help-28

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@celest tapir Has your question been resolved?

[onyx glen]

do you know the general statement of chebyshev's thm

[celest tapir]

Yeah

[onyx glen]

ok can you say it

[glossy valve]

dyxn

[celest tapir]

I guess I have something like P[0 < X < 26] + P[105 < X < 405]

The first probability is P[|X - 13| < 13]

P[|X - 13| < 13] + P[|X - 255| < 150]

@onyx glen something like this?

[onyx glen]

hol up

that seems a bit sus as setup

so X is the number of faulty jackets in the test sample yeah?

X ~ Bin(405, 1/6)

,calc 405/6

[glossy valve]

Result:

67.5

[onyx glen]

mean is 67.5

we want to impose a bound on P(X <= 29 or X >= 106)

,calc (29+106)/2

[glossy valve]

Result:

67.5

[onyx glen]

,calc 67.5-29

[glossy valve]

Result:

38.5

[onyx glen]

@celest tapir the probability we're interested in it $$P(|X-\mu|\geq 38.5)$$

[glossy valve]

Ann

[celest tapir]

Oh right thats what I was missing

[celest tapir]

What's this step about?

[onyx glen]

checking where the midpoint of the excluded interval is

to know if extra fuckery would be needed to apply chebby

[celest tapir]

the problem was set so that it would work out?

[onyx glen]

which in this case it isn't

trust but verify!

[celest tapir]

alright

this makes sense

thanks!

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[hearty ice]

I did b,c, d but stuck on the drawing for a

Is that diagonal vector 145? It seems so small

And not sure where the 20 degrees go and what the velocity triangle is

[drowsy fossil]

I'm assuming the velocity triangle would be the initial velocity, and its respective x and y velocity components

With a low launch angle of 20 degrees, youd have a high x component with a smaller y component. but neither of them would exceed 145 ft/s

You can do some trig to find the x and y components., so if the launch angle is 20, you can do 145sin(20)= y to find the y component and 145cos(20) = x to find the x component

[hearty ice]

I am fine with the algebra

Im just not sure how to draw the diagram

[drowsy fossil]

Vectors are drawn with a start point, direction, and magnitude

so for the initial vector it would just be an arrow with pointing in a 20 degree angle with a magnitude of 145 ft/s

the y and x components are also still vectors so youd draw the correct direction of them and their magnitudes, it would resemble a triangle if you put them all together

[hearty ice]

Would it be like this?

What about the fence and home base

And the end?

[drowsy fossil]

I don't think that would factor into the velocity triangle its asking for, but it would be part of the diagram

But yea that would be the "velocity" triangle

You need magnitudes for the x and y component and it wouldn't hurt to put the direction too, since they're vectors

[hearty ice]

How would i draw the fence and home plate?

Arent the vectors drawn in ft/sec?

The units are not the same

[drowsy fossil]

They are in ft/sec. all of the units are the same despite the sin and cos

if you solved those youd get a number, in ft/s

[hearty ice]

I meant like the 17 foot fence

Isnt that just ft?

And the fence being 380 ft from the plate

[drowsy fossil]

Yea, not vectors, but just dimensions. ill draw something real fast

This essentially tells you everything about the physical components of the problem

[hearty ice]

Huh, I did not know you could draw the diagram like that

Thank you!

[drowsy fossil]

Thats how I did it in physics!

Hope its sufficient for you

[hearty ice]

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[shut prairie]

please don't look at my circles

[shut prairie]

so A, B and C is not possible

but i dont know how to find if its d or e

i thought it'd be 18 because of y=2x

but its 15 and I'm blasting a head on this for the last 15 minutes

wait

i think i see something

nvm it didn't work help

[marble solstice]

the midpoint of OB (the centre of the black circle) is the same distance from C as it is from O

You know O and you know C, so you can find this midpoint

[shut prairie]

oke lets try

oh so my drawing is ugly!!

:D

[marble solstice]

Do you know how to find the distance between two points?

[shut prairie]

oh im supposed to use analytics here?

i tried to solve this with geometry

[marble solstice]

This is basically geometry

[shut prairie]

and i failed

[marble solstice]

I have a picture one second

This is basically what is being asked of you from my knowledge

[shut prairie]

maybe i should study this lesson as a whole

[sudden condor]

(0,x) is a little misleading imo

[marble solstice]

The answer to the question would be 2x

[shut prairie]

oh i thought you were asking for a number

[marble solstice]

So you need x

Start with the distance between (0,0) and (0,x). What would that be?

[shut prairie]

x?

[marble solstice]

The pic I gave

[shut prairie]

oh sorry

im sorry i think i won't be able to solve this question

i dont want to waste your time

I'm a lost cause

[marble solstice]

It's ok

you're not

If you truly give up I can give you a picture of the full solytion for understanding

But I would keep trying

[shut prairie]

okey lets try again then >:3

I FOUND IT

!!!

oh it's flipped

but I FOUND IT

[marble solstice]

CORRECT!

