help-36

[sullen gale]

!show

[final saddle]

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

[old nacelle]

4 = (3(r2)^2)/(r1)^2

i don't think its ritght tho

also i need more equations

maybe there is another approach

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[restive snow]

I know the semi perimeter is 21

i have this graph

x=CN=CP=21-13=8
y=BP=BM=21-14=7
AM=AN=s=21

and this

where N and M are the tangents of the excircle to the extensions of AB and AC

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[south slate]

can u guys explain how did this happened

[south slate]

the graph part

[compact valve]

Hey !

I can try to help you

what do you mean by "how did this happen" ?

Do you mean "how did they find this result" ?

[south slate]

yes

@compact valve

[compact valve]

Ok. Do you have the exact formula for Price in function of QS by any chance, or just this graph ?

(don't hesitate to tag me)

[south slate]

@compact valve i only have the graph

[compact valve]

do you not have a table ? Because this is supply, but the text mentions Demand, and a "Table 2.2" ?

[south slate]

this was the exact page of it i only cropped it

[compact valve]

ok this is weird

it really feels like there are formulas for QD before in this document

[fiery bluff]

Hmm, it seems like there is a typo and the equation should be QS

[compact valve]

oh, that could be too

in which case, now i have all the information i need haha

As you can see, the equation for QD is : $QD(P) = 965 + 0.1 \times P$ (assuming the first equation contains a typo)

[soft zealot]

Grui

Grui

[compact valve]

except here, we want to use that function the other way around

knowing the price, what would be the demand

so it's like solving an equation

you substitude QD(P) with 1250

and solve for P

so you need to solve 1250 = 965 + 0.1x

does this help ?

[final saddle]

@south slate Has your question been resolved?

[south slate]

yes

@compact valve THANKS ALOT

[compact valve]

you're welcome haha

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[gentle cove]

so like theres a minesweeper game with a gravity mode but idk how to solve this
not sure if this counts as math help but yea

[versed crater]

That looks hard what have you tried

[gentle cove]

tbh i'm not even sure where to start

i tried putting it into an equation format thing

like 1.71 = power/(1.8^distance)

but i literally don't know where to go from there

[whole halo]

instructions dont even make clear whether its every mine added together or the closest mine

[versed crater]

Do you know if the gravity have infinite range?

[gentle cove]

no :')

[whole halo]

the 0.04 shows they have too much range

probably whoever made this never thought to cut off the gravity so its a fair assumption

also 1.8^distance instead of distance^1.8 by itself is a rancid decision

man what is this

[gentle cove]

roblox minesweeper game..............

[versed crater]

It also doesn’t say what the power of any particular mine is

[gentle cove]

yea

[versed crater]

Roblox developers

[whole halo]

do you know about different powers of different mines

[gentle cove]

all i know is that the powers are probably randomized too

[whole halo]

the time they finally use a different font from the preset is the time we can treat it as a regular game instead of a roblox one

if the powers are randomized, that sort of ruins the point

imagine if we had stricter instructions

like all the powers were the same

and you had to gauge how close the mines were by these specific numbers

itd be like holding a geiger counter around

[versed crater]

It also doesn’t say if a number means there is or isn’t a mine there

[gentle cove]

maybe i should just go to the games discord server

[whole halo]

is the games discord server going to have a better idea to doing this

[versed crater]

Does 0.04 just mean there is a weak mine on that tile

[gentle cove]

i think the numbers on the board is what i'm supposed to use to find the distance

and minepower

idrk

[versed crater]

How is distance measured?

[whole halo]

dear god theres too many sides to this problem

[versed crater]

Well you’ll have a hard time solving an ill-defined problem

[whole halo]

can you try triggering a mine on a tile thats already "cleared" (has a number on it)?

[gentle cove]

i just died

[whole halo]

was it on a number?

[gentle cove]

uh

the tiles with numbers dont have mines on them

[whole halo]

alr just making sure

[gentle cove]

yea

wait you know how minesweeper works right

[versed crater]

You still have this problem

[whole halo]

well frosst was asking otherwise and I was wondering maybe this minesweeper game was even more broken than usual

[versed crater]

And this problem

[whole halo]

theres also this: look at the 1.71 and the 1.74

[gentle cove]

oh yea

i thought that was weird

[whole halo]

theyre 0.03 off even though theyre on a corner tile which is guaranteed to have a mine

that 0.03 difference indicates its looking at more than one mine then

however its not very much of a difference since 1 / 1.8^distance is exponential

[versed crater]

So perhaps we’re looking at taxicab distance

Still it’s hard to know if it doesn’t say

[whole halo]

oh wait Ive just realized

this is minesweeper

[gentle cove]

yea

[whole halo]

power is likely just the number you usually see

[versed crater]

