help-0

[winter light]

I believe there's a mistake, the very last bit should be dx not dt

Nooo, this is one of the worst mistakes you can do in math 😭

$$\frac{A}{B + C} \neq \frac{A}{B} + \frac{A}{C}$$

[ocean seal]

Alberto Z.

[winter light]

Instead, this rule is correct: $$\frac{B+C}{A} = \frac{B}{A} + \frac{C}{A}$$

[ocean seal]

Alberto Z.

[winter light]

If $y = \frac{1}{2 - x}$, what is $y'$ (or if you prefer $\frac{dy}{dx}$)?

[ocean seal]

Alberto Z.

[winter light]

$$y = \frac{1}{2 - x} \implies y' = \frac{dy}{dx} = \frac{1}{{\left(2-x\right)}^2} $$

[ocean seal]

Alberto Z.

[winter light]

Do you agree with this?

Chain rule

Because you have to multiply by the derviative of 2 - x, which is -1 (and this minus cancels the previous one)

Yes

$\frac{d^2y}{dx^2} = y''$ simply

[ocean seal]

Alberto Z.

[winter light]

It's another notation for the second derivative

Nope

$$\frac{dy}{dx} = \frac{1}{{\left(2-x\right)}^2} \implies \frac{d^2y}{dx^2} = \frac{2}{{\left(2 - x\right)}^3}$$

[ocean seal]

Alberto Z.

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[crimson tundra]

How do I find the quantile value of chi square with 5 degrees of freedom and 0.1 significance level

[crimson tundra]

Using table

[zealous lichen]

find out the p value

then use the table

[crimson tundra]

How do I find the p value?

P = 1 - alpha and that is 0.9

[zealous lichen]

umm

that's for single tail

[crimson tundra]

How would I calculate it for single tail then

[zealous lichen]

so p =1-0.1=0.9

then look for df =5 and p =0.9

[crimson tundra]

In the table I have only degrees of freedom and level of significance

[lone heart]

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[crimson tundra]

.close

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[vapid shuttle]

I'm having trouble understanding this proof

particularly starting at the last line, when it says, U intersect S closure is a subset of U intersect V closure equals the emptyset

Also, I don't see why they ignore this case. How is it demonstrated that neither set intersects the closure of the other?

this is my other confusion

[oak perch]

S in V

Therefore closure of S in closure of V

take intersection with U both sides

[vapid shuttle]

are you explaining the top image cogwheels

[oak perch]

Your last confusion

The one you put why under

[vapid shuttle]

ah

and U intersected with the closure of V is empty why?

because we suppose U and V form a disconnection of S closure

nevermind

[oak perch]

Because closure of S is disconnected

[vapid shuttle]

okay so what about the first confusion then, I understand that part now

[oak perch]

Still I prefer to transform your condition of being connect to a equivalent but more convenient form

What is your version of definition of being connected again?

[vapid shuttle]

it is connected if it is not disconnected, and it is disconnected if it is the union of two nonempty sets that do not intersect the closure of eachother

[oak perch]

So

Intersection (U and cl(V)) is empty

[vapid shuttle]

no

[oak perch]

This gives us U contained in X-cl(V)

[vapid shuttle]

that can't be true

because U and V form a disconnection of S closure

oh sorry you are right

[oak perch]

I mean the condition of disconnectness

Then X-V=U in X-cl(V) in X-V

Therefore U=X-cl(V), meaning U is open

[vapid shuttle]

what is X

[oak perch]

Similarly V is open

The total set X

[vapid shuttle]

total set?

[oak perch]

X is disconnected

Then there exists subsets U,V of X satisfying your condition

[oak perch]

Then this shows U is open in X

Similar V is open

[vapid shuttle]

I'm not following this

I don't know what X is, or why it matters that any of our sets are open

[oak perch]

… never mind then I use your notation then

[vapid shuttle]

okay

[oak perch]

Because if they are both not non-empty, and it isn’t a disconnection of S

Then we have S=(S intersection U) union (S intersection V). S is union of two non-empty subsets. This isn’t a disconnection meaning

Cap is notation for intersection, cup union right?

[vapid shuttle]

yes

[oak perch]

Then

cl(S cap U) cap (S cap V) is non-empty or (S cap U) cal cl(S cap V) is non-empty

Contradiction

[vapid shuttle]

what is that contradicting?

[oak perch]

Because cl(S cap U) cap (S cap V) is contained in cl(U) cap V

The latter is empty

So the former can’t be non-empty

Similar for the another one

[vapid shuttle]

why is it contained in cl(U) cap V

[oak perch]

S cap U in U
S cap V in V

[vapid shuttle]

I see

so that is why neither of those can be nonempty

[oak perch]

Yeah

[vapid shuttle]

so then from this earlier

we conclude that

this means that U must be empty

and that contradicts U union V forming a disconnection for S closure

right?

