help-17

[halcyon cypress]

I am crossing it?

oh

[hushed pewter]

I meant y

[halcyon cypress]

hmm

wait thats true

shiet

[hushed pewter]

,w sin(ln((-1)^2+0^2-2(-1)+1))

[twin meteor]

Results provided by WolframAlpha

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Requested by SWR#8588

[hushed pewter]

[halcyon cypress]

I have done something wrong then I think because when I paint up this function in 3D GeoGebra I can just tell by eye

that I do not cross it

Which is a really bad argument haha

But in your wolfgram example you use x = -1

[hushed pewter]

Oh oops

You had it right

I made an oopsie

out radius is 1

so it should not cross y axis

[halcyon cypress]

Oh nice

Thank you so much really you have helped me a lot

Could you just summarize what we did so I can get a deeper understanding?

[hushed pewter]

you're trying to find all the $(x, y)$ that satisfy $f(x, y)<0$

[twin meteor]

SWR

[hushed pewter]

Or, more specifically,

So you need to know when sin<0, which is in all intervals (2π(n+1/2), 2π(n+1))

So you find

[halcyon cypress]

Yea I understand so far

Upside down A means all right?

[hushed pewter]

"for all"

[halcyon cypress]

Ah and then we have this inequality and we want to "place" it as close to the point (0, 0) and therefore we chose n = -1 because we cannot chose something like n = -1/2 for example

How did you however simplify (x-1)^2 + y^2 = 1

Since you write e^-π < 1 < e^π

omg

wait

dont even answer

bruh im so slow

haha

we chose 0,0

[hushed pewter]

[halcyon cypress]

nvm that its late for me xD

And then we have our intervall where we have the answer

And then we just write out the radius

making a "donut"

Where the donut showcases all the negative values of f(x,y) close to the point (0,0)

@hushed pewter Thank you so much, really really appreciate it

[hushed pewter]

"negative values of f(x, y)" is technically okay, but imo sounds kinda weird. I'd say "all (x, y) near (0, 0) where f(x, y)<0"

[halcyon cypress]

Ah ok

[hushed pewter]

get some rest

[halcyon cypress]

I will thank you for the help

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[atomic marsh]

How do i go about solving this problem? It is my understanding that the normal vectors of the 2 planes should have a dot product equal to zero but i do not know how to find the normal vector of plane 2

i know that my normal vector for plane 1, n1 = (2,-4,6)

but past that i am stuck

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[plush quest]

Can someone help me see what I did wrong

I'm not getting the right terms

I just assumed the center is 0

Cause otherwise idk how id do it

[hushed pewter]

why are the factorials on top?

[plush quest]

Because I have fraction on top divided by a factorial so i move the factorial on top

Like this

[hushed pewter]

There's a difference between $\frac{a/b}{c}$ and $\frac{a}{b/c}$

[twin meteor]

SWR

[plush quest]

I dont understand this

[plush quest]

Thats just the same thing rewriten

[hushed pewter]

no

No it is not

That's what I am saying

They are different values

[plush quest]

Im saying you rewrote the same thing

in both texts i just dont understand what it means

[hushed pewter]

I'm saying I did not

oh

divide 1/2 by 2

what do you get?

[plush quest]

1/4

i think

[hushed pewter]

yes

now divide 1 by 1/2

what do you get?

[plush quest]

2

[hushed pewter]

yes

[plush quest]

I think

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[digital shell]

Can someone understand the math that they did to get y/9

Hell me understand

[sullen thunder]

are you aware of exponent rules?

[cosmic kelp]

Is this channel available

[digital shell]

Yes but I don’t understand this particular one

No

[digital shell]

You have to ask for help on another

Check the help rules

Its an open channel

[sullen thunder]

you can see that $3^{x+2}$ has a sum in the exponent

[twin meteor]

Talent Unlimited

[digital shell]

Btw plug x-2 into x in the functions x

That’s all the answer is

Yes

[queen moth]

x^(a+b)=x^a+x^b

[sullen thunder]

so you can split that into two terms

[digital shell]

Yup

[sullen thunder]

$$ 3^{x+2} = 3^x \times 3^2 = y$$

[queen moth]

So 3^(x+2)=3^x+3^2

[twin meteor]

Talent Unlimited

[digital shell]

Yup

[queen moth]

And 3^2=9

[digital shell]

So I’m confused on how to solve

[sullen thunder]

from this you know that 3^2 is 9

[digital shell]

Yep

[queen moth]

So 3^x=y/9

[sullen thunder]

so just transpose that to the right hand side

[digital shell]

Where is the y/9 coming from

That’s what I don’t get

[sullen thunder]

$$ 3^{x+2} = 3^x \times 3^2 = y$$
$$ 3^x \times 9 = y $$
$$ 3^x = \frac{y}{9} $$

[twin meteor]

Talent Unlimited

[queen moth]

Yes

[sullen thunder]

It's being transposed from the left hand side

[queen moth]

You divide both sides by 9

[digital shell]

Why 9?

