help-39

[iron schooner]

I frogot to add this angle A1 is = to 30 degrees and angle B1 is = to 50 degrees find angle A2; O2 and O1

[sacred python]

chill bro

[iron schooner]

it is

[sacred python]

then 30+a2=50

so a2=20

[iron schooner]

whats the reason

for that

[sacred python]

cus a=b

equal angles of the iso triangle

so a2+a1=b

[iron schooner]

there we go yea

[iron schooner]

i will take that

[sacred python]

so now 30+a2=b

[iron schooner]

wait why we saying a2 + a1= b b is already given

b was 60 degrees

[sacred python]

b is 50

[iron schooner]

whoops

B1 is 50 degrees

not the whole B

[sacred python]

ya

my bad

so a1+a2=b1

so 30+a2=50

[iron schooner]

so a2 is 20 degrees

[sacred python]

a2=50-30

[iron schooner]

hold on

[sacred python]

yep

you got it

[iron schooner]

now what>

[sacred python]

what else do you want help in finding

[iron schooner]

ok we whole of angle A is80 degrees

[sacred python]

no its 50

[iron schooner]

and 03 would be 50

[sacred python]

no

[iron schooner]

how is it 50 A2 is 50

wait..

[sacred python]

it will be 80

[iron schooner]

noo

[sacred python]

a2=20

03=80

[iron schooner]

we said a2 is 20 right

so the whole angle A is 80?

[sacred python]

my guy could i just call you and speak

[iron schooner]

30+50

[sacred python]

it will be easier

[iron schooner]

50+20 is 70

thats angle A right?

[kind rampart]

it looks like youre getting confused with the total measure of angle A and the measure of the two angles that make up A

[iron schooner]

alright u can just speak

@sacred python i need to verify

you can talk

[pearl pond]

@iron schooner Has your question been resolved?

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[iron schooner]

thanks for you help @sacred python

🙌

[pearl pond]

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[modest tartan]

doesnt log(e)=2. somethin

[errant solstice]

I presume the log here is actually a natural log.

or ln, if you use that.

[modest tartan]

but in the question log base 10 is used

[errant solstice]

where's the original question?

[modest tartan]

i originally needed to find e^log(x)

[errant solstice]

and the log is for sure base 10?

[modest tartan]

simplfy this part

in a limits question

i would assume so

[errant solstice]

oh. in that case, you are taking the natural logarithm of both sides. that is fine.

[modest tartan]

wont that give ln(z) instead of log(z)

[modest tartan]

here

[errant solstice]

the log here is actually a ln. some syllabi do that.

[modest tartan]

ah

[errant solstice]

I would probably check back on the text to see what the default base of your log is.

[modest tartan]

i did not know that

[modest tartan]

for us its normally 10

[errant solstice]

in most calculus courses the default base of a logarithm is e, not 10.

[modest tartan]

for log base e we use ln

instead

[errant solstice]

then you may want to approach your lecturer/teacher about this notation.

[modest tartan]

ok thanks

.close

[pearl pond]

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[pearl pond]

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[bitter herald]

,,, this is a test

[bitter herald]

.close

[pearl pond]

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[vestal tapir]

,,, amazing

[jolly parrot]

frowny

[wicked edge]

[bitter herald]

@minor cloud

[pearl pond]

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[bitter herald]

,tex aaa

[north talon]

Let $ABCD$ be a rectangle. Let $K$ be a point on $AD$, let $E,F$ be points on ray AB,DC respectively such that $AE=CK$ and $DF=BK$. Let $P$ be a point between $AB$ and $ CD$ such that $\angle KCD = 2 \angle PAB$ and $\angle KBA = 2 \angle PDC$. Show that $PE = PF$

[jolly parrot]

Copter

[north talon]

uhh from my diagram the lines with the same color are parallel

AP is parallel to the angle bisector of KCD, etc

Let C' be the point on BC such that BC'DK and AC'CK are parallelograms to construct E and F

i dont really have progress on this xd

im guessing this is an angle chasing problem? but idk how id do that

<@&286206848099549185>

[valid nova]

bmo?

[inner granite]

Did you manage to solve the P3 from Balkan?

[north talon]

is it?

[north talon]

yesh

miquel point bah bah

[inner granite]

I see

[north talon]

anyway

[odd kelp]

how'd u make tht

[north talon]

geogebra

• im gud

[pearl pond]

@north talon Has your question been resolved?

[shy wave]

question aside how you made ts

[vestal tapir]

geogebra

guessing

[pearl pond]

@north talon Has your question been resolved?

[north talon]

yea

[inner granite]

Try looking at <BAC1 and <DCK

The same for <CDC1 and <ABK

Fun Fact: ||KP perpendicularly bisects EF||

[pearl pond]

@north talon Has your question been resolved?

[inner granite]

<@&268886789983436800>

[pearl pond]

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[pearl pond]

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[supple charm]

Hi

[supple charm]

I need help

Badly

I've been stuck on this for half an hour

[wooden flare]

What have you tried so far?

[supple charm]

I've tried doing it myself, I gave up after 10 mins so I tried using chat gpt

Even that didn't work

[west sapphire]

well, the good news is that there are already no negative exponents

[supple charm]

True

No matter what I put in, it won't work

[west sapphire]

do you know basic exponent rules?

[supple charm]

Sorta

[west sapphire]

for example, how would you simplify (27)^(2/3)?

[supple charm]

3 to the square root of 27?

Then uh

Idk

[west sapphire]

try thinking of the exponent 2/3 as (1/3) times 2

[supple charm]

Alr

[supple charm]

2/6?

I have no clue

[harsh adder]

27 is 3^3

do you agree with me

[supple charm]

3 x 3 x3

Yes

[harsh adder]

okay

[harsh adder]

now use this

its an exponent identity

(we are only working with 27 right now)

you can try the other 2

its the same thing

got anything?

[supple charm]

9a^6b??