Not what I was thinking, but correct

[shut prairie]

thank you so much like actually

can i see your way too

i want to learn it too

not even my teachers are this patient with me

[marble solstice]

give me a minute lemme jot it down and take a pic

[shut prairie]

okkee

[marble solstice]

Your way is probably faster

[shut prairie]

i want to know as many paths as possible it's not important to be simple for me

[marble solstice]

wait the question was in a different language do you know english

[shut prairie]

yeah heres the translated version

im turkish but I know english a bit

[marble solstice]

great

I'm going to write the solution in english

[shut prairie]

yupp

[marble solstice]

Ok sending in a minute

The solution is complex, but can be more versatile

[shut prairie]

thank you a lot!!

[marble solstice]

See the previous drawing

[shut prairie]

yup okey thank you

now I'll solve it again with this way

have a great day!!!

[marble solstice]

Great, that's how you learn😉

[shut prairie]

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[tulip marlin]

Given a uniformly random n-digit base 4 sequence, what is the probability the sequence contains n - 1 digits which are the same. For n > 2.

I think it is

4*(3*n + 1) / 4^n
=
(3*n + 1) / 4^(n - 1)

Choose from 4 digits to be repeating n-1 times.
Either the nth digit is also the same, then there is only 1 permutation of digits.
Or the nth digit is one of the other 3 digits, and there are n permutations.

Does this look right?

[tulip marlin]

| 3: 0.625
| 4: 0.203125
| 5: 0.0625
| 6: 0.0185546875
| 7: 0.00537109375
| 8: 0.00152587890625
| 9: 0.00042724609375
|10: 0.000118255615234375
|11: 3.24249267578125e-05
|12: 8.821487426757812e-06
|13: 2.384185791015625e-06
|14: 6.407499313354492e-07

code output on that formula

3 is lower than i mightve thought

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[vast breach]

طريقة الحل ؟

[rough tundra]

if possible, could you provide an english translation of what this image is showing?

[cold agate]

never thought i'd see math in a language other than english or greek, but here we are...

[sage osprey]

real

[onyx glen]

the image is low-res and i couldn't make out the symbols in the exponents well but i think it says this:

$5^x + 2^3 = 5^{x+1} \ 5^x + 2^3 = 5^x \times 5^1 \ 4 \times 5^x = 8 \ 5^x = 2 \ x = \frac{\log(2)}{\log(5)} = 0.431$

[glossy valve]

Ann

[onyx glen]

and google translate says the message text translates to "How to solve?"

@vast breach do you speak english?

[vast breach]

yes

[onyx glen]

ok

can you check that i translated your image into western notation correctly

if not tell me where i messed up

[vast breach]

✅

[onyx glen]

anyway, there's nothing to solve -- this is already a fully worked solution.
did you get confused at a specific step? if so, which one?

[vast breach]

haw get 4

[onyx glen]

from the 2nd line, subtract 5^x from both sides

get $2^3 = 5 \times 5^x - 5^x$

[glossy valve]

Ann

[onyx glen]

$5 \times 5^x - 5^x$ simplifies to $4 \times 5^x$

[glossy valve]

Ann

[vast breach]

5^x

تروح

[onyx glen]

sorry, i don't understand

[vast breach]

5^x + 2^3 = 5^x \times 5^1
2^3 = \times 5^1

[onyx glen]

no, that's wrong and doesn't make sense.

you cannot just "remove the 5^x" like that.

especially because ×5^1 is just nonsense -- you can't have a dangling multiplication sign like that!

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[unreal crane]

is there any wrong with this

[steel solar]

The factorization of the second denominator is $(x-2)^2$

[glossy valve]

;(

[unreal crane]

isnt it still the same?

[steel solar]

No

Either I'm completely misreading the plus sign that you have or you miswrote it

[unreal crane]

yeah I've miswrote it thanks

Is this correct

[steel solar]

Yeah that works

[unreal crane]

how we do qn 2c?

[old ember]

you have the same terms with constants on the numerator as in part(a), but you have an additional ax+b/(x+2)^2 term

[unreal crane]

so, the first fraction would be A/(2x-1)
second fraction is B/(x+2)
third fraction is C/(x+2)^2?

[full forum]

@unreal crane Has your question been resolved?

[unreal crane]

<@&286206848099549185>

[sage osprey]

is it one word answer or descriptive?

[unreal crane]

partial fractions?

[sage osprey]

yeah

if its one word i know a trick

and if its descriptive u need to do the whole thing

[unreal crane]

can you teach me both?

[sage osprey]

i will give an example

wait imma write it down

it works only if the numerator is smaller

proper fraction i mean

[old ember]

Instead of C, it is Cx+D

[unreal crane]

actually I'm able to solve and its just A and B

but I cant solve when there's A,B and C also

[unreal crane]

there's a D?

[sage osprey]

write A, B, C when the denominators are with degree 1

and when there is square you need Cx+D as @elliot said

[unreal crane]

how we write that out?

[sage osprey]

Cx+D/ the square part

[unreal crane]

I believe qn2c requires Cx+D in that case

[sage osprey]

like Cx+D/ (x+2)^2

somethn like that

[glad haven]

yes

go further.

[unreal crane]

ok let me write it out and show yall

[old ember]

Wait actually you only need Cx+D if the quadratic is not reducible

[glad haven]

(x+y)/2y

[old ember]

so actually you are correct with this

[unreal crane]

then when we use Cx+D?

