help-33

[proper sleet]

sum of arithmetic and geometric series both have formulas where few terms need calculated

[severe island]

okay but what series is that lmao

if there's a pattern then i can do something yes

i don't see any pattern tbh

that's not arithmetic or geometric

[proper sleet]

one that is given on the test

the hypothetical big set on the test

for the symmetrical set on the flashcard there is no need to see a pattern because the mean and median can be calculated and equated

[severe island]

yeah i get the one in the flashcard part is trivial

my overarching question was how would i get the mean

if they defined a sequence that would generate like 100s of those

others would find the median then say that's the mean because they realize it's symmetric (maybe?)

but i end up doing sum crazy calculus lol

anyway

i think i spent too much time overthinking one trivial detail lol

thanks for the help

[marsh citrus]

@severe island Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[feral forum]

I need help urgently

[marsh citrus]

Please don't occupy multiple help channels.

[summer trench]

can you just stop posting ur question miltiple times in multiple channels and abusing helpers ping

[feral forum]

I need help 😭

I’m also new to the server

So I didn’t know my bad

[summer trench]

pinging them multiple times wont help instead it can end up in situations like u being muted or something

check !help

!help

[marsh citrus]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[feral forum]

!help

[marsh citrus]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[proper zodiac]

I'm going to close your old channel, please just wait patiently in this channel for someone to help you.

[feral forum]

Ok

Thanks

We’re u guys bots

[summer trench]

no

[feral forum]

K

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[proper zodiac]

!15m

[marsh citrus]

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.

[proper zodiac]

please dont abuse the ping

[celest wyvern]

bro dont know what patience is

!status

[marsh citrus]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[feral forum]

2

[feral forum]

Idk there’s so many steps on a discord servers

[celest wyvern]

okay if u have some written work send a pic and i could hint/guide u

[feral forum]

I just need to know the working out and I’m good to go

[celest wyvern]

ofc u will be lol

for congurency we will have to show at least a pair of sides to be equal

[feral forum]

K

[celest wyvern]

how tho

[feral forum]

Wdym

Ik about the congruence letters

And all that stuff

[celest wyvern]

oh okay

[feral forum]

I just want to know how fhm and dfp are congruent

[celest wyvern]

well thats what u will be using

[feral forum]

If it helps I’ll send u an image of the triangles

[celest wyvern]

so which test would u use by showing just one pair of sides to be equal tho

[feral forum]

Idk

[celest wyvern]

ok i want u to find out which pair of sides are equal and why that is first of all

[feral forum]

K

HN and MF

If u mean the whole octagon

It can be

HG and CD

EF and BA

AH and FG

And so on

[celest wyvern]

broooo

okay first

[celest wyvern]

this

do u realize why this is true

[feral forum]

?

[celest wyvern]

just showing 3 angles as equal isnt enough since they could just be scaled versions of each other

one pair of sides MUST be equal for congurency

of two triangles

[feral forum]

K

[celest wyvern]

as such we are going to find that one side we can show equal

in these two triangles

which we are supposed to show congurency in

now try to do so

[feral forum]

Umm

HF and FD

[celest wyvern]

yep

and why

[feral forum]

Gimme a sec

I get why they make sense to be equal

But I don’t know how to mathematically describe it

I found out just using the shape of the regular octagon

[celest wyvern]

these two triangles

can be shown to be congurent

with side angle side

[feral forum]

Wait a minute

[celest wyvern]

since the octagon is regular

[feral forum]

So is fgh the same as fhm?

