help-23

[polar birch]

the relative error will just be given by ds/s or (84/pi)/(84/pi^2)

which is 1/84

or 0.012

For volume V = (4/3)pir^3 which reduced to c^3/6pi^2

so dv is given by (1/2pi^2)C^2dc

so subsituting those values

dv becomes 1/2pi^2 times 84^2 times 0.5

which is equal to

[turbid marsh]

quick question: how do you know how to do this one but need help with distributive property?

[polar birch]

1764/pi^2

i mean you dont need algebra to know calc

[turbid marsh]

fair

[desert juniper]

wait, why are you obtaining derivatives here?

[turbid marsh]

differential uses derivatives

[desert juniper]

uh... you get the surface and volume assuming the circunference is 83.5cm and 84.5cm?

[polar birch]

if u want to find the maximum or relative error most of the time you will need to derive the equation

[polar birch]

whats blud talking about

[desert juniper]

it's a sphere.
You got the circunference.
You can get the radius.
With the radius you get surface and volume.
The minimum radius is gonna be when the circunference is 83.5cm, thus minimum surface and volume.
The maximum radius is gonna be when the circunference is 84.5cm, thus maximum surface and volume

the maximum error in each is the difference between maximum and minimum

[turbid marsh]

vanilla is correct, what he said matches the answer key

[polar birch]

i see how you can use that buts its specifically asking to use differiantials

[desert juniper]

fair enough

[turbid marsh]

alright, thanks for your help.

.close

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[tidal fern]

No idea how to do this. I always end up with equations that are true but don't help in solving for T

<@&286206848099549185>

[dusky crag]

I can try helping

Not sure tho

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[wicked echo]

How do i draw an equation like y=(3x+1)/(3-2x) by using limits? I forgot the concept which my teacher showed us.

[rocky ferry]

by draw do you mean graph/plot?

[wicked echo]

graphed on a cartesian plane

[upbeat ridge]

How do you graph a function using limits

Limits is just an expression where the function is headed towards

[rocky ferry]

well, i assume you mean the asymptotes of the curve.

[wicked echo]

Well yes but you can use it as a guideline to help you draw it...

pretty much

[rocky ferry]

what level math is this

[wicked echo]

it's 13th grade in italy

[rocky ferry]

do you know any calculus yet?

[wicked echo]

If you're referring about derivatives, not yet.

If you're referring about sinx, log, etc. yes

[rocky ferry]

well, regardless you don't really need to know it to understand this concept. whenever the curve is a quotient of two functions, you can determine the asymptote by the degrees of the numerator/denominator

since they are both first degree, we can ascertain it will be a diagonal asymptote.

you're probably wondering, diagonal? but they will tend toward +-infinity. yes, but by diagonal i mean the curve will be reflected along x=y

so we can calculate one side of the discontinuity then just reflect it.

[wicked echo]

Alright... and if i want to find the asymptotes' equations so i could draw it?

[rocky ferry]

x and y intercepts are good starts, then we can examine the discontinuity.

do you know how to look for the asymptote here?

[wicked echo]

No, I can't

[rocky ferry]

so this curve is a quotient. that means division is happening. what is one thing we can't do with division?

[wicked echo]

divide by 0

that means 3-2x is different than 0, meaning x is different than 3/2

[rocky ferry]

well, 3/2 is where the asymptote occurs.

[wicked echo]

and for the other side i guess i'd just have to rotate it 90 degrees which in this case it's simple, meaning that the one for the other asymptote is -3/2?

[rocky ferry]

what? no, 3/2 is an x-value, a vertical line placed at x=3/2

[wicked echo]

I understood that for the x axis, i'm now asking for the y axis, sorry for being unclear there.

[rocky ferry]

well, we would have to determine the limit as x->-infinity

do you know how to do this?

[wicked echo]

We'd have to use the condition of(i can't draw the cursive E so i made it "i"):=======
For every i>0 there exists I(-inf): I(-inf) belongs to the real numbers excluding {-inf} => |f(x) -l| < i?

By the way i was rummaging through my notebook and i saw the example which the professor used: f(x)= (2+x)/(3-x). He turned it into [x(1+2/x)/x(3/x-1)] and from there he found the y and the x asymptotes

I also realised that my thought process was wrong initially

[rocky ferry]

well, the method specifically for limits at infinity involving any rational function is done by dividing the numerator and denominator by their respective degree. (largest power of x)

[wicked echo]

I see. I also applied his method now to the previous equation and i finally got it. Just to recap, this is what i believe the method is that the professor used:

Did i understand it right?

[rocky ferry]

yeah so we have a horizontal asymptote at y=-3/2, and a vertical asymptote at x=3/2. then you can find the intercepts and just connect the dots, then reflect what you drew on the left side of x=3/2 along x=y to draw it again on the right side of x=3/2

[wicked echo]

alright. I have one more question though, out of curiosity.

