help-10

[timid silo]

Let me try again

[knotty crow]

It's fine, let's redo it

[timid silo]

But it’s true that x^2-2x-1 cannot be factored right

Like without quadratic formula

[knotty crow]

no, it's not true

every quadratic for which we have D >= 0 can be factored over the reals

[timid silo]

Ok but how

Without the quadratic formula

How is it possible

[knotty crow]

I mean "factored" means it gives ( ... ) * ( ... )

[timid silo]

Without the quadratic formula, is it possible?

[knotty crow]

maybe not using standard methods, but eventually

yes, it's possible

but it can be problematic

But you know, by saying for polynomial that it's factorable we mean it can be rewritten as product of terms of degree 2 at most

Doesn't matter if we use quadratic formula or not

[timid silo]

What I meant is it is not factorable using standard methods such as how I did here

[knotty crow]

fact that you couldn't factor it using your method doesn't mean it cannot be factored

for example x^2 + 1 cannot be factored

further

using any method

(even quadratic formula)

[timid silo]

@knotty crow why does the highlighted have a multiplicity of 2, I’m not too sure but I’m guessing cuz we sqrt to isolate , can u confirm?

Same thing, different question

[knotty crow]

it's not of multiplicity 2

here all three roots are distinct

or do you mean no. 9?

c)

[obtuse pebble]

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[agile cedar]

In a video game:
Weapon A can kill a monster with either 1 crit shot or 3 non-crit shots. It has a 38% chance to crit.
Weapon B can kill a monster with either 1 crit shot or 4 non-crit shots.
At what crit chance is weapon B more efficient at killing monsters than A?

[tired shell]

do you want lowest expected number of shots?

usually it's either that or retaining the most hp which might depend on the game

[agile cedar]

Yes lowest expected number of shots. So for example lets say each weapon fires 100 shots, how many monsters can each weapon kill?

For Weapon A, I expect 38 out of 100 shots to crit but I don't know how the crit/non-crit shots will be distributed.

If Weapon B has 51% chance to crit, will it outperform Weapon A?

[tired shell]

looks like anything above 45.5% works https://www.wolframalpha.com/input?i=.38%2B.62*.38*2%2B.62^2*3%3Dx%2Bx*(1-x)*2%2Bx*(1-x)^2*3%2B(1-x)^3*4

[agile cedar]

Could you possibly explain to me how you came up with it? Entry level of math here 😅

[tired shell]

yea, if something took 1 shot with 30% and 2 shots with 70%, the expected value would be 1*30% + 2*70%

you just multiply the shots with the chance, and add them up

so I did that with the 3 possibilities for A and 4 possibilities for B, where the possibilities look like (shoot accurately)*(miss past n-1 shots)*n

[agile cedar]

I'm not sure if you understood the problem. Maybe I didn't explain it well enough.
Lets say I generate a string of 62 dots and 38 dashes in random order and another one with 54 dots and 46 dashes.
Each dash represents a crit and each dot a non-crit
As an example: for the Weapon A, "-...--.." this would be 4 kills and for weapon B that would be 3 kills

[tired shell]

yup that's what this math answers, on average

[agile cedar]

Ok great thanks!

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[grim ocean]

hello idk how to do c, forming a generalised formula for this

[meager bolt]

well just try constructing a worst case

hint: you can derive it with a)

[grim ocean]

for a, i did 1*3+1

hm

[meager bolt]

thats correct

[grim ocean]

(n-1)(n+1)+1?

[meager bolt]

i would guess its 4N

how did you optain (n-1)(n+1)+1??

[grim ocean]

using a) as an example, n=2 so its (2-1)(2+1)+1 which is (n-1)(n+1)+1?

[meager bolt]

if we toss the pair 4 times, we are guaranteed to get one outcome 2 times

because there are 3 possibilities

HH TT and HT

so if you want to guarantee getting one outcome N times

[grim ocean]

but doesnt 4N mean 4 * the one outcome occuring N times

wudnt that give more than 4

[meager bolt]

w8 im confused aswell

no i think

it should be 4N

[grim ocean]

but like if i use a) as example, N=2 so 4N is 8 numbers of tosses??

but its 4 tosses in total?

[meager bolt]

what do you mean

i didnt understand

the answer to a is 4

[grim ocean]

yea but if we use 4N in a, its gonna give us 8 as the smallest number of tosses cus its 4*2 (when its meant to be 4)

so.. idk

[meager bolt]

ah

[meager bolt]

i mean is the answer to a 4?

[grim ocean]

mhm

[meager bolt]

its 2

for a

just assume tossing two times HH

[grim ocean]

o crap i forgot to send answers

[meager bolt]

i mean

i get what they mean

but i think the task is formulated wrong

[grim ocean]

oh

[meager bolt]

What is the smallest number of tosses of the pair required to get at least one outcome occurring twice?
Answer: in the best case its 2 tosses

in order to have it guarranteed you need 4 obv

[grim ocean]

yes

[meager bolt]

confusing

3(n-1)+1 aswell

[grim ocean]

ye i was confused by that

[meager bolt]

no now it makes sense

[grim ocean]

??

oh wait actually

i get it nvm

thanks for helping 🙂

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[cobalt walrus]

Is this right?

The answer sheet shows a different answer 5/108π

[obtuse pebble]

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.close

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[drifting flume]

Hi. I have no idea how to show this identity. All I have is that k! S(n,k) is equal to an ordered set of k blocks for [n].

[obtuse pebble]

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@drifting flume Has your question been resolved?

[tired shell]

for a "both sides count this scenario" proof, I see both sides count fubini numbers, which is the number of ways to order n elements with ties allowed

so for the left side, Stirling counts the number of nonempty partitions, and k! counts permutations, so you can say the left side is k=number of partitions between ties or equal signs, and k! orders them which gives you fubini numbers

to finish with the right side, try searching what eulerian numbers count and try to factor in 2^k to give another way of counting orderings with ties

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[lime heath]

hi peeps! trigonometry isnt rlly my strong suit so i have no idea on how to solve this trigonometry pile up, is there anyone available to help?? the answers should have 4 decimal places i believe

[swift sky]

My guess would be to use trigonometric functions

To solve for the needed side

Cos first one

[lime heath]

HELP- how do u even do that chile..

