help-39

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

we want to prove that whatever 13 elements we pick, we will always end up with at least one pair such that $a = 2b$

[jolly parrot]

1 divided by 0 equals Infinity

[proper nova]

so uhm

[slim jackal]

It's impossible

[proper nova]

rewrite each $a_i$ as $2^{k_i} \cdot b_i$ where $k_i$ is either 0 or 1

[slim jackal]

But idk how to explain

[jolly parrot]

1 divided by 0 equals Infinity

[slim jackal]

If I explain it will be more complex

[proper nova]

we only care the $i$ that $k_i = 1$, so with $i$ like those, we want to care that $b_i = a_j$ ($i \neq j$)

[jolly parrot]

1 divided by 0 equals Infinity

[latent glen]

but what can we get from here?

like any groups?

[pearl pond]

@latent glen Has your question been resolved?

[proper nova]

im just trying to lay the foundation

let me think more

[pearl pond]

@latent glen Has your question been resolved?

[latent glen]

do groups like 1 2 4 8 16, 3 6 12, 5 10,.. work?

[foggy scarab]

aight so if we write out all numbers from 1 to 17 and mark up pairs of a and b, we find that 9,11,13,15,17 are not in any pair so we can include them
this leaves us out with 8 pairs and from them we need 8 elements
once we write the pairs , we find that only pairs (5,10) and (7,14) have elements that are not common in any other pair, Hence we can select one element from each.

This leaves us with 6 pairs and we need 6 elements

But these pairs have common elements
In batch (3,6) and (6,12), we see that their is clearly 6 in common, so we can eleminate it and now we need 4 elements in the 4 pairs (1,2),(2,4),(4,8),(8,16)

However, on selecting one, the pairs with the selected elements get removed as we already selected a common element, leading us to getting only 2 elements from the 4 sets, leading to a maximum subset size of 11 elements

hence proved ig

[latent glen]

the max subset size is 12...

[foggy scarab]

ah mb

yeah 12

but proof is valid

just hunt for the biggest set

ig

[latent glen]

okay i can move on to the other problems

[foggy scarab]

🙏

[latent glen]

it's a mix of many diff topics

[foggy scarab]

where did u get this q from?

[latent glen]

last ques in a model test

[foggy scarab]

😮

[latent glen]

anyways

Find the minimum and maximum value of $S = \frac{m^2-2m+2}{m^2+2m+2}$.

[jolly parrot]

Thomas

[foggy scarab]

you could take 1 out and differentiate

[latent glen]

so that gives us $\frac{-4m}{m^2+2m+2}$

[jolly parrot]

Thomas

[foggy scarab]

ye

differentate and equate to 0

[latent glen]

but how can we progress from here

,w min(-4m/(m^2+2m+2))

[jolly parrot]

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[foggy scarab]

you know how to differentiate?

[latent glen]

haven't learnt that yet i think

,w differentiation

[jolly parrot]

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[foggy scarab]

ohhhh

hmmm

[latent glen]

the first subquestion in this was proving that the equations always has two dinstinct roots and now we're here..

[foggy scarab]

am not sure how we do it without differentiation tbh

[latent glen]

<@&286206848099549185>

,w min and max \frac{x^{2}-2x+2}{x^{2}+2x+2}

[jolly parrot]

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[frank violet]

any restriction for m? like m is a positive number or smth?

[latent glen]

i don't think so

i've just proven that every m can be used

[frank violet]

okay that's a bit annoying

[frank violet]

Can you show for any m<0 it's always greater than any m>0

complete the square in the denom, that would help

[latent glen]

(m+1)^2 + 1?

[frank violet]

yeah so the denom is always positive

[frank violet]

let call this T

you can see how for m<0, T>0

[latent glen]

the upper part is also positive tho

[sterile python]

This channel still on the pair thingy?

[frank violet]

and m>0 , T<0

[frank violet]

nope

[sterile python]

Oo

Which question then

[frank violet]

that one

[frank violet]

Now you can focus on m>0 and use some kind of inequalities

We have already shown that min can't occur for m<0

[sterile python]

Have you guys removed the denominator

[frank violet]

Think of each number as a cursed energy flow
😭

[sterile python]

Would be easier to progress

[frank violet]

I'm sure how that help, but we didn't go that way

[sterile python]

Oh okay

Try your way then

Since i think it would be nicer to make denominator eliminated, then set it into an inequality equation

[frank violet]

not sure what you meant by that

[warm patio]

max is not that difficult

just

divide by m

and use AM-GM

just note that after divding by m we need to find the minimum/maximum of m + 2/m

my bad

amgm is only for positive

[frank violet]

right

[warm patio]

but this shoudl be useful

[frank violet]

It does

We have shown that min only occur for m>0

So we can use AM-GM now

idk where OP is rn tho

[warm patio]

oh cool okay

oh so we are actually pretyt much done

just see

that the function is odd

this m + 2/m

[latent glen]

wdym?

[warm patio]

We had that the min occurs only when m > 0, similarly we can show that the max occurs when m < 0 from m + 2/m being odd we get the max by plugging in -x where x is the argument for the min.

[warm patio]

this is just $\frac{-4}{m+2+\frac{2}{m}}$, so its sufficient to minimise/maximise that m + 2/m

[jolly parrot]

k

[latent glen]

which gives us -4/(2*sqrt(2)+2)

2 - 2*sqrt(2) for minimum value

how do i get max tho

@warm patio

[warm patio]

first we need to show that the max is for m < 0

[latent glen]

cauchy for two numbers work tho

[warm patio]

wdym

[latent glen]

m + 2/m >= 2*sqrt(m x 2/m) = 2 sqrt(2)

[warm patio]

yes this works for m > 0

[warm patio]

exactly just dont forget the +1 we put in front earlier

If you plug -x into f(x) = x + 2/x we get f(-x) = -x + 2/(-x) = -(x+2/x)

so if we have some min (x+2/x) it will be the max of -(x+2/x)

our x + 2/x looks like this

maybe its easier to see on a graph

[latent glen]

oooh ok

[warm patio]

sorry, we actually dont need this just the bit about the function

[latent glen]

last question for today

Given a circle with center O and radius R, and a fixed chord BC. A is a point that varies on the major arc BC. CD and BE are altitudes. Let B' be the point symmetric to B with respect to E, and C' be the point symmetric to C with respect to D.
a) Prove that DE has a constant length.
b) The circumcircles of triangles ABB' and ACC' intersect at point K. Prove that K is the intersection of lines BC' and CB'.

c) Prove that K belongs to a fixed line and AK passes through a fixed point.

