help-23

[safe radish]

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[sharp cypress]

i don't understand how we got that

if we factorize cos alpha ^3
with cos^2 alpha
we would get 1/ cos alpha
(if thats what it was done here)

[compact wraith]

$\frac{1}{\frac{1}{cos\alpha}} = cos\alpha$

[flat frigate]

Max..

[compact wraith]

Yes but it's on the denominator so it's reciprocal of a reciprocal

[sharp cypress]

alright i get it

thank you sm

.close

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[tawdry needle]

Hello I have a statistics question...

Let's say I'm trying to calculate damage in game, there are two stats - ATK and Element bonus.

Damage is calculated by multiplying ATK with Element bonus in some way... There is some dropoff tied to the element bonus so it acts ( multiples atk differently ) based on how much you have. For example let's say 1 element bonus will increase final damage by 10% , 100 element bonus will only increase final dmg by 40%. Dropoff occured since 100>1 , and with each new element bonus the overall "multiplication power" of each element kept dropping. I'm trying to figure out this rate of dropoff...

I will list some stats below and the dmg displayed in game , with each higher stat you will see the value of element bonus keeps dropping.

Dmg calculation : (ATK x Element bonus ) / 4
ATK: 3966.1
Element bonus:0
Dmg: 988

ATK: 3966.1
Element bonus: 4
Dmg:1061

Atk:3966.1
Element bonus: 8
Dmg:1124

Atk:3966.1
Element bonus:12
Dmg:1179

Atk:3966.1
Element bonus:16
Dmg:1227

Atk:3966.1
Element bonus:20
Dmg:1270

From above you see that the element bonus dropped off from giving 18.25dmg per 1 element to giving 10.75dmg per 1 element.

I'm curious how to calculate the exact rate of dropoff so it can be included in dmg calculation formula?

Many thanks!

[safe radish]

@tawdry needle Has your question been resolved?

[grizzled fossil]

Plot points

Then you see it logarithmic

So fit that (or transform data and fit linear)

[tawdry needle]

https://cdn.discordapp.com/emojis/853317415956906025.gif?v=1&size=64&quality=lossless

I'll have to try to google this but would be much easier with an example or something

[grizzled fossil]

Its just using any software and have it fit for you

Using either of the 2 methods I said

[tawdry needle]

I'm sorry I actually don't know how to do logarithms... I'm searching for the software xd

[grizzled fossil]

R, excel, desmos, stata, matlab, python,…

[tawdry needle]

Would you mind guiding me through it...

I started up desmos

[grizzled fossil]

Google logarithmic fit desmos?

[tawdry needle]

Ty I found a video let me see

This how far I got with it xd

[tawdry needle]

I inputed the element bonus as x and total damage as y ... I'm not sure if that's how it's supposed to be done?

Probably not... Should I instead calculate how much damage each element bonus gives based on the amount

[grizzled fossil]

Either doesn’t matter

[tawdry needle]

What does this tell me tho?

So far I'm only telling it what I want drawn on screen

I still haven't gotten to having it calculate rate of dropoff , it's only visualized

x shows how many element bonus I have , y shows how much dmg each element bonus will give

[tawdry needle]

Any ideas?

[tawdry needle]

I really don't know.. My question starts here for anyone else tuning up

I'm trying to learn logarithms from YouTube videos in meantime but I've no clue yet

https://cdn.discordapp.com/emojis/853317415956906025.gif?v=1&size=64&quality=lossless

[safe radish]

@tawdry needle Has your question been resolved?

[safe radish]

@tawdry needle Has your question been resolved?

[tawdry needle]

I need help writing out logarithmic function...

I entered the data from above into table

Then I copied logarithmic expression for linear something... Dunno

Basically I have the dots on the screen , I need it connect then with a curved line - so it can predict properly what the value will be as I go further from what I've told it to draw

It's already making mistakes with these 4 dots on screen as you can see , it's not centered on them but it just averaged them and drew the line through... This won't work as the curve starts getting much steeper

This was the orignal question:

Hello I have a statistics question...

Let's say I'm trying to calculate damage in game, there are two stats - ATK and Element bonus.

Damage is calculated by multiplying ATK with Element bonus in some way... There is some dropoff tied to the element bonus so it acts ( multiples atk differently ) based on how much you have. For example let's say 1 element bonus will increase final damage by 10% , 100 element bonus will only increase final dmg by 40%. Dropoff occured since 100>1 , and with each new element bonus the overall "multiplication power" of each element kept dropping. I'm trying to figure out this rate of dropoff...

I will list some stats below and the dmg displayed in game , with each higher stat you will see the value of element bonus keeps dropping.

Dmg calculation : (ATK x Element bonus ) / 4
ATK: 3966.1
Element bonus:0
Dmg: 988

ATK: 3966.1
Element bonus: 4
Dmg:1061

Atk:3966.1
Element bonus: 8
Dmg:1124

Atk:3966.1
Element bonus:12
Dmg:1179

Atk:3966.1
Element bonus:16
Dmg:1227

Atk:3966.1
Element bonus:20
Dmg:1270

From above you see that the element bonus dropped off from giving 18.25dmg per 1 element to giving 10.75dmg per 1 element.

I'm curious how to calculate the exact rate of dropoff so it can be included in dmg calculation formula?

Many thanks!

[safe radish]

@tawdry needle Has your question been resolved?

[lean otter]

200 people stand in a circle in order to 1 to 200 . number 1 has a sword he kills the next person (i.e. No. 2) and gives the sword to the next living person (i.e. No. 3). all people do the same until only 1 survives. wich number surivives to the end?