[queen moth]

Since 3^2=9

[digital shell]

Is there an example that explains this with just plain numbers

Yes but

Why divide by 9

Is this an algebraic rule

[sullen thunder]

because the question is asking to show $3^x$ in terms of $y$

[twin meteor]

Talent Unlimited

[queen moth]

yeah

as 3^(x+2)=y

[digital shell]

So what does the in terms of y even mean

Am I solving for y

I’m confused

[sullen thunder]

@digital shell

[queen moth]

you are solving for 3^x

terms of y means that 3^x = "something with y in it"

[sullen thunder]

"in terms of" means representing a particular thing using a particular variable

[queen moth]

yeah

[sullen thunder]

in this case representing $3^x$ using $y$

[twin meteor]

Talent Unlimited

[digital shell]

Oh

So does this mean to keep 3^x and y

And j algebraic ally simplify the other parts

Algebraically

[queen moth]

it means that one side of the equation will contain $3^x$

[twin meteor]

voiceless

[sullen thunder]

basically yes

[queen moth]

and the other will contain a expression containing the variable y

[queen moth]

yeah pretty much

any other questions?

@digital shell

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[sharp oasis]

Good morning gents. What that sign means?

[sharp oasis]

its Discrete Math btw

[lusty fox]

phi can mean a lot of things

[sharp oasis]

oh to its phi

actually i needed a sign meaning,not a context of it

thanks!!!

[lusty fox]

oh lol np

[sharp oasis]

thanks a lot,solved!

[karmic imp]

Fyi to close a channel, use .close

[urban hamlet]

need help

[karmic imp]

Dude read #❓how-to-get-help

And open your own channel

[sharp oasis]

.close

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[vast shale]

-8/7 = slope , 5/7 = y intercept

[vast shale]

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[vocal sleet]

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.

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[elfin parcel]

<@&286206848099549185>

[sly sierra]

don't ping helpers until after 15 min

[elfin parcel]

Oh okay

[sly sierra]

nw

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[sly sierra]

have you tried drawing a picture? seems like it would be helpful here

[elfin parcel]

Yes

[sly sierra]

oh, i had thought that the 60 and 40 referred to the outer rectangle's sides, not the inner rectangle

i might be misinterpreting the question tho

actually after reading again i think yours is probably right

[elfin parcel]

Lol perhaps

[sly sierra]

i guess try it your way and see if you get one of the answers haha

[elfin parcel]

So I used the equation;

(2x+60)(2x+40)=666-2400

4x^2 + 200x + 2400 = 666 + 2400

4x^2 + 200x - 666 = 0

2x^2 + 100x - 333 = 0

And then I'm stuck

We aren't supposed to use the quadratic formula

[sly sierra]

one moment

what area does (2x+60)(2x+40) represent

in your picture

[elfin parcel]

The entire tablecloth + the lace

[sly sierra]

ok

shouldn't that equal 666 + 2400

not 666 - 2400

[elfin parcel]

Oh yeah I meant that

But I still solved the rest like it was + instead of -

[sly sierra]

ok, let me read over your work now

[elfin parcel]

Just a typo

[sly sierra]

yea so far i agree

and now because we are lazy we appeal to wolfram alpha haha

,w 2x^2 + 100x - 333 = 0

[twin meteor]

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Requested by Bungo#8186

[elfin parcel]

Okay, I can't find the numbers that multiply to -666 and add to 100 😦

Hmmmm

With the level of work I'm doing, it should be at most a fraction or whole number 😭

[sly sierra]

hmm, that doesn't match any of the answers

maybe my interpretation was the right one then

[elfin parcel]

I'll try it

Okay I got

(60-2x)(40-2x)=666+2400

Which simplified to 4x^2 - 200x - 666 = 0

[sly sierra]

wait shouldn't the RHS be 2400-666

[elfin parcel]

Which still have no even factors

[sly sierra]

in that case

[elfin parcel]

Ohhh yes

[sly sierra]

,w 4x^2 - 200x + 666 = 0

[twin meteor]

Results provided by WolframAlpha

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Requested by Bungo#8186

[elfin parcel]

Ahh

Maybe my teacher made a typo... 😭

[sly sierra]

ugh, or we're both misinterpreting

would have been cool if the teacher had provided a picture haha

[elfin parcel]

Yes 😢

This is the type of problems we did in class

[sly sierra]

hmm alas i'm out of ideas

[elfin parcel]

I see

[sly sierra]

hopefully your teacher can clear it up

[elfin parcel]

Thank you for trying

[sly sierra]

cheers

[elfin parcel]

🥲🥲 have a nice day

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[barren dragon]

anyone in calc 3 know how to do this? im not sure where/how to start

also, part b) should say 6a and not 4a

[rugged vortex]

Start by parameterizing the ellipse-disc

[restive compass]

like this?