My guess is that for the 1.74 mine there is a strong mine on the bottom right corner of the adjacent 7.01 tile

[whole halo]

so if there were 2 mines in the neighborhood, the power would be 2 for that tile

[versed crater]

Compared to a slightly weaker mine on the top left of the 6.87

[whole halo]

theyre calling it power because, as usual for roblox games, anything that indicates intensity but cant immediately be described in clear terms is usually called "power"

so power is just a multiple for how many mines are in the immediate neighborhood, its a potential guess

[gentle cove]

probably yea

that would make since

[whole halo]

however that doesnt explain the 5.57

since there can only be one mine next to it but 1 / 1.8 < 1

[gentle cove]

i might just ask the other discord server lol

with the people who made this game

[whole halo]

I hope you get an actual clear explanation in less than a thousand words

[gentle cove]

thanks though

[whole halo]

np

[gentle cove]

i hope so too

.close

[final saddle]

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[nova jay]

a group of people are running in a straight line one after another all in the same speed (V), the length of the line is L, towards them their coach is running at speed of U (U<V) when one of the people that are running reaches the coach he turns 180 degrees and keeps running in the same speed towards the other side, what would be the length of the straight line when the last runner will reach the coach

[desert mantle]

figure out the time until it stops

but as written there isnt enough information for a proper answer

e.g. distance between first and last runner

[nova jay]

but cant we describe the length using vars

[nova jay]

like lets say

coach(t) = Ut + L
lastrunner(t) = lt(t) = Vt + L # assuming he meets the coach at t=0 cause we dont care about the rest

firstrunner(t) = fr(t) = Vt # x=0 when t=0

we can do coach(t) = fr(t) to get the time when they meet and then plug that time in lr(t) to get the location of the last runner when the first one and the coach meets

then subtract no?

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[stuck juniper]

119

[serene panther]

!status

[final saddle]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[stuck juniper]

Not getting

[serene panther]

What did you try

[stuck juniper]

Well

Feynmans popped up

Didnt get

By parts

Substitution

Definite properties

[serene panther]

Oof feynmans looks hard to do here

[stuck juniper]

Question from my mock

[serene panther]

Nah this doesn't seem to be solvable

Did you apply kings rule

Nah even that don’t work here

[stuck juniper]

😨😨

[thin cloud]

I put it in the calculator and it outputs an error lmaoo

[stuck juniper]

[thin cloud]

I also think so

[final saddle]

@stuck juniper Has your question been resolved?

[serene panther]

At best you can approximate it

Assume cos^2(x) <= 1

[stuck juniper]

Maybe we can use inequalities

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[compact bolt]

.reopen

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[tepid creek]

Is this correct? Its writing quad equation in s form and finding the values of a b c

[tepid creek]

I used foil method

[cobalt perch]

where did you get x² from in the 2nd part of equation?

[tepid creek]

x • x

[cobalt perch]

oh wait gulp

I dont know the foil method..

-# bye now...

😔

[tepid creek]

Its first, outer, inner, last

[jagged flare]

x-2x is -x not +x

so it shoudlve been x^2-x-10=0

[tepid creek]

Oh alr srry

THANKYOUU

I mostly get confused on integers

Can i get smth checked too?

[bold turtle]

The goal just being to find some a b and c to put into a quadratic equation, yeah?

[tepid creek]

Yes

Put to standard form (ax2 + bx + c = 0) then find the values of a b and c

[bold turtle]

Then yeah, they're all good

[tepid creek]

Ty

[final saddle]

@tepid creek Has your question been resolved?

[tepid creek]

How to solve this

Is it this?

[paper kite]

yes

3/5 if you simplified it further

[tepid creek]

OH I FORGOT THEY WERE DIVISIBLE BY 3

Oops

Tyy

[tepid creek]

So 3/5?

[paper kite]

you just take the square root

of the numerator and the denominator

[tepid creek]

Okie

[tepid creek]

Ty

[tranquil pine]

@tepid creek Help need msg 🙂

It's ez Exponents..

[tepid creek]

Whats msg

[tranquil pine]

DM or direct chat here

[tepid creek]

Ok

[tranquil pine]

@tepid creek which grade you study?

[tepid creek]

9

[tranquil pine]

Ogg

Ohh*

[tepid creek]

Its easy but i get confused sometimes

[tranquil pine]

Ohh!

Don't worry you're on a right server

[tranquil pine]

You Playin' roblox ..

[tepid creek]

Whyy🥹

Is it bc its 36 hours?

[tranquil pine]

38Hours*

[tepid creek]

Oh

I was afk all night

And im afk rn

While studying math

We have a graded activity tomorrow and a test

Ahsisjisosjd

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[rose hemlock]

Find all natural numbers $n$ that have a divisor $d$ such that $n^4 + d^3$ is divisible by $n^2d + 1$.

