help-13

[proven fable]

so 11

is cd

[fallen moat]

nah, you just didn't remember

no dumb

[proven fable]

so c is (-2,0)

[fallen moat]

done

[proven fable]

now I need to calculate the area of apb

[fallen moat]

first find B

[proven fable]

ok i have two points now

so i can find it

[fallen moat]

why

cand or can?

oh

[proven fable]

I can find it

[fallen moat]

good

[proven fable]

for some reason I cannot come up with an answer

8-0/5+2

[fallen moat]

what have you tried?

[proven fable]

it comes 8/7 as m

[fallen moat]

i see

well, i think it's correct

[proven fable]

it comes out weird

can you help me with that? @fallen moat

[fallen moat]

sorry, was ordering food

now, we have 8/7, then we plug it into
(y-0)=8/7(x+2)

and find what's y when x=0

[proven fable]

I got it y = 8/7+b

ok

[fallen moat]

so, what's b?

[proven fable]

for me it's 0 = 8/7(-2)+b

let me multiply

0 =-16/7 +b

b= -16/7

[fallen moat]

nah

[proven fable]

so what to do/

?

[fallen moat]

do you know how this come from?

[proven fable]

yeah you put y1 in y and the incline which is 8/7(-2) that is x1

no?

[fallen moat]

so we have
y=8/7(x+2)

now since B has y-coordinates = 0

[proven fable]

how (x+2)?

isn't it just -2

?

[fallen moat]

we plug y=0

[fallen moat]

no

[proven fable]

yes

but (-2,0)

[fallen moat]

it's x-(-2)

[proven fable]

you plug 0 from here nort

oh ok

so how we continue

?

[fallen moat]

lemme think a bit

oh

i get it

next

[fallen moat]

we put x=0 into this

to find y

[proven fable]

wait what is the equation

[fallen moat]

and that's y-coordinates of B

[fallen moat]

y=8/7(x+2)

[proven fable]

ok

my bad

so 0 =8/7(x+2)

[fallen moat]

no

y=0

not y=0

it's x=0

[proven fable]

i put 0 in y

oh we need to find y

yeah so 0 in x

[fallen moat]

yes, if we put 0 in y, we can find C

if we put 0 in x, we find D

[proven fable]

so y=8/7(+2)+b?

[fallen moat]

no +b

why +b?

[proven fable]

ok so without

y = 16/7

[fallen moat]

yes

[proven fable]

so how do we continue?

[fallen moat]

Time to find A so that we have baseAB

[proven fable]

A is (0,18)

already calculated

but where does 16/7 plug

[fallen moat]

good

[proven fable]

to b?

correct?

[fallen moat]

AB is base

guess its length

[proven fable]

idk what is b

[fallen moat]

B is (0,16/7)

[proven fable]

My brain just stop wroking after the (-2,0(5,8) fuss

[proven fable]

yes that is what i though

t

so it's 18- 16/7

[fallen moat]

yes

[proven fable]

110/7

?

[fallen moat]

,calc 18*7-16

[wraith dagger]

Result:

110

[fallen moat]

yes

[proven fable]

ok and how can i find the height

[fallen moat]

it's given

P

[proven fable]

8?

[fallen moat]

nah

[proven fable]

5 is x

[fallen moat]

yes

[proven fable]

so 5

[fallen moat]

yes

[proven fable]

550/7

[fallen moat]

550/7 ?

yes

and then

1/2

[proven fable]

yes

so 39 2/7

[fallen moat]

,calc 39*7+2

[wraith dagger]

Result:

275

[fallen moat]

yes

[proven fable]

THANKS so much

.close

[cedar kiln]

Channel closed

Closed by @proven fable

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[marsh glade]

How do you solve this question?

[marsh glade]

I don't get the concept

[hot axle]

@marsh glade are you having trouble understanding the questions ?

[marsh glade]

Yeah

[hot axle]

Honestly I find them pretty unclear as well

[marsh glade]

Oh

[hot axle]

The questions are from 1 to 4

1. to 4.

[marsh glade]

Yeah

[hot axle]

the other numbers are indications ok

so

The concept is you're presented with "axioms" of geometry and the exercise wants you to put them in practice

[marsh glade]

Oh

So do I need to know the "axioms" to solve this question?

[hot axle]

The weird thing is it's not regular geometry, it's a "Fano Plane" so you have to use only the last 3 rules, not your own intuition

[hot axle]

Yeah

[marsh glade]

Yeah its confusing

[hot axle]

Don't worry it would confuse me as well

What level math are you doing ?

[marsh glade]

I'm in Geometry Honors

I'm a freshman

[hot axle]

Ok that explains it haha

[marsh glade]

Yeah haha

[hot axle]

It's very niche geometry topic

[marsh glade]

It is

If this question is too confusing can I ask about another question?

[hot axle]

the weird thing is I think l6 is a circle 😂

[hot axle]

Don't worry I can help

[marsh glade]

Oh ok thanks so much

[hot axle]

So if I understood correctly, you have 6 lines and seven points on the Fano Plane presented

The lines also include a circle which is confusing

[marsh glade]

Yeah

[hot axle]

Yeah

[marsh glade]

That should be right

[hot axle]

So what you need is to find out how the lines and points connect thanks to the drawing on the right

[marsh glade]

Oh

[hot axle]

Honestly this feels more like graphs than actual geometry, are you familiar with graphs a bit?

