help-39

[midnight haven]

here

[whole idol]

Does this help anyhow

[midnight haven]

yeah

it does

lmaoo

[whole idol]

Oh shi yea he left us use this and write notes on the back

[plush moss]

Yea that's just the table but graphically represented

[whole idol]

so 5 over 5 is just 1 over 1 right?

[midnight haven]

lil bro lmao why u need to use calculator then

[midnight haven]

yeah?

[whole idol]

Don’t need it

?

[midnight haven]

u don't

[whole idol]

ngl I go on my phone all day in class

[midnight haven]

...

[whole idol]

Senioritis is real

[plush moss]

I mean if you dont mind not passing thats fine

The class doesn't matter right

If you're on your phone during it

[whole idol]

It doesn’t but I wanna maintain a C atleast

[midnight haven]

🤣

true colors came out

💀

[whole idol]

I already have all my credits and more

[plush moss]

Yea I guess that's fine

Just don't fully tune out

[whole idol]

So on the pic I sent

On the graph

how will I know which to use

[plush moss]

You see the numbers with the ° with them

On each line

That's the angle

[whole idol]

The degrees

yea

[plush moss]

And on the same line you see the corresponding trig ratios

[whole idol]

3/2, 1/2

[plush moss]

The x coordinate corresponds to the ratio for cos

The y coordinate corresponds to the ratio for sin

[whole idol]

So how does my 5 and 53 correspond

[plush moss]

And if you wanted to figure out the corresponding tan ratio

Could you figure out how to do that

[whole idol]

No

[plush moss]

Nah nah

Just think about it

I'll give you 30 seconds

[whole idol]

plug in my numbers instead of those

?

[plush moss]

No no

Act nvm

Just take the sin ratio over the cos ratio

To find the corresponding tangent ratio

[whole idol]

what am I looking at in the paper

[plush moss]

The unit circle

[whole idol]

yea

But like which one

[plush moss]

30°

Look at the one that says 30°

[whole idol]

I am

[plush moss]

Take the y coordinate divided by the x coordinate

What do you get

[whole idol]

So 1 half divided by 3 over 2

Idk

Is my bottom number the y?

[plush moss]

Sorry I gtg

[whole idol]

ok

[plush moss]

But you should've got 1/√3

[whole idol]

yea ok and what after thst

[plush moss]

And that's your tangent ratio

So we know tan(30°)=1/√3

[whole idol]

I domt know that

[plush moss]

So tan^-1(1/√3)=30°

Ah it's fine you got the credit

Just get a C and you're good

[whole idol]

I have. A test on Monday about it

And what if the numbers are different

[plush moss]

I gtg maybe later

[whole idol]

Ok

.close

[pearl pond]

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[south urchin]

how to prove that <BDM = <CDN

[toxic lichen]

that's a lot of moving parts in the diagram

do you have a written problem text to go w/ it

[south urchin]

i do have it , but its not in english so

[toxic lichen]

what language is it in

[south urchin]

vietnamese

[toxic lichen]

ok, can you post that and also a translation into english?

[south urchin]

yeah i just did that

Given acute triangle ABC (AB < AC). Draw the altitude AD and the internal bisector AO of triangle ABC (D and O belong to BC). Draw a circle with center O tangent to AB, AC at M and N respectively.
a) Prove that quadrilateral MDON is inscribed.
b) Prove that BDM = CDN.
c) Through A draw a line perpendicular to BC intersecting MN at I. Line AI intersects BC at K. Prove that K is the midpoint of side BC

[pure rapids]

This geo intresting

[south urchin]

its the hardest part of my homework today

[pure rapids]

Ah

Olympiad?

[south urchin]

no just daily homework

[pure rapids]

damn it's hard

Wait inscribed means cyclic right

[south urchin]

yes

[pure rapids]

k b follows from a easily

Which part u stuck

[south urchin]

im stucking at b , c too

[pure rapids]

Ah

So you did a?

[south urchin]

yes i did

[pure rapids]

Ok

[pure rapids]

Oh wait

K i got it

@south urchin

[south urchin]

im here

[pure rapids]

Because MDON is cyclic

[south urchin]

yes

[pure rapids]

Angle NOD is angle MDB

[south urchin]

yes

[pure rapids]

Then since MO = NO

(radius)

[south urchin]

ok

[pure rapids]

Mb wait

OMN is ONM

[south urchin]

alright

[pure rapids]

Which is NDO

Cos of cyclic

And were done

[south urchin]

alright i understand it

Thanks!

[pure rapids]

Np

Need help with c?

[south urchin]

no thanks man ,

i wont be able to understand that haha

[pure rapids]

Uhh ok

Have a good day 🙂

[south urchin]

you too.

[pearl pond]

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[pearl pond]

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[rapid wagon]

Given f(x) is non-negative and integrable on [a,b]. Use Riemann sums to prove the (f(x))² is also integrable on [a,b]

[rapid wagon]

How ?

[pearl pond]

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[pearl pond]

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[midnight haven]

please tell me I dont have to do this manually

[limpid lily]

Well, there's a simplification.

The top helps determine the bottom.

Let's say you pick ABCD for the top.

The first two letters of the bottom can't be A or B.

The second two letters of the bottom can't be B or C.

The last two letters of the bottom can't be C or D.