[safe radish]

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[dry barn]

Looking for help with a divergence test problem

[stray socket]

Sure

But post the question

[dry barn]

Sorry, forgot that

[dry barn]

Not sure where to begin unraveling this

[stray socket]

Hm

[grizzled fossil]

Wdym unravel?

[stray socket]

Solving

[dry barn]

I dont know how to approach

Yes

[grizzled fossil]

Divergence test?

What does that mean?

[stray socket]

For series

Any tests for divergence

[grizzled fossil]

All other questions he wanted to use a specific one

Is it the same here, if so which?

[stray socket]

He doesn't know which one

He's asking which one he should use in the first place

[grizzled fossil]

Okay

[dry barn]

I believe this tests for divergence only?

[grizzled fossil]

So do that

Yes

[shy temple]

Note that ln 1 = 0, so the first term is undefined. So the series cannot possibly converge.

[dry barn]

So is the lim

[grizzled fossil]

Yes? So wdym how do I unravel this

Clearly you knew smth

[dry barn]

I dont understand the process. Rather check that I'm following the correct steps than get bad habits going

[grizzled fossil]

What process?

[dry barn]

Determining if the series converges or diverges

[stray socket]

Well first evaluate the limit

[dry barn]

So the cubic root is confusing me, not sure what to do with it.

Does the exponent 2 in the denom cancel out with the cube power?

[stray socket]

Well what's the cube root of infinity

[stray socket]

No

You would evaluate the limit of the numerator and the limit of the denominator

So let's look at the numerator

[dry barn]

Oh

[stray socket]

What would the limit, as k goes to infinity, be for the numerator?

[dry barn]

infinity

[stray socket]

Yeah

And then the denominator?

[dry barn]

and I dont think we can do anything with the denominator. Maybe ln^2(k)?

[stray socket]

You can

What's the limit of the denominator as k goes to infinity

[dry barn]

Also infinity?

Yeah, I believe so

I guess infinity^2 but, infinity

[stray socket]

Yes

So you have infinity/infinity

Yikes

But what can you do to manipulate the equation?

[dry barn]

hint?

[stray socket]

Rhymes with Hopital

[dry barn]

l'hopital rule?

[stray socket]

Ye

But youre gonna need to do it a couple of times

Do it, evaluate it, see what you get

[dry barn]

Is this the correct form?

@stray socket

[stray socket]

Yeah that's limit chain rule

[dry barn]

I keep coming back to infinity

[stray socket]

Yeah so use L'Hopitals again

[dry barn]

Am I just going around in circles?

With radical rule

[grizzled fossil]

No you just need to do it two times?

[stray socket]

Just two times I think

[quasi bison]

i would have suggested simply noting that $\frac{n^p}{\ln(n)} \to \infty$ for any $p>0$, but seeing that requires some fluency with calculus (and tying that in with the problem at hand requires some fluency with algebra)

[flat frigate]

Ann

[quasi bison]

l'hop feels gross here tbh

[grizzled fossil]

Which is typically shown using L’H?

[quasi bison]

do you even need l'hop to show that in case you're actually considering it as a sequence and not silently passing from a sequence to a function of a real variable?

[dry barn]

Thanks @quasi bison @grizzled fossil @stray socket I got to go to work.

.close

[safe radish]

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[quasi bison]

don't thank me this time lol i was just rambling

[grizzled fossil]

I mean its 1 line using L’H

[safe radish]

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[tawdry needle]

.reopen

[calm merlin]

Evaluate 32 + (6 − 2) ⋅ 4 − 6 over 3 .

[thin bridge]

do you have a pic of the expression?

i'm interpreting that as
$$32 + (6 − 2) \cdot 4 − \frac63$$

[thin bridge]

what have you tried

[chilly trail]

Use pemdas

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[low swallow]

Not really a problem help, I'm just trying to understand the following statement:
"All functions are continuous in its domain"

[low swallow]

I'm currently in limits, asymptotes and continuity and I don't really understand how I can visualize the statement above or how it helps me

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[lean otter]

I have a recurrence relation i need to solve using matrices

[lean otter]

its this

here's the matrix that satisfies the equation

for n>=2

Im wondering, how can i calculate a_100?

I already diagonalized the matrix and stuff

found P, D

[stoic dune]

So yeah, you want A^100

Which is ez if you have A diagonalized

[lean otter]

Why A^100?

I dont understand the logic

also how would I be able to find an explicit function of n, using this information?

[lean otter]

Also what would i do with this

[stoic dune]

Multiplying by A gives "the next term"

So multiplying by A 100 times should...

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

Yes but if i multiply A by 100 how will that give me an exact answer for a_100?

don't I have to then multiply by x_n?

[stoic dune]

Yes

Well, more specifically,
(A¹⁰⁰)x_0 = x_100

Or, actually I guess
(A⁹⁹)x_1 = x_100

[lean otter]

yup

this is right

i had to flip the initial conditions

and i got it right

thank you so much

[safe radish]

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[urban oriole]

Can someone help me work through this?

[stoic dune]

Need to find a linear function g(x) that has the same value and derivative as f(x) at x = a

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[limber flame]

can someone explais wats going on here?

so in this step lets consider 6f^2(x)-3f(x) as h

then ln(1+h)/h form is eqaul to 1

so we are left with (6f^2(x)-3f(x))/3f(x)

f(x) in numerator and denominator cancel

and then we get 6f(x)-3/3

6f(x) is 0?

as f(a)=0?

[safe radish]

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[lofty sundial]

if something has 5 options,
thing 1 - 69% chance
thing 2 - 10% chance
thing 3 - 10% chance
thing 4 - 10% chance
thing 5 - 1% chance

how many times would i need to do it for a good chance at that 1%

[safe radish]

@lofty sundial Has your question been resolved?