[soft zealot]

Randel_

[rose hemlock]

What I've tried is:

n = d*k

$n^4 + d^3 = d^3(dk^4 + 1)$

[soft zealot]

Randel_

[rose hemlock]

$n^2d + 1 = d^3k^2 + 1$

[soft zealot]

Randel_

[rose hemlock]

$d^3k^2 + 1$ is not divisible by $d^3$

[soft zealot]

Randel_

[rose hemlock]

Therefore, $d^3k^2 + 1$ must be divisible by $dk^4 + 1$

[soft zealot]

Randel_

[rose hemlock]

What I am now trying to prove is that n needs to be a perfect square

aka d = k

I am stuck here

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[rose hemlock]

This is the problem

[scarlet sequoia]

more than that you need to invoke they're coprime

[scarlet sequoia]

in order to assert this

but yes you can do that

[scarlet sequoia]

also isn't it the reverse?

dk^4 + 1 is divisible by d^3k^2+1

[rose hemlock]

Yes

You are correct

mb

[scarlet sequoia]

I'm assuming the divisor must also be positive?

[rose hemlock]

Yes

[scarlet sequoia]

if so, recall that divisors are smaller than the original number when they're not equal to it

[rose hemlock]

Well, it isn't stated in the condition

But I do not think it affects the path of solution

[rose hemlock]

that is true even if they are negative

[scarlet sequoia]

mmh wait

[scarlet sequoia]

yeah but then you can't manipulate inequalities very well

i.e when you square a negative number

it's complicated

here we have to remember that dk = n

[rose hemlock]

so if d < 0, then so is k

[scarlet sequoia]

also, quick thing to notice

do we have an easy case where this is true

[rose hemlock]

if d = k

[scarlet sequoia]

and so that would mean...

[rose hemlock]

that the number must be a perfect square

that's one case

I am trying to figure out whether there's more

[scarlet sequoia]

well, if d and k are not equal

I think we should start reasoning in terms of inequalities

which one out of d and k would be smaller

assume positive divisors first

[rose hemlock]

yes

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@rose hemlock Has your question been resolved?

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[final saddle]

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[thin cloud]

Given function $y=f(x)=2024^x-2024^{-x}+\ln^{2023}(x+\sqrt{x^2+1})$. Find all $m \in \mathbb{R}$ so that inequality $$f(\sqrt{\log_{2}x}\cdot\log_{2}^{2}(2x)) + f[\sqrt{(m+1)\log_{2}x}-(4+\sqrt{m+1})\sqrt{log_{2}^{3}x}] \leq 0$$ has exactly 16 integer solutions

[thin cloud]

I have no idea where to start with

[soft zealot]

Alexis_Fx

[severe verge]

vcl tsa

chú ý hàm f(x) là hàm lẻ

[thin cloud]

Hmmm, okay

[severe verge]

để a xem

thử đặt log2(x) = t

xem có j hay ho

[thin cloud]

Hmm, yeah

Có vẻ hướng đó được

Lâu lắm rr ko làm logarithm h quên hết 😂

Não cá vàng nhanh quên

[severe verge]

chứ ko bt f(x) hàm lẻ có dùng đc ko

ô đợi

f(x) đơn điệu

@thin cloud Cái này ngon

[thin cloud]

Uk, luôn đồng biến

Khoan, sao chắc được phần bên trong ln lớn hơn e

[severe verge]

x^2+1 lúc nào cx lớn hơn x^2

nên căn của nó cx thế

bất chấp x âm hay dương

mé x>0

[thin cloud]

Chờ tí nào, đang k có giấy bút để làm

Chờ e tí

[severe verge]

thua

ko cần đạo hàm

x tăng thì log2(x) có tăng ko

[thin cloud]

Đạo hàm a ms tính đc đơn diệu chứ

[severe verge]

ko cần

[thin cloud]

Em quên rồi

[severe verge]

1/(x.ln(a))

[thin cloud]

Okay, hàm này đồng biến

Em quên hết mấy cái tính chất của hàm mũ rồi

[severe verge]

còn là hàm lẻ nx

[severe verge]

tính chất 1: nó như cứt

h e chỉ cần lấy cái dưới chuyển vế là xong

một bài tương tự từng xuất hiện trong đề thật

năm 2023 hsy

[thin cloud]

Em từng làm 1 bài ntn rồi

[severe verge]

nhìn nó ko phức tạp thế này nhưng ý tưởng y hệt

[thin cloud]

Ý tưởng là hàm lẻ rồi f(x) = -f(-x) đk

[severe verge]

vừa hàm lẻ vừa đơn điệu

[thin cloud]

