help-39

[deep turret]

yep

[junior carbon]

this...........

pre calc 11 stuffs

[deep turret]

uhhh

uhhhh

i know this

[junior carbon]

let me show u the last question

[deep turret]

show me!

[junior carbon]

this was previous

[deep turret]

okay make

make it uhhh

dean 😭

damn

RAGHHH I UED TO KNOW THIS

[junior carbon]

@strong dune

DWKI)DAWKODKWOADKOAWDKO

.close

[pearl pond]

Channel closed

Closed by @junior carbon

Use .reopen if this was a mistake.

[deep turret]

oh just

times it all by root 5a/3

right?

@junior carbon

idk

[junior carbon]

HMMMMMMMMMM

i think so actually

[strong dune]

holaaaa

[junior carbon]

OMG VLAN!!!!!!!!!!!!!!!!!!!!!!

KIYOOOOOOOOOOOOOO

[strong dune]

😁

[junior carbon]

HES ON!!!!

yay

goodmorning!

[deep turret]

do .reopen

[junior carbon]

.reopen

[pearl pond]

✅

[junior carbon]

vlan i have like 15 rationalize denominator questions

i can only solve the super basic ones

u thought me how to solve congajate questions tho

that was amazing

wont forget that

[deep turret]

I GOT IT

I GOT I GOT IT

[junior carbon]

TAF!!!!!!!!

[deep turret]

rewrite it as

a

$6*((5a)^1/2)/(3)^(1/2)$

damn 😭

that looked worse

[junior carbon]

square everything TAF!

Ok imma try

[deep turret]

okay re write it as 6 * (root 5a)/(root 3)

do that ^^^^

[strong dune]

i need to restart this garbage

[junior carbon]

ok i basically rewrote the question taf

[deep turret]

and then do the fraction bit

[junior carbon]

oh ill make the denominators bigger

so i can diivide

right?

[deep turret]

and then times the fraction by root 3

[deep turret]

yessir

[junior carbon]

15 a/3

i always get stuck here

[deep turret]

whats 15a/3

[junior carbon]

5

[deep turret]

5a

[junior carbon]

YES

ok im with u taf

answer is 2 squared 15a

[deep turret]

yessirrr

[junior carbon]

i didnt get that

im at 5a rn TAF LOLZ

[deep turret]

oh shoot 😭

okay okay dont divide it by three

[junior carbon]

LMAO

[deep turret]

cause its root 15a over 3

you cannot do that

so then divide the 6 by 3

[junior carbon]

is that legal taf?!

[deep turret]

yessir

[junior carbon]

ok 6 divde 3

[deep turret]

2 root 15a

[junior carbon]

thank u so much

[deep turret]

np :]

[junior carbon]

rawr

AHHH so scaryyyy

[deep turret]

you got this!

[junior carbon]

ok i try taf!

imma start multiplying top and bottom by 3

[deep turret]

root 3*

[junior carbon]

WHERES MY PEN!:!!??!?

7 squared 9+2

7x2 devided by 3?

or am i dumberson

i have already done what they asked and got the square root out of the bottom of the fraction

but i need to simplfy more prolly

[strong dune]

aqui estoy

que pasó

racionalizar el denominador?

[junior carbon]

Siiiii

[strong dune]

wait a second im opening windows 🤓

[junior carbon]

I actually learnt how to do those questions we did the other day

Thank u so much for helping me

I literally practiced them all day yesterday

One questions by the way 😭

I learnt it tho

I did that question all by myself yesterday thanks to u!

[strong dune]

i did this idk if is okey

is the same thing

[junior carbon]

For a?

[strong dune]

si

[junior carbon]

21+2?

[strong dune]

hmmm

[junior carbon]

i thought im multiplying the top and bottom by the square root of 3

[strong dune]

5.a

[junior carbon]

idek

5.a yes

[strong dune]

maybe if you do this

[strong dune]

yep this

[junior carbon]

hmmm ok

[junior carbon]

i just dont understand why the 3 stays bracket wtf

WOAHHHH

1 question btw

those are the practice ones

i feel like the practice is harder LOL

[strong dune]

3.7=21

you could follow

[junior carbon]

why did u put 3 squared upstairs kiyo

ik its right

[strong dune]

"Rationalizing" in mathematics refers to removing the root from the denominator of a fraction. This process is done to simplify the expression

[junior carbon]

i was just wondering how there ended up being 3 threes on top

[strong dune]

Because you want to eliminate the root in the denominator, when you multiply it by the same number divided by itself (which is equal to 1), you’re not changing the value of the entire expression. (The square root of three divided by the square root of three is equal to 1

[junior carbon]

ok so he put 6 on top and bottom

then the third step he has 3 6s

WHAT?

im just really confused on that

[junior carbon]

cause the square roots on bottom cancels

why do you guys decide to put 2 square roots on top

i thought it was done once

[junior carbon]

omg

that makes sense

[strong dune]

Because it's distributing the product: the first term, multiplied by the square root of 6, plus the second term, multiplied by the square root of 6.