[safe radish]

@lofty sundial Has your question been resolved?

[fair beacon]

The chance of not getting thing 5 is 99%
So, for example, the chance of getting not thing 5 tens times in a row would be 0.99^10
So the chance of getting at least 1 thing 5 is 1-(0.99)^10

In general the chance of getting at least one thing 5 in n times is
1-(0.99)^n
Your answer would depend on what your idea of a “good chance” is

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[dense yacht]

Hi all,
I need a tip on how to prove this:
I am examining the Naturals. So all numbers and parameters are natural numbers.

Linear recursive functions are f(n) = c1 * f(n-1) + ... + c_d * f(n-d). One example is Fib(n) = Fib(n-1) + Fib(n-2).

Claim:
For all m > 1, every linear recursive function of degree d repeats (mod m) at n and k, for some n,k of [d-1, d+m^d)
{// This seems odd: at d-1, f would reference f(-1) which is not something that it should... so i'm also assuming that the possible n,k are in (d-1, d+m^d)}.

The "repeats (mod m)" means:
f(n) ~ f(k) (mod m)
f(n-1) ~ f(k-1) (mod m)
f(n-2) ~ f(k-2) (mod m)
...
f(n-(d-1)) ~ f(k-(d-1)) (mod m)

So far, my intuition is that somehow I will find some multiple of previous f's, but that's not working...

Also, I was searching for patterns in remainders... but that also led me nowhere...

[dense yacht]

(I'll use the symbol ~ for "is congruent" or "equivalent" mod m):

[fiery merlin]

d is the number of terms in the recursive function definition.

[dense yacht]

Yes, thats right.

Fib would be a degree 2 linear recursive function

[fiery merlin]

I'd separate it out into a proof for f(x) = 0 (where d = 0) and the rest, where d > 0.

That way, d - 1 is 0 or greater.

It's easy to show that f(x) = 0 repeats.

[dense yacht]

Yes, it is a constant, so for any k,n i'll have congruence (0 is a multiple of everything)

Oh, i see... I'll try to prove using induction on this "constructive" things (linear combinations of constant functions and functions of constant functions)... Thank you!

[fiery merlin]

No problem.

[dense yacht]

@fiery merlin You are the best!

.close

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[green mulch]

hello

[green mulch]

im trying to figure how to solve these on my own im not understanding tho

<@&286206848099549185>

[sonic spruce]

BCNM and MAND are similar and so are RSFE AND EFTQ i think?

[green mulch]

similar to what

[sonic spruce]

to each other

[green mulch]

oh yeah I think

i thought u were talking about a previous problem someone else did

[sonic spruce]

oh

can you solve it now?

[stoic dune]

Gotta assume that M is a midpoint for AB and N is a midpoint for CD

Then 6x+3 is the average between 21 and 57

[green mulch]

im not really sure what youre talking about this is my first time doing this

its not similar to the previous problems before it so I dont have any reference on how to solve it

[green mulch]

oh wait maybe I understand

is it 39 because that is half between 21 and 57

the equation is the top number plus the obttom number divided by 2?

so 21+57= 78/2 = 39

so then its 6x + 3 = 39

x = 6?

and for number 5

its 24 + 38 = 62 / 2 = 31

3x + 31 = 31

x = 0

<@&286206848099549185>

[safe radish]

@green mulch Has your question been resolved?

[green mulch]

no

can anyone help please

[safe radish]

@green mulch Has your question been resolved?

[sonic spruce]

Your solution is correct

[green mulch]

is the rest corrext

[sonic spruce]

yes

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[tight coral]

Used the average value integral formula

Got 8/pi

Set equal to 2 sec^2 c

Got that and was wrong

[ancient escarp]

it wants answers within the interval

your format implies infinite answers

also i'd assume it wants an exact answer?

not a decimal approximation

[tight coral]

How is that not in the interval

[ancient escarp]

because 327.207 is not between -pi/4 and pi/4

and that's what you get if you let n = 104

[tight coral]

Where did u get 327 from

[ancient escarp]

by plugging in n=104 to your answer

which is just one of the infinitely many answers you can get from that

[tight coral]

Well can u tell me where I went wrong

This what I did

Mean value integral

Got 8/pi

[ancient escarp]

looks good

[tight coral]

I set 8/pi to 2sec^2 c

Then got that number

And another one but I think it’s out the interval

Solving for C

[ancient escarp]

again i'd assume it wants an exact answer

and, again, pi*n implies infinite answers

so i'm not sure why you have that there

[tight coral]

maybe arcsec(2sqrtpi/pi)?

[ancient escarp]

how did you get that

[tight coral]

How do I get the exact answer

Online calculator

[ancient escarp]

write it down and solve

[tight coral]

I get that Long number

[ancient escarp]

use a pencil

write it down

don't put anything into a calculator

[tight coral]

I don’t have one

I’m in Nigeria

[ancient escarp]

you don't have a pencil?

