help-23

[tardy mango]

You can say that $y=a|x-4|-2$

[flat frigate]

Civil Service Pigeon

[tardy mango]

the coordinates (x,y) of any point on the graph will satisfy this

and thus gives you an equation in a that you can solve

(also, minor remark: math is case sensitive, so a and A aren't the same thing)

[mighty totem]

so a doesn't necessarily matter?

[tardy mango]

what do you mean

what I said above is how you determine the value of a

[mighty totem]

Ohhhh okay thank you, how do you solve for a

[tardy mango]

Test a point on the graph

pick a point on the graph

[mighty totem]

(3,2)

[tardy mango]

you're doing this graph right

[mighty totem]

yes

if the the point I already gave you doesnt work then do (4,-2)

[tardy mango]

substitute x=4 and y=-2 into this

what do you get

[mighty totem]

-2= a|4-4|-2

[tardy mango]

mhm

keep going with that

can we solve for a using this

[molten acorn]

I mean, the point (4, -2) always works no matter what a is?

[tardy mango]

I was going to have them figure that out and then generalise

[molten acorn]

Ah I see

[tardy mango]

but anyway the cat is out of the bag

you need to make sure that the point you pick to compute a

doesn't make the term with a equal to zero

[tardy mango]

ex. here, a|4-4|=0a, meaning the term with a vanishes

so we can't use this to solve for a

but any other point that doesn't make the a|x-4| will work

so pick another point

[mighty totem]

okay so (2,-4)

[tardy mango]

alright

what equation do you get from that

[mighty totem]

-4=a|2-4|-2

[tardy mango]

good

can you solve for a using this now

[mighty totem]

yes

[tardy mango]

alright I'll let you do that

[mighty totem]

i got a=-1

[tardy mango]

good

and that's how they got this here

[mighty totem]

ah okay

[tardy mango]

can you do this yourself now

it's the same thing

just a different "type" of function

[mighty totem]

yes thanks for your help

[tardy mango]

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[soft lava]

I’m looking back in pay homework and I’m very confused . How was I able to extend the fraction line without multiplying anything

[tardy mango]

where exactly are you talking about

[soft lava]

5,6,7

The second step I took the first one in pencil I would just extend the line and I know this is correct because my teacher checked it but I don’t undestand how it is

[soft lava]

Yes

[tardy mango]

that is multiplying the fractions

$\frac{a}{b} \cdot \frac{c}{d}=\frac{ac}{bd}$ for $b,d \neq 0$

[flat frigate]

Civil Service Pigeon

[tardy mango]

aka when multiplying fractions, you can multiply the numerators to get the numerator and same for the denominators

[soft lava]

So u can just extend the line?

[tardy mango]

if you want to use that as a shortcut, then sure

[tardy mango]

but this is what is happening

[soft lava]

Kk ty

[tardy mango]

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[orchid ibex]

I have a question regarding the creation of a formula for a puzzle. Its a game where the only move allowed is a jump, where a piece jumps over another piece horizontally or verdically adjacent to it. the piece must land on an empty space. So to the left, right up or down; jumping diagonally isnt possible. The piece that is jumped over disappears. For example the piece in the bottom right jumps over the piece above it and lands two spaces above its starting space. There is now one fewer piece in the game. A next move would be: the leftmost piece jumps over the buck to the right of it and lands two spaces to the right. At the beginning of the game, a certain number of pieces may be places in the white squares in the 'starting position' after all pieces are placed, they can only jump over each other. the goal is to get as far as possible into the gray area. My question is: How can i create a formula which shows the minimal number of pieces needed to go to each grey row.

[peak estuary]

so conways soldiers?

frankly, experimentation

[orchid ibex]

yeah i guess, my teacher presented it as the goat jump game or sm

[peak estuary]

things like this often do not have nice "formulas"

(in fact you cant go arbitrarily far into the gray area, no matter how many pieces you start with)

[orchid ibex]

yeah I know that the 5th row is unachievable, I'm just trying to optimize reaching the 4th row and was wondering if there's an approach that isnt pure trial and error

[peak estuary]

I dont doubt that such an approach exists, but the way you would discover it is by first doing trial and error a lot until you understand the problem well

which again, is very normal for problems like this

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[compact verge]

Can someone help me understand how they derived equation 2 here?

[compact verge]

if you write d^2y/dx^2 = d/dx(y') and you plug in y' you just get d/dx(dy/dt/dx/dt) where nothing is a function of x

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[compact verge]

solved it

[rugged arch]

okay..

[compact verge]

calculus textbooks are very confusing sometimes imo

[rugged arch]

is that thomas?