In that case, the second term in the parentheses has a negative (-2), so the result of multiplying the second term by the square root of 6 will be negative.

[junior carbon]

can i use foil pretty much on any questions?

so far so good

[junior carbon]

im using this

[junior carbon]

im stuck on c

i dont think i can use foil on it anymore

[strong dune]

an example

[junior carbon]

wait vlan

if the virables are being added

it would just stay the same no?

[strong dune]

🤔

I don’t understand it 🤓

[junior carbon]

on question c im ending up with -2srt70

and +3sqrt30

ik im on the right track to the answer

i might factor it here but i get lost in that LOL

SO DUMBBBB

ok so i apprently i dont even need to factor it

i just leave it how it is after multiplying by 10

......

[strong dune]

🤔

I separated the fraction 1/2 to keep only the square root.

[junior carbon]

i appriciate u

I’m on conjugate rationalizing questions now

[junior carbon]

I did 10 times the square roots

And made the 10 cancel so it’s 10 times 2

20 on bottom and the top I left after multiplying them

BUT WHY

they said simply

And I literally have 70 and 30 as a square root

Can’t that be simplified?

I don’t even know what real anymore

[strong dune]

'If the denominator has more than one expression besides the square root, you can apply the same rationalization principle, but by using the conjugate of the denominator.'

[junior carbon]

OMG

ur a super hero forsure

[strong dune]

God, the last time I solved exercises like these must have been over 15 years ago.

🥵

[junior carbon]

God bless u

I hope I can retain all this info for others as well

this is the type of math u need for careers

[junior carbon]

im doing that

i cant seem to get the ending

AHHHHHHH

i get 8 x 3 on top

is that not right

[junior carbon]

yeah im lost asf

LMAO

one sec

its truly a talent how fast i can get lost when it comes to something new

hmmmmmmmmmmmmmmmmmm

ok.....

[strong dune]

exacto

[strong dune]

In this case, in the denominator, you don't have a sum, but rather a product, so I have separated the 2 on the left from the denominator, into 1/2. Then you can continue operating with the fraction

[junior carbon]

i got 7 + 7sqrt 3 on top

with 4 sqrt 3 on tops

the answer is -5 -4sqrt 3

[strong dune]

wait a second

[junior carbon]

i actually think ur right tho

WTF

[junior carbon]

for the bottom its minus 2 apprently

[strong dune]

yesss

[junior carbon]

how to get 4

[strong dune]

8/2

-2

8/-2

[junior carbon]

wow

really

wo

w

thank KIYO!!!!!!

[junior carbon]

thank u sso much

This one has 2 square roots

[strong dune]

🤔

[strong dune]

i think its right

[junior carbon]

oh

so i foil the bottom

interesting

[junior carbon]

damn

i was not close to that

[junior carbon]

wait wtf

i was doing that

idek

i got the plus 8

and end -10

the other 2 brackets have me confused

cause do i do 8x10

or cancel them

[strong dune]

The second and third terms are the same, but with opposite signs, so they cancel each other

[junior carbon]

is it 8 plus 8 minus 20?

[strong dune]

where

?

[junior carbon]

like the 8 cancels and becomes 8 positive right

the square roots come off

[strong dune]

yess

the first one

[junior carbon]

how about the 2nd and 3rd

the 2nd and 3rd got me mind boggled LOL

cause before i would multiply them

[strong dune]

$\sqrt{10}.\sqrt{8}=\sqrt{8}.\sqrt{10}$

[jolly parrot]

vlankiyo

[junior carbon]

i dont mutiply them right

or do i

[strong dune]

It cancels out with the second term

[junior carbon]

wait its 8x8

[strong dune]

where

[junior carbon]

8s cancel and become positive 8

and the ten cancels

theres 2 -10s right

[strong dune]

$-\sqrt{8}.-\sqrt{8}=(-1).\sqrt{8}.(-1).\sqrt{8}=(-1).(-1)\sqrt{8}.\sqrt{8}=\sqrt{8}.\sqrt{8}=8$

[jolly parrot]

vlankiyo

[strong dune]

$-\sqrt{8}.\sqrt{10} +\sqrt{10}.\sqrt{8}=-\sqrt{8}.\sqrt{10}+\sqrt{8}.\sqrt{10}=0$

[jolly parrot]

vlankiyo

[strong dune]

the second and third

[junior carbon]

omg

they equal 0??!?!??!

HOW THO

[junior carbon]

oh

thank u!!

lets see if i can get the answer now

yeah im not doing that question

the answer is ridiculious

this is what im currently working on now

[strong dune]

3?