[celest wyvern]

no

are the triangles i marked not clear

😓

[feral forum]

Is the triangle I shaded fhm

[celest wyvern]

if u have an argument for this we can hear it out ig

[feral forum]

Is the shaded fhm

[celest wyvern]

im not understanding u sorry

[feral forum]

It’s alright

Is the answer SAs

[celest wyvern]

yea SAS can be used to show THESE triangles are equal not fhm

[feral forum]

Ok

[celest wyvern]

from here for the required triangles fhm and dfp

a pair of sides can clearly seen to be equal

[feral forum]

So can you tell me what the full working out would be for this question to get full mark

[feral forum]

I see it now

[celest wyvern]

yea

now we can look for angles to show equal

[feral forum]

So just to clarify

Are both those triangles the same

As in the image

[celest wyvern]

nope

again, if u have an argument for why they would be equal share it

since i myself dont know everything

[feral forum]

Umm

So

I get why it’s not equal now

[celest wyvern]

lets go angle hunting now

hope u with me

[feral forum]

K

[celest wyvern]

do u know what angle dfh is

[feral forum]

Isn’t the entire triangle 180 degrees

[celest wyvern]

no no

the angle

angle dfh

[feral forum]

Oh so the angle of F

I just remembered the angle is usually referring to the middle one my bad

[celest wyvern]

this angle

[feral forum]

90

[celest wyvern]

woah

u did thhis part already

how is it so

[feral forum]

Idk

[celest wyvern]

bro

u dont get marks for guessing

[feral forum]

I see it’s an isocoloses triangle

DF and Hf are equal right?

[celest wyvern]

yea

[feral forum]

So the other 2 sides are 45 degrees

[celest wyvern]

y

[feral forum]

It all adds up to 180

And it’s isocolose

And F is 90

[celest wyvern]

a very rough diagram

[feral forum]

K

[celest wyvern]

there are 3 right angles

i removed the octagon

[feral forum]

K

[celest wyvern]

why is f 90 tho

also dont say f its confusing, its dfh

[feral forum]

K

Idk

[celest wyvern]

do u know the angle of fgh

[feral forum]

No

[celest wyvern]

ok ok

the total internal angle of an octagon

[feral forum]

Can you tell me the answer and then we can go backwards from it

I think it will help me understand

It’s also 8th grade btw

Australia

[celest wyvern]

theres no "answer"

the "answer" is given

those two are equal

[feral forum]

Don’t we have to prove why they are equal

It’s what it says

[celest wyvern]

well look first u need to understand this

total internal angle of a polygon

180 for triangle

what for a quadrilateral

[feral forum]

I think I’ll try figure out on my own and i gotta go

Thanks for all the help

[celest wyvern]

k

[feral forum]

I appreciate it a lot

[celest wyvern]

np gl

[feral forum]

Thanks

[celest wyvern]

umm also

u know how to close right

.close

[feral forum]

.close

[marsh citrus]

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[marsh citrus]

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[cedar briar]

.open

[broken hound]

it's true ?

[marsh citrus]

@broken hound Has your question been resolved?

[marsh citrus]

@broken hound Has your question been resolved?

[broken hound]

how 3

?

xD

[quaint elm]

was a troll

compare the factors of each side

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

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[sterile shoal]

Hello

[sterile shoal]

so

[hidden plaza]

[sterile shoal]

We have to set up a equasion

so from what i see f(0)=30 f(0)=30 f(10)=100 f(10)=100

for both

[hidden plaza]

well one of them is a straight line

Should be easy to find the equation

[sterile shoal]

yes so liniar and exponential

[hidden plaza]

You can assume the equations are in the form $$f_1(x) = mx + b, f_2(x) = ka^x$$

[elfin berry]

$Pure$

[hidden plaza]

m,b,k,a to be determined

[sterile shoal]

so for f2 k would be the starting value no?

[hidden plaza]

Mhm

Which is?

[sterile shoal]

okey so i did it for the f1 i think f(10)=10m+30

f2 would be f(10)=30* a^10 maybe?