[oak perch]

Yeah contradiction

Yeah

[vapid shuttle]

this is very difficult to grasp

but I feel closer

[oak perch]

That’s why I was

Trying to give you a more convenient equivalent definition of being disconnect

But anyway you can prove it yourself

X is disconnected <-> X is disjoint union of two non-empty open subsets <-> X is disjoint union of two non-empty closed subsets <-> X is disjoint union of two non-empty subsets who are both open and closed <-> X has a proper subset that is both open and closed

But anyway, they are all equivalent, it’s just I personally feel these are more convenient to use

[vapid shuttle]

Sorry Cogwheels I was just trying to completely wrap my mind around what'd we just done and get the proof all settled

reading now

ty

[oak perch]

Np

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[vapid shuttle]

@oak perch my completed proof if you were interested in seeing it

ty again for your help

I was having lots of trouble on that one

[lone heart]

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[rancid tartan]

construct a box and whisker plot to display the number of pets owned by a number of randomly chosen students. The data are, 2, 0, 5, 1, 2, 1, 0, 8, 4, 3, 9, 1, 2, 3, and 1

[lone heart]

Please don't occupy multiple help channels.

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[worn hinge]

...

[tropic grail]

Nevermind, we posted a problem at the same time

[worn hinge]

Hi, so why was ls 2 delta r^2 like that? shouldn't it have a pi as well?

when it was distributed?

or there is some rule I don't lnow about this

it should be pi delta r^2

this part

[tropic grail]

Why is the left part delta A / delta x?

[worn hinge]

idk as well

it should be delta r right?

[tropic grail]

I thought so

[lone heart]

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[worn hinge]

<@&286206848099549185>

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@worn hinge Has your question been resolved?

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@worn hinge Has your question been resolved?

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[flat roost]

is this the most simplified form or is there anything more i can do?

[tacit arch]

That's fine as is

[flat roost]

oh ok

i have another problem

The cells in a culture in a lab grow according to the logistic equation dP/dt=0.02P(1-(P/2000)). Initially the population of cells is 300. Estimate the amount of cells after 3 hours.

I got 315.6, but the answer says its 315

here are my steps

,rotate

[lone heart]

@flat roost Has your question been resolved?

[lone heart]

@flat roost Has your question been resolved?

[nimble fern]

looks good

except the part written p=1/2000
which should be B=1/2000

[flat roost]

oops my bad

did you get 315.6 too?

[nimble fern]

,calc 6000*e^0.06/(17+3e^0.06)

[ocean seal]

Result:

315.62340475972

[nimble fern]

yea

[flat roost]

oh alr

thanks

.close

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[faint frost]

i have a lot of problems in geoemtry specifically triangle congruence. i have a worksheet that i have left mostly blank cuz i dont understand how to do anything at all.

[nimble fern]

wow so many

[faint frost]

i just dont get it

for the proof problem 7, i know i have to prove the triangles are congruent then use cpctc

and for everything else i have aboslutely no idea

[nimble fern]

oh okay, i never heard of cpctc

[faint frost]

corresponding parts of congruent triangles are congruent

[nimble fern]

oh that

so, you'll have to prove congruent triangles first 👍

https://www.onlinemathlearning.com/image-files/congruent-triangles.png

[nimble fern]

do you know these? @faint frost

[faint frost]

yes

we learned those in class

[nimble fern]

that's good!

[faint frost]

i know i have to use those

but idk how

[nimble fern]

let's start with Q7

what do we need to prove BD=EA?

(i use = instead of the congruence sign for ease of typing)

[nimble fern]

like which pair of triangles should we use?

[faint frost]

uh

bdc and aec

[faint frost]

sure thas fine

[nimble fern]

so, what info will we need?

[nimble fern]

for this to happen

[faint frost]

i think its one of the angles right

its gotta be the included angle

[nimble fern]

nah, we need to angle between 2 lines

[faint frost]

between the 2 corrosponding congruent lines

[nimble fern]

the given 2 lines

yes

[faint frost]

which one tho

oh wait

nvm

bcd and ace

wait a second

@nimble fern since we already know that the corrosponding lines are congruent, bcd = ace

right

what reason is that tho

not sure what the postulate or theorom would be

[nimble fern]

we need 2 coorespdoning lines

and

the angle between them

[faint frost]

yeah

but what reason would that be on the proof

to prove bcd = ace

[nimble fern]

good question, do you know angle sum of triangle / exterior angle of triangle?

[faint frost]

the exterior angle is the sum of the 2 remote interior angles?

is that what ur talking about?

[nimble fern]

yea

[faint frost]

how would i use that

[nimble fern]

like we know, for now, the base angles of the inner triangle are same

and base angles of the outer triangles are same

[nimble fern]

we can use this to prove
angle BCD = angle ACE

[faint frost]

how do we know that the base angles of the outer triangles are the same

and bcd and ace arent exteroir angles

so i dont get it still

[nimble fern]

ok, I'll go through step by step then

ah, btw can you repost Q7 for easier reading?

angleCDE = angleCED (base angles, isos triangle)
angleCBD=angleCAE (base angles, isos triangle)
angleCDE=angleCBD+angleBCD (exterior angle of triangle)
angleCED=angleCAE+angleACE (exterior angle of trianlge

all good till here?

[faint frost]

yeah i got it

[nimble fern]

good

next step will be

angleCDE-angleCBD=angleBCD
angleCED-angleCAE=angleACE

now that since
angleCDE=angleCED
and
angleCBD=angleCAE

we have
angleCDE-angleCBD=
angleCED-angleCAE

therefore
angleBCD=angleACE

therefore we can finally prove triangle BCD congruent to triangle ACE

all good till here?

[lone heart]

@faint frost Has your question been resolved?

