help-19

[nocturne belfry]

you know what these symbols mean, right?

the angle here

[tacit hound]

no i dont

[nocturne belfry]

they mean angle

[tacit hound]

know what the symbols mean

[nocturne belfry]

$\angle DBA$ means the angle that starts at D, moves to B, then ends at A

[clever fjord]

jan Niku

[tacit hound]

ohh i see

am i supposed to fi nd

the angle

that

goes from

[nocturne belfry]

its written here

[tacit hound]

cbd

[nocturne belfry]

right

[tacit hound]

yea

ok

so

[nocturne belfry]

and im telling you this equation

$\angle DBA = \angle ABC + \angle CBD$

[clever fjord]

jan Niku

[tacit hound]

133*

[tacit hound]

is it that

[nocturne belfry]

which you can maybe see from the picture

[tacit hound]

133

[nocturne belfry]

Green angle = (Blue angle) + (some unknown angle)

you want unknown angle

[tacit hound]

oh 60

i think

[nocturne belfry]

yea 60 sounds good

looks about right too

[tacit hound]

ok

thankyou

[nocturne belfry]

np

[tacit hound]

wait

are u still there

@nocturne belfry

.close

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[last lodge]

help

[true musk]

help about what?

[last lodge]

lim x tends to infinity

wait ill write the eq

tt\lim _{x\to \infty :}\sqrt{x}\left(\sqrt{x}-\sqrt{x-a}\right)

how do we convert this

[true musk]

$\lim _{x\to \infty :}\sqrt{x}\left(\sqrt{x}-\sqrt{x-a}\right)$

[clever fjord]

MetuMortis

[true musk]

what is the question

[last lodge]

you have to calculate the limit of the function

like when x tends to inifnity where does the above function tend to

you just have to simplify it

but idk how

[fast tundra]

so obviously, if you direct substitute. you get 0 * inf. which is an indetermined form.

[last lodge]

nope

ans is a/2

[fast tundra]

what would you do to get rid of 0 * inf ?

[last lodge]

simplify

[fast tundra]

how ?

are you still with me ?

[last lodge]

yeah

[ember oak]

ihapiw was saying that, right now, the multiplication is indeterminate, and you need to do work to ultimately get an answer.

[last lodge]

nvm got it

[fast tundra]

basically multiply it by ( sqrt(x) + sqrt(x+a) )/( sqrt(x) + sqrt(x+a) )

and note that as x approaches infinity, x-a = x

[last lodge]

ye got the answer

[fast tundra]

alright

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[real tusk]

what

[real tusk]

am I doing wrong?

[bitter knoll]

Try rounding

Or by using the Z-table

@real tusk

[real tusk]

how would I use the z table>

?

and if I did round it wouldnt work

[bitter knoll]

There is a formula

[real tusk]

I got 2.6

you cant round that to 2.5

oh the z table

but the whole lesson was about using desmos and geoalgebra

and I tried both and got 2.6

@bitter knoll

[bitter knoll]

Like for the first one it's 49.865

It could be rounded to 49.87

Same with the second one

[real tusk]

yea but the answer says its 49.85

[bitter knoll]

Here it's not really specified if u should round not

[real tusk]

so thats where im confused

because even if I round it wont work

with what desmos

and geo algebra is giving me

so do I have an incorrect setting

or is the course incorrect

[bitter knoll]

I don't thing the settings are wrong

It's probably the answer key

It sometimes has problems with decimal

[real tusk]

ill try some things out

[bitter knoll]

Also the question says approximately

[real tusk]

but if it doesnt work its a problem with the course and im just boned>

?

[bitter knoll]

It's a pretty vague term

[real tusk]

Im not sure what it meant by that yea

I dont want to have incorrect math and then contact customer service

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[lyric dust]

I was wondering if there was a way to convert vector form of plane to cartesian without using the point normal form of a plane

[lyric dust]

its quite trivial to just use determinant to find a normal vector and use pt normal form to convert to cartesian but I was wondering if there is a way that requires less 'prerequisites' since its technically not in the syllabus for my exam to use matrices

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@lyric dust Has your question been resolved?

[lyric dust]

<@&286206848099549185>

[lyric dust]

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[tiny sparrow]

You could use exterior or cross product, since a vector cross itself is 0

.reopen

[lyric dust]

.reopen

[odd edge]

✅

[lyric dust]

the determinant form is just a mnemonic for the cross product

its also not in syllabus unfortunately

[tiny sparrow]

Technically not, if you do it term by term

[lyric dust]

(just going to close since i found my answer elsewhere, thx for responses tho)

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[odd edge]

@obtuse pasture Has your question been resolved?