[tight coral]

No

Why do u think im using a calculator

[ancient escarp]

because 99% of math students would rather make a calculator do the work for them instead of doing it themselves

how do you have electricity but no pencil

??

you're on an electronic device, just use something on there to write

[tight coral]

At the office building

It’s okay I’ll try to struggle by myself thanks for nothing

.close

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[ancient escarp]

.it's hard to guide someone who ignores everything i say

[green mulch]

why couldn’t he use a calculator

[ancient escarp]

he was either inputting it incorrectly or the calculator was giving a decimal approximation

the website he's on most likely wants an exact answer, as it asks for values

[green mulch]

.close

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[fading raft]

So, if I understand the question currently then u = constant would just be a more general case of the first part

Say $u=a$ where a is some constant. then we have $z=e^{a+iv}$

[flat frigate]

azeem321

[fading raft]

In the $x-y$ plane, would this represent a line through x=a and y can take any value

[flat frigate]

azeem321

[sharp island]

The way I understand it is that varying w has the effect of changing z, since the value of z is determined by w. So since w has two variables u and v, in order to get a curve, we need to fix one of them, so I think you are on the right track in setting u=a for some real a and them varying v. Then do the opposite way round for the next part

But I'm not sure about it being a straight line though

[fading raft]

But I am not quite sure how to get this in terms of (x,y). I thought about doing $z=e^{a+iv} \implies \sqrt{x^2+y^2}e^{\tan^{-1}(\frac{y}{x})} = e^{a+iv}$

[flat frigate]

azeem321

[fading raft]

But that doesn't seem right

[sharp island]

It is easier if you don't use complex exponentials here

So

u is a real number, right

exp(u) must also be a real number

So given that v is real too, then z is in the form z=r(cos(t) + i sin(t)), which is just the ordinary complex polar form.

We can see that the r will be exp(u) and t will be v

Since we are varying v, how does this change z on the argand diagram?

In other words, in the z=r(cos(t) + i sin(t)) form, what does t represent?

For any complex number written in this way

[sharp island]

Would it help if I elaborate on anything?

[fading raft]

Nope. Your explanation was brilliant. I've got it now, thanks!

.close

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[sharp island]

Np!

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[marsh walrus]

question is "State the null and alternative hypothesis"

I'm still pretty confused by this, I'd asked earlier and someone had given the advice that 'null always preserves the status quo"

which to me implies that $H_0 : \mu {placebo} - \mu{supplement} \geq 0$

[flat frigate]

jan Niku (Shuri for Honorable)

[marsh walrus]

could someone walk me through the logic of selection on this one?

nvm

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[sick mortar]

I have to find the equilibrium solutions, I'm stuck at this step

[sick mortar]

I tried this but idk if this is correct

[safe radish]

@sick mortar Has your question been resolved?

[sick mortar]

I tried this and am stuck at a different part

<@&286206848099549185>

I've got 5 solutions idk how many I should have or if they are right

I think I got it
Solutions are
0,-1
And
-2,1

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[next crater]

2,3,5 is perimeter

[next crater]

how to find n

[snow robin]

you set up equations for the perimeters of k l and m

so that you can find the sides

[next crater]

i do

$p + l=1 \ p_{1}+l_{1}=1,5 \ p_{2} + l_{2}=2,5$

[flat frigate]

2022 squared

[snow robin]

explain?

[next crater]

K, L, M = 2 ( l + h )

[snow robin]

yes

[next crater]

so 2(l+h) = 2

2(l+h) = 3

etc

[snow robin]

yes but the heights and lengths are similar

$K = 2 = 2(l_1 + h_1)$

[flat frigate]

Katharine

[snow robin]

$L = 3 = 2(l_2 + h_1)$

[flat frigate]

Katharine

[next crater]

h1?

[snow robin]

$M = 5 = 2(l_1 + h_2)$

[flat frigate]

Katharine

[snow robin]

the first bit of the total height

and the second bit of the total height

[next crater]

so just make it l1 = x
l2 = y
h1 = a
h2 = b

?

[snow robin]

that's fine

[next crater]

k thanks ill do the rest

.close

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[lean otter]

Can someone explain to me the basics of soacahtoa?

[novel edge]

Do u have a specific problem you have questions about?

https://www.mathsisfun.com/algebra/sohcahtoa.html

Here is a resource

[lean otter]

like

whats the formula

the equation I mean

and what they do

or what they "find"

They are functions which take input as the angle and spit out the ratios of two different sides as the output. Or at least that is how they are defined. Also, when you mentioned "soacahtoa", I thought it is some theory in higher maths or something lol

[lean otter]

they can be used to find the angle (given the sides) or be used to find ratio of two different sides (given the angle)

I see

[lean otter]

those can be found online?

like the link that person shared above

Ok I see

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[spark mountain]

So I have this quadratic:

[spark mountain]

I think this should have at least one (possibly complex) solution as long as not all of a, b, c are equal to each other

is there some nice way to write that solution?

just applying the quadratic formula works, except for the case where a + c = 2b

[mortal meteor]

then solve this case separately and look for a common form after you solved that case

[spark mountain]

the solution in that case is like x = c / (2c - 2b)

does a common form necesarially exist?

[mortal meteor]

i don't think so

[spark mountain]

guess I'll suck it up and just write some cases lol

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[midnight patio]

why is it C?

[wicked axle]

It loads B.

then second step subtracts A from B

[midnight patio]

yeah

[wicked axle]

so say b >= a then the answer is positive or 0.

otherwise it will always be negative.

[midnight patio]

isn't the first part of A and C the same

[wicked axle]

So the value in the accumulator is positive if and only if b >= a

[midnight patio]

wait the load b and store C is different

[wicked axle]

just listen

[wicked axle]

.

the instruction JMPN

will only jump to label else if the accumulator is negative

which is what we want

[midnight patio]

yeah

[wicked axle]

are you sure it's C?

i think it's D

[midnight patio]

the solution says it's C

[wicked axle]

wait

no

You can check by the LOAD and STORE statements.

in both the if branch and the else branch, you want to load c all the time.

the only option that does that is B

[midnight patio]

I it it is loading c

[wicked axle]

wait

nvm

it's C

i confused what the store statement does.

[midnight patio]

but C is not too obvious

[wicked axle]

store stores the accumulator value that is loaded in to x

so in both cases of accumulator we want to load in a value, and then store it into c

which only c does.

[midnight patio]

Okay I get it thanks and also do you know what part of the cpu stores the results?

what register is it?

