help-23

[south sparrow]

Is 'throwing a fair die, each number has a 1/6 probability chance to be the outcome' a model?

Model is a conjecture

https://en.wikipedia.org/wiki/Mathematical_model

Mactutor has no entry on the history of "model" https://mathshistory.st-andrews.ac.uk/Miller/mathword/m/

There are mathematical models in music!

[safe radish]

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[mystic timber]

I have no clue on this

Like if you read choices backward they doesn't makes sense

[fickle pendant]

oh it took me a minute

the names are written vertically @mystic timber
e.g.
I
D
A
H
O

which, if you flip it, becomes
I
ᗡ
A
H
O

[safe radish]

@mystic timber Has your question been resolved?

[mystic timber]

So the answer is Hawaii supposing all the words are written capital

@fickle pendant

[fickle pendant]

yes

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[quartz wasp]

Trapezoids area is 594 the hight is. 22 the difference between the parralel sides is 6 need to find both of the parralel sides

[quasi bison]

do you know how to find the area of a trapezoid in general?

was asking op

[humble helm]

he is lost

[quasi bison]

you should have let him either try to state the formula or tell me he doesn't know

[humble helm]

he wasn't answering for 3 minutes

[quasi bison]

3 minutes is not a stall

and even if he was not answering for 30 you should not have spoiled it

if op chooses to stall and let the channel time out, that's his choice

or if he has to go for something urgent, idk

[quartz wasp]

Ik the formula

I was making cofee

[quasi bison]

okay then say the formula

[quartz wasp]

A+b/2*h

[quasi bison]

missing parentheses

(a+b)/2 * h

but ok alright fine

you can find a+b from here

and the problem tells you a-b=6

[quartz wasp]

A+b equals 2h no?

[quasi bison]

incorrect.

[humble helm]

no way

Where is area

[quartz wasp]

Wait

[quasi bison]

$S = \frac{a+b}{2}\cdot h \ h = 22, , S = 594 \ a+b = ; ?$

[flat frigate]

Ann

[quartz wasp]

2s/h?

[humble helm]

I guess yes

So find

[quartz wasp]

I can make a system right

[quasi bison]

yes exactly

[quartz wasp]

Ok ill do it

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[safe parrot]

*A tower has four asymmetrical wings.

Determine the relationship for the number of blocks+1 needed for the squares of the height+1 if you know the number of squares of the height.

Then show by induction that an=n(2n−1), n=1,2,3,...*

[safe parrot]

Does someone actually understand how to solve this?

"Determine the relationship for the number of blocks+1 needed for the squares of the height+1 if you know the number of squares of the height. "

Particularly, what does this even mean?

[south sparrow]

What does it mean that a tower has a wing?

[safe parrot]

.close

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[south sparrow]

What is a model?

[final halo]

Context?

[south sparrow]

Throwing a fair die, or tossing a coin.

[final halo]

In those cases a model would be coming like a way to represent the outcomes of those events mathematically

Or the probabilities of things occurring

[south sparrow]

1/6 and 1/2 are models

[final halo]

https://www.usu.edu/math/schneit/StatsStuff/Probability/probModels1.html#:~:text=A probability model is a,probability model with a table.

[south sparrow]

A probability model for a coin toss looks like this.

Probability model for fair die roll:

import random

num_trials = 1000000
num_heads = 0

for i in range(num_trials):
    coin_toss = random.choice(['H', 'T'])
    if coin_toss == 'H':
        num_heads += 1

probability_heads = num_heads / num_trials
print(probability_heads)

Is this Python snippet still a probability model?

[safe radish]

@south sparrow Has your question been resolved?

[south sparrow]

Thanks @final halo , this is great, now I can create a model!

import random
print(sum(random.choice('HT') == 'H' for _ in range(10**6))/10**6)

---

    |       |
  H |  0.5  |
    |_______|
    |       |
  T |  0.5  |
    |_______|

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[covert kiln]

can anyone please explain to me when doing implicit diffrentiation where does the dy/dx come from

[covert kiln]

for example

[empty gyro]

Chain rule on y

[covert kiln]

wouldnt it just be 3y^2

[empty gyro]

Chain rule

3y^2*y'

[south sparrow]

We can think of d as a little bit of something. y depends on x, so if we adjust x by a tiny bit, y is changed by a tiny bit.

This can be written, by adding dx to all the x terms, and adding dy to all the y terms.

[covert kiln]

where does prime y come from

[empty gyro]

Chain rule

[covert kiln]

k i understand that

[south sparrow]

What is prime y?

[covert kiln]

y'

[empty gyro]

Would you know how to find the derivative of [f(x)]^3?

[south sparrow]

Say we have y = x^2, the derivate of this is y + dy = (x + dx)^2, where we adjusted the y and x values by a little bit.

[covert kiln]

3F(x)^2?

[empty gyro]

You're forgetting chain rule. Again

@covert kiln are you familiar with what chain rule is?

[covert kiln]

yh

[empty gyro]

Do you know how to use it?

[covert kiln]

u bring power down and diffrential and times it by the origanl and subtract 1 to the power

[compact pelican]

I think that's the power rule

[empty gyro]

That's power rule..

[compact pelican]

If you mean d(x^n) = n*x^(n-1)

[empty gyro]

@covert kiln read up on chain rule. You need to understand it to understand implicit differentiation

[covert kiln]

isnt that chain rule

[compact pelican]

Where did the x after the 5 come from?