[lone heart]

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[vivid basin]

If a²+1/a² = 786, then a-1/a = ?, I know it can be solved by adding/subtract 2 on both sides, but can it be solved by finding out the value of a, from the first eqn

[ocean seal]

Jshy

[gray isle]

some people respect the order of operations

[vapid shuttle]

I am not one of those people

[vivid basin]

It's a²+(1/a²) and a-(1/a) sry

[gray isle]

no need to apolgise

those () weren't necessary

it'd be pain to solve for a first

[vivid basin]

As I'm good in just solving by linear equations in 2 variables that's why I asked to find a firs

[gray isle]

you could convert the first equation to a quartic equation
for which you'd get 4 solutions, which may not necessaily be nice
and end up doing something like 4 cases

[vivid basin]

I'm not so good in maths so I just know some basic techniques like solving using 2 vars, it would be very helpful if you could solve it using an efficient method and a method which would be in my context of understand like using linear eqns

[tardy stag]

it's a biquadratic so it wouldn't be impossible just annoying

[lone heart]

@vivid basin Has your question been resolved?

[alpine sable]

what part of your question hasn't been resolved

[vivid basin]

I want know the methods by which it can be solved, till now I know just the method of solving it by adding 2 on both sides

[alpine sable]

the most efficient way to solve this problem is subtracting 2 on both sides

[vivid basin]

If I solve the first eqn for a I get √343

Ok np I will use the efficient method

.close

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[vague iris]

Do we always get the same derivative if we simplify the function first before getting its derivative in multiplication and division, and if we use the rules first before simplifying?

[nimble fern]

do you mean like
f(x)=(x+1)⁴
f(x)=x⁴+4x³+6x²+4x+1?

and find f'(x)?

[vague iris]

like f(x)=(x-4)^2 (2x=3)

[nimble fern]

does the 2x=3 a condition for the f'(x)?
I don't get it 😔

[lone heart]

@vague iris Has your question been resolved?

[tardy stag]

i suspect you meant (2x-3)

and yes you'll get the same answer

although you may have to factor one or expand the other, to easily see that they're the same

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[high mauve]

Hey

[high mauve]

Trigger Warning: Arachnophobia

A spider wants to crawl the shortest way from corner a to corner b

So my book says 64 cm

Idk how it got to that

[pulsar laurel]

Is it like a box

[high mauve]

Yes

sqrt30^2+20^2 is 36.0555

[pulsar laurel]

So then it can go up then across diagonally

[high mauve]

yes

That'd be

Sqrt(30^2+40^2)=50

50 + 20

70

But book says 64

[pulsar laurel]

Maybe it can go straight from a to b

[high mauve]

sqrt(40^2+20^2)=44.72

[pulsar laurel]

Pythogoras in 3d

[high mauve]

No it can't, spiders can't fly 😅

[pulsar laurel]

Webs

[high mauve]

Besides, 50^2(diagonal of box) +20^2 (height) is 53.85, so shorter than 64

[pulsar laurel]

64 doesn't sound right

[high mauve]

I have no idea what the book wants me to do

It's a chapter about repeating the Pythagoras

Nothing advanced

[pulsar laurel]

Is there like a solution at the back of the book

[high mauve]

Do you think 64 is an error?

Yep

[pulsar laurel]

Sounds like it

[high mauve]

6.28

Ok I trust you @pulsar laurel thank you for your time!

.close

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[chrome dagger]

How do I do this

brb in 5

dont leave if you come

im back

[alpine sable]

\dm
I think it is to safe to start with the fact that
[
\s{a+b+c} \mathbin{\c r \ne} \s a + \s b + \s c
] in general

[chrome dagger]

ok

ohhh wait

oops

ok but what now

[alpine sable]

\dm
horizontal asymptotes are defined by an asymptotic/convergent behaviour in the function as it approaches $\pm \infty$

[chrome dagger]

yeah

[alpine sable]

\dm i.e. check if $\ds \lim_{x \to \pm \infty} \map f x$ exists

[chrome dagger]

were finding the y value as x aproaches + or - infinity

that's what you said?

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[lone heart]

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[simple saddle]

how interquartile range is calculated here?

[simple saddle]

here in image 1st one denominator is always 365 why so and in second case with replacement approach is used ? what is the reason for this denominator thing

please help!

[heady pollen]

how is interquartile range defined?

or is that the question

hmm

are your messages getting deleted?

[simple saddle]

yeah

[heady pollen]

odd

[simple saddle]

why so

[heady pollen]

idk

never seen auto-deletion before

[simple saddle]

now its ok ig

i thought for a sec i got banned lol

[charred summit]

@simple saddle
https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-data-statistics/cc-6th/v/calculating-interquartile-range-iqr

[simple saddle]

here how they found inter q range i did't understand

[charred summit]

This is a good video for your question

[simple saddle]

no but how to find that if only 2 or 3 terms are there

its understood for 10 and so terms but this i didn't understand

[charred summit]

If there is 2 terms
Then it is simply the difference between them

If it's for 3 terms then it is the difference between the final and the first one

[simple saddle]

oh just like range?

okk

[charred summit]

Yeah

[simple saddle]

can you help with this too @charred summit

[charred summit]

What is the question of the left one?

[simple saddle]

oh wait

[simple saddle]

so instructor showed an exaple for 6 people

[charred summit]

To solve this we need to know how many ways can 2 persons have birthdays

So do you know about combination?