[rugged vortex]

I was thinking more of a direct parameterization, but that also works

Actually that works fine

But uh

Personally I'd parametrize the ellipse-disc directly

Because then the jacobian requires little thought if you're familiar with them

[restive compass]

yeah im familiar with doing that part but im not exactly sure what paramterizing is

[barren dragon]

wouldnt the jacobian be easier the way allarkvark did it

[rugged vortex]

Like parametrics

[restive compass]

this was a different one but similar

[rugged vortex]

Well if you're familiar with Jacobians

Personally I'd rather go with an ellipse

There's a neat trick about Jacobians

If x = au cos(v) and y = bu sin(v), the jacobian is just (a•b)u

That's why I personally like to parametrize an disc

[barren dragon]

how would i turn the ellipse into x=au cos(v) or y=bu sin(v)

[rugged vortex]

a is the radius in the x direction, b is the radius in the y direction. You let u go from 0 to 1 (which should be clear as to why, since you gotta go from 0 to the full length of the axes) and v go from 0 to 2π

[restive compass]

parameterizing when u just set a variable to another or after u solve the integral

[barren dragon]

i figured it out

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[plush roost]

could I say for this when a/b are put into simplest form such that a and b are co prime, and the denominator (b) is a prime number greater than 2, or a multiple of a prime number?

[sly sierra]

a terminating decimal can always be expressed in the form a/10^n for some n

where a and n are integers

[plush roost]

but 1/2 is terminating and the n that satisfies 10^n = 2 is non-integer

right?

[sly sierra]

1/2 can be expressed as 5/10

a = 5, n = 1

[plush roost]

oh i see

is what i said wrong though?

[sly sierra]

well doesn't 2/3 satisfy your criteria?

and it doesn't have a terminating decimal

[plush roost]

oh no i meant if b is not a prime number (aside from 2) and not a multiple of a prime number

my bad

[dark kiln]

so like 1/12

[plush roost]

12 is a multiple of 3 though

[dark kiln]

any number is a multiple of prime number, that's just composite, there's no third type

[plush roost]

i know. So something like 1/12, and since 12 is a composite number that is a multiple of prime number 3, 1/12 must be non terminating

[dark kiln]

i don't get it

then why 1/8 is terminating

[plush roost]

1/8 is terminating because 8 it not prime, and it is not a multiple of any prime number (with the exception of 2)

[dark kiln]

ah ok

[plush roost]

did i just word the criteria poorly?

[dark kiln]

it's not complete

observe how 1/25 is terminating

[plush roost]

yikes lol

now i see Bungo's criteria. Is that like the universal criteria?

oh wait it makes sense actually

that's pretty cool now that i think about it lol

[sly sierra]

sorry, i got distracted... i was going to follow up and say, if you start with a/b in lowest terms (coprime), then if b has any prime factors other than 2 or 5, it's not gonna have a terminating decimal expansion

otherwise it will

[plush roost]

no worries, im just grateful to get any help at all. But now i feel like my original criteria might need to be completely scrapped, but 1/15 is non-terminating

[sly sierra]

yea for 1/15, the denom has 3 as a prime factor

there's no way to get that into the form a/10^n where a is an integer

[plush roost]

oh right lol

forgot 3 was a factor lol

[sly sierra]

basically as long as the denom only has 2's and 5's as factors, you can multiply num and denom by either powers of 2 or powers of 5 until the denom is a power of 10

[plush roost]

right, thank you so much!

[sly sierra]

sure, cheers!

[plush roost]

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[undone aurora]

,rotate

[undone aurora]

what did I do wrong

I checked wit the integral calculator and this is the difference

I cant seem to find the error in my work

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😭

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[halcyon cypress]

for 1d)

Is the answer: As ln > 0 and the quota is not 0, {(x,y) belongs to R2: (x,y) is not (1,0)} , so the function is continous in its domain

?

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[icy plume]

Hey, I am very confused by this factor, can someone tell me what was done between these equations

[vast shale]

put this on a common denominator

[icy plume]

oh, thats simple. Thank you

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[timber gale]

Umm hello i need help evaluating this piecewise function

[timber gale]

Idk how to figure out which one to use in the question 16

[paper depot]

wdym "which one to use"?

[timber gale]

It’s kinda hard to explain but like 15

[paper depot]

which of the three function values are you having trouble calculating? f(2), f(-2) or f(-1)?

[timber gale]

Ah shoot

I thought it was a whole thing

Since there’s no comma

But the answer is just -2..

Whilst question 15 it was 3 different set of answers

[paper depot]

yeah bc you're asked to calculate one number and not three numbers.

but calculating f(2)-f(-2)+f(-1) requires some steps, of course

[timber gale]

Ah shoot bruh

[paper depot]

you need to calculate f(2), then calculate f(-2), then calculate f(-1), then subtract and add them as written.

[timber gale]

Yes I just realised so basically its just algebra oops alright thanks

[timber gale]

Yes I’m dumb I just realised what you said

Nobody saw

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[golden onyx]

do pentagons always have the same angles if all sides are equal in length?