[rocky ferry]

?

[wicked echo]

Not going to lie, I have no idea what to call this monstrosity i've created, but umm...

If for some apparent reason in an alternate universe i'd have seen that equation, how would i be able to graph it? or is it too hard for pre-university?

[wicked echo]

because you stated that at the beginning so i got creative

[rocky ferry]

yeah i was mistaken on that part, i misremembered it has to be a difference of 1 for it to be slanted asymptote. but looking at this, its the same exact procedure.

[wicked echo]

oh, and for x there are 2 different values which inside those is the inner line

[rocky ferry]

yeah i see. the x intercepts can be found with the quadratic formula, and y intercept is straight forward

this is easily graphable by hand

discontinuities and intercepts are the most important parts of drawing your graph

[wicked echo]

the hyperbole part yes, the part in between seems a bit more difficult in my opinion

[rocky ferry]

its not, really, its the same exact procedure.

you just have to use the quadratic formula in this case, which provides 2 answers instead of 1.

[wicked echo]

ah yeah i see now you're right, i'd have to use the top formula to find it the interception point with the x axis, and knowing the boundaries i'd be able to draw it

but how would i know if i were to draw it right? as in if i didn't know how it looked like, how would i know if it's a function which increases or decreases?

[rocky ferry]

Derivative, or finding it through the limit way (x+h)

You can brute force it by looking for certain values too. Checking small intervals for change in y to estimate what direction it's going.

[wicked echo]

alright, thanks for your help!

[rocky ferry]

👍

[wicked echo]

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[hazy sentinel]

i don’t understand this, welp

[summer wigeon]

what don't you get

[hazy sentinel]

should I turn the radian to degrees in order to do this?

[summer wigeon]

no need

one full rotation is 2 pi radians

[hazy sentinel]

i see, then idk where the radian lies

[summer wigeon]

is the angle measure was pi then what would it look like

[hazy sentinel]

idk

[summer wigeon]

it would be a line right

[hazy sentinel]

yeah

[summer wigeon]

okay do you know the unit circle

forget that

forget the unit circle

[hazy sentinel]

Okay

[summer wigeon]

You understand 360 degrees is 2pi radian right

[hazy sentinel]

ohh

yeah, i see

[summer wigeon]

okay so 1pi radians would be 180

and pi/2 would be 90

[hazy sentinel]

yeah

[summer wigeon]

so now approximate 5pi/6

[hazy sentinel]

5pi is 810

and divided by 6 is 135

[summer wigeon]

okay but without doing that

if you know pi is 180

and pi/2 is 90

is 5pi/6 less then pi or more then pi

[hazy sentinel]

more

[summer wigeon]

okay is 5/6 more then 1 or less then 1

[hazy sentinel]

less

[summer wigeon]

so is 5pi/6 less or more then 1pi

[hazy sentinel]

more

[hazy sentinel]

according to chatgpt

but it’s less

[summer wigeon]

🙂

good

and so is it more or less then pi/2

[hazy sentinel]

5pi/6 is more than pi/2

[summer wigeon]

so what quadrant would it be in

[hazy sentinel]

[summer wigeon]

like where would the angle be

would it be between 0 and pi/2

pi/2 and pi

pi and 3pi/2

[hazy sentinel]

[summer wigeon]

or 3pi/2 and 2pi

[reef estuary]

😐

[hazy sentinel]

what

imma watch a video

[reef estuary]

is your question which quadrant 5pi/6 will be in?

[summer wigeon]

yes

[hazy sentinel]

yea here

[reef estuary]

ok look

compare 5pi/6

to pi

both have pi, so when comparing, you dont need to think about the pi

just compare 5/6 to 1

pi is 180 degrees, so completely flat on the left would be pi

and you have 5pi/6

5/6 is almost one

but not quite

it is less than 1

so which option do you think it could be

[hazy sentinel]

gimme one moment

C

[reef estuary]

yes

[hazy sentinel]

thx

thx yall

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[tough briar]

Where do you get N = 2 from?

Also, can you show your work for the integral?

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[heavy glade]

is this right?

[rancid sand]

I have no idea what ur writing...

Pls write it better

[heavy glade]

also theta is angle between include and x

[hard crest]

you've got a naked T in the first and a Tcos(theta) in the second

[heavy glade]

is this better

[hard crest]

clearly one of those isn't right

[heavy glade]

yes

wait why

the second t is refering to the vertical one

[hard crest]

what are we summing up here

the forces on the upper box, right?

[heavy glade]

forces

the system

[hard crest]

T is pulling parallel to motion

[heavy glade]

in the parallel axis and the perpendicular axis

[heavy glade]

yes

[hard crest]

so why do you have a perpendicular component?

[heavy glade]

for the T pulling the other box up?

is that wrong?