[latent glen]

<@&286206848099549185>

[slim vigil]

You should start a new channel for a new question

[latent glen]

oh ok

getting late now

.close

[pearl pond]

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[pearl pond]

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[somber night]

im sorry dude i think no one rly understand the question

im not good in this field of math so i cant rly help im sorry , i hope someone can help u though maybe ping mods

[merry carbon]

No, don't ping us for this please

[slow oak]

the mods are not picked on the basis of mathematical ability

[merry carbon]

And don't encourage people to do that mods are for moderation issues, not for solving problems

[lilac jackal]

pls ping us for server issues or queries, not math help

[merry carbon]

Not your fault just remember that for next time

[pearl pond]

@odd prism Has your question been resolved?

[pearl pond]

@odd prism Has your question been resolved?

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

[supple hill]

Can you explain more where is your question and what you tried?

[pearl mauve]

What????? I thought all mods were uncs who have a phd in math

[supple hill]

Think about the fact that your figure is not a complete circle, but a sector of a larger regular polygon.

We would better use this formula since in your problem, that curve is made of n_2 straight segments of length s. The line below has n_1 segments of length s.

Yes.

Yes, but to find it you need to isolate theta.

You are right, you have it!

Okay, now that we've solved the problem, here are the formulas I would use for a problem like this:
Find $\theta$: $1 = \sin(\theta) / \sin(\theta/2) \implies \theta = 120^\circ$. Then, Calculate X: $X = 180^\circ - 120^\circ = \mathbf{60^\circ}$. Finally, Calculate Y: $Y = 90^\circ - \frac{(2-1)120^\circ}{2} = 90^\circ - 60^\circ = \mathbf{30^\circ}$.

I used the coefficients given in this problem.

[jolly parrot]

Ga³¹Br³⁵I⁵³9000✞

[supple hill]

We can find both, and we use theta as a centre angle.

It's complex to explain, the point is that as we have observed, if you use the formula, you find theta and you can then find the angles.

Do you understand it better?

Yes. How we can get that?

Right. Use that.

[supple hill]

With the angular change formula.

Do you remeber?

interior angle=180∘−Δθ

Yes.

No, because 90*4 are: 360, right?

Sorry, you're right, it's only for each individual vertex.

[somber night]

Oh im sorry i didnt know that

[brisk lynx]

square root of 156?

[brisk lynx]

im 8th grade

whats a prime actor and factor

[robust oar]

https://studymathnow.com/natural-numbers/prime-numbers-and-factors/

[pearl pond]

@odd prism Has your question been resolved?

[sand dagger]

Prime factor is a factor that can't be split into any more factors other than 1 and itself

And a factor is...just any factor

😢 Mb bro

[sterile python]

Just go to other channels btw

[pearl pond]

@odd prism Has your question been resolved?

[jolly parrot]

Civil Service Pigeon

[dense jasper]

are you sure that this is well posed

If it wasn't obvious, this configuration is for n1=1 and n2=7

You can scale it up so that all of the segments have length 1

my point still stands

oh I'm a muppet I connected them in the wrong order

hold on

fixed

So did you mean to say that the segments that make up "line" B cannot be self-intersecting

These angles are all between 0 and 180 degrees. So I'm asking again - are you trying to say that the segments in line B cannot be self intersecting?

because your phrasing is very hard to follow at the moment

ok

please try to make sure that the questions you ask have all of the relevant details

\textbf{Updated Question Statement}:
\begin{tcolorbox}
Consider two paths in a plane composed of segments of equal length. Path $A$, consisting of $n_1$ segments, remains perfectly straight. Path $B$, consisting of $n_2$ segments, bends such that every interior angle between adjacent segments is a uniform value $X$. \

The endpoints of the longer path ($B$) coincide with the endpoints of the shorter straight path ($A$). The interior angles formed at the two points where these paths intersect are also uniform, denoted as $Y$. \

Given that $n_1$ and $n_2$ are positive integers such that $n_1 < n_2 < \infty$, and assuming Path $B$ does not intersect itself, determine the values of angles $X$ and $Y$ in terms of $n_1$ and $n_2$.
\end{tcolorbox}

[jolly parrot]

Civil Service Pigeon

[dense jasper]

So is this what you meant

[dense jasper]

I don't feel like culling through the entire chat, so I'm just going to give my commentary from the start: \

Without loss of generality, assume that each segment has length $1$. Let the turning angle of path $B$ be $\alpha \coloneqq 180-X$. Because all of the interior angles of $B$ are equal, each step of the chain turns by the same amount $\alpha$. Since the two endpoint angles with the straight path $A$ are equal to $Y$, we have that
$$2Y=(n_2-1)\alpha \implies Y=\frac{n_2-1}{2} \alpha.$$
The distance between the endpoints is precisely $n_1$, so
$$n_1=\frac{\sin \left(\frac{n_2 \alpha}{2} \right)}{\sin \left(\frac{\alpha}{2} \right)}.$$
Equivalently, in terms of $X$ and $Y$,
$$n_1=\frac{\sin \left(\frac{n_2}{n_2-1} Y \right)}{\sin \left(\frac{Y}{n_2-1} \right)}, \qquad X=180^{\circ}-\frac{2Y}{n_2-1}.$$
However, except for very special cases (such as when you get a regular hexagon), the values are not "nice" and cannot be easily found in closed form because the equation is transcendental (the equation is not algebraic). In cases like this, you would normally resort to numeric approximation. If you have a decent starting guess, Newton's method works. You could also do bisection on $\alpha$ if you don't have a good initial guess since
$$\frac{\sin \left(\frac{n_2 \alpha}{2} \right)}{\sin \left(\frac{\alpha}{2} \right)}-n_1$$
has the relevant root in and is well-behaved on the interval $\alpha \in (0,2\pi/n_2)$ (note that the total curvature $n_2 \alpha$ must be less than $360^{\circ}$ for path $B$ to not intersect itself).