[simple saddle]

yes i do

but i was thiking it would be like (365/ 365) *( 364/364)... like that as without replacement it is? right?

[charred summit]

Nice
So to get all the ways for their birthdays
We do a little easy equation

,calc 365 * 365

[ocean seal]

Result:

1.33225e+5

[charred summit]

,calc 365^2

[ocean seal]

Result:

1.33225e+5

[simple saddle]

but why then the denominator is all 365?

[charred summit]

I don't understand anything about the solution you gave to be honest

So now we want to know how many ways they can have the same birthday which will be
365 ways

[simple saddle]

this they did after that using the logic

[simple saddle]

ya the first person

[charred summit]

Getting the probability of two having the same birthday is
$\frac{1}{365}$

[ocean seal]

Sherif Player

[simple saddle]

bro my doubt was why denominator was not altered like it is when no replacemnet case is there in the second q that happend

[charred summit]

Which gives us the percentage 0.274%
Meaning to get a 100%
That there is at least 2 people have the same birthday

We do

,calc 2 / 0.274%

[ocean seal]

Result:

729.92700729927

[charred summit]

So we need 730 people just to get 100 percent probility to have 2 people of the Same birthday

[simple saddle]

bro

its 23 people

[charred summit]

Hmm

So I can't solve it ?

[simple saddle]

you tried man

but the ans is 23 thats what

i wasn't asking the ans just the denominator concept

noo need to solve

[charred summit]

Hmm I think that it is 365 which represents the year days

[simple saddle]

no why not like this (365/ 365) *( 364/364)...

[charred summit]

Do you mean the way he said
$$\frac{364}{365} \times \frac{364}{365}$$?

[ocean seal]

Sherif Player

[charred summit]

Look I think I understand it now
When
To calculate the probability that bunch of people doesn't share any birthday is easy
We will just take one out of them then say that
The second one needs to be different than this person so it will be
$$\frac{364}{365}$$
Then the third one needs to differ from both of these guys so
And as both of the conditions needs to be applied we will use multiplication
$$\frac{364}{365} \times \frac{363}{365}$$

[ocean seal]

Sherif Player

[charred summit]

And so on

Now to get that at least 2 of them has the same birthday we will just say
1 - the result

So we get
$$1 - \frac{364!}{(365-K)! \times 365^{k-1}}$$

As K is the number of people

[ocean seal]

Sherif Player

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@simple saddle Has your question been resolved?

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[vague iris]

Where did I go wrong?

[hot snow]

honestly looks alright to me

[vague iris]

i think there was a mistake in the answer key. The denominator was raised to positive 2.

Thanks!

.close

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[lethal belfry]

I would like some advice on how to solve the following diophantine equation $x^2+y^2+x-y= 2021$( The question is actually to find the number of diophantine roots, but I would like to find the roots first

[ocean seal]

physicsrocks

[mellow grail]

dont take my advice too seriously, but solving for either variable in terms of the other using quadratic formula should allow u to place a few bounds on the values

[lethal belfry]

Yeah, I'm aware of that, but I was hoping for an alternate solution

I have tried a few tricks. Like$(x+0.5)^2+(y-0.5)^2= 2020.5$ and $(x+y)(x+y-1)-2xy=2021$ But neither seem to help

[ocean seal]

physicsrocks

[mellow grail]

in this case its possible your only option is my way

reason being its not factorizable

[wind cloak]

I mean isn't this just the equation to a circle

[mellow grail]

u sure?

[wind cloak]

You can find the length of the intercept made on the x axis and calculate the distance from the x coordinate of the center

[wind cloak]

well yeah it clearly is

[mellow grail]

+x-y doesnt matter?

[wind cloak]

He's even got it in center-radius form here

[wind cloak]

why would it

[mellow grail]

hmm

[mellow grail]

well then u have radius as sqrt(2020.5)

[lethal belfry]

oh yeah, I don't know how I missed that . Thanks!

[wind cloak]

square root of 2020.5

Is the radius

[lethal belfry]

.close

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[frosty swift]

I have doubts in this question's solution I found online.

[frosty swift]

Somebody did this:

Where did x + 1/x come from?

This is ref page I found from that person.

[alpine sable]

they're using the fact that if f(x)=x+(1/x) then the range of f(x) is [2,infinity)

that's smart actually damn

[frosty swift]

But why though?

[alpine sable]

using the information from the four equations, you can see how they got to l+(1/l)=q-2

[frosty swift]

Yeah.

[alpine sable]

so they see that l+(1/l) is f(l) in this case and its range is [2,infinity)

[frosty swift]

Ok I see.

[alpine sable]

and then they're substituting for l+(1/l) in that inequality to get the range of q

[frosty swift]

But wait, shouldn't the range by -1 to infinity taking only natural numbers?

Or I'm mistaken?

(probably).

[alpine sable]

you can find the range of x+(1/x) by completing the square

[frosty swift]

Oh ok.

[alpine sable]

wait

i think im massively mistaken

[frosty swift]

No.

You're right.

[alpine sable]

wait yeah

im right

ok no i was half right the range also includes the interval (-infinity,-2]

[frosty swift]

But wait, you can't complete the square for this I don't think so.

[alpine sable]

yeah that part was wrong

if l+(1/l) was less than -2, it would imply q-2 was less than -2 or q was less than 0

which isnt possible because q is given to be natural

[frosty swift]

Yeah I got that.