[marsh glade]

Kind of

Not too well though

[hot axle]

https://en.wikipedia.org/wiki/Graph_(discrete_mathematics)

it looks like this

[high coyote]

It's the typical exercise of projective geometry

[marsh glade]

Oh

Its just even if I try to use everything I learned it doesn't work

[hot axle]

Your Fano Plane is basically a graph where lines can have more than 2 points on them

[high coyote]

It's not graph theory

[marsh glade]

I can't seem to solve it

[marsh glade]

Oh

[hot axle]

Ok I haven't heard about it thanks!

[hot axle]

I mean this as a way to picture it, it's not really a graph ok

[marsh glade]

Oh

[high coyote]

@marsh glade what have you tried for 1.?

[marsh glade]

I got some answers

So for points 1 and 2 I did line 1 contains both, for points 2 and 7, I did line 5 contains both, and for points 5 and 6, I did line 2 contains both

I still don't get the violations of the first axiom part

[hot axle]

The first axiom being that only one line goes through 2 points, none of the examples you gave violate it right?

[marsh glade]

Yeah

It can't violate it

[hot axle]

Well that's the point

It means the drawing seems to not violate axiom 1

[marsh glade]

Because theres no multiple lines that goes through 2 points at least from what I see

[marsh glade]

Oh

[hot axle]

Right, exactly

[marsh glade]

But then the thing is question 2 starts to get me confused

[high coyote]

Why should you? The axiom 1 holds

[marsh glade]

I don't get question 2: Which point is the intersection of line 1 and line 2? 3 line 4 and line 2? line 3 and line 6? Are there any violations of the second axiom?

Because there is no point that intersects line 1 and 2

[marsh glade]

Oh

[marsh glade]

I don't think there is any points unless I'm wrong

[hot axle]

I would assume you get the intersection of l1 & l2 ; as well as l4 & l2. In practice they look like regular lines and the drawing shows they do in fact intersect right ?

[hot axle]

Oh you don't see a point where they cross ?

[high coyote]

Look at the picture and see which lines 1 belongs to.

[marsh glade]

Yeah

Ohhhh

I get it now

[hot axle]

Awesome

[marsh glade]

Now question 3 and 4

Is making my life difficult

Q3: Why is it that regardless of which four points you choose, one point will not be on the same line as the other three?

Wait

[hot axle]

The weirder one is l3 & l6 for question 2

[marsh glade]

Isn't the point where they intersect point 3

Not sure

[hot axle]

oh wait I thought l6 was the circle my bad

Yes you're right

[marsh glade]

Oh ok

Wait for question 3

[marsh glade]

would the answer be that 4 points are only coplanar but not colinear?

So since there is only 3 points for side of the triangle,

4 points cannot be all on the same line

[hot axle]

They are ALL "coplanar" it's a Fano Plane haha

[marsh glade]

Oh

But no four points would be collinear right?

[hot axle]

Yeah you're right that's the idea

[hot axle]

exactly

[marsh glade]

Oh ok

I get it now

But whats a projective plane?

[hot axle]

I have never seen projective planes in geometry I'll check

[marsh glade]

Oh ok

[hot axle]

Or maybe @high coyote can help

[marsh glade]

Oh ok

Thanks so much guys for all of your help

[hot axle]

Oh wait I'm an idiot

The definition of a projective plane is the 3 axioms we saw

I thought that was the definition of the Fano Plane

[hot axle]

You're welcome <3

[marsh glade]

Oh

[hot axle]

So what you really want is to verify all 3 axioms on the Fano Plane, that way you will know that it is in fact a projective plane

[marsh glade]

Oh

But then would question three verify the third axiom?

[hot axle]

Yes precisely

Fano Plane basically overkills the 3rd axiom by having all sets of 4 points non colinear

[high coyote]

(1) (2) (3) is a possible definition of projective plane

I'd say it's a projectivization of a third-dimensional afine space tho. 😋

[marsh glade]

Oh

[marsh glade]

Oh ok thanks a lot

[hot axle]

cool words :)

[marsh glade]

yeah

Oh ok I get it now

[hot axle]

Awesome !! You did it

[marsh glade]

Yeah!!!!

Now I get what this question is

Thanks a lot guys

[hot axle]

And about this - that's precisely the point of the exercise, so don't worry

[marsh glade]

Oh okay

[hot axle]

Next time try to treat something new like you don't know anything and try to piece it out from scratch, then you should find similarities to other things you've done

[marsh glade]

Oh ok

Thanks

Also I have one more math question I want to ask

[hot axle]

That way you're sure to not go in blindly thinking you know stuff you don't

[marsh glade]

Oh yeah

[marsh glade]

Should I ask it here?

[hot axle]

sure

[marsh glade]

Its this

All of the questions I do are complicated

The thing is I feel like I can use a protractor, and at the same time, I feel like I don't need it

[hot axle]

Yeah you're not supposed to

[marsh glade]

Oh ok

[hot axle]

All the info is in the runway numbers

[marsh glade]

Oh

I get question 3

[hot axle]

Yeah

What about 1 and 2 ?