Once you figure out the number of combinations for ABCD on top, you can multiply it by the permutations of ABCD.

[midnight haven]

can you give me an example

maybe in a 2x2

[limpid lily]

Well, ABCD/CDAB works.

[midnight haven]

no I know that but what do you mean multiply it by the permutations

and the number of combinations on top you mean like abcd acbd cbad dbac all of these?

[limpid lily]

No, I mean the allowable bottom arrangements.

Like ABCD on top stays constant.

How many ways are there to arrange the bottom?

Then multiply that by the number of permutations of ABCD.

[midnight haven]

what about all the different arrangements on top?

[limpid lily]

That's what the permutations of ABCD are for.

[midnight haven]

or do we squaer the combinations that we get in the bottom?

[limpid lily]

No.

Like with ABCD/CDAB, you can change that to ACBD/BDAC.

Do you know how permutations work?

With that, I went ABCD -> ACBD on top, which means A changed to A, B changed to C, C changed to B, and D changed to D.

So, I made those changes in the bottom as well.

And that produced another correct result.

You can make ABCD/CDAB into another correct answer simply by making a change like that.

Like let's say we do A -> C, B -> D, C -> B, D -> A.

ABCD/CDAB becomes CDBA/BACD, which is a correct answer.

[limpid lily]

Does that make sense?

[pearl pond]

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[pearl pond]

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[sterile pecan]

Find the volume of a figurę limited by oxy plane, oxz plane z=3x and y^2+x^2=4

[pearl pond]

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[pearl pond]

@sterile pecan Has your question been resolved?

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[wild mantle]

I know about general equation of circle plus we have 2 point on x-axis one is negative and other positive with that we have y = 0

If you please explain me the question or how to solve it I will be very thankful

[dawn sand]

Try completing the square to get the circle into standard form

by which I mean the form (x-a)^2+(y-b)^2=r^2 for reals a,b,r

[wild mantle]

okay

but it is already is standard for

[dawn sand]

Ok then whatever the name of (x-a)^2+(y-b)^2=r^2 is. I don't know all the names

[wild mantle]

noo I got you point

you are saying convert it into this for right (x - h)^2 + ( y - k)^2 = r^2

[eager jewel]

i dont think converting it in that form will help

[slow grove]

it will

[slow grove]

yes

[eager jewel]

huh

[slow grove]

stick around and youll see haha

[pure rapids]

Completed square form

[wild mantle]

No offense but how I convert it into that form like I just know x point with + - a and y = 0

@slow grove sorry for mentioning you but can you help me

<@&286206848099549185>

[frozen lantern]

by completing the square

[wild mantle]

he man what's up we meet again

ap ki kon si University ha?

[frozen lantern]

giving entrances

[wild mantle]

Jee Clear ho gia

Jee Advance

[frozen lantern]

no

anyway complete the square lo

lo

lol****

[wild mantle]

bhai gusa na kari kasa karon

ya hi to samaj nahi a rahi standard form ha already or g,f ki value bhi nahi pata

[frozen lantern]

well

at least complete the square no

we can see baaki ka later

[wild mantle]

okay

( x + g ) ^2 + ( y + f ) ^2 = -c -g^2 -f^2

Is it right

.close

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[slow grove]

nearly

[pearl pond]

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[dense iris]

quick question

[dense iris]

what do they mean by normal reaction

[flat cedar]

Normal force?

[dense iris]

wait my b i j forgot for a sec

u j draw a line going the opposite way from the 60N force at point A innit

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[dense iris]

I get everything here except

[pearl pond]

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[dense iris]

.close

what am i missing

why do they have an m

[supple mural]

wdym

[pearl pond]

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[dense iris]

like y is m a coefficient there

[dense iris]

i thought m just equals 4gcos70

[eager jewel]

m is the mass brother

[dense iris]

ah i think i overthinking this

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[orchid token]

i need help with C

[orchid token]

i have done a distribution table

like x = 0, 1,2

and P(X=x) for x = 0,1,2

and my probabilities are 0.81 for x = 0, 0.29 for x =1, and 0.02 for x = 2

where x is number of times a red is achieved

what do i do next lol

i know expected value is just E(X) = n(p)

but i dont think that is the expected "winning"

[pearl pond]

@orchid token Has your question been resolved?

[woven matrix]

your probabilities are wrong since the total isn't 1

[orchid token]

yes i found

out

[woven matrix]

P(x = 1) should be 0.17

[orchid token]

ya

i got

so

then

[woven matrix]

also, expected winning = sum of gains * their associated probability

[orchid token]

ohh

i see

kk

so

(0.17)2 + (0.02)10

0.54

[woven matrix]

Idk if you have to count the $1 you spend to play, the exercise isn't clear about that

like if you win 0.54 on average by paying 1, technically your average "winning" is a loss of 0.46

[pearl pond]

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[pearl pond]

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[eager jewel]

can someone explain the last line of the question

i didnt understand what they want us to find

[placid geyser]

It's asking for the focal length of the eye's lens

[eager jewel]

thats what i thought as well but they are not asking for that

according to the solution they are asking for the image distance

[pearl pond]

@eager jewel Has your question been resolved?