Oh wait.

1/x is a rational function.

So the minimum val. is 1 for natural numbers.

for f(x) = x + 1/x => f(1) = 2

So that's why they did that.

Ah that's acutally smart.

(That's what you said in the beginning).

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[final mural]

Hello, is this the right drawing of the situation

[final mural]

Orthogonal means independent here

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[humble leaf]

Let v1, v2 and v3 be distinct elements of a vector space V over a field F and let c1, c2, c3 ∈ F . Under what conditions is the subset {c2v3 −c3v2, c1v2 −c2v1, c3v1 −c1v3} of V linearly independent?

[humble leaf]

i think the answer is never cause c1(c2v3 −c3v2)+c3(c1v2 −c2v1)+c2(c3v1 −c1v3) is always 0

so they're always a linearly dependent set

someone pliz sanity check, thanks!

[oak perch]

$\begin{pmatrix}v_{1}&v_{2}&v_{3}\end{pmatrix}\begin{pmatrix}0&-c_{2}&c_{3}\-c_{3}&c_{1}&0\c_{2}&0&-c_{1}\end{pmatrix}$

[ocean seal]

Cogwheels of the mind

[oak perch]

So only when v_i are linear independent and that matrix is invertible

Calculate its determinant

Oh seems like you did

[oak perch]

You are correct

[humble leaf]

thats where i got the idea yea

cool thanks 🫡

.close

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[oak perch]

Np

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[alpine sable]

hi can anyone help me with this probabilities question

[alpine sable]

hi

i know this is related to pascals triangle and ncr but i dont fully understand how to do it

[waxen flame]

I think the setup was a bit confusing.

The first sentence basically tells that you are using combinations of four fruit so 4C1 + 4C2 + 4C3 + 4C4.

The next line does not clearly define if Olivia is just adding one fruit to each of the possible combinations or there is an additional fruit to consider in the combinations above. The explanation given would indicate they intended the former.

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

thank you sm

i just got one more question, it might be dumb

but what i did was i looked at pascals triangle and i included 4c0 and 5c0 in my calculations so i did 32-16=16, why did you leave out 0?

4c0?

so i did (1+5+10+10+5+1)-(1+4+6+4+1)

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[viscid lynx]

how do you do this starting with dy first? (It will be two integrals added together), what are the limits for dy?
https://cdn.discordapp.com/attachments/689915677431169069/1146997301240397834/image.png

[oak perch]

I would do it like this:

y=rcos(u), z=rsin(u), 0<=r<=3, u from 0 to 2π, x>=0
The 3x-y+z=6 condition becomes sqrt(2)rsin(u-π/4)=6-3x

dxdydz=rdxdrdu

[viscid lynx]

starting with dy my friend

not dx

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[raven coral]

Hi, I have a question about the definition of a gradient vector field. I know that $for a C^{1}$ vector field F, we have that $F = \nabla f$ (i.e F is a gradient vector field), then for any simple closed curve $C$, we have that
$$\int_{C} \textbf{F} \cdot d\textbf{S} = 0$$ . However; what if F is not $C^{1}$ continuous? For example, consider $$\textbf{F} = \frac{-y}{x^2 + y^2} \textbf{i} + \frac{x}{x^2 + y^2} \textbf{j}$$. When $C$ is the unit circle, $\int_{C} \textbf{F} \cdot d\textbf{S} = 2\pi \neq 0$. And yet, ... F is the gradient of -arctan(x/y).

[ocean seal]

Maria

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[winter current]

Hi, guys! If someone could help me with this, that’d be amazing I’m so confused.

[vapid shuttle]

What have you tried?

What part(s) specifically are confusing you?

[winter current]

I thought I was supposed to use the contribution margin ratio but I’m not sure 🥲

[vapid shuttle]

When you tried labeling the fixed costs

what crossed your mind

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@winter current Has your question been resolved?

[winter current]

I’m not even sure where to start if I’m being honest

[lone heart]

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[weak ridge]

PLease how do you solve

[oak perch]

It’s the same as (2r+4)+(r-2)=18

[weak ridge]

how

[oak perch]

Because C(n,a)=C(n,b) iff a+b=n

[weak ridge]

What

[weak ridge]

Why 19

[oak perch]

n=18 here

Wait sorry

n

Editing

18 not 19 my bad

Edited

[weak ridge]

Why do you have to add

[twin nimbus]

You familiar with pascals triangle and the relationship between it and the binomial coefficients?

[weak ridge]

What do you mean

[twin nimbus]

https://en.wikipedia.org/wiki/Pascal's_triangle

[weak ridge]

i know it

but tell me what it has to do with the question

please

[twin nimbus]

So the coefficients of (1+x)^n are the nth row of this

[weak ridge]

ok

[twin nimbus]

For instance, the third row is 1 3 3 1, and (1+x)^3 is x^3 + 3x^2 + 3x + 1

Each entry of Pascal's triangle is C(n, k)

Where n is the row and k is the column

And it's symmetric

Meaning that C(n, k) = C(n, n-k)

So we know n = 18, because (1+x)^18

So we just need to find the appropriate k

[weak ridge]

why n-k

[twin nimbus]

Symmetry

[weak ridge]

from where

[twin nimbus]

Pascal's triangle

[weak ridge]

What do you mean

[winter light]

Or simply by definition of binomial coefficient, you can easily show C(n,k) = C(n, n-k)

[weak ridge]

how

[winter light]

Using the definition of binomial coefficient (if you know it of course, otherwise forget it for now)

The one with the factorials

[weak ridge]

which one

[twin nimbus]

n!/(k!(n-k)!)