[verbal narwhal]

cosine = adjacent/hypotenuse. The adjacent side of the 34 degree angle at the bottom is 8 cm. So cosine(34°)=8/hypotenuse. Solve for hypotenuse. Then subtract 2.5 cm to get that unknown length, and add it to 3.6 cm to get the hypotenuse of the next angle. Repeat until you get to the top.

[lime heath]

oh like this?

[verbal narwhal]

yes, that works

[lime heath]

then next is this?

is this right OML

[verbal narwhal]

yep, that's right

[lime heath]

how abt this omg IDK

OOP

[verbal narwhal]

yes, that's correct

[lime heath]

then next is the blue one right?

[verbal narwhal]

yep, all of that looks right to me. Seems like you have a good handle on this, I don't think I need to check your work every step of the way lol.

[lime heath]

HAHA, wait pls i might be wrong huhu

wait imma send the whole thing

js tell me if theres a wrong computation huhu

cz i want to make sure that its correct huhuhu

[verbal narwhal]

just a sec

yep, I don't see any errors, so it's safe to say you're correct

[lime heath]

thank you smmmm!!!!

[obtuse pebble]

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[cobalt pelican]

I need to find the angle BAC

[rapid summit]

is this all the information you were given?

[obtuse pebble]

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[merry girder]

11

[merry girder]

idk how to do it

[high lily]

sub in the points, solve the system of equations to determine k and c

[merry girder]

ok thanks

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[vocal shell]

What am I doing wrong? My answer doesn't line up with the graphing calculator.

[vocal shell]

Thanks for not helping but you made me notice the mistake

I didn't distribute the 2 at the end correctly

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[rigid sky]

wolfram alpha is nto giving an answer

[cunning burrow]

What techniques have you tried?

liebniz rule?

[timber pine]

does using ln rules help

[rigid sky]

um

f(a+b-x) rule

[high lily]

wolfram should be able to give an answer for that

[rigid sky]

the answer i got is 0

[high lily]

that is correct

,w integrate cos(x)log((1+x)/(1-x)) from -1/2 to 1/2

[warm shale]

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[rigid sky]

okay thanks

[rigid sky]

damn

,w integrate x/(1+cos2x+sin2x) from 0 to π/4

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[rigid sky]

wut

i got negative 😭

[rigid sky]

can someone help

,w integrate ((2x+3)/x^2(x+3))

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[timber pine]

This?

expand (2x+3)(x+3) into 2x^2+9x+9 then divide by x^2

2+ 9/x + 9/x^2

[high lily]

show your work

[obtuse pebble]

@rigid sky Has your question been resolved?

[obtuse pebble]

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[lost otter]

i have a question

[lost otter]

so like

integral of (1/1+e^-x) dx

[normal osprey]

?

[lost otter]

i got it as 1/e^-x ln(1+e^-x) + c

looks like its wrong

idk why tho

[normal osprey]

looks right to me

[lost otter]

yeah

but its not

,w integrate 1/1+e^-x

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[lost otter]

eh

,w integrate 1/(1+e^-x)

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[lost otter]

see

why is this the case

[severe reef]

How did you arrive here

[lost otter]

Well

Um

How do i show u

One sec

this

[severe reef]

Well that's...

Too many violations

You can not move anything containing x just out of the integral

[lost otter]

o

h

[timid silo]

yeah

if it was 2 * something

then yeah sure

2 integral of ....

[lost otter]

uhuh

so what do i do

[severe reef]

Look at e^-x

[timid silo]

yeah

what can you do to him

[severe reef]

Try writing it as 1/e^x

[lost otter]

Eh?

[severe reef]

Did you try this

[lost otter]

how do i write e^-x as 1/e^x

like, the numerator right?

[severe reef]

a^-b = 1/a^b

Basic laws of exponents

[lost otter]

Yeah

Thats true

What then

[severe reef]

Cross multiply and simplify

You'll see a beautiful numerator for the denominator

[lost otter]

oh

bet

Wait

Wouldnt the numerator be

-1/e^x

[severe reef]

We could have also just multiplied both numer and denominator by e^x to get the same thing

[lost otter]

Instead of 1/e^x

[severe reef]

What

[lost otter]

Yah?

[severe reef]

No

After doing one of the two methods I suggested

[lost otter]

Where did the e^x come from anyways

[severe reef]

What do you mean

[lost otter]

Wait

In the numerator

How did u ge

t

e^-x

in the first place anyways

[severe reef]

We never got it?

[severe reef]

Do this

And you'll see why

[severe reef]

In the original expression

[lost otter]

Oh

[severe reef]

It seems you have trouble simplifying fractions so let's just multiply both numerator and denominator by e^x instead of writing e^-x as 1/e^x

[lost otter]

okay

so that'd be

$\int\frac{e^x}{\times{e^x){1+e^x}}$ ?

[warm shale]

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

[lost otter]

oh

[severe reef]

You should be getting e^x/(1+e^x)

Yes

[lost otter]

yes

wait no

i got

e^x/e^x(1+e^-x)

[severe reef]

Yes distribute that e^x

In the denom

[severe reef]

Getting this

[lost otter]

Ohh

Yah

Mhm

Then?

Oh wait

[severe reef]

Now do you see something beautiful

[lost otter]

OH

Wait a min

[severe reef]

What is the derivative of the denominator

[lost otter]

if u do this

You get

e^x/e^x + e^(x-x)

that becomes

[severe reef]

.

[lost otter]

e^x/e^x+1

[severe reef]

Yes

[lost otter]

OH

YA

H

WAITTT

soo

isn't that

ln(e^x+1)

[severe reef]

Yes

[lost otter]

so the ans is

ln(e^x+1) + C?

[severe reef]

Yes

Gj

[lost otter]

But

Howd u know

to multiply num and denom

by e^x

@severe reef

[obtuse pebble]

@lost otter Has your question been resolved?