[flat cedar]

Ray diagram would be something like this:

So focus won't be the point where the parallel lines focus (i.e intersect and form image)

[eager jewel]

huh oh

so by focus they mean the point the parallel lines intersect not R/2

[flat cedar]

Yh, "they will come to focus" means the point they will meet or intersect

[eager jewel]

oh ok thank u

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[steady wolf]

for this question, i checked the signs for f''(2). the signs changed, BUT f'(2) is not defined.. is it still consduered a point of inflection??

[fickle schooner]

I'm not 100% on this but I believe that's okay if f(2) is defined

[sharp vigil]

a point of inflection is defined as a point in the function where the concavity changes. so even if the function is not differentiable it should still count

[thick raven]

Relevant

[sharp vigil]

it depends of course on how your calculus course defines point of inflection. here's thomas calculus, stewart calculus, and paul's notes (note the thomas definition is more strict than the other two, but both allow this type of point as a point of inflection)

[pearl pond]

@steady wolf Has your question been resolved?

[steady wolf]

okay thanks all, ill just stick to that definition then

[pearl pond]

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[dim temple]

I'm having trouble trying to solve this matrix, have I committed an arithmetical error or have I done something wrong?

[eager jewel]

dude ur handwriting 😭

i cant understand

can u give the original question

[tiny lintel]

Y'a pas une couille ici ? Image 2 dernière ligne

C'est pas [1 -2 1] la dernière ligne de la matrice droite ?

[dim temple]

Ah ouiii t'as raison 😅

Probably what ruined the matrix

[tiny lintel]

Je veux pas juger, mais si tu prends le temps d'écrire proprement sur tes brouillons, je te jure que ça te faciliterais de ouf la relecture et d'éviter les "mini-couilles". Je te parle en connaissance de cause, parce que j'étais comme ça et ça m'a couté cher trop souvent

[dim temple]

C'est-ce que j'ai commencé à faire dernièrement comme tu peux voir aux alentours de la deuxième page, par contre parfois je fais un peu trop confiance à ce que je fais et je fini par foirer quelque part

Et merci!

[tiny lintel]

Je t'en prie

Attend j'ai autre chose à te montrer

@dim temple

Il peut t'aider en vrai pour ce genre de choses

Quand tu as une erreur de calcul mais que tu vois pas où

Il est nul par contre pour les raisonnements

[dim temple]

Much appreciated!

L'écricome est dans deux jours donc il faudrait que j'arrête de commettre des fautes comme celle-ci

[tiny lintel]

Courage, c'est normal tu bosses beaucoup donc tu fatigues

[dim temple]

Les maths me fatiguent plutôt vite donc il se peut que ça m'arrive au jour du concours

Mais bon, j'espère que je ne commettrais plus de couilles d'ici là

[pearl pond]

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[pearl pond]

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[spring dove]

Don´t know how to solve this thing

[modest widget]

thats gnarly

should be 330

if im not overthinking it

[spring dove]

why? Do I need trigonometric ratios?

[modest widget]

i

thought

sums of areas on opposite lots are equal?

and then run through all the options

[spring dove]

if that's rigth should be 350

[modest widget]

yeah 350

my bad

is it 350?

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[pallid forum]

Could anyone help me with this one

[pearl pond]

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[pearl pond]

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[acoustic barn]

How would I find
Sin(2x) given Sec(x) = 5/4 and 0°<x<90°

[jovial kiln]

use the fact that secx = 1/cosx and draw a right angle triangle to find all sides of the triangle

and sin(2x) = 2sinxcosx

[acoustic barn]

It has to be the exact value
Would 2Sin(3/5)Cos(4/5) be valid?

Doesn't seem nearly simplified enough, that's as far as I got

[toxic lichen]

no

sin(3/5) is not something that should ever come up for you

you are NOT taking the sine of 0.6 radians 😭

nor of 0.6 degrees for that matter

[acoustic barn]

Secx is 5/4, it's hypotenuse over adjacent. Remaining side of the triangle would be 3 unless my sleep deprived brain is making me extra stupid

[toxic lichen]

you should have gotten to

sin(x) = 3/5
and
cos(x) = 4/5

which are not the same thing as the expressions sin(3/5) and cos(4/5)

[acoustic barn]

Ah

Well then

[toxic lichen]

and then sin(2x) = 2 * 3/5 * 4/5

[toxic lichen]

if you are sleep deprived then you should NOT be doing math

[acoustic barn]

Tough luck for me, not much of a choice rn :D

[toxic lichen]

are you facing literal execution and death if you do not get this homework done

[acoustic barn]

Ty though that helps, haven't had a problem where we had to solve with a given equivalence rn

[pearl pond]

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[midnight haven]

heloooo

first time

r u real people

i want new international friends

hmmmm

[pearl pond]

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[honest spear]

Prove by mathematical induction that
$$2^n > n^2 \text{ for } n \in \mathbb{N} \text{ where n is sufficiently large}$$

[jolly parrot]

Nyxzore

[honest spear]

How do I find out with values of n are sufficiently large?

[toxic lichen]

try some

[honest spear]

The thing is it works for n=1 but breaks at n=2,3,4 but then works again at n=5,6,7 but how can say for sure that it won't break again for larger n

[violet tundra]

perhaps graph it and find the intersection point(s)

[mortal spade]

I think proving 2^x -2x -1 is increasing for n>5 can work

[pearl pond]

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[pearl pond]

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[brave lava]

Finding the area inside the circle r = 4cos(θ) and outside the circle r = 2

[brave lava]

I’m confused as to why the bounds of the integral are -pi/3, pi/3 in the textbook answer

[pearl pond]

@brave lava Has your question been resolved?

