help-10

[copper latch]

Yes but you see thats not correct

So that’s incorrect statement

Hence x is not = 2

[dense pier]

but if i take 0 then -2sin(0)=0 and it is correct

[copper latch]

Yes

[timid silo]

A more systematic way to do this would be by saying -2sin(x) is strictly decreasing over the domain while x is strictly increasing

So only one value for x satisfies the equation that's the intercept

[copper latch]

So now find if there’s any other solution for x other than 0

[timid silo]

=>x=0

[copper latch]

Is it strictly decreasing??

[timid silo]

Calculate the derivative

Wait it isn't

Lol

[copper latch]

What about x between (π/2,π)

[timid silo]

I am dumb

I mean between (0, π/2)

[copper latch]

Yes

The argument should be
-2sinx < = 0 for x between [0,π]

[timid silo]

x on that domain is greater than π/2 and -2sin(x) can't be greater than π/2 on that domain namely (π/2, π)

For all real numbers for x atleast

[copper latch]

So only one value for x satisfies the equation that's the intersection

[timid silo]

ya so only value for x is 0

[copper latch]

Lol but he needs to say this not us

[timid silo]

in real numbers at least

[copper latch]

He needs to get

[timid silo]

Ya

[copper latch]

@dense pier ig you get it

Right?

[dense pier]

im just calculating lol

well yes but

[timid silo]

It's obvious for the most part

[dense pier]

how do i fit in this answer

[timid silo]

I already said it all

Lol

[copper latch]

Oh sure verify the statement written above

[dense pier]

[-2;0]

[copper latch]

But -2 isn’t in the domain

[timid silo]

He meant output domain

Probably

[copper latch]

Huh what?

[dense pier]

i mean
the correct answer is C - [-2;0]
how could it be correct if -2 isnt in the domain

[timid silo]

I think we wants to say 0>f(x)>-2 for values of x in the domain from 0 to π

[copper latch]

No its wrong

Care to translate first two options?

[dense pier]

a) Empty set
b) none of the options is correct

[copper latch]

Option b seems correct then

[timid silo]

Ya but does @dense pier get why

[copper latch]

Idk @timid silo can you take over i gotta go

[dense pier]

i do because there's nothing in common between A&B except for 0

[dense pier]

thanks for your time and have a good day🙏

[copper latch]

So doesn’t option b seem correct to you?

[copper latch]

Np

[dense pier]

well it does, but im really confused due to option C being marked as the correct one

[copper latch]

Do you think 2sin(2-π)=2?

[dense pier]

no i definitely dont

[copper latch]

,w 2sin(2-pi)

[warm shale]

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[copper latch]

…

[obtuse pebble]

@dense pier Has your question been resolved?

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[marsh crow]

How much is x?

is it 4,20m?

[lucid flame]

we need more context

[timid silo]

5

[marsh crow]

btw its about enlarge

[timid silo]

if same shape scaled

[marsh crow]

enlargeing

[naive owl]

are they proprtional

[marsh crow]

what does that mean?

[naive owl]

ok so
100cm/x = 80cm/4

[timid silo]

x/1m = 4m/0.8m

[marsh crow]

20

0.20?

m

[lucid flame]

if they're the same thing but scaled up then the ratios of the sides are the same

[timid silo]

5m

[marsh crow]

Btw i am swedish

[naive owl]

,w solve 100cm/x = 80cm/4

[marsh crow]

so i dont know what they are called in english

[naive owl]

ye

u got the concept

[marsh crow]

So the answer is 20 cm

[naive owl]

just cross multiply lol

[marsh crow]

I mean m

[naive owl]

$$100cm/x = 80cm/4$$

[warm shale]

Chunkin

[timid silo]

,w solve 100cm/x = 80cm/4cm

[warm shale]

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[naive owl]

5

[marsh crow]

But they didnt learn us in school

Bruh

[naive owl]

bro

[marsh crow]

School doesnt learn anything

[timid silo]

this looked so misleading

[naive owl]

u should know how to solve fractions

[marsh crow]

It does

[marsh crow]

Tell that to the book

But i dont think the book can speak

anyways

thank you guys

i thougt it was 4,20m

but why is x 0.20

Because on the left one

its 1m

ANd right one

20 cm less

So

why is the right side more than the left side?

On the second one?

[marsh crow]

Thats not fractions

Its enlrage

[marsh crow]

wait

then

how do you solve this

[timid silo]

12/8 = x/10

,w x = 12*10/8

[warm shale]

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[marsh crow]

Then isnt this 100/4= 80/4=

Why are they differnet

[timid silo]

huh?

[marsh crow]

Like

you solve both probelms in a diff way

[timid silo]

well u can just shift around the terms

they are all the same

[marsh crow]

look

[marsh crow]

look

diff way

why?

[timid yacht]

look

[timid yacht]

same way

[marsh crow]

yeah

but he didnt multiply

like he did in the other one

why?

[timid yacht]

multiply what

[marsh crow]

WAIT

BRUH

MY MIND

ITS DEAD

I UNDERSTAND NOWW

Im so dumb

Thank you guys so much

.close

[obtuse pebble]

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[lucid flame]

plug in the gradient and (1,5) then solve for c

[timid yacht]

draw a graph

no, its not that kind of channel

[lucid flame]

it's a help channel

[timid yacht]

XD

maybe elsewhere

no

its not useful for your understanding

read #❓how-to-get-help

<@&268886789983436800>

[obtuse musk]

.close

[obtuse pebble]

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[obtuse musk]

thanks

[hardy merlin]

Lol

[obtuse pebble]

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[grim bison]

Hello, does someone know if graph k5 and k3,3 are homeomoprhic?