[sage osprey]

when u have (SOMETHING)^2

[old ember]

If e.g. you had x^2 + 1 on the denominator

because you cannot break this up into linear factors

[unreal crane]

so its like x^2+2?

[restive geyser]

This is not quite right - you can split this up into two fractions, one with a constant over (x+2), the other a constant over (x+2)^2

[unreal crane]

this is used for Cx+D?

[sage osprey]

i just specified that part

[restive geyser]

We use this if the denominator is a quadratic that can't be reduced (which for all intents and purposes in this case just means we can't factorise it)

[unreal crane]

so like x^2+2?

[sage osprey]

like look at here

[restive geyser]

That would be an example, yes

,rccw

[sage osprey]

pardon my potato camera

[unreal crane]

but what if its the case like qn2c?

does Bx+C still apply?

[restive geyser]

Then it is exactly this

[old ember]

No, because everything is broken up into linear terms, you just have repeated terms

[sage osprey]

yeah but you need to add num/den if the fraction isnt proper

[old ember]

Sorry meant to reply to this

[sage osprey]

break it down and checks if den is greater than num

i mean factorize

[unreal crane]

Like this?

[sage osprey]

yes

factorize the numerator

it will be easier

[restive geyser]

The numerator doesn't factorise

[unreal crane]

I dont think we can factorise

[restive geyser]

You can't

[sage osprey]

mb i noticed it now

[restive geyser]

It's not factorisable

Nor is it necessary to

[restive geyser]

Multiply both sides by the denominator of the fraction on the left

[sage osprey]

equate the coefficients or put the value of x

[unreal crane]

you mean A/2x-1?

[restive geyser]

"on the left" meaning "on the left of the equals sign"

You have an equation here

You should get $7x^2 + 11x - 1 = A(x+2)^2 + B(2x-1)(x+2) + C(2x-1)$

[glossy valve]

Waes (Wires)

[unreal crane]

wait I dont get this

[restive geyser]

I just multiplied by the denominator of the fraction on the left of the equals sign

(cancelling out common factors on the right)

[old ember]

multiply every term in the equation by (2x-1)(x+2)^2, and some things cancel

[unreal crane]

oh so for e.g (2x-1)(x+2)^2(2x-1)
we cancel out 2x-1

[old ember]

the 2x-1 is on the denominator, and the other terms appear on the numerator now, but yes they will cancel

It's more like $$ \frac{(2x-1)(x+2)^2}{2x-1} = (x+2)^2$$

[glossy valve]

elliot

[unreal crane]

Like this?

[old ember]

No, your aim is to have no denominators at all once you multiply through

[sage osprey]

multiply the denominator on both sides

[old ember]

On the left hand side, we are doing
$$ \frac{7x^2 + 11x -1}{(2x-1)(x+2)^2} \times (2x-1)(x+2)^2 = 7x^2 + 11x -1 $$

[glossy valve]

elliot

[unreal crane]

Like this?

[old ember]

yes exactly

do you know how to solve for A,B,C from here?

[unreal crane]

not really

[sage osprey]

now its a real problem

enter into a voice chat with @elliot if he/she agrees

it cant be taught properly via texts

[old ember]

you want to start by seeing if you can make some terms on the right disappear, and we are free to choose different values for x

the equation must hold for every value of x

e.g. what happens if I plug in x = -2 into the entire equation

[sage osprey]

its somethn like this

[unreal crane]

so you mean x+2=0
x=-2

[old ember]

yes

[sage osprey]

shouldn't we take it from the given equations?

[unreal crane]

but isnt it (x+2)^2 not (x+2)?

[old ember]

at this point this is not so relevant, we can just work with the expression we have arrived at

[unreal crane]

so we ignore the ^2?

[old ember]

Well if (x+2) = 0, then (x+2)^2 = 0, yes?

we are just trying to figure out values to pick for x

[unreal crane]

yeah

[old ember]

so what do you get if you plug in x = -2 into the equation?

[old ember]

+7 rather than -7 on the left hand side

[unreal crane]

-51=-5B-5C?

[old ember]

Does the B term survive?

[unreal crane]

yeah?

[old ember]

Also remember it is +7 on the left

[old ember]

You have B(2x-1)(x+2), right?

Which means $B \times (2x-1) \times (x+2)$

[glossy valve]

elliot

[unreal crane]

yeah

[old ember]

so what is the value of this when x = -2?

[unreal crane]

oh I see it now

so its 5=-5C

[old ember]

yes

so now we know what C is

[unreal crane]

yeah C=-1

[old ember]

so now can you do something similar to work out A?

[unreal crane]

B has two expressions tho unlike A?

[old ember]

what is the concern?

[unreal crane]

like initially its just x+2 right but now we got (2x-1)(x+2)

[old ember]

yeah but we can still continue to work out coefficients

to work out C we subbed in a value for x when x+2 = 0

what should we try now?

[sage osprey]

hows it going guys?

[unreal crane]

2x-1?

[old ember]

yeah

[sage osprey]

cant we equate the coefficients?

nvm

[unreal crane]

do we sub in the value of C also?