[weak ridge]

then

what please

[weak ridge]

What i do then

[weak ridge]

what am i doing with this

[winter light]

Write, using that definition, what C(n, n-k) is equal to

[weak ridge]

what definition

the formula

[winter light]

This one! This is the definition of C(n, k)

[weak ridge]

2 and 18

[winter light]

Don't think about numbers for now

[weak ridge]

Then what am i doing

please

[winter light]

This...

Btw I'm going to bed now cause it's quite late

There is for sure someone available to help you

[weak ridge]

wait

please

how you got the n and n-k is what i need help in

[weak ridge]

how does symmetry deal with this

[winter light]

I was saying that in few minutes I'm going sleeping and I wanted to let you know that you aren't left alone, but instead a lot of helpers are available to help you

[weak ridge]

no sorry wrong reply

[weak ridge]

this

[winter light]

I thought @twin nimbus wanted to make you note that C(n, k) and C(n, n - k) are equal to eachother

And what I was telling you is one way to prove that

[weak ridge]

How is this possible

how are they equal

What does the k even mean

[winter light]

You have to prove that they are equal if you don't know it

[weak ridge]

but how

[winter light]

And I'm trying to tell you how to do that 😅

[winter light]

Oh my god

[winter light]

This...

[weak ridge]

What does th k mean

[winter light]

I don't know how to explain you if not in this way

[weak ridge]

is it 18

[winter light]

@weak ridge

[winter light]

No, n is 18

[weak ridge]

but why are we talking about symettry here

how does have to deal with equality

[winter light]

@twin nimbus can you explain it again please? I don't know how to help him

[weak ridge]

please explain the relationship between the equality and the symmetry

[winter light]

<@&286206848099549185>

[winter light]

I don't think I'm good enough to explain that

[weak ridge]

just try please

[winter light]

The only thing I can suggest is going back to the messages and restart to read carefully what @twin nimbus told you

[weak ridge]

i did that

i came with this question

going back wont help

[winter light]

Starting from here

[junior citrus]

this?

[weak ridge]

yes

[junior citrus]

do you know binomial theorem?

[weak ridge]

yes

[junior citrus]

then whats the issue?

[weak ridge]

this

[junior citrus]

if the term you are expanding is 1+x, then the coefficient would not change from either the a or b terms

therefore you could just set the nCr terms equal to each other and solve for r

[weak ridge]

why

so they are the same

;but term 2 and 18 are the same

but they dont give me same r

[junior citrus]

well each term, according to the binomial theorem, could be written as nCr (1)^r*x^n-r

and you want to find out the coefficients of the (2r+4)th term and the (r-2)th terms

since no matter what 1^r * x^n-r will not have a coefficient

you can ignore those

and you just focus on the nCr terms

so you would set them equal to each other

18C2r+4 = 18Cr-2

[weak ridge]

18C1 is the same as 18C18

[junior citrus]

yes

oh

[weak ridge]

Hello

[weak ridge]

Are you there

[junior citrus]

if the binomial is raised to the nth power, the coefficient of the rth term will be equal to the coefficient of the (n-r)th term, given that r starts at 0

[weak ridge]

so that means

what do you mean r starts at 0

[junior citrus]

like

[weak ridge]

why is that a condition

[junior citrus]

x^4 + 4x^3 + 6x^2 + 4x + 1

x^4 would be 0

4x^3 would be 1

[weak ridge]

ok

but how is thata useful

[junior citrus]

i mean then if you want r to start at 1 you could have the coefficient of the rth term to be equal to the coefficient of the (n-r+1)th term doesn't really matter

[junior citrus]

i mean you wanted the relationship between equality and symmetry of the coefficients didn't you?

[weak ridge]

why does be n-r+1 be equals to r

[weak ridge]

why does this have to be so please

[weak ridge]

why are the equal please

[junior citrus]

nCr = nC(n-r) => n!/r!(n-r)! = n!/(n-r)!(n-n+r)! => n!/r!(n-r)! = n!/r!(n-r)!

[weak ridge]

why are we substracting r from n

[weak ridge]

why

[junior citrus]

i just proved that nCr is equal to nC(n-r)

[weak ridge]

where

[junior citrus]

here

[weak ridge]

please explain

[junior citrus]

$nCr = \frac{n!}{r!(n-r)!}$

$nC(n-r) = \frac{n!}{(n-r)![n-(n-r)]!} = \frac{n!}{(n-r)!(n-n+r)!} = \frac{n!}{(n-r)!r!} = \frac{n!}{r!(n-r)!} = nCr$

[ocean seal]

PyroShock

PyroShock

[junior citrus]

@weak ridge

[lone heart]

@weak ridge Has your question been resolved?

[lone heart]

@weak ridge Has your question been resolved?