But, aside from that, correct

[covert kiln]

that times

[covert kiln]

yh taht what i said

[compact pelican]

OK, really not great to use x as both a variable and to mean times

[empty gyro]

That's an example, yes

[empty gyro]

But you don't seem to understand how chain rule applies here. Which tells me you need a thorough review of it before we can proceed

[covert kiln]

with y^3 it will be 3(1)(y^2)

where doe y' appear

[south sparrow]

Can you explain it in simple terms please, because these concepts show something dead simple in complicated language.

[compact pelican]

True, that's an example of using the chain rule with the power rule. In general, the chain rule applies when you compose two functions f(g(x))' = f'(g(x))*g'(x)

[covert kiln]

the question taht im asking?

[covert kiln]

isnt that prodcut rule?

[compact pelican]

If it helps, remember that y is a function of x, so you can write it as y(x)

[south sparrow]

To explain things to you, if someone knows the simple thing behind the complexity.

And to me!

[compact pelican]

No the product rule is (f(x)*g(x))' = f(x)*g'(x) + f'(x)*g(x)

[covert kiln]

yh mb but i was only taught chain rule via power rule

[compact pelican]

OK, you're going to need it in other cases too like sin(x^2)

[empty gyro]

This is incorrect. Again. You are ignoring chain rule

[empty gyro]

$[f(g(x))]'=f'(g(x))g'(x)$

[flat frigate]

SWR

[covert kiln]

can u give an example with numbers please

[covert kiln]

if we diffrentiate x^3 we just get 3x^2

[plucky elk]

Try finding
(sin(x^3))'

[covert kiln]

3x^2cos(x^3)

[covert kiln]

?

[plucky elk]

Yes that's right

,w diff sin(x^3)

[flat frigate]

Results provided by WolframAlpha

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

Requested by rie.mann

[covert kiln]

how does this help with my orgianl question

[plucky elk]

You asked for it?

[covert kiln]

oh ok

i dont understnad how it used for y^3 tho

and how it ends up being 3y^2 dy/dx

[plucky elk]

Let y=f(x) then apply chain rule

[empty gyro]

Let $f(x)=x^3$ and $g(x)=y$

[flat frigate]

SWR

[covert kiln]

f(X)^3 =3F(x)^2

[empty gyro]

Try this one.

[covert kiln]

im lost on the fact that there 2 diffrnet functions

[empty gyro]

And apply as written here

[covert kiln]

kl il try

[covert kiln]

[f(y)]' = f'(y) (1)

[empty gyro]

Nope

[covert kiln]

im so lost on that formula

[empty gyro]

$g'(x)=y'$

[flat frigate]

SWR

[covert kiln]

y'=1

[empty gyro]

I would review it in your book if I were you

[empty gyro]

False

[covert kiln]

oh yh taht is false

what is y'

[empty gyro]

That would only be true if $y=x$ (plus some constant but I'll ignore that)

[flat frigate]

SWR

[covert kiln]

our teacher didnt teach us that formula

[empty gyro]

Unknown. That's why we leave it as just y'

[covert kiln]

is it equivalent to dy/dx

[empty gyro]

I highly doubt that. It's necessary for this stuff. Ask your classmates if they have seen it

[empty gyro]

Yes. It's just shorthand notation

[covert kiln]

😭

our teacher shows us shortcut

[covert kiln]

why isnt there dy/dx when we diffrentiate x^3

[empty gyro]

[empty gyro]

Because there is no y in x^3

[covert kiln]

but there no x^3 in y

[empty gyro]

You are differentiating y^3 though, not y

Okay, try this:

What is derivative of $[\sin{x}]^3$

[covert kiln]

if we diffrnetiate y^3 we get 3y^2 dy/dx but there no dy/dx for diffrentiating x^3

[flat frigate]

SWR

[covert kiln]

that what 1 of my classmate said

[empty gyro]

It's mostly right. But it's kind of a word soup

[empty gyro]

Correct

[covert kiln]

wdym word soup

[covert kiln]

yh i dont understand why tho im copying that from the anwser

[empty gyro]

It rambles on a bit too much

[covert kiln]

3cossin^2

[covert kiln]

can u give me a simple explaination

[empty gyro]

This is correct. Suppose y=sin(x). What is y'?

[covert kiln]

cos

what does dy/dx mean does it mean diffrnetiate everything

[empty gyro]

dy/dx means "differentiate y with respect to x"

[covert kiln]

what does that mean tho

can i also diffrentiate y terms?

im confused with the wording with things0=sdo[kdasdasd

[empty gyro]

It means to find the derivative of y with respect to x

[covert kiln]

so to diffrentatie what y is equal to

y= 2x

so i would diffrrentiate 2x

[empty gyro]

Yes

[covert kiln]

y=2x + 6

6 would disappear or = to 0

cause it not an x

[empty gyro]

Yes

[covert kiln]

y= y^3 +2x

what happens to the y^3

does it disappear

[south sparrow]

dy/dx also means a tiny change of values, it is a slope value. ∆y/∆x

As we change x a little bit, y changes a little bit with it.

[covert kiln]

is the statment i said true

[south sparrow]

And I think in your original equation we write x^3 + x + y^3 + 3y = 6 as x^3 + x + u = 6

[south sparrow]

Sorry, which statement?

[safe radish]

@covert kiln Has your question been resolved?

[south sparrow]

https://www.youtube.com/watch?v=WUvTyaaNkzM&pp=ygUTZXNzZW5jZSBvZiBjYWxjdWx1cw%3D%3D

There is a video on chain rule as well, in this series.

https://archive.org/details/calculusmadeeasy_202001/page/n3/mode/2up

[safe radish]

@covert kiln Has your question been resolved?

[safe radish]

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[violet geode]

yo
if we have a^k = x mod t
and we have a and x and t
when do we know that this will cycle
for example(every fifth k)?

[safe radish]

Please don't occupy multiple help channels.