[river minnow]

No

I believe that applied to any n-gon with n > 3

Because for quadrilaterals there are rhombi

They have equal sides, but angles are not always equal

[golden onyx]

so like this

but it wouldn‘t be possible without having angles greater than 180

[river minnow]

You can come up with many such shapes, yeah

Right

[golden onyx]

so if the angles are less than 180, it is a regular polygon

[river minnow]

Can't think of a counterexample, so yes, I suppose

[golden onyx]

k thanks

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[river minnow]

Actually wait

.reopen

[vocal sleet]

✅

[river minnow]

One came to my mind

Draw a square and an isosceles triangle on top of it

[golden onyx]

oh right

thanks

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[plucky path]

can i get some help

[strange crater]

2 to the what power equals 128?

[plucky path]

7

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[fluid notch]

The denominator of a fraction must be equal during addition or subtraction for two fractions to be added or subtracted in the first place

The denominator does not change in addition or subtraction. Only the numerator

Maybe an example will help?

So if we take 2 fractions like the following: 3/5 and 4/7
How would you approach adding these two?

I wanna see how you'd approach this first before going into further detail

[loud walrus]

Do you know why are you making the denominators equal?

[fluid notch]

So basically what you've explained in that is exactly why you make the denominators equal

lcm of the denominators works or just multiplying either fraction by the denominator of the other in order to make the denominators equal

Depends what you mean by that?

[loud walrus]

Maybe you need something visual

[fluid notch]

If you add 3/5 and 4/7 without making the denominators equal you'd get 7/12. 7/12 is not the same as 41/35

[loud walrus]

This denominators are already equal, so you can add or substract right away

[fluid notch]

Additionally you can look at it through decimals

[civic drift]

Let's say we have 1/4 of a pizza and 1/2 of the same pizza if we add it it becomes 3/4 of the pizza rather than 2/6 of the pizza

[loud walrus]

Here you have one example where you can't just add without making the same denominator, and you will see the reason why looking at the size of the parts

1/3 of a whole is bigger than 1/4, so to add them both, you must make them to be equally big

You can do them equally with a bigger denominator

for example: 1/2 is equal to 2/4

so if you have to add 1/2 + 1/4, you can make 1/2 = 2/4 and add them up

2/4+1/4 = 3/4

you are multiplying by 1

for example

2/2 is = 1, right?

so if you multiply 1/2 * 2/2

you get 2/4

and because 2/2 is = 1, that means 1/2 * 1 = 2/4

so 1/2 = 2/4

because multiplying by 1 is like doing nothing

it's just a transformation of your number into something you want

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[old niche]

i am very confused on how to do the following q

[old niche]

.close

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[crimson grove]

So what is p if not 6?

[hot ledge]

It would be 2

Because $\sqrt{n^6} = n^3$

[twin meteor]

black_couscous

[rugged orchid]

$\sqrt{n^6} = |n|^3$

[twin meteor]

Frosst

[hot ledge]

But here n is positive

[rugged vortex]

n is positive so the absolute value doesn't matter

[rugged orchid]

i am blind i better sleep

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[arctic panther]

how do i

[arctic panther]

find the T value?

do I use excel?

[strange crater]

you can use whatever software/method you are comfortable with

[arctic panther]

oh hiii

okay what is the function

i use

[strange crater]

I don't use excel, so I don't know sorry

[arctic panther]

okay no worries i will scour my notes

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[carmine kettle]

How is this looking

I am not sure if simplifying the power is a legal step like I did

(Getting rid of the 2/n)

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[vocal sleet]

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[split wind]

you might need to repost since this channel is closed

[narrow stone]

Okay

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[tidal light]

how do i figure these out from the top determinant

[karmic imp]

Do you know how each row operation affects the det?

[tidal light]

i forgot how the sign changes

but do i take the 2 out

multiply by 3 and take the negative out

[karmic imp]

That's useful

If I'm not mistaken, column operations do the same too, so you can take the 2 out of the colum, instead of the row

[strange crater]

they do

[tidal light]

what’s the rule for multiplying c2 by 3

[karmic imp]

It's the same as row operations, if you multiply a row by a constant, it scales by that, same logic with columns

[tidal light]

.close

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[lyric grove]

Hey, is this proof right? My professor gave an easier proof, but this is what I came up with. I feel like something is up with the substitution.

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[muted wyvern]

is the remainder thing for series ( ∫n+1 < Rn < ∫n) the same for alternating series?

[muted wyvern]

i know the first term is the biggest

but how do we do the remainder

[flat whale]

do you have an actual problem you're working on?

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[magic belfry]

does this look right

that's valid innit

I think it is

[hot ledge]

It is

Just be careful with the notation

It is a bit confusing

[magic belfry]

wym

[hot ledge]

k=2k²

You could write m=2k² so that n²=2m

But it's a detail

[magic belfry]

ooh ok

gotcha

.close

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[vast shale]

hello, I have a problem. for context, i am writing a small programming project that involves a little bit of angle work.

into the question. I have been trying to find a function to "convert" "unit circle" degrees into "compass" degrees. (see diagram below). I've tried a ton of things but my brain is seriously fried. I've been thinking of a solution since sunday. please help!

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@vast shale Has your question been resolved?

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@vast shale Has your question been resolved?