[grim bison]

can't finda nything on the internet

[obtuse musk]

homeomorphic?

[grim bison]

Homomorphism of Graphs: A graph Homomorphism is a mapping between two graphs that respects their structure, i.e., maps adjacent vertices of one graph to the adjacent vertices in the other. A homomorphism from graph G to graph H is a map from VG to VH which takes edges to edges

[obtuse musk]

thats not a homeomorphism though

a homeomorphism is between topological spaces

anways, they have different number of vertices

[grim bison]

graph can be homeomorphic too tho

[obtuse musk]

what does this mean

[grim bison]

<@&286206848099549185>

[obtuse musk]

😵‍💫

please dont ping helpers before 15 mins

also i asked you to elaborate and you just ignore me?

[grim bison]

srry

what do you want me to elaborate

[obtuse musk]

what homeomorphic graphs means

you just posted that definition of homomorphism, which is a) different from a homeomorphism and b) doesnt answer the question

[grim bison]

ill try my best, english is not my main language

imagine you have a triangle with 3 vertex and 3 edges, you can add a new vertex to the graph and make it a square, which means that a square graph is homeomorphic to the triangle

[obtuse musk]

subdivision?

well

[grim bison]

yes

is it not called homeomorph?

[obtuse musk]

i looked it up, it is

its a bit more complex than just homomorphisms

[grim bison]

i see

[obtuse musk]

you need an isomorphism between subdivisions

anyways, they shouldnt be

[grim bison]

so they are not homeomorphic?

[obtuse musk]

k_5 has vertices of degree 5

and subdivisions only add vertices of degree 2

k_3,3 has no vertices of degree 5

[grim bison]

ooooh

ty so much didn't think about that

sorry for ghosting you there i was thinking😅

[obtuse musk]

np

so the question is answered?

[grim bison]

yes

[obtuse musk]

.close

[obtuse pebble]

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[slow wadi]

I have no idea where to start

[obtuse pebble]

@slow wadi Has your question been resolved?

[slow wadi]

<@&286206848099549185>

[obtuse pebble]

@slow wadi Has your question been resolved?

[timid silo]

Can you tell the answer for part (i)

[slow wadi]

i dont know it

[timid silo]

Ok i found an answer but I'll have to check

[slow wadi]

could you walk me through it please

[timid silo]

it might be wrong tho

[slow wadi]

oh ok

maybe theres something that can get me started?

[timid silo]

Ok here's what I think we should calculate the probability of turning on the correct switch in hallway and in the kitchen seperately

If a guest changes the state of the switch and we have to calculate the probability for the light turned on then all the switches in the hallway must be turned off

[slow wadi]

not necessarily

the correct switch in the hallway could be on but the one in the kitchen has to be on aswell

[timid silo]

yes but it's necessary that a guest changes the state of a switch in the hallway

[slow wadi]

yes but what im saying is that the guest can choose the wrong switch in the hallway (and turn it off) whilst also turning the correct switch on in the kitchen

[timid silo]

they have to be turned off from before
We are calculating the favorable outcomes

[timid silo]

But that won't turn on the light in swimming pool

[slow wadi]

why not?

the correct switch is on in both the hallway and the kitchen

[timid silo]

Oh

Right

You are right

[slow wadi]

yeah but im not sure what to do now

[timid silo]

Two more cases then

All switch turned on in the hallway and then all turned off

[slow wadi]

ok

[timid silo]

If all are turned on then he has to switch off the wrong one

[slow wadi]

yes

which is 3/4

[timid silo]

3p/4

[slow wadi]

oh yes

[timid silo]

then the kitchen

[slow wadi]

would it be 1/8?

wait no

[timid silo]

Wait there are two more casew

If he turns on the correct one or if the correct one's turned on and he switches off the wrong one

Like you said

[slow wadi]

yeah

if the correct one is turned off, he has a 1/4 chance of picking it

correct?

[timid silo]

Yes and 1/2 it's turned off

so 1/8

[slow wadi]

surely its 3/4

since it doesnt matter if the other switches are off

as long as he doesnt pick the correct one if it is on

[timid silo]

I was talking about first case only

[slow wadi]

oh my bad

[timid silo]

The first case the correct switch turned off that's 1/2 he picks it 1/4 so 1/8

[slow wadi]

but yes

you are right

yes

[timid silo]

Second case is the correct switch is turned on so it's 1/2 and he doesn't pick that so 3/4 that's 3/8

Right ?

[slow wadi]

yes

[timid silo]

So total 1/2

[slow wadi]

so its 4/8 he picks correct switch in the kitchen

yes

[timid silo]

Then 3p/8

For first case

[slow wadi]

wait what?

[timid silo]

First case in the hallway all switch turned on

[slow wadi]

oh yes

that is 3p/8

as you said

[timid silo]

That's 3p/4 and then 1/2 in kitchen

[timid silo]

Yes

[slow wadi]

yeah

so second case

[timid silo]

All turned off

Picks the correct one

[slow wadi]

1/4

[timid silo]

(1-p)/4

[slow wadi]

yeah

so 1-p/4 multiplied by 1/2

?

[timid silo]

and the kitchen one will be 1/2 only

[timid silo]

Tes

Yes

[slow wadi]

1-p/8

so final answer is 1+2p/8?