[oak perch]

It’s simply C(n,a)<C(n,a+1), strictly increases for a<[n/2], because the latter is the former multiplied by (n-a)/(a+1)>1. And C(n,a)=C(n,n-a). So increases then decreases, So C(n,a)=C(n,b) we have only a+b=n

[lone heart]

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[terse fable]

Can someone explain how I get from point a to b

[terse fable]

a being where the arrow starts and b the equation its pointing to

or if anything how I get (a^2-2a-2) from combining the two fractions

oh wait never mind

.close

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[terse fable]

how does factoring the denominator end up with (2a+5)(a+5)?

I'm referring to the second denominator btw

nevermind

my brain is being dumb today

.close

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[lavish kelp]

Ok ok this is a little dumb

[lavish kelp]

Me and my friend have been so confused on what to do

EVEN THOUGH WEVE DONE STUFF ABOVE OUR GRADE LVL

cnr+r/n

Can you reduce that into

Cr+r/n

[near apex]

c, n, r are variables?

[lavish kelp]

Yes

[near apex]

Then no. Not unless n is 1 which then makes division by n in second term redundant.

Wait...

c and C are same. Right?

[lavish kelp]

Yes

Capital was mistake

[fallen verge]

wheres your parentheses?

$\frac{cnr+r}{n}$ or $cnr+\frac rn$

[ocean seal]

GarlicB

[near apex]

Lol. True. Maybe, they had the first one and ended up with cr + r/n.

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[alpine sable]

hay

[alpine sable]

i wanted do dicusses about roudning to the next dollor but it doesnt add up

[median oar]

What is the issue?

[fierce prairie]

send your problem?

[lone heart]

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[pale granite]

i need help

[pale granite]

i got a calculus problem but idk if yall can do it in this section

this is the very last problem of the summer hw and me and my classmates (7 of us) dk how to even approach this

[tardy stag]

what are some observations you can make about the limit on the left?

[pale granite]

its approaching 0

i got to the point where i canceled out the x, but idk where to go from there

[tardy stag]

well the overall limit should be approaching sqrt3

but do you mean the denominator is approaching 0?

[pale granite]

mb

the x is towards 0

which means that we have to somehow cancel out the x in the denominator with another x in the numerator otherwise this limit would be DNE

[tardy stag]

well... it means that the numerator has to approach 0 as well

or else yes it would be DNE (or infinite, and certainly not sqrt3)

[pale granite]

what i did was multiply by the opposite thingy

so instead of a+bx- root 3

i did + root 3

lemme send a picture of my messy work

[tardy stag]

i thought about doing that but i wasn't sure if it'd be useful

[pale granite]

what would be better?

[tardy stag]

i just used l'hopital's rule but your way is probably better

[pale granite]

i didn't solve for a and b tho

we arent allowed to use l'hopital's rule

because we havent learned it yet

[tardy stag]

so you ended up with $\lim \frac{b}{\sqrt{a+bx}+\sqrt3}$ if i'm reading that right?

[ocean seal]

hayley!

[pale granite]

uhh lemme check

yea

then the bx goes away

cuz x=0

[tardy stag]

yeah

[pale granite]

then that is equal to root 3

and idk where to go from there because there isn't a system of equations i can use to solve for a or b

after that im just rearranging the equation

[tardy stag]

yeah you only found one equation so far

i used the fact that the original numerator had to = 0 to get another equation

[pale granite]

could u pls explain why it has to =0

the numerator specifiaclly

specifically

w9

w8

a+bx=3

a=3

then solve for b

and you get b -6

b=6

[tardy stag]

imagine if it didn't

then what would our limit be

[pale granite]

w8

this got me confuzzled

lol

but wouldnt that be the function

aka the y-value

and the y-value has to equal root 3

[tardy stag]

hmm?

[pale granite]

the thing is = to f(x)

aka y

which is root 3

so why does the numerator of it have to = 0

[tardy stag]

well

[pale granite]

is it a special rule idk

[tardy stag]

what's $\lim_{x\to 0} \frac{x^2+1}{x}$?

[ocean seal]

hayley!

[pale granite]

idk lol

um

DNE

[tardy stag]

how do you know that?

[pale granite]

cuz you cant get rid of the 0 in the denominator

[tardy stag]

why not?

[pale granite]

can you?

x(x+1/x)

over x

then x+1/x

its 0

i feel stupid

lol

[tardy stag]

you are correct that it's DNE

[pale granite]

bet

[tardy stag]

and you'll note that if you plug in 0

you get 1/0

[pale granite]

yea

if i plug in 0 to the originial one its also gonna be DNE

but we know its not

[tardy stag]

right, we have to stop it from being DNE

[pale granite]

yea

[tardy stag]

thankfully we have some knobs to turn

[pale granite]

wdym

im just gonna resend the pic so i dont have to keep looking up

[tardy stag]

right, we have to stop that limit from not existing

we know the denominator is going to 0, we can't do anything about that

and if the denom goes to 0 but the numerator goes to some number that isn't 0, then the limit will not exist (or will be infinite)

[pale granite]

w8 no thats only if there is an x in the numeraot

numerator

when the function is heading towards positive or negative infinity

w8

im so lost

is this a rule idk?

that the numerator and denominator have to approach the same number?

cuz how cany they both approach 0 if it says = to 3

[tardy stag]

well i mean

what's $\lim_{x\to0} \frac{3x}{x}$?

[ocean seal]

hayley!

[pale granite]

3

obv

[tardy stag]

yes obv

but

the numerator and denominator are both approaching 0

[pale granite]

w88888

thats when you plug in for x

then it is approaching 0

is there a yt vid i can watch

[tardy stag]

probably

[pale granite]

so

basically

just write the numerator = 0

solve for a

plug into other one for b

rightttttttttttttt???