[marsh glade]

I still don't get

[hot axle]

I think in this situation you need to get out a paper a pencil and draw out examples so you know you get it

[marsh glade]

Oh ok yeah I have a paper and pencil out

Wait for question 1, will the runway number be 21?

[hot axle]

first you need to draw North because everything is based off of that

[marsh glade]

Oh

[marsh glade]

Okay

[hot axle]

yeah I think you got the main idea

Try to draw 1, 36 and 27 for example

and 18

[marsh glade]

Oh ok

[hot axle]

I'll give you an answer on my magnicent paint drawing 🙃

[marsh glade]

Okay haha

Thanks

[hot axle]

Are you good ? should i send my png ? @marsh glade

[marsh glade]

Yeah it would be great if you could

[hot axle]

im so proud of my drawing lol

[marsh glade]

lol

Its very good

[hot axle]

Thanks

[marsh glade]

I get it now

but I'm now stuck on question 2 and I have to get the value for angle 3 and 4

I got 1 and 2

which are 30 degrees and 150 degrees

Not sure if I am correct

[hot axle]

What's the easiest way to calculate them with what you know already ?

[marsh glade]

You get the runway number and then you multiply it by 10 and then subtract it from 360?

Not sure

[hot axle]

Yep I agree with those answers as well

[marsh glade]

Thats how I got 1 and 2

But still don't get 3 and 4

[hot axle]

Yeah so what's so different about 3 and 4?

[marsh glade]

Wait

I still can't solve 3 and 4

[hot axle]

Ok

What is it you don't know that you need to know to answer? @marsh glade

[marsh glade]

I don't get how the runway number will work using the same method I did for the other runways

The numbers are bigger

so like runway 33

[hot axle]

yeah

Ok I see you're intuition, you're careful to not make a mistake by overgeneralizing

[marsh glade]

Yeah

[hot axle]

So here it's fine as long your angles don't go over 360 and reset to 0

[marsh glade]

Oh

[hot axle]

Because that's the only way your angle substractions will not work

[marsh glade]

Oh okay

Then for 3

like angle 3

will it be like 50?

50 degrees?

Because if the unknown runway number is 21,

[hot axle]

the runway numbers are 15, 21 and 33 right ?

[marsh glade]

Yeah

To solve angle 3 and 4

[hot axle]

So if you put it all clearly like that you're unlikely to mess up

[marsh glade]

Oh

[hot axle]

Was it really 50 ? between runways 15 and 21

[marsh glade]

I think I'm getting confused of finding the degrees

[marsh glade]

Wait no

[hot axle]

Yeah it seems that way

[marsh glade]

60 degrees for angle 3 and 120 degrees for angle 4?

Because if you subtract 330 from 210 you get 120, and if thats angle 4 then angle 3 is 60?

[hot axle]

(it seemed like you're measuring discrete intervals like between 1 and 3 strictly you have one integer but 3-1 is 2)

[marsh glade]

Oh

[hot axle]

Yup perfect

[marsh glade]

Ohh okay

[hot axle]

You need to crosscheck things when you get the chance, it lowers your amount of mistakes

[marsh glade]

Oh okay

Question 5 is just a lot of drawing: Airports are designed with more than one runway so that pilots landing planes will not have to deal with the difficult situation of landing in a strong crosswind. Design a runway system with the fewest number of runways so that a pilot will never have a crosswind angle of more than 30 degrees. Number each of the runways in your plan

I have to make my own runway system

[hot axle]

Yeah the hard part is to interpret the question

[marsh glade]

Yeah I agree

[hot axle]

"the fewest number of runways so that a pilot will never have a crosswind angle of more than 30 degrees"

[marsh glade]

Hmm

[hot axle]

what spacing does that suggest you try

[marsh glade]

30 degrees spacing or less?

[hot axle]

yeah

perfect

you can try and see if it works, and if you can prove it's the fewest amount of runways you won

[marsh glade]

Oh ok

Give me a minute

Let me try it out

[hot axle]

Honestly it's been so fun helping you out, I'm really happy to do this

[marsh glade]

Thanks so much

I learned so much from today

[hot axle]

That's awesome, my pleasure

[marsh glade]

Would this work?

[hot axle]

yes that's the idea

[marsh glade]

Oh okay

[hot axle]

but what about wind coming from the West?

[marsh glade]

Oh yeah

Then I would need to make one more?

[hot axle]

yep you need more runways

Do you see where this is going ?

[marsh glade]

Yeah

Would it be like this?

Or more like

this

[hot axle]

Prove to me that it works for wind coming from the west

ok

[marsh glade]

Oh wait there is still a area

[hot axle]

what about, wind at 61 degrees from North

[marsh glade]

that is larger than 30 degrees

[hot axle]

yeah exactly

[marsh glade]

Would this one work ?

[hot axle]

wait

I just realised something

[marsh glade]

yeah?

[hot axle]

this one actually works

try 61 degrees wind, which track would that be?