[plucky elk]

.close

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[plucky elk]

!1c

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[warm oyster]

Find the inflection point in the curve f(x) = | x³ - 1 |

[warm oyster]

When I differentiated again the function became x6

So x6 =0

Then x = 0

But when I looked at the answers it turns out there are 2 values

{0 , 1}

How did x become 1?

[peak estuary]

what

dont ping me. in general dont ping specific people

[warm oyster]

Ok should I ping helpers role?

[devout shale]

after 15 minutes of waiting you can ping the helpers role once

[warm oyster]

Ok

[warm oyster]

.

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[visual shell]

I don't know how to find the range of values for this problem y=1-sin²(x)cos²(x)

[fallen grotto]

start by finding the range of values for sin(x)cos(x)

Hint: sin(x)cos(x) = 2*sin(x)cos(x)/2

[visual shell]

It's y=[-0.5;0,5]?

[fallen grotto]

yeah good job

should be pretty straightforward from there

[visual shell]

Thank you

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[pallid glacier]

hi. can someone explain why the circled part of the solution is the case.

oh

[stoic dune]

That's a second derivative test

[pallid glacier]

yes. but specifically those values

x=L,L/2

[stoic dune]

Solutions to part a

[pallid glacier]

nono like

how is it certain that when plugging in those values to the second derivative that the value of the second derivative is >/< 0

is it to do with 0<=x<=L

[stoic dune]

/<?

[pallid glacier]

when u put L/2 into the second derivative

how do u know that the second derivative is negative in that case

[stoic dune]

You calculate it

Get the second derivative, plug L/2 into it

They did calculate the second derivative above your circled part

[pallid glacier]

oh ye brainfart sry

i missed the carry through of L on 6xL which made it impossible

thanks

.close

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[white umbra]

OP got muted

.close

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[wheat apex]

I'm having trouble with the substitution method, i think i just have a misunderstanding of something simple

please could you hlep me with this:

oh wait

i think i get it

.clsoe

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[wheat apex]

.reopen

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✅

[wheat apex]

i didnt lol

oh wait

maybe i am

sorry

one second

was getting confused with this now writing it out

slowly making sense

is this the correct preliminary steps for the question?

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[loud bone]

I wanna solve this problem but when I try to solve it my answer does not match the answer given by the Google

√x +√x-64 = 16
(√x +√x-64)^2= (16)^2
x+x-64=256
2x-64=256
2x=256+64
2x=320
X=320/2=160

The answer given by Google is 100 where did I went wrong?

[desert pasture]

sub x=u^2 probably

[dreamy elk]

$(\sqrt{x}+\sqrt{x-64})^2\neq x+x-64$

[flat frigate]

Crystopher

[humble helm]

what a hell

[exotic sky]

tf

[humble helm]

I guess you need to square both sides

nah?

[exotic sky]

oh wait

he didnt

nvm

[bleak tapir]

You forget 2ab in (a+b)^2

[humble helm]

He did wrongly

[exotic sky]

its going to be hectic, id recommend you take a term to RHS and then square

[humble helm]

that's fucking square formula a²+2ab+b²

[humble helm]

? Just say x>0 x>64

Idk

[loud bone]

It's getting too hard

[humble helm]

anyway

It's the only way

[humble helm]

First of all square both sides correctly

[loud bone]

Let me try it

[twilit spindle]

$\sqrt{x}+\sqrt{x-64}=16\ x+2\sqrt{x^2-64x}+x-64=256\ 2\sqrt{x^2-64x}=320-2x$ do you think you can take it from here?

oh it got cut off

[flat frigate]

PajamaMamaLlama

[humble helm]

Square both sides again nah?

[exotic sky]

its going to worsen

as i said

take a term to the rhs

then square

[loud bone]

Can you do what @exotic sky said

[exotic sky]

this is wrong

you wont get solution

the root is going to be annoying

[exotic sky]

take the rootx or the other term to the RHS

then square

2 times

you'll get your answer

[twilit spindle]

that is the exact same thing I did above?

[humble helm]

Agree

[exotic sky]

i am blind ig

yes iam

[loud bone]

Okay like this

√x-64=16√x

[exotic sky]

im totally sorry

i read that as

oh my god

[exotic sky]

16 - rootx

[loud bone]

Oh yes

I forgot that

[exotic sky]

now square

2 times

[exotic sky]

like this

[loud bone]

√x-64=16-√x

(√x-64)^2=(16-√x)^2
(√x)^2-2*√x64+(64)^2=(16)^2-216*(√x)+(√x)^2

=x-2x*64+4096=256-32√x+x

[bleak tapir]

√(x-64)≠√x-64

[humble helm]

But he did correct nah?

Ah whatever

[loud bone]

So it is a whole value in itself?

[humble helm]

Ah ye

Didn't you see

It is given

[loud bone]

(64x+4096)^2=(224√x+x)^2

.close

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[dull stratus]

https://tutorial.math.lamar.edu/classes/calciii/eqnsofplanes.aspx#:~:text=This is called the vector,a vector for the difference.&text=where d%3Dax0,y 0 %2B c z 0 .

[dull stratus]

What does Paul mean by normal vector not having to touch the plane?

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@dull stratus Has your question been resolved?

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@dull stratus Has your question been resolved?

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[fluid holly]

How do I start

[lean otter]

what have you tried

[fluid holly]

Well

My favourable out come is 3c12?

12c3

Mb

[safe radish]

@fluid holly Has your question been resolved?

[fluid holly]

<@&286206848099549185>

[lean otter]

Not sure if my method is right but i think u have to use begger here and then apply probability multiplication theorem

[sonic spruce]

whats that?

i think you gotta consider what happens to the other 9 balls. how they are put into boxes

[lean otter]

begger os probability multiplication theorem?
Which 1 u talking about?