[harsh adder]

yep

thats your answer

[harsh adder]

you wrote b in the power

thats why you got it incorrect

[supple charm]

OMG

I can't believe it

This is what happens when you're sleep deprived bruh

Lmao

@harsh adder Tysm bro. I still got one more

[harsh adder]

sure

[harsh adder]

hmm

you try that 1

same thing as last question

[supple charm]

Alr

[harsh adder]

if you get stuck anywhere @ me

[supple charm]

4^8b^9?

[harsh adder]

4^8?

[supple charm]

uh

4 with the exponent being 8

[harsh adder]

what happened to a

[harsh adder]

nope

write 4 as 2^2

and same thing as last question

[tulip ore]

<@&268886789983436800> @willow epoch

[supple charm]

Ofc

[tulip ore]

most intelligent discord user:

[harsh adder]

this has gotten boring now

[supple charm]

How hard is it not to fall for those scams bro

[supple charm]

Sorry

[harsh adder]

no not your q

the scam

[tulip ore]

yea the scam

[supple charm]

Oh alr

[harsh adder]

we need a smth new

[tulip ore]

maybe they dont need to refer to mrbeast

[harsh adder]

i guess

[supple charm]

@harsh adder I'm lost

[harsh adder]

which part

[supple charm]

2^2

I probably look stupid

I'm just tired af

[harsh adder]

nah dw

mistakes happen

[supple charm]

What next after the 8?

is it

8a/9b?

[harsh adder]

can you show your working

if possible

ill tell you where exactly you're going wrong

that will help you better

[supple charm]

Alr one sec

@harsh adder

[harsh adder]

4^-8 is kinda incorrect

we wrote 4 as 2^2 and then simplified it

to make it 8

so 8 is the first term here

like in the prev q it was 9

[supple charm]

Damn it

Alr

[harsh adder]

also wheres the a

the 8 is in the power of a

not 4

[supple charm]

I keep forgetting the a

Fuck

@harsh adder Ok I don't know what I'm doing

I think I need to go back to step one

OK

I GOT IT

YES

[harsh adder]

o

[supple charm]

TY!!!

[harsh adder]

yw

tho you solved it yourself

[supple charm]

Yea but you helped me so much

I was stuck on that god awful question for over half an hour

[pearl pond]

@supple charm Has your question been resolved?

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[pearl pond]

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[viscid dew]

Pairwise distinct points A, B, C, D, E, F, G, W1, W2, W3, W4, W5, X1, X2, X3, X4, X5, Y1, Y2, Y3, Y4, Y5, Z1, Z2, Z3, Z4, and Z5 on a plane satisfy AW1⟂W1X1⟂X1Y1⟂Y1Z1⟂Z1G, AW2⟂W2X2⟂X2Y2⟂Y2Z2⟂Z2G, AW3⟂W3X3⟂X3Y3⟂Y3Z3⟂Z3G, AW4⟂W4X4⟂X4Y4⟂Y4Z4⟂Z4G, AW5⟂W5X5⟂X5Y5⟂Y5Z5⟂Z5G, BW1⟂CX1⟂DY1⟂EZ1⟂FG, BW2⟂CX2⟂DY2⟂EZ2⟂FG, BW3⟂CX3⟂DY3⟂EZ3⟂FG, BW4⟂CX4⟂DY4⟂EZ4⟂FG, BW5⟂CX5⟂DY5⟂EZ5⟂FG, and AG=2026. Find the product of the lengths of AW1, AW2, AW3, AW4, and AW5.

[wicked edge]

Just send a picture.

And what have you done?

[pearl pond]

@viscid dew Has your question been resolved?

[split void]

give the actual problem in a pic.

and ofc, a diagram.

[viscid dew]

there's no picture / diagram

[quaint arrow]

Hlo guys

[viscid dew]

wha, what?

[wicked edge]

Please provide a screenshot of original question.

[sweet junco]

ok so from here we can tell that E, Z2, Z5 are collinear
I can't read all that

[sweet junco]

ok that makes it more clear

have you drawn a diagram?

[viscid dew]

no, not yet

[viscid dew]

i did realize what this meant B, W1, W2, ..., W5 are collinear, and similarly for the others

[sweet junco]

weull uh, since B, C, D, and E don't change as it goes through W X Y Z 1-5 and are perpendicular to FG, we can assume W1-5, X1-5, Y1-5, and Z1-5 are linear

[viscid dew]

(C, X1...5; D, Y1...5; E, Z1...5)

right...

yah

[sweet junco]

ok can you imagine those rn?

they're kinda like a square grid

[sweet junco]

now what do the lines W1X1, X1Y1, Y1Z1 tell us

[viscid dew]

W1X1 and Y1Z1 are parallel?

[sweet junco]

ignore this

the lines are perpendicular with common points

so they're kind of either zigzags or squares or smth like that

the lines specifically

[sweet junco]

also nvm they are

[viscid dew]

mhm, right

[sweet junco]

assume W1-5, X1-5, Y1-5 and Z1-5 are arranged like in a square grid

like 5 by 5

4 by 5 woops

draw them out

[viscid dew]

hmm

like this?

[sweet junco]

yeah

now what can we tell about the general direction of A

[viscid dew]

well, but actually the W's would be on a horizontal line, and then the X's would be on a vertical line, right?

[sweet junco]

honestly

dunno

my brain fried

u can ask sum1 else mopre imaginative than me I'm out of ideaas

[viscid dew]

ok

its fine, thank you!

[sweet junco]

yep the Z would also be on there

[viscid dew]

would it look something like this

[sweet junco]

so like draw 2 columns and 2 rows W1-5&B Y1-5&D and X1-5&C Z1-5&E

[sweet junco]

yeah prolly

[sweet junco]

swap the W1-W5/mirror it and the bottom

and you should get what ur prolly looking for

[viscid dew]

this?

[sweet junco]

yes

so now wait

W1X1, X1Y1 yeah that should be riht

right

what it looks like

now place A hey wait a minute

[viscid dew]

A is like, non-existent 😭

[sweet junco]

wouldn't A be in line with the Ws and G in line with the Zs?