[marsh glade]

Hmm

[marsh glade]

like which runway?

[hot axle]

I mean which runway yes

Which runway has least wind

[marsh glade]

the horizontal runway

?

[hot axle]

ie smallest angle from wind

[marsh glade]

Or

[hot axle]

Yeah because 90-61 = 29

[marsh glade]

Yeah

Wait so it works?

[hot axle]

so you get wind at 29 degrees

[marsh glade]

Oh yeah

[hot axle]

Yeah ! Weird right?

[marsh glade]

Yeah!

Wait so it will still work if the angles are 30 degrees right?

[hot axle]

So maybe our first idea for a solution wasn't the minimum amount of runways

[marsh glade]

Oh

[hot axle]

Yeah of course, but is it minimal?

[marsh glade]

Maybe?

[hot axle]

Well

[hot axle]

this all 30 degree angles

[marsh glade]

Yeah

[hot axle]

This has some 30 and some 60 degrees

And both work right?

[marsh glade]

Oh yeah

Yeah they do work

[hot axle]

So which one is better, ie minimal amount of runways?

[marsh glade]

the minimal amount of runways?

[hot axle]

Yeah which has less runways

[marsh glade]

This one

[hot axle]

Yep

[marsh glade]

Ohhhhh

[hot axle]

So then the other can't be our final answer yeah

[marsh glade]

Yeah

[hot axle]

It's disproven to be our final answer

[marsh glade]

Yeah

Because while it works theres too many runways

[hot axle]

Exactly

[marsh glade]

Ohh okay

I get it now

[hot axle]

Any ideas to make our current best one even better? Is it really the best it can be?

[marsh glade]

I don't see any which can work to be even better

There might be a way

[hot axle]

All right

[marsh glade]

But you need to consider wind from all four directions

[hot axle]

That's good thinking

[marsh glade]

👍

[hot axle]

Actually, is there only just 4 directions of wind?

[marsh glade]

No theres a lot more

From like southeast, northwest, southwest, northeast

[hot axle]

Right

In this model we can just think of the wind as an angle

[marsh glade]

Yeah

[hot axle]

So any angle can be a wind angle

[marsh glade]

Yeah I agree

[hot axle]

right

Honestly it's the kind of question where if you can find the best answer immediately you can justify that it's the best one

And the hard part is finding it

[marsh glade]

Oh yeah

Because even with our current runway system I feel like there is a way to get even less runways

[hot axle]

It's kinda hard to brute force your way, but you can probably do it analytically

[marsh glade]

Yeah

Because what I drew is it all intersects at one point, but what if they intersect at different points?

[hot axle]

It's just going to be more complicated

[hot axle]

Nah we don't do that, remember runways are just angles and wind is just an angle so it doesn't matter

[marsh glade]

Oh yeah

[marsh glade]

Oh yeah

[hot axle]

if you take another runway that runs parallel it will still have the same angle with the wind

[marsh glade]

Oh yeah so that won't work

[hot axle]

Yeah

I can either give you a solution

or help you do it analytically (takes alot more time)

[marsh glade]

Uh

Which one will be better?

I feel like the analytical one is better

[hot axle]

Ok

Let me write out my model in LaTeX

[marsh glade]

Thanks

[hot axle]

So let n be the number of tracks, you have tracks $t_1, t_2, \cdots t_n$ integers between 1 and 36 that represent the track numbers (note : you can't have more than 36 tracks)

[wraith dagger]

Mathospondylus

[marsh glade]

Yeah

I think even if you just give me a solution I can try to figure out how you got to that solution

[hot axle]

The tracks are already in growing order so our problem translates to having the space between tracks ie $t_k -t_{k-1} \quad \forall k \in {2, \cdots n}$ to verify an equation

[wraith dagger]

Mathospondylus

[hot axle]

ok if you want

[marsh glade]

Sure but after you give me the solution can you just explain it to be simply?

[hot axle]

Ok

I think you don't really like the analytical approach haha

it is quite arduous

[marsh glade]

Oh yeah haha

I would take the analytical approach if I had more time

But I need to go shower and to go sleep soon

haha

Thanks

[marsh glade]

I agree

[hot axle]

Ok no problem

[marsh glade]

Oh

I see now

[hot axle]

actually it's algebraic my bad

[marsh glade]

Oh

[hot axle]

because it uses letters haha

Nice !

[marsh glade]

Oh yeah haha

I get it now

Thanks so much!

I learned so much today

[hot axle]

Sorry I gave you a red herring, I would have made the same mistakes on my own haha

starting from 30 degrees and realising it's not actually minimal

[marsh glade]

Oh no its fine haha

[marsh glade]

Oh yeah

[hot axle]

Perfect! I'm glad I could help

[marsh glade]

I think thats all the questions today!

Before we go can I just send a friend request

[hot axle]

Nice! Don't hesitate to reach out to me if you want

Yes

Awesome

.close

[marsh glade]

Thanks!

.close

[cedar kiln]

Channel closed

Closed by @marsh glade

Use .reopen if this was a mistake.