[junior carbon]

doing 3 a couple times until i know how to do it exactly step by step

then doing 4

[strong dune]

😎

[junior carbon]

im on 4 >:)

the last 2 is like HUH?

uhm

oh do the 5s cancel

........

😉

LOL

its time .

[pearl pond]

@junior carbon Has your question been resolved?

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[sudden badger]

If this function has a vertical asymptopte at -2

why does it touch it

it only seems to not touch 1

[pulsar lark]

-2 is not a vertical asymptote

You can simplify the function

[pearl pond]

@sudden badger Has your question been resolved?

[sudden badger]

what bt this one

[pulsar lark]

Simplify more

[sudden badger]

its x-4

however it has a hole in -4

right?

[pulsar lark]

Why hole?

[sudden badger]

no?

[pulsar lark]

No

[sudden badger]

wow

that's what my teacher said

[pulsar lark]

Test it

[sudden badger]

then

this says in the book

its wrong?

[pulsar lark]

Test in the calculator graph

Desmos

[sudden badger]

i did

[pulsar lark]

Hmm strange

[sudden badger]

yes it has a whole

hole*

[pulsar lark]

Does -2 have a hole too in origin question?

[sudden badger]

in this function?

yes

[pulsar lark]

Ok

[sudden badger]

i mean

that's what i thought

i dont really understand why

there's a hole but its not an asymptote?

[pulsar lark]

https://xaktly.com/MathLimits.html

The difference between holes and asymptotes come down to limits I guess

If you are looking for an algebraic explanation

[sudden badger]

yes

i got it

basically if the factor repeats in num and denom, its a hole

if its only in the denom, then ts an asymptote

[sudden badger]

.solve

.solved

[pearl pond]

Channel closed

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[pearl pond]

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[cinder portal]

im a bit sick in the head n i can't get the answer to no.3 , im stuck asf

[pure rapids]

okay

[pure rapids]

do you know how to factorise quadratic expressions?

[cinder portal]

i do nw its js the process

[pure rapids]

oh

then we split 32 into factor pairs

and see which pair sums to 12

[cinder portal]

isnt that 4 n 3 then or

?

[pure rapids]

32 = 4 x 8

and 4+8 = 12

[cinder portal]

oh OH

LMAO

[pure rapids]

there's our pair lol

k then its pretty easy

[cinder portal]

thanks i legit js did multiplication on both

no wonder

[pure rapids]

💀

rip

[cinder portal]

you can close it nowbtw

[pure rapids]

uh you need to do .close

.close

[pure rapids]

i cant since its ur channel

[next estuary]

💀💀💀

[pure rapids]

💀 indeed

[next estuary]

She her he him they them how is this dude all of the above

[pure rapids]

wait what

'all of the above' frfr

[next estuary]

The other dude

[cinder portal]

KNAI

[pure rapids]

yeah

[cinder portal]

LMAO*

ok wait.

[pure rapids]

'knai' 💀

[cinder portal]

shh

isnt it .solved

or

?

[next estuary]

.close

[pure rapids]

do .close

[cinder portal]

.close

[pearl pond]

Channel closed

Closed by @cinder portal

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[pearl pond]

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[agile ridge]

here x is in [a,b] f(a)=f(b)=0

[agile ridge]

and idk why it's a+b/2 here

[dusty flame]

mean value theorem?

[dusty flame]

what do u want to prove here?

[agile ridge]

if general(let along this q), if x is in [a,b] then x-a must be in a, a+b/2?

[dusty flame]

hmm'

[agile ridge]

wait wait

[dusty flame]

im thinking if we can prove that

wait

so u want to prove this inequality?

[agile ridge]

wait ok lemme translate the orginal q

it'll take a min

[dusty flame]

yes plz

[dusty flame]

ok

gimem a moment

oh

u just invoke mean value theorem

take absolute value

[agile ridge]

I kinda understand the a+b/2 stuff but

[dusty flame]

which part are u most confused abt

[agile ridge]

they pluged |f(x)| when a<xi<x in the interval a,a+b/2??

[agile ridge]

no i mean like if x is in a, a+b/2, then I understand, but it's not

[dusty flame]

ah

im thinking of using rolle's theorem someway somehow

the lower one assumes that the function can be split in the middle

but what im assuming is

u use this here IF x is bigger than a

this here IF x is smaller than b

in which case where x changes from being closer to one boundary than the other is

(a+b)/2

[agile ridge]

I think they splited the integral into two integrals So the range of x values in these two integrals should be different, right?

The range of x in the first integral should be between a and a+b/2, and the range of x in the second integral should be between a+b/2 and b

[dusty flame]

where:
one integral is for the points closer to a

the other is for points closer to b

[agile ridge]

Isn't that right?