[viscid dew]

i mean, AW1 is perpendicular to W1X1

[sweet junco]

what if you shift W5-2 lower kinda forming like a diagonal thing

do the same with Y- yeah no that doesn't wok

work

[pearl pond]

@viscid dew Has your question been resolved?

[pearl pond]

@viscid dew Has your question been resolved?

[pearl pond]

@viscid dew Has your question been resolved?

[wicked edge]

Are you still stuck?

[viscid dew]

yah

[wicked edge]

Where?

[viscid dew]

i mean, i haven't really tried more, since i don't know what to do

[viscid dew]

basically just labelled the points

the W points, X points, Y points, Z points are colinear (in each set), and are like perpendicular

[wicked edge]

Well, I read from the question.

For any index i = 1,2,3,4,5, the conditions described that 2 orthogonal paths.

What if we identifying these 2 parts?

Maybe try from vertical - horizontal chain.

[viscid dew]

oh, i forgot to include the given information that AB=1

[viscid dew]

.close

[pearl pond]

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[pearl pond]

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[bright nacelle]

I need assistance with doing 10-20 questions really fast, they are very basic, but neither me nor my friends can do them fast enough. I am not completely sure how I would get assistance with them in this channel as I would need to screenshare them to you though.

[plush bramble]

just screenshot one problem at a time and send your work with them

[plush bramble]

like it says here

[bright nacelle]

The issue is that this is an interesting scenario.

[plush bramble]

are you taking a test

[bright nacelle]

No I swear I am not that kind of person.

It is merely a game

[plush bramble]

this still applies

[jolly ice]

Sounds suspiciously like a test😭😭

[bright nacelle]

There is a boss that has its health attached to math problems, I am doing a strat with friends, I can do the math fast enough, but they cannot.

The math problems are split across 3 people and its been an issue.

One of them is on console so they cannot send images of the math problems.

He can only screenshare them.

[plush bramble]

where is the actual math question

[bright nacelle]

This image is only used to display that I am not taking a test.

[plush bramble]

then just follow this

[bright nacelle]

I can't, they are extremely basic and are during a time constraint. If we were to type them out, we would lose by the time we get the answers.

[plush bramble]

can't help if you can't ask the math question ¯_(ツ)_/¯

[bright nacelle]

That is why I wished to screenshare them to you.

[plush bramble]

.

[bright nacelle]

I have already explained as to why that is not an option.

[feral olive]

Just give a couple examples so we know what kind of problem we deal with

Not even a screenshot

Just state an example

[bright nacelle]

Here is one taken from the pool. (This is the most difficult type of question that it gives.) with a 4-7 second time limit. The rest can be as easy as 4+5+8.

I do not have an issue with them, but my friends do, and they need assistance.

[feral olive]

practice

Actually that makes the game rly good cuz if you do this over and over again you get better

[bright nacelle]

This game has absolutely nothing to do with math, It is a random throw in enemy that has no reason to be here other than this singular boss.

[feral olive]

A good way to practice then might just be to spit out examples to each other and get them to answer as quickly as possible maybe

You can also definitely automate making these kinds of problems

[bright nacelle]

An issue is they also have to focus on micro-managing their units while doing problems which is an issue

Well it is clear I am not getting assistance here.

.close

[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

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[zenith rapids]

i know that the number must be odd and not divisible by any primes up to 7, but i cant figure out anything more than that
ive set up a series of representations of these divisions and remainders using modular arithmetic but i dont know where to go from there

[errant solstice]

my first instinct is to do something about the system of congruences using CRT.

[flint basalt]

please show what you have tried so far for other helpers to know where to start :)

[zenith rapids]

well obviously the integer in question is odd since it has a remainder of 1 when divided by 2
and ive basically written
n ≡ 9 (mod 10)
n ≡ 8 (mod 9)
n ≡ 7 (mod 8)
etc.

also i tried thinking about like

doing 2^2 * 3 * 5 * 7

and then that minus 1

thats what my intuition told me but it didnt work

[errant solstice]

hint: this system can be written another way such that the congruences end up having the same remainder mod each divisor.

[zenith rapids]

only works for the division by 10 case

[wicked edge]

-# Chiaki you are green!

[zenith rapids]

ohhhh do you mean like writing n+1 divides 10, n+1 divides 9, n+1 divides 8, etc.

[errant solstice]

close!

you are indeed going to have to do something with 1. but it might not be a +1 that you're looking for.

well I sense another helper is about to come in and take over, so I'll leave this to the other helper.

[flint basalt]

another way you can think of this is since n+1 has to be divisible by all numbers 2-10, what's the smallest number thats divisible by all of those?

[zenith rapids]

yeah thats what i was saying sorry i mixed up n+1 divides 10 and 10 divides n+1

[formal wedge]

can someone help me understand how to solve this? its on my test tomorrow morning

[flint basalt]

!occupied

[pearl pond]

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

[flint basalt]

and how would you go about getting that number?

[zenith rapids]

ig its just 2^3 * 3^2 * 5 * 7 right? since that covers all factors

[errant solstice]

-# crossing out this if you're following Krish's method.

[zenith rapids]

and then

you do that\

-1

[flint basalt]

yeah that works perfectly i think

[zenith rapids]

2519?

[flint basalt]

was just going to say n+1 equals that number but you got it!

[flint basalt]

yep

[zenith rapids]

yeah thats pretty cool

its like system of equations but with modular arithmetic

[flint basalt]

imo thats the most simple way to do it but if you want to try to learn chiaki's method you can go ahead and do that

[zenith rapids]

learning this actually helps with a problem i got stuck on before that which i think is another system

so imma go do that one rq and see if the same method works

yep works with that too

[errant solstice]

!done?

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[zenith rapids]

yep

.close

[pearl pond]

Channel closed

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[pearl pond]

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[fallow quarry]

I have a desire to know what every math symbol means. (I am running short on time but will return in the near future).