I should say that you can also write the answer in terms of the definition of the Chebyshev polynomial of the second kind: If you let $\theta \coloneqq \alpha/2$, then your equation becomes
$$U_{n_2-1} (\cos \theta)=n_1$$
meaning that your angle is determined by the real root $x \in (-1,1)$ of $U_{n_2-1}(x)=n_1$, and then $\alpha=2 \arccos x$.

[jolly parrot]

Civil Service Pigeon

[dense jasper]

anyway that's all I rlly have to say

imma go eat now

[pearl pond]

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[pearl pond]

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[warm patio]

<@&268886789983436800>

[pearl pond]

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[wooden flare]

whats the way to like "show" a function is a one-to-one function or not?

[supple hill]

What did you try?

Think about x.

[wooden flare]

the way the textbook has us do it is to show: f(x) = f(y) and end up with x = y

[green aurora]

Assume f(x1)=f(x2)->x1=x2

[wooden flare]

so if i end up with anything but x = y, the function is not one-to-one right?

[supple hill]

Yeah.

[green aurora]

Example : 3x_1+2=3x_2+3 -> x_1=x_2

[wooden flare]

x = x^2?

[green aurora]

I made it clearer

[wooden flare]

that does look better

or just use x and y

[green aurora]

Yes

[wooden flare]

just preference

[green aurora]

Yes

[supple hill]

Yes, you can use what you prefer.

[wooden flare]

wait. so is f(x) = 1/(x+3) one-to-one?

do the asymptotes mean anything?

bc i do get x = y, but is doesn't look right idk

[supple hill]

Try it.

[wooden flare]

i did

[supple hill]

And what did you get?

[wooden flare]

x=y

[supple hill]

Then, that’s the answer.

It’s one-to-one.

[green aurora]

Its correct

[wooden flare]

yea, but idk it doesn't "feel" right bc of the asymptotes

[green aurora]

,w graph 1/(x+3)

[jolly parrot]

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[green aurora]

Graphically function one-to-one : it must pass the horizontal line test, intersecting only once.

[wooden flare]

doesn't it intersect at pos infinity and neg infinity?

so it has two intersections?

[green aurora]

Infinity Is not a Number

[supple hill]

No, because infinity it’s not a number.

[green aurora]

Then

A + and - ∞ the function ->0

[wooden flare]

i mean yea, but if it doesnt intersect the x axis then what do you mean by

[supple hill]

For example, if we draw a horizontal line in the function, we would need to draw two to intersect both. That's another indication that it's one-to-one.

[wooden flare]

that makes more sense

[supple hill]

To avoid a one-to-one interaction, you would need to be able to select two points on the graph.

[wooden flare]

bc it never equals 0, so a line drawn horizontally will never cross the "other side of the function"

[supple hill]

Right.

[wooden flare]

that makes sense

[supple hill]

Do you have more questions?

[wooden flare]

and now how would you show that: f(x) = x^n + x for n>0 is not one-to-one. would you state a case for n where it is not true, and hence it is not one-to-one for all n

because i dont believe showing f(x) = f(y) => x=y is possible for this question?

in general case for n rather than n = some constant

[supple hill]

First, you should analyze the parity, since depending on the value of n, it would be either even or odd. Then you would give it an even value and you just need to calculate the counterexample.

[grave mirage]

If it's always increasing or decreasing ig

[supple hill]

You will have two answers. Depending it's n pair or impair. Try it.

[grave mirage]

If it's continuous that is

If it's piecewise check the ranges of individual fxn

[grave mirage]

You'll always end up at x=y

But if you end up anything besides x=y, it's not one to one

[supple hill]

In this case, try n=2 and n=1, for example, and see what happens.

[grave mirage]

Also it's also not a good way to do things

try doing this with f(x)= e^x+x

[pearl pond]

@wooden flare Has your question been resolved?

[wooden flare]

sorry, had to do a thing rq

[supple hill]

Np, do you have more questions about that?

[wooden flare]

im going through it atm

if n=1, its oto. i assume if n=2 its not oto

for for n=2k+1, f(x) is oto, and for n=2k, f(x) is not oto

is my assumption, but ill see

[supple hill]

Yeah, you're right.

You get it.

[wooden flare]

hm, i mean. i still gotta prove it tho no?

so for n=2, i end up with: x(x+1) = y(y+1) how do i progress from here

[supple hill]

You can, but you are right.

[grave mirage]

Try to factor (x^2k)-(y^2k)

[supple hill]

Yes, factor and move to one side.

[wooden flare]

x^2 - y^2 = y - x?

so (x+y)(x-y) = y-x

[grave mirage]

No what?

[grave mirage]

How did you reach this conclusion?

[wooden flare]

x^2 + x = y^2 + y ==> x^2 - y^2 = y - x

[grave mirage]

Right

I'd advise to bring everything to lhs

[wooden flare]

soo x^2 - y^2 -y + x = 0

[grave mirage]

also use completing the squares

[wooden flare]

ok ill try from here

[grave mirage]

But here it works either way

[supple hill]

Yeah, when you have that, you can continue.

[wooden flare]

idk how do this with 2 variables...

[wooden flare]

how do i go about this?

[grave mirage]

It's not necessary for this example

[wooden flare]

i know how to CTS but with 2 variables idk what im doing

[grave mirage]

What's your general f(x)?

[wooden flare]

x^n + x

[grave mirage]

Okk

Yeah don't complete the square then

[grave mirage]

Just do it one at a time

[pearl pond]

Channel closed

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[wooden flare]

.reopen

[supple hill]

You can have on one side the letters with n and on the other the letters without.

[pearl pond]

✅ Original question: #help-39 message

[grave mirage]

Ok do you know x^n-y^n is always factorable by x-y?

[wooden flare]

so: x^2 + x = (x+1/2) ^2 - 1/4

[grave mirage]

Yeah

But this will only be usefull for n=2

[wooden flare]

yea

[grave mirage]

Even then it's just better to factor

[grave mirage]

...

Do you know differentiation?

[wooden flare]

yea

[grave mirage]

Ok so

[grave mirage]

If you differentiate a fxn

And the derivative is always greater than or equal to 0

OR

always less than or equal to 0

[supple hill]

That's it.

[grave mirage]

It is one to one

Use this for x^n + x

[wooden flare]

... wha

[grave mirage]

Visualise

[wooden flare]

how do we come to this conclusion?