Em chx kịp lấy giấy ra làm

Bài bx trc em làm cũng có hàm dạng f(x)= a^x + a^(-x)

Hình như hay ra dạng ntn

[severe verge]

a chịu

[thin cloud]

Được 15p chx

@helper

Nhầm

<@&286206848099549185>

[severe verge]

[thin cloud]

This is the question

Given function $y=f(x)=2024^x-2024^{-x}+\ln^{2023}(x+\sqrt{x^2+1})$. Find all $m \in \mathbb{R}$ so that inequality $$f(\sqrt{\log_{2}x}\cdot\log_{2}^{2}(2x)) + f[\sqrt{(m+1)\log_{2}x}-(4+\sqrt{m+1})\sqrt{log_{2}^{3}x}] \leq 0$$ has exactly 16 integer solutions

[soft zealot]

Alexis_Fx

[wet sentinel]

thử đặt t= log2(x) thì x = 2^t xem thử

[final saddle]

@thin cloud Has your question been resolved?

[final saddle]

@thin cloud Has your question been resolved?

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[lofty wing]

I don’t really understand how to fact check the middle statement, I just would like someone to explain to me the steps to confirming it or not 😓

[serene panther]

Do you know definition of increasing and decreasing functions

[tired walrus]

i mean...

[brave galleon]

increasing and decreasing functions are referring to the slopes. positive slope = increasing, negative slope = decreasing

knowing how many zeroes you have and the general end behaviors of polynomial functions will help greatly in giving you a very rough sketch of the function, which would be more than enough to get visual idea of what g(x) looks like

[lofty wing]

Wait idk I’m probably just dumb but how do I know what specific points it’s decreasing and increasing on?

[serene panther]

Where derivative is 0

Do you not know wavy curve method

[lofty wing]

I don’t know anything about derivatives 😓😭

Sorry 😢

[serene panther]

You don’t know how todifferentiate or you don’t know what it is

[lofty wing]

What it is

[serene panther]

I see

In that case

[lean imp]

Your gave to differentiate it on x

[serene panther]

I suggest you learn differentiation first

[lean imp]

Yep

[serene panther]

Before going into increasing and decreasing functions

[lofty wing]

Okay 😅 thanks

[serene panther]

Khan academy is good

https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new

[lofty wing]

I’m just confused because this is a pre calculus equation and my syllabus doesn’t talk about derivatives or differationation 😓😭 sorry about being confusing

[serene panther]

Pre Calc you say?

Then drawing a graph will help immensely

Unfortunately to draw the graph of cubics, you have to know about derivatives as well

[lofty wing]

Aw okay I’ll figure it out then

[lean imp]

Rudy you have good knowledge of mathematics, where are you from bro?

[serene panther]

India

Figures

[lean imp]

Oh nice me too

[serene panther]

Nice

[lean imp]

So are you a college student?

[serene panther]

Let’s move the discussion somewhere else

[lean imp]

Ok fine

[serene panther]

[lean imp]

Nice to meet you

[final saddle]

@lofty wing Has your question been resolved?

[final saddle]

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[soft zealot]

Mathematics (268882317391429632)

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[tranquil pine]

sin^2 theta+cos^2 theta = ?

[final saddle]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[blissful zenith]

1

[tranquil pine]

I need proof for that

[terse crypt]

Pytahgorus

[tranquil pine]

I have the answer as 1, I don't have the proof

[severe verge]

Ptolemy's

[blissful zenith]

What tool assumed?

[terse crypt]

Use definition of cosx and sinx for the unit circle

[tranquil pine]

Just basic trig

[blissful zenith]

Okay

[thin cloud]

draw a unit circle

[tranquil pine]

Is it this?

[magic coyote]

.

[tired walrus]

!msgdel

[final saddle]

The original post of this help channel has been deleted, and it will abruptly close and possibly lock. (This is irreversible.) Please claim a new channel, and don't delete the first message of any future channel you claim.

[terse crypt]

Interesting

[severe verge]

indeed

[tranquil pine]

Alright

[terse crypt]

Haven’t seen that one

[tranquil pine]

uh

[terse crypt]

This constitutes a proof ye

[tired walrus]

this channel is about to implode

[blissful zenith]

I just wrote, this is basic identity and direct algebraic manipulation

Or am I not allowed for the manipulation then

[tranquil pine]

Not sure

[tranquil pine]

I used this

but I'm not quite sure about how this transition works

[tired walrus]

yall

[severe verge]

sin x = (1 - cos^2 (x))^2 ???

[blissful zenith]

1/2

[tired walrus]

this channel is about to close, why are yall still typing in here

[tranquil pine]

When you divide all by H^2

sin^2(t) + cos^2(t) = 0

right?

[final saddle]

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.