[sand orbit]

Did you try a piecewise function y=90-x for [0,90] and so forth for each quadrant

[tacit flame]

are you allowed to import packages?

there should be something in almost every language to do that for you

i think i could also come up with a solution in Java you could build your solution off of let me take a look

could you give me an example of a conversion on a piece of paper or soemthing im not sure what you are trying to convert from/to

[half imp]

(450-x)%360

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[gusty pelican]

ANOVA: Does it have to be categorical?
My understanding is, that it can be used to analyze any model right?
Just simply replace the $\bar{y}_i$ with the model prediction, right?

[twin meteor]

BeatriceBernardo

[gusty pelican]

I'm using this as a reference https://www.itl.nist.gov/div898/handbook/prc/section4/prc421.htm

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[astral pilot]

,rotate

[empty frigate]

,rccw

[empty frigate]

well
you know from (a) that it's 125/4 m above the ground
you know the acceleration (due to gravity) is 10m/s^2
you should know how fast it's initially moving vertically
that should be enough information to work out how long it will take for it to hit the ground

@inner lodge

[inner lodge]

i know

i have the information

but im not sure what to do with it

its moving 15m/s initially i worked out

im not sure what to do as it "bounces"

[empty frigate]

huh, that's... not what i expected
is that 15m/s up or 15m/s down?

[inner lodge]

upwards i think

I did 30sin30

[empty frigate]

i think you just ignore any bouncing and work out how long before it hits the ground for the first time

[empty frigate]

oh wait you mean at the time when it's thrown?

[inner lodge]

yep

[empty frigate]

i meant "initially" as in "at the start of it falling after hitting the wall"

[inner lodge]

ohh

so would i solve for the vertical displacement eqution equal to zero first

i got that it hits the ground again at t=4 in that case

[empty frigate]

...i don't know which equation that is

[inner lodge]

um i worked out another equation

so it gave -10 m/s^2 initially for acceleration

integrating i got -10t+c

where i subbed in the initial conditions

where i got the velocity as -10t +15

and similarly doing the same but subbing initial conditions with t=0 and y=20 since it is starting 20m above origin

giving -5t^2+15t+20

[empty frigate]

...yep that sounds right

although the question they asked was how long does it take after rebounding from the wall, not from after it's thrown, so you'll need to adjust that time a bit

it hits the wall at t = 1.5 i think?, so just subtract that

[inner lodge]

would it be 2.5?

t=2.5?

since time of flight is 4 seconds

and it hits the wall at 1.5

[empty frigate]

yep
which also agrees with what i got by a slightly different method

i just thought "well, $\frac{1}{2}at^2 = \frac{125}{4}$ where $a = 10$" and solved that and you get $2.5$

[twin meteor]

bee

[inner lodge]

right

so with this time what should i do next?

or wait

is that the answer

[empty frigate]

well for (b) this is just the answer

[inner lodge]

it is right..

okayyy

well the next question should be fairly straight forward

[empty frigate]

yep
now that you know how long it took between hitting the wall and hitting the ground, and how fast it was moving during that time... :)

[inner lodge]

thank you so much

.close

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[plucky kernel]

$(1+\frac{1}{k(k-1)})^k>1+\frac{1}{k-1} \to \sqrt[k]{\frac{k}{k-1}}<1+\frac{1}{k-1}-\frac{1}{k}$

[twin meteor]

Quasar

[plucky kernel]

Is the first inequality somehow famous?

cz someone just used it to solve the below question.

[\sum_{k=2}^n\sqrt[k]{\frac{k}{k-1}}<n.]

[twin meteor]

Quasar

[slim prism]

i mean thats just comparison test no?

[plucky kernel]

$(1+\frac{1}{k(k-1)})^k>1+\frac{1}{k-1} $

[twin meteor]

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

[slim prism]

you kth root both sides to show this

and apply comparisson test on the series

or i guess the order perserving property of sums

[plucky kernel]

I meant this one.

[plucky kernel]

The problem is to prove this

The question is that and below is the solution lemme write wait

[slim prism]

i think they see that kth root and come up with a way to prove its always less than one

[plucky kernel]

$(1+\frac{1}{k(k-1)})^k>1+\frac{1}{k-1} \to \sqrt[k]{\frac{k}{k-1}}<1+\frac{1}{k-1}-\frac{1}{k}$

[twin meteor]

Quasar

[plucky kernel]

$\sum_{k=2}^n\sqrt[k]{\frac{k}{k-1}}<\sum_{k=2}^n (1+\frac{1}{k-1}-\frac{1}{k})=n-\frac{1}{n}$

[twin meteor]

Quasar

[slim prism]

yea

[plucky kernel]

oh how did you get idea of that?

looking at kth root right?

[slim prism]

yea

[plucky kernel]

cool.

[slim prism]

like its fairly standard i think

[plucky kernel]

Thanks. I was overthinking 😦

sorry bad question hope I didn't waste your time 🙂

[slim prism]

thats not always the worst thing

overthinking good

[plucky kernel]

Thanks .

how to close the channel?