[pearl pond]

@brave lava Has your question been resolved?

[brave lava]

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[pearl pond]

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[chilly jay]

[discrete mathematics] can anyone please recommend me a good book to study this stuff from, which also has a lot of practice questions?

[tender ingot]

Intro to math logic

[plush bramble]

#book-recommendations

[chilly jay]

.close

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[stiff forge]

ok. I am struggling with this problem. im not sure exacrlt how much I should round this. i have 0.0197 as my answer atm

[stiff forge]

for #1

[jovial kiln]

this is already rounded off nicely

use the decimal precision of your Z score table if you have one

[stiff forge]

wdym? like for rounding

[jovial kiln]

yes for rounding

im guessing you have a Z score table right?

if not then just do 3 significant figures

[stiff forge]

nah

im using a calculator and using those valuesx

[jovial kiln]

yeah then 3 sig figs

is usually the standard

[stiff forge]

even with 3 sig figs its still 0.0197 at least is what i

i use 23.3/11.3

[jovial kiln]

yes

thats why i said 0.0197 is correct rounding

[stiff forge]

0.0197 is wrong though

[jovial kiln]

okay then try one more sig fig

[jovial kiln]

😭

[stiff forge]

i was jus showing my answers for

this lol

ill start with number 1

4 sig figs is 0.0196

[jovial kiln]

all your answers is correct btw

[stiff forge]

0.0196 is right

onto next..

[jovial kiln]

try 0.149

[stiff forge]

0.1499 you mean?

yep that was correct

okay now these ones are whole numbers ? or

1499 is right

maybe 196?

[jovial kiln]

perhaps

it would be funny if they werent whole numbers

[stiff forge]

yep

there we go lmfao

that is so weird bruh idk it saays no where the sig figs

[jovial kiln]

ask your teacher to lock in and specify their demands

[stiff forge]

@jovial kiln last one rq

63.67 and 63.66 both wrong

i have other 2 correct so

just need the final part

[jovial kiln]

im also getting 63.67

i did 55.25+1.03643(8.12)

,w calc 55.25+1.03643(8.12)

[jolly parrot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by _mia_34

[stiff forge]

exactly what i did

(number 2 is 0.4618)

[jovial kiln]

try 63 itself

maybe they dont consider age to be continuous

[stiff forge]

nope

[jovial kiln]

okay then you are cooked i think

[stiff forge]

LOL

[jovial kiln]

brute force

[stiff forge]

could probably email

maybe they qwant 4 decimals?

[indigo pewter]

did you try 64?

[stiff forge]

yep

[stiff forge]

@indigo pewter

[indigo pewter]

bruh what

its def 63.67

or 64

ts cooked

did u try 4 dp?

63.6658

[stiff forge]

just did

also wrong

[indigo pewter]

try 46.83

[stiff forge]

thats right??

goated bro

[jovial kiln]

we might be cooked bro

its time to drop out

[indigo pewter]

yeah younger than 85% means bottom 15% 💀

[indigo pewter]

so its the invnorm of 15%

[stiff forge]

you guys are both legends thank you

[indigo pewter]

do you need part two or ru good

[stiff forge]

would give rep if I could

nah im good i had that one

[indigo pewter]

alrr

[stiff forge]

Thanks guys :))

.close

[pearl pond]

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[pearl pond]

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[hazy glade]

Hi! My research examines how the COVID-19 pandemic affected Belgian start-ups, focusing on three main areas:

1. Start-up creation: Did more start-ups emerge during and after the pandemic? Were trends different across sectors?
2. Financial health: Did COVID-19 affect start-up profitability, liquidity, and solvency, and did this vary by sector?
3. Survival: Did start-ups survive better during/post-COVID, and were severely hit sectors impacted differently?

I use data with multiple firm-year observations (used firm younger than 6y as definition for start-up, so some firms will be 6 times in dataset, some 3, others once etc.).

Given this setup, what statistical tests should I perform?
I thought I had longitudinal data, so I did GEE's and the basic descriptives. But I'm not sure anymore, should i just do the descriptives, correlation matrices, T-tests and an OLS instead?

Thanks in advance if you want to help 🙏🏻

[gloomy quiver]

For problems Д) and Е)
How do you know what to multiply with to rationalize the deniminator?

[gloomy cliff]

the conjugate

[gloomy quiver]

For example when multiplying trinomials with binomials I usually do
(a + b) × (x + y + z)

I do ax + ay + az + bx + by + bz

But this method is so inconvenient now

[gloomy cliff]

(√2 - 3) has a conjugate (√2 + 3)
our goal is to exploit the identity:
(a-b)(a+b) = a² - b²
this can be used multiple times for expressions with more than 2 terms

for example:
for, 1/(1 + √2 + √3) you can multiply and divide first by (1 + √2 - √3) giving you (1+√2+√3)/((1+√2)² - (√3)²)
if you further simplify it, you will get an expression with only one square-root which can easily be solved similarly

[solid ivy]

For e) you exploit this identity https://math.stackexchange.com/questions/11618/algebraic-identity-an-bn-a-b-sum-limits-k-0n-1-akbn-1-k

good one to memorize tbh. There's also one for a^n + b^n when n is odd. Just let b = -b in a^n - b^n

[jovial kiln]

!occupied

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[jovial kiln]

guys this is someone else's channel

[gloomy quiver]

I'm so sorry, I didn't mean to do that

[jovial kiln]

its okay, just open a new channel

[solid ivy]

Ah woops, didn't scroll that far up lol mb

[gloomy quiver]

This is RDF's question

[jovial kiln]

this is the original question for helpers

[gloomy quiver]

@hazy glade could you resend your question?