[hidden plaza]

Yeah

That’s just true but what is f(10)

[sterile shoal]

100

100=10m+30 and then m=7

and a is 1,27

thanks .close

.close

[marsh citrus]

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[marsh citrus]

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[undone summit]

I'll tell you steps to solve

first get area of whole circle

second area of sector by formula

[still temple]

4436.5 mam

Mum

Mm

[undone summit]

what's this

[still temple]

Area of circle

[undone summit]

ok

good

now area if sector

*of

[still temple]

Is it 844

[undone summit]

now subtract the area

to get

required

area

[still temple]

3593.5

I got it wrong

Now the question has changed

I got it

.close

[marsh citrus]

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[humble river]

wait nvm

.close

[marsh citrus]

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[waxen vortex]

A packaging company that manufactures Kraft cardboard boxes wants to reduce associated costs
to raw materials in the manufacture of a line of products with low circulation, but with orders
large quantities (more than 1000 units per order); one of these products is composed
of self-assembling die-cut boxes, with the dimensions presented in the figure at the end of the
document and whose volume is $1008𝑐𝑚^3$

Since it is a die-cut box, there is waste between the cardstock sheet and the template.
the box at the time of manufacturing, so a new template is desired that meets the
following conditions:

1. The constructed boxes must have a lid and must have a volume of $1000 ± 10𝑐𝑚^3$.
2. The template should reduce the amount of material used to manufacture the box
3. The template should reduce the amount of waste generated in the die-cutting process

(reference image)

[elfin berry]

ElPanaArturo

[waxen vortex]

I really don't know how to start, I would like to know or at least have a guide on how to start with this exercise :c

[marsh citrus]

@waxen vortex Has your question been resolved?

[marsh citrus]

@waxen vortex Has your question been resolved?

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[harsh rain]

any help i have no idea what to do because sin has x^2

[karmic inlet]

I dont think this is enough info to help out, im not in calculus but what is goal here? Get derivitive?

[harsh rain]

find the limit?

[karmic inlet]

oh so what Y tends to or something

2pi = period here

[harsh rain]

dont mean to be rude but i am trying to get some help here so i am not really here to explain to you what this is

this is a help channel

[stark trail]

Striker, do you know Lhopitals rule?

[harsh rain]

i know it but cant use it here

[karmic inlet]

no problem

[stark trail]

Also, is this sin^2(x) or sin(x^2)

[stark trail]

Why not?

[harsh rain]

sin(x^2)

it is weird to explain

havent been taught that at school but i have learned it at the extra classes i am taking in order to get good results for my exams

this exams determine to what uni i go

just no l'hopital rules

?

[marsh citrus]

@harsh rain Has your question been resolved?

[harsh rain]

thank you for the help guys appreciate it

[marsh citrus]

@harsh rain Has your question been resolved?

[marsh citrus]

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[polar ledge]

this tends to what

[marsh citrus]

Please don't occupy multiple help channels.

[polar ledge]

can i do like this

[main idol]

closed your other channel

[polar ledge]

--

[polar ledge]

ok

[main idol]

no

you're overcomplicating it

[proper zodiac]

you are thinking of $\lim_{x\to 0}\frac{\sin(x)}{x} = 1$

[elfin berry]

ΣΑC

[proper zodiac]

but your x is not going to 0

[polar ledge]

ohh

so its just sin2/4 or not

[proper zodiac]

yes

[polar ledge]

oh thank you

[marsh citrus]

@polar ledge Has your question been resolved?

[marsh citrus]

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[iron meadow]

my suggestion is to multiply the right side by 3/3 to have same numerator, then consider at what values denominators are changing signs and think about the couple of different intervals independently a little, it should be clear what happens

@viscid radish

[marsh citrus]

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[tough abyss]

here we're solving an equation using fourier transform and some dirac delta integral proprities

i would like to know why we added that identity?

i've never heard of a step function

then they devided them i don't know why

[main idol]

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

[main idol]

which part of the image exactly are you talking about

[tough abyss]

this one i don't know why they insert it

[main idol]

they use it later probably

yea to simplify the delta function

follow the delta

[tough abyss]

we could have done that without that identity

or we did that for the K0 >0 condition ?

[main idol]

did you try this

if you can simplify their calculations, you should show your professor

[tough abyss]

alright

thank you sm

.close

[marsh citrus]

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[simple quiver]

Need help with 8, 9, 10

For 8 it seems like I did it right but for 9 and 10 I’m not sure

[simple quiver]

@ helpers

<@&286206848099549185>

[marsh citrus]

@simple quiver Has your question been resolved?