[old ember]

yes we can, but I think it won't matter in this step because I imagine the last two terms will die

[unreal crane]

A=1

[old ember]

how did you get it?

[unreal crane]

6.25A=6.25

[old ember]

yeah that should be right

So now we know what A and C are, we just need to find B

To summarise, we have $$ 7x^2 + 11x -1 = (x+2)^2 + B(2x-1)(x+2) - (2x-1) $$

[glossy valve]

elliot

[old ember]

this holds for any value of x

so how could we figure out B?

[unreal crane]

Do we sub in the values of A and C?

[old ember]

well we have already done that once we got here

[old ember]

this holds for e.g. x=0, x=1, x=2, x=1/2, x=17,.....

[unreal crane]

then how we find B tho

[old ember]

to find A and C, we subbed in a particular value of x

why not just sub in something else?

as long as its not 1/2 or -2 (the ones we used up before) it should work regardless of what we choose

[unreal crane]

so any value works?

[old ember]

yeah, we just want a particular equation that lets us solve for B

[unreal crane]

so 100 also can?

[old ember]

yeah or -1000000

0 is usually easiest if you are able to

[unreal crane]

so B =3?

[old ember]

yeah looks like it

[unreal crane]

ok thanks a lot

by the way, can you give a qn on an example of the Bx+C?

[old ember]

Yeah, try expressing $$ \frac{x^2-6x+8}{(x-1)(x^2+1)} $$ in partial fractions

[glossy valve]

elliot

[old ember]

I can't guarantee that the values will be nice, I just made this up

[unreal crane]

Its ok, I found a qn

is the idea the same as what we did earlier?

[old ember]

Yes, but it is slightly harder

There is a clear value of x that we can use, right?

[unreal crane]

wdym

[old ember]

So we do the whole process before, write it in terms of A,B,C, multiply through by the denominator

[old ember]

get to something like this

can you see already what value of x we will be able to sub in further down the line?

[unreal crane]

but this time round there's only two fractions on the right hand side?

[old ember]

Yeah so you get $$15x^2 + 5x + 2 = A(1+2x^2) = (Bx+C)(2-x) $$

[glossy valve]

elliot

[old ember]

should be a plus at the end

[unreal crane]

Like this?

[old ember]

yeah that's right

so now what to do?

[unreal crane]

1+2x^2=0
2x^2=-1
x^2=0.5

[old ember]

no that doesn't work, this is why this one is harder

[unreal crane]

oh so we have to use 2-x?

[old ember]

note that $1 + 2x^2 \geq 0,$ so it definitely has no roots, so you cannot use this one

[glossy valve]

elliot

[old ember]

yes

[old ember]

should be $\geq 1$

[glossy valve]

elliot

[unreal crane]

I got 52=9A-8B+4C

[old ember]

But shouldn't B and C die?

[unreal crane]

nvm I got A=8

[old ember]

yeah that should be right

now we have just one other nice value of x that gives us some cancellations

how else could we kill B?

[unreal crane]

Bx+C?

[old ember]

our goal is to figure out B and C separately

How could I kill Bx?

[unreal crane]

other than (1+2x^2) and 2-x I dont see any expression of x

[old ember]

what happens if x=0?

[unreal crane]

oh yeah

do we sub in value of A?

[old ember]

yeah that is known information now

[unreal crane]

I got C=-3

[old ember]

It's $2 = A + 2C$ right? So yeah that looks correct

[glossy valve]

elliot

[unreal crane]

so to find B, we sub in A and C?

[old ember]

we should do that yes, but remember what else we had to do before

[unreal crane]

the x value

we still do x=0?

[old ember]

try it and see

(it won't work but you should try and see why)

[unreal crane]

yeah it kills off B

B(0)=0

I guess x=1 should work

[old ember]

that's what I mean when I say values have been 'used up'

[old ember]

yep it should

[unreal crane]

I got B=1

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[raven beacon]

You need to find the circles raidus I got 5.9 and the answer is 6.7

[indigo frost]

,rccw

[raven beacon]

I did y/24 = 26-y/24

That’s probably where I made the mistake

Not sure how to find y

[cyan walrus]

Area of triangle = semi perimeter * radius of incircle

radius of incircle = area / semi perimeter

[raven beacon]

What do you mean by semi perimeter

[cyan walrus]

A/ S where s = (a+b+c)/2

[cyan walrus]

half of perimeter

sum of 3 sides divided by 2

[raven beacon]

Can you show what you mean

[cyan walrus]

mmm wait

[raven beacon]

Alright

[cyan walrus]

@raven beacon

here's the derivation as well

you can directly use this as a formula

[cyan walrus]

^

[raven beacon]

I’m pretty sure that’s above my course

[cyan walrus]

then follow the derivation

[raven beacon]

We are supposed to use this

[cyan walrus]

it's 2 extra lines of work

I can't read that but, similar triangles?

[raven beacon]

Yep

I’ll translate

Real quick

intercept theorem

[cyan walrus]

hmm, using this would be complicating the question unnecessarily

[raven beacon]

What’s the best way to solve it

Then

[cyan walrus]

are you sure you HAVE to stick to it?