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[zinc mantle]

hello

[zinc mantle]

how do I find the slope of tangent

I can't use a graph for this one

3/5 is correct?

[plain hound]

Perfect

you right

The answer is - 1 so do the one that is close to - 1

C or D

[zinc mantle]

ok

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[dapper canyon]

doppler affect?

not sure how temperature factors in

does the speed of sound remain the same

what's the frequency at 20 degrees

sound is a wave

speed = frequency * wavelength

idk

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@crystal kernel Has your question been resolved?

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@crystal kernel Has your question been resolved?

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[random wing]

i revised my work based on prior feedback. Can someone please look over my work

[random wing]

<@&286206848099549185>

[warped grove]

Please don't ping Helpers before 15 minutes have passed

[random wing]

my apologies

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[river osprey]

may someone help me

[river osprey]

where does the part in the red come from

[empty bough]

what's the question?

[river osprey]

its a super long question but basically its partial fraction decompostion with integrals

so the original question is this

[empty bough]

I can't help u sry (it's too advanced for me lol)

[river osprey]

thats ok! no worries

<@&286206848099549185>

[warped grove]

Please don't ping Helpers before 15 minutes have passed

[river osprey]

oh sorry!

[ember oak]

You can integrate $\frac{1}{(u^2+1)^2}$ with the substitution $u=\tan v$. Maybe that will help you get to the red.

[clever fjord]

SWR

[odd edge]

@river osprey Has your question been resolved?

[river osprey]

1 second..

[river osprey]

dude, you are seriously the best.. i dont even know how you had an eye for that, thats awesome thank yo so much @ember oak

AHHHH THAT MAKES SO MUCH SENSE OH MY GOOODNESS

[ember oak]

thank yo so much @am1r

[river osprey]

IT WAS ACCIDNET

LOL

i @'d myself cuz i wanted to send my work on my iapd but i could not fgindf it

[ember oak]

hah

yeah whenever you see x²+1 in the denominator, it's almost always x=tan(u)

[river osprey]

man who knew one little trig substitution could do all that

THANK GOD.

[river osprey]

alrighty will take note

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[mystic saffron]

i have no idea how to do this.

[ember oak]

you want to isolate y to be on just one side of the equal sign

[mystic saffron]

oh ok

writing it down rn

[glad osprey]

Yeah so add both sides by 13

Then multiply both sides by 11

[mystic saffron]

ohh

i have never been taught this so im really bad at it

[ember oak]

I was hoping to guide OP there rather than give it all away at once...

[glad osprey]

Sorry

Maybe give her another question then?

[mystic saffron]

im still solving this one

almost done

[ember oak]

Looks like you're learning it now, so there's that

[mystic saffron]

thank you

ok for both sides I got 33 and 260

if that makes sense

[summer cradle]

omg hi

[wanton bison]

i was about to ping ya

[mystic saffron]

helloo

oh i did it wrong

lemme restart]

got it right

now its this one.

hmm

[wanton bison]

In order to solve such equations you need to first know about arithmetic operations

You can add, subtract, multiply and divide by both sides

The goal is to manipulate the equation with arithmetic operations in such a way so that we isolate the variable

The only things you are not allowed to is to multiply and divide by 0

That being said

Your goal now is to isolate c

Your first question in mind should be how can I get rid of the +29 and the factor -3

Any ideas?

[mystic saffron]

yes

thank you

its making sense

i will try to figure it out from here

igtg soon

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[quartz pier]

why is there a 5 at the front of y

shouldnt it have been

yz^2(5x)^(yz^2-1)

used a calculator online and it gave me an output with a single 5 aswell

oh shit i am braindead its (5x)^ not 5x^

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[quartz pier]

.reopen

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✅

[quartz pier]

wait but even then it should have been

5^(yz^2) * yz^2 (x)^(yz^2-1)

right?

[dawn tiger]

since you're differentiating wrt, you can isolate the functions of x and lump everything else together as a constant if that helps

(5x)^(yz^2) is kind of like (5x)^C in the x world

[weary pelican]

you just rewrote what was already written in a different way

[dawn tiger]

which becomes C * (5x)^(C-1) * 5 when differentiating

[weary pelican]

in any case, (f(x))^k gets differentiated as kf'(x) (f(x))^(k-1) as per the chain rule

[quartz pier]

oh shit ur right

🤦

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[delicate pilot]

Calculate the area of an equilateral triangle inscribed in a circle of radius 6 cm.

[delicate pilot]

Is the answer is 27 sqrt3?