[compact verge]

ye

[rugged arch]

😭

[compact verge]

chapter 10

cause i understand that if y=h(x) and h is C^1

then we can use the chain rule

and then from algebra we get dy/dx

and then i was like what we differentiate both sides and nothing was a function of x anymore on the right since we had dy/dt/dx/dt

but if we just apply equation 1 without differentiating anything

then we basically plug in y' instead of y

[rugged arch]

yeah well it was just chain rule to switch from wrt x to wrt to x

d/dx = d/dt * dt/dx

[compact verge]

and then h has to be C^2

same condition for dx/dt =/= 0

so i think its solved

[rugged arch]

[compact verge]

i noticed that thomas / stewart kind of calculus books are a bit boring and very long with examples and things

im hyped to get to some shorter books that include trickier examples so i dont have to read as much but think more

[rugged arch]

yeah i guess you're supposed to skim those books

😭

because nobody's wants to work thru 1000 pages

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[polar whale]

tried solving it with the concepts of telescopic series but couldn't solve it.

[quasi bison]

this one does not telescope

[polar whale]

how come

I mean yeah it didnt

but how can you tell this doesn't telescope

[quasi bison]

there won't be any cancellation

3n+1 never equals 3m+2

[polar whale]

oh yeah well

how do you solve it

[fiery stratus]

OMG

This question

Its not telescopic

[polar whale]

it was kinda stupid i didnt get that lol

[fiery stratus]

Its integration

[polar whale]

no way

wait let me try of my own

[fiery stratus]

Ye crazy ass question

[ornate wyvern]

can b done by converting infinite sum over 0 to 1 of x^(3n+1) and so on

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[lone void]

ye

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[polar whale]

it was kinda long but finally did solve it

crazy fr

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[half sorrel]

i would like to find the demonstration of this. i've been looking for it on the internet and some books and i dont find it anywhere, does anyone know where can i find it or could do it please?

[compact plover]

it mainly comes from the derivation of e^iz=cosz+isinz

e^(-iz)=cos(-z)+isin(-z)

recall some trig rules

cos(-z)=cosz
sin(-z)=-sinz

so, (e^iz+e^(-iz))/2=cosz

we can also get sin by substituting in the cos component for e^iz=cosz+isinz

so, e^iz=(e^iz+e^(-iz))/2+isinz
(e^iz-e^(-iz))/2=isinz
(e^iz-e^(-iz))/2i=sinz

[half sorrel]

okay

thaank you

got it

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[south lynx]

the answer key says that it is an euler path and not an euler circuit, hence the starting vertex is different from the ending vertex. can someone please tell me why my solution does not work?

[south lynx]

(a minute, uploading my solution)

[glacial cairn]

Your solution works

[quasi bison]

can you show the answer key too for reference

yeah your solution seems to work

[glacial cairn]

The graph only has even vertices so yes it does have an euler circuit

[south lynx]

sure, here

so it is incorrect, right?

[glacial cairn]

Those degrees don't match

[south lynx]

oh

true

well

all right, maybe it is a misprint

thanks for the help

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[fathom yacht]

What have you tried?

[spring wasp]

nothing yet tbh

my mind has gone blank

[fathom yacht]

Okay

[spring wasp]

all i can really see is the fact 1/(nlnn) goes to 0 as n goes to inf

[fathom yacht]

Does it look like a familiar series?

Like something you have seen?

[spring wasp]

um i dont think so maybe

[fathom yacht]

If i take out the ln

what does it look like?

[spring wasp]

just the sum of 1/2+1/3+1/4....

[peak estuary]

what tests do you have available

[spring wasp]

i think only comparison at the moment

i cover more later in the year

[peak estuary]

hmm ok unlucky

would be really nice if you had the integral test

[spring wasp]

theres this result i can use?

[peak estuary]

thats the integral test, yes

use that

[spring wasp]

but how can i use that to show its divergent cause even if the integral is divergent since the sum will be less than or equal can i say for sure its also divergent

[peak estuary]

Then int f(x) dx is convergent if and only if sum f(n) is convergent

[spring wasp]

ahhh

[peak estuary]

infinity = int f(x) dx <= sum f(n) implies that sum f(n) = infinity aswell

[spring wasp]

ok yeah i got it thank you

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[vague rain]

Help

[mental salmon]

hii i need help with this (9th grade maths btw)

[lofty kelp]

can someone help me in solving it ?

[vague rain]

How i can solve thus

[lofty kelp]

is the answer 4 ?

[vagrant ice]

!occupied

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@vague rain Has your question been resolved?