[wicked axle]

That isn't part of the question.

[wicked axle]

accumulator @midnight patio

[midnight patio]

okay thanks .close

.close

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[digital mural]

what do i do after i find the area of the triangle?

[digital mural]

i dont know what this means or how should i solve

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[pseudo copper]

Hello math friends :). Does anyone know how to prove this inequality?

[stoic dune]

What is $||x||^2_2$?

[flat frigate]

Kaynex

[stoic dune]

I'm assuming this is a distance function and obeys the triangle inequality?

Then again that wouldn't explain the 3

[light shoal]

wait never mind, it's norm squared, not norm

try writing $|x+y+z|^2 = \langle x+y+z, x+y+z\rangle$ and expanding out the RHS

[flat frigate]

OurBelovedBungo

[safe radish]

@pseudo copper Has your question been resolved?

[safe radish]

@pseudo copper Has your question been resolved?

[safe radish]

@pseudo copper Has your question been resolved?

[pseudo copper]

I try and I get $x^2 + y^2 + z^2 - 2xy + 2yz - 2zx$ but I don't understand where the "3" comes from

[flat frigate]

Juan_Duran

[pseudo copper]

<@&286206848099549185>

[light shoal]

this one is tricky but here is a solution

I'm too lazy to latex it

||x+y+z||^2 + ||x-y||^2 + ||x-z||^2 + ||y-z||^2

= <x+y+z, x+y+z> + <x-y,x-y> + <x-z,x-z> + <y-z,y-z>

= ||x||^2 + ||y||^2 + ||z||^2 + 2<x,y> + 2<x,z> + 2<y,z>
+ ||x||^2 + ||y||^2 - 2<x,y>
+ ||x||^2 + ||z||^2 - 2<x,z>
+ ||y||^2 + ||z||^2 - 2<y,z>

= 3(||x||^2 + ||y||^2 + ||z||^2)

therefore:

||x+y+z||^2 = 3(||x||^2 + ||y||^2 + ||z||^2) - ||x-y||^2 - ||x-z||^2 - ||y-z||^2

which is obviously <= 3(||x||^2 + ||y||^2 + ||z||^2)

the second to last line also shows that equality is achieved if and only if x = y = z

[safe radish]

@pseudo copper Has your question been resolved?

[cursive spindle]

Instead of x,y,z I recommend just renaming as $x_{1}, x_{2}, x_{3}$, and just use indexing

[flat frigate]

pepper

[cursive spindle]

Then the repetition is also done using $\forall$ ...

[flat frigate]

pepper

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[bronze moth]

Hi guys

I forgot

how to do this qs

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[viscid sparrow]

heLLO im kinda stuck on a lesson and cant rly seem to figure out the answer, but is the graph of k12 euler, hamiltonian, both, or neither?

[viscid sparrow]

wait nvm got it

.close

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[autumn locust]

"how long is a chord of a circle with equation (x+2)^2 +y^2=169 which is at a distance of 5 units from centre"
Umm here how are we supposed to find the length of the chord from the given information?
I believe we can find the centre of the circle from the equation but after that what are we supposed to do?

[potent roost]

first step would be finding the chord

do you have an idea of where the chord would be, or its orientation

[autumn locust]

Hmm I don't think so as the question only stated the equation and the distance from the centre

[potent roost]

yes using that you can figure out the its orientation

when we are told the distance of a chord from the center it means the perpendicular distance

[autumn locust]

Oo

[potent roost]

from the midpoint of the chord

one sec

there can be two chords that satisfy that

[autumn locust]

Aren't we given the equation of the circle so cant we find its radius? if the distance is perpendicular to the chord and bisects it can't we make a right angle triangle and find half the length of the chord?

[potent roost]

wait nvm me, I misread the question

I read it as 5 units from the origin

💀

[potent roost]

yes

[autumn locust]

Xd

[potent roost]

so you think you can do it now?

[autumn locust]

Ye let me try

[potent roost]

aight good luck

[autumn locust]

Thanks alot

Ye did it. Thanks again

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[lean otter]

A+B=8

B=2A

So A+2A=8

3A=8

A=2.66 or A=8/3

Is this the right way of doing this?

Or

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[fiery spoke]

How do I prove this?

[shy temple]

You can't. It's not true for all n. For example, for n =2, you get 1 - cos(2pi) = 1 + cos(pi/2)

[fiery spoke]

For some reason I was told it can be proved

[shy temple]

You were lied to.

[safe radish]

@fiery spoke Has your question been resolved?

[fiery spoke]

<@&286206848099549185>

[shy temple]

The statement is false. Note that for n = 2, you have that
1 - cos(4pi/2) = 1 - cos(2pi) = 1 - 1 = 0
1 + cos(pi/2) = 1 + 0 = 1.
And 1 =/= 0.

[safe radish]

@fiery spoke Has your question been resolved?

[fiery spoke]

What about for 2pi?

Like on RHS you have 1+cos(2pi/n)

[shy temple]

You mean n = 2pi? Then you get 1 - cos(4pi/2pi) = 1- cos 2.
The right side is 1 + cos(pi/2pi) = 1 + cos (1/2).

[fiery spoke]

no

I mean this

[shy temple]

that's what I did

[fiery spoke]

You said n=2pi

[fiery spoke]

Instead of 4pi on LHS you have 2pi

[shy temple]

because you said what about for 2 pi? here

[fiery spoke]

Sorry, that's my bad

[fiery spoke]

What about for this

Is what I meant

Sorry for badly expressing myself

[shy temple]

Okay, typo then.
1 - cos(2pi/2pi) = 1 - cos 1
1 + cos(pi/2pi) = 1 + cos(1/2)

,w plot cos x

[flat frigate]

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[shy temple]

From the graph you can see cos 1 is not cos 1/2, so 1 - cos 1 is not 1 + cos(1/2)

[fiery spoke]

Hm, yeah alright

Thank you

For your help

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[green mulch]

hello

[opaque prism]

hi

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[frosty cairn]

sorry for the messy writing, i've done this question 4395345 times and i'm frustrated 😭
i'm not sure what i did wrong, though i missed my previous 2 classes and i'm just winging it

[safe radish]

@frosty cairn Has your question been resolved?

