help-19

[mystic saffron]

--- ( 20i + 8j )

[odd edge]

@modest void Has your question been resolved?

[buoyant hinge]

no taking the negative would not change the direction

because its still there

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[frigid canopy]

Prove that gcd(5a+2,7a+3)=1

[frigid canopy]

so $k_1(5a+2)+k_2(7a+3) =gcd(5a+2,7a+3)$

[clever fjord]

ƒ(Why am. I here)=I don't Know

[frigid canopy]

I then let $k_1=7, k_2=-5$

[clever fjord]

ƒ(Why am. I here)=I don't Know

[inland rover]

Hmm didn't you say you were gonna do LA this month?

[frigid canopy]

which gave me a gcd of -1 somehow

[frigid canopy]

both

I'll start lin alg after lunch

[mystic saffron]

and what is this? proofs?

[frigid canopy]

Number theory

[mystic saffron]

oh ok

[lavish kite]

well that's true for some values of k1 and k2, it isn't true for all of them

[lavish kite]

this pair of values works in the sense that you can use it to solve the problem, but not in the sense that you literally get the gcd out

[frigid canopy]

but isn't there a therom that states $\frac{a}{d}+ \frac{b}{d}=1$ implies the two are co prime

[clever fjord]

ƒ(Why am. I here)=I don't Know

[lavish kite]

no there isn't, $\frac 26 + \frac 46 = 1$ but none of the numbers involved are coprime to each other

[clever fjord]

bee [it/its]

[lavish kite]

but anyway, looking at this problem

basically the point is that we got -1, and indeed that can only happen if they're coprime

more precisely their gcd must divide -1, and the only numbers that do that are 1 and -1

[frigid canopy]

hmm,ok

got it

[lavish kite]

and in that case we take the gcd to be the positive value, so the gcd is 1

[frigid canopy]

got it

thanks

.close

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[pulsar elbow]

not me though, i’ve picked a terrible system of representatives ;) (hope ur doing well)

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[mystic saffron]

help. I need to know something related to logarithms

[thin kelp]

ok

[mystic saffron]

For which range of numbers is value of a log negative. I mean to say that for a number let's say log x to the base n where n is a natural number, how should 'x' vary so that value of log becomes positive or negative?

[thin kelp]

when x is less than 1

[mystic saffron]

I know that if the base is fractional and number a positive integer, then the logarithm value is negative.

[mystic saffron]

Oh

[thin kelp]

thats also true

lets say we have log base 'a' of 'b' or log_a(b)

if a and b are on either side of 1, then the logarithm is negative

[mystic saffron]

Ohhhh

[mystic saffron]

I get it now. i forgot what our teacher told us😅

[clever fjord]

Bomb

[mystic saffron]

So this logarithmic value will be negative right @thin kelp , because sqrt(13) - 3 is positive and the number will be a positive fraction, whereas e is a positive integer and both will be on either side of 1

[thin kelp]

either side of 1

but yes'

[mystic saffron]

oh yeah

my bad

thankss

[thin kelp]

all good

[mystic saffron]

.close

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[rapid rivet]

I have a 40 by 2024 lattice that contains 80960 points. Some of them are coloured red. Four red points are not allowed to be placed if, when joined together, can make a rectangle with sides parallel to the axes of the lattice. What is the maximum number of red points

[rapid rivet]

i got 820 but im not sure if it's correct

my reasoning is that

if i place a red point in the corner

and if i want to maximize no of points

then i am not able to place a point each on the adjacent axes

so i choose an axes

and then

in 1 column i place the red point in the corner

then in the 2nd column i put it on the axes and then the point below is also coloured red

etc

[tepid pelican]

Im unsure on how to solve it, but just by placing 2024 points in one row, you would get more than 820

[rapid rivet]

oh yeah

💀

uhhh

<@&286206848099549185>

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@rapid rivet Has your question been resolved?

[mystic saffron]

please occupy only one help channel l

as others also need one..

[rapid rivet]

i closed the other one

[odd edge]

@rapid rivet Has your question been resolved?

[static totem]

maybe it's like this

i didn't think much, it's just three lines idk

i mean i thought much, but there was nothing clever

[odd edge]

@rapid rivet Has your question been resolved?

[rapid rivet]

ty

i mean

i had to submit my work

so yeah

i think i got it wrong but oh well

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[foggy ibex]

Hello guys, I am just about to join college doing a major in electronics, But I still want to continue learning math and going in depth, Is there any books that you recommend for self study?

[odd edge]

@foggy ibex Has your question been resolved?

[amber current]

go to #book-recommendations

[foggy ibex]

my bad then

!close

.close

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[gilded flower]

Last question on this page

[gilded flower]

i keep getting 1:1

area of shaded region = 4pi and the bigger circle is 4pi

is the question wrong or are my answers wrong

[stark pike]

what does it mean by big circle? as in the we have to compare the area of the circle of radius 2 and the shaded area of the circle of radius 3?

[gilded flower]

yeah

[thin kelp]

weird i dont get any of those solutions

[gilded flower]

wait

[stark pike]

well then the answer is 1:1

[gilded flower]

im not sure actually

[stark pike]

you are correct

[thin kelp]

am i misinterpreting the question?