[grave mirage]

If a fxn is both increasing and decreasing

There has to be a point

Where the fxn went from increasing to decreasing

Do you get that idea?

[wooden flare]

a point of inflection right?

[grave mirage]

Critical point

[wooden flare]

or that

[supple hill]

Yeah.

[grave mirage]

A function doesn't go from increasing to decreasing on an inflection point

[grave mirage]

Basically a minima or a maxima

[wooden flare]

so set f'(x) = 0?

[grave mirage]

Yeah

That works

No

Actually

Set

f'(x)≥0

Or f'(x)≤0

[wooden flare]

hmm ok, lemme try

[grave mirage]

And see if it holds

If it does, you have an one to one

If it doesn't, it isn't a one to one

[wooden flare]

this looks harder to solve no? x^(n-1) >= -1/n

[grave mirage]

Nope

[grave mirage]

Take cases

When n is even or odd

What's the domain on n?

[wooden flare]

wait 1sec

[grave mirage]

Huh?

[wooden flare]

n {R\{0}}

[grave mirage]

It's in the set of real numbers?

[wooden flare]

yea

i dont have the stuff to type it out atm

[grave mirage]

I thought it was natural

[wooden flare]

cant it be any number?

[grave mirage]

What's given

[wooden flare]

its not a discrete function right?

its just n>0 is our condition for this qustion

[grave mirage]

Well it wouldn't be either ways

x is not discrete

fxn is in terms of x not n

[grave mirage]

Is it a self made question?

[wooden flare]

no

[grave mirage]

Can you show the original question?

[wooden flare]

question 2

[grave mirage]

x^3+x is definitely one to one what do they mean 🥀

[supple hill]

Now it's clearly.

[grave mirage]

Tell the teacher it's wrong

[wooden flare]

yea it is wrong

like this textbook is cooked 😭

thats def not one-to-one (trust)

[grave mirage]

🥀

[wooden flare]

lemme check the answer rq

[grave mirage]

Also it becomes veryyy complicated if you take n in R

[supple hill]

No, because if n=a non even it will be non.

[grave mirage]

Atleast at a starting level

[wooden flare]

guys guys

this textbook is ragebait

the answer is proof

[grave mirage]

😭😭😭😭😭😭😭

[supple hill]

Oh.

[grave mirage]

Proof by proof

[wooden flare]

so would we just prove, if n is odd f(x) is oto, and if n is even f(x) is not oto

[grave mirage]

Yeah

But n is given in R

[wooden flare]

R+ \{0}

[grave mirage]

So you'd have to first see if it is rational or irrational

[wooden flare]

i assume its only discrete n

bc what is x^2.7 even mean

[grave mirage]

So {N}

[wooden flare]

more than likely yea

[grave mirage]

(x^27)^1/10

[supple hill]

Yeah, it's a good option.

[grave mirage]

Ok ye then

We just look at natural numbers

[wooden flare]

ive already shown for n=1, x=y. for n=2 im stuck on (x+1/2)^2 - (y+1/2)^2 = 0

[grave mirage]

Use difference of squares

[wooden flare]

oh yea that does work here

[supple hill]

Good.

[wooden flare]

ok so

x=-y -1 or x=y

[grave mirage]

So if x+y=-1, the equality still holds, thus it is not one to one

[wooden flare]

since it has one case where its not oto, then the entire thing isnt oto right

[grave mirage]

Yes

[grave mirage]

It's really infinite cases

But yes

It it's even one

It's not

[wooden flare]

so thats for odd and even, now we extend it to general cases? if n=2k+1 and n=2k

or do we not need to do that in this case

[supple hill]

It's not necessary because, if it fails in at least one case, it will be non-one-to-one.

[wooden flare]

that makes sense

[supple hill]

Do you have more questions?

[wooden flare]

the others seem doable

i hope

tysm for the help!

[supple hill]

You're welcome! Have a nice day.

🙂

[wooden flare]

.close

[pearl pond]

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[pearl pond]

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

[graceful marsh]

This is a solved example from a textbook and i dont quite understand the solution given in the textbook

and i don't know where to begin

[wary bane]

well since they're rationals, my approach would be to use that

although that might not help

without further context / book solution its hard for me to tell what you are or aren't allowed to use

[crimson veldt]

think that any relation lets me reduce everything to degree <=1 forces x rational

the key fact is that { 1, p^(1/3),p^(2/3) } are linearly independent over Q

which followes from the minimal polynomial of p^(1/3)

since p is not a perfect cube

x^3-p is is irreducible over Q

if factored it would have a rational root

[graceful marsh]

that sounds like very advanced math

[crimson veldt]

thus some rational q with q^3=p which contradicts p not being a perfect cube

can u show the solution from the textbook?

[graceful marsh]

This is the solution given in the textbook

[wary bane]

so we need more context to determine what you're allowed to use or not

so you need to use viete's identities to solve this

[graceful marsh]

sorry i am unfamiliar with those

[wary bane]

that's what that solution looks like its using to me? maybe not

oh no its not, mb

I mean - I follow what the solution is doing, but are you stuck trying to understand it
or worried you wouldn't be able to come up with it on your own

Which bit don't you understand

[crimson veldt]

so (1) is the og equation ,multiply through by p^(1/3) to get (2). so it's shifting each term up by one power
hen bx(1) - cx(2) is an elimination.it removes the p^(2/3) terms and leaves you with only p^(1/3) and rational numbers in equation (3)
since sine p^(1/3) is irrational the only way (3) holds is if both coefficients are 0 independently..so b^2 - ac =0 and ab = c^2p

[graceful marsh]

i am not understading the solution

[graceful marsh]

oh

[wary bane]

which part

[graceful marsh]

well now i do understand it

[crimson veldt]

so we got
c^4p^2=a^2b^2=a^3c, if c!=0 you can divide through and get p^2=a^3/c^3, meaning p^(2/3)=a/c is rational, which forces p^(1/3) to be rational too

[graceful marsh]

i think the b*(1) - c*(2) part was confusing me

but it is irrational, so every cofficient must have been 0?