[cedar kiln]

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.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[simple bane]

how to simplify

[violet brook]

theres a few ways u can simplify

[simple bane]

$\sqrt3^4$

[wraith dagger]

张元英

[simple bane]

and then i cancel the threes?

[violet brook]

the other way i can think of

is

u can split 8^(4/3) into 8^(3/3) * 8^(1/3)

which becomes 8 * 8^(1/3)

8^1/3 is the same as cuberoot of 8 which is 2

8*2 = 16

[simple bane]

$8^\frac{4}{3}=8^\frac{3}{3}*8^\frac{1}{3}$

[violet brook]

i think u meant to write 8^(3/3) lmao

[wraith dagger]

张元英

[violet brook]

there we go

yep

u get why that works right

[simple bane]

uhm, $\frac{3}{3}*\frac{1}{3}\neq\frac{4}{3}$

[wraith dagger]

张元英

[simple bane]

its more like

$\frac{3}{3}+\frac{1}{3}=\frac{4}{3}$

[wraith dagger]

张元英

[violet brook]

$a^m * a^n = a^{m+n}$

oops

[simple bane]

oh

[wraith dagger]

taro

[violet brook]

yk this index law right

when the bases r the same

you add the powers

[violet brook]

anyway using this method ur thing would be

3sqrt(8^4)

[simple bane]

i kinda forgot i got the unstable foundation

[violet brook]

think of it this way

[simple bane]

does this work?

[violet brook]

$2^3 * 2^4 = (2 * 2 * 2) * (2 * 2 * 2 * 2) = 2^7$

[simple bane]

ah

[wraith dagger]

taro

[violet brook]

thats why we add the powers

[violet brook]

idk why u cubed the 2 tho

its sll good exxept for the 2^3

wait no

ur right

wait

now im confusing myself

[simple bane]

so what is $$\frac{8^{\frac{4}{3}}{8^{\frac{3}{3}}$$

[violet brook]

oh right ur right

u just changed 8 to 2^3

yeye that works

👍🏻

so u can turn it into a root

or play around w index laws

until u get smth easy to work with

up to you

[wraith dagger]

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

[violet brook]

hmm?

[simple bane]

so what is (8^(4/3))/(8^(3/3))

[violet brook]

$\frac{8^{4/3}}{8^{3/3}}$

this?

[wraith dagger]

taro

[violet brook]

fgs good enough

[simple bane]

yes

[violet brook]

when ur dividing powers

with the same base

u subtract the powers

[simple bane]

o

[violet brook]

so it becomes

$8^{4/3 - 3/3}$

[wraith dagger]

taro

[violet brook]

= 8^1/3

which equals 2

as cuberoot 8 is 2

[violet brook]

have u learnt these index laws before?

[simple bane]

$2^3^2$?

[wraith dagger]

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

[violet brook]

${2^3}^2$

[wraith dagger]

taro

[violet brook]

this?

[simple bane]

yesyes

[violet brook]

so w this

(a^m)^n = a^mn

[simple bane]

2^6

[surreal cave]

Is it $2^{(3^2)}$ or $(2^3)^{2}$? Parentheses matter when doing these problems

[violet brook]

$({2^3})^2 = 2^3 * 2^3 = 2^6$

[wraith dagger]

taro

[simple bane]

2^5

[violet brook]

im hoping he means the second one

[wraith dagger]

MrFancy

[simple bane]

whats the diffrence

[violet brook]

if theres brackets

[simple bane]

2^9

[surreal cave]

parentheses first :)

[violet brook]

yep

[simple bane]

o

oh ok

thanks

for the help

[violet brook]

npnp

[simple bane]

great help

[violet brook]

those r called the index laws if u need to learn them all

[simple bane]

o

.close

[cedar kiln]

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[cedar kiln]

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[peak bough]

is the inner function here (x^2+1) or sqrt(x^2+1)?

[mighty shuttle]

Wdym by inner function? If you mean under the root it's $x^2 +1$

[wraith dagger]

physicsrocks

[peak bough]

is that what I should substitute the u?

[mighty shuttle]

What's the question? Is this an integral?

[peak bough]

I have to solve for the derivative using the chain rule i think

.close

[cedar kiln]

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[hot sigil]

Hello! I have a question in discrete calculus. For a finite set of natural numbers S, where |S| is the number of elements in S and σ(S) is the sum of all elements in S, find

[hot sigil]

Where T={1,2,…,10}

[wintry pier]

!status

[cedar kiln]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[hot sigil]

2

[wintry pier]

Then !show

!show

[cedar kiln]

Show your work, and if possible, explain where you are stuck.

[hot sigil]

I started finding the possible partitions of the sums and find the ones where it would cancel out to eventually find a pattern

For example {1,2,3} and {1,5} would cancel out in that expression

Since they have the same sum but the parity of the number of elements is different

I looked for the sums of the first 12 numbers but it’s very random

[vagrant elbow]

what the hell is discrete calculus

Sounds oxymoronic

[hot sigil]

The ones where the sums don’t cancel out are 2,5,7,11, and 12

[hot sigil]

It’s calculus for integers

I know it doesn’t add much lol

[wintry pier]

Where did you find this exercise ?