[lone elbow]

radius of the shaded is 3

[stark pike]

i dont know why the answer isnt mentioned

[gilded flower]

this happens a lot in this book because it's translated 💢

[stark pike]

ah

[gilded flower]

yeah it means compare area of shaded to the circle with radius 3

[stark pike]

anyway the important part is that you get the concept

[gilded flower]

now it makes sense

[gilded flower]

yeah yeah

[stark pike]

i suggest to move on and solve other questions

great job

[gilded flower]

ok yeah

thx everyone

[thin kelp]

isnt the area of the bigger circle 9pi

[gilded flower]

yeah

the answer is A

4:9

thx

.close

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[thin kelp]

ah right, shaded area is red

mb

yeah

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[copper quarry]

im not sure what to do for this , the only idea i have is finding the stationary points but dk what to do after

[dawn scarab]

It should be periodical, so find one point and add the period? Idrk

[copper quarry]

this is part of the series and sequences topic so

sumtin to do with that

[dawn scarab]

What does being stationary mean exactly here?

Maybe I‘m out of my depth

[copper quarry]

x' = 0 ig

find the 50th stationary point? i have no clue

[dawn scarab]

Well my thought is use the periodicity of the thibg, but if you tell me it‘s not that, I can‘t really help you, sorry :(

[copper quarry]

its ok typo

youll think of something

[thin kelp]

yes

[thin kelp]

this should come into play

[copper quarry]

$x'\left(t\right)=\frac{3\pi}{4}\cos\left(\frac{\pi}{4}t\right)e^{3\sin\left(\frac{\pi}{4}t\right)}$

[clever fjord]

tox was here

[thin kelp]

right

and this must = 0 as u said

[copper quarry]

yes

but it will have infinite solutions unless im given a domain restriction

[thin kelp]

yes but infinite solutions is kinda what we want here

the first solution wiill be the 1st time it is stationary

and the 2nd solution will be the 2nd time its stationary, etc.

[copper quarry]

im looking for $\cos\left(\frac{\pi}{4}t\right)=0$

[clever fjord]

tox was here

[copper quarry]

but how do i find the 50th

[thin kelp]

remember that cos is periodic

so u can keep adding the period and that will be a solution each time

[copper quarry]

but doing that 49 times surely theres a better way right

[thin kelp]

add 49 * period

[copper quarry]

hm

will that work?

[thin kelp]

yes

[copper quarry]

oh wait

i see where the sequence comes in

this is an arithmetic sequence

[thin kelp]

ah yes

that is correct

good spot

[copper quarry]

2,6,10....

$T_{n}=a+\left(n-1\right)d$

[clever fjord]

tox was here

[thin kelp]

yes

but the common difference is the period

so you'll still end up adding 49 lots of the period

anyways

[copper quarry]

$T_{50}=2+\left(50-1\right)4=198$

[clever fjord]

tox was here

[thin kelp]

yeah this sounds about right

[copper quarry]

thats a clever question

[thin kelp]

hmm yeah

[copper quarry]

anyways thanks, btw shouldnt u be studying for ur ext 2 test or sleeping 🤣

[thin kelp]

yeah i most definitely should be doing one or the other

[dawn scarab]

Common Galaxy W

[copper quarry]

lmao

[copper quarry]

i finished all of my tests so now its just the trials next term, hbu

[thin kelp]

2 more to go, maths ext 2 on wednesday and a software design test on thursday

[copper quarry]

ah GL

i just had my last assessment today which was a phys report

[thin kelp]

yeah gl for the trials man

[copper quarry]

u too

[thin kelp]

ah nice, think u did well? I think i flunked my last physics test

[copper quarry]

i hope so 😂 3.2k words and it was on the wave model of light

[thin kelp]

holy

[copper quarry]

oh damn what was it on

[thin kelp]

all the theory for the nature of light, excluding special relativity

[copper quarry]

sheesh

pretty big test then

u guys have started mod 8?

[thin kelp]

yeah my teacher somehow managed to pull a 1 part 15 mark question

[thin kelp]

nah, we're behind as hell

[copper quarry]

lmfao damn

[copper quarry]

rip

i gotta finish mod 8 by end of holidays to prep for trials too

[thin kelp]

bruh we'll be learning content after the trials 💀

[copper quarry]

i think same for us as well

[thin kelp]

oh

[copper quarry]

but im just studying ahead of the class

cuz things are slow and they arent a good teacher lol

[thin kelp]

amen to that

[copper quarry]

my phys teacher is doing mod 8 rn

but

they skipped special relativity and energy mass equivalence

lolololol

[thin kelp]

ohh

that aint good

[copper quarry]

she said she will come back to it later but kinda weird still

anyways

go study for ur ext 2 😂 or go sleep

sleep is a good option

im gonna go sleep now

[thin kelp]

aight gn, i cant sleep im in too deep

[copper quarry]

lmao alr

gn and gl

.close

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[lean saddle]

In an math competition there were 10 exercises.If the student in one exercise answers right he gets 10 points.If wrong -2 points and if he lets it blank he gets 0 points from it.If in the end he got 36 points how many blank exercises he let?

[fickle coral]

What if he is brave and takes risk of solving all and gets 5 right and 7 wrong

[lean saddle]

There are 10 exercises

[fickle coral]

Right I didn't see

[vivid girder]

a mix of both may be used, you can make an equation system with 3 variables and 2 equations, then you try to set some bounds for $y$ and find it using the system. Since the system has more variables than equations you have guaranteed infinitely many solutions for $y$, among them you choose the most adequate.