[tranquil pine]

Find the angles between the line root3x - y +6=0 and y+3=0

[magic coyote]

Yes, we can see that

What have you tried

[tranquil pine]

I can't think of a thing

I don't know nothing

I want to learn

I have a urge to learn everything

I quit all my addictions

No games no nothing

[tired walrus]

your root symbol

[magic coyote]

Do you know of the formula $\tan(\theta)=\mod{\frac{m_1-m_2}{1+ m_1 m_2}}$

[tired walrus]

mod fail

[tranquil pine]

What's mod

...

[magic coyote]

Absolute value

[tranquil pine]

U mean

[tired walrus]

did you mean $\sqrt{3}x$ instead of $\sqrt{3x\phantom{ZZZZZZZ}}$?

[soft zealot]

Ann

[tranquil pine]

So I need to use tan theta = m1-m2/1+m1m2 ?

And I'll find the angle

[magic coyote]

absolute value

but yes

[tranquil pine]

Yes

[magic coyote]

two angles*

[tranquil pine]

So which one is m1 and which ones m2

[magic coyote]

any can be m1 and any can be m2

hence the mod

[tranquil pine]

Can u text it down

Like

My question in formula

[magic coyote]

Just find slopes of both lines

and plug them in

[tranquil pine]

So I can get the answer

So I find slope

[magic coyote]

yes

[tranquil pine]

And fuse it into formula

That's it ?

[magic coyote]

yes

yes

[tranquil pine]

Alr ima search how to find slope

[magic coyote]

[tranquil pine]

Your a genius thanks man

[magic coyote]

y=mx+c

[tranquil pine]

This server is really helpful

[tranquil pine]

How do I write it down with my question

[magic coyote]

Isolate y

and whatever the coeff of x is

is your slope

[tranquil pine]

I'll try to chatgpt once aswell

[magic coyote]

k

[tranquil pine]

Can I keep this open till then ?

Or do I close for now ?

[magic coyote]

ok

keep it open

[tranquil pine]

Alr

@magic coyote

Are you a mathematician +

?

[magic coyote]

No

[tranquil pine]

Ur very smart

I wana absorb every mathematical knowledge from u like a sponge

Yo

1 thing

How does this happen.

[dusty quarry]

how does what happen

[tranquil pine]

I understand that we compared line 1 with Ax+Bc+C=0

But

In line 2

ChatGPT showed me line 2 = 0

Y+3=0

Y=-3

...

Doesn't makes sense to me.

[dusty quarry]

it does

seems like your second line doesn't have an x

which means the value of y doesn't depend on x

[tranquil pine]

Ahhh anyone help

[dusty quarry]

that's a big clue already

[tranquil pine]

Oh

So

[dusty quarry]

if y = -3 everywhere, what kind of line do you think y + 3 = 0 is?

[tranquil pine]

Since we can't compare it with Ax+Bx+C.

It is 0 ?

[dusty quarry]

we don't need to technically speaking

it has no slope, correct

[tranquil pine]

mhmm

[dusty quarry]

so m = 0 is not surprising

[tranquil pine]

So is Ax+By+C a universal formula ?

[tranquil pine]

Sorry I'm still learning

[dusty quarry]

i've never heard of that formula before

i prefer y = mx + c

[tranquil pine]

How do we use that

Can u tell me

[dusty quarry]

rearrange both equations to say y = {something}

then compare with the general form y = mx + c

[tranquil pine]

Can we be friends since I can dm u if ur a friend?

[dusty quarry]

why do you want to dm me?

[tranquil pine]

So I can learn

I wana be sponge.

[dusty quarry]

aren't you learning here lol

[tranquil pine]

Learn everything.

Well

I could anony u in dms

[dusty quarry]

maybe look at my bio lolz

[tranquil pine]

Is y=mc+c universal formula ?

[dusty quarry]

y = mx + c is the slope-intercept form of a linear equation

[tranquil pine]

Only dm if urgent business or a friend

I wana be friend

[tranquil pine]

In this coordinate geometry can it be used everywhere to find angles ?

[dusty quarry]

so-called because m is the slope and c is the y-intercept

i don't think you can use it directly, but it exposes the slope

[tranquil pine]

Mhm

[dusty quarry]

so you can use the slope in trig calculations to find the angle

[tranquil pine]

Once can u covert root3x -y+6 =0 into y=mx+c ?

[dusty quarry]

that depends

is x also part of the root?

[tranquil pine]

Yes

[dusty quarry]

because if it is this is gonna be a pain in the ass

[tranquil pine]

...

So what's the best

To use

U got any other formulas ?