[hot sigil]

It’s on an online math platform

It’s in french

[wintry pier]

I think I know exactly which exercise it is

In your online platform

[hot sigil]

Did you solve it?

Btw I’m not asking for a solution I’m just trying to find a hint to move forward

[wintry pier]

Yes

Try what doing what the clue ask you

[hot sigil]

Yeah I tried to reason using the clue but should I do it on sigma(S)?

[wintry pier]

What did you find with the clue

[hot sigil]

Well there would be 10 derivatives and since the maximum degree is also 10 it means that I should only reason on x^10

All the other ones with a degree less than 10 would just disappear

[wintry pier]

Aren’t you copying the forum ?

[hot sigil]

I checked the forum later on but I kinda got that on my own

I already have a good background in calculus

I know how derivatives work

I’m just really new to discrete calculus

[wintry pier]

You find the degree of the polynomial try finding the dominant coefficient

Think of Ponchammer polynomial

[hot sigil]

Alright one second

I think I have something

[wintry pier]

So ?

Did you find anything

[cedar kiln]

@hot sigil Has your question been resolved?

[cedar kiln]

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[crimson sedge]

Triangle ABC is an isosceles triangle in which AB = AC. Side Ba is produced to D such that AD = AB. Show that angle BCD is a right angle.

[cedar kiln]

Please don't occupy multiple help channels.

[deep oriole]

please delete either this channel or the other channel you are occupying

[crimson sedge]

<@&286206848099549185>

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[crimson sedge]

Bruh

[nimble veldt]

name the angle at B as alpha. write the angle BAC in terms of alpha, writhe angle CAD in terms of alpha, write angle ACD in terms of alpha. then it is obvious.

[crimson sedge]

What is term of alpha

[silent finch]

note A is circumcenter of ABC and midpoint is only circumcenter when triangle is right by Thales

[nimble veldt]

if angle aBC = alpha what is angle BAC?

[crimson sedge]

Bruh this is the Triangle lesson question

[nimble veldt]

and what do you want to tell us with this?

[crimson sedge]

That I don't know who is alpha ; - ;

[nimble veldt]

name it x, or s or whatever you want.

[crimson sedge]

Ok then

What to name x

[nimble veldt]

the angle at B

[crimson sedge]

I don't understand

Anything

[nimble veldt]

ABC is isoceles, right?

[crimson sedge]

Yes

[nimble veldt]

what does this mean?

[crimson sedge]

Abc and acb are equal

[nimble veldt]

ok. and what else?

[crimson sedge]

Don't know

[wintry pier]

What about triangle ADC

[crimson sedge]

Acd and adc are equal

[nimble veldt]

isoceles means: two sides are equal, and therefore the opposite angles are equal. and its a triangle. su the sum of all angles is ...?

[wintry pier]

What else ?

[crimson sedge]

180

[nimble veldt]

so the angle at BAC = ....?

[crimson sedge]

What?

I think it will be 90 but don't know how

[nimble veldt]

angle ABC = angle BCA (as isoceles) and angle ABC + angle BCA + angle BAC = 180, so you get angle BCA = 180 - 2 x angle ABC

[wintry pier]

I think you also have angle bac=angle dac ?

[crimson sedge]

No

Not given

[crimson sedge]

Wat

[nimble veldt]

what do you not understand?

[crimson sedge]

But it is coming bac is equal to 180 - 2x

[nimble veldt]

thats what i said.

[crimson sedge]

Oh

Ok

So wat now?

[nimble veldt]

now, what does this mean for angle CAD?

[crimson sedge]

Angle cad is equal to 180 - 2y

[nimble veldt]

what the fuck is y?

[crimson sedge]

Variable

[nimble veldt]

forget it. BAC = 180-2x, BD is a straight line, so CAD = 180-(180-2x) = 2x

[crimson sedge]

Ok ok got

Wait

I think I can solve it now

Gimme 5 min

@nimble veldt hey

I cant

Find it

[nimble veldt]

you have one angle in CAD (2x), triangle CAD is isoceles, so two angles (the 2 unknows) are equal, and the sum is 180. so the angle ACD is ,,,,

[crimson sedge]

X?

[nimble veldt]

CAD + ACD + CDA = 180, now use CAD = 2x and ACD = CDA, so you get 2x+2ACD=180 -> ACD = ...

[crimson sedge]

90 - 2x

[nimble veldt]

no

[crimson sedge]

45 - x

[nimble veldt]

no, 2x+2ACD = 180 -> 2ACD = 180 - 2x -> ACD = 90-x

[crimson sedge]

Oh shit I got it

Ohhhhhh broooo

[nimble veldt]

and for the final step. sum up ACD and ACB.

[crimson sedge]

Thats what

Thanks mannn

Damn

.close

[cedar kiln]

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[mint viper]

My lecturer asks us to "Discuss" theorems such as the Mean value theorem. I am not sure what this means, but he says he does not want us to define it or calculate the MVT for a function. What must we do when he asks us to "Discuss"

[nimble veldt]

think about what it means. try to find interpretations. try to understand for what each of the prerequisites is necessary.

[tropic oxide]

so you asked him what he wants you to do, but he ONLY told you what NOT to do?