[timid silo]

Yes

[slow wadi]

ok thank you

that took too long lmao

tysm tho

[timid silo]

Did you do part (ii)

[slow wadi]

no

i dont need to do that tho

[timid silo]

It's fun doing probability without studying it

[slow wadi]

yes

thank you again

[timid silo]

Welcome friend

[slow wadi]

.close

[obtuse pebble]

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[arctic panther]

hi, i'm looking for a function f(x) such that f(f(x))=x^2 and inverse of f, let's call it g such that g(g(x))=sqrt(x)

[arctic panther]

or for a general way to make functions such that f(f(x))=g(x) and inverse of f (h) such that h(h(x))=inverse of g

if it's even possible

[warm canopy]

assume $f(x) = x^a$, then figure out what you need a to be to get $f(f(x)) = x^2$

[warm shale]

iCaird

[arctic panther]

oh

i was expecting something more complicated than x^a so i didn't even think of that

thanks a lot

.close

[obtuse pebble]

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[teal mango]

How do i reverse engineer this so that i can use the final result to find n?

[obtuse pebble]

@teal mango Has your question been resolved?

[teal mango]

<@&286206848099549185>

[warm shale]

LF

[coral nest]

you can get a quadratic in terms of n

which can be solved using the quadratic formula for a given output

[obtuse pebble]

@teal mango Has your question been resolved?

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[split wraith]

I need to list all of the partitions in the set A={a,b,c} and I am having trouble understanding the concept. I really need help asap, it is due very soon.

[timid silo]

a partition of a set is a set itself, whose elements are taken from the bigger set.

Let's say A={a}

what are the possible partitions

[split wraith]

a..?

[timid silo]

"a" written like this is not a set, all partitions are sets

[split wraith]

so then {a}?

[timid silo]

Yes, correct

Now there is another partition of the set {a}

[split wraith]

{}?

[drifting wraith]

you;re describing subsets

are you sure they used 'partition" to mean susbet?

[split wraith]

the question is from the book of proofs, it is not about subsets because that was chapter 1 material, this is chapter 11 or 12 I believe

[timid silo]

Ah you're right, should've said that the elements are sets

[drifting wraith]

yeah

[timid silo]

So you said that {a} and {} are subsets

[split wraith]

yes..?

I need partitions though

[timid silo]

a partition of a set is a collection of subsets of that set.

[split wraith]

I thought that was a power set?

[timid silo]

the power set contains all the subsets, it is the maximal partition

but a partition can contain only some subsets

[split wraith]

so then how do I find the partitions for my set?

[timid silo]

Okay, so we know that : {a}, and {} are subsets. what is the biggest partition ?

[split wraith]

{a} because it has a value

[timid silo]

that not a partition, that's a subset

because its element is not a set

it's a

and a is a number, not a set

{{a}} is an example of partition

[split wraith]

okay cool

so then for a set of {a,b,c} how do I find the partitions? is it just {{a}},{{b}},{{c}}?

and {{}}

[timid silo]

nope, you also have {{a}, {b}} and {{b}, {c}} and {{c}, {a}} and {{}, {a}, {b}} and {{}, {b}, {c}} etc.

[split wraith]

how do I know how many there are?

[timid silo]

I misspoke here, a partition doesn't contain the empty set.

[split wraith]

okay

[timid silo]

the number of partitions is known as the bell number, it doesn't have a simple formula

your example above A={a,b,c} has exactly five partitions

[split wraith]

okay

I think i kinda got it now

[timid silo]

Forget everything above, I was explaining how to get the power set. The partitions are,a way to group elements of a set. In your example :
A= {a,b,c} can be regrouped like this { {a,b}, c} or like this {a, {b,c}}

can you give other ways to group these elements

[split wraith]

so its like associativity with sets in a way?

[timid silo]

yes kind of !

{ {a}, {b} } is NOT a partition because none of its blocks contains c

each element of A has to be in at least one block

So the partitions are :

{{a}, {b,c}}
{ {a,b}, {c}}
{{a,c},{b}}
{{a},{b},{c}}

there is one remaining

can you guess what it is

[split wraith]

hmm.

[timid silo]

notice that in the first three partitions, each time we grouped two elements together

and left one

in the fourth partition

we took each element in isolation

what's left

[split wraith]

not the empty set right?

[timid silo]

nope the empty set in never in the partition

[split wraith]

Hmm

[timid silo]

you have three elements, you can either take 1 element in each block , or 2 together, or ...?

[split wraith]

{{b,c},{a}}?

o wait

[timid silo]

correct but already given

[split wraith]

{{a,b,c}}

[timid silo]

haha correct

so we have 5 in total

[split wraith]

i know that it was a stupid struggle but seeing the little correct mark made me smile

i promise im not stupid, im just not smart 😅

[timid silo]

haha

[timid silo]

i don't think anybody is smart when it comes to math lol

[split wraith]

im sure plenty of people are lol

[timid silo]

they just hide their struggles, i know that from experience.

but abilities differ evidently

and surely, everybody struggles

[split wraith]

well, i cannot thank you enough for this

it rly helped me understand

[timid silo]

no need to

glad it helped

[split wraith]

I know I was being a little difficult and I apologize for that

[timid silo]

not at all!

[split wraith]

such a kind soul 🥺

thanks again (:

.close

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[worn ocean]

@timid silo Hey, I asked a question yesterday about the combinations of subsets but no one really solved it/explained how to do it, can you help? ;-;

[worn ocean]

This is the question ;-;

[violet sentinel]

Oh geez, give me a second to think on that

[worn ocean]

Please, take all the time you need.