[pearl pond]

@hazy glade Has your question been resolved?

[hazy glade]

Hi! My research examines how the COVID-19 pandemic affected Belgian start-ups, focusing on three main areas:

1. Start-up creation: Did more start-ups emerge during and after the pandemic? Were trends different across sectors?
2. Financial health: Did COVID-19 affect start-up profitability, liquidity, and solvency, and did this vary by sector?
3. Survival: Did start-ups survive better during/post-COVID, and were severely hit sectors impacted differently?

I use data with multiple firm-year observations (used firm younger than 6y as definition for start-up, so some firms will be 6 times in dataset, some 3, others once etc.).

Given this setup, what statistical tests should I perform?
I thought I had longitudinal data, so I did GEE's and the basic descriptives. But I'm not sure anymore, should i just do the descriptives, correlation matrices, T-tests and an OLS instead?

Thanks in advance if you want to help 🙏🏻

[pearl pond]

@hazy glade Has your question been resolved?

[pearl pond]

@hazy glade Has your question been resolved?

[pearl pond]

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[shadow fable]

where does x=y come from?

[sleek frost]

x = y is a hypothesis in this case

[shadow fable]

i see

so we're seeing where x=y?

[indigo pewter]

since the argument is pi/4, x=y

[shadow fable]

ohhhhhh

thank you

.close

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[frigid saffron]

i can't understand algebra

[pearl pond]

Please don't occupy multiple help channels.

[frigid saffron]

ok

[plush bramble]

.close

[pearl pond]

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[frigid saffron]

.reopen

[pearl pond]

✅

[plush bramble]

you're already in #help-28 stop claiming multiple channels

.close

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[rapid wagon]

in how many ways can you organize 17 students, 40 lecturers and 60 randoms in a line such that between every 2 students theres exactly two lecturers and three randoms (u can organize randoms and lecturers at the edge of the line too)

[rapid wagon]

can someone help with this one

[toxic lichen]

so like:

• arbitrary number of randoms and lecturers
• student #1
• 3R, 2L in some order
• student #2
• 3R, 2L in some order
• student #3
• ...
• student #16
• 3R, 2L in some order
• student #17
• rest of the randoms and lecturers

like this?

[rapid wagon]

Yes

That makes it easier actually

[toxic lichen]

yeah i mean then you have like... some selections of randoms and lecturers to place into "sections" demarcated by the students

times the permutations of those within each section

times the permutations of the students among themselves

[rapid wagon]

Yes

I get it

but the final answer

Is not like what i think of

Like they did extra stuff

For the permutatiins of students among themselves u have 17! And then also for all the randoms and lecturers 40!60! Okay. Then between every student there's 16 gaps in total and in each gap there's 5C2 options for the lecturers and randoms so put it 5C2 to the power of 16 and now im left with the 8 lecturers and 12 randoms how many ways can they be ordered in the line ?

The final answer is 17!40!60!5C2^16* 21* 20C8

But i dont understand where 21 and 20C8 come from

[coral holly]

im not good with permutations and combinations but i'll see if i can do something

[toxic lichen]

oh there is a good explanation for that

so like

you have this entire midsection in the line between the 1st and 17th student inclusive

and then you have to arrange the other 20 combined lecturers and randoms, and this midsection considered as a single unit

21 objects in all

[coral holly]

so does randoms mean that you can use combinations on them no need of permutations or something like that?

[rapid wagon]

I dont understand what u mean

[coral holly]

by saying "random" people does it mean that in what order they are organized matter?

like if it doesnt matter you can use combination otherwise you oughta use permutation

[rapid wagon]

Yeahh it matters "random" is just their type

[coral holly]

ok thanks

[rapid wagon]

No, order matters

I still dont understand why the answer has 21*20C8

I understand that when midsection is counted as 1 and we have the leftovers we get 21 objects but then why is that (21*20C8) the number of ways to order them i dont get it

[coral holly]

sorry man i didnt get it....i dont get permutation and combination that well

[rapid wagon]

Np its okay

[pearl pond]

@rapid wagon Has your question been resolved?

[rapid wagon]

Pls

[rapid wagon]

So i figured 21 is the number of different places the big unit can be placed like in between every lecturer and random left which is 20 theres 21 places between them

Still have to figure out why its 20C8 now

[pearl pond]

@rapid wagon Has your question been resolved?

[pearl pond]

@rapid wagon Has your question been resolved?

[fringe robin]

the number of ways you can permutate the rest of the lecturers and the randoms

[pearl pond]

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[minor ibex]

is this right? i js took the diagonals and then used the rhombus area formula but im still paranoid i did it wrong 😞

[dense jasper]

You can check this yourself in #bots

,w area of quadrilateral (-1,-5), (6,-1), (7,7), (0,3)

[jolly parrot]

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[minor ibex]

thank ya i did not know this!