[dusty quarry]

if the x is under a root as well

sorry, that is not a linear equation

and finding an angle between a quadratic and a linear makes little sense

[tranquil pine]

Wait

Lemme send u question from

My question paper

[dusty quarry]

should have done that earlier

i suspect the x is not under the root

[tranquil pine]

Mhm

[dusty quarry]

if it is $sqrt(3)x - y + 6 = 0$ it would be trivial

[soft zealot]

Seia

[dusty quarry]

ok guess that's not the way to do roots in texit TIL

[dusty quarry]

where am i even supposed to look at

[tranquil pine]

The first question under coordinate geometry

It's written short question

[dusty quarry]

the x is not under the root

[tranquil pine]

Ohhh

So

Can u

Put it in

Y=mc+c

Lnce

Once

I got maths exam tmrw..

[dusty quarry]

y = mx + c, yes

[tranquil pine]

Put that root3 x-y+6 in y=mx+c please

[dusty quarry]

the equation is $\sqrt{3}x - y + 6 = 0$

[soft zealot]

Seia

[tranquil pine]

So

[dusty quarry]

try it yourself

[tranquil pine]

How

Mhm

Idk how

I learned different teachings

[dusty quarry]

rearrange the equation so that you have y = {something}

[tranquil pine]

My book says

This

[dusty quarry]

that works too i suppose

[tranquil pine]

Mhmmmmmmmmm

[dusty quarry]

if that's the case just follow your book and plug your values in

you already have the equations in the right format

[tranquil pine]

But y+3 =0 no a only b...

B also 0/1

1*

[dusty quarry]

did you know multiplying something by 0 makes it disappear

[tranquil pine]

Yeah ik...

Why is everyone here smarter than Mr.

[dusty quarry]

so who's stopping you from introducing 0x into your second equation?

[tranquil pine]

Me*

Is grade 10 high-school or ?

Middle school

My education system is different

[dusty quarry]

dun ask me, i'm not from one of those countries that have a high school system

back to your question

[tranquil pine]

I'll try to solve

Gimme 5 mins

Bro

My m1 is-root3

And tan inverse-1 -root 3 is -60 degree

[dusty quarry]

nothing wrong with that

[tranquil pine]

And answer shows 60 degree

[dusty quarry]

it is 60 degree

you're asked to find a magnitude

the sign only shows direction

so the raw angle between those two lines is 60 degrees

[tranquil pine]

Wasn't I supposed to find angle.

[dusty quarry]

an angle is a magnitude

[tranquil pine]

Ohhhh

I forgot formula

[dusty quarry]

that's why the helper before me told you to "mod" the formula you were given to find the angle with

[tranquil pine]

Tan theta =+- m1-m1/1+m1.m2

[dusty quarry]

because you only wanna know the magnitude

[tranquil pine]

YES

Mod

It means +- right?

[dusty quarry]

technically it's not called a mod

[tranquil pine]

So

[dusty quarry]

it's called an absolute value

mod is another thing entirely

[tranquil pine]

Mhm

[dusty quarry]

that's a whole other can of worms

[tranquil pine]

So

If have yo find another value

Do

I do

180-first value ?

Taking -

?

If we take + it's 60

[dusty quarry]

idk if you studied physics before

[tranquil pine]

If I take -its 180-60 +

?

[dusty quarry]

but we are talking about total angles here

[tranquil pine]

I'm in grade 10 bro 😭😭😭

[dusty quarry]

or an angular difference

the sign doesn't matter

[tranquil pine]

WHATS THAT.

Don't we have to find 2 angles according to question ?

[dusty quarry]

i'm not sure why you think i know what you are or have studied. that's why i said if

[dusty quarry]

if you need two angles, the other is just 180 - that angle you found earlier

that's all

no negative signs

again, absolute values

[tranquil pine]

Lemme show u smth funny

[dusty quarry]

if you get a negative, toss it out

[tranquil pine]

I just found

[magic coyote]

No

[tranquil pine]

Damn it my slow network

[dusty quarry]

not wrong. but do you see a negative angle anywhere?

[tranquil pine]

No.

But

[dusty quarry]

exactly

[tranquil pine]

I had to take

[dusty quarry]

the tangent can be negative

but not the angle

[tranquil pine]

Mhmm

I get it

Yall are super smart

Teach me yall ways

I gotta pass atleast

Atleast

Tmrw

Or my mom will be disappointed

I want to learn everything aswell

May I continue with another question ?

[dusty quarry]

yes

[magic coyote]

\sqrt{3}

[dusty quarry]

yea i found it after some google

[dusty quarry]

there my first texit usage

am proud

[magic coyote]

[tranquil pine]

The damn photo won't load

The damn photo won't load

Now my text isn't loading

.

2 photos

Very good

Very clear

[dusty quarry]

nice, now i'm doing geometry

turning head 90 degrees

this is simple. you are given the big bombshell that the two lines are parallel

the first line already has its equation in y = mx + c form

should tell you everything you need

[tranquil pine]

So

I need slope ?