The answer is 40 btw, no clue how though :(

[limber quartz]

there's 63 subsets, not including the empty set

2^n - 1, n = 6

[timid silo]

Hi, real headscratcher you have here, sounds like the pigeonhole principle is involved, not sure though how to get the result

[worn ocean]

Permuting the sets?

[fair pulsar]

Ok so here is my thinking: First we decide which combinations of 1,2,3,4,5,6 result in a multiple of three. Then using nchoosek we see how many subsets of the total are left

[worn ocean]

Adding 123456 is 21
So
12456 and 12345 right?

Thats just the 5 ones

It can also be 3456

And so many others ;-;

[violet sentinel]

Okay so the sum of the digits are a multiple of 3

So we first can start with the sets that are of one element: {1}, {2}, {3}, {4}, {5}, {6}. The only ones that are a multiple of 3 is {3}, {6}, so from the single elements we have 2 total.

For subsets with 2 elements, we have {1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {2, 1}, {2, 3}, {2, 4}, {2, 5}, {2, 6}, {3, 1}, {3, 2}, {3, 4}, {3, 5}, {3, 6}, ... {6, 1},{6, 2},{6, 3},{6, 4},{6, 5}

this is getting too much for hand counting, so there's got to be an easier way

[worn ocean]

You're bruteforcing, that's kinda not... efficient

[timid silo]

there are 64 subsets, there has to be another way than brute force

[worn ocean]

I get 120 seconds to solve this question

[violet sentinel]

well yes, but you have to start somewhere to discern a pattern

[fair pulsar]

@violet sentinel the sum of the digits must not be multiple of three

[violet sentinel]

fixed the edit, thank you

[fair pulsar]

oh ok

nvm

[worn ocean]

If you subtract from total, u would get the not multiple

[limber quartz]

*subsets, and 63, b.c. the problem specifies non-empty subsets.

[fair pulsar]

@worn ocean ordered or unordered subsets? Ie does (3,4) and (4,3) count?

[limber quartz]

that;s a good question

[worn ocean]

unordered

they both count as one

its implied i think

[fair pulsar]

ok, so lets try find all ordered ones first that satisfy the requirement (should not be too difficult)

[worn ocean]

think of it as balls numbered 1 to 6 and we have to make many non empty bags of balls

[worn ocean]

6!?

6 factorial?

[fair pulsar]

6! would be all the subsets no?

[worn ocean]

yeah all ordered ones right

thats what you said we should find ;-;

[fair pulsar]

I was referring to all ordered ones that satisfy the conditions of non-multiplicity

[worn ocean]

Ah

[fair pulsar]

but now im thinking its too tidious

there must be a smarter way

[timid silo]

the number of all the subsets is equal to the cardinal of the power set, that is $2^6$, if we count the empty set as disorganized pointed out

[warm shale]

AimaneSN

[worn ocean]

yeah because we would have change the sets

[worn ocean]

Hmm, this would include empty sets

but if we subtract one from 2 power 6, it should be fine right?

[limber quartz]

We can leverage the fact that any set whose sum of entries is a multiple of 3 can be a subset of a larger subset whose sum of multiples are 3

...spoken stupidly

so we can just start small.

{1,2}

[worn ocean]

yeah thats one

[fair pulsar]

Ok lets try something else to see if we are on the correct path: assume we have counted all ordered sets that satisfy our requirement (say they are n). How would we derive the answer? We would have to see how many permutations all n sets have. But this would require us to know how many elements each of the n sets has. Therefore the approach I proposed earlier is not enough we need more information

[limber quartz]

{{{1,2},3}6}

but also {2,4} 🤔

[worn ocean]

12 45 would also

yeah

45

36

[limber quartz]

oh, just ascend

you know

this is an algorithm though, not a formula

[worn ocean]

yeah yeah but i cant sit there making sets

;-;

[fair pulsar]

Perhaps we should start with a smaller problem: replace {1,2,3,4,5,6} with {1,2} and then {1,2,3,4}

[worn ocean]

If I give you guys the solution can u try reverse engineering something?

[worn ocean]

sure

[limber quartz]

{{{2,4},3},6}

[fair pulsar]

ok so for {1,2} we have {1},{2} but not {1,2}

[timid silo]

and I was thinking about pigeonholes

[worn ocean]

But this is an algorithm like disorg did

[fair pulsar]

I dont think they did

[worn ocean]

which is not helpful when im giving the test

i only get 120 seconds ;-;

[timid silo]

still, lots of trial and error

[limber quartz]

neat

[worn ocean]

can u pls explain ;-;

[limber quartz]

🧠

[fair pulsar]

I think this sentence is the key

I dont understand it yet but lets see

[worn ocean]

[timid silo]

that's because the remaining subset is 21-(the subset that is not a multiple of 3)

[fair pulsar]

I think I am getting somewhere: Suppose x is a multiple of three and y isnt. Is (x-y) a multiple of three?

[timid silo]

if x isn't a multiple of 3 then 21-x isn't either

[timid silo]

nope

[worn ocean]

yeah but total sets arent 21

the sum of digits of the sets is 21

[timid silo]

I mean the sum of all digits

[worn ocean]

obv not ;-;

[timid silo]

yes, so the sum of the digits of ANY subset is equal to 21 - (the sum of the digits of the complementary subset)

[fair pulsar]

Ok so, suppose our elemets are a,b,c,d. If {a} works then since {a,b,c,d} doesnt work, {a,b,c,d}{a} = {b,c,d} also works

[worn ocean]

okay makes sense

but then you would still have to hand count the subsets, is my point

[worn ocean]

knowing this fact just helps you count half of them, instead of all of em

[fickle turret]

So I think I found an easy method to bruteforce it without actually using maths. May I share my idea?