[plucky elk]

should translate the question if you want help

[vague rain]

I need find limit

[vague rain]

Is a whole part function

Lim x_>0 x[1/×]

[vague rain]

You understand?

[plucky elk]

you just need ton consider values of x in (0, 1) and x in (-1, 0) separately

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[long igloo]

what is the theorem called where i can split a double integral up into 2 single integrals multiplied

[long igloo]

i cannot find it in my notes

i just need the name

i keep thinking fibonacci but that isnt it lol

[peak estuary]

fubini

[normal moss]

Fubini

[long igloo]

is it fubini?

[normal moss]

[peak estuary]

well, its not integrals multiplied

its iterated integrals

you can sometimes pull those apart

[long igloo]

yeah like dydx = dxdy

[peak estuary]

but not always

[long igloo]

okay

so what i am thinking of is fubini

thanks all

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[pulsar notch]

Hi, are these s.s correct

[ornate wyvern]

yes everything is right except in step 4-(one should always put modulus after taking a square root) hence here the correction would b to put a modulus arounf sqrt(3)/2

check with various 'theta' for even betr confirmation

[pulsar notch]

Alright thankyouu

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[real imp]

how to do this

[astral glacier]

I'm guessing you're meant to calculate the areas

What's the formula for the area of a sector

[real imp]

bro

[astral glacier]

?

If you don't know I'll help you dw

I'm just trying to gauge what you know

[real imp]

they dont mean find the literal area bro

[astral glacier]

How else would you "justify"?

"oh it just looks like grey is more"

"proof by using my eyes"

[real imp]

yeah but

bro

you arent doing area of sector for this question

you have to calculate the sector of blue x 6

and grey x 6

and then see which one has a greater area

[astral glacier]

...and you do realize you can drop the x6 right

[real imp]

no

[astral glacier]

...why?

[real imp]

i didnt realize

[buoyant umbra]

Just calculate both the area of sector and multiply by 6 and compare

[real imp]

ok ty

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[frozen quarry]

How to do equation out of this?

[thin bridge]

there are multiple ways to do this,
depending on what the question asks for it may be better to approach it differently

did it the question ask for a specific form they wanted the equation in?

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@frozen quarry Has your question been resolved?

[frozen quarry]

No, out of pure curiosity

[drowsy slate]

i know this

i can help u

[thin bridge]

depending on what's given
you'd usually start with
factored form
or vertex form
for a parabola

here there is sufficient info for both, so you could start with either depending on personal preference

[drowsy slate]

vertex is (1.5, 4)

hence proving tha

idrk

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[slender depot]

hi gang, so this is basic trig but i dont get whats with the apostrophe at the end of the angle?

[slender depot]

the base number is the same as i get on the calc but they changed the decimal and added the apostrophe

[proper zephyr]

Need help with this question

[severe pond]

DMS

[proper zephyr]

For a math test on quadratics algebra 1

[glacial cairn]

That's the symbol for "arcminute"

https://en.wikipedia.org/wiki/Minute_and_second_of_arc

[slender depot]

oh i see

so they converted the decimal into this then

[glacial cairn]

Presumably

20' = 20/60 º

So it's like 37.333...º

[slender depot]

yeah thats right

wait so how would the conversion work

[glacial cairn]

1' = 1/60º

[slender depot]

mm

so if i had this same 37.33

one sec

would i multiply the decimal by 1/60

no that cant be right

[glacial cairn]

No, by 60

[slender depot]

oh right

yeah there we go

thatll be it then thanks tons

have a wonderful day

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[cunning kindle]

Got stuck on 72

.close

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[thick sedge]

Pls help

Given the sequence 3, 6, 12, 24, 48, determine the sum of the first 10 terms.

I think this is the formula

Sn = An × r - A1/ r -1

[quasi bison]

your formula is rather poorly typed

maybe you could write on paper what you meant?

[thick sedge]

Right, my bad

Its either the first one or the second one

Idek

[quasi bison]

the first one just gives the n'th term

the second one gives the sum of the first n terms

and in discord you have to type it like this:

S_n = (a_n * r - a_1)/(r-1)

anyway you have the formula in front of your eyes. so what's the issue?

[thick sedge]

ChatGPT has given me a result that does not match what I have calculated, and when I show it my calculation, it also tells me that it is correct...

[empty gyro]

!nogpt

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[thick sedge]

HMM GTK TY

[thick sedge]

I just dont know what to do

[mortal sandal]

the first step is to figure out what each of the parameters are and then the second step is to plug them into the formula

[quasi bison]

do you know what each piece in the formula stands for

and secondary to that, do you know what the VALUE of each of those things should be in your question

[thick sedge]

Ifinally understand tysm everyone

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[regal sage]

Where did i go wrong?