[pale granite]

.

[tardy stag]

that is what i did yes

[pale granite]

ok perf

what is the thing u were talking abt

what is it called

so i can search it up

[tardy stag]

what thing

uhh look at indeterminate forms

[pale granite]

ok thx

i appreciate it

have a good night

[lone heart]

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@trail summit Has your question been resolved?

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[rapid gate]

c

[pulsar aspen]

!status

[lone heart]

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

[pulsar aspen]

@rapid gate

[lone heart]

@rapid gate Has your question been resolved?

[rapid gate]

so

so either

i dont understand the english part

or the math part

but when it saids 13-19 inclusively

does it mean numbers 14-18 including 14,18

[lone heart]

@rapid gate Has your question been resolved?

[lone heart]

@rapid gate Has your question been resolved?

[rapid gate]

,w x^2-6x-27=0

[ocean seal]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by fwgx

[rapid gate]

,w log_3 (x-6)+log_3 (x)=3

[ocean seal]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by fwgx

[rapid gate]

.close

[lone heart]

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[waxen seal]

is my proof for a and b good enough?

<@&286206848099549185>

[sour verge]

Give me 15 minutes, Ill come back to it if nobody answered

[waxen seal]

ok thank you

here is number 3c

aw i think i read the question wrong

[sour verge]

I'm still doing something, but at first glance it seems you have the preimage definition a bit wrong for a)

(g dot f)-1 (E) = {a in A | g(f(a)) in E}

[waxen seal]

how about this? is my definition correct?

[sour verge]

That seems about right. Now I'd advise trying to pick an element from the first set and show that it must be in the second.

[waxen seal]

yes im doing it right now ill send it

here it is haha

[sour verge]

I get what you're trying to write down, but there's some things a bit iffy with it. I'll annotate it just give me a minute

[waxen seal]

ok thank you ill try proving b and c atm

[sour verge]

I think this might get you going a bit further

[waxen seal]

ok let me try yo fix it

i use g^(-1)(e) as elements of set g^(-1)(E)

[sour verge]

Again, x is not in g-1(E) since x is in A and g-1(E) is a subset of B.

[waxen seal]

ohhhh

[sour verge]

However, you can say that if (g dot f) (x) = e in E, then you have g(f(x)) in E, so f(x) is in g-1(E)

[waxen seal]

f(x) is in g(-1)(E)

[sour verge]

Yeah

But then that means this x is in A

With f(x) in g(-1)(E)

So it's also in f-1(g-1(E))

Makes sense?

It's not as short a proof as it could be, but it should do the trick.

[waxen seal]

f(x) is in B so there must be an x in A which is inside f^(-1)(g^(-1)(E))

[sour verge]

Yeah the x you picked to begin with is that specific x actually

But that's the reason yes

[waxen seal]

is it correct now?

[sour verge]

Yeah looks good

[waxen seal]

i can draw it or visualized it but i cannot say it mathematically

can i send more proofs?

[sour verge]

Yeah it takes some practice

Ofc

[waxen seal]

ok thank you very much im proving b and c now

[lone heart]

@waxen seal Has your question been resolved?

[waxen seal]

here is my attempt at 3b and 3c

[lone heart]

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[wind hill]

hello guys

[wind hill]

i want to learn, is there a way to answer rational function inequality fast?

like is there a shortcut?

[vale wigeon]

shortcut compared to what

[wind hill]

compared to the conventional, just a curious question, im fine with the conventional it's just that it takes me so much time when answering quizzes and test

for example let's have this

and this

[vale wigeon]

what's "the conventional"?

everything to one side then factorization?

[wind hill]

I just watched organic chemistry, i thought that was conventional sorry ^_^

any method you do?

[vale wigeon]

no, there is no shortcut around that.

[wind hill]

do you ahve a method u can share with me?

[tardy stag]

well like in this case you can use the fact that a square is always ≥ 0

[wind hill]

ok ok thank you

im sorry if im asking too much can u show how that works, i do promise i'm just advance studying, this isn't out homework!

[tardy stag]

well like so for that second one

$\frac{(x-7)(x-3)}{(x-5)^2} > 0$

[ocean seal]

hayley!

[tardy stag]

you can just multiply both sides by (x-5)^2; because it's always positive we don't have to change the sign of the inequality

so we get $(x-7)(x-3) > 0$ and we can easily solve that

[ocean seal]

hayley!

[tardy stag]

and remember that 5 can't be a solution since it's not valid in the original expression (because it would be dividing by 0)

[wind hill]

thank you!

[gray isle]

that's applicable more generally too

[wind hill]

how did the answer went to this?

[gray isle]

e.g. in that first example you can multiply both sides by
(x+5)^2(x+6)^2

after multiplying both sides by (x-5)^2,
you'd get
(x+7)(x-3) > 0, (make sure to remember to exclude 5 from the solution set)
and the most intuitive way to identify the solution to that would be the wavy method where you draw a rough sketch of y=(x+7)(x-3)

[alpine sable]

ah this is wrong

[gray isle]

where the main features you care about are just the direction and x-intercepts

[alpine sable]

it's x + 7

[wind hill]

ok ok i see

[gray isle]

just a typo, idea still stands