[raven beacon]

Yep sadly

[cyan walrus]

using areas ig

since you said incircle formula is out of syllabus for you

[raven beacon]

You have to solve it using similarity

It’s written on the side

[cyan walrus]

hmm

idt I can help man, sorry 😔

[raven beacon]

Its all good

[cyan walrus]

good luck though,
ping helpers if you don't get a response for too long

[raven beacon]

I’ll Ask My teacher tmrw

Thanks for

Doe

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[thick ivy]

any hints welcome

i tried the following:
we take some pair.

if it is alone in a circle then there are R(n-1) ways,

but if the pair is not alone in the circle, then how many ways ?

[potent shale]

what do you mean by alone in a circle?

if they are alone, isn't it n=1?

@thick ivy

[thick ivy]

its a typo

sorry

i meant

lets give each pair a number from 1 to n
assume we are looking at pair number one, and we want them to be alone in a circle.

to do this there are R(n-1) ways (where the pair number one is alone in a circle)

[potent shale]

Oh, i see, I misread the question.

[thick ivy]

now i need to figure out the number of ways for when our pair is not alone in a circle

well to arrange n-1 pairs there are R(n-1) ways, now if we want to include our pair in this arrangement ?

[potent shale]

We can forget the twins and think about one of them

[thick ivy]

if we can do that then it is R(n-1) * (n-1)

because the "pair" has n-1 people it can be near

but what i dont understand is why we can look at them as the same person ?

[potent shale]

because the second twin position is determined by the first

so you should only choose the position one of them

[thick ivy]

oh

and also switching between the twins doesnt add any possibilty since they look the same

[potent shale]

yes

[thick ivy]

R(n-1) + (n-1)R(n-1)

is my answer

[potent shale]

right

[thick ivy]

simplified:

n * R(n-1)

in closed form n1

n!

[potent shale]

yes, you can think about it as if you select where to put them between the other twins

and if you select the first place, you create a new circle for them

so n options

n * R(n-1)

[thick ivy]

yup

thanks a lot

[potent shale]

yw

[thick ivy]

this is what i missed

thanks a lot

< 3

[lethal spoke]

@potent shale

did you read my sol though

[thick ivy]

shashank you complicated it

[lethal spoke]

what did i do wrong

can't understand

[lethal spoke]

i guess

lol

[potent shale]

I did not

where is it?

[lethal spoke]

can't find it

[thick ivy]

the closed form solution is n!, @lethal spoke you got something with a fraction

[lethal spoke]

hmm

[thick ivy]

in help 9

[lethal spoke]

actually i didn't thought of it recursively

i misinterpreted

i calculated only the number of ways

that's why

[thick ivy]

its ok, they specifically asked for a recursive func

[lethal spoke]

hmm

[thick ivy]

yeah but still we got n!

you got smth with a fraction

[lethal spoke]

no no no, that's different

i calculated circular permutation

[thick ivy]

oh

yeah yeah right

[lethal spoke]

fraction came because there were identical an repeated cases

nvm

[thick ivy]

you understood the question in a different way i think

[lethal spoke]

question is solved

[thick ivy]

alr i think i am gonna close this

[lethal spoke]

yup lol

[thick ivy]

its alr thanks anyway

[lethal spoke]

i later on googled it

something came up and then i gave up

didn't understand a single thing

using my approach

yours was way better and straightforward lol

recursion is not my forte

[thick ivy]

haha its alr

goodluck thanks

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[short dock]

can someone pls tell me if my answer is correct?

[short dock]

i am unsure if they are equivalent, it doesnt look it to me. where did i go wrong?

[wise gate]

$\frac{2u}{u^3+u}$ does not equal $\frac{2}{u^2}+2$

[glossy valve]

d

[short dock]

oh silly me 😭😭 thanks sm, rookie mistake

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[gilded thorn]

Help

[gilded thorn]

Is this correct

[gritty rose]

you can use calc to check

,calc 8 + 9

[glossy valve]

Result:

17

[full forum]

@gilded thorn Has your question been resolved?

[gilded thorn]

Progress

how to simplify it on calcu

[gritty rose]

use ^2 for squaring

,calc 23^2

[glossy valve]

Result:

529

[gritty rose]

,calc 23 * 23

[glossy valve]

Result:

529

[gilded thorn]

can you show me?

if you can

[gritty rose]

right here

copy paste it but with your numbers

[gilded thorn]

simplifify in calcu?

[gilded thorn]

i get 529

so how to put this on calcu

[gilded thorn]

I'm done so is this good

Guys how do I do the last one

<@&286206848099549185>

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[gilded thorn]

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[gray orbit]

How do I make two tangents between two arbitrary circles?

[gray orbit]

I'm trying to figure out what what point along one circle to start, so that it is also tangent with the other circle. They will have arbitrary size and position

[torn jolt]

Start at any point such that the tangent meet the circles

Simple

[gray orbit]

what

[gray orbit]

I do not know what this point is.

I am trying to do math to find out what that point is

[torn jolt]

The tangent must satisfy the equation of both of your circles

[gray orbit]

could you explain why this new line goes from the top to the bottom?

[gray orbit]

by "the tangent" you mean the tangent you just drew?