[sonic spruce]

begger

[lean otter]

its a topic from Permutation and combination, Number of ways to distribute "n" items among "r" ppl
$${n+r-1}C{r-1} $$

[flat frigate]

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

Snow

[safe radish]

@fluid holly Has your question been resolved?

[safe radish]

@fluid holly Has your question been resolved?

[safe radish]

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[covert kiln]

when diffrentiation cant i just rerannge and make y the subject

then diffreentaite and p[lug in 2

[dim meteor]

if thats possible then do that yes

[covert kiln]

i did it but i got a wrong anwser

i think it cause i didnt include -3 somehow

[dim meteor]

show your work

[covert kiln]

k but its kinda messy

[dim meteor]

alright

[covert kiln]

gimme a sec

[covert kiln]

that the anwser

[covert kiln]

if u cant read it i can rewritre

[dim meteor]

just that last part where you differntiate it please

[covert kiln]

is that better

i pluged in 2 and i got -11.296

[dim meteor]

you sure $v'=2$?

[flat frigate]

The Great D

[covert kiln]

oh minus 2

[dim meteor]

yes

[covert kiln]

still got the wrong anwser 😭

[dim meteor]

and why do you have it multiplied by 6^x?

[covert kiln]

oh

cause before i did 2x3^x which equals 6^x

but that was wrong

forgot to change it

let me retry

[dim meteor]

careful there 2*3^x is not the same as 6^x

[covert kiln]

yes i got it right

tu

ty

.close

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[wooden oyster]

sqrt(x^3) = x^(3/2)

[wooden oyster]

why is this not true

[halcyon carbon]

It is true.

[wooden oyster]

ok good

.close

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[mellow mountain]

how would i do this

[safe radish]

@mellow mountain Has your question been resolved?

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[flint nebula]

I'm trying to solve for x but I completely forget how

[slender juniper]

well

first thing u can do is multiply 39/50

[eager torrent]

not really, first add the stuff up

[slender juniper]

yea ofc

but i was thinking she have already done that mb!

[eager torrent]

the multiply the added numbers by 39/50 and also multiply the part related with x or sqrt(x)/4 with another 39/50 and add them together

[flint nebula]

167*39/50?

[slender juniper]

yea so u have ((167+sqrtx/4)x39/50)+10=170

and then just as i said multiply the 39/50

[eager torrent]

167*(39/50)+(sqrt(x)/4)*(39/50)

btw stubtract ten from both sides too as we dont want that plus ten there for nothing so we just got 167(39/50)+(sqrt(x)/4)(39/50) = 160

[flint nebula]

Sorry I'm confused how did you get that

[eager torrent]

so we multiplied teh parts right?

oh i see what you mean

the 167 is oviously the added numbers 103+14+50

[flint nebula]

Why are we doing 167(39/50)

Oh wait I get it

[eager torrent]

here there is this that can help you: (a+b)*N = aN+bN

[flint nebula]

If you're multiplying stuff in brackets it's like you're multiplying it with everything inside separately

[eager torrent]

yup

[flint nebula]

Okay yeah

[eager torrent]

now we subtract ten from both sides as we got that plus ten at the end meaning that 167(39/50)+(sqrt(x)/4)(39/50) +10-10=170-10

those plus tens cancel out, and 170-10 = 160

now we multiply the 39 with 167 and then divide by fifty for one of theh parts

[flint nebula]

Okay

[eager torrent]

167*39=6513

if we divide it by 50 we get 130.26

now we can subtract frm both sides

and the +130.26 and the minus 130.26 cancel out for the left part while we get 29.74 for teh right side

now, we multiply the sqrt(x) and 39 on the top and 4 and 50 on the bottom

[flint nebula]

Yep

[eager torrent]

now since we got divide on one side, we can multiply both sides by the 200

[flint nebula]

Alright

[flint nebula]

Then we divide both by 39?

[eager torrent]

yup

[flint nebula]

Okay it's like 152.51

[eager torrent]

now since we see that the fraction is infinite/very long we just let it be, dont simplify it to anything just let it be

[flint nebula]

So √x ≈ 152.51

I don't have to be exact by the way

[eager torrent]

yup, but we wantv exact answers, in math the more precise you are the btter you are

[flint nebula]

Ok

[eager torrent]

but since we got sqrt(x) not just x , we can square both sides

[flint nebula]

Yes

x = 23260.160420775

[flint nebula]

Yes I jumped the gun

[eager torrent]

just leave it there, dont simplify, tahts what you get when you square 5948 and 39

[flint nebula]

Okay cool

I have another kind of algebra question but should I make a new help thing?

[eager torrent]

no we can finsih it here its fine

[flint nebula]

So I should post my other question here?

[eager torrent]

why not, keep it running, if it tells you to stop we can try another help channel

[flint nebula]

Okay

[eager torrent]

ohh, webpages

[flint nebula]

Nope sry

One sec

It's kind of the same.
This is the formula for calculating HP in my game and it has five variables and I know that if you have four variables you can calculate the fifth variable so I wanted help getting one of the variables to be like what it's equal to.
I also tried simplifying the formula 🙂

((B + I + (√E/4) + 50) * L / 50) + 10 = F

There

[eager torrent]

my question is, whats te context on this, what do you mean by hp/ hpiv/hpev

horsepower???

[flint nebula]

Right so that's why I simplified it to this
((B + I + (√E/4) + 50) * L / 50) + 10 = F

Base HP is one variable and HP IV is one variable and HP EV is one variable and Morty Level is one variable

[eager torrent]

abs value is noot needed when we got square roots, so remove those,

[flint nebula]

What sorry?