[regal sage]

(I know its nit the most efficient way of solving)

[tardy mango]

,w factor 1-x^3

[flat frigate]

Results provided by WolframAlpha

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Requested by civil_service_pigeon

[regal sage]

Ah ok ty

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[south lynx]

i need help with the second part of the question. is there an easy way to figure it out, without assigning a colour to each vertex manually?

[south lynx]

please ping me if responding

[cedar widget]

Identify cliques

This is a lower bound on the number of colours needed

The maximal size of a clique

[south lynx]

sorry, what is a clique?

[zealous tundra]

show your graph

and yes theres actually a direct formula

[cedar widget]

Full graph

[south lynx]

if this is how you're supposed to do it

[zealous tundra]

why is it directed

[south lynx]

my bad

[zealous tundra]

instead of names of cities. just put a dot (vertice) and make it undirected

[south lynx]

okay, give me a couple of minutes

[safe radish]

@south lynx Has your question been resolved?

[safe radish]

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[umbral bobcat]

okay so this is the question and the formula for the first question is
TiWiLi=TfWfLf
(i: stands for initial, f: stands for final)
but my question here what is the initial? is it before going thru the first step or is it the first step or is it the step before the last one?

[safe radish]

@umbral bobcat Has your question been resolved?

[safe radish]

@umbral bobcat Has your question been resolved?

[hoary ibex]

im assuming the initial dimensions of the slab are the ones stated at the beginning of the problem

though given the equation it shouldn;t matter whether you use the ones from the beginning or towards the end since they're all equivalent

you could express the final thickness and width of the slab as t=5cm(0.75)^3 and w=25cm(1.03)^3 and then solve for the length

[safe radish]

@umbral bobcat Has your question been resolved?

[safe radish]

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[fair goblet]

Can someone tell me how Im suppose to mathc and match

[fair goblet]

by js looking at the equation

cause i dont have that mind muscle connection

wondering if some1 can guide pls

ping if u can hep

[twilit spindle]

try holding onle variable to be 0, suppose y=0 and change x around (so we're moving along the x-axis) then see how the vector evolves as you do so

[fair goblet]

yea

but how am i suppose to know what it like looks liek andstuff cause like

ok look

vector field for x, -y as x should go right when its positive and y should go down when its > 0

right

for the first one

[pure agate]

If (x,y) = (1,1), what does (x, -y) equal?

[fair goblet]

1,-1

right down

IV

[pure agate]

Yes, so at point (1,1), there should be an arrow pointing right and down.

[fair goblet]

yes

but the other ones are too hard to do

they are being rly weird for no reason

yk what i mean

[molten acorn]

Your technique is already pretty good, which is observing when it is going up/down and left/right

Let's check number 14 first

[fair goblet]

okay

y. x-y uhhhh

[molten acorn]

Right

[fair goblet]

when y is positive it arrows should point ->

i think

for x

[molten acorn]

No, you shouldn't think about x

[fair goblet]

oh

x-y = 0

[molten acorn]

You need to think about x-y

[fair goblet]

x= y

[molten acorn]

What happens if x-y is positive?

[fair goblet]

uhhh

0

it gets smaller towards x -y

oh

ok

[molten acorn]

So uhm

If x-y is positive, will the vector point up or down?

[fair goblet]

up

up right

how would i even know

do i just plug stuff in

[molten acorn]

Well the vector is <x, x-y>

If x-y is positive, that means it would go up wouldn't it?

[fair goblet]

yes

[molten acorn]

Because x-y here is the vertical part of the vector

[molten acorn]

Okay, now when or where in the graph is x-y positive?

[fair goblet]

above y=x

[molten acorn]

If it's below, the vector will be pointing down

[fair goblet]

yes

[molten acorn]

Now, with these clues and this one too

Try to figure out which graph it is

[fair goblet]

V

[molten acorn]

Sorry but that isn't correct

And I gtg, sorry Light

[fair goblet]

it ok

thank of you

[safe radish]

@fair goblet Has your question been resolved?

[safe radish]

@fair goblet Has your question been resolved?

[fair goblet]

,

[safe radish]

@fair goblet Has your question been resolved?

[tardy mango]

,w vector field (y, x-y)

[flat frigate]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by civil_service_pigeon

[tardy mango]

I feel like it should be that tbh

Check to make sure there’s not another one that matches better

Cause my eyes can suck

[safe radish]

@fair goblet Has your question been resolved?

[molten acorn]

Oh wait I just realized this is wrong

It should be below

That means yeah, it should be V

[safe radish]

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[safe radish]

Available help channel!