[wicked edge]

In which field?

Sets, or..?

[flint basalt]

can you send some examples of ones you don't know?

[fallow quarry]

Sure thing.

One sec

[versed mica]

it would be much a better usage of your time to just learn the math you want to learn

there is nothing deep about the symbols

you'll learn the symbols as you go

[versed mica]

gamma isn't really a "symbol"

[fallow quarry]

Yes... but I am unable to work the math currently (I can mentally) but not on paper. So DOING the math becomes more difficult.

[versed mica]

just a variable name that's used for a function mostly

[flint basalt]

well for the E, it says what it is in the title: "is an element of" and it refers to sets.

[wicked edge]

The first one is just an element that belongs to a set.

[fallow quarry]

Ok... what does an 'element' mean. Or 'set'?

[wicked edge]

The Pi one is for products, right?

[flint basalt]

the capital pi is similar to sigma notation, except instead of summing it means a product

[versed mica]

not sure what you mean

what math have you learned so far?

[fallow quarry]

I have no paper nor pencil at this moment. Not any surface where I can work complex problems out.

[versed mica]

you can just read actual math texts and try some of the problems

[flint basalt]

and of course, lowercase pi is just the ratio of a circle's circumference to its diameter

[wicked edge]

Hmm. For the Gamma one, I don't think it is used much in math, but logically it's stand for basis.

[timber cape]

thats gonna take forever

[fallow quarry]

It might, but an edge is an edge.

[wicked edge]

Which is "already knew some conditions".

[flint basalt]

doesnt gamma(x) = (x-1)!

[versed mica]

an edge?

[fallow quarry]

To get ahead of others.

[wicked edge]

Never heard of that unfortunately.

[versed mica]

brother

[timber cape]

for only x belongs to positive integer

[versed mica]

😭

[flint basalt]

ive heard of it once or twice but never really used it

[versed mica]

learning math symbols does nothing for you

there is no substance to it

[wicked edge]

Well Gamma mostly used in CS.

[fallow quarry]

It does if we see them...

[timber cape]

,w gamma function

[wicked edge]

Or in Lambda Calculus.

[flint basalt]

that really won't help you if you don't do any problems involving it to understand it

[jolly parrot]

Results provided by WolframAlpha

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

Requested by nemdt

[timber cape]

bruh

anyways

[timber cape]

oh yeah true

[versed mica]

people that have "an edge" over you in math aren't wasting their times stroking their egos thinking that learning symbols makes them smart

[fallow quarry]

Well, I see it commonly displayed on papers that I aspire to be able to write

[versed mica]

they are doing problems and learn whatever symbols show up

[wicked edge]

For, example: $\Gamma \vdash B: C$.

[jolly parrot]

Mercury (ヤフォダ)

[flint basalt]

right, but you won't be able to write a paper if you don't know how to use the symbols

[versed mica]

glorifying symbols is such an elementary thing

it's like glorifying the alphabet

[timber cape]

damn i completely forgor all of lambda calc now 😭

[fallow quarry]

That's why I am trying to learn.

[flint basalt]

knowing the definition is one thing and knowing its application is another

[wicked edge]

You want things to be clear, not too confusing by applying symbols.

They avoid using too much symbols for some reasons.

[flint basalt]

more symbols in a math problem =/= more knowledge on the topic

[versed mica]

or complexity

[fallow quarry]

Can we start with Element for now?

[olive otter]

Just help the guy already

[flint basalt]

in what context? we don't know what you have covered and what you haven't

[versed mica]

there is nothing to help him with

go to wikipedia

[timber cape]

-# except the lagrangian model for particle physics

[wicked edge]

What math are you doing now?

Know your level.

[olive otter]

Ouch

[fallow quarry]

Assume I have never seen this symbol in my life... (Because I have no idea its application nor definition)

[wicked edge]

But do you need it yet?

[flint basalt]

well, have you seen any sets before?

[fallow quarry]

No, I have not.

But, I have seen it

[versed mica]

have you taken algebra?

[flint basalt]

have you ever seen any of these?

[versed mica]

youtube video?

[fallow quarry]

I have seen R, N, and X.

[flint basalt]

or anything in notation {x1, x2, x3, ...}

[timber cape]

[fallow quarry]

I am unsure I am well versed enough to distinguish a 'set'.

[worldly jewel]

U and W tryna sneak in

[timber cape]

fahhh

[worldly jewel]

insert the who is he gif

[timber cape]

https://tenor.com/view/komik-gif-14211273463614358620

[fallow quarry]

Ok, so what is a 'set'?

[flint basalt]

https://en.wikipedia.org/wiki/Set_(mathematics)

[wicked edge]

I mean U is legit for some cases.

[fallow quarry]

Ok, how would I notice this or apply it?

[flint basalt]

apply it to what?

[fallow quarry]

You know more than I do...🥀💔

I don't know. When do we use this?

[flint basalt]

well its used in a variety of classes in different ways. discrete math, calc, linear algebra, stats, computer science

[fallow quarry]

Alright, would it be better for me to ask: How do we use it in algebra?

[flint basalt]

uhh well you would probably see it in like a set of solutions to a given function, or domains of a function

for example if you had a function:

$f(x) = \frac{1}{x-2}$

[jolly parrot]

Krish

[flint basalt]

the domain of this would be ${x \in \mathbb{R} : x \neq 2}$

[jolly parrot]

Krish

[flint basalt]

and this reads as follows: x is in the set of Real Numbers, such that x is not equal to 2.

or "x is an element of the set of Real Numbers, ..."

[fallow quarry]

Ok, I see now.

[flint basalt]

i guess you could also make a solution set for solving an equation for a variable

$x^2 = 9 \implies x \in {-3, 3}$

[jolly parrot]

Krish

[fallow quarry]

Oh, so Element refers to (in this case) what the 'x' could be?