[eager torrent]

you know those straight lines around the hp ev part?, yea we dont need those

[flint nebula]

Right

Ok

[eager torrent]

so now we multiply all parts by morylevel/50 which we can call m/50

[flint nebula]

Yup

Actually one second did I tell you what I was trying to do

[eager torrent]

k why not

[flint nebula]

I'm trying to flip the equation around so we can solve for HP EV

[eager torrent]

so hp-ev is not hp*ev, from what ur saying

[flint nebula]

Correct

[eager torrent]

oh ok make sense

[flint nebula]

HP EV is one number

[eager torrent]

that make it way easier foor me to follow

[flint nebula]

It stands for health point effort value

[eager torrent]

you doing pokemon?

[flint nebula]

Kind of

It's like pokemon set in Rick and Morty

You use different universes mortys as Pokémon

[eager torrent]

I would rarely dought you would need specific values even in compeetivve or coding

anyways lets continue

[flint nebula]

Ok

[eager torrent]

if thats the case we gotta restart, so its easier

[flint nebula]

Yep

[eager torrent]

ill write it down in desmos just wait

[flint nebula]

By the way what did you use to write these

[eager torrent]

desmos

btw what do you mean by morty level, im familier with gen 1-3 but i dont remeber a morty level?

[flint nebula]

Well it's not Pokémon it's a different game in the style of Pokémon it's just it's level

[eager torrent]

ok i see but can that morty level be negative by any chance

[flint nebula]

No

It's 1 to 100

[eager torrent]

what about the hp evs

[flint nebula]

That can be between 0 and 65535

[eager torrent]

ok, that amkes it way easier, and heck this game is completly ripping off old pokemon

[flint nebula]

It's called pocket mortys

It's set in the Rick and Morty universe

Multiverse

And there's only three types rock paper and scissors

Lol

[eager torrent]

heh, anyway

[flint nebula]

Lol

[eager torrent]

i believe thats it, migh need some checking though

[flint nebula]

Oh Base HP is all one thing sorry about that

I did take the formula and make them all single letter variables if you'd like that

[eager torrent]

I don't get you, what do you mean by all one thing?

[flint nebula]

"Base HP" represents a single number

[eager torrent]

oh, liek its constant like :2, and is always something like 2

[flint nebula]

Are you familiar with base stats?

[eager torrent]

yup

am a competetive player after all

[flint nebula]

So base HP is it's HP base stat

[flint nebula]

I used to do that a little

[eager torrent]

yup, i see, but i stilld ont get the problem

[flint nebula]

Over on the left you have just "hp" and I'm not sure what that's representing from the original image

[eager torrent]

this?

[flint nebula]

Yes

Give me one second I'll be right back

[eager torrent]

oh by hp its means what its hp is at what lvl, what evs, while base stats represent its orignal values before calculuation, a morty charector with higher base p, will have more hp than a one with lower base hp, if the evs and ivs and moretylevek are the same

[flint nebula]

Okay there are five variables
"Base HP"
"HP IV"
"HP EV"
"Morty Level"
"HP" (the final HP stat)
The three instances of HP does not mean those are the same number

Base HP is 170 but the HP IV is 14

Each circle is one variable separate from the rest

[eager torrent]

yup i just put the base(hp) as its easier to write not cause i thought it was a function

[flint nebula]

Oh I thought you meant "base" multiplied by "hp"

As if they were two separate things

[eager torrent]

anyways have a nice day

[flint nebula]

Okay

Thanks bye

.close

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[pure grail]

.close

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[torpid pumice]

s2 = 20, s4=200 in a geometric sequence

[torpid pumice]

s2 = 20, s4 = 200

20=a(1-r^2)/1-r
200=a(1-r^4)/1-r
20(1-r)/(1-r^2) = a
subs
200=20(1-r)(1-r^4)/(1-r)
200=20(1-r^3)
200-200r^3=20
200r^3=180
r^3 = 9/10
is that right?

[shy rover]

why is the second eqn 400?

s4=200

[dim meteor]

I dont understand the sub you did to combine them into one equation. Can you please explain?

[safe radish]

@torpid pumice Has your question been resolved?

[torpid pumice]

oh right my bad

[torpid pumice]

yeah, I put the values i got for a into the other equation

[torpid pumice]

<@&286206848099549185> is this right? im pretty sure its still wrong

[dim meteor]

ok in the sub...

where did the (1-r^2) part go

its part of the expression for a

it should be $\frac{20(1-r)(1-r^4)}{(1-r)(1-r^2)}$ no?

[flat frigate]

The Great D

[torpid pumice]

oh shit im so stupid

ok wait ill try the whole thing from the beginning

[dim meteor]

ok

[torpid pumice]

20(1-r)/(1-r^2) = a , 200=a(1-r^4)/1-r

200=2(1-r)(1-r^4)/(1-r^2)(1-r)

200=2(1-r^2)

200-200r^2=2

198=200r^2

r^2=99/100

<@&286206848099549185>

[safe radish]

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[torpid pumice]

.reopen

[safe radish]

✅

[torpid pumice]

20(1-r)/(1-r^2) = a , 200=a(1-r^4)/1-r
200=2(1-r)(1-r^4)/(1-r^2)(1-r)
200=2(1-r^2)
200-200r^2=2
198=200r^2
r^2=99/100
<@&286206848099549185>

[sage idol]

these are two diffrent eq right'?

[torpid pumice]

ya

[flat frigate]

CosMo

[vapid relic]

I believe this is right

[torpid pumice]

i

do not understand wtf

[vapid relic]

This is your question right? (s2=20)

[red delta]

send the original question

[torpid pumice]

yup

[torpid pumice]

geometric series where s2=20, s4 = 200

find r and a

[vapid relic]

So then where are you stuck right now?