[hard crest]

this doesn't seem right

if you have one component you should have the other component too

[heavy glade]

wait your right

[hard crest]

$T + F_{g_1}\sin\theta$ accurately describes the forces on box 1 parallel to the slope

[heavy glade]

for perpendicular I left out a Tsin(theta)

[flat frigate]

hayley.tiff

[heavy glade]

for the top one

[heavy glade]

would it also accurately describe the system in the parallel axis since box 2 has no parallel component

for perpendicular?

[hard crest]

what are these forces being applied to...

[heavy glade]

box 1 and box 2 in the perpendular direction

[hard crest]

box 2 is very simple, it has T countering Fg2

[heavy glade]

for box 1?

[hard crest]

yeah

[hard crest]

box 2 doesn't need a perpendicular or parallel component

or rather everything is parallel to the rope

[heavy glade]

I understand no parallel but why no perpendicular

[hard crest]

the forces on box 2 are very simple

there is tension pulling it up

and there is gravity pulling it down

[heavy glade]

yes but since im summing up the system perpendicularly, i need those components because they are not perpendicular to the slope I thought

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[gilded shell]

Need help with this problem. I keep getting 4.961

[torpid fable]

Can you show your work?

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[cunning kelp]

For part a, why multiply by 5!

[cunning kelp]

Since the rooks are indistinguishable shouldn't the order matter, does not needing to multiply by 5!

[quasi bison]

say you have a rook on each of the 1st through 5th ranks

(a.k.a. rows but it's chess terminology)

it matters whether you have rooks on a1, b2, c3, d4, e5 or b1, a2, d3, e4 and c5

even though the set of ranks and files occupied by the rooks is the same in each case

[cunning kelp]

But isn't this part why we do (8P5)^2?

[quasi bison]

8C5.

we do (8C5)^2 to determine the set of ranks and files that the rooks will go in

[cunning kelp]

The answer key uses permutation not combination though

[dapper venture]

the answer given equals ${}_8C_5\times{}_8P_5$

[flat frigate]

WhereWolf(ping if needed)

[quasi bison]

$\binom{8}{5}$ is a combination not a permutation

[flat frigate]

Ann

[cunning kelp]

Hmm okay thank you, I don't really get it but I'll try again

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[cosmic grove]

$\frac{4}{x-2} + \frac{x}{2-x}$

[flat frigate]

Herels

[obsidian vine]

How do you solve this

Is iti 8 + x^2 -6/(x-2)(2-x)

[cosmic grove]

do you know how to add fractions with different denominator ?

[obsidian vine]

LCD

[cosmic grove]

yea

[obsidian vine]

What's the LCD, is it not (x - 2)(2 - x)?

[cosmic grove]

it is

[obsidian vine]

So this is correct?

[cosmic grove]

(8-4x + x² - 2x) = x² -6x +8

[obsidian vine]

So that's all?

[cosmic grove]

maybe, but its better to simplify

$\frac{4}{x-2} + \frac{x}{2-x} = \frac{4}{x-2} - \frac{x}{x-2}$

[flat frigate]

Herels

[cosmic grove]

faster

[obsidian vine]

Ohhh

Nevermind I get it

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@obsidian vine Has your question been resolved?

[obsidian vine]

Wait I just asked my friends, they said x-4/2-x is wrong. They said that the correct answer is x-4/x-2. Why is this? <@&286206848099549185>

[obsidian vine]

Where did you get x-2 on the second fraction from?

[reef estuary]

(x-4)/(x-2) is wrong. tell your friends they are wrong

the right answer is (x-4)/(2-x)

[waxen sapphire]

ur friend is correct bro

[lean otter]

What to do?

[reef estuary]

?

[obsidian vine]

Where did x-2 come from in the denominator of the second term

[reef estuary]

multiply top and bottom by -1

[waxen sapphire]

wait wait
​

tell me the question again ps

[reef estuary]

.

[obsidian vine]

.

[waxen sapphire]

k

[cosmic grove]

2-x = -(x-2)

[reef estuary]

@obsidian vine ans is (x-4)/(2-x)

= (4-x)/(x-2)

[obsidian vine]

Is it not LCD?

[cosmic grove]

??

[reef estuary]

??

bro just multiply by -1 to get the same denominator

[obsidian vine]

Huh

Oh

[cosmic grove]

^

its the same as (x-4)/(2-x)

[waxen sapphire]

here

hmm

yup

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[broken flame]

Can somebody check if it is correct?

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@broken flame Has your question been resolved?

[broken flame]

<@&286206848099549185>

[broken flame]

<@&286206848099549185>

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[plucky elk]

.close

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[normal tartan]

I just want to get this vocabulary clear:

Local extrema, global extrema, local maximum, local minimum, stationary point, inflexion points, turning points

[normal tartan]

from what I know, the local extrema of function are its turning points?