[crimson veldt]

so a contradiction since p isn't a perfect cube so c=0 & then the og equation collapses to a+bp^(1/3)=0 which immediately gives a=b=0 since p^(1/3) is irrational

[wary bane]

they are equating coefficients on 2 sides of a polynomial equation

since the rhs is the 0 polynomial, all coefficients on the lhs must equal 0

[graceful marsh]

hmm

thanks yall

🙏

.close

[pearl pond]

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[tawdry charm]

i need help understanding this task i have due

[tawdry charm]

like what math i should apply

[sterile python]

Maths methods here? Wow

Basically derivatives

[tawdry charm]

yes

i dont get the 25-30s bit?

and the "within the allowable ranges" part, what are the ranges?

[sterile python]

So you find heights of infernos to find total distance travelled

[tawdry charm]

mhm

am i basically sketching a time-displacement graph

[sterile python]

Yes

You need to find h(t) and derivatives

Acceleration and velocity too

[tawdry charm]

no integration?

[sterile python]

No i dont see anything that need integration

I havent done a method assignment ever, i did double specialist only

[tawdry charm]

it says vertically up and down twice each, uh how am i supposed to do that? with like four different functions thata re going to differ

idk im really dumb at math ahhh

[sterile python]

First you might need to find max height and min height

Note that 10% taller sth and 20% lower sth there

You could construct a function on desmos that starts at lowest point (lowest height?)

Go up twice, down twice? What does this suggest about polynomial structure? Quartic, cubic, or more?

Include that in function formula as well

It also reach highest peak (highest height) only once

And total time between 25 -30s, which indicates an interval

[pearl pond]

@tawdry charm Has your question been resolved?

[pearl pond]

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[crimson haven]

Question 3 pls

[somber night]

u dont need to calculate tough

[supple hill]

!nosols

[pearl pond]

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.

[supple hill]

What have you tried?

[somber night]

he sent it

on the right

[supple hill]

Yeah, I didn-t see, excuse me.

🙂

[crimson veldt]

b3

[somber night]

do u have to calculate it

[crimson haven]

Uh no we use the Cartesian plane

[somber night]

cz if not just imagine the unit circle going 180 degrees to the left

like a pendelum

[crimson veldt]

uhm

[somber night]

?

[crimson veldt]

why do i get a different result

[somber night]

show ur work

[crimson haven]

Mines wrong

[crimson veldt]

wait a sec cause im on phone

[cosmic sinew]

how did that turn to that?

[crimson veldt]

they got the demoninator wrong i think

[crimson haven]

Yes I just realized

Tysm

Pls help me with question c

I found Ira angle but idk what else to do

[rough forge]

nice bunny

[prisma kernel]

have you solved (i)?

[agile hedge]

check if this matches

[crimson haven]

No

[crimson haven]

Won’t it be in the third quadrant?

[prisma kernel]

well, !nosols

wait

!noans

[pearl pond]

The purpose of this server is to help you learn; please don't ask for direct answers. Ask for guidance, explanations, or feedback instead.

[prisma kernel]

damn it

!nosols

[pearl pond]

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.

[agile hedge]

for ii,

sin^2(180-theta) = sin^2 theta (identity)

and 1-sin^2 theta = cos^2 theta

and using i

cos theta = 2/rt 5

so cos^2 theta = 4/5

[prisma kernel]

dude, guide her to the answer

[agile hedge]

since we are using pythagorus, we cannot expect it the length to be in negative

[prisma kernel]

don't just show it at once

[agile hedge]

ah

mb

[prisma kernel]

no problem man

[crimson haven]

I got 5

Oops wrong questions

[crimson haven]

So I put the values given in to cos and 1-sim^2

How do I do d2)

[prisma kernel]

well sin^2(pi + theta) = cos^2(theta), and 1 - cos^(theta) = sin^2(theta)

[prisma kernel]

okay let me see

[prisma kernel]

yeah you did it correctly here

[crimson haven]

But it says 17

[prisma kernel]

wait, you squared it, so it should be just -8 - 9

[prisma kernel]

can you confirm if it's -17?

[crimson haven]

Yeah

[prisma kernel]

yup, then this should do it

(-3)^2 becomes 9, so it's just -8 - 9, and not -8 - (-9)

[crimson haven]

Oh I did it wrong

[crimson haven]

Tysm

How do I do question f pls

[prisma kernel]

god okay let me think properly

okay okay, so sin(alpha) is positive and lies in second or third quadrant, but sine is positive so we know it lies in second quadrant

similarly, cos(beta) is negative and lies in first or second quadrant, but cosine is negative so we know it lies in second quadrant too

so, alpha and beta are both in second quadrant

now, what is cos(pi + alpha)?

@crimson haven

[crimson haven]

Huh

Wdym cos (pi +alpha)

[prisma kernel]

ahhh, you can say cos(180 + alpha)

[crimson haven]

-cos

[prisma kernel]

okay nice, so it's just -cos(alpha)

and tan(180 + beta) [hint : if you don't know it directly, convert it into sin/cos and then find it]

[crimson haven]

Positive tan

[prisma kernel]

correct

okay, so it's just tan(beta) - cos(alpha)

[crimson haven]

Oh

Wow

I sunk at thinking

[prisma kernel]

where they both are negative because alpha and beta are both in second quadrant, and cos and tan are negative in second quadrant

so yeah, just convert sin(alpha) into cos(alpha) {use the negative value of cos}, and convert cos(beta) into tan(beta) {use the negative value of tan, convert cos into sec, and then apply tan^(beta) = sqrt(sec^2(beta) - 1)}

[prisma kernel]

don't we all sometimes

[crimson haven]

Tysm

.close

[pearl pond]

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[pearl pond]

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[frank urchin]

How do we see invariant subspace in rotation?

[pseudo oxide]

invariant = doesn't change

[frank urchin]

x axis

y axis

[pseudo oxide]

i.e., after rotating it, the subspace stays the exact same

[frank urchin]

??

[pseudo oxide]

nope

try again

[frank urchin]

(0,0)?

[pseudo oxide]

it needs to be a 1 dimensional subspace

are there any other vectors, apart from the 0 vector, that are invariant under rotation

[frank urchin]

I guess there is none

[pseudo oxide]

correct!

[frank urchin]

a line which is rotated 90°

[pseudo oxide]

therefore how many 1 dimensional subspaces are there

[frank urchin]

0

[pseudo oxide]

won't remain the same

yes,

there are none

precisely

[frank urchin]

If i change the degree

180 or 360 like

[pseudo oxide]

?