[dawn tiger]

yes

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[wispy vine]

I walk at 3km/hr and miss the bus by 2 minutes. If I walk 4km/hr, I reach 3 minutes before the bus arrives. How far do I walk to reach the bus stand?

[wispy vine]

How do you do this

how i tried to solve it was assume it takes an hour to walk to the bus stop

for walking 3km/hr

so he is at a distance at 2900

metres

however if he walks 4km/hr

then he only needs to be at a distance of 3800m to arrive

nvm

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[polar hazel]

coulld someone explain this

[polar hazel]

i dont understand how the sqrt 3's went away into one sqrt 3

[green elm]

they had 1/2 powers on them

[supple mural]

x^1/2 * x^1/2 = x

[polar hazel]

but isnt it .5 + .5 makes 1

wait

[runic swift]

exactly

power of 1

[polar hazel]

since they are being multiplied

u add exponet

so 1

can someone agreee

[runic swift]

yes indeed

[polar hazel]

tahnks

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[signal trout]

describe k recursively and create an FSA that accepts K

[latent scaffold]

What do you think?

[signal trout]

∅ ∈ Σ, K ⊂ Σ*, a^3b ∈ Σ

[latent scaffold]

Sigma is just the alphabet you're given. The language is K

What's the smallest string you can have in K?

[signal trout]

∅?

[viscid flint]

wait is K a language or a set of languages

[latent scaffold]

It's a subset of Sigma*, so I assumed it was a language, although I agree that it is written weirdly.

[viscid flint]

yeah

ok

well ∅ isn't a string

[signal trout]

so then a^3b as the base?

[viscid flint]

idr whether computer science tends to have 0 as a natural number but yes

[signal trout]

no i think it starts as 1

[viscid flint]

kk

[signal trout]

actually im not sure i dont think that was ever mentioned in this class but in my other class it starts at 1

[latent scaffold]

It won't change the rest of the exercise much. Maybe an extra state in the FSA if anything.

Now that you have a base string for K, if I give you any string in K, call it w. Can you give me a string that will also be in K?

[signal trout]

a^3bw

because concatenation

[latent scaffold]

Well if w is in K to begin with, it should have a^3b in there in the first place. Rather, what could you add to the end of w, and yet still get a string?

[signal trout]

another b?

[latent scaffold]

Indeed. So we can define K as :
aaab is in K, and for any w in K, wb is in K.

[signal trout]

K ⊂ Σ*, a^3b ∈ K, w ∈ K logically implies wb ∈ K

i think i might know the fsa but im not 100 percent confident

i wasnt sure whether to put an a in front of the start state or not but i figured that since it is certain that there are 3 a's, then i can just do 3 states to add the a's and then the last one is to add as many b's as are in the word

[viscid flint]

seems legit

[signal trout]

thank you

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[rich wharf]

when you integrate tan/sqrt tan wrt some variable, is there some trig rule you can get to derive arcsec?

[rich wharf]

im working through a physics problem and it involves that integral and i looked at the solution and it somehow gets from that integral straight to arcsec

i literally don't know how on earth that integral was solved when other sources i look at find nasty looking solutions to integral of sqrt tan

i can send a photo, not looking for answer btw, I just would like insight on how this integral is tackled

as it is just a small part of solving the overarching problem

[odd edge]

@rich wharf Has your question been resolved?

[shy sundial]

whats the integral?

[magic girder]

is this it?

[magic girder]

but yeah the solution is pretty huge

[rich wharf]

Right. I agree

So this is where I am confused.

I am clueless as how they went from what is effectively just square root of tan to arcsec

[shy sundial]

what is the original expression?

[rich wharf]

sure

[shy sundial]

or what is the physical problem in words

[rich wharf]

ah

I don't care necessarily about the physics. It's basically asking that when F = 0 that some particle should move in a straight line.

I just am having trouble with the math between the beginning and the solution

And i don't understand how the solution got freakin arcsec

[shy sundial]

i think there's a typo

maybe tan^2 ?

[rich wharf]

ohhh was it supposed to be tan^2?