[clever fjord]

Crystopher

[vivid girder]

in this case I let $y$ be the amount of blank answers.

[clever fjord]

Crystopher

[lean saddle]

And?

[fickle coral]

I would try to make equations

x right

y wrong

z left

10x-2y=36

[lean saddle]

Bro ik that but idk how to make the equation

[fickle coral]

x+y+z=10

[lean saddle]

I did this too

Yeah

I stopped here

Couldnt do more

[fickle coral]

5x-y=18

I divide by 2

y=5x-18

Put in x+y+z=10

x+5x-18+z=10

[lean saddle]

6x+z=28

[fickle coral]

Yes

Z=28-6x

But z is positive Integer

28-6x>0

6x<28

This gives choices for x

Because x is also positive Integer

[lean saddle]

1,2,3,4

1 no way

2 no way too

3 no way

So only answer is 4

[fickle coral]

I think so

[lean saddle]

10x3-.......≠36

And so on

[fickle coral]

4,2,4?

[lean saddle]

Exactly

[fickle coral]

Oh ok

[lean saddle]

Thanks

👍

[fickle coral]

[lean saddle]

.close

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[slate spear]

Need help

[odd edge]

@slate spear Has your question been resolved?

[slate spear]

bruh

[slate spear]

.close

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[charred yarrow]

HOW is this wrong?

[charred yarrow]

<@&286206848099549185>

[elfin zodiac]

How did you get to the answer 0.001?

[mystic saffron]

So?

[elfin zodiac]

@charred yarrow

[charred yarrow]

umm

so

i millisecond has 0.001 s

so i kept it as a answer

[elfin zodiac]

But the question is not what is a millisecond in terms of seconds

Read the question again

[charred yarrow]

degrees per seconds

so 1 / 0.001?

[elfin zodiac]

If the speed is 7 degrees each millisecons

When a second happens, how many degrees?

[charred yarrow]

this is true in the question?

oh yeah

[elfin zodiac]

That is the first statement

[charred yarrow]

um 7 divided my 0.001?

7000

[elfin zodiac]

That looks better

[charred yarrow]

oh

for the last 3 days im stuck on questions like these

Thx

can I dm u if there are related questions? @elfin zodiac

[elfin zodiac]

No, i have blocked all private messages

[charred yarrow]

why

[elfin zodiac]

Just write here

[charred yarrow]

ok

[elfin zodiac]

In other channel

!done

[odd edge]

If you are done with this channel, please mark your problem as solved by typing .close

[charred yarrow]

ok

bye!

.close

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[marsh hound]

hello

[marsh hound]

im working on intervals

so A is clearly an interval

c and d are more exclusion restrictions

so that leaves b as the other interval

and the solution says the interval is the set of all real numbers

im assuming this is because x can be any value greater than -1, or any value less than or equal to 2, and so there will be an overlap and the value of x can take any real value

is this correct?

also, here are you able to make a distinction between open and closed sets?

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@marsh hound Has your question been resolved?

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@marsh hound Has your question been resolved?

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@marsh hound Has your question been resolved?

[marsh hound]

.close

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[uncut arrow]

how to solve the following problem:

Let $ T(R^3) \leftarrow R^2 $ be a linear transformation defined as,\
\begin{equation*}
T(a,b,c) = (a+b,2a-c).
\end{equation*}
Then find out $T^{-1}(1,11) $.
\end{document}

[clever fjord]

shanks44

how to solve the following problem:

Let $ T(R^3) \leftarrow R^2 $ be a linear transformation defined as,\\
\begin{equation*}
  T(a,b,c) = (a+b,2a-c).
\end{equation*}
Then find out $T^{-1}(1,11) $.
\end{document}

[sand horizon]

a+b + 0c = 1
2a +0b -c = 11
Then you put it like matrices

[tulip niche]

hmm, this should read $T : \mathbb{R}^3 \to \mathbb{R}^2$ be a linear transformation

[clever fjord]

Pseudonium

[uncut arrow]

I tried to solve it in the following way :
a + b + 0.c = x, and
2a + 0.b -c = y

[uncut arrow]

yes yes

[uncut arrow]

yes but then another equation can be obtained - 2b + c = -9

[sand horizon]

Yeah and from this i let u cook i gtg

[uncut arrow]

but I think I am lacking the basic concepts

[tulip niche]

oh?

what do you mean?

[uncut arrow]

there is a confusion, is it normal for a system of linear equations to have yet another linear equation that was not mentioned previously

[tulip niche]

why not?

[uncut arrow]

was it because there are 3 unknowns and 2 equations ?

[tulip niche]

im not quite sure what you mean by another equation not mentioned previously

like

$a + b = 1$

[clever fjord]

Pseudonium

[tulip niche]

if you double this, you get $2a + 2b = 2$

[clever fjord]

Pseudonium

[tulip niche]

would that count as another equation not mentioned previously?

[uncut arrow]

actually 2b + c = -9 was obtained

[tulip niche]

yes or no to this?

[uncut arrow]

I donot understand

[tulip niche]

does $2a + 2b = 2$ count as another equation not mentioned previously?