[worn ocean]

Please do

[fair pulsar]

@worn ocean a lot more than half since as you increase the size its gonna get more and more difficult

[worn ocean]

It will always be half, since its complementary in nature o.o

if the set was 1234567, then too, it would help us count the other half automatically

[fair pulsar]

If the answer is 50 and the initial set has an even number of elements then you will have to count 25. But what I am saying is that if you were to try to brute force all the number of elements in subsets (ie not stop at 3 number of elements) it would become a lot more difficult to count from 20 (in our example where the answer is 40) and upwards. Therefore by the proposed method you are saving quite a bit of time compared to doing it all brute force. So it should be appropriate for the exam. However it still is interesting to see if there is a smarter way (ie lets see what @fickle turret is typing)

[fair pulsar]

I am saying should be appropriate for the exam because it is their proposed solution

[worn ocean]

Ah yes true, its hard to keep track of stuff after that point

[worn ocean]

The people asking and solving are different entities ;-;

[fair pulsar]

what do you mean

[worn ocean]

The question is taken from somewhere else

[fair pulsar]

ohhh

[worn ocean]

Then asked in the test

[fair pulsar]

so you found the solution indendently of the provided material, I see

[worn ocean]

That's my teacher's solution, which is very slow, but I'm guessing that's the best he knows

[fickle turret]

So as we already discuessed there are 63 combinations in total. My idea would be to subtract every possible combination that is divisable by 3. Doing that we could first focus on the set S={1,2,4,5} and than reconsider the 3 and the 6. The maximum value a number in S can get is 12. So we could just focus on searching the combinations in S to make 3,6,9,12 (I found 5 of them). Than we need to consider the 3,6 which could be added while not changing the result modular 3. So each of this 5 combinations needs to be multipled by 4 (the normal version, +3, +6, and +3+6), which gives us 20 possible ways to get a number divisable by 3. The last thing we need to consider is the empty set combined with 3 and 6 so just {3} {6} {3,6} leading us to having 23 possible combinations to get a number being devided by 3

Sorry for my bad english and hopefully it helps

[limber quartz]

...is the answer 46?

[worn ocean]

The answer is 40

[fickle turret]

63-23

[limber quartz]

grr

[fair pulsar]

he arrives at the correct answer 63-23 = 40

[limber quartz]

oh, we don't count the complements?

[worn ocean]

We do

[limber quartz]

why

[violet sentinel]

lol I love how many people are involved in this right now haha

[worn ocean]

mb, we actually do count em

you can see the double from each subset complements

[worn ocean]

What about 1236

[worn ocean]

I dont see you account for sets other than 3 and 6

[limber quartz]

^23

[fickle turret]

I don't understand the question?

[worn ocean]

6 and 3 are divisible

in the first row

so 21 left, there has to be one more set that u counted extra

[fair pulsar]

@fickle turret I really like your solution, which works fine here but I think there is a catch: you say that first we remove elements already divisible by 3 and then add all the remaining digits to get say X. The we see how many divisors are smaller than X, (ie x1,x2,x3,...). But how do we know that we can combine our elements (generally) to get all x1,x2,x3,...?

[limber quartz]

I'm missing shit

[fair pulsar]

This would need a bit more work but still I find it a nice different point of view

[worn ocean]

yes all the 4 and 5 subsets

[limber quartz]

I was skipping those because we should be able to just double the subsets of size 1,2,3

to account for the complements

[worn ocean]

yeah thats what im saying, just list 20 sets

and then double em

u have 21 sets rn

[limber quartz]

so am I over or under?!

[worn ocean]

you must have made one extra

[limber quartz]

crazy

[fair pulsar]

@worn ocean I think the answer you are looking for is what @fickle turret provided. If I have understood it correctly its pretty fast

and I think you do not need to count anything

like nonthing

nothing*

[worn ocean]

I don't really fully understand it tbh

Can you try to word it better please ;_;

[fair pulsar]

Sure, lets try

[limber quartz]

🧠 🧠 🧠 🧠

[fair pulsar]

Ok, so for a start the idea is to not find how many sets satisfy our requirement but find the total number of subsets and then the number of subsets that dont satisfy our requirement and subtract. I think you got it thus far

[limber quartz]

I'm with Toby now.

[fickle turret]

So, sry that I can't explan it better. The language barrier makes it more difficult for me

[limber quartz]

you did
A M A Z I N G .

[worn ocean]

What language do you speak? :D

[limber quartz]

🌼

[fickle turret]

German

[worn ocean]

I see I see

I have a lot of german friends if you wanna talk to one of them and explain in german

and then they can explain it to me in english

[worn ocean]

yep

[fair pulsar]

Now, the first Idea is that suppose a subset x satisfies our requirement. x union y where y is divisible by 3 will also always satisfy our requirement. So for our example the y's are 3 and 6. So if we know that x satisfies our requirement then {x} {x,3} {x,6} {x,3+6} also satisfy our requirement. @worn ocean do you understand this?

@fickle turret Wie geht's

[worn ocean]

Yeah but I don't understand why we only take 3 and 6

[fickle turret]

Danke, mir geht es gut 🙂

[worn ocean]

Thanks Toby for the solution though ❤️

[fickle turret]

You're welcome

[fair pulsar]

@worn ocean you cant take anything else that the reason. Suppose you took some y that is not divisible by 3, where x is divisible by 3. Then {x,y} is always not divisible by 3.