[flint basalt]

yes

also, you may see:

$2 \in {1, 2, 3}$

[jolly parrot]

Krish

[flint basalt]

and this would evaluate to true since 2 is in the set

or

$4 \in {1, 2, 3}$

[jolly parrot]

Krish

[flint basalt]

which would evaluate to false.

theres also an "opposite" of this symbol, which means "is not an element of"

$4 \notin {1, 2, 3}$ evaluates to true.

[jolly parrot]

Krish

[brittle carbon]

Impossible

[fallow quarry]

So, how does $4\in {1, 2, 3, }$ evaluate to true but the other not?

[flint basalt]

because this is saying "4 is an element in the set {1, 2, 3}"

[jolly parrot]

Kira1sJustice

[flint basalt]

this is saying "4 is not an element in the set {1, 2, 3}"

[fallow quarry]

So how did the one with 2 work?

1,2,3

[flint basalt]

because 2 is in the set {1, 2, 3}

[fallow quarry]

Ok, I understand now.❤️

[flint basalt]

simply its asking:

you have the set {1, 2, 3}. do you see a 2 in there? yes? then its true.

[fallow quarry]

And, in the instance with a variable all it requires is solving, and then finding that number to the variable in the set. Right?

[flint basalt]

yes that is usually the idea

however like i said that's not the only way sets are used, there's different uses depending on what class.

[fallow quarry]

Well, thanks for helping. I have ran out of time today.❤️

.close

[pearl pond]

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[flint basalt]

ofc!

[pearl pond]

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[golden wave]

Not really sure how to attempt the inductive step

[errant solstice]

for future helpers, please show your current progress.

[errant solstice]

you're good so far.

now, you have to show that the RHS of your current inequality is \geq the RHS of the original inequality given n = k+1.

[golden wave]

so would be have to compare them?

[errant solstice]

mhm.

[golden wave]

and subtract

[errant solstice]

subtraction is the right move here, but depends on what you subtract. best to continue your work and we'll see.

[golden wave]

so do 6k^2 + 3 - (2k^2 +4k +3)

lemme try it rq and ill paste it here

[errant solstice]

sure.

[golden wave]

heres what i did

i attemtped the full thing

not sure its right though

[errant solstice]

yup!

also good eye to include the (true for k >= 1) part.

[golden wave]

Thank you!!!

quick question tho

is this like the method i should always use for these type of inductive questions? (proving inequalities)

[errant solstice]

I guess it depends on the inequality.

yes for this one in particular though.

[golden wave]

hmmif u can could u like explain maybe a case where it could be different

if not its fine dw

[errant solstice]

I can't think of one off the top of my head, but I also don't want to nail it down completely lest there is an edge case I never thought of. I'd rather not spread misinformation.

sorry.

[golden wave]

np, thanks for the help on the questions!

.close

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[wild fable]

so this is what i have so far

$\sum_{k=2}^{\infty}k\left(k-1\right)a_{k}x^{k-2}+\left(1-2x^{2}+\frac{2x^{4}}{3}\right)\left(\sum_{k=0}^{\infty}a_{k}x^{k}\right)$

so im not sure what to do next im supposed to mulply 1 - 2x^2 + 4/3 x^4 by the terms from a_k x^k up to k=4?

[jolly parrot]

water beam

[pearl pond]

@wild fable Has your question been resolved?

[wild fable]

.close

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[warped violet]

<@&268886789983436800>

[trail falcon]

List to make 20 biscuits

260g of butter
500g of sugar
650g of flour
425g of rice

How to find the mass of rice as the percentage of the mass of sugar

[pearl pond]

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[warped violet]

[warped violet]

You gotta make a new channel, sorry

This one will close

[mystic geode]

he a mathematician 😢

[merry carbon]

No mathematicians in my server

[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

[wraith jacinth]

<@&268886789983436800>

here too!! ^^

[pearl pond]

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[tight cobalt]

hey how can i solve for every $c_k$ the system of equation $c_1 + c_2 x_k + \cdots + c_n x_k^{n} = 0 $

[jolly parrot]

akbar_m3gaz

[honest oyster]

Given the value of roots?

[lusty nest]

Using vector and matrix?

[honest oyster]

Like solve for c(k) using what info?

[lusty nest]

I think he is asking for a general solution of any polynomial ☠️

[tight cobalt]

yes

[tight cobalt]

the polynom $$P(x) = \prod_{i=0}^{n} (x - X_i) = \sum_{i=0}^{n} a_k x^k$$ i want every $a_k$ expressed with all the $X_k$'s

[lusty nest]

Ohhhhh

[jolly parrot]

akbar_m3gaz

[lusty nest]

Vietas relation

[tight cobalt]

thanks

[honest oyster]

Ye just find it on Google there's many sources explaining it

[lusty nest]

Yeah it is very interesting and long

[tight cobalt]

thanks

.close

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[pearl pond]

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[pearl pond]

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

[stoic imp]

1

[wicked edge]

A translate would be much appreciated.

[jolly parrot]

Renato

[summer imp]

You should have a formula for the average value of a function in a region in your notes.

What’s causing you trouble then?

[stoic imp]

I am scared of double integrals

this the formula

[summer imp]

Have you sketched the region?

[visual kernel]

Hlo

I have a doubt

[stoic imp]

its a disc and x >= -y and y >= 0

[visual kernel]

Umm i have a doubt regarding math and mind

[honest oyster]

#❓how-to-get-help

[visual kernel]

Can I ask?

[honest oyster]

!occupied

[pearl pond]

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

[summer imp]

It is. If you draw it you should see that it’s just a sector of the disk

[stoic imp]

help please

[viscid shale]

Polar coordinates 🥀

[stoic imp]

care to elaborate?

[viscid shale]

Youll have to solve w/ polar coordinates

[stoic imp]

why

[viscid shale]

mostly cause this is a slice of a circle.

[stoic imp]

why

[summer imp]

Polar coordinates just behave really well with circular regions

It makes the actual computation more manageable (and sometimes possible in the first place).