[clever fjord]

Pseudonium

[tulip niche]

there’s not a right or wrong answer here

[uncut arrow]

no

[tulip niche]

ok!

so now we have $2a + 2b = 2$

[clever fjord]

Pseudonium

[tulip niche]

and $2a - c = 11$

[clever fjord]

Pseudonium

[tulip niche]

and then you subtract them, right? to get $2b + c = -9$

[clever fjord]

Pseudonium

[uncut arrow]

yes

[tulip niche]

so - if 2a + 2b = 2 didn’t count

why would this count as a new equation?

again, not a right or wrong answer here

[uncut arrow]

variable a is absent ?

[tulip niche]

hmm i see

that’s fair

i guess - for me, a new equation just means one we didn’t write down at the start

in which case it’s perfectly normal to get new equations as you solve something

so for me, $2a + 2b = 2$ is a new equation

[clever fjord]

Pseudonium

[uncut arrow]

but how ?

[tulip niche]

as is $2b + c = -9$

[clever fjord]

Pseudonium

[tulip niche]

they’re both new

at least, the way i see things

but this is more a matter of language

your manipulations are correct

[uncut arrow]

so now the system can be represented as -
[
\begin{matrix}
1 & 1 & 0 &1 \
2 & 0 & -1 & 11 \
0 & 2 & 1 & -9 \
\end{matrix}
]

[clever fjord]

shanks44

[tulip niche]

sure if you want

[uncut arrow]

oh but writing the 3rd row is not needed

sorry

[tulip niche]

yep

it’s ok

[uncut arrow]

it comes from the first two rows

[tulip niche]

mhm

[uncut arrow]

the following equivalent matrix is derived

[
\begin{matrix}
1 & 0 & \frac{-1}{2} & \frac{11}{2} \
0 & 1 & \frac{1}{2} & \frac{-9}{2} \
\end{matrix}
]

[clever fjord]

shanks44

[uncut arrow]

but from here how to reach the solution

[tulip niche]

try expressing a and b in terms of c

[uncut arrow]

yes then a = 11/2 + c/2
and b = -9/2 - c/2

and there are infinite solutions

[tulip niche]

yep!

[uncut arrow]

for c = 0, a = 11/2 and b = -9/2

answer given is :
\begin{equation}
T^{-1}{(1,11)} = { (11/2 , -9/2 , 0) + t(1 , -1 , 2) : t \in R }
\end{equation}

[clever fjord]

shanks44

[tulip niche]

that should work!

[uncut arrow]

the first part is understood for c = 0

what is the meaning of the 2nd part ?

[tulip niche]

ah well

you know (a, b, c) = (11/2 + c/2, -9/2 - c/2, c)

[uncut arrow]

yes

[tulip niche]

you can write this as (11/2, -9/2, 0) + c(1/2, -1/2, 1)

here $c \in \mathbb{R}$ is arbitrary

[clever fjord]

Pseudonium

[tulip niche]

you can also write c = 2t

[uncut arrow]

but why the particular format is mentioned

ok, that may be a convention

thanks

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[shut turret]

is this right? r is 2/7 which is less than one so i think converges
then 8/(1-2/7)
bc 8 is first term and then the a1/(1-r) formula

[shut turret]

oh wait the first term is 51/7

is the answer 51/5?

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[undone crypt]

is this right?

[undone crypt]

I was asked to traced the first derivate using the original graphic, but I dont think tha flatline should be like that

I made it a little rounder in the extremes but still, that flatline is not something im sure about

This is the original Graphic

I have seem some examples of how to do it, and i must say im lost

<@&286206848099549185>

[plain falcon]

Doesn't look right

[undone crypt]

yeah, i think so too

[green sage]

for one, you should get a cubic

[plain falcon]

There are increasing and decreasing sections in-between the minima and maxima

Should look like this

Every hill should turn into a downwards slope and every valley into an upwards one

[undone crypt]

arent the "hills/slope" were my graphic should go to zero in theirs m? edit: nevermind, I understand now

how do i know where new, "hills/slopes" should be make?

allright i think i see some more tutorials and see if a understand them better now

Thanks

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[steep mantle]

can you get away without using trig on this

[odd edge]

@steep mantle Has your question been resolved?

[muted stirrup]

hii

[steep mantle]

heloo

[muted stirrup]

I'dont understand english

[steep mantle]

bruh

[muted stirrup]

Can you send me the extract? on discord, so I can put it in the translator?

the text question

[viscid flint]

not easily for sure

[steep mantle]

ok

how do i do it with trig

[steep mantle]

use google lens

[steep mantle]

i can get AC with trig, but idk what to do next

[viscid flint]

do you have the height of A as well?

[steep mantle]

uh

idk how to

[muted stirrup]

uh

[viscid flint]

use the area of ADC and the angle C

[steep mantle]

uhhhhh

i need dc no?

[viscid flint]

oh yeah hmm

[steep mantle]

wait can i usd angle b and AB

[steep mantle]

ok

[mystic saffron]

ye

[steep mantle]

um

AB is 2

since a=1

[mystic saffron]

ye

[steep mantle]

wait

ac is a+swrt3 a

[mystic saffron]

ye

[steep mantle]

height of a/sin(75)=2

[fluid tundra]

wdym "without using trig"

[steep mantle]

idk i wanna knoe if i can do it without trig

but its fine

[mystic saffron]

you dont nesd to do thaf bro

[steep mantle]

huh

[mystic saffron]

u know a so u know the area of abc

n adc is given

so abd is abc - 2

[steep mantle]

fight

[fluid tundra]

(for reference, you don't need the 15-75-90 triangle at all for this problem)

[mystic saffron]

(the shortest distance)*(ab)*1/2 = abd

[steep mantle]

ok

[mystic saffron]

ye

[steep mantle]

wait do i use herons?