[slim prism]

.close

[plucky kernel]

.close

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[hybrid marlin]

<@&286206848099549185>

ok, how do i cometo that tho

wait no i cant do that i think

[zealous bluff]

Ye you're right

[hybrid marlin]

or can I

[modest dirge]

cant you choose $\delta=\frac{2\varepsilon}{|x+c|}$

[twin meteor]

llspacebarll

[hybrid marlin]

I thought I cant just leave it like that

[hybrid marlin]

idk i think i can

[modest dirge]

then you jsut work backwards

and you haev the required implication

this just means that the function is not uniformly continuous

its just regular continuous

[hybrid marlin]

wait why is it 2epsilon

[modest dirge]

cause you multiply 2

[hybrid marlin]

isnt it suppoed to be epsilon/2(x+c)

[modest dirge]

you have a typo

[hybrid marlin]

oh

u right

[modest dirge]

i dont think that this works actually

i think delta cant depend on both c and x

just on c

[hybrid marlin]

ye thats what i was thinking

[modest dirge]

use the triangle inequality on |x+c|

[hybrid marlin]

idk what that is

[modest dirge]

|x+c|<|x|+|c|

very important tool for these kinds of proofs

[hybrid marlin]

what after here

@modest dirge

[zealous bluff]

choose delta=min{1 , 2epsilon/1+2|c|}

Cant choose delta depending on x

[hybrid marlin]

why

[zealous bluff]

You want 1/2 |x-c||x+c|<eps

|x-c|<delta
|x+c|=|x-c+2c|≤|x-c|+2|c|

Ande then you say |x-c|+2|c|<1+2|c|

[hybrid marlin]

why is |x-c| < 1 tho

[zealous bluff]

Because we chose delta=min{1,2eps/1+2|c|}

Means Delta ≤1 and delta ≤ that other thing

[hybrid marlin]

ye why did we choose this tho

from what conclusion

[zealous bluff]

Usually when we have a quadratic in delta like here we choose delta to be min{1,something} that something you figure it in the end

It can be really any number not 1 but it's easier if 1

https://youtu.be/GBuULJ_m-mM

Watch this

[hybrid marlin]

ok ty

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[sharp vault]

Construct an acute triangle ABC (AB < AC). Construct altitude AH and median line AM. Draw D and E
so that H is the midpoint of AD and M is the midpoint of AE. From M construct MN so that it is
perpendicular to DE (N belongs to DE). Prove: MN, BE, CD all intersect one point

[sharp vault]

I have an idea

Call the Intersection of BE nd CD O

And prove that angle BOM = EON

but i dont know how to approach it

or maybe prove thatt ON is perpendicular to DE

but

i still have no idea

Note: no pythagorean, circle, quadrilateral, trig theorems

pure triangle theorems are only allowed

and thats the difficult part rlly

[solemn storm]

trig is triangles what?

[sharp vault]

what?

[solemn storm]

trigonometry is the study of trigons, i.e. three sided shapes

[sharp vault]

what do you mean by this

Does this like relate>

?

[solemn storm]

I mean that "pure triangle theorems" excluding the pythagorean and trigonometric theorems makes no sense

[sharp vault]

i mean

i dont know english very well

pure theorems as in

congruency, altitude intersection, median line intersection, bisector intersection, etc

idk how to explain

[solemn storm]

it looks to me like the intersection point is a center of triangle DME.

[sharp vault]

which center

because i cannot spot orthocenter, center of gravity, etc

[empty frigate]

...no it isn't, it's the study of trigonon metron (τρίγωνον + μέτρον), "triangle measure"
which mostly refers to stuff like sin and cos, not just the study of triangles in general (i don't think there's a specific word for that)

[solemn storm]

dunno, that's what needs to be figured out

[sharp vault]

Maybe I can prove O is an orthocenter

oh wait i dont

[solemn storm]

the word for it was historically trigonometry. The reduction to the angle functions we're used to is a recent phenomenon, but is still based in triangles.

[sharp vault]

bc it requires ON to be perpendicular to DE or BC

[sharp vault]

is this... exactly related?

[solemn storm]

no, sorry.

[sharp vault]

its not your fault really

[solemn storm]

you may be able to do so. You know that triangle AHM is right, as is ADE. When in doubt, just construct as much new information as you can.

[sharp vault]

hm

I can prove BE is perp to MD and CD is perp to ME if that was the case

but it seems a bit unfeasable from what I have seen

[solemn storm]

you may want to use the fact that AD and AE are both bisected by BC

this implies that AHN~ADE

(you might need some more rigorous steps to get there, but that's a basic idea)

[sharp vault]

Wait

I have an idea

Maybe prove N is the midpoint of DE

=> N1 = N2 (the angles of N)

=> N1 = 90

by hopefully proving MDN is equal to MEN

by s-a-s

[solemn storm]

I think I have a direction as well, prove triangle MDE is isosceles, then, since M is the midpoint of BC, BE=DC by symmetry. Also by symmetry, they should meet on the altitude of MDE (which is MN). I don't remember the theorems that get you through this, exactly, since it's been a long time since I've done this, but it should work out