[hardy lion]

Yes, they are interchangeable

[normal tartan]

ok, and then

local max and min are types of local extrema

[hardy lion]

Yes

[normal tartan]

local max is a concave down extrema and min is concave up

[hardy lion]

Yup

Not really when you would use vice versa though lol

[normal tartan]

min is not concave up?

or vice versa means something else

[hardy lion]

vice versa means something else

[normal tartan]

ok so local extrema are a part of stationary points

but i think a stationary point also includes inflexion points on top of that

?

inflexion points are points where the gradient of the derivative changes its direction of approach as far as i know

not too sure about global maxima though

also not sure if what I said about stationary points is true...

pls ping if anyone answers

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[cunning ember]

hey yall

[cunning ember]

the question and the solution are both in the pic
edit (the solution is wrong)

idk shit about trig

so can someone tell me what went down

where did 1 + sin11/cos11 come from

explain to me like youre explaining a baby lmao

[twilit spindle]

factor out the cos11 :)

[cunning ember]

like dividing cos 11 on both numerator and denominator?

[twilit spindle]

$\cos(11)\cdot1+\cos11\cdot\frac{\sin11}{\cos11}$

see if that helps

[flat frigate]

FancyBredFries

[cunning ember]

huh

[cunning ember]

ok its the same as cos11 + sin 11

but rewritten in terms of idk

[twilit spindle]

see if you can correlate it with this:

$ab+ac=a(b+c)$

[flat frigate]

FancyBredFries

[cunning ember]

yeah

[cunning ember]

ok i understood this part

what next?

[twilit spindle]

that's how they get 1+cos11/sin11 lol

[cunning ember]

so why is there 1 minus sin11/cos 11 in the denominator
shouldnt be 1 plus sin11/cos 11 as well?

[cunning ember]

this is what im talking about

the denominator

[twilit spindle]

oh

I see, my bad I misinterpreted your question

well cos11+sin11 will just cancal out, no?

[cunning ember]

how?

[twilit spindle]

because they're the same?

[cunning ember]

wha

[twilit spindle]

$\frac{a}{a}=1$

[cunning ember]

hows cos and sin the same

[flat frigate]

FancyBredFries

[cunning ember]

but its cos 11 and sin 11

not sin 11 and sin 11 or cos 11 and cos 11

[twilit spindle]

cos11+sin11=cos11+sin11

[cunning ember]

so how did this become

this

[twilit spindle]

by factoring

[cunning ember]

ok yeah

but what about the denominator

like how did this become

https://i.imgur.com/6Qkgo75.png

this

[twilit spindle]

looks like you added a negative where there shouldn't've been one?

[cunning ember]

probably idk

[cunning ember]

ok i have one more solution to that problem

this

[cunning ember]

https://i.imgur.com/p9ts8Fz.png

where did (tan pi / 4) come from in the denominator

.close

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[noble solar]

So the DV (diavolume) is calculated by P(Total Permeate)/CF(Concentration Factor) * 9
and the value of P is constantly increasing as time goes by.
The process will end if the DV reaches 9.

I need help with the mathematic equation where the value of CF changes let's say from 2 to 3 after certain time.
So the value of DV should be the amount of DV while CF was at 2, plus the amount of DV while CF was at 3, not just whole value of DV changes to value of 3.

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@noble solar Has your question been resolved?

[safe radish]

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[tender scaffold]

confused on how to format my answer

[old veldt]

well, your format should be in

a <= x <= b

for A and B

for C it should be in the format

p, q, r

[old veldt]

and as the question states

You either use set notation (U) or separate it with "," I think

[tender scaffold]

THANK YOU

🙏

.close

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[radiant jetty]

Could someone tell me why this is incorrect?

[errant rain]

there’s no help channels left 💀

[radiant jetty]

rip 💀

[shut wigeon]

cant they just print more lmao

[lean otter]

inflation

it's autumn

[errant rain]

i call next help channel

[lean otter]

send it

[errant rain]

actually can i just use this one

[lean otter]

ohh

[radiant jetty]

lmao alr 😭

[buoyant gazelle]

mystifine taking one 4 the team

[errant rain]

alr so

[lean otter]

soooo

[errant rain]

Is that correct

[lean otter]

you have fantastic penmanship

[errant rain]

ik

[lean otter]

math checks out

it's correct

[errant rain]

alright @radiant jetty have ur channel back

[lean otter]

can you show off some more of that penmanship pls

[errant rain]

i’ll write ur user

[radiant jetty]

thanks

[lean otter]

do you use any special equipment?

[lean otter]

impressive

[errant rain]

i use a bic pencil 0.5mm

[lean otter]

but the tidyness with the matrices is really something

[errant rain]

yea i kinda have ocd so that may explain why

[errant rain]

but thanks for confirming

[lean otter]

my pencil/pen is a lot thicker

[errant rain]

i used to have that pencil

that silver one

[lean otter]

ah really?

it's not bad

[errant rain]

graphgear?