[severe reef]

Kinda comes with experience

It was kind of intuitive to write it as 1/e^x tho

The more integrals you solve the more things you'll 'predict' about the expression

[obtuse pebble]

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[crisp marten]

Does this formula for a derivative of x^m work when m is a fraction?

[crisp marten]

Of only if it's an integer?

[median dome]

Yes it does

It works for all real values of m

[crisp marten]

Got it

Thanks

.close

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[kind hawk]

.reopen

[obtuse pebble]

✅

[kind hawk]

nvm.

.close

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[median dome]

Did I say something wrong?

[kind hawk]

nah

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[dusk fractal]

hi, could you check my answers? I got 1. B 2. E 3. C and 4. F. Although i feel pretty confused, because at first i though that everything below the x- axis is negative. So i dont understand why A2, is a postivie number.

[tender stratus]

they mean the magnitude of area is 50 prolly

I feel the first one should be -31

oh it seems you've taken int a to b f(x) as +50

I do think it should be -50

[dusk fractal]

yeah but the thing is, if you start to consider 50 as -50 then the problem is that, once you get to the third equation you realize that the 3. one is -31 too.

and that cannot be..

[tender stratus]

no?

that would be 17 - (-50) + 2 = 69

[dusk fractal]

._.

damn those minuses

i see, then 1. A 2. C 3. B and 4. is D

[ember frost]

!1q

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[dusk fractal]

.close

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[ionic sandal]

hey i have a problem where i have to vary x from 1 to k in the 2 expressions [ (k-x)/2, x ] and then take the smaller value out of the two expressions [ (k-x)/2, x ], and check for which x the smaller value is maximum

[ionic sandal]

how do I do that exactly?

I've thought of minimizing the expression | (k-x)/2 - x| and then finding the corresponding x, is that valid?

<@&286206848099549185>

[crimson jetty]

So if im getting this correctly we want (k-x)/2 < x and (k-x)/2 maximum at the same time?

[ionic sandal]

not really, more like we want to find an x' such that min( (k-x')/2, x') ≥ min( (k-x)/2, x ) for all x such that 1 ≤ x ≤ k

[crimson jetty]

ah shi

[ionic sandal]

any ideas?

[crimson jetty]

can we just compare all the elements?

so (k-x')/2 > (k-x)/2 and > x

x' > ...

[ionic sandal]

they won't hold together though

[ionic sandal]

i guess so but i was trying to look for a "neater" way

[crimson jetty]

this would also work

tho

find k in both cases

im still not fully getting the question but yeah

[ionic sandal]

alright

[obtuse pebble]

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[heavy badger]

Can someone tell me how we came out with (1, x)
And why not the versa (x, 1)
In the graph

[last pilot]

cos(arctan x), yes??
well, write out cosine in terms of tangent

hint: sec^2 x = tan^2 x - 1

oh wait mb

i messed up

[heavy badger]

I think this require me to memorize all the converting laws

[last pilot]

no

[heavy badger]

:/

[last pilot]

look, this is the thing i want you to look at

[warm shale]

ren

[last pilot]

$$\frac{1}{\cos^{2}x}=\frac{\sin^{2}x}{\cos^{2}x}+\frac{\cos^{2}x}{\cos^{2}x}=\tan^{2}x+1 \rightarrow \cos^2 x = \frac 1{\tan^2 x + 1}$$

[warm shale]

ren

[heavy badger]

Alr alr
But where's the point..
My q was why we put x and 1 as x, y..

[last pilot]

my man

try to solve for cos x

then plug x = arctan t

;-;

u'll see what i mean

im not a fool

[heavy badger]

Im still wondering..
In the video.. He jumped into writing x, 1 after knowing which quad to work in
So idk what to do after getting the quad :/

[heavy badger]

I get artan(x) for x and solve for cos
x<0 so cos is in the quad ii
What next..

[last pilot]

what???

bro

[last pilot]

USE THIS

[heavy badger]

Ah

[heavy badger]

Is this what u meant?

Its not the final answer tho

[last pilot]

bro

tan^2 t = (tan t)^2????

u can solve??!?!?!?

hello1??!?!?!?

[heavy badger]

Ah yeah I messed

[last pilot]

;-;

!done

[obtuse pebble]

If you are done with this channel, please mark your problem as solved by typing .close

[heavy badger]

I think I will just start from the zero
Not attending the class made me forget almost everything :/

Alr, I wont come back here I any sooner..

.close

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

im trying to work on this equation for overdamped current. how would i go about choosing values of A and B?

[timid silo]

i'm trying to plot cases of over, under, and critical damping, so i suppose i have to use some sort of thinking when choosing the values.

this is regarding LRC circuits

[tired shell]

this is when you factor out an e^ct that makes the exponential sum Ae^ikt+Be^-ikt to let you convert to cosine with this formula

I think at the end critical damping occurs at a certain ratio of the constants? I'd have to see what it ends up as

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

the image i provided was for overdamping. critical damping occurs when there is two equal roots to the equation ' Lk^2 + Rk +1/C = 0', and you solve it like any other ODE.

i havent heard or read anything about the constants being at any specific ratio

[crimson jetty]

Don't A and B arrise from +C from integrals?

as in A and B have to be set by the enviroment

[timid silo]

right, its an initial value problem.

but this circuit exists on papper. im not actually measuring anything.

[crimson jetty]

hold on i actually have notes on that somewhere with me

[timid silo]

that would be great, thanks.

[crimson jetty]

nvm its just for derivation

where A,B are independed on starting ODE

[timid silo]

i can get to i = some function of time for over, under, and crit. i can find the values for k. just not A and B as im unsure how i am supposed to choose them. im pretty sure i'm supposed to choose them

[crimson jetty]

oh

1 sec

you mean that kinda stuff?