[simple quiver]

<@&286206848099549185>

[marsh citrus]

@simple quiver Has your question been resolved?

[simple quiver]

<@&286206848099549185>

[simple quiver]

.close

[marsh citrus]

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[neat pewter]

i dont understnad what the red boxes mean ik that converge means its tending towards something for eg the first box tells us the function f(t) = e^at is converging to something at s-a > 0 then not sure what Re stuf mean

[glass silo]

If you have s and a being complex, then the real part of s is greater than the real part of a (for the first red box)

[neat pewter]

where does it say stuff about if s and a are complex?

[glass silo]

Assumedly they wanted to imply it

[neat pewter]

Oh okay and how did they find out that the real part of s will be greater than the real part of a in the first condition

[glass silo]

If you break s and a into the real and imaginary parts, then it's similar to how you'd show convergence in the purely real case

in particular, the imaginary bits basically trace out the unit circle so those are bounded, and you're left only considering what happens to the real part

damn it autodelete

[elfin berry]

@glass silo

[marsh citrus]

@neat pewter Has your question been resolved?

[marsh citrus]

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[inland bramble]

Hi

[inland bramble]

How to get factors of this. Sir directly wrote it as (t+1/2)² in the next step

Anyone free to help me?🥲

[hasty ruin]

complete the square

[inland bramble]

I am dumb

[inland bramble]

Like 1/4 you mean?

I doubt we should do that. cause in next step it's 1/2 itself

There's an identity or something. I just couldn't recall it

@hasty ruin bro you there? =3

[inland bramble]

<@&286206848099549185> is somebody free to help guys =)

[ocean ruin]

Hey

Real quick

So yk the form of a binomial squared

(a+b)²

@inland bramble you here?

[ocean ruin]

=a²+2ab+b²

So you can notice that we have t²+2(½)(t)+(½)²

Which makes it (t+½)²

@inland bramble got it?

If you have any more questions please do @ mention me

[inland bramble]

Yup know that

[ocean ruin]

So you got how I did this?

[inland bramble]

Oh ok. That's what chatgpt also said but I wasn't convinced. But now I am. Thanks mate

[ocean ruin]

No problem 🙂

[inland bramble]

Yeah bro thanks

So like now I am suppode to do .close? I am new here

I guess yeah lemme try

.close

[marsh citrus]

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[inland bramble]

Yup it worked

[ocean ruin]

By sending the message "yup it worked" you reopened it 😂

. close

[marsh citrus]

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[sick sedge]

im not sure what else it can be besides DNE

[wise jackal]

have you considered the domain being x >= -1?

[sick sedge]

hmm

[sick sedge]

ok so plugging in -1 does the trick

but can you explain why it works

[wise jackal]

have you drawn it?

may help to see it

[sick sedge]

isnt it just a linear line

[wise jackal]

yeah, but the restriction is what's key here

it begins at its maximum at x = -1

[sacred kestrel]

absolute extrema can be found at interval endpoints is what he's trying to get at.

domain of the function is [-1, inf)

thinking about it logically, we know its a constantly decreasing function, so we can make a conclusion from that where as x grows smaller, the function grows larger. and following that thought, we might say "f(x) will be largest at the smallest x in its domain."

[sick sedge]

i just woke up and my brain not work

wouldn it be x <= 5

why graph no make sense

oh wait

nvm im dumb

im talking about y

[sacred kestrel]

x >= -1 is given as the domain of f(x), where f(x) = y.