So this is why you should draw the region and figure out how you’d express it in terms of r and theta (instead of x and y)

[stoic imp]

how

[summer imp]

Draw the boundary curves of the region

[stoic imp]

x = -y

[summer imp]

y=0

[stoic imp]

y = 0

x^2 + y^2 = 1

y = -x

[summer imp]

Then shade the appropriate region. It should make it clearer that it is a sector of a circle.

[stoic imp]

oops

i made a mistake at the first one

[summer imp]

Then you can either try to delimit it in r and theta visually, or you can use the equations x=rcos(theta) and y=rsin(theta) for polar coordinates and see what the inequalities for your region translate to.

[stoic imp]

@summer imp

[stoic imp]

this?

[summer imp]

That’s not the portion with x >= -y.

You can check that (-1,0) doesn’t satisfy that.

[stoic imp]

yeah

[summer imp]

Why is x=0 delimiting anything ?

[viscid shale]

Try graphing the 3 conditions as regions, and find the intersection of all three.

[stoic imp]

easy to say

[viscid shale]

pretty easy to do too

[stoic imp]

x >= -y
-x >= y

y <= -x

[summer imp]

Careful with signs and inequalities

[viscid shale]

Not quite

[stoic imp]

fuck my life

x >= -y
-x <= y
y >= -x

[summer imp]

You need y>= 0 though.

You want the region which is above the line y = -x, above the x axis, inside the circle.

[stoic imp]

I CANT

[summer imp]

Well just draw the three inequalities and take their intersection like laplace said then.

If anything just indicate with an arrow the side of the lines you’re keeping

[stoic imp]

can u help

[viscid shale]

Can you graph the region x² + y² =< 1?

[stoic imp]

how?

[summer imp]

If you just take the time to do it instead of trial and error you;ll understand.

Draw the three inequalities y >= -x, y >= 0 and x^2 + y^2 <= 1

[stoic imp]

i have trouble with y >= - x

help

i can draw y = -x

but idk y >= -x

[viscid shale]

You can usually consider that the line y = -x divides the region which doesnt belong to that criteria from the one which does

So, try a value above the line and one below

[stoic imp]

what

@summer imp

[summer imp]

Just test a point on one side of the line to see if it satisfies the inequality

[stoic imp]

like what

point

[summer imp]

Any point that isn’t on the line.

[stoic imp]

0-1 and 10

[summer imp]

Right so which one of those satisfies the inequality

[stoic imp]

10

[summer imp]

So that indicates which side of the line your inequality is describing

The side which contains the point (1,0)

[stoic imp]

ye

i got it now

you make it look easy tho

[summer imp]

It should be easy to graph inequalities if you’re doing multivar calculus

Anyways

So now you have your region.

If you can guess right away from the graph the way you would express this with r and theta, then you should do that.

If not, then consider the inequalities y>= 0, x >= -y and x^2 + y^2 <= 1 along with the polar coordinate equations x = rcos(theta), y= sin(theta).

[stoic imp]

don't bully I'm new to calc

[stoic imp]

how?

[stoic imp]

sin(t) >=0, cos(t) >= -sin(t), r <= 1

[summer imp]

It was not a dig at you. I'm just saying that this will come up a lot, first with 2D regions, then with 3D regions, and you should either be already comfortable with dealing with them, or get used to them urgently.

[stoic imp]

r is less than or equal to 1 and angle is 90 + 45 deg

[summer imp]

Yep

[stoic imp]

what about it

[stoic imp]

anal is hard

[summer imp]

What about sin(t) >= 0?

What does that say?

[stoic imp]

angle is pi/2 + pi/4

angle is [0, 3pi/4]

r is [0,1]

[summer imp]

Yes

[stoic imp]

unsure

[stoic imp]

i don't know

@summer imp

[summer imp]

Well that's it. sin(t) >= 0 just means your angle should be in [0,pi].
sin(t) >= -cos(t) just means your angle should be in [-3pi/4, 3pi/4]
r <= 1 tells you r is in [0,1].
So taking the intersection for the angles you need it to be in [0, 3pi/4].

[stoic imp]

how?

trig scary

[summer imp]

They're the same equations as your region. It should make sense that the angle in the disk is restricted to [0,pi] by y >= 0 and that it's also restricted to [-3pi/4, 3pi/4] by y>=-x.

[stoic imp]

yep

nice

[stoic imp]

now what

?????

@summer imp

[viscid shale]

If youre tasked with using polar coordinates you should start practicing for trig.

[summer imp]

You're looking for the average value of f(x,y) = x+y in this region.

[stoic imp]

don't bully

[summer imp]

So now you need the area of the region first.

[summer imp]

Can you compute that (without an integral)?

[stoic imp]

ye

2/3 × pir^2

@summer imp

[viscid shale]

2/3?

[stoic imp]

angle is [0, 3pi/4]
r is [0,1]

[stoic imp]

3/4

@summer imp

right(?

[viscid shale]

Do you remember the formula for the area of a sector?

[stoic imp]

3pi/4 is A(R)

[summer imp]

To get the area of a sector you multiply the area of the disk by (sector angle)/2pi.

[summer imp]

Idk why you would expect this to be the area

[viscid shale]

,, A_s = \frac\theta2 r^2

[viscid shale]

For clarity

Here r = 1

[stoic imp]

lets just use integrals please..

[summer imp]

You don't need integrals to find the area of the sector of a disk (not that it's particularly hard)

[viscid shale]

Can you set up the integral to calculate the area?

[stoic imp]

yes

[viscid shale]

go ahead

<@&268886789983436800>

[stoic imp]

i forgot how

[summer imp]

Can you write down the general double integral for the area of any region?

[stoic imp]

,, \int \int_R f(x,y) dydx

[jolly parrot]

Renato

[viscid shale]

Thats for the cumulative value of f(x,y) over R

we want the area of R

Notably, theres a really specific """function""" that gives you the area of a region.

[stoic imp]

just 1

[viscid shale]

btw, \iint exists

,, \iint_R\ dxdy

[stoic imp]

,, \iint_R 1 \ dydx

[viscid shale]

Now, what happens if you translate this to polar coordinates?