This is because sum({x,y}) = x+y which isnt divisible

because y is a single element

[worn ocean]

Yeah but the set {1,2} should be included in here, but by this logic we won't ever count it here

[fair pulsar]

should be included as x?

ohhh wait no

I see what you mean

you will

let me explain hold up

[worn ocean]

You can also explain on call if you want ;-;

[fickle turret]

So maybe I can share my paper

[fair pulsar]

@worn ocean I think that would be better. Can you set it up because this is like my second discord session

[worn ocean]

Please do tobias

[worn ocean]

Sure

[fickle turret]

Well, I think it is not really orderly

That is how I came up with 5

[fickle turret]

.

[worn ocean]

@fickle turret Got it!!!

@fair pulsar Actually explained your method super well on call

Thank you @limber quartz @violet sentinel @winter crest for helping too! ❤️

[limber quartz]

np

[fickle turret]

Thank you @fair pulsar for explaining it. And good night everybody .

[worn ocean]

Goodnight bud

[worn flame]

So I’m supposed to do these 10 questions for homework, and if I don’t turn it in correctly I will fail. Anyone care to help?

[obtuse pebble]

@worn ocean Has your question been resolved?

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[bright palm]

Hi just wondering if someone could help tell me what topic this is?

[swift root]

im having an issue at the end of the problem

[obtuse pebble]

@bright palm Has your question been resolved?

[bright palm]

<@&286206848099549185>

[obtuse pebble]

@bright palm Has your question been resolved?

[bright palm]

<@&286206848099549185> really sorry to ping again but I need to try and figure this out in the next few hours

[uncut ridge]

simplify 15/42 and 21/49

Note that 15 = 3x5 and 42= 3x2x7

and also 21=3x7 and 49=7x7.

Then cross multiply the simplified fractions

[bright palm]

Ah thank you!!

.close

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[deft kestrel]

how can i show that if $z\in\mathbb{Z}[\sqrt{-3}]$ divides $2$ and $1+\sqrt{-3}$ then $z=\pm 1$

[warm shale]

Beous

[deft kestrel]

$\mathbb{Z}[\sqrt{-3}]={a+b\sqrt{-3};|;a,b\in\mathbb{Z}}$

[warm shale]

Beous

[obtuse pebble]

@deft kestrel Has your question been resolved?

[obtuse pebble]

@deft kestrel Has your question been resolved?

[deft kestrel]

nvm i figured it out

.close

[obtuse pebble]

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[obsidian isle]

The center is (0,0)

[worn jackal]

oh, so it would be x^2+y^2=10?

instead of in the (h,k) format

[nocturne minnow]

How did you calculate r?

[worn jackal]

distance formula

[obsidian isle]

Try again

[nocturne minnow]

Can you show your work?

[worn jackal]

sqrt[(-7-0)^2 + (3-0)^2]

sqrt[49 + 9]

7+3

[nocturne minnow]

No

You can't do that

[obsidian isle]

Haha

[worn jackal]

oh, any suggestions on what to do instead?

[glad totem]

Hello

[obsidian isle]

Exponents don't distribute over addition

[nocturne minnow]

$\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$

[warm shale]

dldh06

[obsidian isle]

Exponents distribute over multiplication

[glad totem]

Can I get Some help

[nocturne minnow]

#❓how-to-get-help

[worn jackal]

wait so

then it would be sqrt(58)

[obsidian isle]

Yes

That's r

[worn jackal]

this still shows as wrong for some reason

this is also wrong

[obsidian isle]

r^2

[nocturne minnow]

Because $r = \sqrt{58}$

[warm shale]

dldh06

[nocturne minnow]

The equation of a circle is this

Notice how it's r^2

But you just have r

[worn jackal]

oh, so its 58?

[nocturne minnow]

Try it

[worn jackal]

thank u so much, it worked

.close

[obtuse pebble]

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[wind field]

#5, solving limits
just wondering why #5 the limit does not exist

[gilded needle]

what do you get if you plug x=1 into the two expressions

10-x^2 and 7-x

[wind field]

9 and 6

[gilded needle]

right

so that says that the left hand limit and the right hand limit do not agree

[wind field]

ah okay okay

thanks a lot!!

[gilded needle]

pleasure

[wind field]

.close

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[dreamy ember]

I need help with it

[dense imp]

Any idea how to start if you need to get the variable out by itself on the left and it’s inside the tangent function like this?

[obtuse pebble]

@dreamy ember Has your question been resolved?

[obtuse pebble]

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[polar siren]

Can anyone help me for the hypothesis part, only Question B

[obtuse pebble]

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[still cairn]

Hi I have a question about sequences

[still cairn]

the problem was to find the formula for {-3, 2, -4/3, 8/9, -16/27,...}

^ here's my answer

https://cdn.discordapp.com/attachments/903430332324405288/971985145089388564/931030786541502515.png

^ here's the books answer

Both formulas give out the same sequence, but I wonder if my answer is acceptable on let's say an exam. If these formulas arent equal please tell me.

I'm starting at n=1 btw

[fierce lagoon]

I would prefer the book's answer for the sake of it being geometric

But like

Yeah I'd rather have it in geometric form

Rather than alternating

But they're mathematically equivalent so I think they shouldn't make a difference

[dense imp]

theyre the same sequence but the book's version is simplified a little

[still cairn]

I looked up a solution and saw how they got it, but I would've never guessed it first try.