[rich wharf]

oh yeah

its supposed to be tan^2

so it cancels out

and you just get a constant

making the function constant

icic

yeah this book has a lot of typos in the solution so i always read them w a grain of salt but for this problem i had no idea how to integrate that

tfw they forget 1 exponent and makes me waste 3 hours 🥲

[shy sundial]

im also kind of lost in the notation

i can see where it is going but its a mess lol

but yeah it was a typo

ok now i understand forget the last two. i just noticed he just replaces phi back from r=Asecphi and -phi_0 is convertend into +constant

[rich wharf]

yepp

that is exactly whats done

i got that part

[shy sundial]

but i think the constant should be outside sec^-1

[rich wharf]

phi-phi0 gets subbed in by Asecphi or whatever it was

[shy sundial]

maybe sec^-1(something+constant) still equals sec^-1(something) + another constant so that step is valid, im not sure

anyways yeah lol i think thats it

[rich wharf]

yeah now the rest of the problem is just the physical interpretation

now, I just solve to get r on its own to find the motion of the particle

show that it's a constant

implying that when the force F = 0, particle moves in straight line

so convoluted lmao

thank you though, sv

svc

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[short scaffold]

Can someone check why is it wrong?

[desert marlin]

Have you tried an online derivative calculator

[short scaffold]

more importantly, the second derivative one

[short scaffold]

ive graphed the 2nd der in desmos and my answer doesnt match up

[desert marlin]

does it by chance differ only by a constant

[short scaffold]

also i found that the second term on my answer should be positive for it to match up

[short scaffold]

^

so maybeee I messed of on the sign of my 2nd term?

this is confirmed by an online der calc

oh shoot i noticed why!

apparently i used a negative sign on the product rule lmao

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[exotic blade]

I don't fully understand the concept of "n" and how it is applied here. If someone could explain "n" in terms of the question and walk me through it thatd be great.

[sand horizon]

Ok so you know that trig function are known to be periodic

[exotic blade]

yea like between waves

[sand horizon]

Like sin is the same every 2pi

[exotic blade]

mhm

[sand horizon]

Sin(0) = sin(2pi)= sin (2024pi)

[exotic blade]

so for every multiple of 2pi?

[sand horizon]

You have a value, you add a multiple of 2pi and its the same

Sin (x) = sin(x+2pin)

And that n comes from here

[exotic blade]

so the period for the equation is 2pi/12 right

[exotic blade]

oh and that means the identity is the same is 2pi n

because multiple of 2pi

ohhh

[sand horizon]

So pi/6 indeed

So we gonna solve it

Rq

[exotic blade]

yea !!!!

[sand horizon]

Cos(x) = -1 what are the values ? Possible

[exotic blade]

could you rephrase that sorry

[sand horizon]

What are the value of x such that cosx = -1 ?

[exotic blade]

180?

[sand horizon]

Yeah pi [2pi] indeed

So we have cos(12x) = cos(pi [2pi])

The method gives us two possibilities,
12x = pi[2pi]
Or 12x = -pi[2pi]

For the first one what we get ?

[exotic blade]

just to clarify its pi * 2pi

right

[sand horizon]

Pi + 2pin

Its the mod notation

[exotic blade]

oh okay

x=

pi/12 +pi/6

[sand horizon]

pi*n/6 indeed

And for the other one ?

[exotic blade]

-pi/6?

[sand horizon]

• pi*n/6

Indeed

So

We can look closer to the answer

And compare

[sand horizon]

-pi/12 so

[exotic blade]

i made a typo sorry

so its -pi/12

and -pin/6

[sand horizon]

So now we look at the answer

We get

We have pi/12 + pi*n/6, so its pi/12, 3pi/12, 5pi/12 etc..

And on the other hand we have -pi/12 + pi*n/6
So -pi/12, pi/12, 3pi/12 ....
Something like to be the same right ?

[exotic blade]

yea because its an even function

it has the same sin reguardless

[sand horizon]

Indeed

So what is the answer ?

Pi/12 = 15° and pi/6 = 30°

[exotic blade]

ookok

but in the answer set

i dont understand why its written like that

[sand horizon]

I dont know, id honestly prefer works with radian

[exotic blade]

it makes sense for this one since pi*n

but there should be two answers at least

i feel

[sand horizon]

Since its even

[exotic blade]

yea

[sand horizon]

There is not

Or there is the same actually

[exotic blade]

hm

it was pi/12 -pi*n/6

[sand horizon]

That minus is irrelevant since the n can compense it by taking the value n = -1

n is an integer

[exotic blade]

yea

but that still only leaves one answer

[sand horizon]

Two identical answer

[exotic blade]

is it D?

[sand horizon]

No its C

[exotic blade]

so theres only one answer?

oh what if it was

pi *2/12

does that justify D

[sand horizon]

Pi*2/12 = pi/6 so no

[sand horizon]

Ah you mean if student make an error ?

Well yes, its a trap

[exotic blade]

oh no

it was just me being paranoid

c is the right answer thanks !!!