[lean otter]

that's right

i've had this one for a few years now

those black pens are also real great

they're japanese

uni ball

[errant rain]

i just use this nowadays

with this lead

[lean otter]

i use 2b as well

but 0.7mm

I have a deathgrip on my pencil

[errant rain]

yea i feel like 0.5 is a little too thin

always breaks

[lean otter]

Yeah, and even I snap 0.7mm all the time

[errant rain]

damn

[lean otter]

I was that kid in middle school who abused the fuck out of their pencils

so my grip is accustomed to that

@errant rain how far into linear algebra are you?

[errant rain]

just started the 3rd week

[lean otter]

really? I started 4 weeks ago

[errant rain]

we're covering linear transformations and linear independence rn

[lean otter]

linear combinations?

[errant rain]

which topic are you guys currently on

[lean otter]

I'm just self-studying

[errant rain]

oh really

[lean otter]

Right now I'm learning about determinants

Inverse matrices

Using inverse matrices to solve systems of linear equations

[errant rain]

this is our calender

[lean otter]

I see

fairly comprehensive

[errant rain]

yep

[lean otter]

https://www.engineer4free.com/linear-algebra.html

I'm using this as my main reference (path)

[errant rain]

no way

im saving that

[lean otter]

you want more?

[errant rain]

how is that free lmfao

[errant rain]

sure

[lean otter]

https://tutorial.math.lamar.edu/Classes/Alg/AugmentedMatrix.aspx

https://tutorial.math.lamar.edu/Classes/Alg/AugmentedMatrixII.aspx

https://tutorial.math.lamar.edu/Classes/Alg/NonlinearSystems.aspx

good problem sets

https://www.youtube.com/playlist?list=PLX2gX-ftPVXVuWEutmvA163NoDCAhiWv2

https://www.youtube.com/playlist?list=PLX2gX-ftPVXUT6cBuDAjO2Z2XURDgdp8f

https://www.youtube.com/playlist?list=PLX2gX-ftPVXWjzkAjEu4Sd16R4wz55Bwe

[errant rain]

very helpful

i saved them

[errant rain]

thanks for sending

[lean otter]

np

[safe radish]

@radiant jetty Has your question been resolved?

[safe radish]

@radiant jetty Has your question been resolved?

[radiant jetty]

https://cdn.discordapp.com/attachments/903481106819612714/1161062514516308038/image.png?ex=6536eeba&is=652479ba&hm=f201866cf1037b98272c329efd4b7a9db0a111de1051ae4c06cc8cf3ed5db3da&

<@&286206848099549185> Why is this incorrect?

[safe radish]

@radiant jetty Has your question been resolved?

[sinful zodiac]

TO GET EXPECTED VALUE, YOU NEED TO INTEGRATE THE PRODUCT Y*F(Y)

RN, YOU'RE ONLY INTEGRATING F(Y)

[radiant jetty]

Expected value is the mean ah okay

Ok so I just have to re-integrate thanks

.close

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[radiant jetty]

waited 3h for that 🙂

[safe radish]

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[fickle grove]

discontinuous at -5 but how do i know if it's point, jump, or infinite discontinuity ?

[gritty glacier]

use LHL RHL

[fickle grove]

wahts that

oh

how do i do that

[gritty glacier]

,, \lim_{x\to 5} \frac{x^2 -25}{x+5}

[flat frigate]

Bettim

[fickle grove]

do i plug in -5?

i got -10 as the limit with factorin

factoring

[gritty glacier]

wait

so at 5 its continous

[fickle grove]

yeah

[gritty glacier]

we can ignore that

now for -5

[fickle grove]

i js wanna know if its gonna be point, jump, or infinite discontinuity at -5

or like how to find out

[gritty glacier]

if LHL and RHL is equal it is jump discontinity

no wiat

point

[fickle grove]

can u explain how to find the lhl and rhl

i photomathed it and it said its infinite but idk how

[gritty glacier]

wait

so

for LHL

x is slightly less than -5

so we represent it as

$\lim_{x \to -5^-} \frac{(x+5)(x-5)}{x+5}$

we can cancel the term

[fickle grove]

why x-5 in the denominater

[gritty glacier]

ooops my bad

[flat frigate]

Bettim

[gritty glacier]

$\lim_{x \to -5^-} (x-5)$

[flat frigate]

Bettim

[gritty glacier]

is this right?

[fickle grove]

yeah thats hwat i did

[gritty glacier]

now sub in -5 in the place of x

[fickle grove]

yeah i got -10

[gritty glacier]

do it same for the rhl

[fickle grove]

wouldnt it be the same

[gritty glacier]

yes

if it is the same then you find f(x)

if all 3 are equal then its continous

[fickle grove]

wait what do you mean

-5 is discontinuous right

[gritty glacier]

if lhl and rhl are equal but f(x) is not equal then its discontinous

[fickle grove]

do you plug it into x-5?