[mystic saffron]

idk is it solving this cube?

[mystic saffron]

no

[steep mantle]

how would i find abc then

[mystic saffron]

U know ac n the height

[steep mantle]

oop

is the shortest distance (3+sqrt3)/2

ok then thanks yaal ill take that as a yes :3

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[lapis palm]

How do I find the asymptotes of f(x)=2/x-3

[lone elbow]

for vertical

asymptote

you take the limit from both sides as x approaches 3

[lapis palm]

what is the limit?

how do I find it?

[lone elbow]

you havent learnt limits tho?

[lapis palm]

nope

its college algebra

I havent done algebra in 10 years

I really sorry I know I am inept

[lone elbow]

well if I am not mistsken we will be needing limits here though

so what are we

[lone elbow]

whats ur age

[lapis palm]

28

I was in the army for 9 years and am getting a degree in cyber security

am I screwed?

[latent scaffold]

I mean you can also just look for places where the denominator is 0.

You want it to be (some nonzero constant) over 0

[lone elbow]

u are not screwed u are good and ready to go

[lapis palm]

so how do I find it?

[latent scaffold]

What is the denominator in your expression?

[lapis palm]

x-3

[latent scaffold]

Ok, so you want to find values of x where x-3 = 0

Can you solve for x?

[lapis palm]

so 3

[lone elbow]

when x is 3, i wanted to know if this would be a horizontal or vertical or oblique asymptote

[lapis palm]

horizantal right x-axis?

[desert marlin]

I've seen not much of your work but from what I've seen I think you'll be just fine. Looks like you're open to learning and you're here working so if you just keep that up I think you'll do great

[latent scaffold]

Realistically, if you've not been introduced to limits (I suppose you're not doing calculus, just learning rational functions), you will likely mostly deal with vertical asymptotes for now, like this one.

Horizontal asymptotes are a bit harder to explain without taking limits and such, although they might be taught as a canonical formula for rational functions, or "graphically".

[lone elbow]

okay mb

[lapis palm]

yeah we should have any horizontal or diaginal

[lone elbow]

so at x = 3 we have a vertical asymptote then

?

[lapis palm]

I guess? XD

[lone elbow]

lets see the graph

[lapis palm]

is it one or 2?

[lone elbow]

when you solved

for x

did you got one solution or two

[lone elbow]

.

[lapis palm]

1

[lone elbow]

,w solve x-3=0

[clever fjord]

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[lone elbow]

only one

,w plot 2/(x-3)

[clever fjord]

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[lapis palm]

so for the option of 1 asymptote it wants me to give an equation

[lone elbow]

option?

[lapis palm]

yeah

[lone elbow]

x = 3

[lapis palm]

yup

now it wants me to identify horizantal asymptotes

same problem

so how do I find the horizantal ones?

[lone elbow]

for the same function?

[lapis palm]

yup

[lone elbow]

you need limits

tho

[lapis palm]

@<@ what are limits?

[lone elbow]

this is the formulas for finding oblique

[lapis palm]

T_T never seen that even once

and I have attended class every day and taken thuro notes

[lone elbow]

show your notes on horizontal and oblique

!show

[odd edge]

Show your work, and if possible, explain where you are stuck.

[lone elbow]

vertical asymptote was what we did first

[lapis palm]

I just dont know how to do it from reading my notes and I dont have that formula

T_T im sorry

[lone elbow]

but

[lapis palm]

I'm retared I'm sorry

[lone elbow]

oblique asymptotes

is not present in your pictures

only vertical and horizontal

[lapis palm]

I dont think we are doing oblique

[lone elbow]

.

u sure?

[lapis palm]

I was but now I am not

[lone elbow]

https://media.discordapp.net/attachments/903463355619110912/1257569329382428782/image.png?ex=6684e234&is=668390b4&hm=294325653ddba987a67573e5d3fade5f7ef69a945ac7061480f92ffc4f8c3949&

[lapis palm]

@<@

I am retarded

[lone elbow]

[lone elbow]

.

[lapis palm]

this is embarasing maybe I am just bad at reading

[lone elbow]

but

do u covered oblique or no

or u dont remember?

[lapis palm]

I guess so if oblique also means horizantal

is it y=0?

[lone elbow]

wdym

[lapis palm]

cause 2<x-3

?

[lone elbow]

no

leading coeficient top / leading coeffi bottom

[lapis palm]

so 2/1

2

[lone elbow]

lesding cefficient top is 0

[lapis palm]

how would I know that without seeing the graph?

[lone elbow]

0*x + 2

[lapis palm]

how is the coefficent 0 why wouldnt it be 1?

[lone elbow]

leading coefficient of a polynomial for a homographic function

a homographic function is the quotient of two polynomials

or something

[lone elbow]

f(x) = 0*x + 2 = 2

,, \frac{f(x)}{g(x)} = \frac{2}{(x-3)}

[clever fjord]

938c2cc0dcc05f2b68c4287040cfcf71

[lone elbow]

using limits this is more understandable imo

[lapis palm]

please

[lone elbow]

well

but do you know the basics of limits?

,, \lim_{n \to \infty} = 1/n =?