[sharp vault]

oooh

well

i think its like the same to MDN = MEN

just more convenient

[solemn storm]

it should be

[sharp vault]

wait a few mins

also i already proved AC // BE

[sharp vault]

And AB//CE

also BD = CE and CD = BE

these kind of rigorous prooofs are asians staple

[solemn storm]

they're common in US geometry classes as well

[sharp vault]

i mean

grade 7,8,9 is ALL proofs

we dont learn pythag until late grade 9

they gotta make life insanely hard for us

[solemn storm]

arguably that makes things better. Learning these proofs first makes trig easier once the pythagorean theorem is learned. You get special triangles naturally from these kinds of constructions

[sharp vault]

couldnt they just like move pythag down one class

it makes calculation problems easier

anyways

its going offtopic

[solemn storm]

just use these proofs to construct the pythagorean theorem, then you're allowed to reference it as a lemma

[sharp vault]

in vietnam there is no such thingss as a lemma (until g10)

we only are allowed a set of rules

you have this, now follow it to tooth and nail

[solemn storm]

you do have lemmas, you just don't call them that. You proved that BD=CE, that's a lemma (thing proven from the given rules that is used to prove other things).

[sharp vault]

hmmm
how do you even say it in english

like BD = CE is only specific to this problem

and proving general theorems dont get applied

[solemn storm]

yes, but it holds for all possible cases where this problem applies. Lemmas are usually more general, but they're just stepping stones

[sharp vault]

ah i understand

i was going to ask another question after this but this most likely answered it

so

heres my proof

we have AH = AD and BH perp AD => ABH = DBH

AC // BE => DBC = ACB; AB // CE => ABC = BCE

=> ABH = DBH = BCE

Evaluate Triangle BMD and MCE:

BM = MC (midpoint)

MBD (DBC) = MCE (BCE)

AD = CE (proven above)

=> BMD = MCE

=> MD = ME

=> MDE is isoceles => MDN = MEN (angle)

Evaluate triangle MOD and MOE

MD = ME

DMO = OME (corresponding angles of equal triangles)

MO: Shared length

=> MOD = MOE

=> MDO (angle) = MIO (angle)

Because MDN = MDO + ODN; MEN = MEO + OEN

=> ODN = OEN

=> OD = OE

And Triangle MDN = MEN (similar to MOD and MOE)

=> DN = NI

Because OD = OE and DN = NI =>

ON perpendicular to DE (theorem)

BC MN perpendicular to DE and ON perpendicular to DE

=> MON is on a straight line

Because O is the intersection of CD an BE

=> the three lines intersect at O

@solemn storm correct?

[solemn storm]

be careful, AH=/= AD, they are colinear, not equal.

[sharp vault]

oh

i meant AH = HD

[solemn storm]

something here isn't matching up in the first half of the proof. I'm not strong enough in this to exactly say what though

[sharp vault]

really

can you tell me like the zone of the first half

like is it the triangle proof

or the parallel proof

[solemn storm]

something about AC || BE doesn't seem obvious to me. It may be a typo

[sharp vault]

oh

I proved it in question a0

A)

this is D)

i think i said it to you earlier

[sharp vault]

.

[solemn storm]

ok, if you've proven it (therefore it's a lemma), then that seems good enough

[sharp vault]

ah ok

ill send this to the prof soon

the class hasnt finished this yet

.close

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[brittle zealot]

hi is it possible if I can ask math questions involving finances here

[brittle zealot]

I’m just going to send the question here anyway

If I’m finding the capital gain, it would be:

RM 900000 - RM 54000(10% down payment) - RM 15000 - RM 8000 - RM 18000 - RM 666000

But why can’t I include the original price in the deduction?

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[prisma viper]

Hi, can anyone help me with this question please

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[split wind]

do you recall what is three-figure bearing?

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@prisma viper Has your question been resolved?

[prisma viper]

yh, i do, but im confused about how it works with vectors

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[near hill]

since these digits are just 0-9

im not quite sure why we can be sure that y isn't equal to an already defined xn

[worthy citrus]

because, by construction, y has atleast one different digit to every xn

[near hill]

ohh lmao i see

thanks!

.close

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[vast shale]

why is d/dx wrote as a fraction is there a reason why its written in this specific way?

[worthy citrus]

in the most simple interpretation its meant to be the infinitesimal analogue of "change in function/change in x", usually written $\frac{\Delta y}{\Delta x}$

[twin meteor]

ΣAC

[worthy citrus]

its not really a fraction, but for nearly all cases you can treat it as such

and then eventually you discover that in fact it really is a fraction, of things called "differentials", that are a fairly advanced thing

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[slender quarry]

henlo,

[slender quarry]

$\nabla u \cdot \nabla v = \nabla v \cdot \nabla u$

[twin meteor]

fuli

[slender quarry]

this is true right?

[worthy citrus]

dot product is symmetric yes

[slender quarry]

ah yah omg the gradient symbol threw me off

thanks xD

.close

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[torpid sequoia]

I apologize

[hushed pewter]

@fresh wing what have you tried?