[timid silo]

yes

[crimson jetty]

when discriminant < 0 we have x12 = A * exp(-gamma*t) * exp(+-i omega t)

where the 1st part is a new amplitude

[timid silo]

yes

so i need to figure out N1 and N2, the amplitudes. i have the exponents sorted.

i feel like theres something obvious im not understanding. if i had initial values for i(t) and di/dt (t), i could find these using simultaneous equations.

i can say i(0) = 0, so my plots will start at (0, 0)

[crimson jetty]

you said k^2 here

is that actually k^2 or did you mean 2nd derivative

from the general formula:

L dI/dt + RI + q/C = 0

[timid silo]

k raised to the second power

k*k

[crimson jetty]

aight

so we have Lk^2 + Rk +1/C = 0

[crimson jetty]

which has to satisfy this

then k^2 = dk/dt

?

also the q in the formula is 1

[timid silo]

this is where k is comming from

often lambda is used

[crimson jetty]

ffs im too dumb to make sense of this

i'd be better if it was a mechanical wave

[timid silo]

if you can make an analogy to a mechanical wave, please do lol

[crimson jetty]

if we dont have a starting condition then A doenst depend on k

any kinda A solves the ODE

so what am i missing

[timid silo]

just a second, i cant tell if im stupid

[crimson jetty]

nah its probably me being stupid

if its overdamped it means the discriminant somewhere above is > 0

gamma^2 - omega^2

[timid silo]

well yes, any value of A will solve this. but how do i choose one sensibly

[crimson jetty]

idk what else this shit gives us

[crimson jetty]

ur usually just given it in starting conditions on a test

say i(0) = 5

[timid silo]

yeah, thats what's throwing me off.
because this isnt for a test. its for a lab, where i have to make some sort of model for over, under, and critical damping

[crimson jetty]

oh

yeah then A and B doesnt matter

;-;

its still gonna be an overdamped wave

[timid silo]

true true

[crimson jetty]

naww dont tell me we've spent 15 minutes on this

i was tryna make some sense and doubt myself

[timid silo]

useful experience maybe? confidence in yourself?

[crimson jetty]

sure

ill take that

so your models for different damping will only depend on the dampening constants

or dampening constant and stiffness

which in your electricall case is whatever

R and C

[timid silo]

i think what i'll do is set i(0) = 0, i'(0) = 0 to get some sort of consistency

because at the moment, i get order of magnitude differences between models for magnitude of current

[crimson jetty]

careful with that

if ur using exponential and not trig form

i(0) means A * e^0 = 0 => A = 0

[timid silo]

yeah, so my graphs will simply start at 0

[crimson jetty]

no like

amplitude would be 0

everywhere

unless ur using A*sin(whatever)

[timid silo]

i(t) = Aexp(kt) + Bexp(Kt)

i(0) = Aexp(k0) + Bexp(K0)
i(0) = A+B = 0

di/dt(0) = kAexp(k0) + KBexp(K0) = 0

0 = kA + KB.

i know k and K, so i can solve for A and B using simultaneous equations. i have set the initial conditions, and these values of A and B will satisfy the initial conditions

[crimson jetty]

yeah sure that works

[timid silo]

awesome

i'll leave this for a couple of minutes just in case i've done something alarmingly stupid, but after that, ill close it. thanks

.close

[obtuse pebble]

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[full cloud]

I need to prove that X (the reflection of H, the orthocentre about BC) and Y (reflection of H about E, the midpoint of BC) lie on the circumcircle of ABC

If I'm able to prove AXYC is a cyclic quadrilateral, I believe that should be enough to prove that X and Y lie on the circumcircle

So I know HXY is 90 as triangle HXY is similar to HDE

I need to prove ACY is 90

I can get to there by proving that angle HCY is 90. I can clearly see that triangle EYC is similar to EHB and if I'm able to prove angle EHB is 90, then angle HYC would also be 90

Till now all I could figure out is that BHD = 90 - HAC = ACB

[obtuse pebble]

@full cloud Has your question been resolved?

[full cloud]

<@&286206848099549185>

Ok it turns out, I have made a mistake

[full cloud]

This is incorrect, HCY doesn't have to be 90 for ACY to be 90

Somehow, I need to prove ACY is 90

<@&286206848099549185>

Aha

I think I got it

[surreal forge]

is this the ultimate task?

[full cloud]

ACY = EBH + ACB (EYC ~ EHB)
And clearly by observing triangle CBJ, EBH + ACB = 90
Hence, AXYC is a cyclic quadrilateral

[full cloud]

No, I have mentioned the ultimate task following this image

[full cloud]

Here

@surreal forge sorry, but does proving AXYC is cyclic enough to say that X and Y lie on the circumcircle of triangle ABC?

[surreal forge]

that alone is insufficient

[full cloud]

But why?

[surreal forge]

you'd also have to establish at least one of X and Y actually lies on the circumcircle

[full cloud]

Oh

Yeah that makes sense

If I can prove that AYXB is also cyclic, does that help or no?

[lucid glade]

It's easier to separately prove statement for each point (X, Y)

[full cloud]

Maybe, but I want to continue what I did

[full cloud]

I'll try doing this after

[obtuse pebble]

@full cloud Has your question been resolved?

[full cloud]

@surreal forge

I'm able to prove this

[surreal forge]

that should prove ABXYC are concyclic

[full cloud]

Right, so that would mean X and Y lie on ABC's circumcircle correct?

Because the circumcircle is unique

[full cloud]

Here BHCY is a parallelogram (diagonals bisect each other) hence angle HCB is equal to angle YBC, similarly angle YCB is equal to angle HBC. It's easy to show angle HCJ is equal to HBA. Hence, angle ABY must be 90. Since angle ABY = angle AXY = 90, ABXY is cyclic with AY being the diameter of (ABXY)

[full cloud]

@surreal forge sorry for mentioning you again 😅

[obtuse pebble]

@full cloud Has your question been resolved?

[obtuse pebble]

@full cloud Has your question been resolved?

[obtuse pebble]

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[craggy heart]

So the NBA has 30 teams that play a season of 82 games. The league is split in two conferences with 3 divisions in each.
4 games against the other 4 division opponents (4×4=16 games)
4 games* against 6 (out-of-division) conference opponents (4×6=24 games)
3 games against the remaining 4 conference teams (3×4=12 games)
2 games against teams in the opposing conference (2×15=30 games)

• A five-year rotation determined which out-of-division conference teams are played only 3 times.
  How would I use this formula for 256 teams?