[sick sedge]

ok yeah i get it now

my lowesr

one of my lowest moments

.close

[marsh citrus]

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[twin fulcrum]

I need help

[twin fulcrum]

I can’t seem to know where to put the 5th point

<@&286206848099549185>

[marsh citrus]

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[flint jolt]

231

[sacred kestrel]

!status

[marsh citrus]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[flint jolt]

1

[marsh citrus]

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[weary kettle]

how to solve sec x = 2 csc x

[weary kettle]

(with calculator)

[abstract knot]

express them in terms of sin and cos, manipulate till you get a single trig function, take inverse

[weary kettle]

yeah

I got to tan(theta) = 1/2

and took inverse

[abstract knot]

where'd theta come from

[weary kettle]

tan(x) sorry

[abstract knot]

also are you sure about this

[weary kettle]

no lol

[abstract knot]

whats sec

in terms of sin and cos

[weary kettle]

1/cos

[abstract knot]

right

whats csc

[weary kettle]

1/sin

[abstract knot]

right

[weary kettle]

so 2/sin in this case

[abstract knot]

so we have 1/cos=2/sin

[weary kettle]

yeah

[abstract knot]

so what is tan = to

[weary kettle]

I got 2tan = 1

[abstract knot]

hmmm

is tan sin/cos or cos/sin

[weary kettle]

sin/cos

[abstract knot]

right

[abstract knot]

so if we multiply both sides of this by sin, we get?

[weary kettle]

sin/cos = 2

ohh

[abstract knot]

yeah

[weary kettle]

ok I see

thank you

[abstract knot]

np

[marsh citrus]

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[tepid breach]

When proving that A ' (x) = f(x), do we say that A(a) = 0 is a 'boundary condition' or just a 'condition' ?

[marsh citrus]

@tepid breach Has your question been resolved?

[tepid breach]

<@&286206848099549185>

[marsh citrus]

@tepid breach Has your question been resolved?

[marsh citrus]

@tepid breach Has your question been resolved?

[blazing robin]

What

[tepid breach]

A(a) = 0 is a boundary condition and A(b) = the whole area, boundary condition #2

[blazing robin]

Your question lacks context

[blazing robin]

Ah

So A is a function such that

A(x) gives area under graph of f between a and x

The conditions for this are
A'(x)=f(x)
A(a)=0

I would just call it a condition
I doubt there is an exact mathematical definition of what counts as a boundary condition

[tepid breach]

But isn't it implied by the so-called condition set forth that A(a) = 0 etc?

[blazing robin]

I have no idea what you mean

[tepid breach]

Nvm. our teacher just said that we need to call it a boundary condition for whatever reason.

But why do we actually draw the A(x): "A(x) gives area under graph of f between a and x"

When we want to comment on the orange area?

[blazing robin]

What

I gtg

Ask someone else

[tepid breach]

Fair enough

.close

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[marsh citrus]

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[fleet sable]

can someone explain how the answer at the bottom is recieved

[fleet sable]

.close

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[junior sonnet]

hey

[junior sonnet]

u0 = 2; v0 = 1; y0 = [u0;v0]; % Initial conditions
% Setup the ODE solver, these are default tolerances:
options=odeset('RelTol',1e-3,'AbsTol',1e-6);
tspan=[0 75];                 % Time span
solver=@ode45;                % Chose the appropriate ODE solver
% Solve the equations:
[t yout]=solver(@fun1,tspan,y0,options,parameters);
u=yout(:,1);  v=yout(:,2);    % more intuitive variables
% Plotting results: time series
figure(1); plot(t,[u v],'o-'); %axis([0 20 0 1.25*max([u;v])])
legend('Prey','Preditor');
title('Time Series'); xlabel('Time (non-dimensional)'); 
ylabel('Population (non-dimensional)');