.close

[odd edge]

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[thick fiber]

Identity

Sen = sin

[sand horizon]

U want to show this ?

[thick fiber]

Yes

[sand horizon]

Well tg = sin/cos

Sin/cos - sin = (sin -sin*cos)/cos

(Sin - sincos)/cossin^3(x)

Sin(1-cos) / cos sin^3

[thick fiber]

Wait that last step

Nvm nvm

[sand horizon]

1-cos / cos sin^2

[thick fiber]

is that what I have to prove

[sand horizon]

(1-cos)/cos(1-cos^2)
1-cos /cossin
Since secx = 1/cos and 1 + cos = 1/sin
1-cos /cossin
sec / 1+cos

[odd edge]

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[arctic dawn]

i don't get why the volume of the cone is subtracted here? would appreciate any help!

[trim owl]

There’s an additional line in the problem statement

[arctic dawn]

i don't understand

which one exactly?

[trim owl]

The line “3y-x=3”. Based on the solution, I’m guessing the curved sqrt function is the inside

Maybe it isn’t. I take that back

[arctic dawn]

hmm still don't get why the volume of the cone is subtracted

[arctic dawn]

i get the former part of the line i've underlined here, cuz u rotate the curve but why subtract the volume of the cone from it?

[ancient patrol]

u subtract the volume if the line is outside the curve in which case the volume is the difference between the lines/solids (don't trust me on this)

[arctic dawn]

wait should i construct the line they've given?

the y = x/3+1 thing

that slipped my mind

[trim owl]

You definitely should lol

[trim owl]

This is true, but I assumed that he already accounted for the line

Or she

[arctic dawn]

ohhh got it now

[arctic dawn]

i totally ignored the line part lol, thanks!

[ancient patrol]

thar we go

[odd edge]

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[scenic vessel]

Need help solving this qu

[astral mirage]

does the answer is 7 ohms ?

[boreal crag]

start by finding a set of resistors which are all in series

[boreal crag]

!nosol

and no

!nosols

[odd edge]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[boreal crag]

that solution is not correct, and again

!nosols

[odd edge]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[astral mirage]

why ? where did it went wrong ?

[astral mirage]

what is no solution

[boreal crag]

read

can you not see the bot messages or something

[astral mirage]

sorry my bad

new here

[scenic vessel]

the ans is 14/3 ohm but i dont get how

[boreal crag]

can you find a set of resistors which are all in series in the problem

[scenic vessel]

the 2 6 1 2 and 3 ohm ones?

[boreal crag]

that one doesn't count because there are other resistors branching off as well there

you want to find something like

wire leads in

resistor

resistor

resistor

wire goes out

without any branches

[scenic vessel]

Oh so the 1 2 and 3 ohm ones

[boreal crag]

yea exactly

so those can be thought of as one big resistor

since they form a series

[scenic vessel]

yea

so it would basically be a 6 ohm resistor right?

[boreal crag]

yeah exactly

so now you can kind of redraw or reimagine the setup as looking a bit simpler

[scenic vessel]

alr what abt the other redistors

[scenic vessel]

yea

[boreal crag]

and the next challenge will be finding resistors which are in series here

not series sorry

parallel

[scenic vessel]

4 and 3 ohm?

[boreal crag]

not quite because again there's another resistor in between them

[scenic vessel]

oh…

Uhh im not sure

[boreal crag]

it'll involve the new 6 ohm resistor we just created

[scenic vessel]

wait remind me what sre the conditions for parallel again?

[boreal crag]

wire goes in, branches to 2 paths which each have a resistor in them, then reconnects at the end of the 2 paths

[scenic vessel]

ohhh ok

then wouldnt the parallel be 2 6 6(new) 3 and 4?

[boreal crag]

only 2 of those are in parallel

the 3 ohm and the new 6 ohm

[scenic vessel]

Oh

[boreal crag]

maybe there's a better way to describe the requirements for parallel

they both have to start at the same place and they both have to end at the same place

by the same place i mean there's no resistance between the start of one and the start of the other (or the end of one and the end of the other)

for series on the other hand, you want one to start where the other one ends, and you also want there to not be any other branches in the middle

[scenic vessel]

Umm

ok…

i think i get it

[boreal crag]

all right nice

so next is 2 resistors in series

[scenic vessel]

my doubt is why isnt the 4 in the parallel

[boreal crag]

because there is a 6 ohm resistor between it's starting point and the starting point of the new 6 and the 3

[odd edge]

@scenic vessel Has your question been resolved?

[odd edge]

@scenic vessel Has your question been resolved?