[frosty cairn]

<@&286206848099549185> pre calc

[safe radish]

@frosty cairn Has your question been resolved?

[safe radish]

@frosty cairn Has your question been resolved?

[marsh walrus]

take the change and divide it by the time

what makes you think to use cosine?

[frosty cairn]

that's what the teacher said, use cos if it's the high or low and use sine if it's the middle

[marsh walrus]

use sine if its the middle okay

let me check out what you have

so its a little concerning you dont have an x in your function

typo?

[frosty cairn]

yeah i was just trying to get through it, like i said i had already done similar problems 57395 times

[marsh walrus]

so the problem is really just translating a cosine function?

your scaling looks good

i think its just translation thats off

lmc

lets start with your translated cosine

$\frac{39}{2} \cos (x) + \frac{59}{2}$

[flat frigate]

jan Niku (Shuri for Honorable)

[marsh walrus]

this has the correct amplitude and range

the problem is frequency

i think the correct order of fixing things is to first fix the period, then fix the initial point

can you tell what the period of this thing is now, and what it should be?

[marsh walrus]

youre honestly so close!

[frosty cairn]

thanks! i missed the lesson for this so i just picked up scraps off the internet lmao
i'm not sure why the period is wrong, unless you're just counting the x? if not, i honestly don't know ;-;
i think it'd be 12 months because it's during rhe year, but i'm not sure bc it starts in decenber

[marsh walrus]

no, youre exactly right

it should be 12

we want it to start really low in december right

then in the middle of the year its high

then next winter its back low

so 12 full months to get back to where it starts

you should know a trick to make cosine period 1 is to multiply it by 2 pi

we have the translation on the outside, lets try to make a cosine that has period 12

how would you do this, knowing that $\cos ( 2 \pi x)$ has period 1?

[flat frigate]

jan Niku (Shuri for Honorable)

[frosty cairn]

2 pi x 12?

[marsh walrus]

not quite, but close

if you multiply the argument (by a number bigger than 1), it will speed up

but with cos(2pi*x), we have period 1, and we want 12

so we need to slow it down

[frosty cairn]

so divide?

so it'd still be pi over six?

[marsh walrus]

yea

but with x

$\cos (\frac{x \pi}{6})$

[flat frigate]

jan Niku (Shuri for Honorable)

[marsh walrus]

this is a cosine with period with 12

but whats wrong with it?

wheres the peak?

[frosty cairn]

hmm i've never seen it written like that
i'm not sure what's wrong with it

[marsh walrus]

try plotting it

[frosty cairn]

i don't know how 😭

[marsh walrus]

use desmos

[frosty cairn]

alrighty, one second

[marsh walrus]

https://www.desmos.com/calculator/p69wfugerr

heres our function so far

its close

but it has a minor problem

[frosty cairn]

does it need to be negative?

[marsh walrus]

yea

how can we make it negative?

[frosty cairn]

-6

[marsh walrus]

wheree?

did you get it?

[frosty cairn]

wdym

[marsh walrus]

where does the 6 go

and did you figure out your answer

[frosty cairn]

sorry, i meant -39/2, i was multitasking

[marsh walrus]

no

[frosty cairn]

i don't know then

[marsh walrus]

this about translating cosine forward

[frosty cairn]

i'm not sure what that means

[marsh walrus]

https://www.desmos.com/calculator/l7frahlvca

try sliding a around

[frosty cairn]

is A not 39/2?

[marsh walrus]

in what?

[marsh walrus]

what is the relationship between cos(x) and cos(x+pi)?

[frosty cairn]

the horizontal shift?

i think?

[marsh walrus]

its a horizontal shift but what does it do

what do they look like compared to each other

[frosty cairn]

moved to the left or right

[marsh walrus]

be more specific

you are describing what happens for any horizontal shift

what about specifically its if its by pi

[frosty cairn]

i don't know ;-;

[marsh walrus]

graph it

[frosty cairn]

i don't know what to do to graph that

i don't know i didn't go to the lesson i don't know how to do this at all

and i've been stuck on this one problem for 5 hours

[marsh walrus]

you need to use desmos

type in cosx and cos(x-pi)

[frosty cairn]

this?

[marsh walrus]

how do they compare

,w plot cosx and cos(x+pi)

[flat frigate]

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[frosty cairn]

one is negative

[marsh walrus]

what do you think that means

can we use this?

[frosty cairn]

i thought it would be negative 39/2

i don't know what else it would be

[marsh walrus]

can you tell why that doesnt work

if not you should graph it

you should be graphing everything, really

[frosty cairn]

well i havent been graphing cause we arent allowed graphing calculators on the final, but i'll check one second

[marsh walrus]

then you need to learn some identities

or the unit circle

[frosty cairn]

i put a negative in front of everything and the graph didn't change

[marsh walrus]

?