[frank urchin]

Rotation when 180° or 360°

[pseudo oxide]

ah, yes

then there are indeed invariant subspaces, but notice that rotation by 360 is just the identity matrix

and rotation by 180 is the negative of the identity matrix

[frank urchin]

By 180° can we make any invariant subspaces?

R^2 is an invariant subspace

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[crimson haven]

Question f

How do Ik which too use

[somber night]

wdym by wich

[crimson haven]

There sin and cos given

Idk if I’m to use the values

And how

[toxic remnant]

can you simplify cos(180+alpha) and tan(180+beta)

[crimson haven]

No?

Wait wdym

[supple hill]

<@&268886789983436800>

[crimson haven]

Oop

[wary bane]

??? is that necessary

[toxic remnant]

you have been given alpha and beta, not 180+alpha or 180+beta

no but it'll be easier to see

[wary bane]

the point of the question is to surely use CAST to figure out the sign

/ correct quadrant

you're told sin a = 3/5

what quadrant must angle a be in?

given this information

[crimson haven]

1st or 2nd

[wary bane]

yes

[toxic remnant]

i mean that is what i mean by simplifying

[wary bane]

but then you're also told this

so what quadrant is angle a?

[crimson haven]

2nd or third

[wary bane]

yes but combining the 2 pieces of information together

[crimson haven]

Second

[wary bane]

ok do the same for beta

[crimson haven]

1st

[wary bane]

doubt it

cos is not negative in the first quadrant

[crimson haven]

Oh yeah mb

End of third quadrant

2nd

[wary bane]

its in the 2nd quadrant

[crimson haven]

Ok

[wary bane]

so they're both in the 2nd quadrant

[crimson haven]

Oki

[wary bane]

that should then help you simplify the expression then.

[crimson haven]

Not rl
Y

Not rlly

Is implied the cow(180+a) +tam(1+b)

Bit idk what else to do

[wary bane]

if you add 180 degrees to an angle in the 2nd quadrant

which quadrant does it end up in

[crimson haven]

Third

[wary bane]

4th

[crimson haven]

Huh

[wary bane]

adding 180 degrees is like going from 10 o clock to 4 o clock

on a clock

?

10 11 o clock are in the 2nd quadrant

4 5 o clock are in the 4th quadrant

[crimson haven]

My teacher taught as to start at 180 and add by one

[wary bane]

???

No: I'm asking

given angle alpha and beta are in the 2nd quadrant

[crimson haven]

Huh

[wary bane]

if you add 180 degrees to either of them

where is the angle after?

[crimson haven]

Why are we adding 180

[wary bane]

because look at the question

[crimson haven]

I still don’t get it

[wary bane]

do you understand why i'm asking you which quadrant 180 + a is in?

or not.

you need to know in order to figure out whats the sign of cos(180 + a)

[cosmic sinew]

hm.. do you agree alpha lies in second quandrant, second quadrant starts from 90 degree and ends at 180 ?

[crimson haven]

I do

[wild wren]

So first we establish where α and β are, because that gives us an idea about their sign
And then we want to establish where (α+180) and (β+180) are

Because +180 is just moving by two quadrants

Aka half a cycle

We know that the unit cycle has four quadrants, each quadrant has certain properties i.e cos and sin being positive/negative

That's usually the visual approach, it's also the easier to understand

||You can also think of it as cos (α+π) which should be an identity, but I suggest following the previous approach||

Try to draw a unit circle and plug random values of α and β that follow the condition, and then add 180°

One way to do that is to draw a line connecting that angle α or β with the center of the circle, and check where it intersects with the other side

[crimson haven]

The thing is

Is don’t understand moving by 180

It’s different to what I was taught

[cosmic sinew]

what do you not understand?

[cosmic sinew]

again can you answer this?

[crimson haven]

Yes

But what I mean is that my teach made as start from 180

And move forward once

So I’d move form the second to the third quadrant

[wild wren]

That's pretty much what it means to move 180°

[wary bane]

(from the first quadrant)

[wild wren]

Yes that's just a clarification of what moving 180° means, regardless of where your initial angle is

So in this example, α was in the 1st quadrant, we can see that (α+180) is in the 3rd quadrant

Clear enough?

[crimson haven]

No

Why did it mov 180

[cosmic sinew]

its simply wherever you are you reach the opposite quadrant

[wild wren]

We were told to compute for α+180, so we looked into it

[wild wren]

Very good

[crimson haven]

It moved backwards

[wary bane]

aka if theta is in quadrant 1

[wild wren]

Quadrant 1 θ

But what if θ is not in the first quadrant?

How can we adjust

[wild wren]

@crimson haven I think a good way to go about this is: try to think why this was made

[crimson haven]

Ok

[wild wren]

Why theta +180 was there

And why theta - 180

Oh and remember that those are just degrees, and that adding a degree is basically a rotation

[crimson haven]

For these my teacher makes as start on the lines

[wild wren]

So adding 180 is just turning around

[cosmic sinew]

theta is acute, so it lies between [0,90] if you add 180 + theta it will range from 180 +0 <= 180 +theta<= 180+ 90
180<= 180+theta <=270 which you can see is the 3rd quadrant
consider case where theta is >90 but < 180
as you can see the bounds change again and you moveto the 4th qyadrant

180 + 90 < 180 +theta< 180+180
270 <180 +theta< 360

[crimson haven]

Is that a Ute a right angle?

[cosmic sinew]

UTE?

[crimson haven]

acute

[cosmic sinew]

acute angle means theta lie in 1st quadrant or 0< theta < 90

[crimson haven]

Ok

[pearl pond]

@crimson haven Has your question been resolved?

[midnight haven]

ts looks so hart

hard

😭

[crimson haven]

It is

I have to go tho

[pearl pond]

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[blissful brook]

yo

[blissful brook]

not sure how to do b

this is how far i got

[terse root]

is this an A-level paper by any chance?

[blissful brook]

yes

this is the only question in this paper that i didnt understand

and its 3 marks

[terse root]

is this normal or futher? which exam board?