% Phase plane:
f=figure(2); f.Position=[200 200 600 300];
subplot(1,2,1); plot(u,v,'b-','LineWidth',2); 
title('Phase Plane');xlabel('Prey (non-dimensional)');
ylabel('Preditor (non-dimensional)'); axis equal; 
%
hold on
ur=get(gca,'XLim');             % Get x and y ranges off previous plot
vr=get(gca,'YLim');
number=15;                      % Number of subdivisions for plot
[um vm]=meshgrid(linspace(ur(1),ur(2),number),...
                              linspace(vr(1),vr(2),number));
% du/dt and dv/dt (right hand sides) evaluated at (xm,ym):
dudt=um * (1 - 0.2*um - vm); 
dvdt=-vm*(1 - um);
quiver(um,vm,dudt,dvdt,'k')
hold off
%
subplot(1,2,2); plot(u,v,'b-','LineWidth',2); 
title('Phase Plane');xlabel('Prey (non-dimensional)');
ylabel('Preditor (non-dimensional)'); axis equal; 
%
hold on
ur=get(gca,'XLim');             % Get x and y ranges off previous plot
vr=get(gca,'YLim');
number=15;                      % Number of subdivisions for plot
[um vm]=meshgrid(linspace(ur(1),ur(2),number),...
                              linspace(vr(1),vr(2),number));
% du/dt and dv/dt (right hand sides) evaluated at (xm,ym):
dudt=um * (1 - 0.2*um - vm); 
dvdt=-vm*(1 - um);
scale1=1./sqrt(dudt.^2+dvdt.^2);
quiver(um,vm,scale1.*dudt,scale1.*dvdt,'k')
hold off

function [ydot]=fun1(t,y,parameters)  
  % Set up the output data structure (assures it is a column):
  ydot=zeros(size(y)); 
  % Differential equations:
  ydot(1)=y(1)*(1-0.2*y(1)-y(2));
  ydot(2)=-y(2)*(1-y(1)); 
end

anyone know why my vector field isn't following my phase plane for the whole thing?

!close

?close

/close

RESOLVED

/close

.close

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[upbeat harness]

Hey I'm trying to find what this distribution is, given a pdf

[upbeat harness]

the pdf is as follows:

And my first thought is a beta distribution, but I have trouble transforming it into that form

it's continuous btw

[marsh citrus]

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[marsh citrus]

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[upbeat harness]

.reopen

[marsh citrus]

✅

[main idol]

What's $I_{(0, 1)}(x)$

[elfin berry]

riemann

[upbeat harness]

It's an indicator function

so if x is in (0,1) then it returns 1 and if not 0

basically it bounds the function between 0 and 1

That's what made me think that it could be a beta distribution, cause not a lot of other distributions are bounded between 0 and 1 like that

[main idol]

did you plot it in desmos for different theta

[upbeat harness]

Yeah, I got this kind of curve

[main idol]

beta distribution has two parameters, so you can try to do some transformation of theta to two params or vice versa take two parameters from beta and make them relate to each other

in this formulation for beta distribution, alpha and beta are independent, but maybe make one equal to theta, and solve for the other

[upbeat harness]

Yeah I was thinking that alpha could be like 1 or something like that

Is the Beta function symmetric like that?

[main idol]

the formula for beta distribution is above, so try playing around with different parameters in desmos

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

[upbeat harness]

I get really stuck trying to do the algebra, I just have no idea what to do with $(2x-x^2)^{1-\theta}$

[elfin berry]

Karl

[upbeat harness]

I mean I can factor out an $x^{1-\theta}$

[elfin berry]

Karl

[upbeat harness]

But after that I can seem to cancel it out with anything

Maybe I need a change of variables?

Okay wait I don't even need to think about this

What the question is really asking is

And I can just use mgfs for that

[marsh citrus]

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[severe whale]

Guys I have a question, why is the sum of all integers till n-1 equal $^nC_2$

[elfin berry]

diaas_(yt)

[severe whale]

Is it a coincidence or is there an argument

[sweet pawn]

let me make a spreadsheet

[proud ice]

Do you know the formula for $\sum_{k=1}^n k$?