,w plot cosx and cos(-x)

[flat frigate]

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[marsh walrus]

sorry, wrong plots

,w plot cosx and -cosx

[flat frigate]

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[frosty cairn]

i don't get this

i'm too stressed out to try to wrap my head around this right now

i don't understand what i need to do

[marsh walrus]

we just need to shift our argument by pi

thats all

here is the answer

if you feel too stressed you should maybe take a break but

oops, 1s

https://www.desmos.com/calculator/wtg2fr3csy

heres the answer

id suggest starting here, and seeing if you can understand what each piece does

try deleting it piece by piece until you can get back to cos(x) alone

if you cant do it with desmos youre gonna have a hard time without

just focus on trying to understand generally what each element is doing

[frosty cairn]

alright

thank you

i'm sorry for getting snappy, today has just not been the day

[marsh walrus]

lol for me either

you can definitely get it

[frosty cairn]

thank you again, ik it's frustrating trying to teach someome who doesn't get it 😭 hope you have a better day tomorrow :)

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[lunar gazelle]

If the Central Limit Theorem suggests we can get a normally-distributed sample of a dataset by using no more than 10% of the data, why would a machine learning algorithm need to be trained on 20% of the dataset?

[plucky elk]

depends on your application

[lunar gazelle]

As in?

[plucky elk]

depends on "need"

there is no singular percentage of dataset to train on that is universal for all applications and all business needs

[lunar gazelle]

So the 20% is just a completely arbitrary rule of thumb?

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[lean otter]

what is the author’s mindset when it comes to solving this problem? I completely understand the proof but it seems over complicated.
the author ends his proof with 1 + (1/2)+ (1/(2^2)) + …. < 2 which seems like he figured by manually adding up a few values from left to right untill
1/(2^n) > 2 - (s(n+1))
(inequality satisfied when n >= 6)

if there exists an n such that the inequality is satisfied than it is implied that it’s impossible for the sequence to be bounded above by 2

s_(n) := the n-tail of sequence s

my point is, we could have used that strategy in the first step via the inequality 1 + (1/(2^2)) + (1/(3^2)) + …

maybe the author had reduced the denominator to a number with base 2 for a obvious reason? idk

[plucky elk]

if it's over complicated do you have a simpler solution?

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[lean otter]

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✅

[lean otter]

I already proposed it

s := 1 + (1/(2^2)) + (1/(3^2)) + …
1/(n^2) > 2 - ((n + 1)_tail_of_s)

if this inequality is satisfied for n >= some arbitrary N that’s a member of naturals then the series converges to a value less than 2

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@lean otter Has your question been resolved?

[lean otter]

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[royal iris]

why can't anything be done to 1-sin(theta)/cos(theta)

[royal iris]

when simplifying*

[quasi bison]

do you mean $\frac{1 - \sin(\theta)}{\cos(\theta)}$?

[flat frigate]

Ann

[quasi bison]

if you do, then you are missing a pair of parentheses.

and if you really want to, you can rewrite 1 - sin(θ) as 2cos^2(θ/2 + π/4) perhaps

[safe radish]

@royal iris Has your question been resolved?

[royal iris]

no its just that and I have to prove why its wrong

[royal iris]

and yes

[mortal meteor]

there is no claim that can be wrong. There is just an expression.

[royal iris]

srry caps

sorry wrong language

prove that it cant be simplified

[mortal meteor]

If i write the expression differently, what criterion is there to decide which expression is simpler?

[royal iris]

honestly no clue man sorry

[mortal meteor]

if its just an opinion which expression is simpler, one can't prove that you can't simplify it. I mean maybe i think after rewriting that it looks simpler now and you think that is not simpler...

[royal iris]

fair, idk this is just a question that got my brain hurt

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[proven surge]

what is this question?

[fallen hamlet]

3,25

[proven surge]

'What are the reciprocals of the fopllowing. also expressed in SF wherever appropriate?'

then it asks me that

[fallen hamlet]

https://youtu.be/xY_Ha_Vc2pg

[proven surge]

thanks

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[proven surge]

thank you again Asger, have a nice day

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[tall sphinx]

I want to find the gradient of a function with multiple variables

[tall sphinx]

as of now, that is the work i have done. Is all i do apply the variables P(1,4,2)

[whole geyser]

Are you trying to find the gradient at a point? If so then yea

[tall sphinx]

I have another question

the partial derivative of the function with respect to x

that should be 1

[whole geyser]

Yes

[tall sphinx]

so do i have \vec{j} 1

i wont bother with latex

[whole geyser]

The partial of any constant is 0, which means that you’re left with the partial of x which is just 1

[tall sphinx]

yes, and when i get the gradient of g

is that correct

how i did it

[whole geyser]

Yes

[tall sphinx]

Wonderful

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[upper pewter]

if I want to express a graph with y axis being intensity and x axis being time, should I say "intensity over time graph"?
like is that the correct terminnology? I need for a research paper

[safe radish]

@upper pewter Has your question been resolved?

[upper pewter]

<@&286206848099549185>

[raw wharf]

language is scary 😭

[upper pewter]

yes.. unfortunately i get marked down if I dont use correct terms 😦

[plucky elk]

depends on your domain. do a lit review

[upper pewter]

what's a lit review?

[plucky elk]

if you are contributing to a research paper, that's one of the first things you do so that you know you're not reproducing work

[upper pewter]

oh ok

[plucky elk]

https://www.scribbr.com/dissertation/literature-review/

[upper pewter]

the research paper is literally a high school assignment, its not supposed to be an academic publish though

[plucky elk]

oh hahaha then what you said is fine

[safe radish]

@upper pewter Has your question been resolved?

[upper pewter]

oh ok cheers 🙂

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[hollow moon]

yo

[hollow moon]

I need help

I tried this

U sub

U= e^x+1

Shit did not work like

I dont understand

[ancient escarp]

can you show work

[hollow moon]

yes

here

[patent thorn]

it 1^3 =1^5
according to law of exponent bases same so pwer same same this means
3=5

[ancient escarp]

bro seriously

[ancient escarp]

your problem here is pulling the 1/e^x out in front

[ivory meteor]

that's smart howd you do that

[hollow moon]

can you not do that

[patent thorn]

you are really disappointing me you are not answering and ......