[nova ginkgo]

@scenic vessel hello

[subtle sleet]

hi

[nova ginkgo]

not you

[odd edge]

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[tawdry agate]

Is this shape geometrically possible? where 3 hexagons which aren't on the same plane meet?

[sick plaza]

no

not regular hexagons

afaik

[tawdry agate]

so the angles are slightly skewed?

[sick plaza]

the angles in this aren't 120 per hexagon

[tawdry agate]

Yea I guess that makes sense.

Could it be constructed with regulars on a non-euclidean plane though?

[sick plaza]

if the space the object occupies is non euclidean

wouldn't be a plane

but a space

hyperbolic space

then it could be possible because then any point in that space has more than 360 degrees required to be "flat"

for example on a sphere

a triangle has angles adding up to 270

specifically

the right angle triangle

with all right angles

idk actually if all triangles do

imagine taking those hexagons

and putting them together in that shape

and having some angle left not covered by a hexagon

in such a space a shape like this might be possible

[tawdry agate]

Right. I am gonna see if it could work for my application later

.close

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[alpine laurel]

Anyone help me simplify: 5 sqrt(12) • sqrt(18) - 3 sqrt(8) • sqrt(48)

[sand horizon]

Sqrt(12) = sqrt(3x4) = sqrt(4) x sqrt(3)

=2sqrt(3)

Apply the same to the others

[alpine laurel]

Sqrt(18) = sqrt(2 x 9)

Sqrt(8) = sqrt(2x4)

[jovial gate]

ez

[alpine laurel]

Sqrt(48) = 4 sqrt(3)

[jovial gate]

5.root12x18 - 3.root8x48

[alpine laurel]

Btw here are the choices

[jovial gate]

you can multiply root

[alpine laurel]

5 sqrt(216) - 3 sqrt(384)

Now what

[jovial gate]

lcm

[alpine laurel]

5√12 x 18 - 3√8 x 48
5√216 - 3√384
5√36 √6 - 3√64 √6
5 x 6 √6 - 3 x 8 √6
30 √6 - 24 √6
6√6

Like this right?

Thanks

.close

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[broken sail]

Can someone explain how we solved the last 4 steps? The ones highlted in black and indented to the right a bit

[broken sail]

I believe this question asked to get the third deravative

[nimble blaze]

the work so far is just the first derivative

[broken sail]

Then how was it done?

[thin kelp]

factorising is the first step

[nimble blaze]

separated terms with dy/dx from terms without
factored out dy/dx, essentially solving/isolating like a linear equation

[thin kelp]

common denominators for step 2

[nimble blaze]

combined terms into a single fraction

[thin kelp]

then division to get dy/dx

[nimble blaze]

then division, (again like solving a linear equation)

[broken sail]

Pardon my idiocy but how did we go from step 2 to 3? Like in the denominators how did we get rid of ln(5)?

[nimble blaze]

multiply both sides by ln(5)

[broken sail]

Makes sense, what about the last step?

[nimble blaze]

then division, (again like solving a linear equation)

the same way you'd solve
pq = r
for p

[broken sail]

Yes, but the division itself confuses me

[nimble blaze]

division by a fraction is the multiplication of its reciprocal

$\frac{a}{\br{\frac bc}} = a \cdot \frac cb$

[clever fjord]

ℝαμΩℕωⅤ

[broken sail]

Ahaaaa

Amazing!

Thank you so much man!

[odd edge]

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[quartz adder]

Hii

[hallow wind]

I'm confused has to how I'm supposed to do this. I don't typically ask for help, so I'm sorry if I'm not doing it right. I can't exactly show my work since I don't know how to do this in the first place, so I just need a head start with how to do it. Thanks

[hallow wind]

Sorry, I didn't know you were here first-

[sand horizon]

Hum, he did not ask any questions tho

[hallow wind]

Oh alright, I just saw the room was claimed by him so I thought I was interrupting something..

[sand horizon]

Wait ima ask

<@&268886789983436800> sorry to bother you all

Someone claimed by saying hi, without question and connor drop a question at the same time

[viscid flint]

better if you get a new channel so that the pings go to you and you have the option of closing it

[hallow wind]

Okay, thank you then :]

[hallow wind]

Thank you for helpin' me out

[quartz adder]

Can i ask my doubt..will anybody judge me

??

[sand horizon]

Ask your question

No place for judgement here

[quartz adder]

If a = c^z , b = a ^x and c=b^y , prove that xyz = 1

Its class 9 question

[quartz adder]

ok

[quartz adder]

@sand horizon

[sand horizon]

Do you have an equality for a,b and c ?