[lapis palm]

not at all

[clever fjord]

938c2cc0dcc05f2b68c4287040cfcf71

[lapis palm]

what is the variable n represent?

[lone elbow]

is an variable n that grows up to infinity

,calc 1/999999999999

[clever fjord]

Result:

1.000000000001e-12

[lone elbow]

,w 1/999999

[clever fjord]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

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[lapis palm]

I dont really follow?

[lone elbow]

is a really small positive number

[lapis palm]

ok

[lone elbow]

0.0000000000000000000000000000000001 or something

anyways the limit of 1/n as n goes to infinity is 0

[lapis palm]

ok so what does that mean in relation to the equation?

[lone elbow]

sure

,, f(x) = \frac{2}{x-3}

[clever fjord]

938c2cc0dcc05f2b68c4287040cfcf71

[lone elbow]

if we take the limit for x as it goes to infinity for negative infinity and positive infinity both sides give 0

I mean

we already knew the horizontal asymptote was y=0

due to top coefi / bottom coefi

but we were just making sure with limit

,w limit x to infinity 2/(x-3)

[clever fjord]

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[lone elbow]

,w limit x to -infinity 2/(x-3)

[clever fjord]

Results provided by WolframAlpha

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[lapis palm]

how do I write that?

[lone elbow]

y=0

[lapis palm]

and oblique = horizontal?

[lone elbow]

?????? wdym

[lapis palm]

nvm sorry

they are different

[lone elbow]

but you lowkey dont have the notes for oblique

[lapis palm]

I know

[lone elbow]

i mean

so what

<@&286206848099549185>

[lapis palm]

.close

[odd edge]

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[lapis palm]

Thank you boss

I appreciate it I will continue to apply this

Im gonna try a few on my own now

[lone elbow]

keep doing exercises

u got this

[lapis palm]

Thanks man I hope if I botch this test I am gonna fail

[odd edge]

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[torn rune]

i'm legally blind visually imapred or rather low vision would be a better way to put it so it's hard for me to read all the rulses and guuide lines but i was wondering if someone or someones wouldnt mind video chatting with me for a bit so i can share some of hypothsothysysys and to learn how to do video and audio on on discord

[torn rune]

?

[odd edge]

@torn rune Has your question been resolved?

[torn rune]

no

[ember oak]

Most people won't want to video chat. Sorry

[torn rune]

thankyou

[ember oak]

What is your hypothesis

[odd edge]

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[empty cosmos]

I know the answers here but I don't know how to express them mathematically

[steep mantle]

venn diagram might help

[empty cosmos]

Idk if that counts but I could try it

[steep mantle]

it should

[empty cosmos]

From a "showing work" perspective not a "getting the right answer" perspective

[steep mantle]

huh

um

idk then, the only way i know is venn diagram but i know very little about set theory

maybe wsit for a helper who knows set theory well

[empty cosmos]

Wait it just occurred to me that I don't actually know the answer for B

Because I don't know how much of the 0.3 would be taken out of set A or B respectively

[steep mantle]

venn diagram might help tbh

[empty cosmos]

I tried that but I'm having trouble visualizing it

Is it just 0.1?

[steep mantle]

could you use venn diagram to show your work?

[steep mantle]

i think so?

[empty cosmos]

I don't have an easy way to post one here

I am shit tired and really need to go to bed this homework is not worth it 💀

[empty cosmos]

Does that mean the P(A U B) is actually 0.6 and not 0.9?

[steep mantle]

afaik ye

[empty cosmos]

Ok

[odd edge]

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[lucid sonnet]

$y = \ln \cosh v - \frac{1}{2} \tanh^{2} v \ $
$\frac{dy}{dv} = \frac{1}{ \cosh v} \cdot \sinh v - \cdot \frac{1}{2} 2\tanh \cdot\sech^{2} v$

[lucid sonnet]

is this correct so far?

<@&286206848099549185>

[mystic saffron]

nope

theres a slight mistake

[supple zealot]

-sinhv

[lucid sonnet]

Isn't the derivative of cosh positive sinhv I think it differs from how it'd work with just sin.

[supple zealot]

Derivative of cosx is -sinx

Derivative of sin is cos

[lucid sonnet]

those are hyperbolic functions above

[spring nacelle]

You are correct

[lucid sonnet]

so is this correct so far?

[spring nacelle]

Sech instead of sec

[supple zealot]

No there's slight mistake

Why is there a tanh²v

It's correct if take out that part

[lucid sonnet]

ah yes I shouldnt have that

[clever fjord]

Tomi

[supple zealot]

Yes

This is correct

Now it's just 2 steps

Tanh³x

[lucid sonnet]

$\frac{dy}{dv} = \tanh v - \cdot \tanh v\cdot\sech^{2} v$

[supple zealot]

Tanhv that side

[clever fjord]

Tomi

[supple zealot]

Yes

[lucid sonnet]

wait how is this tanhv to the power of 3?

[supple zealot]

Tanhv(1-sech²v)

Tanhv(tanh²v)

Tanh³v

Got it?

[lucid sonnet]

yes, thank you

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[mystic saffron]

I am supposed to simplify similar terms
But idk where to start

[shut wedge]

Try grouping the terms inside the brackets first

[mystic saffron]

-6/6A?

[shut wedge]

Whats your thought process behind this

Just curious

[mystic saffron]

I mean
I wanna turn the fractions in same denominator

[mystic saffron]

So, -A=-6/6A ?