.close

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[sonic hound]

Draw the graph of f(x).

[sonic hound]

Where f(x) = ax^2 + bx + c.

How can this be done?

This is so hard.

a > 0

[placid geyser]

Depends on the signs of b and c

[raw anchor]

^

[sonic hound]

Really?

[placid geyser]

Do you know what a parabola is?

[digital carbon]

depends on Δ as well

[toxic lichen]

do you know what the graph of y = x^2 looks like

start from there

[sonic hound]

Yes, the shape of the curve of a quadratic equation.

How? It's so hard. To draw. To plot every point. On the graph.

[latent quail]

what's the original question tho

[sonic hound]

Plot f(x).

[latent quail]

I mean full content

[sonic hound]

To draw the exact shape of the curve.

For x^2 + y^2 = 1, we can use a compass.

For y = mx + b, we can use a ruler.

[placid geyser]

I don't think they need an "exact" version if they didn't even give b and c values

[placid geyser]

Oh, u mean constructing the graph?

[sonic hound]

Drawing.

Constructing.

[toxic lichen]

ok hold on a second

[sonic hound]

Same thing?

[raw anchor]

where is this question asked?

[toxic lichen]

are you expecting a compass and straightedge construction for a parabola

cause no, you can't really do that

[raw anchor]

yea

[sonic hound]

Some sort of construction

[raw anchor]

just make it look neat

[placid geyser]

Kinda, but constructing is drawing a more accurate one, is what I'd like to think

[sonic hound]

So yes, here it is to be constructed

[toxic lichen]

if you need to graph one by hand, the usual strat is to pin down a few key points such as vertex and/or axis intercepts and then freehand it through those

otherwise:

!xy

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[oak ivy]

For ax^2+bx+c>0

[toxic lichen]

if it is a product of your imagination then tell us so still

[sonic hound]

What if they look at x = 500 and cut the marks?

😦

[toxic lichen]

who are "they"?

marks on what?

[sonic hound]

The gods.

[toxic lichen]

[raw anchor]

where were you asked this question ?

[toxic lichen]

be serious now.

[raw anchor]

a school test?

[placid geyser]

Bruh, you won't have enough paper

To draw it

[oak ivy]

ax^2+bx+c>0 and if a>0 D<0

It's an upward parabola not touching x axis

[latent quail]

The result varies, and it depends on the value of a, b, and c.

[sonic hound]

Yes

[oak ivy]

What does that hourglass indicate ann

[raw anchor]

just draw a neat upward parabola above the x axis

they'll give whatever marks u want

[sonic hound]

So there ain't no way

[placid geyser]

Nope

[sonic hound]

Ok

[raw anchor]

whats ur motive tho

[placid geyser]

It's just a wild guess, but are you an Indian by any chance?

[sonic hound]

Yes

[raw anchor]

do u wanna master the art of drawing incomplete parabolas?

[placid geyser]

Oh lol

10th?

[sonic hound]

yes

[placid geyser]

Thought so

[raw anchor]

10th

icse?

[sonic hound]

you?

no

[placid geyser]

11th

[sonic hound]

k

[raw anchor]

bro just byheart the formula

[placid geyser]

Passed 10th just a month ago

[oak ivy]

Suika didn't you ask about some surds question?

[raw anchor]

u don't need to know this lmao

[placid geyser]

Take a few points and make sure to draw a graph passing through it

[placid geyser]

Yes

[oak ivy]

You learn about that in 11th wow

[raw anchor]

they won't ask you in puc also

[toxic lichen]

ok like literally

where is this concern coming from

is it coming from your imagination only?

[oak ivy]

yes

[placid geyser]

Actually they taught it to us back in 8th

[oak ivy]

..

[toxic lichen]

ok right but like no

[placid geyser]

For jee prep

[oak ivy]

Ok good luck next year

[placid geyser]

Thanks

[toxic lichen]

if you graph a parabola by hand it's literally completely fine to freehand it through anchor points

[raw anchor]

did you just start 11th?

[placid geyser]

Yep

[raw anchor]

nice

[sonic hound]

that's what I wanted to know

thanks

[placid geyser]

ICSE or state board?

Or cbse?

[sonic hound]

Cbse

[placid geyser]

Oh okay

[raw anchor]

you wrote ur boards?

[sonic hound]

no

[oak ivy]

There is nothing involving parabola like this?

[sonic hound]

Yes, nothing

[raw anchor]

wait so then who just wrote their boards

[oak ivy]

suika

[sonic hound]

Suika

[raw anchor]

right

[oak ivy]

Did you get 95%

[placid geyser]

Didn't get the results yet

[raw anchor]

u wrote boards as well?

[raw anchor]

icse?

[placid geyser]

But I expect ~95%

[placid geyser]

State

[sonic hound]

damn

[oak ivy]

bruh

[raw anchor]

i dont remember state release date

[oak ivy]

state 95% should be easy

[placid geyser]

Hehe

Ik

[oak ivy]

ANyways

mathematician

[raw anchor]

what was urs

icse?

[oak ivy]

Just know about the formula for roots

and D>0 and D=0 and D<0 properties

[sonic hound]

So three things need to be indicated with coordinates on the graph sheet:

1. The vertex
2. The x-intercept(s)
3. The y-intercept
   right?

[oak ivy]

And that sum of roots product of roots

[raw anchor]

sure

[sonic hound]

Okay. Thanks y'all.