[torn jolt]

Not really

[gray orbit]

cuz I don't think that is the case

...

[torn jolt]

The line I drew is also a tangent though

Should you draw 2 tangents or 1 common tangent ?

[gray orbit]

that is very cool, but I'm focusing on the lines I drew

those are the lines I need

they must be two tangents

[torn jolt]

1st , find the slope of the lines

[gray orbit]

each as depicted

I think the slope is also an unknown variable

[torn jolt]

Then , you see they have the same equation

[torn jolt]

Do you have the circle’s equation ?

[gray orbit]

I have the diameter, and their positions relative to each other

[torn jolt]

Because they both satisfy the 2 circles

[torn jolt]

Good

Now , construct a standard eqn

[gray orbit]

not sure what that is

[torn jolt]

(X-h)^2 +(Y-k)^2 = a^2

Where a’s the radius

[gray orbit]

hmmm are there not two radii? two circles?

[torn jolt]

You get 2 equations

You solve these equations

Or you solve these circles each and the lines each

Then you get 2 more linear equations

Now , you solve these 2 linear equations to get a common point

Or there’s another way

Don’t get confused

You first solve the red circle and the lower tangent

Then you find 1 point of contact on the red circle

You do the exact same thing with the green circle

And tada , you have a second point of contact on the green circle

Those are the points you draw your tangents from

@gray orbit

[gray orbit]

hmmmm sorry

actually chatgpt figured it out ;-;

gave me the formula tho so i can test it :)

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[tame elbow]

I just finished a math Olympiad comp and what the hell is this 😭😭😭

[tame elbow]

Pls help 😭

All I could find out is guess and check can’t work since a1 to a5 are real but probably like irrational or something but I could find an alternative method

[gritty rose]

Why do you think a_k can be irrational

You probably need to do prime factorization of 1183 and guess and check the permutations of each term in parens since the b-a_k terms are all distinct

[green merlin]

@tame elbow smo junior also cooked

i did 24 carelessed 1

anyway @tame elbow what I did was prime factorise 1183 and realised that the factors only have 1 choice

Good luck for getting into round 2

i carelessed q25 and put 6 😭

@tame elbow

blyatttt i got q19 wrong

@tame elbow u there

[onyx glen]

,w factorize 1183

[glossy valve]

Results provided by WolframAlpha

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

Requested by ann_dec

[green merlin]

||6 is what i got in the competition||

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[calm burrow]

https://codeforces.com/problemset/problem/1438/D

[calm burrow]

what I've thought :

/* 
    Find invariants 
    -> property or something that doesn't change after transformations 
    it can be used to prove the correctness of an algorithm 
    a, b, c -> change all of them to (a^b^c)

    after applying this operation, xor of this triplet is still the same 
    what does this tell us ? 
    This means total xor of the array remains constant always. 

    can I group the array in distinct triplets ? 
    suppose 3 | n 
    then is it always possible to make the array same -> one number at the end ? 
    yes
    now, 3 doesn't | n 
    n mod 3 = 1, 2 
    Does it depend on the parity of the number of p's we have at the end ?
    what is invariant here ? 
    what type of tranformations or operations should I apply ?

    if we know it is possible to make the array numbers equal -> what does that mean ? 
    n even 
    if totxor = 0 - yes

    n odd 
    equal number = totxor 
    cnt[totxor] >= n mod 3 
    but I just guessed it from the samples -> wrong 

    after that, I know n mod 3 = 0

    Do we ever have to apply coinciding operations ? 
    xor of the triplets can still be different and still produce the prescribed totxor
    
 */

can someone give a hint ?

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@calm burrow Has your question been resolved?

[rain vale]

Is this a doubt on coding or smthn?

[calm burrow]

it's just math

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@calm burrow Has your question been resolved?

[cursive pier]

@calm burrow what is the qyestion?

definition of invariance principle?

can you define what do you mean by distinct triplet

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@calm burrow Has your question been resolved?

[calm burrow]

https://codeforces.com/problemset/problem/1438/D

[calm burrow]

that's just my thoughts on the problem

[calm burrow]

triplets with different xor's

[sullen hinge]

@calm burrow lets think about just one bit across all the numbers, say the 2^0 bit

consider the number of 1s that appear in the 2^0 across the array

when we do an operation what are the possible outcomes that can come from this number (split in cases, e.g. three numbers with 0 in the 2^0, 2 with 0 and one with 1, etc)

i dont think ive fully solved the problem because of the 'at most n operations' condition but i think this provides a good start and insight into the problem

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[sudden oasis]

trying to find this maclaurin series, i'm almost there but missing a factor of 1-x^2 outside of the sum and a few other things that im pretty sure are just change of index.

[sudden oasis]

this is the answer

also ignore the last line on my page i didnt finish it

[gritty rose]

try writing out the first two terms: n=0 and n=1 separately from the summation

[sudden oasis]

k

like from my summation?

yeah i got nothin

tried both

[sudden oasis]

got anything else

[leaden ermine]

You need to split the sum I think.

[sudden oasis]

wdym split the sum

[leaden ermine]

Then the right sum you would index shift it so that it starts at n = 2.

Combining both then again. What a weird way to construct a solution.