[lone niche]

the simplest answer is very simple don’t overthink just try some stuff

*some rationals

[hushed pewter]

1/3+1/4+1/(12/5)

[dark kiln]

nice lol

[lone niche]

ez

gg

[fresh wing]

It says to get the maximum amount though

I got 43 values so far

[hushed pewter]

where does it say that

[fresh wing]

Someone told me that

[hushed pewter]

[fresh wing]

No, seriously lol

[hushed pewter]

No one told us

[fresh wing]

Yeah

You're right

[peak matrix]

lol they didnt mention that x has to be integer

[lone niche]

dang
find how many sets of nonrepeating integers from 2 to 2023 have inverses that sum to 1

hard q

[hushed pewter]

What is this from?

[lone niche]

true but that doesn’t make it possible to cheese bc integers are just as easy

[hushed pewter]

It reads like a #competition-math problem, but it's pretty amateur hour to neglect mentioning x be an integer

[fresh wing]

Idk, one of my friends said it's an interesting problem so I thought I'd try it

[hushed pewter]

Try #competition-math math too if you haven't

[fresh wing]

I don't think it's a competition one

[hushed pewter]

Best I can think of is the one easiest solution, and then solving computationally

[hushed pewter]

The 2023 screams competition math

[fresh wing]

Yeah, we are allowed to use a computer and program

[peak matrix]

Thats quite important info

[fresh wing]

Well my friend told me he used one

[sturdy pumice]

probably an olympiad problem

[hushed pewter]

You need to tell us these things

[peak matrix]

brute force

[fresh wing]

Lol

[hushed pewter]

I can read minds, but you're out of range. So just tell us

[fresh wing]

I tried programming and I got 43 values

[peak matrix]

it should be around 2023*2023 loop so it should execute in no time. Brute force is best approach

[hushed pewter]

2023!

[fresh wing]

I'm not that good at programming

[hushed pewter]

You can have more than two fractions

[peak matrix]

oh dang, im idiot

[hushed pewter]

Brute force is checking all possible fraction combinations

This sounds like dynamic programming

I hate dynamic programming. They take forever to set up

[fresh wing]

The answer is a solution strong?

[peak matrix]

wait wouldnt it actually be 2^2023? For each number you either include or dont include it

[fresh wing]

String

[hushed pewter]

Ah I believe you are right

[peak matrix]

still too slow tho

[lone niche]

if you have 2 sets of coprime integers that have a sum of 1, where the greatest values have a product less than 2023, you can multiply them

bc 1*1 = 1

might help

[worthy citrus]

you can also prune in the sense that if you tested a set of integers that reciprocal summed to something bigger than one, you wouldnt test any sets that contain that set as a subset

sounds annoying though

[fresh wing]

What would that look like?

[hushed pewter]

painful

[worthy citrus]

yeah i dont know how to implement it, or if its even useful

[hushed pewter]

What's annoying is need to do fraction math in programming without floating point error

So you gotta make a fraction struct

And, there's a possibility to go above 64-bit, so you gotta add a BigInteger struct too

There's a lot of work in this problem

What's the time frame?

[fresh wing]

Sometime in may

Well there's no deadline

[worthy citrus]

alternatively if for example you're looking at 3 integers, $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1 \iff yz + xz + xy = xyz$

[twin meteor]

ΣAC

[worthy citrus]

so you could avoid floating point

[lone niche]

it’s definitely less than 1280 because the lowest 1280 values sum to >1
not that helpful

[fresh wing]

Won't we need a loop inside another for loop, an array to keep the correct rationals in, and some common sense?

[lone niche]

we will need some common sense

I also claim there are no multiples of primes greater than 404 because their inverses have to add up to something with a numerator not divisible by the prime and 1 + 1/2 + 1/3 + 1/4 + 1/5 (or some subsum) has denominators less than 404

not 100% sure on that

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@fresh wing Has your question been resolved?

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[vague oriole]

How do i solve these?

[ornate seal]

Sum of exterior angles?

[vague oriole]

if ur asking if thats the type of problem idk, i think?

the assignment is js labeled lesson 12 so 💀

[ornate seal]

Sum of exterior angles is always 360 degrees no natter the size of the polygon

[vague oriole]

so then i'd have to add up all the other numbers and the number thats missing is the answer?

[soft walrus]

the sum of all the exterior angles, i.e., the ones given, is 360º

[vague oriole]

ohh okay ty

wait so for the third one how do i solve it since theres 2 x's, do i have to give 2 answers or? 😭

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[karmic imp]

No

[tacit flame]

solve for the exterior angles as normal by using 360- all the angle values and then set that equal to the missing value, say for example if the missing angle is 40 set it equal to 2x and solve dy dividing both sides by the coefficient of x (in this case its 2).

[vague oriole]

oh got it, ty!

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[vast shale]

number 56. Im confused on how to write the summation because the deposit period isnt the same as the intrest period

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[vast shale]

What am I doing wrong again

[vast shale]

🌟

Ehat am