[obtuse pebble]

@craggy heart Has your question been resolved?

[zenith raft]

i don't think there is one particular "correct" way to generalize it to 256 teams, especially with no other info (e.g. do you want each team to play 82 games still? do you want 2 conferences? etc)

[craggy heart]

2 conferences 128 teams each 8 divisions 16 teams each division

thats the thing I dont know how many games should be played

I was assuming this formula maybe could help with the games played

[obtuse pebble]

@craggy heart Has your question been resolved?

[obtuse pebble]

@craggy heart Has your question been resolved?

[obtuse pebble]

@craggy heart Has your question been resolved?

[hallow prairie]

i dont think ur question is complete, cos like there's isnt enough info to go off of

[obtuse pebble]

@craggy heart Has your question been resolved?

[obtuse pebble]

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[humble storm]

To prove the two are Continuous

[obtuse pebble]

Please don't occupy multiple help channels.

[humble storm]

Helps

[pastel wren]

can you write down those functions? i dont use desmos often

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

What is the x angle

[swift sky]

Use the corresponding angle

Wait

Nvm

I can’t tell

Can you show the original problem

[obtuse pebble]

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[sweet pond]

How do you do this

[cunning breach]

Make them to one fraction.

[sweet pond]

oops not that

this one

[cunning breach]

The same, find the common denominator.

[sweet pond]

erm

so lik

[pastel wren]

add fractions

[sweet pond]

how do you do that

[pastel wren]

dont you know how to add fractions?

[sweet pond]

no i do

but like

oh wait oops

5+6?

[pastel wren]

5+6????

its 11 bro

[sweet pond]

no im asking if im supposed to do it like that

[pastel wren]

like what

[sweet pond]

solving the numerator first then the denominator cause dont you need to solve the lcm for the denominators

nvm

.close

[obtuse pebble]

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[midnight gorge]

wait can you help me

@pastel wren

[obtuse pebble]

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[whole dock]

Looks correct

$\pm8$

[warm shale]

ColdTee

[obtuse pebble]

@open walrus Has your question been resolved?

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[pastel wren]

@open walrus its not +-8 its just 8 you did it right

[obtuse pebble]

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[frail depot]

How can I prove this by induction?

I have done the base case

And assumed the induction hypothesis (n=k)

But I am stuck on the n= k+1 part

[obtuse pebble]

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[fallen arch]

$\int[x^2cos(nx)]dx = \frac{(-1)^n4\pi}{n^2}$

[warm shale]

ElPanaArturo

[fallen arch]

from $ {-\pi} to {\pi}$

[versed stratus]

you want this verirfied ?

[fallen arch]

so

the intervale is a=-pi b=pi

help

help

help

[obtuse pebble]

@fallen arch Has your question been resolved?

[obtuse pebble]

@fallen arch Has your question been resolved?

[full cloud]

What's your question

@fallen arch

[fallen arch]

solve and evaluate the integral

[cinder basalt]

n es un entero o cualquier numero?

[fallen arch]

entero

[cinder basalt]

,w integral from -pi to pi of x^2cos(nx)

[warm shale]

Results provided by WolframAlpha

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

Requested by svc20060505

[cinder basalt]

esta casi bien pero falta algo

fijate que cos(pi n) puede ser negativo o positivo dependiendo de si n es par o impar

estoy ciego olvidalo

si lo pusiste

no entiendo que te falta ya lo tienes hecho

[fallen arch]

q

la demostracion

[hazy spoke]

1+1 2

[obtuse pebble]

@fallen arch Has your question been resolved?

[fallen arch]

No

[obtuse pebble]

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[timid silo]

i know how to like do the y = (1-r)^t thing

but we didnt even learn what to do if we had a variable in the exponent

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[tired shell]

you can get rid of that 6 in the exponent with this rule

so that it becomes 1700(0.75^6)^x

the logic is that it's like applying 0.75 six times each

[obtuse pebble]

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[north anchor]

How does one belong to the vector space?

Given the vectors:•
v1 = <4, 8>•
v2 = <3, -5>•
w = <1, 31>•
z = <-1, 31>•
Given the scalars:•
k1 = 2•
k2 = -3• |
Find the vector v3•
Determine if the vectors w, z belong to the vector space of v1 and v2•
Explain.

I need ; Determine if the vectors w, z belong to the vector space of v1 and v2•

[obtuse pebble]

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[obtuse pebble]

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[grand crater]

How do i find the derivative of this using product rule fgh

[fallow reef]

sqrt(x) = x^(1/2)

[grand crater]

is it fine to do it without multiplying the x terms first? or would that make it too hard to simplify later

[swift sky]

It’s better to write it as x^(3/2) easier to simplify

The multiplication is also easier to see for that

Then the derivative should be normal

Product rule would be an unesassary pain

[obtuse pebble]

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[sharp ridge]

how to prove the lateral surface area of a cylinder is 2πrh

[latent walrus]

what do you mean by prove it?

[sharp ridge]

like deriving the formula

[latent walrus]

it would come from the circumference of a circle, then just multiplying it by the length, but then you have to say where the circumference comes from

if you unravel the cylinder
then youd get a rectangle of side length 2pir and width h or vice versa

[sharp ridge]

im trying to do deriv the formula from that rectangle

[latent walrus]

if you have the rectangle its just length * width, im not sure where it would be a problem from there

[sharp ridge]

So this, using the surface area formula

2π∫y sqrt(1 + f'(x)) dx

like im not sure if it's correct if i put my y as r

[latent walrus]

drawing it like this is a bit confusing, not even centered on the origin or starting at the origin

if you just ignore the bit in the negative side then its fine except that r should be 2r

[obtuse pebble]

@sharp ridge Has your question been resolved?

[sharp ridge]

so since it is with respect to x, would r(x) = x - 0 and f'(x) be....2?