[dusty flame]

yes

[agile ridge]

But we don't know if x is within this specified range about the expression we plug in

they only have this

x can beyond a+b/2 but still satisfies these stuff

[dusty flame]

they apply mvt for these invertals

[jolly parrot]

k

[agile ridge]

but look this case, x is not in the a,a+b/2, then how can we plug in the first integral

[dusty flame]

how is it not in (a, a+b/2)

we stop using the top definition

[agile ridge]

x>a+b/2

[dusty flame]

because since a+b/2 onward the point will be closer to point b

[dusty flame]

then it will be closer to b, yes?

[agile ridge]

yea

[dusty flame]

so u use this mvt

[agile ridge]

but if x>a+b/2, then how can we plug it in the integral whose integral interval is a,a+b/2

[dusty flame]

we plug into this one instead

what im thinking is

[agile ridge]

shouldn't it be like this way

[dusty flame]

the author wants to apply mvt for interval for $(a, \frac{a+b}{2})$ and another mvt for ${\frac{a+b}{2}, b}$

[jolly parrot]

k

[dusty flame]

the right one should be ${\xi_2}$

[jolly parrot]

k

[agile ridge]

which one?

[agile ridge]

Yea I labeled it as xi2 already

just a bit ugly

[dusty flame]

oh wait

im high

[dusty flame]

this diagram is right ye

but x can be anywhere

[agile ridge]

Yes so this is just one case

[dusty flame]

remember when u integrate

u move x from a to a+b/2

u dont keep x stationary

think of the sweeping motion

[agile ridge]

I know, but you see this diagram, in this situation x is in the second range, so it should be in the second integral while a<xi1<x should fit in (x-a) f(xi1) which should be plug in the first integral

[dusty flame]

a<xi1<(a+b/2) should be in the first integral

since we are evaluating it till (a+b/2)

[agile ridge]

So actually it has nothing to do with where x is located?😵‍💫

[dusty flame]

if it is between a<xi1<(a+b/2) we use the first integral

if it is between a+b/2 and b, we use second integral

[agile ridge]

what about this situation

[dusty flame]

well

we dont care about this no more

since its not between a and (a+b/2)

[agile ridge]

Then the integral in this interval is....?

[restive condor]

hey i'm new what does 1+2+3+4+5+...+100=?

[dusty flame]

5050

[restive condor]

cool ty

[dusty flame]

evaluating the first function; the one for x closer to a

using the integral from a to (a+b)/2

[agile ridge]

I mean in this situation😭 where x is closer to b

[dusty flame]

then we use the second function for the case where it is closer to b

and evaluate using the second integral

[agile ridge]

but then in this case what the first integral will be?

[dusty flame]

u dont assume x to be stationary 😭

like

[agile ridge]

<@&268886789983436800>

[dusty flame]

think of the first one as for x moving from a to (a+b/2)

and the second as for the x moving from (a+b/2) to b

[unborn abyss]

stop trolling and go away.

[dusty flame]

u dont assume x to be a certain numberthat is nonmoving

[agile ridge]

I think that's the reason behind

x is changing

Darn😭

[dusty flame]

damn

[agile ridge]

Thank you so much for your patience lol

Fxck that's what confused me for almost an hour😭

I'm the biggest idiot in the world

Anyway thanks, finally understood

.close

[pearl pond]

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[abstract quiver]

can I assume d divides c in the last line....I tried to understand with an example

12....18

2×2×3....2×3×3

now c will be {2,3,6} ..??..and d{6}...because gives condition c belongs to z ?

Am I correct?

[pearl pond]

@abstract quiver Has your question been resolved?

[pearl pond]

@abstract quiver Has your question been resolved?

[abstract quiver]

.close

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[surreal light]

I need help with part ii

[surreal light]

<@&286206848099549185>

[latent quail]

,rotate

[pearl pond]

@surreal light Has your question been resolved?

[surreal light]

<@&286206848099549185>

[pearl pond]

@surreal light Has your question been resolved?

[pearl pond]

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[chilly terrace]

whats the formula of calculating by how many % does a price low?

[chilly terrace]

does a price decrease

[hollow cobalt]

Divide the difference by the original price

[chilly terrace]

i did original price-new price divded by new price * 100

[hollow cobalt]

Right

new - original will be more intuitive though

[midnight haven]

Hmmm why divide by new?

Isn’t it divide by old?

[chilly terrace]

no

[midnight haven]

👍

[chilly terrace]

.close

[pearl pond]

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[junior vapor]

how do i do this?

[supple knoll]

a is slope

[cosmic garnet]

and |x-b| + c shifts the origin of |x| from (0, 0) to (b,c)

[junior vapor]

oh wait sorry i meant how do i do b

[cosmic garnet]

i think it means range of possible values of m

because theres infinite possible values of m

[junior vapor]

oh so do i find the points of intersection?