.close

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[fervent wagon]

.close

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[hoary mirage]

hey guys, Im stuck on this accounting problem

[hoary mirage]

can anyone guide me

I just cant understand problem 2c

<@&286206848099549185>

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@hoary mirage Has your question been resolved?

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@hoary mirage Has your question been resolved?

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@hoary mirage Has your question been resolved?

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[pallid void]

Original function is 1/x

I said that the shift was a reflection across the y axis and a 2 unit shift to the left

but apparently its a 2 unit shift to the right...

why is this the case, whenever you add n units shouldn't it shift n units to the left?

[quasi bison]

reverse the order of those and you would be right.

[pallid void]

hmmm why can't I use the original order?

[quasi bison]

cause it gives you something different

reflection across the y axis gives you 1/(-x) and then shifting 2 left will give you 1/(-(x+2))

or -1/(x+2)

[pallid void]

so 1/(-(x-2))?

So its always a shift before reflection?

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@pallid void Has your question been resolved?

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[rare hornet]

Hi

[frigid locust]

hello

post ur question?

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[stable pollen]

So far I’ve done (2/3)^4 and (1/3)^3 and I was gonna assume that I should find the lowest common multiple of the two but I’m fairly certain that’s incorrect here…

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@stable pollen Has your question been resolved?

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[fierce wind]

Was trying to solve this problem from Bertsekas' book, but I am confused with why it is a problem? We can always have a mapping for the optimal pis subsequences right and then set PI to be a set containing the pi sequence? Can someone help here?

[safe radish]

@fierce wind Has your question been resolved?

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@fierce wind Has your question been resolved?

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[dark mesa]

red ones are what i tried

[somber sail]

it's the area of the triangle

x2

[dark mesa]

what is the area of what triangle?

also what is x2

[somber sail]

the area of the triangle can be got by two ways

first. use the right angle

1/2 * a * b = triangle area

[dark mesa]

yes

[somber sail]

second. use a side and height

1/2 * c * h = triangle area

so therefore

1/2 * a * b = 1/2 * c * h

a * b = h * c

[dark mesa]

1/2 * c * h = triangle area can you explain this?

[somber sail]

well,

consider a rectangle

that you drew with red lines

[dark mesa]

yes how can we be sure that is half of it

[somber sail]

do you know triangle's congruence

[dark mesa]

yes

[somber sail]

when we are young, we just learned that it is 1/2

but it could be proved by the way

uh...

can i let

(what was the name)

[dark mesa]

i think i see it

[somber sail]

ok

[dark mesa]

a*b is 4 triangles right?

4 equal triangles

[somber sail]

there are 4 equal triangles

like you said

[dark mesa]

thank you so much

.close

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[somber sail]

hmm.

[dark mesa]

.reopen

[safe radish]

✅

[dark mesa]

what happened?

[somber sail]

wait not exactly equal because a and b isn't confirmed equal

there are two equal triangles on each side

did you understand?

[dark mesa]

it's hard to imagine from text

what each side?

[somber sail]

like this

we don't know if a and b is equal or not

[dark mesa]

and why that matters exactly?

oh

[somber sail]

you cannot say there are four equal triangles

[dark mesa]

okay

[somber sail]

.close

[dark mesa]

yeah that's right

wait do not close that means i still couldn't get it

[somber sail]

uh

you could get it

[dark mesa]

wym

[somber sail]

let's consider left side triangle

a

[dark mesa]

oaky

[somber sail]

and right side triangle

b

the total area

[dark mesa]

a*b

[somber sail]

is 2a + 2b right?

[dark mesa]

what

yeas

yes

a is area

[somber sail]

i took the other variable

to explain this

ok

2a + 2b is the rectangle area

and a + b is the triangle area

[dark mesa]

yes

[somber sail]

so it's half

due to congruence

[dark mesa]

oh yes

nice

[somber sail]

now you got it

[dark mesa]

thanks

.close

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[royal wolf]

hi guys, can you help me?

[bold aurora]

Which one are we doing?

[lean otter]

any would be fine, you can start from the first one

he just needs a thorough explanation

[royal wolf]

Yes

[safe radish]

@royal wolf Has your question been resolved?

[bold aurora]

Let's do the first one, any ideas?

[royal wolf]

Well, we need to solve these equations

and I'm bad at math

[bold aurora]

Well ok, if you have no idea then I can give you a tip to start with:

• The top equation looks very close to (x + y)^2
• We can manipulate the bottom equation to get a x + y

[royal wolf]

Do you know how to solve this yourself?

if you can solve, please solve, I just don’t understand anything

[bold aurora]

We don't do solutions here, sorry

[royal wolf]

Okey

[stray warren]

maybe

try working on

doing the questions

as a first step to be pro

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@royal wolf Has your question been resolved?