[red delta]

where the sum of the first 2 terms =20
and
the sum of the first 4 terms =200?

[torpid pumice]

yup

[red delta]

$S_2 = a(r^2 - 1)/r-1 $
$S_4 = a(r^4 - 1)/r-1 $
when |r|>1
$(S_4/S_2) = (r^4 - 1)/(r^2 - 1)$

[flat frigate]

ishigamisenku_

[red delta]

@torpid pumice

[torpid pumice]

omfg

im

so fucking sorry

i didnt even write the q properly

im so sorry

wait no its right

imlosing my mind

[torpid pumice]

YES

[red delta]

sub value of $S_4 and S_2$

[flat frigate]

ishigamisenku_

[torpid pumice]

ok so 200/20 = (r^4-1)/(r^2-1)

10 = (r^4-1)/(r^2-1)

[red delta]

don't forget its division

[torpid pumice]

ohr yeah

[red delta]

$10 = (r^4-1)/(r^2-1)$
$10 r^2 - 10 = r^4 - 1$

[flat frigate]

ishigamisenku_

[red delta]

collect like terms

[torpid pumice]

okay makes sense

10r^2 - 9 = r^4

[red delta]

then $r^4 - 10r^2 + 9 = 0$

[flat frigate]

ishigamisenku_

[torpid pumice]

okayy

[red delta]

factorize

continue from here

[vapid cypress]

or just assume r^2 = x

and use quadratic formula

[torpid pumice]

OH

[vapid relic]

Isn't r^4-1=(r^2+1)(r^2-1)?

[torpid pumice]

i kept trying to figure out where a went n all you just ignored it

[red delta]

[red delta]

yes

[vapid cypress]

bro it makes it slightly easier

[red delta]

when you divide a at top cancels out a at bottom

[red delta]

ye ik

[vapid relic]

Then isn't it just r^2+1=10 (r!=1)?

[calm brook]

then $r^4 - 10r^2 + 9 = 0$ \
let\
$r ^2 = x\$

$x^2 - 10x + 9 =0$\

$=> \frac{10 \pm \sqrt{100 - 4 (9)}}{2}$\
$= \frac{10 \pm 8}{2}$\

$r^2 = \frac{18}{2} = 9 $\
$r = \pm 3$\
or\
$r^2 = \frac{2}{2} = 1 $\
$r = \pm 1$\

r is either 3, -3, 1, -1

[red delta]

expand the top dif off 2 squares

[calm brook]

[red delta]

r^2 + 1 = 10

|r| = 3

r = 3

[red delta]

yes r = 3

[calm brook]

bru

[red delta]

yes

[calm brook]

±

it doesn't know how to write ±

[vapid relic]

Can you tell me if this is wrong

[flat frigate]

realtime

[red delta]

its unclear

[red delta]

\pm

[calm brook]

I got r^2 = 9 and 1

[red delta]

if its 1 then (r^4 - 1)/(r^2 - 1) = 0/0 or -1/0

so its r^2 = 9

[calm brook]

ah

[red delta]

r^2 = 9
(r^4 - 1)(r^2 - 1) = 10

[calm brook]

wait how is it 0/0?

if r=1, -1 in

(r^4 - 1)(r^2 - 1) = 0

(1 - 1) (1 -1) = 0
0*0 = 0

[red delta]

oop i forgot "/"

[calm brook]

I dont get you

[red delta]

i omitted the divide sign

this is what i mean

if its 1 then (r^4 - 1)/(r^2 - 1) = 0/0 or -1/0
so its r^2 = 9
r^2 = 9
(r^4 - 1)(r^2 - 1) = 10

[calm brook]

ahh

im sry

yeah

s oit only has 2 solutions

[red delta]

yes

[calm brook]

@red delta

what about this equtiona

$r^4 - 10r^2 + 9 = 0$ \

[flat frigate]

realtime

[calm brook]

for this -1 and +1 do happen to reach 0

1 - 10 + 9 = 0

[red delta]

hmm

[calm brook]

also 3^4 - 10 (3)^2 + 9 = 0

so -3 and +3 work

[safe radish]

@torpid pumice Has your question been resolved?

[safe radish]

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[tardy mango]

@stable sedge #help-4 message

[tardy mango]

Why are you multiplying the x coordinates by 2?

Just leave it as such

and then substitute into the equation of the line

and that gives you the coordinates of P and Q

[stable sedge]

oh wait frick

so $y = x+ k = \frac{(-k \pm \sqrt{-k^2+8})}{2} + k = \frac{(k \pm \sqrt{-k^2+8})}{2} $ if i can add right

$y = x+ k = \frac{(-k \pm \sqrt{-k^2+8})}{2} + k = \frac{(k \pm \sqrt{-k^2+8})}{2}$

[flat frigate]

ISAVAGE

[stable sedge]

ok

then would i add the two and divide by two?

please say yes

ok gud

am i trippin $\sqrt{-k^2+8}} $???

oh wait

M(-k, k)???

but i then that would be wrong 😭

hjhmmm

i redid got (-2k, 2k) but also wrong oof

[stable sedge]

@tardy mango wait how is this right?

isnt it $(x1+x2)/2, (y1+y2)/2$?

[flat frigate]

ISAVAGE

[stable sedge]

isnt the 1/2 over the fraction?

wut

oh i forgot k

uh

im trippin lemme look again

[tardy mango]

nvm it's $\left(\frac{\sqrt{8-k^2}}{2}, -\frac{1}{2} \sqrt{8-k^2} \right)$$

[stable sedge]

wut u get -k/2??