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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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[fervent pulsar]

help

[fervent pulsar]

tried

solving this

.close

[safe radish]

Channel closed

Closed by @fervent pulsar

Use .reopen if this was a mistake.

[fervent pulsar]

.reopen

[safe radish]

✅ Original question: #help-23 message

[fervent pulsar]

i tried

sovling this

kept running in circles

[forest sonnet]

I think you should cool down and wait for a helper to arrive

[fervent pulsar]

yh

[left gyro]

you here?

[fervent pulsar]

yep

[left gyro]

show what youve tried so far

[fervent pulsar]

erm

litr alot

im on pc though

so cant take a pic

i jst tried simultaneous equations

in anyway

[left gyro]

what thing is the work on

[fervent pulsar]

book

[left gyro]

Im talking about the work you wrote down where you said you kept running in circles

if you can show all of the work you tried, there is a chance you tried something very good

it doesnt look like you can value how much youve done so far, because you havent said anything more specific about where you are or what you've figured out already

for one, you can use equation 1 and equation 2 to solve for c

you didn't say if you've reached the "I've solved for c" step or not

[fervent pulsar]

ok

well

it ried to write c in terms of logs

ive tried to make things equal each other

and then

try make it the same base

and susbistute

in the end

bc=c/log base a c

i had also reached

log base c a= a^c

or log base c b=bc

[left gyro]

do you know what the change of base formula is

[fervent pulsar]

erm yh

its like

e,g

if i had log base 3 5

and i make it log base 2

it becomes

log base 2 5/ log base 2 3

[left gyro]

you can use these log rules for more than just calculating numbers you know

it could make the algebra easier

you could simplify all the logs to use the same base instead of three different bases

[fervent pulsar]

yh i used it

in this

case

i didnt rlly end up anywhere

[left gyro]

it almost sounds like you should show your work

it does in fact end up somewhere

now how am I supposed to tell you why it ended up nowhere?

[fervent pulsar]

ok let me try again

ill try send it if i can

idk if it ends up readable

[left gyro]

you dont have to worry

your text is already barely readable

but I can still understand it

it is readable enough

[fervent pulsar]

trust me

my handwriting is buns

[left gyro]

I trust you

you can type it if you need to

[fervent pulsar]

ok

i did get a quadratic

in terms of c

c^2 +3c/2 -1 =0

ill solve

[left gyro]

thats the correct quadratic

[fervent pulsar]

c must be 1/2

[left gyro]

yes

[fervent pulsar]

so surely

that automatically makes it a

wait nvm

it cld be a,b,c

[left gyro]

you should solve it and be sure about it

youve now used one aspect of equation 1 and equation 2

you should try out equation 1 on its own and equation 3 on its own next

[fervent pulsar]

i get a=2^-b

[left gyro]

knowing c = 1/2, its bound to simplify them

[fervent pulsar]

b=a^1/2

[left gyro]

also correct

[fervent pulsar]

b^2=a

but ill solve for b but ill need to make sure

which one is a real solution

cos i jst created another by squaring

[left gyro]

if you square both sides, you know the other solution would just be negative

[fervent pulsar]

yh

[left gyro]

so no, you didnt create another solution by squaring

this is because the next step after this isnt to solve for b

[fervent pulsar]

but i jst end up with b^2=1/2^b

[left gyro]

yes, the next step after this isnt to solve for b

you now need to consider what the question is asking

[fervent pulsar]

i need to see

if its unique

or not

[left gyro]

no cheating with a graph

you need to think about this

[fervent pulsar]

ok

wait

[fervent pulsar]

with this eq

i need tos ee

whether

if multiple solutions to this

[left gyro]

although you cannot cheat, you can still do the next best thing and picture what the general shapes of b^2 and 1/2^b are

keep in mind we have b > 0, so its not the entire graph

[fervent pulsar]

yh exactly

2^-b

wld be like

ok

theres only one solution

for b

cos

[left gyro]

thats correct

[fervent pulsar]

i kinda cheated

cos of the graph

in my head

[left gyro]

fortunately that is not counted as cheating

[fervent pulsar]

cos half of the quadratic from 0,0

so

its a

[left gyro]

yes

[fervent pulsar]

i also have another question

[left gyro]

you cannot solve for b unless you have more operators to use

[fervent pulsar]

i think this ones harder

[left gyro]

alr

[fervent pulsar]

like

gives alot of info

[left gyro]

this is interesting

heres a question

can you find an example of a straight line with length 1/2?

[left gyro]

we know √3/2 happens in that side

now what about 1/2?