[junior vapor]

true

[cosmic garnet]

nah

range of m starts from here (this isnt included)

and it ends when the y=mx line is parallel to the right part of the graph because then there will only be 1 intersection point between the line and the graph

[junior vapor]

how can it be parallel when its moving in the opposite direction from the right part of the graph?

[cosmic garnet]

when negative slope

orange line (of y=mx form) parallel to right part of the graph

[junior vapor]

I'm confused again. Why do the two orange lines have different slopes? I thought I had to find one line that cuts the graph is two places?

[junior vapor]

But you drew two orange lines that each cut the blue graph in one place?

[cosmic garnet]

yes

thats the possible range of values of m

[junior vapor]

Oh, I see now🤡

Hm ok I think I got it from here

Thanks

.close

[pearl pond]

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[whole silo]

Hey 👋
I have a question regarding linear algebra.
I am missing something fundamental in my understanding.
Given the plane 2x + 3y -6z = 0.
Find the orthogonal projection of the vector [1 0 0]^T on the plane.

[whole silo]

If the vector is orthogonal to the plane then the projection would be the 0 vector.
In the solution they use the gram-schmidt technique, but what I don't understand is how finding the ON-bases is even required to find the projection.
I would understand finding only the vector that is in the plane that is closest to the vector and then simply project it

[pearl pond]

@whole silo Has your question been resolved?

[coarse harbor]

gram-schmidt is only one of the methods that can be used

And you correctly mentioned that the closest vector from the plane would be the projection

[whole silo]

It was the one given in the solution so I would like to know how it works

I have been looking up explanations and I don't quite get what they're doing by using this method

[coarse harbor]

The idea is that given two orthonormal vectors in your plane, you can explicitly construct the projection.
Let $v_1, v_2$. The geometric intuition is as follows. Imagine extending $v_1, v_2$ to the orthonormal basis by choosing a proper $v_3$. Then in this basis first two coordinates of any vector will correspond to the projection of that vector to our plane

[jolly parrot]

EQUENOS

[coarse harbor]

We can look at our basis "from above" so v_3 looks like a point. Then any vector from that point of view will look exactly like its projection

[whole silo]

I am following thus far

[coarse harbor]

Mathematically we just calculate first 2 coordinates of some vector $u$ from $\mathbb{R}^3$: $v_1 \cdot u,\quad v_2 \cdot u$

[jolly parrot]

EQUENOS

[whole silo]

v1 and v2 being the vectors in the plane yeah

[coarse harbor]

Yes, and being orthonormal

We can check that $(v_1 \cdot u)v_1 + (v_2 \cdot u)v_2$ is indeed the closest vector to $u$ from the plane

[jolly parrot]

EQUENOS

[coarse harbor]

We want to minimize the following function: $|u - (x v_1 + y v_2)|^2$ by choosing proper $x, y \in \mathbb{R}$

[jolly parrot]

EQUENOS

[coarse harbor]

Actually I'm not sure if explaining this particular property is useful

But we can discuss it anyways, it's up to you

(I mean the property of orthogonal projection being the closest vector from the plane to our vector)

[whole silo]

I understood your explanation, it was helpful to help me better understand the gram-schmidt

wait I am gathering my thoughts

[coarse harbor]

One more thing: I forgot to mention that in order to choose such v_1, v_2 we need gram-schmidt

Basically GS is an algorithm that takes in arbitrary linearly independent vectors and turns them into an arthonormal system

It does it as follows. Imagine that we want to orthonormalize n vectors and we already did that with k vectors. In order to get (k+1)-th vector, we subtract from it the orthonormal projection to the space formed by first k vectors, and then normalize the length

[whole silo]

Are we projecting a dot onto the plane? Then everything makes sense to me all of a sudden, I imagined the projection would be a line

[coarse harbor]

Yes, we're projecting a dot (vector) onto the plane

[whole silo]

Arr then I get the solution and all the videos I watched 😩

[whole silo]

I can't get an intuition of what is actually happening here, I only imagine that we have one vector, then with math magic we get orthogonal vectors to it

and divide by it's value

[coarse harbor]

Given a subspace, we can uniquely represent any vector from the space as a sum of 2 vector orthogonal to each other, one of which is from the subspace

[whole silo]

That's a crystal clear explanation

Thank you sir, it was helpful

[coarse harbor]

Yep, so here on this pic yellow vector is the orthonormal component of the blue vector:

Red vector is the orthonormal projection

So yellow = blue - red

And the we normalize the result by dividing it by length

[pearl pond]

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[shadow cargo]

Studying for an exam from last years tests to understand
anyways we have to turn something in2^result and i am wondering why what i underlined(2^20+3) turned into (4)

[shadow cargo]

where did the 3 go and what happened to 2^20

[sinful nebula]

looks like they took 2^20 common

[shadow cargo]

what do you mean by that

?