[dense imp]

your answer should be acceptable on an exam, unless it's specifically asking for some form of simplification, but it's helpful to see how they arrive at that version in case you'd have to pick it out on say a multiple choice exam

[still cairn]

they put the 2nd term over the 1st term and figured out the pattern from there. I would've never saw that.

[dense imp]

the idea is to just make the exponents match, so for example you can multiply the top and bottom by -3 / -3, then the -3 in the denominator would have it's exponent go up by 1 and you're left with an extra -3 on top, does that make sense?

$$\frac{(--1)^n \cdot 2^{n-1}}{3^{n-2}}\frac{-3}{-3}$$

oh boy i made a mess of that

$$\frac{(-1)^n \cdot 2^{n-1}}{3^{n-2}}\frac{-3}{-3}$$

[warm shale]

Sooshon

[dense imp]

so just your version multiplied by -3/-3

[still cairn]

hahah yeah i think my teacher would be forgiving

[dense imp]

$$\frac{-3 \cdot (-1)^n \cdot 2^{n-1}}{3^{n-1}}$$

[warm shale]

Sooshon

[dense imp]

so you keep the -3 on top, multiply the -3 in the bottom and now you can combine the 2 and 3 bases under the same exponent

[still cairn]

yeah I see now

[dense imp]

the reason you would think to do this is that the exponents are so close

[still cairn]

oh okay

I wouldve never guessed that because I wasn't taught that trick

[dense imp]

it's really just to make the expression more concise

[still cairn]

i see

thanks for helping

[dense imp]

but your answer is perfectly correct

(good luck telling that to online answer submission forms)

[still cairn]

online answer submission forms?

[dense imp]

lol just a joke ragging on those online homework submission things, they are endlessly frustrating to students because they have to mold their answer into some very specific expression they are designed to accept

hopefully you don't need to use such things : )

[still cairn]

ohhh that

yeah I turn in my homework as a pdf

its all good

I cannot imagine using that for calc 1 and forward

I had to use an online submission system for precalc, but there's usually only one answer for precalc problems.

.close

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[undone vector]

can sme help me pls

[obtuse pebble]

@undone vector Has your question been resolved?

[tardy epoch]

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

[undone vector]

idek where to start lmao

[nocturne minnow]

Plug in the values

[undone vector]

do ive to take the antiderivative

[obtuse pebble]

@undone vector Has your question been resolved?

[marble shuttle]

You need to integrate

[obtuse pebble]

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[radiant yew]

how do I do this

[lucid flame]

do you know law of sines? (if not then that's fine)

[radiant yew]

no

I don't

[lucid flame]

ok

look at each triangle

(ABC and ABD)

[radiant yew]

yes

[lucid flame]

let CB be x

and AB be h

then whats sin(38 degrees)

actually tan(38 degrees)

not what i meant

[radiant yew]

0.78

wat

[lucid flame]

in triangle ABD

its a right triangle with one angle 38 degrees

[radiant yew]

ye

[lucid flame]

which ratio will be tan of that angle

[radiant yew]

the 38 degree

tan 38=h/BC+23?

[lucid flame]

yes

tan(38) = h/(23+x)

and then look at trangle ABC

create another equation for tan(52)

[radiant yew]

tan 52=ab/bc

which is x/h

[lucid flame]

h/x

[radiant yew]

oh ok

wait ye

typo

[lucid flame]

so now you have 2 equations

[radiant yew]

yes

[lucid flame]

you can solve for h

[radiant yew]

is it tan52-tan38

tan14=23

wait I'm wrong

[obtuse pebble]

@radiant yew Has your question been resolved?

[radiant yew]

hoe do I solve

aaa

how

<@&286206848099549185>

[radiant yew]

Let BC = x
Let AB = h
tan 38° = h/(x+23)
tan 52° = h/x
how do i substitute the values of tan 52° and tan 38° and use substitution method to get value of x and h

[high lily]

you'd substitute one of your variables into the other equation

tan(52°)and tan(38°) are known constants

[obtuse pebble]

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[rugged vine]

Can i ask a question about work word problem?

[earnest elk]

go ahead

[rugged vine]

Alex can paint a house in 6 days. WIth Bryan's help, they can paint the same house in 3 days. How many days would Bryan take to paint the house alone?

i know how to do if they give how long alex and bryan take and ask to find how many days needed

but i dont know how to do the question when the question ask for amount of time bryan take

[royal basin]

in one day, Alex paints 1/6 of the house

while Alex and Bryan together paint 1/3 of the house per day

so what fraction of the house does Bryan alone paint per day?

@rugged vine

[obtuse pebble]

@rugged vine Has your question been resolved?

[rugged vine]

1/3?

so its 3 days right?

[obtuse pebble]

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[sterile trellis]

pls help

[sterile trellis]

😃

[royal basin]

have you made any progress so far?

[sterile trellis]

i calculate like

i can just get this

[royal basin]

ok

my instinct here is to substitute t := x - pi/2

so that the limit becomes $\lim_{t \to 0} \frac{\cot(\pi/2 + t)}{t}$

[warm shale]

Ann

[royal basin]

which in turn can be written as $\lim_{t \to 0} \frac{-\tan(t)}{t}$

[warm shale]

Ann

[sterile trellis]

oh yes

let me see

thank you so much i did it!