[elfin berry]

SWR

[severe whale]

k(k+1)/2

[proud ice]

n(n+1)/2

but yea

[severe whale]

Right yeah

Yeah but I'm talking about n-1 objects

[glass perch]

One way this is intuitive is because $^nC_1 = n$

[elfin berry]

Error_5506

[proud ice]

in n(n+1)/2, replace n with (n-1). You end up with n(n-1)/2

[severe whale]

Yes and that's $^nC_2$

[elfin berry]

diaas_(yt)

[sweet pawn]

it's a running tally

[severe whale]

I'm asking why that's true

[severe whale]

Can you explain. WHT this is

[hidden plaza]

the intutive way is thinking about n people in a room and everyone shakes hands with everyone else and count how many handshakes occured

[sweet pawn]

start with 1+2. then add 3. then add 4...

look at the spreadsheet

[severe whale]

Hmm k

[severe whale]

OK thanks man this makes sense

[sweet pawn]

the intutive way is thinking about n people in a room and everyone shakes hands with everyone else and count how many handshakes occured

[severe whale]

the intutive way is thinking about n people in a room and everyone shakes hands with everyone else and count how many handshakes occured

OK Ill think about this later thanks guys

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[tame talon]

I have to sketch the following set of numbers

I think I was able to find |z|=0,5

But I am stuck with the Re()

[night mica]

hint: $\frac{1}{z}=4\bar z$

[elfin berry]

everg

[night mica]

since $\bar zz=|z|^2=\frac{1}{4}$

[elfin berry]

everg

[marsh citrus]

@tame talon Has your question been resolved?

[tame talon]

ok i am not sure how this should help me:(

i am just really consued by this notation. idk what does the argument mean

[marsh citrus]

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[marsh citrus]

@tame talon Has your question been resolved?

[marsh citrus]

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[tulip folio]

This is just e^-1 right?

[marsh citrus]

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[polar oracle]

Hii

[polar oracle]

Can someone help me with this?

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.close

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[marsh citrus]

@still temple Has your question been resolved?

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[slate blaze]

im a little confused

[old dagger]

with

wot

[slate blaze]

like do I need to find the vertex?

[old dagger]

i guess you could

did you try any of them

do you understand what standard form tells you

ax^2+bx+c

[slate blaze]

oh wait it's standard form?

I thought it's factored form

[old dagger]

well the first one is in standard form

Second one too

[slate blaze]

how do you know

if it's standard or which

[old dagger]

Well it’s in the form of ax^2+bx+c

Vertex is a(x-h)^2+k

[slate blaze]

oh

[old dagger]

Factored is a(x-r)(x-s)

So just look at it ig

But you understand how to graph it right

[slate blaze]

i know how to do factored is it like the same way?

[old dagger]

Factored is not the same

Factored gives you the zeros right

Standard and vertex don’t

Do you know the graph of x^2

[slate blaze]

oh

[old dagger]

x: -2,-1,0,1,2
y: 4,1,0,1,4

Have you seen that

On a table of values or sum

for values of x and y

[slate blaze]

the teacher tells us to use tov for standard form

[old dagger]

what’s tov

oh table of values

So u understand how to use it yea?

How the c impacts the x value

Oops I mean y

Not x

[slate blaze]

yes

[old dagger]

Alr

[slate blaze]

.close

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[carmine terrace]

Could somebody please check this proof? I don't think it's well articulated even if it might be correct so feel free to point out how that could be better as well

[marsh citrus]

@carmine terrace Has your question been resolved?

[carmine terrace]

<@&286206848099549185>

[marsh citrus]

@carmine terrace Has your question been resolved?

[carmine terrace]

Ok I've been informed that I can't use an infinite partition. Now I'm wondering if there is any way to make this approach work?

[stark trail]

Hello Nash

[carmine terrace]

Yo what's up

[stark trail]

There is a way to make the approach of bounding the upper sums below arbitrary epsilon positive work

Likely it will have to do with rationals, so we are going to Let $\varepsilon>0$ and then we can always find a natural number $q$ such that $\frac{1}{2q}<\varepsilon$

[elfin berry]

Austin

[stark trail]

right?

unrelated to p/q in reduced form from the problem statement btw

[carmine terrace]

Are you saying that uou're trying to bound the output 1/q below some epsilon?

Or am I misunderstanding?