[sinful nebula]

$2^{20}\cdot(1)+2^{20}\cdot(3) = 2^{20}\cdot(1+3)$

[jolly parrot]

Alaska

[shadow cargo]

ooooh

that makes sense

thank you

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[shadow cargo]

Problem 2. Given the six-digit positive integer A =
2023xy, where
𝑥, 𝑦 digits of the decimal numbering system.
Determine the digits 𝑥, 𝑦 so that the number A is divisible by
number 17. can somebody explain to me how to solve this

[shadow cargo]

i know the solutios is there ( in greek) but i dont understand it

[lofty aspen]

alright, i will

So, the number 2023xy can be written as 200000+00000+2000+300+10x+y

[shadow cargo]

alright

[lofty aspen]

So, then adding the constants, you get.. 202300 + 10x + y

if it is divisible by 17, it should be expressed by a multiple of 17

lets say 17k

now 17K - 202300 = 10x + y

But 202300 is divisible by 17…

202300=11900*17 (i looked in the solution😅 )

so you can write it as 10x + y is a multiple of 17

agree?

[shadow cargo]

why is it 10x+y instead of 10xy or 10x+100y

[lofty aspen]

10xyu?

Oh-

well, its because the number 23 is expressed as 20 + 3 and not 20 + 300… here x and y are digits…

Well, do you know number expansions?

[shadow cargo]

if i had to guess

i probably dont

[lofty aspen]

oh, in my country, its taught in lower grades😅

alright.. lemme give you a breifing

So a number 4723 can be expanded as 4x1000 + 7x100 + 2x10 + 3x1

[shadow cargo]

oh yes

makes sense

i know that

but arent x and y supposed to be decimals?

[lofty aspen]

Yes, the digit multiplied by its place value

[shadow cargo]

oooh

now i get it

[lofty aspen]

no, if you can see in the question.. they are the digits of demical number..

[shadow cargo]

whats the difference

[lofty aspen]

well.. digits are 10, they are 0, 1, 2, 3, …, 8, 9

decimals are any positive real numbers like 1.5 or 2.71282 or etc

so can you agree to my previous statement now?

[shadow cargo]

yeah

[lofty aspen]

ask any question if you dont have any clarity..

[shadow cargo]

i thought they were decimals

[lofty aspen]

oh well… that clears the question then! Do you want me to solve it further?

[shadow cargo]

yeah if you could

[lofty aspen]

sure!

We want the number in question to be divisible by 17

[shadow cargo]

oh never mind i got it but whats this symbol ∈

[lofty aspen]

so 10x + y should be divisible by 17

[lofty aspen]

Oh alright

the sumbol there is

”belongs to”

[shadow cargo]

can it be skipped but written with words?

[lofty aspen]

What do you mean?

[shadow cargo]

can i write that 10x+y has to be 17,34...85 instead of 10x+y ∈ 17,34...85

[lofty aspen]

sure, they both mean the same thing…

but in deeper studies like calculus… we would use that sumbol..

so id prefer you using it and getting comfortable around it

[shadow cargo]

yeah but this is like 9th grade so i doubt that could matter a lot

[lofty aspen]

Haha, i guess.. but if the foundations are laid right, building a building is a piece of cake

[shadow cargo]

yeah i guess

[lofty aspen]

Trust me on this one..

😁

[shadow cargo]

yeah i do

you helped on this one

[lofty aspen]

Alright, any more questions?

[shadow cargo]

maybe the last line of the solution?

[lofty aspen]

that means a lot.. thank you for trusting mw haha!

[lofty aspen]

sure!

[shadow cargo]

why do we use ,

[lofty aspen]

its to distinguish two different elements in a set

the symbol “,” pronounced as comma, here is not needed to be pronounced, but can be indicated in speech with a small gap

basically a short full stop

[shadow cargo]

yeah

cant we write x+y ∈(10+7,30+4)

as well

is that an option?

[lofty aspen]

yes we can!

based on how they ask us to write…

both answers are absolutely true!

[shadow cargo]

ok thank you i think i got it.although probably gonna have to come back later for a lataer problem

[next estuary]

Oops

[lofty aspen]

sure!

[next estuary]

Wrong place

[shadow cargo]

anyways see ya

.close

[pearl pond]

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[lofty aspen]

if you want to chat me about it, you can friend request me…

😁

[shadow cargo]

Sure you make it easy to understand maybe even more than some teachers

[lofty aspen]

oh, that means a lot! Thank you!!