[obtuse pebble]

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[north anchor]

Given the vectors:•
v1 = <4, 8>•
v2 = <3, -5>•
w = <1, 31>•
z = <-1, 31>•
Given the scalars:•
k1 = 2•
k2 = -3• |
Find the vector v3•
Determine if the vectors w, z belong to the vector space of v1 and v2•
Explain.

[north anchor]

I need ; Determine if the vectors w, z belong to the vector space of v1 and v2•

[tired shell]

I'd check if (<4,8> , <3,-5>) is linearly independent. if so, then with 2 basis vectors, this vector space would be everything, all of R2 and every vector would be in it

[north anchor]

how

[north anchor]

v3 is <-8.7> right?

[tired shell]

I'm not sure what k1, k2, and v3 mean from the question

[north anchor]

are scalars

i am new with space vectors

[tired shell]

ah okay, to multiply numbers by a vector you multiply everything inside by the number

so that equation becomes <2*4,2*8>+<3*3,3*-5>

[north anchor]

How I know if a matrix has a solution , no solution or infinite solutions?

[tired shell]

when you row reduce, if it's a full diagonal of 1's (pivots), that's one solution

infinite is when there's a row of 0's and none is when it's a row of 0's with an actual number at the end

[north anchor]

If i Use addition/elimmination method?

[tired shell]

yea that's another name for row reduction

[obtuse pebble]

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[rustic moss]

Just a quick question, when do I know to use either law of sine or law of cosine?

[thick oracle]

law of cosine when you have 2 sides and the angle between them

law of sine when you have 2 sides and 1 angle

or 2 angles and 1 side

[rustic moss]

oooh ok ok

thank you

.close

[obtuse pebble]

Channel closed

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[thick oracle]

np

sometimes you can use both

just see which one fits best

[rustic moss]

so its kinda like the quadratic equation but only sometimes?

[thick oracle]

wdym

#❓how-to-get-help

[tall ocean]

Sorry wrong channel

[rustic moss]

oh shoot

quadratic formula

sorry if i confused you

yk how it can be used for any quadratic equation? Im thinking about it like that

correct me if im wrong

[thick oracle]

it can

but law of sine and cosine cant always be used

[rustic moss]

oooooooh ok

thanks for clearing that up

[thick oracle]

np

[rustic moss]

.close

did i break it ;-;

[thick oracle]

its already closed

[rustic moss]

alright

thanks again

[thick oracle]

np

[obtuse pebble]

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[edgy needle]

excuse me how do i do this

[edgy needle]

question bi

[obtuse pebble]

@edgy needle Has your question been resolved?

[modest ingot]

Lol where's the math in this

[obtuse pebble]

@edgy needle Has your question been resolved?

[edgy needle]

well it says calculate LOL

bi

[obtuse pebble]

@edgy needle Has your question been resolved?

[torpid herald]

im joking

[edgy needle]

RAAAAAAAAAAAHHHHHHHHHHHHHH

[drifting wraith]

Divide 500 million by .....03

By 0.4 too I guess

[sage dagger]

hi

[edgy needle]

hi

[obtuse pebble]

@edgy needle Has your question been resolved?

[cyan oxide]

Ramanujan summation

[sage dagger]

ramanujan summation

[obtuse pebble]

@edgy needle Has your question been resolved?

[obtuse pebble]

@edgy needle Has your question been resolved?

[eager niche]

Did you solve it?

[edgy needle]

yeah

im in my school toilet

LOL

brooo my stomach hurts too bad

ok bye got to hide my phone in my locker

DOING MY TEST NEXT PERIOD WISH ME LUCK

[swift sky]

Tf pause

[urban lichen]

🤣

what is this

[hollow solstice]

💀

[obtuse pebble]

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[cunning burrow]

Are there any similar questions in a textbook you could follow?

[obtuse pebble]

@vagrant spear Has your question been resolved?

[obtuse pebble]

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[mighty wraith]

are tere any voice channels

[fathom flicker]

No, if you have a math question post it here, if you are looking for a discussion channel use #discussion or #chill

[last pilot]

that's... not a question you ask here

the "math help" channels are to ask math questions

[mighty wraith]

buhhhh. if x = voice channel and y = here. does x = y

jk thx

[last pilot]

what...?

do .close

[mighty wraith]

.close

[obtuse pebble]

Channel closed

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[mighty wraith]

how much homework in math should i do if im doing the hardest math subject in school

actual question

.close

[last pilot]

what???

that... doesn't make much sense

[mighty wraith]

sry i though i was in discussion

[last pilot]

ohk

[obtuse pebble]

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[dusk fractal]

.clsoe

.close

[obtuse pebble]

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[uncut pebble]

Each afternoon the probability my cat sleeps is 0.6 and the probability that my dog sleeps is 0.7. The probability that the dog sleeps given that the cat is sleeping is 0.9. Find the probability that both sleep in the afternoon. Give your answer to three significant figures.

[uncut pebble]

<@&286206848099549185>

[unique gust]

have you tried something or you don't know how to start?

[last pilot]

!15m

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[uncut pebble]

i tried subbing into the equation but it was wrong

[unique gust]

what's the equation you was trying to sub it into?

[uncut pebble]

i got P(A n B)= P(A given B) x P(B)

so i got P(A n B) = 0.9 x 0.6 = 0.54

[unique gust]

ok, so in this case, which are the events A and B?

[unique gust]

isn't that the answer?

[uncut pebble]

cat sleeping (0.6) and dog sleeping (0.7)

[uncut pebble]

no it was wrong apparently

[unique gust]

what's the answer supposedly?

[uncut pebble]

im not sure its on transum so u cant see the answer

[unique gust]

weird

[unique gust]

I think this is correct

let's see if someone else helps here

because that was my answer lol

[uncut pebble]

lol nw ty anyways

.close

[obtuse pebble]

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[obtuse pebble]

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[graceful coral]

I'm having a math problem, it is about allocating resources with a condition. Where should I post this question? Thanks! 🙂

[graceful coral]

I'm trying to fairly allocate foodstamps between 4 centers. There are 10,000 requested and only 5,000 available. I want to give centers that have less % requested so far more "right" to available stamps. I found a way, which seemed to work as long as C64 is smaller than E72. See first picture

However, as soon as C64 becomes larger than E72, The correction % becomes more than 100%, which causes overallocation. See second picture

I could seet the max to 100% and manually distribute the remaining stamps. But This model should also work for 100+ centers, therefore, I'm looking for a mathematical solution. Can anyone help?