Is that possible?

[shut wedge]

Yep

Very possible

[mystic saffron]

So, the brackets would be 3/4B-6/6A-6/6B-1/6A
Right?

[shut wedge]

Remember, A and B are different terms

Make the B into the/4

[mystic saffron]

Right, so -B=-4/4B

So, now should I simplify them

-6/6A-1/6A

[shut wedge]

Yep

[mystic saffron]

-5/6A?

Or 7/6A because of the - and -

[shut wedge]

You add two negatives

So yes it would be -7/6A

[mystic saffron]

So, is it [-1/4B-7/6A]

[shut wedge]

Mhm

Then what do you think you do next

[mystic saffron]

First, the symbols, right?

There is a negative before the brackets

So it would be 2/3A+1/4B+7/6A-2/5B?

[shut wedge]

Yep

[mystic saffron]

So, I got 33/18A-3/20B

I think I did something wrong there

Oh, nevermind

I got 11/6A-3/20B, that's the answer

[shut wedge]

Nice

Gj man

[mystic saffron]

Thank you, I've got another one

[shut wedge]

Trust yourself more you basically did this by yourself

[mystic saffron]

Yeah, I just needed some guidance and self trust

[mystic saffron]

Can I get help w that one?

[shut wedge]

Let's go

[shut wedge]

What do you think you do first

[mystic saffron]

The parenthesis, so it would be -11b+4a^3?

[shut wedge]

Mmmm

Not quite

If there are two pronumerals together like the a^2b, thats a different term than a^2

Just like how a^3 is different to a^2

[mystic saffron]

Isn't 4a^3=4a^2*a?

[shut wedge]

Yea, but you wouldn't add 4a^3 and a^2 together

Anyway, the a^2b is its own term, and the a^2 is its own term. So inside the bracket,
there are no like terms

[mystic saffron]

So the only thing I can do is operating the symbols, right?

a^3-a^2-11a^2b-4a^3+a^2+3a^2b

Is that correct?

[shut wedge]

Yep

[mystic saffron]

The answer would be -3a^3-8a^2b

[shut wedge]

Yep

[mystic saffron]

Is the final result -17x?

[shut wedge]

Should be

Gj

[mystic saffron]

I think this one is gonna be difficult

[shut wedge]

Hmm

Whats your first step

[mystic saffron]

First, the parenthesis

But, the thing is the exponent of the -3b^1/3

Which is kinda scary

[shut wedge]

its fine

Dont pay attention to it

[mystic saffron]

I can add the 12a^4+4a^4

3/8ab^6 can be operated w -1/2ab^6, right?

[shut wedge]

Yep

[mystic saffron]

So I got -1/8ab^6

[shut wedge]

Whered the b^1/3 go

We didnt do anything to it but its still there

Also the a^4

Oh is this only your ab^6 term?

[mystic saffron]

Yeah, in the final parenthesis would be:
(16a^4-1/8ab^6-3b^1/3)

[shut wedge]

Yep

Looks good

[mystic saffron]

Yeah I just realized I can -1/3ab^6-1/8ab^6

Right?

Actually, it's plus

[shut wedge]

Mhm but theres a negative on the bracket si be careful

[mystic saffron]

Because of the symbol

[shut wedge]

Nice

[mystic saffron]

yeah

so the result of that would be -5/24ab^6

[shut wedge]

Yep

[mystic saffron]

So the final result is -5/24ab^6-16a^4+3b^1/2

[shut wedge]

Typo? +3b^1/3

But yep

Seems good

[mystic saffron]

Oh, yeah

1/3

[odd edge]

@mystic saffron Has your question been resolved?

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[mystic sail]

Hello : D, in the first screenshot you can see what the exercise is asking me to do (to find the range of k ).
First of all I do not understand from where 1/5 comes from (watch the solution), moreover I need help to understand the range of k given the graph ( 1 sol means that the straight line only hits one time the elipse) .
Steps I did:
Found A and B points of Elipse.
Found K(s) based on those points.
Found tuition based on those K.
Then I solved the system and found another K
Then I found another tuition based on this K.
Then I have put everything in the graph

[mystic sail]

I hope y'all can understand my handwriting, ik it's pretty bad

Thanks in advance

<@&286206848099549185>

[devout silo]

!15min

[odd edge]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[devout silo]

anyway, could you tell me what happened here?

[mystic sail]

I just rewrote the elipse equation in a better form

" means same as before

[devout silo]

well the question gave you an eclipse but you somehow transformed it into a circle without changing the bounds

[mystic sail]

I did not transform it into a circle, it's still an elipse from what I can see

a^2 is 1 and b^2 is 1

no?

[sweet dagger]

No one:
The question: 🌑🌍☀️

[devout silo]

well, your result should have been x^2 + 16y^2 = 16 instead of the x^2 + y^2=1 written in your work

[mystic sail]

I have transformed sqrt(16) into 4 and then moltiplied with the fraction

[mystic sail]

😶

[devout silo]

ok does this mean sqrt(16-x^2) or sqrt(16) - x^2?

[mystic sail]

2nd

[devout silo]

yeah that's the wrong equation

the question gave you y = 1/4*sqrt(16-x^2)

[mystic sail]

ooooh

you are right

epic fail

[devout silo]

also if you are writing y = 1/4*sqrt(16) - x^2, you still can not get x^2+y^2=1

these two equations are not equivalent

[mystic sail]

yeah cause it may be +-4 I guess right?