[flat frigate]

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

[tardy mango]

I'm multitasking so I'm doing this in my head lol

[stable sedge]

i was too lazy but ??

[tardy mango]

your intersections are wrong

[stable sedge]

i thought we are finding the midpoint?

[tardy mango]

...

[stable sedge]

cant u simply do that???

[tardy mango]

read what you wrote carefully

and compare.

[stable sedge]

the intersections divided by 2?

[stable sedge]

are they not the same?

ok imma just rewrite everytghing i have

$x = \frac{-k\pm\sqrt{-k^2+8}}{2}, y = \frac{k\pm\sqrt{-k^2+8}}{2} \
\therefore, \
P(\frac{-k + \sqrt{-k^2+8}}{2}, \frac{k+\sqrt{-k^2+8}}{2}), \ Q(\frac{-k - \sqrt{-k^2+8}}{2}, \frac{k - \sqrt{-k^2+8}}{2}) \$
The midpoint $M$ of $PQ$ is
$M(\frac{x1 + x2}{2}, \frac{y1 + y2}{2})$

[flat frigate]

ISAVAGE

[stable sedge]

is that right?

like on the first line?

arent they just solutions to the line when substituted?

ok for the P and Q is it the same?

tho

cuz yours rewords the numbers

if thats the case imma divide by 2 again to find the midpoint hopefuly it works

[tardy mango]

yeah ignore everything I said earlier

I read the output incorrectly

it's (-k/2, k/2)

[stable sedge]

o

oh

wait then i could have just did -b/2a

[stable sedge]

hey i just wanted to say thanks for the help

tho

🙏

sorry for being really dumb

[tardy mango]

massive faults on both sides lol

let's just move on from this

[stable sedge]

yea aha

[tardy mango]

anything else?

[stable sedge]

well tbh i dont really understand the follow up question

but its ok

[tardy mango]

what's the follow up question?

[stable sedge]

when the value of k changes in the range found in Question (1) i.e. $-2\sqrt{2}<k<k2\sqrt{2}$, find the locus of Midpoint M like bro i thought we just did that

[flat frigate]

ISAVAGE

[stable sedge]

like did we not just do that

[tardy mango]

wait send the entire question

cause I kinda forgot it tbh

[stable sedge]

ok

1. Given that circle $x^2 + y^2 = 4$ and line $y = x + k$ intersect at two different points $P$ and $Q$, solve the following questions. \
   (1) Find the range of the values of constant $k$. alr did, i.e., $-2\sqrt{2}<k<2\sqrt{2}$
   (2) Express the midpoint $M(x,y)$ of line segment $PQ$ in terms of $k$. we just did this one $M(-k/2,k/2)$
   (3) When the value of $k$ changes in the range found in question (1), find the locus of midpoint $M$.

OMG

[tardy mango]

it's fine

ik what it says

[stable sedge]

ok

[tardy mango]

I think it means to find a geometric figure that all M lie on

[flat frigate]

ISAVAGE

[stable sedge]

ok nvm

[tardy mango]

ex. a circle centered at the origin w/ radius 1

obv this isn't the answer to this question

[stable sedge]

like i know locus is like a line, circle, parabola etc

[tardy mango]

yeah

so which of those figures will all possible locations of M lie on?

[stable sedge]

figures meaning the locus?

[tardy mango]

yeah

[stable sedge]

well

maybe when the line is perpendicular?

cuz when i move the line down M is like diagonal

[tardy mango]

Hold on

are you japanese

[stable sedge]

nah lol

[tardy mango]

oh nvm

💀

[stable sedge]

ok

[tardy mango]

I was about to switch languages cause the (1), (2), (3) is smthn I've only seen in Japan to represent subparts

[tardy mango]

what did you mean here then

[stable sedge]

like when u change the value of k

[tardy mango]

perpedicular to what?

[stable sedge]

to y = x + k

or maybe i could use points that pass through M

?

[tardy mango]

the line is in fact perpendicular

but idt you rlly need that to figure out the locus

there's a much simpler way

[tardy mango]

yeah this a good strat

[stable sedge]

ok i would use k = 2, -2

lemme see

i wait im using desmos

so it would obv be (1,-1), (-1, 1)

then i would do
$y - 1 = \frac{-1 - 1 (x - (-1))}{1 - (-1)}$

[flat frigate]

ISAVAGE

[stable sedge]

uh

weird format

bruh my pencil is gone

nvm

ayo its $y = -x$

[flat frigate]

ISAVAGE

[tardy mango]

yeah

[stable sedge]

dam

[tardy mango]

an easy way to recognise this is that the y coord is the negative of the x coord

tho to be safe, you might want to restrict the domain

cause there are certainly k where there's no intersections b/w the two graphs

[stable sedge]

like u mean $-2\sqrt{2} < k < 2\sqrt{2}$

[flat frigate]

ISAVAGE

[stable sedge]

wait

thats in k

[tardy mango]

good catch

[stable sedge]

uh

[tardy mango]

Recall the midpoint is (-k/2, k/2)

[stable sedge]

hm

[tardy mango]

The x value is -k/2, so the range is ||-sqrt 2 < x < sqrt 2||

[stable sedge]

sorry i dced

wait the midpoint lies within the range

[stable sedge]

ok so i figured tjat u divided by 2

cuz then it would make the x value -k/2

alr thank you

for the help

🙏

math is a struggle fr

[safe radish]

@tardy mango Has your question been resolved?

[safe radish]

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[cinder oxide]

Use identites to simplify the following expressions
-tan(-x)cot(-x)

[cinder oxide]

Any ideas on how to simplify this expression?
I had tried but I honestly don't know if it's 1 or -1

[rustic goblet]

show what you've done

[cinder oxide]

This is what I have done.