.close

[pearl pond]

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[rich jolt]

how to solve this

i have no idea how to start

i could find Gravitational intensity and use pressure formula idk doesnt make sense to me

[fringe robin]

you'll need to open a new channel

[toxic lichen]

!msgdeel

!msgdel

[pearl pond]

The original post of this help channel has been deleted, and it will abruptly close and possibly lock. (This is irreversible.) Please claim a new channel, and don't delete the first message of any future channel you claim.

[fringe robin]

Ann you might as well be the bot

[rich jolt]

.close

[pearl pond]

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[restive wedge]

how do you do this

[restive wedge]

i do not know where to begin

how

witha rea

how do i

turn that into sides

[digital carbon]

circumference is wild

[digital carbon]

draw a diagram for the equilateral triangle

then you can use pythag to find the side length

[raw anchor]

i was thinking

a=4b/3

area of the triangle = root3*a/4

equate that to 4*root3

[raw anchor]

you got a, now find b with this

and do b*b

[raw anchor]

is the answer 9?

[restive wedge]

yes

what

[raw anchor]

so the circumference is equal right?

[restive wedge]

yeah

[raw anchor]

so let a = triangle side, b = square side

[restive wedge]

the total of all sides of both trianle and square is equal

[raw anchor]

3a = 4b (maths)

[restive wedge]

ok

[raw anchor]

now area of triangle (equilateral) is root3*a^2 /4

they gave 4*root 3

you get a = 4, find b, you get b=3

and do b square

[restive wedge]

is this a formula or

[raw anchor]

yes

[restive wedge]

ohhh

is there a bunch of these

or

[restive wedge]

is it only equirateral triangle

thats special

[raw anchor]

works only for equilateral triangles

[restive wedge]

oh ok

yeah i got it

ty

once i get the a

i jsut plug it into the 3a=4b

.close

[pearl pond]

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[pearl pond]

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[old salmon]

Hello since the other fricking channel isn't working I need help on this please

[old salmon]

My teacher did not explain anything correctly at all

And I got too much stress thats why I am freaking out

[green dew]

the equation of any straight line on a graph is: y = ax + b

ur goal is to determine a and b

[old salmon]

A and B?

[green dew]

yes

[old salmon]

My teacher said like y=mx+c

Man I am so confused

I am cooked

[green dew]

u can even say y = Ωx + ρ

it doesnt matter

all u need to know is that the equation of any straight line in general

[mint gazelle]

u can use any variable insteade of a and b

[old salmon]

Oh

[mint gazelle]

yuhh

[green dew]

is y = ax + b, as in something multipled by x

and u add to it another number

[mint gazelle]

yupp

[green dew]

by word, its hard, so we symbolize it

y = ax + b, where a and b are real numbers

[old salmon]

So it can be used by any?

But how can I work out A?

Line a

[green dew]

so say that line a has an equation of: y = ax + b

[old salmon]

Yeah

[green dew]

b is always the y cordinate of intersection of line a and the y axis

so check at which point line a intersects the y axis

[old salmon]

Alright

[mint gazelle]

@old salmon

[old salmon]

3

[mint gazelle]

if u r confused how does these graphs work

https://www.desmos.com/calculator

go on such a website and play with graphs a bit

[old salmon]

Alright

[green dew]

if ur talking abt the green line yes

[drowsy fjord]

Use different values to see what they do

[green dew]

so the green line's equation is y = ax + 3 then

whats left is a now

[old salmon]

Yeah

[green dew]

you would need two points on the green line

which yk the coordinates of

to find a

[drowsy fjord]

a is the slope

[old salmon]

So x is -3?

[green dew]

coordinates

not x value

[mint gazelle]

he prly meant x-coordinates only

[green dew]

coordinates is (x,y)

not just x

[mint gazelle]

yuhh yuhh

[drowsy fjord]

Just tell him what a means to not confuse him with points

[green dew]

a straight forward formula wont help

[drowsy fjord]

No what it means on the graph

[green dew]

the slope of the function

[drowsy fjord]

Right, if the graph is going down, a is negative, if it goes up, it's positive. Give him an idea what a does

[mint gazelle]

@old salmon Just change the values of a here, u will understand it very neatly

[green dew]

a just tells you how much the line is inclined

[mint gazelle]

make a function y= ax + b there

[green dew]

and depending on the sign, it will also tell u its variation

on the whole interval

[drowsy fjord]

On number 1 a) the right value for a would be 1, because for each box it goes to the right the line goes 1 box upwards. If the line would be steeper, a would be a higher number

[mint gazelle]

the 1 who needed help isn't online ig 😐

[drowsy fjord]

Gave up ig

[mint gazelle]

i don't think so

🤷

but maybe

i always learned about graph using desmos like stuff

where u can play with values oyo

it's what makes u learn it the ezy way it seems to me

but i still can't understand a Point slope form graph, i am yet to learn them 😅

https://www.desmos.com/calculator/onfy0dzx2o
these ones

[drowsy fjord]

Point slope form?