Or ?

(3 -4) (-2 a) use +

?

[magic coyote]

,rotate exists

[dusty quarry]

oh.

see, told you i'm dumb : P

[magic coyote]

its bot commands kek nothing to be smart about

[dusty quarry]

huh

[magic coyote]

you just pick up from reading other ppl use it

[dusty quarry]

you need the slope of the other line, yes

the other line can also be written in y = mx + c form

but you already know the m of the second line

[tranquil pine]

Wait

Can I turn y+2x+3=0 into -2x-y+3=0 ?

[dusty quarry]

you don't need to

[tranquil pine]

Why is everything so hard when I'm trying.

[dusty quarry]

you can if you want to, but why not just kick the x and the constant term to the right

[tranquil pine]

Can u tell me steps what what i need to do.

[dusty quarry]

again, rearrange the equation into y = {something}

that's all i am going to say

[tranquil pine]

Y=-2x-3 ?

AHHHH

THIS IS HEADACHE

WHY IS IMPROVING SO HARD

SCROLLING INSTA AND SLEEPING AND PLAYING GAMES IS MUCH EASIER

[dusty quarry]

you will need some work to improve, so

[dusty quarry]

but yes, this is correct

and now you have the slope of that line

and you know the second line is parallel to the first

what does that tell you about the slope of the second line?

[tranquil pine]

That means

First line + second line =-1?

[dusty quarry]

that's perpendicular

[tranquil pine]

Eait

[dusty quarry]

parallel is even easier

[tranquil pine]

First line + second line = 0

?

I wish worked hard from earlier

[dusty quarry]

no

if they are parallel

means they never meet

[tranquil pine]

And realised this anime gaming shit isn't worth it

Only costed me my vision

ik that.

[dusty quarry]

so if the lines never meet

visually, what do their slopes look like?

[tranquil pine]

But.

Idk hro

[dusty quarry]

draw

[tranquil pine]

My damn internet

Wait

Lemme draw

[tranquil pine]

How long are you free btw

So ik u will teach me till then

...

[dusty quarry]

i'll tell you when i'm leaving

[tranquil pine]

Reply ?

[dusty quarry]

chill? i have other stuff to do too from time to time. relax.

[tranquil pine]

Alr I'll try to utilise my time fully as long as I can

I got 1 hour before I sleep

Even if I fail this first terminal I'll try better next terminal

I'll just try to learn

Ig this is the only way to think so I can learn quicker

[tranquil pine]

Smth like this?

[dusty quarry]

mhm

what do you think the slope of the second line has to be if they don't meet (aka are parallel)

[tranquil pine]

Mhmmm

Let me think

So 2nd line is (-2 a ) ?

[dusty quarry]

that's a point, not a line

ignore those two points first

[tranquil pine]

Wait.

Can

[dusty quarry]

they will come into use later

[tranquil pine]

U help me

Convert (3 -4) and (-2 a) into a line

So i can use this

[dusty quarry]

if you can do that you might as well have answered the question

you don't have a

that's why you need to get the equation of the second line from the fact that the two lines are parallel

[tranquil pine]

Oh.

So ur teaching me

How to make

This two point

Into a line

So I to make this two line into a equation

I meed to use the fact that they are parallel

[dusty quarry]

no

[tranquil pine]

.

Oh

Then ?

[dusty quarry]

you need to get the equation of the second line from the parallel part

like i said, ignore the points first

you need the gradient of the second line

or the slope

[tranquil pine]

Ok I ignored it

Now ?

I need slope?

From 2nd line

[dusty quarry]

^

[tranquil pine]

From y=-2x-3 ?

[dusty quarry]

my good sir. you are already staring at the answer

[tranquil pine]

SORRY MAN

IM TRYING MY BEST

...

So

Slope of first point

=slope of 2nd point ?

Pr

Or.

....

[dusty quarry]

slope of first line = slope of second line because they are parallel, yes

[tranquil pine]

Mhm

So now

I meed to find slope

[dusty quarry]

you could have seen it here. look at the slopes of the two lines you drew. aren't they the same?

[dusty quarry]

you alrd have it

[tranquil pine]

Howwww

OMG

[dusty quarry]

this form of the equation already exposes the m

[tranquil pine]

I DONT UNDERSTAND A THING

[dusty quarry]

that's why i told you to rearrange to this form

[tranquil pine]

Mbis -2 ?

M is -2 ?

[dusty quarry]

yes.

[tranquil pine]

So

The

[dusty quarry]

so your other line also have a gradient of -2

[tranquil pine]

Slope of the line and 2 point are

-2

Oh.