[ancient escarp]

you cannot, since e^x is not constant

[hollow moon]

OH OK

[ancient escarp]

it changes with respect to x, and x changes with respect to u

consider the fact that you set u = e^x+1, though

you can solve that for e^x

and get everything in terms of u

[hollow moon]

But it's not integratable

Ohhh

U have to play wit it

[idle basalt]

hi sorry to interupt but this serv is only for algebra or geometry too ?

[hollow moon]

It's for both dawg

If no one responds to u

[ancient escarp]

it's for every single mathematical concept you can think of

#❓how-to-get-help #rules

[hollow moon]

It's cuz they dont know how to do ur question

i looked at ur question

[idle basalt]

oh ok

[hollow moon]

I couldnt do that shit

i didnt know what wuz going on

srry

guys

Is ln of e^x = x

Ln of e to the anything is that anything

Ok nvm

[safe radish]

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[hollow moon]

.close

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[cerulean edge]

What's the equation of the circle?

[cerulean edge]

|z| = r?

What's the equation of the circle?

[feral linden]

x^2+y^2=r^2

[queen parcel]

That looks good to me

[queen parcel]

We're in the complex plane though

[feral linden]

Then he said it himself

[feral linden]

This

[cerulean edge]

@queen parcel How would you answer? It says ”write the equation of the circle”

[queen parcel]

You're basically right

[cerulean edge]

Do I just write |z| = r

[queen parcel]

The magnitude of the complex number is the radius of the circle

[queen parcel]

I would

[cerulean edge]

Thanks

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[whole stone]

I'm learning Trigonometry in my math class, and he gave some homework. I'm doing fine, but I am stuck on this type of problem. I tried dividing the tangent and multiplying, but I failed.
The Problem:
10= x/tan(35)

[queen parcel]

Think of tan(35) as just another numnber

How would you solve for x in something like x/5 = 10

[whole stone]

X would be 2 in that, right?

Cause the 5*x would be 10

wait

no?

[winter pivot]

yeah that's right

so just multiply both sides of your equation by tan(35)

[whole stone]

I got 8.19

Alright, thank you

close.

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[loud island]

So i've been learning logarithims and this question has stumped me. Ive tried simply substituting the letters with the numbers above but it keep saying im wrong when i input the anwer i get

For a) I've been getting ln(8/(5/3))

[queen parcel]

Do you have work for each problem?

If so I can check it

[loud island]

Yes one second. I rewrote it out so its easier to understand

[queen parcel]

Log properties (although I'd google them):

1. ln(xy) = ln(x) + ln(y)

2. ln(x/y) = ln(x) - ln(y)

3. log_x (y) = ln (y) / ln (x)

4. ln(x + y) ≠ ln(x) + ln(y)

5. ln(x - y) ≠ ln(x) - ln(y)

[queen parcel]

a ≠ 2, b ≠ 3, c ≠ 5

[loud island]

Do i substitute the whole ln(a) for 2 then?

[stable ruin]

@loud island if ln a = 2

Do you know how to solve for a

[empty marten]

||is it e^2?||

[loud island]

wouldnt it be e^2? Sorry this is literally my second day learning logs

[stable ruin]

Yes

So can you do the same thing to solve for b and c

[loud island]

ok so then would i put e^2 in place of a in the equations? and do the same with b and c?

[stable ruin]

Yes

[loud island]

Thank you! So then when I put e^2 in my first equation does the law of exponents apply and I would multiply the ^2 and the ^3?

[stable ruin]

Uh

The exponents would be added but y’a pretty much

[loud island]

woops lol

thx

Sorry im still confused. So i understand when a^2 * a^3 you would add them. But when I put e^2 into the initial question. Wouldn't it be (e^2)^3 so i would multiplied them for e^6?

[stable ruin]

If it’s that then yes e^6

[loud island]

So would this be correct?

[stable ruin]

Looks good

[loud island]

Okay sorry to ask. But does this all look good?

[stable ruin]

Double check on d @loud island

[loud island]

oh woops multiplication. d =-35?

[stable ruin]

Should be good

[loud island]

omg thank you soooo much! I can finally go to bed! Its almost 3 am

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[errant brook]

can someone help w/ this truth table?

Is it supposed to have just 4 letters: X,Y,A,B?

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[lean otter]

Please could someone tell me if I have done this right?

[vagrant acorn]

Yes. It's correct.

[lean otter]

Amazing! thanks for your help

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[lean otter]

Hi, can someone help me? My math exams is in like 3 days and I know nothing

[lean otter]

Idk lol is it my fault for having like a shitty math teacher??!?!?

we'll never know

nvm I'll just try my best to recall what I learnt ❤️❤️

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[sonic spruce]

@lean otter what do you need help with?

[safe radish]

@carmine oxide Has your question been resolved?

[sonic spruce]

ok, so notice how AOBD is a cyclic quadrilateral

Consider Chord AB, the angle subtended by a chord on the centre is always double the angle subtended anywhere else on the circle

so, 1/2 * Angle ACB = ADB
Also, Angle AOB=Angle ACB

Then 1/2 * Angle AOB= angle ADB

Sum of Opposite angles of a cyclic quadrilateral is 180 degrees

now can you solve it?

@carmine oxide

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[rustic hound]

how can I find solutions to 3rd degree polynomials? I usually try to factor out stuff or divide the polynomial by another one if I already have a solution and most of the times it works, but in cases like this, those don't work.

[rancid wagon]

add + 0x^2 into equation

should be fine for u