[mint viper]

Exactly

[tropic oxide]

sounds like a purposefully unhelpful teacher who wants you to fail.

[mint viper]

Thank you this helps, but I feel like he still might want more than this.

[nimble veldt]

he wants you to learn and understand the mean value theorem. do you have qeustions? ask them in a group with your colleagues. and discuss them.

there are no qeustions for you? try to explan the questions from your colleagues.

[mint viper]

Alright thanks

.close

[cedar kiln]

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[tacit mulch]

How to solve the equation set 2x + 3y = 2, -3x+2y=10?

[tacit mulch]

do I begin like 5(2x+3y)=-3x+2y or?

because that becomes "13y=-13x", thusly y=-x

well i found the solution, x=-2 and y=2

😅

.close

[cedar kiln]

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[misty notch]

am i correct?

[misty notch]

ping me when you reply plz

[umbral dew]

csc is cosec?

[umbral dew]

.

[cobalt granite]

yes

[umbral dew]

yes, youre correct

[misty notch]

yes

[misty notch]

alr ty

[cedar kiln]

@misty notch Has your question been resolved?

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[crimson sedge]

Why is (b) is not a partial order and (c) is not a total order ?

[crimson sedge]

I found that b is reflexive, transitive and anti-symmetric thus it shall be a partial order ?

[radiant mulch]

help i need to submit tomorrow

[crimson sedge]

There are free channels

why are you writing in an occupied channel ?

[cedar kiln]

Please read #❓how-to-get-help

[south tundra]

Are you sure about the antisymmetry of b?

|x| = |y| doesn't mean x = y

[obsidian coral]

Because people don't read server instructions

[tropic oxide]

some people read the server instructions and actively choose not to follow htem

[crimson sedge]

I need some time to process your answer

[south tundra]

Could you show your reasoning behind why the relation in b is anti-symmetric?

[crimson sedge]

Well if for instance |1| =< |2| then not |2| =< |1|

The only case when that happens is when a = b

For instance |1| = |1|

[south tundra]

How about the fact that |-1| <= |1| and |1| <= |-1|?

[crimson sedge]

Oh you're right

I didn't see that

[south tundra]

Yeah so in b if you have xRy and yRx then you have |x| <= |y| and |y| <= |x|

Meaning |x| = |y|

But that is as far as you can go

|x| = |y| means x = y or x = -y

[crimson sedge]

Yes I see

For c I see that it is transitive, reflexive and for symmetry...

Well it looks anti symmetric

[south tundra]

For anti-symmetry assume xRy and yRx meaning (|x| < |y| or x = y) and (|y| < |x| or x = y)

[crimson sedge]

And I am trying to see if it's a total order

[south tundra]

Now, we can't have |x| < |y| because it would contradict |y| < |x| and x = y at the same time

Meaning x = y is forced

[south tundra]

Generally try to come up with a rigorous proof

[crimson sedge]

Ok I am gonna do that next time

[south tundra]

Hm, the answer sheet says that the relation in c is not a total order?

[crimson sedge]

Yes which is strange

[south tundra]

Ah I see

[crimson sedge]

I found

[south tundra]

You can't compare the nonzero numbers with opposite signs

E.g. you can't verify whether (-1)R(1) or (1)R(-1)

So it's not total

[crimson sedge]

Yes

Thank you for your help !

.close

[cedar kiln]

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[crimson sedge]

I don't understand how to progress at all

[pliant wedge]

We want to take notice to "If the least amount of material is to be used", which means we will be needing the surface areas of a cylinder as well

[crimson sedge]

OH

lemme try

@pliant wedge I did this.. abomination

this was my idea initially

i got it through getting the area of each can

adding it all up together

[pliant wedge]

right

[crimson sedge]

then

substituted

h1 and h2 in respect to r

with these

I got those off of the volume formula

[pliant wedge]

I'm not quite sure how to get to here now

But I think it might have something to do with min values of the function

[crimson sedge]

i mean

im guessing i derive

then i get critical values

see local minimum

[pliant wedge]

Yeah that makes sense

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[misty venture]

find a linear transformation T:R^3 -> R^3 that maintains
I don't know how to solve this please help

[misty venture]

ty
btw where did u get this from

.close

[cedar kiln]

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[crimson sedge]

Pick a number between 1 to 1000 at random. Find the probability
that is neither divisible by 12 nor 15.
How i solve this? or learn to solve this

[prisma gull]

So we have to first find which all are divisible by 12, 15, 12 and 15 then substract from the total

does that sound good?

[toxic bison]

Every 4th and 5th multiple of 3 is a number divisible by 12 or 15

You have to find out how many such multiples there are, and then subtract that value from 1000. Dividing that value by 1000 would give you your answer

[crimson sedge]

@toxic bison I found this youtube video, let me try that, send you the pic of the solution.

[toxic bison]

Sure

[prisma gull]

No of multiples of 15= $\lfoor{1000/15}$

[crimson sedge]

oh okaqy

[toxic bison]

Yes, but that will not work. This is because there will be common multiples of 12 and 15

For example 60, 120, etc

[prisma gull]

I didn’t complete

[toxic bison]

My bad

[prisma gull]

You have find multiple of 15

similarly for 12

and substract their common multiples

To find common multiples= floor(1000/lcm(15,12))

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

@prisma gull @toxic bison

[toxic bison]

The LCM of 12 and 15 would be 60

[crimson sedge]

so?