[fervent pulsar]

1/2

is midppint of a side

[left gyro]

yep, you can draw the line horizontally

well no nvm thats not true

the center O isnt halfway up the triangle

its 1/3 up the triangle instead whoops

sorry

[fervent pulsar]

wdym

[left gyro]

centroid

the center of the triangle is closer to the base than the tip

[fervent pulsar]

isnt the cords of O

1/2, root 3 /4

[left gyro]

not really

[fervent pulsar]

How dyk that

[left gyro]

(1/2, √(3)/4) looks like this

you are supposed to memorize where the center of an equilateral triangle is at

one way to find it, if you dont remember it, is to draw three lines
each line starts at a corner and ends at a midpoint

if you do that, the center looks more like this

[fervent pulsar]

yh

i got that

so

what do u think we do

shld we

graph it

make 3 lines

y=0

y=x

[left gyro]

its not y = x

[fervent pulsar]

why not

[left gyro]

theres an easier way to go about this

first off, you can see the slope of the white line is not 1

a slope of 1 looks like this

also a slope of 1 would mean this is a 45-45-90 triangle

that is not true

[fervent pulsar]

oh yh

ur right

[left gyro]

there is an easier way to go about this

[left gyro]

take a look at this picture

it is clear the six smaller triangles are all congruent 30-60-90 triangles, right

[fervent pulsar]

yh

er,

erm wait what

ok

i see

yh yh

[left gyro]

yep

you can see that because the shape is symmetric

so all three of these Ts should have the same angles

and since those Ts are isosceles and have a perpendicular in the middle,

[fervent pulsar]

yh

[left gyro]

that must apply for all of them

ok now for something harder

pay attention to the center line

[fervent pulsar]

yh

[left gyro]

notice where the center is at

can you prove that the center is 1/3 along this line?

[fervent pulsar]

maybe

cos u know 1 side

is 1/2

and u knpw angles

[left gyro]

be more specific

[fervent pulsar]

like

u see

this triangle

[left gyro]

and the other triangle is at?

[fervent pulsar]

wldnt

that be enough

cos

2 sides

i mean

2 angles

1 side

[left gyro]

it sounds like you havent had all the words fully thought out yet

heres a more enticing picture

[fervent pulsar]

like yk one side

but like

ok wait

i see it

like

cant we find

[left gyro]

thats not an ideal solution

[fervent pulsar]

lengths

and then match it

[left gyro]

you dont need to

[fervent pulsar]

with our first triangle

[left gyro]

you just need to use 30-60-90 triangles

and that some segments are equal in length

[fervent pulsar]

wdym

is it cos

the height of all triangles

make up one big triangle

so the height of the bottom one aka green line

must be 1/3 of total height

[left gyro]

why 1/3?

[fervent pulsar]

cos

3 triangles

make up the one big triangle

[left gyro]

first off, no

you can see the second triangle doesnt contribute any height to the center line

only the bottom and the top triangle do that

second off,

you need to use that these are all congruent 30-60-90 triangles

you said you only needed one 30-60-90 triangle to solve this

this is correct

look at the 6 segments that eminate from the center

how many different lengths do you see in these segments?

[fervent pulsar]

2

3

3

[left gyro]

2

[fervent pulsar]

howis it 2

its 3 no?

[left gyro]

lengths that combine other lengths do not count

[fervent pulsar]

oh ok

2

[left gyro]

yes, 2

and is one of them twice as long as the other one?

[fervent pulsar]

erm

idki

nee phytag

[left gyro]

30-60-90

you dont

30-60-90, what do you remember with 30-60-90?

surely it cant just be the angles

[fervent pulsar]

nthn much tbh

i dont really rmbr

doing this

[left gyro]

what about the sides? the ratios?

[fervent pulsar]

i assume

the sin

values

sin 30 = 1/2

sin 60 is root 3 /2

so

[left gyro]

sin 30 = cos 60 = 1/2

now here,

[fervent pulsar]

yep

[left gyro]

what about sin 30 or cos 60 can tell us that the purple length is 2 x the green length?