[sudden oasis]

what the fuck i can do that

well i guess theres no reason i cant

[sudden oasis]

right???? some of these solutions are set up really weird

[leaden ermine]

Well the series converges.

[sudden oasis]

idk why they didnt just leave it like how i had it

it was good the way it was

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[left rivet]

how to i get a normal number 😔 in geogebra ive been trying and i have no please help

[stiff musk]

wdym normal number?

[left rivet]

something like this

[stiff musk]

you want to solve g'(x) = 0?

[left rivet]

i wanan find the growth rate so yh to the graf :> and i gotta make a presentajion about all this and i cant really do that with that number

bc doing just normal points gives me the number :>

idk if im going mental

[stiff musk]

well it's a function of x, so if you want numbers you do have to plug in x = something

[left rivet]

g'(x)=0?

[stiff musk]

g'(x) is never zero for your function

because the numerator is never zero

[left rivet]

o

wait sooooo do i gotta paste the number on the x line here to get the point

[void nova]

What is your actual task? Like, what did your teacher tell you to do?

Not but your own words, but the original

[left rivet]

find the growth rate of the graf

if i translated that right

im not going mental am i it is g'(x)

idk with my geogebra is going insane bc its big numbers or im messing something up

[stiff musk]

!original

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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.

[left rivet]

idk if this also helps but i choose a logistic regression model ?

[stiff musk]

chatgpt's attempt at a translation to english:

let me know if that doesn't look right

[left rivet]

looks good

[stiff musk]

now can you show the original problem statement for the one you're working on

[left rivet]

all this?

[stiff musk]

the question as originally stated to you

[left rivet]

why cant i get a normal number when trying to find the growth rate to the graf

[void nova]

Wdym with "normal" number?

[void nova]

I mean, this is just... the derivative

As a function of x of course

[left rivet]

i want the derivative in the spot where the growrate is highest

is it not just putting in g'(x)

and it pops out a number

like 2300101

[void nova]

Then you have to find the max of g'

[left rivet]

thats just putting it to inf

6402878.257413

right

[void nova]

Huh?

[left rivet]

thats the max aint it 😭 or am i going insane

thats like the Vertical asymptote where it maxes at

right

[void nova]

Suppose you wanted to find the max of a function (let's call it f)
What steps would you do?

[left rivet]

put the limit to inf g>inf

[void nova]

Nope

[left rivet]

https://tenor.com/view/2-gif-7749019966943875056

[void nova]

(whatever that means)

[left rivet]

im confused now.

[void nova]

Does the word derivative ring you a bell?

[left rivet]

yh

the thing i hate most rn

do i do the g''

but the graf is a logistic

aint you suppose to make it go to inf

max value is always L

[void nova]

Where have you heard about this inf lol?

[left rivet]

um school did they mess up

or something

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[glad drum]

for b) how did they get that? why does the ln have to be less than 0?

[shut thorn]

,, x = \frac{-1}{k} \ln(\frac{3}{k})

[glossy valve]

Bettim

[shut thorn]

what do you know about the domain of logarithm function? @glad drum

[glad drum]

x has to be larger than 0?

[shut thorn]

yes

so here $\frac{3}{k}$ has to be larger than 0

[glossy valve]

Bettim

[shut thorn]

which means we know that k is also larger than 0

since we know that

we can conclude the term $\frac{-1}{k}$ will always be negative

[glossy valve]

Bettim

[shut thorn]

so for x to be positive, the term $\ln(\frac{3}{k})$ must be negative

[glossy valve]

Bettim

[shut thorn]

$\ln (\frac 3k) < 0$

[glossy valve]

Bettim

[shut thorn]

got it?

[glad drum]

i still dont get this part

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@glad drum Has your question been resolved?

[shut thorn]

x = (some negative number) (logartihm)

if x needs to be positive

should the log term be positive or negative

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[glad drum]

ohhhh negative

[glad drum]

okk thank you!

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[full forum]

Please don't occupy multiple help channels.

[ember solstice]

My other occupied channel is a completely different question.

Why does this approach not work for the given problem?

[lime ether]

36 = 9 * 4

[ember solstice]

Sorry, I don't understand how this disproves my approach.

I am NOT saying that my approach is definitely correct.

[lime ether]

because you can have a perfect square that factors into two factors that aren’t equal

[ember solstice]

Okay, so you're saying some of factors of a square number might "bleed into" (for lack of better term) one of the other factors.

[lime ether]

sure

[ember solstice]

Thanks!

Anything else you need to add, or can i close?

[lime ether]

a perfect square need not be factored into the product of its square roots

[lime ether]

you can close

[ember solstice]

.close

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[hushed stone]

how to do 14

[hushed stone]

did i do it right?
something feels off

[analog sand]

hm you've definitely done something wrong if you considered a to be the first term of the gp and got r>1. Did you?

[hushed stone]

yep i considered that

[hushed stone]

but wasn't mod r suppose to be less then 1?