[pearl pond]

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[sand coyote]

I started off with the parameters

And at the end when I compared the two equations with parameters 0=1 so no intersection

<@&286206848099549185>

[winter elbow]

Anyone can explain this to me

<@&286206848099549185>

[acoustic belfry]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[sand coyote]

Do you know if I did it right

[acoustic belfry]

no this is beyond my level sry

[sand coyote]

All good

[pearl pond]

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[pearl pond]

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[midnight haven]

I know how put both into exponential form but I do not know how to keep the angle in the range above

@lofty aspen

You replied earlier

[lofty aspen]

heyyy

oh its simple!

well, when you raise a complex number to a power n, the argument of the number gets multiplied by n

and the magnitude is raised by the power n

well, this is the proof obtained from demoivre’s theorem…

I’d rather ask you to check out videos made by 3blue 1 brown or some other experts on youtube…

but since we’re here, ill help you!

[midnight haven]

Yeah thanks for the help I acc figured it out just now

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[iron basin]

What do those AM-GM questions mean when they ask something like\\"Find minimum of $\frac{a^5+b^5+c^5}{3}$ if a, b, c are positive real numbers"\\ and the answers themselves are in terms of a, b, c? Wouldn't it be easier to say that the minimum is close to 0?

[iron basin]

The options are $\left(\frac{a+b+c}{3}\right)^5$, $a+b+c$, $a^2+b^2+c^2$, and $\frac{1}{3}[abc(ab+bc+ca)]$

[pearl pond]

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[pearl pond]

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[next estuary]

help

[next estuary]

i did this so far

,rcw

[dusty flame]

try letting u = 9-x^2 instead of x^2

[dusty flame]

this substitution is wrong, it should be 9-u not 9-u^2

[next estuary]

wait what

oh bruh

i see

Square root -3😞😞😞

This can’t be right

Nvm nvm nvm NMV

I got just 1 as the answer

That seems too suspect

[dusty flame]

lets see chat

,w integral form -3 to 1 of (x/sqrt(9-x^2)) dx

[jolly parrot]

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[dusty flame]

i thought u fixed it

but

${du = -2x \neq 2x}$

[jolly parrot]

k

[next estuary]

😞

[pseudo oxide]

just sub u = -x^2?

[pseudo oxide]

and incorrect, du = -2x dx

[next estuary]

Yeah

[pseudo oxide]

which is very, very important

[dusty flame]

mb mb

[next estuary]

I also skipped a step here

Hold on I got this

I got -sqrt8

That’s -2sqrt2 right

Ok Ty guys this was a simple question I just made stupid errors

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[pearl pond]

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[uneven river]

Hey guy does anyone know how to make inverted sine in graph program

[uneven river]

something like this but inverted

[dire flume]

do u mean reflected along x axis?

[uneven river]

same as the first one but inverted by 180 degrees

like what do i have to type in ?

[dire flume]

-sin(ax)

[uneven river]

thanks

[dire flume]

np

[uneven river]

@dire flume sorry to bother you but do you know how to make this in that program ?

[dire flume]

in which program?

[uneven river]

symbolab would be best

[dire flume]

also r u an IB student?

[uneven river]

what is IB ?

[dire flume]

the board ur school follows

does ur school follow an ib board?

international baccalaureate

[uneven river]

im on high school

[dire flume]

also sorry, not sure how to do it, because the graph u have is in terms of radian, im not sure how to bring that to numerical values on platforms

[uneven river]

oh okay no problem and do you where you can make like wave forms ?

[dire flume]

no, sorry

[uneven river]

because im doing voltage waveform

oh okay

thanks for the help anyway

[barren arch]

This looks like the graph for sin^2(x) or |sinx|

[pearl pond]

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[deep turret]

can someone help me find the inverse function of:
(3x+1)/(2-x)

[deep turret]

with steps :]

[dusty flame]

Hint: 2-x/2-x = 1

[deep turret]

thank you!

i can try that?

but

how tf 😭

do you get to that

[fickle schooner]

$$ y = \frac{3x + 1}{2-x} $$

[deep turret]

is the bot down?

[jolly parrot]

StrangeQuarkAL

[fickle schooner]

Try to make x the subject of the equation

[west vault]

The general process for inverting a function $y = f(x)$ is just to replace all the $y$s and $x$s, like $x = f(y)$, and then solve for $y$.

[jolly parrot]

jewels!

[deep turret]

i know how to find the inverse

i just get stuck on like

uhhh

2x-xy=3y-1

[fickle schooner]

Factorize the x in the left hand side

[deep turret]

x(2-y)=3y-1

[west vault]

you need to be solving for y here

you replaced them right?

[deep turret]

yeah

i swapped x and y

[west vault]

yeah so take xy to the other side and factor y out

[fickle schooner]

Oh interesting, I usually do it the other way round

[deep turret]

x=3y-1/2-y ?

[deep turret]

where did you get -x?

[dusty flame]

I do it like this :P

[dusty flame]

Multiply 1/-1 on both sides

[fickle schooner]

After this, solve for y

[deep turret]

thats where i started 😅

[deep turret]

why?