[quartz adder]

meaning

[sand horizon]

Nvm

Ok

So

[quartz adder]

yes

[sand horizon]

We have a^x = b
So a^xyz = b^yz
a^xyz = (b^y)^z
a^xyz = c^z
a^xyz = a ...

And now compare the power, and there you go

[sand horizon]

All good ?

[odd edge]

@quartz adder Has your question been resolved?

[odd edge]

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[ember shell]

uhmmm i have a basic idea where i can use 1/2bh

and i got stuck there

i went through the problem a few times and then i even looked on the answer key

but the explaination given was just way too confusing for me

i understand everything but the formula given for the tip of the angle

i dont see why it should be x-s

[orchid torrent]

It doesn't have to be

you can pick almost anything wlog tbh

[ember shell]

huh

what does that mean

😅

im extremely new to this kind of stuff

[desert marlin]

Without loss of generality

[ember shell]

...

i dont know what that means

😓

uhm

this was supposed to be a geometry problem

supposdely

supposdedly

god dammit

supposedly

[ember shell]

i actually didnt really understand that part tbh..

this is so confusing

[orchid torrent]

In essence, if you draw a cevian from a vertex down to the opposite side, the ratios of the original shaded regions and the parts of the regions cut off by the cevian will be the same

[ember shell]

...

An auxiliary line?

[orchid torrent]

A cevian is a segment drawn from one of the vertices of a triangle to the opp. side

[ember shell]

Oh..

Can you show me how I'd draw that??

This entire problem confuses me beuh

[orchid torrent]

https://mathworld.wolfram.com/Cevian.html

see the picture

[ember shell]

Yes

Uh

But like

...how do I draw that here

I don't have any points in the middle

[orchid torrent]

"as dividing the angle doesn't change the problem" refers to drawing a cevian

tbf the line could lie outside the triangle but

same idea tbh

just treat the original side as a cevian

[ember shell]

Oh

Uhm ok

So then

Back to the angle

Why is it y=x-s

I can't really understand that too much

[orchid torrent]

that's what we've been talking abt the whole time tbh

the slope of the line (which defines how much the angle is divided by) doesn't matter (see earlier discussion)

we just introduce an intercept to make sure the line passes through A

[ember shell]

....maybe this solution is too hard for me 😭 I have never learned any of what you guys have been saying

Idk what a cevian or any of that is

im pretty sure this might be the solution she wanted

but that one kind of confuses me

everything is so confusing

[orchid torrent]

it just saying that the ratio of A and B

would be the same as the areas, of say, these regions

[ember shell]

oh ic

i just

AH :FKSSDHIUFLK>JDLSDK

im confused about why the formula is like that

i think i understand what that is now

ok wait

uhmmm

sorry

[orchid torrent]

region C is just this

minus this

rinse and repeat the logic for B

[ember shell]

oh

so

i can understand that

so

y im assumign is teh tip??

so x-s is like

1-s?

1x-s

[ember shell]

does it not mean that the parallel lines are 1 2 3 4 etc etc

?

[orchid torrent]

that's ... exactly what it's saying tbh

[ember shell]

yeah

[orchid torrent]

I'm just showing you where the solution says it

[ember shell]

oh ic

right so the tip will be. y=x-s thats very confusing to me

i dont understand where the equation comes from

i can understand the meaning of it

[orchid torrent]

that's what we've been talking abt the whole time tbh
the slope of the line (which defines how much the angle is divided by) doesn't matter (see earlier discussion)
we just introduce an intercept to make sure the line passes through A

did you not read this?

[ember shell]

afkdahfa;

yeah i did

im still confused about that

i dont really get it sorry

[orchid torrent]

yk how the tangent of the angle b/w a line w/ eq. y = mx + b and the x axis is m

[ember shell]

Yeah

Wait

What

[orchid torrent]

draw a picture and draw the classic triangle that's used to give visual intuitition behind slope if you want to

the pf is trivial

[orchid torrent]

but the pt is that the slope of the line is going to define the angle b/w the x axis and the cevian (which is the line here)

[ember shell]

?