[timid silo]

Is it needed tho can't we just use lhoptal rule from here

differentiate top and bottom I mean

[royal basin]

fuck off with l'hôpital's rule lmao

there's no need for l'hôpital

[timid silo]

[royal basin]

there is no need for it when the same can be accomplished by less nuclear means

[timid silo]

Its just one way I mean

[royal basin]

if you want to l'hop-bash everything like a limit problem speedrunner then go ahead and do it but don't inflict it on others

[timid silo]

It's not so nuclear ig we all know derivative of cot is -csc^2 plug in pi/2 it's -1

[timid silo]

Ok ok chill

[obtuse pebble]

@sterile trellis Has your question been resolved?

[obtuse pebble]

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[covert zealot]

stuck here

[royal basin]

multiply both sides by $\log(k)$ and divide both sides by $b-2$ to get $$\log(k) = \frac{\log(a)}{b-2}$$

[warm shale]

Ann

[meager raptor]

hey can u recommend me a studying playlist?

[teal prawn]

#❓how-to-get-help

[meager raptor]

is there any music help

[covert zealot]

@royal basin how do i get from that to

[royal basin]

raise e^(both sides)

or 10^(both sides) or whichever base you consider your logs to be in

[obtuse pebble]

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[obtuse pebble]

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[strange wing]

A rectangular room measuring 7 meters by 5 meters and 2.5 meters in height is re-papered every year (we leave out any windows and doors). One does not remove the old wallpaper, just glues the new one over it. The wallpaper is 1.2 mm thick. The area of wallpaper needed gets smaller every year. The number in m^2 of wallpaper needed in the n-th year is called $b_n, n=0, 1, 2, ....
It holds: bn=bn-1-0.024, n=2, 3, 4, ....$

a
Explain that.

[warm shale]

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

[strange wing]

how can this be modelled?

[obtuse pebble]

@strange wing Has your question been resolved?

[obtuse pebble]

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[stark ether]

can this be somehow simplified further?

[obtuse pebble]

@stark ether Has your question been resolved?

[timid silo]

is it tan(x)

If so, I don't think so

[obtuse pebble]

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[timid silo]

q5

[timid silo]

ik it's a simple sum but somehow im going wrong in the calculations wit the formula ( SD = PW + TD )

<@&286206848099549185>

[velvet siren]

Is rs a currency?

[timid silo]

yes it is

[velvet siren]

So what i'm getting is the presents price is 11,660 but after 9 months it decreased to 11k?

Or did you buy it for 11k and it increased after 9 months

@timid silo

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

the price rn is 11,660 , so 660 is a sort of interest if paid after 9 months would be written off

so i gotta find r

sorry for the late reply

[velvet siren]

So you began with 11k?

[timid silo]

yes

[velvet siren]

Ok do you have a clue how the formula would be?

[timid silo]

yes sum due = present worth - discount/interest

[velvet siren]

You want the rate of interest

Mostly thats a product

[timid silo]

yes

[velvet siren]

11,660=11,000×X

You have to find X

[timid silo]

the answer seems to be 8% although im not sure of the calculations

( from my answer key )

[velvet siren]

Solve this

11,000x6.25=11660?

11000×6.25 is like 66k

[timid silo]

tru

1.06?

[velvet siren]

Yes

Thats the rate of interest for 9 months

[timid silo]

ah

alrighty

[velvet siren]

But

You probably want the rate of interest for 1 month

[timid silo]

i want it for 9

[velvet siren]

Well for 9 months its 1.06 then yes

[timid silo]

alrigh thanks mate

[velvet siren]

👍

Don't forget to close

[timid silo]

will close now

.close

[obtuse pebble]

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[teal prawn]

the lines of 2 linear equations will be parallel if they have 0 solutions

and not coinciding

[timid silo]

or more simply

if there slopes match

[teal prawn]

ratio of co-efficient of x = that of y, neither of which should be equal to ratio of co-efficients

[velvet fossil]

so it would be D?

[timid silo]

how?

[copper latch]

Yes

[velvet fossil]

ok

[teal prawn]

sorry yeah

[velvet fossil]

ty

[timid silo]

yes D

because there slopes match

[velvet fossil]

ok

[timid silo]

so they are parallel

[velvet fossil]

ok

ty

.close

[obtuse pebble]

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[hushed cairn]

How would I do something like this?

[hushed cairn]

For other calculations

I add one to the power

and then divide it by the new exponent

but it doesn't seem to work here

[pine sail]

What did you get?

[hushed cairn]

5x^-2/-2

or if you wanna bring the x^2 down 5/-2x^2

[pine sail]

+c yeah.

[hushed cairn]

Yea

wait did I do it right?

[pine sail]

Seems like it.

[hushed cairn]

Ok I have another question

What about this?

How should I approach this question

[pine sail]

Well you could try calculating derivatives of the options to see if you get f(x).

[hushed cairn]

Right

If I didn't have the options though

What would you do?

[pine sail]

Quit.

Lol I'm kidding.

[hushed cairn]

I want to

Our teacher started integrals 2 days ago

and we have a full exam on it

in 2 days

so I'm not even sure anymore

[pine sail]

That's sad.

[pine sail]

Integrate by parts.

u = x+5

[hushed cairn]

Yea.

Ok

I'll try that out

[obtuse pebble]

@hushed cairn Has your question been resolved?

[obtuse pebble]

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[scenic lynx]

hello, how can I prove that 5^n is congruent to 25 mod 100 if n >= 2

[tardy epoch]

5^3=125 and induction

[scenic lynx]

oh I am really stupid didn't think abt induction at all