But I think I maybe have made a mistake

Not sure about the identities

[rustic goblet]

you have -(-tan)(-cot) = tan*cot

when it should be -tan*cot

[cinder oxide]

Well isn't the identity is that.
if theta is negative. then tan is positive and vice versa?

same thing with cot

If cot(-x) OR -cot(x)

yk?

[rustic goblet]

what do you mean?

tan is an odd function

so tan(-x) = -tanx

same with cotx

[cinder oxide]

That's what I kinda meant.

[rustic goblet]

so your boxed identities are true

[cinder oxide]

Okay. then I don't see the mistake though

Like is it 1 or -1?

[rustic goblet]

see here

[cinder oxide]

Hold on

I honestly have zero clue on this matter

The identites are a bit confusing.

I do know that both are odd functions

But your telling me that it should be -tan*cot

[autumn drum]

anyone please help me with this question-Consider that a, b, c, d are positive real numbers satisfying (a + c)(b + d) = ac + bd.
Find the smallest possible value of S=a/b+b/c+c/d+d/a

[rustic goblet]

in your work, you have three negatives

[cinder oxide]

Go to the "Math help (available) section to open your own ticket my friend

[cinder oxide]

Here?

[rustic goblet]

yes

but your next line

[rustic goblet]

is written as if there were only 2 negatives

yes

[cinder oxide]

I didn't write any negatives

Oh..

Right right

so -tan(x) cot(x)

correct?

[rustic goblet]

yes

[cinder oxide]

Okay..
so this would mean that the answer is -1?

[rustic goblet]

yes

[cinder oxide]

Okay so the work is correct. just the final answer

[rustic goblet]

yeah you made a simple sign error

[cinder oxide]

And a negative

Yeah

Alright thank you

I shall open another ticket incase I have another inquiries.

@rustic goblet 🫶

[safe radish]

@cinder oxide Has your question been resolved?

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[flat kiln]

In order to compute the area of a particular circle, Juan first measures the length of its diameter. The actual diameter is 20 cm, but Juan's measurement has an error of up to $20%$. What is the largest possible percent error, in percent, in Juan's computed area of the circle?

[flat frigate]

!ｓａｔｏｒｏ !

[neon summit]

20 percent of 20 is 4

i think i misread the question

[flat kiln]

uhh

yeah I think so

so 20% is either 16 or 24 right?

[neon summit]

yes

[flat kiln]

so do we just take one of them into account?

[buoyant shadow]

yeah

[neon summit]

so the actual area is pi10^2

[flat frigate]

Result:

1.44

[flat kiln]

so the correct one should be 100pi, the other one should be 144pi right?

ohh I see

1.44

so 44% ?

is that how you do it?

oh I see, got the question correct

thanks guys

.close

[safe radish]

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[spare forum]

Anyone mind quickly double checking this?

For a number X to be in A the following rules must be met: If the prime number p divides n, then 2p+1 also divides n.
Find the smallest member of A

[spare forum]

I got 10 but im scared its somehow wrong it feels to easy

[left karma]

10 definitely seems to work since 2 is prime and divides 10, and 2*2+1=5 also divides 10. The initial statement says "number X" and then uses n instead which is weird to me, but what you got seems valid. And numbers less than 10 don't seem to work

[spare forum]

sorry x and n is a mistake on my part im used to n for intergers

[left karma]

Oh wait, the trick might be that 5 is also prime, but 11 doesn't divide 10

[spare forum]

hmm so for all primes that divide it then 2p+1 must divide it. I interpreted it as if theres only one p because of the word “the”

[tiny wraith]

Hmm, I've found that $$2\cdot5\cdot11\cdot23\cdot47\cdot19\cdot\left(3\cdot13\right)\cdot\left(7\cdot9\right)$$ seems to satisfy the constraints, but it might not be the smallest

[flat frigate]

Jelle

[tiny wraith]

Maybe you can argue that 2 has to be a prime divisor and work from there actually

Yeah, I think that works

[spare forum]

2 actually doesnt have to be i think since if you start with a prime the next number 2p+1 will always be odd. And to find the LCD you would have multiply two odd numbers together which is always odd

[tiny wraith]

Lol, I messed up parity my bad

[spare forum]

dw i interpreted the problem completely wrongly so just a partity mistake is fine lmao

can we say that n is always equal to the product of the primes that divide it and the 2p+1’s

[tiny wraith]

Yeah, but some of the p's might be 2p + 1's as well

7 for example

(if 2 is in the factorization at least)

so you get these chains

[spare forum]

yeah excluding the overlap then

[tiny wraith]

I'm kinda afraid that it's impossible to prove that a certain prime is the factorization of n though

[spare forum]

Hmm there seems to be something with sophie germain primes maybe

Theyre primes p such that 2p+1 is also prime

[tiny wraith]

What kind of class is this probem from btw?

[spare forum]

highschool challenge problem

So probably 11th 12th ish

[tiny wraith]

okay, I should know enough theory then

[spare forum]

I have side tutoring for some early uni stuff but that probably wont be needed

yeah

[tiny wraith]

(I'm in last year of high school, but I have done olympiad stuff, so I'm familiar with nt)

[spare forum]

Nice

You think this is somehow useful

[tiny wraith]

Is this fermat's last theorem?

[spare forum]

its to do with it but its a stand alone theorem

[tiny wraith]

I've tried a couple things and it seems like starting with any prime below 30 will give a sequence like the one I sent, but without 2

[spare forum]

just conjecturing maybe the solution could be the first non sophie germain prime since that would only need to divide itself?