[analog sand]

yes it was

so you've done something wrong, let me check

[hushed stone]

ok

[analog sand]

it is +6 not -6

[hushed stone]

oh

[analog sand]

@hushed stone

yeah lol

[hushed stone]

ye I see it

man i did all the hard stuff but fucked up subtraction

[analog sand]

hits hard ik

[hushed stone]

kinda sucks

well ima go do some more

thanks

[analog sand]

alr cya

[hushed stone]

cya

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[rich basalt]

1 A wood turner carves out a bowl according to the formula d = 1/3 x^2 − 27, where d cm is the depth of
the bowl and x cm is the distance from the centre of the bowl.
a Sketch a graph for −9⩽x⩽9, showing x-intercepts and the turning point.
b What is the width of the bowl?
c What is the maximum depth of the bowl?
please help me with graph especially

[analog sand]

the graph part would just be a parabola

[rich basalt]

ya i know that bit but idk how to set up the graph since the examples set it up weird

[analog sand]

so for the graph x^2= 3(d+27)

taking y axis as d, the vertex of the parabola would be at (0,-27)

[rich basalt]

how would i label the points?

[glossy zodiac]

hii

[analog sand]

now you've to sketch it from [-9,9]

[rich basalt]

hihi

[analog sand]

well the point where x=0 is the axis of the center of the bowl

so the vertex of the parabola

[glossy zodiac]

i think

[analog sand]

is the center of the bowl

[glossy zodiac]

is parabola symmetrical

[rich basalt]

so likw -9 and then some space then 0 then 9? on the x axis

[analog sand]

about?

its symmetrical about y axis

[glossy zodiac]

vertex will be

-b/2a

yeh

o is vertex

[analog sand]

yeah see because you need to notice that when x is 9 or -9

d becomes 0

[glossy zodiac]

since b = 0

[analog sand]

so those are the edges of the bowl

[rich basalt]

wait so hwo would i label the y axis then

[analog sand]

i'll draw a basic diagram so you can understand

[glossy zodiac]

are

[rich basalt]

ok ty

[glossy zodiac]

roots are

-9 + 9

vertex is 0,-27

i guess

[analog sand]

drew it freehand

so its ugly af

[glossy zodiac]

thats beutiful with a mouse lol

[analog sand]

lol ty

[rich basalt]

omggg im actually dumb i didnt know it wanted me to do that wtf 😭

ty guys

[glossy zodiac]

nah

[analog sand]

yw

[glossy zodiac]

that question is interesting

[rich basalt]

😂

how to close

[glossy zodiac]

.close

[analog sand]

.close

[rich basalt]

.close

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[covert heath]

not sure how to do this.

[covert heath]

i dont recognise it as any form i know

its homogenous but not zero degree

[glossy zodiac]

i dunno

just dericate the options

it will be lenghty

but it kinda helps to see the answer

wait

take

x- y as u

as substitute

[covert heath]

i assume i can do stuff for homogenous DEs even if its not zero degree but idk how to

[glossy zodiac]

try this

1- du/dx = u^2

du/ 1- u^2 = dx

integrate

use partial fractions

@covert heath is option 1 correct

[covert heath]

yeah

[glossy zodiac]

yeh

try that way pecy

percy

take x -y as u ig

[covert heath]

hmm

okay

yeah

[sour pine]

yeah that works

$$ x - y = u $$

[glossy valve]

grey

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[blazing folio]

Hey guys I need help to understand this exercise, if im right, this is a quadratic function where the parabola is crossed by a line(btw the Spanish text means "solve the exercises, factorize and simplify if required"

[steel solar]

Uh, for part a)?

Or for all the parts in general

[blazing folio]

If you can all but with the first part it enough

[steel solar]

I think you just need to factor all of it

For b, c, and d, then for a you just expand

[blazing folio]

What do you mean?

[steel solar]

e.g. for b

You will need to find common factors between x^3+8 and x^2+2x

[blazing folio]

Oh, thanks you so much

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[thorn marten]

Find the 6537th digit after the decimal point in:
the sum from i = 1 to i = 58474 of 1/i^2

[brave ruin]

[onyx glen]

i would be very surprised if there's any sort of nice trick for that beyond just doing the summation with that stupid amount of precision

what all do you need a decimal digit that far deep for, anyway

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@thorn marten Has your question been resolved?

[gritty rose]

Waste your time

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@thorn marten Has your question been resolved?

[analog sand]

bros trollin

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[sour pine]

it's identity based

[winter escarp]

i hate

this

i cant understand

[sour pine]

do you remember sin ( a + b )

[winter escarp]

why sin(a)cos(b) + sin(a)cos(b) is

2sin(a)cos(b)

[sour pine]

huh

[winter escarp]

why isnt it just 2cos(b)

[sour pine]

it's not

[winter escarp]

BUT IT IS

if u forumalte it out

u get to

sin(a)cos(b) + sin(a)cos(b)

and u left with that

to = 2sin(a)cos(b)

[sour pine]

$$ \sin(a+b) = \sin(a)\cos(b) + \cos(a)\sin(b) $$

[glossy valve]

grey

[sour pine]

this is the identity

[winter escarp]

yeah but u gotta add the other identity to it

so it cancels out

cos(a)sin(B)

then u left with

[sour pine]

bro put - in place of + here

then add both

[winter escarp]

this

[sour pine]

then you get this

[winter escarp]

YUES

so why isnt

it

2cos(B)