

[quasi bison]

are you in a dark room

[quasi bison]

show us what error you're getting

[junior smelt]

There’s usually a smaller (-) button (or something) that’ll allow you to select negative numbers

[rapid lantern]

Yes!

At least i believe it is a negative symbol

[junior smelt]

Yea what you probably want is something like the purple (taken from my phone calc) and not the blue

[rapid lantern]

Oooh hold on

Yeeeeah

I thought it was the negative symbol but i guess it was subtraction

Jeez

[junior smelt]

Yea that can trip people up sometimes, does it work for you now?

[rapid lantern]

Ah yes

Tysm:)

Little things really help me

I get the value i wanted now

[quasi bison]

mind you this is not about math this is about the calculator specifically

[rapid lantern]

Yay

[rapid lantern]

But i now have a math related question

Instead of calculator

So if you look at the unit circle,

And look at this problem

Value of c is approximately 133 degrees

But if you go down to about 225 degrees on the unit circle, doesn’t it still give you the same cosine value?

[lean otter]

but a triangle cannot have a 225 angle

[quasi bison]

sure, but you know C lies between 0 and 180°

[rapid lantern]

Ooh

Im dumb

Makes sense now

[quasi bison]

given that it is an angle in a triangle

[rapid lantern]

Lol

Gotcha

Forget it lol

But anyways thank you so much for your helps!

Im closing now

.close

[safe radish]

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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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[dark egret]

i'm literally so done

why am i getting different answers for the exact same problem

is it a calculator issue?

whats going on

i need to solve for x in the problem 1 - sin^2 (x) - cos (2x) = 1/2

i should be able to turn that into cos^2 (x) - cos (2x) = 1/2

which i can turn into sin^2 (x) = 1/2

where are all of these answers coming from

the ones in the picture

[final halo]

From solving sin(x) = ±1/sqrt(2)

[random star]

trig eq have infinite solutions 💀

[dark egret]

yeah ik but i'm asking why they're different for the two equivalent equations

look at the pictures

[final halo]

,calc pi/4

[flat frigate]

Result:

0.78539816339745

[final halo]

They're the same

[dark egret]

............

its too late for this

myb

thank you

.close

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[random star]

CAS use lookup tables for solving equations therefore you might end up with different solutions

[safe radish]

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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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[novel pivot]

i’ve drawn this out but idk what to do more

the answer is 39.2 btw also ignore the bit below different question

[empty gyro]

,rotate

[safe radish]

@novel pivot Has your question been resolved?

[stray jasper]

i dont have much time to help but, the only force acting down the slope from the object is its weight

which is 4/sin45 = 55.493....

Then with that weight component acting parallel to the slope, that must be countered exactly by the force P to ensure no movement

so then Sin 45 = O/H, O is P and H is 55.49 which you calculated before. so H x Sin45 = P = 39.2

which is the same as 4g

as its at a right angle and the angle of action is 45 they have to have the same value as the component acting parallel to the slope is in the same exact proportion for both

also sorry if that is confusing explanation but yeah. im not a teacher

[novel pivot]

thanks that makes sense

[safe radish]

@novel pivot Has your question been resolved?

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