[orchid torrent]

as we've gone over earlier, how we split the angle doesn't matter

so we want to make our life easy

make the slope 1

However, we don't know for sure that y = x will pass through A

so we just shift the line by some arbitrary amount, namely s, to force that

[orchid torrent]

preferably in coordinate space

[orchid torrent]

now find the horizontal and vertical displacements in terms of the slope and some arbitrary scalar

[ember shell]

.....idk what that is

Wait

Is this still geometry

😅

im so sorry dude

genuinely.

i think that the only solution i recognize is the second one but the first one is a lot simpler

but i dont get the concept of the first one

and the second on also confuses me

😭

im really struggling rn

[orchid torrent]

I was referring to this arg

[ember shell]

i erm...

i also have not learned that

😅

ok ok ok

how about we take a step back

[ember shell]

i deffinitely dont think this is something i can do without spending an hour on i know that the second solution is simpler if it doesnt bothe ryou can you help me with that

im really sorry dude

i didnt expect it to take this long

genuinely

[orchid torrent]

uh

I'm too lazy to scroll

can you send it again

[ember shell]

oh sure

its uhm

its proportions

i dont really get why you need to square it

tbh

[orchid torrent]

ratio of areas = (scale factor)^2

you can probably look a proof of this if you want

but it's not hard to derive yourself

[ember shell]

what is a scale factor

[ember shell]

ah ok

[orchid torrent]

also known as ratio of similitude

basically the ratio of corresponding side lengths b/w two similar figs

[ember shell]

oh ic

so i think the second one

would be

((x+2)/(x+1))^2

?

right

[orchid torrent]

b/w this triangle

and this triangle?

yeah

[ember shell]

huh

yeah

i think thats what it is

wait no

uhm

the first one is x+1/x which means that its the second over first

so then

the second should be x+2/x+1 which means third over second right?

or should it be third over first?

i think its second

but my real confern is

what does this mean

[orchid torrent]

order is off but it's whatever

[ember shell]

oh uhm ok

[ember shell]

do you multiply by x or something

i mean x^2

[orchid torrent]

yeah

[ember shell]

so then

do you multiply by (x+1)^22 in the second one then?

[orchid torrent]

yeah

[ember shell]

ohh

and then we subtract by x^2+1

or i mean

(x+1)^2

[orchid torrent]

yeah yeah

[ember shell]

so then that's shaded region b?

so now all we need to do is find c

which would be uhm

(x+4)/(x+3) squared?

[orchid torrent]

that's the ratio, but you have the right idea

[ember shell]

okk

so its 2x+7

meaning

uhm

[orchid torrent]

yeah

quick aside

you can be a bit lazy

The area of the regions is given by $(x+2n)^2-(x+2n-1)^2$

[clever fjord]

Civil Service Pigeon

[orchid torrent]

the coefficient of x is clearly 2 from this

and the constant term is 4n - 1

but you don't need to find that directly

just recall that the first differences of a quadratic sequence form a linear sequence

hence the constant terms of the expressions of A, B, C, D form a linear sequence

ok that's my tangent over

[orchid torrent]

back to you

[ember shell]

ok...

that confused me a lot :D

so uhm

should i use the same logic with uhhh

a and d

technically

well

uhm since technically a is just x^2

it should just be 1/6 right

1/6 squared

mb

so then we need to find d?

[orchid torrent]

uh

you're throwing around some relevant expressions and numbers

but there's no coherence to your thoughts

so imma just let you try it and we'll see what happens

[ember shell]

huhh

uhm ok

thanks

[orchid torrent]

normally once ppl try it

their thoughts become more coherent since they actually know what's happening instead of just trying to communicate a rough plan

[ember shell]

oh ic

[orchid torrent]

yeah being able to see how everything will play out instantly is nice but

don't expect it lol

[ember shell]

i think from what i got was ((2)(1/6)+11)/(1/6)^2

[orchid torrent]

that matches with this

[ember shell]

oh

i forgot about that for a second

okk

thanks a lot man

[orchid torrent]

🍻

[ember shell]

im really sorry about that

😅

[orchid torrent]

I'm here to kill time

dw abt it

[ember shell]

ur very patient

oh ok thx a lot

😄

[orchid torrent]

I'm waiting for my food to slow cook 💀

[ember shell]

nah thats torture tbh

[orchid torrent]

c'est la vie

🤷‍♂️

[ember shell]

lol

ok thanks again i sent a friend req cuz ur nice

😄

[orchid torrent]

being honest here, I generally don't do friend requests just from helping ppl

[ember shell]

ohh thats ok

[orchid torrent]

if I interact w/ you more and we're chill, that's when I open up

[ember shell]

lmaoo

its fine

we all have preferences

on how to make friends

but thanks again

:D

[orchid torrent]

[ember shell]

.close

[odd edge]

Channel closed

Closed by @ember shell

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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[wide drift]

I absolutely HATE word questions

Like theres only 3 numbers

So this question is probably super simple

But I just have no idea what to do

I dont even know what im looking for, the length of the purple line right?

[mystic swan]

Yes