[fervent pulsar]

cos

u have like

CAH

and SOH

so like

they both match

so like sin30= green/purple

and cos 60 = green/purple

[left gyro]

alr, so sin 30 = cos 60 = green/purple = 1/2

[fervent pulsar]

purple=2green

[left gyro]

there we go

and with that youve seen the ratios

the ratios of a 30-60-90 triangle are 1 : √3 : 2

in other words, the sides are always:
shortest side, √3 * shortest side, 2 * shortest side

in this case you can see we're comparing the shortest and longest side

the ratios tell us the longest side is 2x the shortest side

and with that,

we know that the center must be 1/3 up the triangle

[fervent pulsar]

i dont get it

[fervent pulsar]

how

did we get this

[left gyro]

this is something you are supposed to memorize

[left gyro]

for a 30-60-90 triangle,

hypotenuse or purple side = 2 * green side
white side = √3 * green side

you can figure out that this is true, because

green/purple = cos(60) = 1/2
white/green = tan(60) = √3

[fervent pulsar]

ok yh

i get it

[left gyro]

now with that, the ratios of the sides are 1, √3, 2

often written as 1 : √3 : 2 to represent that theyre in proportion to one another

doubling one side requires that you also double the rest too

2 : 2√3 : 4 for instance

[fervent pulsar]

yh

[left gyro]

ok with purple = 2 x green,

we know the center is 1/3 of the way up the triangle

instead of 1/2

in general youd memorize this information

a centroid is "1/3" up the triangle

in that if you drew the line from the midpoint to the top, the longer segment from the center is always 2x the shorter segment from the center

[fervent pulsar]

yh

[left gyro]

ok now with that,

consider this line again

what do you think the length would be now

[fervent pulsar]

i tried to find that distance initally

[left gyro]

well it should be obvious if you look at the picture the right way

think about how you got the 1/2

if you need a hint, heres a different picture of the same line

[fervent pulsar]

i dont see it

im sorry

[left gyro]

youve only looked at the picture for 5 entire seconds

please look at the picture for a more substantial amount of time before giving up

heres a more enticing picture

[fervent pulsar]

is it some sort of ratio

[left gyro]

theres a ratio here

can you see it

its not related to 30-60-90

[fervent pulsar]

its getting smaller

as height increases

[left gyro]

look at the height

are the lines evenly spaced?

[fervent pulsar]

yh

it is right?

[left gyro]

yes

[fervent pulsar]

so

is the same

that way too

/equally spaced

[left gyro]

you need to be more specific

[fervent pulsar]

its

a

the height

[left gyro]

the vertical spaces between the lines are evenly spaced

[fervent pulsar]

we

jst

rotated the triangle

[left gyro]

rotated?

[fervent pulsar]

like

[left gyro]

all I did was cut it in half

I didnt rotate it

[fervent pulsar]

as in

like

the height we knew

the ratio

was

1:@

1:2

it must be the same

[fervent pulsar]

here

too right

[left gyro]

you need to be more specific

that isnt true if you phrase it wrong

shouldnt the number 3 have something to do with this?

its three evenly spaced horizontal lines

the 3 is from 2 + 1

and we need to find the ratio of middle line / bottom line

[fervent pulsar]

yh

thats

what i was trynna say

shldnt the spacing between the horizontal lineds

wait nvm

idk what im saying

[left gyro]

theres something more interesting here

back when you thought the centroid was halfway up the triangle,

[fervent pulsar]

occupied

[left gyro]

you figured the length was 1/2

[novel root]

In neat handwriting on paper 😭

[left gyro]

now where did you get the 1/2 from?

and this time, think about scaling

[fervent pulsar]

bro

it was jst

the midpoint of 1

[left gyro]

and youre saying

you cant just do the same thing but with 3 instead of 2?

its not a midpoint anymore

but its still of 1

[fervent pulsar]

oh i see

but

[left gyro]

can you find the ratio of top horizontal line : bottom line?

[fervent pulsar]

that doesnt give us

horizontal distance though

jst the distance

between each individual line

1/3

[left gyro]

thats correct

so the top horizontal line is 1/3 the length of the bottom line

you just said that

[fervent pulsar]

yh

[left gyro]

so the middle horizontal line : bottom horizontal line, what ratio is that

[fervent pulsar]

2/3 ?

but

[left gyro]

correct

[fervent pulsar]

i guessed that

[left gyro]

no you didnt

[fervent pulsar]

i jst knew

it had to do sm

with 3

[left gyro]

its the literal number between 1/3 and 1

[fervent pulsar]

and it cant be 1/3

[left gyro]

put it this way

we agreed the lines were equally spaced

now looking at the picture, that equal spacing means more than just vertically

[fervent pulsar]

yh

[left gyro]

it has to mean horizontally too

[fervent pulsar]

yh

[left gyro]

so given the bottom line is 1,

the other lines must be 1/3 and 2/3

[fervent pulsar]

yh

i see

[left gyro]

if you need to use proof,

consider drawing these lines through the triangle

then count the number of red segments on each horizontal line

1 2 3

the green lines are made to be parallel to the left diagonal

this also works if the triangle isnt equilateral
equally spaced red lines also have equally spaced lengths

if you want a boring general reason, its because you can form similar triangles that can show that the lengths scale like that

@fervent pulsar you here?