[safe radish]

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[lean otter]

in a group where all elements have finite order that means that the group has finite order?

[light shoal]

not necessarily

[lean otter]

thanks

[peak estuary]

you should really try finding an example of such a group

[lean otter]

yea we need the condition that every proper subgroup has finite order

the subgroups generated by the elements are not sufficient

[lean otter]

and finite number of proper subgroups

[clear blade]

you should also try proving that the converse is true

[peak estuary]

thats still not enough

[lean otter]

with this it is enough

[light shoal]

even this is not sufficient

https://mathoverflow.net/a/120830

[lean otter]

ik bro

.close

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[potent rain]

How to do

[potent rain]

This one

<@&286206848099549185>

[vagrant jolt]

Is the 76% a part of the q?

[potent rain]

what?

[potent rain]

what does that mean?

[vagrant jolt]

The 76% at the top

Next to similarity

I’m guessing it’s not but just making sure

[potent rain]

ah no no no

its just saying the probability a similiar question comnes in the exam

[vagrant jolt]

Ah alrighty then

[potent rain]

yea

[potent rain]

so we have to find some sort of scale factor

[vagrant jolt]

Indeed what do you think you have to do first

Or anything you notice which you can use?

[potent rain]

we got a sort of ratio

[vagrant jolt]

Yh precisely so the question says PR divided by AC is equal to 2/3

Hence we can see that PR is equal to 2

And AC is 3

[potent rain]

ok

so we just assume one of 2cm and the other is 3cm

@vagrant jolt

[vagrant jolt]

Mb assume was the wrong word

We know it

Because if PR divided by AC is 2/3 then PR must be 2 and AC must be 3

[potent rain]

ok

understood

so we have the area of one shape

area of bigger shape / area of smaller shape = k^2 if im not mistaken bro

[vagrant jolt]

Let’s do it bit by bit first

Let’s work out everything we can

Wait have you worked out length of PQ already?

[potent rain]

no mate

[vagrant jolt]

Yh so let’s work out everything first

Got a pen and paper with you?

[potent rain]

yea i do

[vagrant jolt]

So this is what we have rn

Because we know PR is 2 and that AC is 3

Using that we can now work out the scale factor

What do you think it is?

[potent rain]

length scale factor?

[vagrant jolt]

Mhm

[potent rain]

so if we do bigger shape / smaller shape

it will be AC / PR

[vagrant jolt]

👍👍👍👍

So what’s the scale factor?

[potent rain]

3/2

[vagrant jolt]

Yup

It’s 3/2 smaller

Or 1.5 times smaller

[potent rain]

got it

so therefor

9 x 3/2 to get PR?

wait nahhh thats wrong mate

[vagrant jolt]

carry on

So what do you think we do

Keep in mind that triangle ABC is the bigger one

[potent rain]

what if we divide

[vagrant jolt]

👍👍👍

[potent rain]

so PQ is 6

[vagrant jolt]

Because we want to find out the side of the smaller triangle

[vagrant jolt]

Yup

If we had it the other way around and we had the smaller one then we multiply by 3/2

[potent rain]

next question says Area of PQR

[vagrant jolt]

Mhm

[potent rain]

so area scale factor

@vagrant jolt

[vagrant jolt]

No

Be careful here

We can’t just apply the same scale factor to the area

What’s the formula for area of a triangle?

[potent rain]

1/2bh

and 1/2absinc

[vagrant jolt]

This

We can’t rlly use 1/2absinC because we don’t have any angles given and can’t work them out

[potent rain]

alright

but we dont have h

[vagrant jolt]

But can we work it out?

Because we have the area of the first triangle

And have the base

And then we can use the scale factor on the height to find out the second triangle height and work out area

[potent rain]

how?

area scale factor is your length scale factor squared?

[vagrant jolt]

This works too but let’s try it this way

[potent rain]

ohhhh wow

so we got height of first triangle

[vagrant jolt]

Yup

Now we apply the scale factor to get height of second triangle

And then do 1/2bh

Because we know the base of second triangle

Which is 6

Do this and see what you get

[potent rain]

same scale factor as 3/2

[vagrant jolt]

Yup

Because the 3/2 scale factor is used for lengths

[potent rain]

4 divided by 3/2??

[vagrant jolt]

Mhm

And then what?

[potent rain]

its 8

[vagrant jolt]

👍👍👍👍

[vagrant jolt]

But yh this formula works too

But I wouldn’t always rely on it

Use it if you’re rushing in your exam and don’t have time or to check

[potent rain]

ok i get it now bro

i got my final tomorrow. Any advice?