[dusty flame]

It makes it easier imo

I like variable first

But it’s not necessary

[fickle schooner]

First of all: $x = \frac{3y + 1}{2- y}$

That's after swapping.

$$ x(2-y) = 3y + 1 $$
All you have to do is expand this out and factorize the y

[jolly parrot]

StrangeQuarkAL

[deep turret]

thank you i will try that

[deep turret]

how do i get the y’s on the same side?

[fickle schooner]

Take the -xy to the 3y

[deep turret]

mhm :]

OHHH

I GET IT NOW

OKAY THANK YOU

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[fickle schooner]

No problem

[pearl pond]

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[glass finch]

Okay so I understand that if the coefficient of friction from the car moving is greater than 0.7, the car will slide out of the turn, but I'm unsure how to figure that out without the mass of the car given... And I can't figure out how to find the mass either.

[pearl pond]

@glass finch Has your question been resolved?

[wintry gate]

f=ma

[glass finch]

Do I have force??

I mean I can do a=v^2/r to get 32^2/120 to a=8.53 but then I'm missing mass and force

[warm current]

Let m remain a variable. I assume it will cancel out in the end.

[glass finch]

Okay but what about force? Force is not equal to force of friction, right?

[warm current]

What are all the forces acting on your car?

[glass finch]

Do we just consider the velocity force? Other than that there's friction, gravity, and normal force I believe

[warm current]

Yup. From there, work out your free body diagram. Calculate, algebraically, what each force should be. The sum total is the net force, which, algebraically, you at least know. Work with your unknowns

[pearl pond]

@glass finch Has your question been resolved?

[glass finch]

So it does slide out of the turn?

[warm current]

What do you have for $F_f$? How are you getting $m=3.8$?

[jolly parrot]

SWR

[glass finch]

I used F=ma to solve for m 🤷‍♀️ I have no idea

[warm current]

How did you get F then?

[glass finch]

Velocity:(

[warm current]

Okay, we'll just move past that then.

You know $F_g$ and $F_N$, so that's good. Can you find $F_f$?

[jolly parrot]

SWR

[glass finch]

Do I just use the .7 for coefficient of friction? I mean f_f is mew/coefficient of friction times normal force

[pearl pond]

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[fringe sorrel]

I just don't know where to go from here, should I treat the complex numbers as one variable? or a+bi?

[rough forge]

Treat it as a+bi

Having it just z won't help much into finding out the real and imaginary part

[fringe sorrel]

ok, is the part with a+bi ok?

[rough forge]

yes

[fringe sorrel]

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[grave parcel]

HI

[grave parcel]

Can i get help finding the equation of this hyperbolae?

Please 🙂

[plush bramble]

are those two points all you know about the hyperbola ?

[grave parcel]

Yes

Y = k/x is formula correct?

[plush bramble]

that's one type of hyperbola yes

[grave parcel]

That’s the formula I’ve been using

I got the answer of y = 10/x

Is that correct?

@plush bramble

[plush bramble]

that's one possible answer yes

[grave parcel]

Whats the other?

|

Also my teacher gave me this one i said the answer was y = -4/k but apparently the answer is -4 = k/k= -4

@plush bramble

Hello u there?

[plush bramble]

yes this is incorrect

,w plot y = -4/x

[jolly parrot]

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[plush bramble]

x refers to the variable in the horizontal axis

[grave parcel]

So the equation is y = -4/x

[plush bramble]

k is the number

[grave parcel]

This was the example

@plush bramble

[plush bramble]

you don't need to keep pinging me

i don't know what you're asking

[grave parcel]

Sorry

Im so sorry

It’s fine never mind

Thanks you so so much for trying to help explain

.close

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[misty marlin]

Question 23 idk if this is a simultaneous equation or a max value or what

Ive done xtan45=(x-1)tan60 but idk what do to from there i tried tan/tan but idk what to do with that number

[pearl pond]

@misty marlin Has your question been resolved?

[misty marlin]

<@&286206848099549185>

[zenith cipher]

velocity is change in position per change in time.
So here, plane's velocity is given to be 600 miles/hour
So it travels 600 miles in one hour.
Find the distance traveled in 1 min.

l is the distance traveled in 1 min.
Then take tan on both the triangles.

[pearl pond]

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[pearl pond]

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[earnest crane]

the angle of point P(4, -2) should be -26.57°, right?

[solemn linden]

Yup you are exactly right

you can realize this by analyzing the angle of (4,2) and then inverting it

tan^-1 (2/4)

[earnest crane]

thank youu

oh interesting method

makes more sense that way

.close

[pearl pond]

Channel closed

Closed by @earnest crane

Use .reopen if this was a mistake.

[pearl pond]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• 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.

[orchid tapir]

helloi

[orchid tapir]

is that supposed to be 0 or 1?

[tall flint]

what's "that"

[orchid tapir]

the answer

the limit

[solemn linden]

0

[tall flint]

the limit is 0