[jolly parrot]

Renato

[stoic imp]

unsure

[summer imp]

Do you have notes on polar coordinates?

[stoic imp]

r = 1

[viscid shale]

Mostly, how to move from cartesian -> polar

[summer imp]

You should have seen in class how to integrate regions over polar coordinates.

[stoic imp]

what about it?

[viscid shale]

You should prob have some notes on |J|, the jacobian determinant of the transformation

[stoic imp]

the jacobian is r

in polar coordinates

@viscid shale @summer imp

[viscid shale]

So, what happens when you move from dxdy -> dθdr to our previous ""function""

[stoic imp]

1 -> r

[viscid shale]

Write down the integral

[stoic imp]

$\iint_R r \ d \theta dr$

[jolly parrot]

Renato

[viscid shale]

And what are the bounds for "R"

[stoic imp]

hard

[viscid shale]

Its what weve been discussing at the start.

[stoic imp]

,, \int_0^1 \int_0^{3\pi/4} r \ d\theta dr

[jolly parrot]

Renato

[viscid shale]

And now solve

[stoic imp]

how?

[viscid shale]

As any other double integral?

Go from the inside to the outside, compute the antiderivatives and evaluate, etc...

[stoic imp]

@summer imp @viscid shale

[summer imp]

Good.

Now similarly compute the cummulative value for the function f(x,y) = x+y you were given.

[stoic imp]

how

[viscid shale]

use polar coordinate transformations x = rcosθ, y = rsinθ to get the function you wanna integrate alongside with the same bounds

[stoic imp]

i dont follow

[summer imp]

You just did $\iint_{R} \dd{x}\dd{y}$. Now you need to compute $\iint_{R} (x+y) \dd{x}\dd{y}$.

[jolly parrot]

Azyrashacorki

[summer imp]

Same method with polar coordinates. x=rcos(theta), y=rsin(theta).

[stoic imp]

is there any tricks to skip that and go to the result

[summer imp]

Well the point is for you to compute this integral.

So no

[viscid shale]

You forgot the jacobian

[stoic imp]

@viscid shale @summer imp

[viscid shale]

Nobody has introduced you to Fubini's Theorem/Rule?

[summer imp]

Sign error for -1 here in any case.

[stoic imp]

wdym

[summer imp]

Should be -cos(3pi/4) - (-cos(0))

[stoic imp]

fuck this shit broo

[viscid shale]

Ill spoil it to you

,, \int_0^1\int_0^{\frac{3\pi}4}(r\cos\theta + r\sin\theta)\cdot r\ d\theta dr = \int_0^1\int_0^{\frac{3\pi}4} (\cos\theta + \sin\theta)\cdot r^2\ d\theta dr =\ \int_0^1 r^2 dr \cdot \int_0^{\frac{3\pi}4}(\cos\theta + \sin\theta) d\theta

[stoic imp]

is illegal

[summer imp]

It's not. It's just a consequence of Fubini's theorem you've learned about.

[stoic imp]

care to elaborate

[viscid shale]

ngl i dont recall what are all the conditions on Fubini's

[viscid shale]

Big corollary which will prob be corrected:

If a function $f(x,y)$ can be written as the product of two functions, $g(x)\cdot h(y)$, and you have a region $R$ where the bounds are constant\par --- $(a\leq x\leq b) \wedge (c \leq y \leq d)$ --- then:
$$\iint_R f(x,y) dx dy = \int_a^b g(x) dx \cdot \int_c^d h(y) dy$$ 

[viscid shale]

There are a few ifs and buts, but in general this is the idea.

In reality Fubini's Theorem is this huge thing that allows you to draw a lot of conclussions for calculus & real analysis.

[stoic imp]

@viscid shale @summer imp

[summer imp]

You should know what sin(3pi/4) and cos(3pi/4) are.

[stoic imp]

trig scary

[viscid shale]

.

[stoic imp]

,w 8 /(3pi) * 1/3 * (sin(3pi/4) - cos(3pi/4) + 1)

[jolly parrot]

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Requested by renatoxbz

[stoic imp]

.solved

[pearl pond]

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[pearl pond]

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[untold latch]

anyone here

[untold latch]

xan anyone help

[pearl pond]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

@untold latch Has your question been resolved?

[wicked edge]

A specific question is needed here.

[pearl pond]

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[tired radish]

I keep getting this wrong, where did I mess up ?

[dense jasper]

Your setup here means that you essentially calculated the integral of -|2-x| instead

[dense jasper]

Dropped negative

[pearl pond]

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[pearl pond]

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[frank parrot]

Hi does anyone have free

[frank parrot]

Time

And could go over a amc 10 prep with me

Lowk being lazy and I hate it

[dire mortar]

simply ask the question

[frank parrot]

Ok

[dire mortar]

dont say, can anybody help me with…

😉

[frank parrot]

Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are $3$ cm and $6$ cm. Into each cone is dropped a spherical marble of radius $1$ cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level in the narrow cone to the rise of the liquid level in the wide cone?

[asy] size(350); defaultpen(linewidth(0.8)); real h1 = 10, r = 3.1, s=0.75; pair P = (r,h1), Q = (-r,h1), Pp = s * P, Qp = s * Q; path e = ellipse((0,h1),r,0.9), ep = ellipse((0,h1s),rs,0.9); draw(ellipse(origin,r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill(origin--Pp--Qp--cycle,gray(0.8)); draw((-r,h1)--(0,0)--(r,h1)^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(0,Qp.y),Arrows(size=8)); draw(origin--(0,12),linetype("4 4")); draw(origin--(r*(s-0.1),0)); label("$3$",(-0.9,h1s),N,fontsize(10)); real h2 = 7.5, r = 6, s=0.6, d = 14; pair P = (d+r-0.05,h2-0.15), Q = (d-r+0.05,h2-0.15), Pp = s * P + (1-s)(d,0), Qp = s * Q + (1-s)(d,0); path e = ellipse((d,h2),r,1), ep = ellipse((d,h2s+0.09),rs,1); draw(ellipse((d,0),r(s-0.1),0.8)); fill(ep,gray(0.8)); fill((d,0)--Pp--Qp--cycle,gray(0.8)); draw(P--(d,0)--Q^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(d,Qp.y),Arrows(size=8)); draw((d,0)--(d,10),linetype("4 4")); draw((d,0)--(d+r*(s-0.1),0)); label("$6$",(d-r/4,h2*s-0.06),N,fontsize(10)); [/asy]