[vagrant jolt]

If you don’t know how to do a q don’t stay on it

If it’s taking long and you don’t know it then write anything (take an appropriate guess) and move on

[potent rain]

hopefully the paper is easy bro

[vagrant jolt]

Don’t worry and get good sleep

Make sure to get atleast 8hrs

Otherwise you’ll make silly mistakes or wouldn’t be able to think straight

[vagrant jolt]

Also if you did this then we would do 3/2 squared

Which is 2.25

And divide 18 by 2.25 to get area of other triangle which would still be 8

But like I said only do it like this if you’re in a rush

[potent rain]

oh alright man

[vagrant jolt]

Also close the room if you’re all good

🫡

Good luck

[potent rain]

thanks man

.close

[safe radish]

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[potent rain]

Can you help with this @vagrant jolt

[vagrant jolt]

Alrighty

So this is what your diagram should look like

So the only combinations where we can take a green button and one without is if:

We first take a green button and then not a green button

Or if we take a not green button and then take a green button

If you’re confused when to add or multiply think of it like this

When in your head you say or then you add

When you say and you multiply

Or when you go across the diagram you multiply, when you go down you add

Everything good?

[potent rain]

Ok alright thanks bro

[vagrant jolt]

Anything that doesn’t make sense? Or is it okay?

@potent rain

[potent rain]

Yea it makes sense

Thanks bro

.close

[safe radish]

Channel closed

Closed by @potent rain

Use .reopen if this was a mistake.

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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.

[safe radish]

Please don't occupy multiple help channels.

[sturdy swallow]

@vagrant jolt do uk how to do this?

[safe radish]

Please do not ping individual helpers unprompted.

[sturdy swallow]

alr

[quiet juniper]

What's a pronumeral

[sturdy swallow]

a

[quiet juniper]

Oh nvm I see it now

[sturdy swallow]

u might need to draw this out tbh

[sturdy swallow]

got an idea?

[quiet juniper]

nah still thinking

[sturdy swallow]

yep ag

[lean otter]

@sturdy swallow Picture a vehicle going along the path

[sturdy swallow]

nw

[lean otter]

Say it starts at angle 0° and the angle increases clockwise

[quiet juniper]

oh solid idea

[sturdy swallow]

hmm

[lean otter]

I could kiss you

oops sorry, force of habit

[quiet juniper]

lol

[lean otter]

At each turn, calculate its new angle

@sturdy swallow For example, at turn 1, what would its angle be?

[sturdy swallow]

90 degrees?

wait but were r we starting from?

[lean otter]

90° would imply it turned to the right exactly

[lean otter]

Top line

We're saying it starts with an angle of 0°

[sturdy swallow]

ohhh

[lean otter]

By first turn I mean first angle

What would its angle be after that turn?

[sturdy swallow]

im not sure bro

[lean otter]

How about now?

[sturdy swallow]

135

ahhhh okee

[lean otter]

And after the second turn?

[sturdy swallow]

what would be the second turn?

after the second turn?

[lean otter]

If you picture a vehicle going along the path, it'd just be the second angle it encounters

[sturdy swallow]

60 degrees?

[lean otter]

That's how much it turns

But what would its angle be after that turn?

[sturdy swallow]

70 degrees

[lean otter]

Solid

Aight

3rd turn?

[sturdy swallow]

thx g

ah

290

[lean otter]

The angles add up

Which is why I'm confused

Because 70+290=360

OOOOH wait

Right, I made the same mistake as you

[sturdy swallow]

lol ag brother

do u wanna know the answer?

i have the answr

but no idea

how to get it

[lean otter]

This is how much it should turn left

[sturdy swallow]

ohhhhh

120

[lean otter]

Yep

Using that same logic, how much does the vehicle turn on the 3rd turn?

[sturdy swallow]

110

[lean otter]

That's right, now add all the turns up until now

[sturdy swallow]

365

[lean otter]

The 120° was a turn left

So it removes from the total angle

[sturdy swallow]

ohh alr

245

[lean otter]

135-120+110=245?

Oh I see

You took your previous answer and removed 120

[sturdy swallow]

oh mb

[lean otter]

You should've removed it twice though

[sturdy swallow]

125 degrees

[lean otter]

What would be its angle after the 4th turn?

[sturdy swallow]

hmm

im not sure bro

[quiet juniper]

you don't need this at all but for what little it's worth I figured out how to do it with 0 eyeQ insight now (i.e. just drawing extensions of lines and lotso angle chasing)

[sturdy swallow]

lol

is there an easier way doing this?

[lean otter]

That's basically what we're doing

[sturdy swallow]

perhaps creating parallel lines in the middle or smth

im not sure tho

[lean otter]

I feel like you can't get much simpler than this

[sturdy swallow]

tru

[quiet juniper]

Yeah

[lean otter]

Just calculate turns and add up total angle

[sturdy swallow]

wait so what would 4th turn be?

[lean otter]

Continue the line and draw the angle if you're uncertain

Just like I did

[sturdy swallow]

i dont understand bro

is it ok if u can give that same diagram

but for 4th turn

15° (alternate angles in parallel lines)

this was the answer btw

idek