[dusty quarry]

so you now have y = -2x + c

you don't know your c

[tranquil pine]

Mhm

C

[dusty quarry]

but that's fine

[tranquil pine]

C

What do I do

[dusty quarry]

you have a point

[tranquil pine]

Yeah

[dusty quarry]

the one that doesn't have the a

[tranquil pine]

How do I use that point

...

[dusty quarry]

throw it into the equation

[tranquil pine]

Yes

Hpw

How

[dusty quarry]

substitute the x and y values into the equation and solve for c

[tranquil pine]

Substitute from?

The y=-2-3 ?

2x*

[dusty quarry]

no

[tranquil pine]

So.

From where

[dusty quarry]

one of your points is (3, 4)

points are always given in (x, y) format

so use that information

[tranquil pine]

OH

So

-4=-2×3+c ?

[dusty quarry]

oh there's a negative sign for the 4?

then yes

[tranquil pine]

-4=-6+c.

Yes

Yes sir.

Sir ?

R u there ?

[dusty quarry]

finish your working?

[tranquil pine]

Wha.

U said that's a -4 ?

Isn't that

A mistake

Or did u genuinely ask ?

[dusty quarry]

yes, i'm asking

i'm not the one with the question and you posted the question 90 degrees

if there is a - sign before the 4 then yes your calculation thus far is correct

but finish your work at least before asking if i'm here lol

[tranquil pine]

ALR

NET

Bet

Lemme cook

C=2

Now I'll substitute value in another point

[dusty quarry]

keep going

[tranquil pine]

6

Les goo

A=6

Is this correct sir ?

[dusty quarry]

yes

[tranquil pine]

YES

YEA

IM A FKING GENIUS

[dusty quarry]

also if you couldn't tell already, i'm not a "sir"

[tranquil pine]

Wdym

In respecting u.

Sir is respecting ?

Isn't it

[dusty quarry]

do you call females "sir"?

[tranquil pine]

UR A FEMALRE

Didn't know

But thanks mom

Mam*

[dusty quarry]

and you think the red diamond next to my name is for no reason or smth

lol

remember to .close the channel when you're done

so others can use it

[tranquil pine]

Is mam better or mom ?

Ma'am right ?

I need to use it more

...

[dusty quarry]

or, you know, you can just call me by name

i have a name

and you're staring at it

but that's a story for another channel

[tranquil pine]

Seia ?

Sprry

Sorry

Thanks for teaching me

Are u still free ?

I need to learn more

[dusty quarry]

just post your next question

there are many helpers here

[mellow iris]

lmao

[tranquil pine]

Alr

[bold glacier]

Hello I need help with this question

[thin cloud]

!help

[final saddle]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[dusty quarry]

!occupied

[final saddle]

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).

[tranquil pine]

My question isn't uploading

Finally

[bold glacier]

Sorry I didn't know this guy is my friend and I needed help too thank you for informing me

[tranquil pine]

Idk if my question is even right or wrong.

@bold glacier

Go to the help-no.

And

Post question there

And that's ur room

Where they help and tutor u

[thin cloud]

I can't even see what you wrote due to the light

[tranquil pine]

If the lines 11x+m1y+n1=0 and 2x+m2y+nz=0 are perpendicular to each other prove 11/2+m1.m2=1

There

[dusty quarry]

perpendicular
you already know the relationship of the slopes of two perpendicular lines

you even quoted it to me when i asked you for the slope relationship between two parallel lines

[tranquil pine]

M1.m2=-1

But

I'm not sure

If the question is correct

The question paper I got has a strange handwriting of our sir

[tranquil pine]

I thought u went off

[dusty quarry]

i really think you shouldn't be hanging on to the concept of a single helper a lot

[dusty quarry]

ask him then

thought this kind of stuff should be clarified with him already

[tranquil pine]

I jsut thought

[tranquil pine]

That gayass hits us with pipe

[dusty quarry]

?

what pipe

[tranquil pine]

Ofc a wooden one

Once I have a red scare

[dusty quarry]

so asking him is a no go or what?

[tranquil pine]

No.

I don't like him

Bearded motherfuckers

I bet he even beats his children same way

With a wooden stick

Can anyone teach...

[dusty quarry]

you already identified the concept. continue

do be patient, my dude

[tranquil pine]

Wdym

[dusty quarry]

it's not even 2 minutes since my last message and you're already sending stuff like this

[tranquil pine]

U meant Impatient ?

Scam ?

The user looks sus too

[tranquil pine]

Didn't u say not rely on a single helper urself

[dusty quarry]

and again it's been less than 2 minutes since i sent my last message

won't hurt to wait a bit more for others to come and read the long thread so they know what your question is about

anyway

you have m1 x m2 = -1

[tranquil pine]

I'll try to do it ma'am

I'm sorry

I'll patiently do