Let me know if I asked at the wrong place!

<@&286206848099549185> Hopefully someone can point me in the right direction, or knows who to ask

[narrow vault]

if C74>E72 does it just mean you have enough to fulfill all demands?

[graceful coral]

In case you mean C64>E72: No, because in this example, demand is 10,000.

Thank you for taking the time to help me out! 🙂

[narrow vault]

ok i see, probably needs some iterations

main problem is you end up with col H values greater than col C values

the straightforward way i see resolving this is to wrap everything in column H with min(stuff you already have , corresponding col C cell)

and then if any centres gets their entire demand fulfilled remove them and the corresponding total from available qty and iterate

without vba it probably gets ugly, but with vba you can just write a macro

[graceful coral]

Yes but that would mean that I have to manually distribute the excess amount (in this example 156) to the other centers right?

[narrow vault]

it means that those 156 would be distributed among the other 3 in the next iteration (where you only have suppliers ABC and total availability 4000)

misread

[graceful coral]

So like multiple recalculations to get to the desired result? The problem is that I would want to do this with 2000+ centers as well..

[narrow vault]

i mean it’ll run pretty fast and only gets a bit slow (like a few seconds maybe) if almost all of them can have their demand fulfilled

but your problem is begging to be solved in a proper programming language, i’d throw a python script at this if you can’t write vba

oh, maybe you can do it with volatiles

[graceful coral]

I was wondering if there would be a different way to allocate which would make sure that the requested amount would never be exceeded

[graceful coral]

I'd love to be able to properly program something, but I'm afraid that I'm limited to excel..

[graceful coral]

Care to elaborate? 🙂

[narrow vault]

bleh, actually probably no

anyways without vba you theoretically need up to (number of centres - 1) columns to simulate what i described in pure excel, in the case where 1 center gets their demand filled, and then the overflow barely gets another centre to overflow, etc
repeat for n-1 iterations

due to this limitation it’s probably hard to do much better than that, use hidden cells to hide the ugliness away i guess

[graceful coral]

Ahw..

[graceful coral]

Yeah, I don't think 2000+ columns would work anyways.. It sucks that I have 0 VBA or python knowledge

[narrow vault]

you can cut down the work significantly if you only want to maintain absolute fairness for a few rounds, say one or two

then when you’re done just give out the rest by fulfilling centres one by one sorted by column D or something

but then yeah no more absolute absolute fairness

[graceful coral]

Distribution of stamps would be done twice per year, with around 2000 centers participating..

But you don't see any posibility in the way they get allocated right now? Perhaps a different approach on that?

I'm open to anything more than 50% available means everyone gets 50% of their requested amount

[narrow vault]

ok it’s actually somewhat interesting when you’re limited by pure excel and you want to save space, there probably very weird optimisations you can do by sorting on column D then doing modifications directly in column F

[graceful coral]

I don't completely follow you but am curious what you mean

I really appreciate your help by the way, thank you 🙂

[narrow vault]

no worries i kinda like pure optimisation sometimes

[graceful coral]

Hmm I get what you mean, like sorting to sort out the 100%+ cases?

[narrow vault]

so the idea is to sort on column D from high to low

and since the overflowing centres will always be at the top we can use a formula that references {all rows including this one and below} to figure out allocations one by one

it should be computationally equivalent with almost no extra space required

[graceful coral]

And possible with 2000+ cases?

[narrow vault]

running time was never a problem lol

outside of extreme edge cases where it might take like a few seconds to crunch

i would expect it to almost be instant

unless the machines is from like 1997 running on 95

[graceful coral]

Is there a chance there would be a way to solve this mathematically? Like find a way to distribute the available stamps in a way that there will never be overdistribution?

There are two possible ways to solve this: 1. Find a way and hopefully automate the reallocation of overallocated stamps to others that havent reached their max yet. OR 2. Find a way to adjust the original allocation in a way that it never exceeds the requested amount, but still takes the % of "free space" into account.

It does seem easier to first try route 2, but I don't know what other way to do the allocation besides the way we do it right now..

[narrow vault]

1 is basically my idea the entire way and achievable easily in python and slightly harder in pure excel

2 if you’re willing to sacrifice absolute fairness is very easy lol

the fun thing is i actually don’t know where this problem lies, probably somewhere in operational research

[graceful coral]

I feel like you do enjoy the challenge of option 1 haha

But what would the solution be if we do option 2?

[narrow vault]

2 would be allocate by your rule then process the overflow just by blindly fulfilling the remaining centres with some rule that’s easy to follow

say just give all of them to the remaining center with the highest column D value

which is very obviously unfair but also very easy to implement

[graceful coral]

And isn't there a way to push values in G between 0 and 1?

[narrow vault]

yeah should be able to do it in any one of the columns

[graceful coral]

Would that still allocate all available stamps tho?

[narrow vault]

yeah it’ll work

anyways ive been replicating the sheet to figure out how to save space and i have an implementation that only needs 1 column

the limitation is you have to sort it by column D from largest to smallest first

[graceful coral]

Could you show me?

[narrow vault]

top cell

2nd cell onwards, you can drag this down to autofill

eh reference cell numbers are a bit off but the highlights should tell you where things should be

[graceful coral]

What we also can do is sort G from big to small and let excess droop down right?

Which would give the same result?

Damn I feel like there still should be an easier way by finding a mathematical way to distribute the stamps that would never exceed 100% allocation.. Or don't you think there is a way?

[narrow vault]

you have to sort it somewhere between D and G, i just did mine on D because it’s the first column where this matters

i mean the easy way would just be to say to D enjoy your extra 156 stamps

but when you want a cap the problem usually gets harder and i think this is comparable to the stuffs ive seen in operational research

[graceful coral]

Fair enough. Let me ask my friend who I'm doing this for what he thinks of this solution

Thank you for your help! 🙂

And if you think of something in the mean time, let me know!