[blissful brook]

edexcel normal

[terse root]

ah

[light helm]

apply that the line l passes through the origin

[terse root]

so y = mx and not +c ig

m = the thing u got in part a

ig

[light helm]

use the non-derivative way to get the slope

[terse root]

to show something is the solution i suppose u need the factor theorem?

[light helm]

no

[blissful brook]

wdym? like y1 - y2 / x1 - x2 ?

[light helm]

(), but yes

basic slope formula

[blissful brook]

using P and 0,0?

[light helm]

yes

[blissful brook]

ok

lemme try

ok i got it ty

.close

[pearl pond]

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[frank urchin]

I asked the question yesterday but i went to market

How do I find the congruent circles radius?

[pearl pond]

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

[frank urchin]

for circle i can see

tan(theta/2)=r/a

tan(theta/2)=r/(25-x-2r)

I don't know why i am thinking about this angle

[pearl pond]

@frank urchin Has your question been resolved?

[pearl pond]

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[safe prairie]

im finding expectation given a geometric distribution, and i got my hands on a summation of a^x/x. how do i proceed?

[safe prairie]

im unsure what limits to put into the highlighted part if i int wrt x, or if i just use t instead then after swapping int and sum idk how to change the limits of summation to fit the needs of sth thats written wrt t

[supple hill]

I recommend you use that identity: $$\frac{1}{x+1} = \int_0^1 t^x , dt$$

[jolly parrot]

Ga³¹Br³⁵I⁵³9000✞

[pearl pond]

@safe prairie Has your question been resolved?

[safe prairie]

so now its this

[pearl pond]

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[safe prairie]

NO

.reopen

[pearl pond]

✅ Original question: #help-39 message

[safe prairie]

i dont know how to swap summation and integral in this form

well idk even in the form of sum(int(x dx)) neither when it comes to the integration limits but thats for later

[supple hill]

Yeah, do you have any ideas how to continue?

[safe prairie]

no

first one looks like typical geometric sequence but i cant use it

cuz id need to separate the front part from the back then

[supple hill]

The objective is have: $\sum r^n$

[jolly parrot]

Ga³¹Br³⁵I⁵³9000✞

[supple hill]

Right?

[safe prairie]

if this happens itd be nice but idk if thats what we are looking for

[supple hill]

Yeah, because we can apply the sumatory distributive.

[safe prairie]

idk what that is

taking out a constant?

to take out a constant id need a - or + to appear in my summation

[supple hill]

Yes, it is.

[cosmic sinew]

alternately you can sub (1-p) = u then use the taylor series for ln(1-u) change the index for n = x+1

[safe prairie]

wait no by then itd have already been separated

[supple hill]

Yes you can use that.

[safe prairie]

should i have remembered the taylor series form for lnx then

i never remembered any

[cosmic sinew]

not really

you can use the way they suggested

[safe prairie]

i dont understand their way yet

[supple hill]

You can apply that if you don't want the Taylor serie.

$$\frac{1}{x+1} = \int_0^1 t^x , dt$$

[jolly parrot]

Ga³¹Br³⁵I⁵³9000✞

[safe prairie]

what i meant to say is idk how to proceed after this

[supple hill]

Ok.

[safe prairie]

working on it

ok yeah i do its just power rule

[supple hill]

You can move the integral out and the sum inside.

[safe prairie]

whereas here i think i make the mistake of assuming its power rule but its actually int a^x instead

[supple hill]

Yes, it's correct.

[safe prairie]

like this?

[supple hill]

Exactly, it is the exchange property.

[safe prairie]

how does that work

the mess i put before int is not a constant

wait i should check my work again

i got here, which is cleaner then my first photos i think

but with taylor series i got this what i think should be a final answer

[safe prairie]

anyways i dont know how to adjust the bounds of the operators

[safe prairie]

deepseek says this is close except its -lnp so i guess i got it

[safe prairie]

my devices are running low on battery so i think ill stop working for now but i still wanna hear how this works on my phone

[cosmic sinew]

this does not feel right what happened here?

[supple hill]

Np.

I think you missed a negative sign in Taylor.

[safe prairie]

Subject is (1-p) instead of t
Or must I use t to make swapping possible

[safe prairie]

Yup

[supple hill]

Yeah.

[cosmic sinew]

this is the expansion of ln(1-x) what you have is -ln(1-(1-p))

[supple hill]

Do you see it better now?

[safe prairie]

No cuz even if it’s t the summation concerns x instead

[cosmic sinew]

ok, no it will work if you dont take t

[safe prairie]

So it’s fine
But still from double int I learnt that swapping order means redefining bounds

Uh afk

[supple hill]

Yes, it's fine.

[cosmic sinew]

hm.. i do not know multivariable calculus much lol i step back

sorry

[safe prairie]

cuz at this stage int wrt t and sum wrt x does feel like two unknowns

[safe prairie]

so how would redefining bounds work in this context?

[supple hill]

Here it's not necessary since x and t are totally independent. Right?

[safe prairie]

oh alr

[supple hill]

Do you understand better?

[safe prairie]

better now

my first mistake is making int t^x wrt x instead of wrt t, and swapping like this doesnt need to change bounds

[supple hill]

Exactly.

[safe prairie]

but last question, now ik sum(int(x)dx)) can be swapped, what about sum(x int(x)dx))?

[supple hill]

To be valid, the x that is outside the integral must be treated as a constant by the integral.

[supple hill]

Do you have more questions?

[safe prairie]

sounds confusing

sum(t int(x dx)wrt t) works, but sum(x int(x dx)wrt x) doesnt?

[supple hill]

Yes, because the order.

[safe prairie]

Maybe sumt(t int(x dx)) is clearer, so this can become int(t sumt(x)dx)??

Then sumt(x) just multiplies x by range of t, and int whatever dx would look off

Then looks like it’s still impossible

[supple hill]

The sum cannot depend on the same variable as the integral.

I recommend you use the x for the sum and t for the integral.

[pearl pond]

@safe prairie Has your question been resolved?