[safe radish]

@fervent pulsar Has your question been resolved?

[left gyro]

I gtg
you have that 2/3 is horizontal and √3/2 is vertical
proving this more rigorously is not easy to expect, if you want to try reading my mind you can refer to this picture
but for now, you just choose whichever of 2/3 or √3/2 is smaller and thats good enough

[safe radish]

Channel closed

Closed due to timeout

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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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[acoustic zealot]

How do i calculate this.
I know theres a rule, where u can splitt them in different prime numbers.

$&\phi(15) = \phi(3) * \phi(5) \
&\phi(77) = \phi(7) * \phi(11)$

Can someone calculate one and explain me what you did.
Imma try to solve the rest on my own.

Also if you know a source for these tasks, id also appreciate it

[flat frigate]

Darknesslion5|Christian
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[safe radish]

@acoustic zealot Has your question been resolved?

[hoary ibex]

youcan use l’hopital’s rule

[acoustic zealot]

Wrong question

[acoustic zealot]

This is the question

[maiden egret]

No the question isnt incorrect it is correct I have written it correctly and also solved now

[obsidian basin]

!nosols btw, on top of answering the wrong OP

[safe radish]

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.

[hoary ibex]

oh thats my bad im sorry

[acoustic zealot]

Still you cant just overwrite my question

[maiden egret]

What do you mean ?

[obsidian basin]

it's okay, he got his own channel already, and has his channel answered there

[acoustic zealot]

Ok

[obsidian basin]

but as a heads-up, do not post your question in an occupied channel next time please.

if you see the channel has a "| {someone's name}" on it, don't ask your question there.

[maiden egret]

Sorry I was new so I didn't knew

[acoustic zealot]

How do i calculate this.
I know theres a rule, where u can splitt them in different prime numbers.

$&\phi(15) = \phi(3) * \phi(5) \
&\phi(77) = \phi(7) * \phi(11)$

Can someone calculate one and explain me what you did.
Imma try to solve the rest on my own.

Also if you know a source for these tasks, id also appreciate it

[flat frigate]

Darknesslion5|Christian
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[winged flare]

if $n = p_1^{a_1}p_2^{a_2}p_3^{a_3...}$ then\
$\phi(n) = n(1-\frac{1}{p_1})(1-\frac{1}{p_2})(1-\frac{1}{p_3})...$

[winged flare]

so for phi(16) it would be 16(1-1/2)

off context

[acoustic zealot]

What is p_1^a_1 aso

Ah prime

[maiden egret]

I have solved your question hope YOU got it ...

[obsidian basin]

!nosols

[safe radish]

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.

[acoustic zealot]

Ahh ok

I think i got it

$\phi(80) = 80[1 - 1/2] = (80*1)/2 = 40$
Right?

[flat frigate]

Darknesslion5|Christian

[safe radish]

@acoustic zealot Has your question been resolved?

[safe radish]

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[wind halo]

Does anyone know how to simplify a radical

[wind halo]

.reopen

[charred jetty]

Yes, what's the question?

[wind halo]

Radical 67

[charred jetty]

I can feel the beard growing slowly on my chin.

What is your main question?

[rugged arch]

lol

[quasi bison]

but on the off chance you're serious,

67 is a prime

[smoky nacelle]

radical 67 = 67 who believes in harsh change

[quasi bison]

so sqrt(67) is not simplifiable

[wind halo]

Yes that’s the question

[smoky nacelle]

$\sqrt{67}=\sqrt{67\cdot 1^2} = 1 \sqrt{67}$

[flat frigate]

Περσυ

[wind halo]

Oh ok

Thxxxxx

[charred jetty]

Yo, you should go do it with a calculator

If it's allowed.

[wind halo]

Yes it is

[rugged arch]

what’s there to do tho

[wind halo]

It’s kit a review question

For a test

[charred jetty]

I think it could also help you radicalize 69 too

lol

[rugged arch]

[wind halo]

lol

Well thxx

[quasi bison]

do you need help with anything else for now

[wind halo]

No

If I do I’ll come back

[smoky nacelle]

sounds fair

[quasi bison]

close this channel for now then

[safe radish]

@wind halo Has your question been resolved?

[safe radish]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

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.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• 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.

[north dragon]

i dont understand how to get what the question is asking

[north dragon]

i kinda didnt understand it well i think

[floral sequoia]

Draw a circle and a line tangent to the circle

The circle is the path the horse is running on

The line is the fence

[north dragon]

yea i got that bit

the horse is covering 1/8th

so the angle is pi/4

[floral sequoia]

There is a light source on the center of the circle