[toxic bison]

Wait nvm

My bad

[crimson sedge]

this driving me nuts, i am gonna go outside after this and get some food.

[toxic bison]

I think I can help

See

[crimson sedge]

have i solved it correctly?

[toxic bison]

Just understand the logic first please

[crimson sedge]

ok

[toxic bison]

I do not think so

So every 12th number is a multiple of 12
Every 15th number is a multiple of 15
Every 60th number has both factors

Makes sense?

[crimson sedge]

yes

[toxic bison]

Number of 12 multiples: 1000/12 (Only the quotient)

Same applies to 15 and 60

[crimson sedge]

60?

[toxic bison]

Number of 12 multiples: 83
Number of 15 multiples: 66
Number of 60 multiples: 16

[toxic bison]

It is the LCM of 12 and 15

[crimson sedge]

ok

got it

[toxic bison]

So total number would actually be 83+66-32

We are subtracting 16 because when we add the multiples of 12 and 15, we are adding 60 twice

[crimson sedge]

arent we going to multiply 83 by 12?

[crimson sedge]

got it

[toxic bison]

I don't see what that will get you

Can you share the question again please

[toxic bison]

This is your solution right. I need the question

[crimson sedge]

Pick a number between 1 to 1000 at random. Find the probability
that is neither divisible by 12 nor 15.

[toxic bison]

Yes

[toxic bison]

So 117 of them are not divisible

Which means the answer is 117/1000 or 0.117

Did you understand the solution?

[crimson sedge]

So, we are going to multiply the 12 and 15 by 83 and 66 respectively. Then subtract 60x16=960 which is the lcm of (12 and 15)

[toxic bison]

Why do you want to multiply them lol

See, we just need the number of factors right

We don't need to multiply for that

[crimson sedge]

yes

so 12 and 15 minus 60?

[toxic bison]

Every 12th number is a multiple of 12

How many such "12th numbers" are there in a 1000? Well, it is 83

[crimson sedge]

yes

[toxic bison]

Number of factors of 12 + Number of factors of 15 - 2(number of factors of 60)

[toxic bison]

The same you can do for 15 and 60

That gets you the number of multiples

[crimson sedge]

83 + 66 = 154

[toxic bison]

Then you can use this formula

Yes

[crimson sedge]

154 - ?

[toxic bison]

2*16

[crimson sedge]

I am sorry, i am just dumb.

Why are we multiplying 2 by six?

[toxic bison]

We are multiplying 2 and 16

See

It depends on the question

Every multiple of 60 is a multiple of 12 and 15

Right?

[crimson sedge]

yes

[toxic bison]

You need to make sure that the number is neither divisible by 12 or 15

Because if it is divisible by 12 and not 15, it does not fulfill the condition of the question

Because it states "neither"

Makes sense?

[crimson sedge]

yes

[toxic bison]

The only time a number is divisible by 12 and 15 is when it is divisible by 60

True?

[crimson sedge]

okay

yes

[toxic bison]

So you have to remove all 60 multiples

[crimson sedge]

and how we do that?

60 and 16 multiple

[toxic bison]

When you add the multiples of 12 and 15, multiples of 60 come with both of them

[toxic bison]

Leave 16 for now, I'll get to that part

[crimson sedge]

yes

okay

[crimson sedge]

yes

[toxic bison]

But you do not want mutliples of 60, so you have to subtract them twice

[crimson sedge]

yes

[toxic bison]

Coz one for 12 and one for 15, because they both bring the 60 multiples with them

[crimson sedge]

yes

[toxic bison]

Now how many multiples of 60 are between 1 and 999?

[crimson sedge]

16

[toxic bison]

Well, the answer is 16

Yes

So you multiply 16 and 2, and then you subtract that value from the number of multiples of 12 and 15

Makes sense?

[crimson sedge]

yes, it does 🙂

[toxic bison]

Yayy

You can close this now, if your question has been answered

[crimson sedge]

83 + 66 - 32 = 117 is the answer?

[toxic bison]

Yes

But you need to find the probability right

So you need to divide that by 1000, because it is the total possible outcomes

[crimson sedge]

yes

[toxic bison]

Wait wait wait

One more thing

You need to subtract 117 from 1000

[crimson sedge]

why

[toxic bison]

117 is the number of values that are divisible by both 12 and 15

We want the rest of the values

So we subtract it from the total

Get it?

[crimson sedge]

117 divided by 883 is my answer

[toxic bison]

No no no

See

1000-117/1000

See the number of values we want is 1000-177

[crimson sedge]

number of favourable outcomes

i get it

[toxic bison]

Total values is 1000

Yeah

[crimson sedge]

yes

[toxic bison]

That should be your answer

[crimson sedge]

thank you kush

thank you so much 🙂

[toxic bison]

Anytime!

[crimson sedge]

okay 🙂

.close

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[swift fossil]

I am having trouble with questions C F and G