[devout silo]

no, sqrt(16)=4, I have no problem with that.
If y = 1/4*sqrt(16) - x^2 then y^2 = (1-x^2)^2

[mystic sail]

oh

I forgot the ^2

ok so

before I close the issue

could I get your opinion on a similar problem?

thanks btw

[devout silo]

you can keep this open if you still have any questions.
If you do decide to close here though, please note that when squaring both sides, it will give you extraneous solutions.
You are not considering the ellipse as a whole, but only a portion of the ellipse.

and yes, you can ask questions on another problem as well

[mystic sail]

no ok yeah, if it's ok for you I would like to try to do the calcs and then see if the result is the right one

while keeping the thread open

[devout silo]

sure

[mystic sail]

thx

[odd edge]

@mystic sail Has your question been resolved?

[mystic sail]

Alright so seems like it's right but what I do not understand is: Why don't we take solutions under the orange line?

[mystic sail]

Feel free to ping me when you will be able to answer back

(thx)

[odd edge]

@mystic sail Has your question been resolved?

[devout silo]

@mystic sail sorry for the late response, had something do
Results looks correct (assuming you got 0.2<=k<2). I'm too lazy to go through the solving bit.

The reason is because the bit under the orange line don't actualy 'exist'

Remember that the question gave you y=1/4sqrt(16-x^2). These are the points (x, y) that satisfies the equation

notice that y is non-negative here, that is because square root function is defined to return non-negative values.
But remember when you squared both sides of the equation to get x^2+16y^2=16, without any constraint, you are allowing y to be negative. And thus when you graph it out, you see the whole ellipse.

[mystic sail]

ah alright, got it. Thanks you, one final question before closing the issue

Here there were no square root. Altought why don't we take values under blue line

nvm

I got why

[devout silo]

👍

[mystic sail]

tysm

have a great day

[devout silo]

np you too

[mystic sail]

.close

[odd edge]

Channel closed

Closed by @mystic sail

Use .reopen if this was a mistake.

[odd edge]

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[lethal wharf]

Can anyone help me integrate this as quickly as possible:

[clever fjord]

Baby Einstein

[lethal wharf]

Which is equal to

[clever fjord]

Baby Einstein

[lethal wharf]

Can i use this ?

[clever fjord]

Baby Einstein

[mystic forge]

consider ∫csc(x)csc^2(x)dx

then use identities to get two separate integrals

[lethal wharf]

.

Can i use this

[mystic forge]

idk i havent tried that

is that a substitution?

[lethal wharf]

Is it the parametric formula of sin? If you know

[mystic forge]

how do u derive it?

[lethal wharf]

For now lets try this

What do I have to do ?

[mystic forge]

do u know csc^2(x) identity with cotx?

[lethal wharf]

Mm no

I do not remember

[mystic forge]

this one

[lethal wharf]

Ah ok

So csc x •( 1 + cot ^2 x)

[mystic forge]

yeah remember the brackets

now distribute the integral

u shud get ∫csc(x)(1+cot^2x) dx = ∫csc(x)dx + ∫csc(x)cot(x)^2 dx

[lethal wharf]

ah csc x + csc x cot^2 x

[mystic forge]

yups

now lets rewrite cot^2 x as cotx * cotx

integral of cscxdx is standard

u shud know that right

[lethal wharf]

(-) log (| csc x + cot x|)

[mystic forge]

yeah

now

∫csc(x)cot(x)^2 dx focus on this

we have ∫csc(x)cot(x)cot(x) dx

[lethal wharf]

Yes

[mystic forge]

do integration by parts with u=cotx and dv=csc(x)cot(x)dx

and rewrite the integral

[lethal wharf]

Wait

Im back

@mystic forge

[mystic forge]

did u find the new integral?

[lethal wharf]

i write u = cot x , du = -(csc x)^2

dv = csc x cot x dx

v = -csc x

Now what should I do?

[mystic forge]

we get that ∫csc(x)cot(x)cot(x) dx=-cscxcotx - ∫-cscx*-(cscx)^2dx

so we get -cscxcotx - ∫csc^3(x)dx

have u heard of cyclical integrals?

[lethal wharf]

Why ?

[mystic forge]

by parts formula

∫udv = uv - ∫vdu

[lethal wharf]

👍

And now ?

[mystic forge]

Lets define a new function: I(x) = ∫csc^3xdx

so we have I(x) = -ln(cscx+cotx) - cscxcotx - I(x)

so 2I(x) = -ln(cscx+cotx) - cscxcotx

now solve for I(x)

[lunar fulcrum]

true, although there is an easier way by finding the function

the new function can be simplified by this

l(x)= ∫csc^3xdx is a sufficient method

but

[lethal wharf]

How

[mystic forge]

u add I(x) on both sides

from I(x) = -ln(cscx+cotx) - cscxcotx - I(x)

[lethal wharf]

Oh Sorry

I mean this

[mystic forge]

divide 2 on both sides

[lethal wharf]

Ok

Then ?

[mystic forge]

then boom

u have I(x)

u have ur answer

[lethal wharf]

eeeeeeasy 👍👍👍👍

Thanks !

.close

[odd edge]

Channel closed

Closed by @lethal wharf

Use .reopen if this was a mistake.

[odd edge]

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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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.