[obtuse pebble]

@graceful coral Has your question been resolved?

[graceful coral]

@narrow vault This is what i've done now, agree?

It trickles down as you can see

[graceful coral]

Let me know if you still want to help, I have a new idea..

[obtuse pebble]

@graceful coral Has your question been resolved?

[narrow vault]

i don’t know what processing you’re doing them in, although that’s not my output when you switch to 9000

ok it’s the trickle down thing

yeah if you’re fine with that you can use it

[graceful coral]

So what is the difference between my method and your method? I'm confused. Thank you for your help!

I cannot seem to duplicate your results..

Got it, thank you

What if we transfor allocation % of requested in such a way that it shrinks around the mean, so that the values in the picture become in the range of for example 12%-100%?

[graceful coral]

Could you please explain the theory behind this?

I got it in my excel now, but I'm trying to get the rationale behind it

[narrow vault]

column D is a metric that i'll describe as "fulfillment percentage weight" (im also bad at naming things)

it represents the percentage of request that should be filled if you have just enough stamps equal to your E72

so if you have 7700 center A gets the full 4k, B gets only 100, etc

a useful property of this metric is that the center with the highest weight will always overflow first, because suppose center B has lower weight and it overflows, then A has a higher percentage of its need fulfilled, meaning it also overflows

this means when sorting by this metric you already automatically sifted all the centers that will overflow to the very top

[graceful coral]

So you basically guarantee that the overflowing goes from top to bottom?

[narrow vault]

which means by calculating row by row while sorted that way you automatically get what the overflow amount is
which means when doing the next row (say B) you can repeat the calculations for the remaining rows by subtracting the demand from the one above, and only calculating with weights from B and down

ye

[narrow vault]

the formula here effectively says:

subtract all distributed stamps to centers above this one from the total, and distribute the remaining stamps among the remaining centers (for that row 3 centers, A,C,B) based on your preferred system

and if this center overflows again we only distribute an amount equal to its demand (the min() wrap)

and if it doesn't overflow, this is equivalent to just your original system

[graceful coral]

So you basically recalculate the distribution for A, C and B with the remaining available shares?

[narrow vault]

yes
it's more overhead and there are ways to speed it up (by adding more ugly cells) but that's the minimal version that basically needs no additional space

[graceful coral]

But with this way, no more columns have to be added if we have lets say 2000+ centers right?

[narrow vault]

no
in fact if you want to paste that directly on top of column G it should be fine, 0 extra space needed :p

[graceful coral]

This I do not understand yet haha

[narrow vault]

oh, i can't count
i meant column H

[graceful coral]

So now we can see the increase of % of requested which is allocated. Is there a way to all have the same increase in percentage points?

[narrow vault]

i don't know what column J and K are doing

but i checked column H/column E and all non-fully fulfilled centres should have the same ratio

[graceful coral]

You see that all non-fully fulfilled centers get their percentage increased by 4.05% (column L), the percentage point increase (column K) is different for all of them

[narrow vault]

it's physically impossible to keep both K and L equal
so choose your favourite i guess

[graceful coral]

Yes I understand

But would there be a way to allocate the overalocated amount evenly over the rest?

[narrow vault]

you can do that, much easier too

[graceful coral]

Could you show me how?

[narrow vault]

just keep redistributing evenly until none of them overfill

[graceful coral]

Could you please show me how I'd do that in Excel?

So I mean percentage point increase

Wait now I'm confusing myself haha..

[narrow vault]

what i gave you is already the same percentage increase

if you check column H/column E for 8000 it should give the 3 non overflowing centers 108% fulfillment wrt your percentage weights

if you want to evenly distribute the overflowing stamps by amount (i.e for the 150 stamps just give 50 to each of the other 3) it's a bit different

[obtuse pebble]

@graceful coral Has your question been resolved?

[obtuse pebble]

Channel closed

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[obtuse pebble]

Available help channel!

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[gilded pumice]

Let f : R → R defined by -pic-
prove that f is continuous at 1 and discontinuous at 2.

[gilded pumice]

any pointers? i literally have no idea how to start😂😅

[tired shell]

it's essentially showing the limit of x and x^2 at x=1 is 1, but the limit at x=2 is different for x and x^2. when you write out the definitions with |f(x)-f(c)|<epsilon you always have to look at both cases x and x^2, since there'll always be a rational and irrational in the interval (it's called Q and R\Q being dense in the reals)

[gilded pumice]

any chance you could give me a working example?

[tired shell]

uh not really, I'd have to know what the proofs in your class tend to look like

[gilded needle]

if you don't like fiddling with the two cases, you could do something like:
|f(x) - 1| is either |x-1| or |(x-1)(x+1)|, and that's bounded by 3|x-1| for x reasonably near 1
then you can use the squeeze theorem

[obtuse pebble]

@gilded pumice Has your question been resolved?

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[halcyon horizon]

Trying to simplify a partial derivative. Not sure how they work, though. How can I show the expression I need to simplify?

[halcyon horizon]

I can describe the expression in words maybe

its the partial derivative of (y-(y hat))^2 with respect to W, where we have a relation between y hat and W, but y can be treated as a constant

y hat = U * tanh(Wx + b) + c

ok I figured latex out

,tex $\frac{\partial({-2y\hat{y}})}{\partial{W}} + \frac{\partial(\hat{y}^2)}{\partial{W}}$

[warm shale]

Tathya

[halcyon horizon]

and

,tex $\hat{y} = U \tanh({Wx + b}) + c$

[warm shale]

Tathya

[halcyon horizon]

<@&286206848099549185>

[oak plover]

You don't need to put ,tex behind the latex

[halcyon horizon]

oh

sorry I've never really engaged in this server before

[oak plover]

Have you tried getting -2y out of the partial derivative and computing dy/dW

[oak plover]

No it's fine I was just informing

[halcyon horizon]

unfortunately I don't understand a single thing you're saying

im 14