Is it P + v

Point plus some vector

In this case it would be (0; 0) + k<1; 1>

Or just the vector part

Which theoretically is just y=mx+c wrote down differently

Where for the point you would use the y-intersept, k = x and the vector reduced to its unit vector

@mint gazelle

[mint gazelle]

@drowsy fjord

this thingy

[drowsy fjord]

Well it's y=mx +c but with the point distributed so it's gives us the slope

[mint gazelle]

it's nearly same ig, just with one more var

[mint gazelle]

Yuss!!

[drowsy fjord]

It's the slope formula

I have never seen it before tbh

[mint gazelle]

ohh rly!?

[drowsy fjord]

Used as a way to display a line

[mint gazelle]

yup it's usually not told in any book or anything

i found it online only and got a bit confused

[drowsy fjord]

OK, don't use it

[mint gazelle]

Mhmm

[drowsy fjord]

You could also use a polar form

[mint gazelle]

uhh, ik I can, i just wanted to know how this thingy worked

just a lil curious yk

😅

[drowsy fjord]

Yeah it's a good thing ig

[mint gazelle]

Indeed it is

i learn alot bcz if it

Cyaa

[drowsy fjord]

Cya

[pearl pond]

@old salmon Has your question been resolved?

[pearl pond]

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[pearl pond]

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[pearl pond]

Channel closed

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[pearl pond]

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[vapid valley]

Why is there integral of 1? Where did it came from?

[toxic lichen]

$\int 3 \dd{x}$ but they pulled the 3 out

[jolly parrot]

Ann

[vapid valley]

Why tho

[oak ivy]

Because constants are pulled out of integral

Same with derivatives

[toxic lichen]

eh doesnt make much of a difference

[vapid valley]

I think i misunderstood it, the first integral is jist the xdx and then the 3 is outside of the integral of the other?

[toxic lichen]

WA likes to do it this way i guess

$\int 3 \dd{x}$ and $3 \int 1 \dd{x}$ are the same even in terms of how difficult they are to calculate

[vapid valley]

Kinda my first time using wolfram

[jolly parrot]

Ann

[toxic lichen]

(which is to say: not very)

[vapid valley]

Aight. Thanks

.close

[pearl pond]

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[pearl pond]

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[frozen lantern]

how can i think of two line equations added together?

like $A_1x+B_1y+C_1 + \lambda (A_2x+B_2y+C_2) =0$.

im tempted to think of this vectorially, because what i see when graphing it is that changing $\lambda$ just rotates the resultant line, but i cant think of a nice way to make vectors work.

[jolly parrot]

Percy

[snow sail]

maybe parametric

?

[rustic gate]

https://en.wikipedia.org/wiki/Pencil_(geometry)

[frozen lantern]

okay ill read this and see if it helps

[frozen lantern]

can you elaborate?

[frozen lantern]

okay so this does tell me what it does but i still dont see why

[rustic gate]

graph z = ax + by + c in 3d

get 2 planes

setting = 0 is just intersection with xy-plane

when you tilt one plane up higher itll rotate where the sum intersects the xy-plane

[pearl pond]

@frozen lantern Has your question been resolved?

[frozen lantern]

@rustic gate so basically when graphed in 3d, the lines correspond to the intersection of the planes with xy plane, and multiplying the planes by any factor just rotates them.

i follow till here.
i cant quite see how that one point is fixed though

[rustic gate]

0 + 0 = 0

[frozen lantern]

uh huh

i dont really get it

[frozen lantern]

are you basically just saying 'this is definitely satisfied when both are zero, so their intersection (the point where both are zero) must always be included'?

[rustic gate]

no i'm saying that's the only point where both planes are 0

so it's the only point that's fixed as you tilt the planes

because 0 + 0 = 0

[frozen lantern]

okay but why is fixed though

[rustic gate]

and generally speaking s*nonzero + t*nonzero isn't 0

[rustic gate]

quite literally because 0 + 0 = 0

[frozen lantern]

why is the point fixed as i rotate the planes

desmos 3d super isnt helping here

[frozen lantern]

also can i think of it like that

[rustic gate]

cuz it's s*plane1 + t*plane2 = 0

[frozen lantern]

yes

[rustic gate]

which is 0 when both plane1 = 0 and plane2 = 0

i.e. the intersection

[frozen lantern]

mm

[rustic gate]

and as you vary s and t

the other points will generally not be 0

[frozen lantern]

oksy i think thats what i said here?

[rustic gate]

is it though

maybe it is

[frozen lantern]

hmmm

i think i get it yeah

.close

[pearl pond]

Channel closed

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[rustic gate]

it's easy to prove that there's only one point that appears in all lines

[pearl pond]

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[midnight haven]

.

[midnight haven]

why did we permutate

[past perch]

D1 E L V R D2
is different from D1 L E V R D2

[midnight haven]

alr

[brave hawk]

basically you have 2 D's so picking a D in a given position is more likely

cause this is probability

[midnight haven]

oh

so if it didnt say randomly

[brave hawk]

as they say, we're not counting the number of different arrangements that satisfy the condition

[midnight haven]

we would do combination

[brave hawk]

we're looking at how likely

[brave hawk]

if we were counting the number of distinct arrangements yes we would treat the two D's as interchangeable

so we wouldn't need the 2! there

[midnight haven]

wait im kinda lost

[tropic saddle]

frankly, "randomly chosen" is imprecise wording and both interpretations can be argued for

[midnight haven]

can i have an example of combination and one for permutation

why didnt we multiply 2!

the C on both ends