$\textbf{(A) }1:1 \qquad \textbf{(B) }47:43 \qquad \textbf{(C) }2:1 \qquad \textbf{(D) }40:13 \qquad \textbf{(E) }4:1$

Solution

[jolly parrot]

Hate

[frank parrot]

Oh it cant make the image

And these

<@&286206848099549185>

[livid goblet]

for q12 if all roots are integers they must be factors of 16
and that their product is 16

you also know that they add up to 10

[pearl pond]

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[pastel valley]

I dont get where the 3 and 4 come into this equation, its very annoying

[kind rampart]

can you send the full original expression? it appears to be cut off in this picture

[kind rampart]

i assume you got the LCD correctly, right

[pastel valley]

Yeah the lcd is right, then I always get the top part wrong, i dont get where I get it wrong though

[kind rampart]

can you send your rough work?

[pastel valley]

Lemmie put it to paper, 1 sec

[kind rampart]

ah i see i can probably guess what you did

you did 2 ⋅ 12 and 7 ⋅ 9 and got 24 + 63 = 87

[pastel valley]

Yeah I did that exactly

[kind rampart]

alright

so you want to have both of the fractions have the same LCD (36) so you can add them directly

right ?

[pastel valley]

Yeah

[kind rampart]

lets work on the first fraction. whats the denominator in 2/9 ?

[pastel valley]

18?

[kind rampart]

not quite

in $\frac{2}{9}$, what is the denominator?

[jolly parrot]

insult

[pastel valley]

3?

This is breaking my head

[kind rampart]

can you work me through how you got to 3?

[pastel valley]

No i would be 4 no?

Because 2 can fit into 9 4 times?

[kind rampart]

no, that isn't what i'm asking

do you know what a denominator is ?

[pastel valley]

The lowest both can fit into

Like 4 and 3 is lcd 12

[kind rampart]

not sure what this means, exactly

for example, in $\frac{1}{2}$, what would the denominator be?

[jolly parrot]

insult

[pastel valley]

2?

[kind rampart]

alright yes, its 2.

how about in $\frac{2}{3}$?

[jolly parrot]

insult

[pastel valley]

6?

[kind rampart]

why 6?

[pastel valley]

Multiples of 2 are 2, 4, 6
Multiples of 3 are 3, 6, 9

Lowest thats both multiply into

[kind rampart]

you're getting LCD mixed up with denominator

[pastel valley]

Ohhh

[kind rampart]

besides, you shouldn't even be finding the LCD of that fraction ... because 2 is a numerator and 3 is the denominator

how about in this fraction, $\frac{2}{5}$? what is the denominator here?

[jolly parrot]

insult

[pastel valley]

5

[kind rampart]

alright

how about in $\frac{3}{10}$?

[jolly parrot]

insult

[pastel valley]

10

[kind rampart]

okay

[kind rampart]

so whats the denominator in 2/9, in the previous question

[pastel valley]

9

[kind rampart]

alright

so we want to make that denominator into 36, the LCD, so we can add them directly, right

[pastel valley]

Yes

[kind rampart]

alright. so we want to turn $\frac{2}{9}$ into some fraction with denominator 36, $\frac{?}{36}$

[jolly parrot]

insult

[kind rampart]

what do you think should we do?

[pastel valley]

We find that by also finding the denominator of the fraction we are adding

So the other fraction denominator is 12, so we simply find the lcd of both

And we see its 36?

[kind rampart]

alright, thats how to find the LCD. but we've already done that

[kind rampart]

what do you think we should do to get 2/9 to have a denominator 36?

[pastel valley]

How much we need to multiply 9 to get there to the denominator of 37

36 mb

[kind rampart]

yup, so how much?

[pastel valley]

4

Holy shit I get it now

It clicked

[kind rampart]

alright, four. now since we multiply the bottom by four, we have to multiply the top by four to keep it equal

$\frac{2}{9} \cdot \frac{4}{4}$

[jolly parrot]

insult

[kind rampart]

what does this equate to?

[pastel valley]

8 over 36

[kind rampart]

alright, keep that in your mind for later. now, for the second fraction, we have $\frac{7}{12}$. what do you think we should do here?

[jolly parrot]

insult

[pastel valley]

Do the same, multiply by the amount needed for lcd, 3 so
7 times 3 is 21
So 21 over 36

So its 8 plus 21 so 29
The answer would be 29 over 36 since the denominator always cancels out

[kind rampart]

alright, so now we have two fractions with the same denominator (you may have called these like fractions). $\frac{8}{36} + \frac{21}{36}$

[jolly parrot]

insult

[kind rampart]

what does that equate to?

[pastel valley]

29 over 36

[kind rampart]

great, you've added them. before answering however, you should check if you can simplify it further. does 29 and 36 have any common factors?

[pastel valley]

No the top one isn't more than the denominator

So its just like that

[kind rampart]

yes, it is unsimplifiable. however, this is the wrong reasoning

for example, look at fraction $\frac{6}{4}$.

[jolly parrot]

insult

[pastel valley]

1 and 2 over 4

[kind rampart]

ah you're using mixed Numbers

[pastel valley]

Huh

[kind rampart]

dont mind that. its just that later youll find that its often easier to work with fractions that aren't in this form

anyways, ill just leave off saying that you can simplify $\frac{6}{4}$ as $\frac{3}{2}$.

[jolly parrot]

insult

[pastel valley]

Ohhhhhh I see what you mean