[pearl pond]

Channel closed

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[sterile python]

What have you done

[cosmic sinew]

im about to post the same gif

[sterile python]

Please no

I am asking what have he done, not to teach him now

[subtle crow]

Ohh sorry

[sterile python]

We need to know what he up to yet

[lament eagle]

Thank you

[cosmic sinew]

you good im jk

[sterile python]

Thank you, what have you done?

https://tenor.com/view/the-batman-what-have-you-done-gif-25280767

[lament eagle]

I'll use the Pythagorean theorem to do this

[lament eagle]

So far I've found the volume of the top cone

AD² = 3 -> AD √3cm

[subtle crow]

No no

That would make it lengthy

You see, finding the area of the rest of stuff will be complex

I'd say you're better off doing this

First find the initial volume of the cone (before stuff was removed). I believe you can do that. Then, find the volume of the cylinder removed. Now,
Final volume = initial volume - volume of cylinder

[lament eagle]

Understand it now

Thank you

[sterile python]

No he did correct, he got AD

[lament eagle]

.close

[pearl pond]

Channel closed

Closed by @lament eagle

Use .reopen if this was a mistake.

[subtle crow]

Oh did he? Well mb then

[pearl pond]

Available help channel!

Send your question here to claim the channel.

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

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

[smoky gull]

A person has 6 friends and during a certain vacation he met them during several dinners. He found that he dined with all 6 exactly on one day, with every 5 of them on 2 days, with every 4 of them on 3 days, with every 3 of them on 4 days, with every 2 on 5 days. Further every friend was present at 7 dinners and every friend was absent at 7 dinners. The number of dinner he had alone is equal to?

[sterile python]

What do you think? What have you done?

[smoky gull]

!status

[pearl pond]

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

[sterile python]

Let N be total number of dinner

Sk is sum of the number of days he dined with a specific combination of k friends

[smoky gull]

no i think number of dinners is given in question i just know what it may be

[smoky gull]

but sure okay

[sterile python]

The number of ways to choose k friends from 6 is 6ck right

[smoky gull]

hm

[sterile python]

But 6 friends

[smoky gull]

@sterile python i appreciate it but please stick to 1-2channel at a time else other helpers might assume you are currenty helping me

[sterile python]

I am bro

[smoky gull]

K sorry then

[sterile python]

So you need at least 2 friends to dine with

[smoky gull]

no only 1 is also fine

[sterile python]

So you take 6C6+6c5+...6c2 is total of ways to find k

[cosmic sinew]

you are underestimitating his power

[sterile python]

N=7+7 where absent and present meals

[smoky gull]

okay and?

[sterile python]

Now find s1

[smoky gull]

wait are you sure abut this tho

[sterile python]

Yes

We know each of the 6 friends was present at 7 dinners right

[smoky gull]

yeah

[supple hill]

And absent in 7.

[rough forge]

I thought you were @rigid mist

[sterile python]

Then s1 is sum of dinners of each people attend

6*7=42

[smoky gull]

but that doesnt mean there cant be more righ?

oh wait nvm yeah okay go on

[sterile python]

Then use Principle of Inclusion-Exclusion

[smoky gull]

ik thqt dude but sry not him

[cosmic sinew]

confusing wording he dines with every 5 on 2days so with all set of 5 people on 2days or what how many times is he dining

[smoky gull]

doesnt come out

[rough forge]

You are the good yajat dw

[smoky gull]

tried this

[sterile python]

Try

[smoky gull]

I tried

[supple hill]

And what did you get?

[sterile python]

P=S1-s2+s3 -s4+s5-s6

[smoky gull]

so i gto 13

just checked again

[sterile python]

Correct

[smoky gull]

yeah butanswer isnt 1

[sterile python]

Now find dinners alone

Huh

Thats bs

[smoky gull]

yeak ikr

[supple hill]

It's right.

[sterile python]

Then idk theoretically its 1

[supple hill]

Yeah I calculated and it is.

[smoky gull]

?

[vestal tapir]

i don't get it

[sterile python]

Get what

[vestal tapir]

like i can assume 2 days includes the 6-friends day, so there was only one day where he met 5 specific people

it still means some person went to 6 dinners already

how can it add up to 7

[supple hill]

Can you send us your process?

It can be clearly to see what happened.

[smoky gull]

yeah gimme a sec to type it out in notes using latex it

[supple hill]

👍🏻

[frank violet]

hmmm, every 5? exactly 5 or just 5?

if it's not exactly 5

Then when you meet all 6 of them you also meet 5 of them and that's counted as 1

[vestal tapir]

it's just nonsensical

[smoky gull]

Let At,t=1,2,3,4,5,6 be the set of days on which the friend is present at dinner & Bt be the set of days on which friend is absent at dinner
Then
|At|=|Bt|=7

Also
∣∣Ai∩Aj∣∣=5,∣∣Ai∩Aj∩Ak∣∣=4∣∣Ai∩Aj∩Ak∩Aℓ∣∣=3,∣∣Ai∩Aj∩Ak∩Aℓ∩Am∣∣=2
& |A1∩A2∩A3∩A4∩A5∩A6|=1
Where i,j,k,ℓ,m vary from 1 to 6& distinct number of days/dinners at which at least one friend was present =|A1∪A2∪A3∪A4∪A5∪A6|
∑|Ai|−∑∣∣Ai∩Aj∣∣+∑∣∣Ai∩Aj∩Ak∣∣−∑∣∣Ai∩Aj∩Ak∩Aℓ∣∣+∑∣∣Ai∩Aj∩Ak∩Am∣∣−∣A1∩A2∩…..=6C1×7−6C2×5+6C3×4−6C4×3+6C5×2−6C6×1=13
Total number of dinners =|At|+|Bt|=14

Aw frick

one sec lemme find a latex software

[sterile python]

https://tenor.com/view/alien-extraterrestrial-movie-scene-gif-6746292148918175301

Fuckass alien language

[supple hill]

Yeah, better.

[vestal tapir]

oh wait so
with every 4 of them on 3 days is just covered already we ignore that

i don;t get it anyway

[smoky gull]

sorry could you not curse😅

[sterile python]

Skibidi

[smoky gull]

Tryna keepit family friendlu,y

[supple hill]

It's correct.

[smoky gull]

😭😭

[supple hill]

What do you not understand?

[smoky gull]

Sorry?

[vestal tapir]

i don't understand anything

[supple hill]

What's your question about the problem?

[smoky gull]

the answer is wrong😭😭

its not 1

what?

[supple hill]

Nothing, I was thinking.

[vestal tapir]

basically there was 7 dinners, and everyone missed one dinner
so then he had 1 one on one with each one, so everyone missed 7

so it's 0