[stark trail]

We are trying to bound the upper sums above by any epsilon positive

likely it will be nice to work with a rational upper bound

so since we can always find a 1/2q < epsilon

if we bound our upper sums by this 1/2q

we are also done

this q, is unrelated to our functions output as of rn

we can use a different letter if it is confusing

[carmine terrace]

So you're saying $U_Pf = \frac{1}{2q} < \epsilon$ for any $q$?

[elfin berry]

Nash

[stark trail]

not for any q

No

that's not what I am saying

I am saying

Let eps>0

there exists (going to use a new letter instead of q)

z

such that

1/2z < epsilon

and now we want to form a partition, such that the upper sum is < 1/2z and hence < epsilon

[carmine terrace]

Ok I see what you're saying

Sorry that took me a while lol

[stark trail]

no worries

and the reason for the 2

is because

we can make like

two different upper sum partitions, that collectively sum to an entire upper sum

if you can bound each of those upper sums independently by 1/2z

then over all they are bounded by 2*1/2z = 1/z < epsilon

I introduced the 2 too early actually

Mind if we just restart @carmine terrace

?

[carmine terrace]

Yeah fs

[stark trail]

Let $\varepsilon>0$ there exists $z\in \mathbb{N}$ such that $\frac{1}{z}<\varepsilon$. Now our goal is to make two seperate upper sum partitions, each independently bounded by $\frac{1}{2z}$, that in combination sum to a complete upper sum of our interval, and hence are bounded by $\frac{1}{2z}+\frac{1}{2z}=\frac{1}{z}<\varepsilon$

[elfin berry]

Austin

[stark trail]

that's our overall goal. lmk if that makes sense atleast

[carmine terrace]

Yeah I understand

[stark trail]

Okay, so I'm not sure how much help you want, I can walk you through it, or just give a hint here now

what would you prefer

[carmine terrace]

I'll just take the hint

[stark trail]

okay

there are finitely many rationals in [0,1] of the form p/k with k<=2q

so those could form one of your seperated upper sum partitions

and then of course you'd have to account for the other rationals in the other upper sum partition

[carmine terrace]

Ok great

Thank you for your help

[stark trail]

[carmine terrace]

.close

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[wide gulch]

Why is this wrong

[wide gulch]

If Z=2x-y

then why is the polar conversion not 2rcos(theta)-rsin(theta)

please ping if you reply

.close

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[coral lagoon]

How do you derive concavity using f' being neg, pos, or zero

[floral linden]

if from the graph you see that f' is increasing (so f' goes from -ve to +ve) then f'' > 0

[cloud iron]

You use the second derivative to determine concavity

In a similar way to how the first derivative can tell you whether the curve is increasing or decreasing based on the zeros of f'

[coral lagoon]

yeah i was triyng to do it lie that but i kept messing up, i also tried drawing but that didnt help. 1 sec

so its going from negative to zero to negative

so it goes from concave up to concave down?

for a to b

[cloud iron]

At a certain point between a and b it does, but that doesn't necessarily have to do with whether f is positive or negative

[coral lagoon]

im using f' though not f

[cloud iron]

The points where f' changes sign are the critical points of f

Or the maximums and minimums

In order to determine the inflection points (where f changes between concave up or down) you need to find the second derivative f'' and identify its zeros

[floral linden]

but the points where f' is increasing give you the points where f'' > 0

[floral linden]

bruh they dont have the function how will they find f''

you just have to use the graph

[cloud iron]

My bad, reading is difficult

[coral lagoon]

ik how to find cc up and down using a function, im just confsued on how u get that from f and f' with a graph

you said that when f' is increasing, f'' > 0 but when f = 0 and f' = negative, its positive (i can see the answer) im confused on how it got positive when it seems lie itd be cc down

[floral linden]

look at the point A. is the slope of the curve becoming less steep?

so it's becoming more positive?

i.e. it is increasing

[coral lagoon]

yes

so for b

its becoming less and less positive, so going negative?

[floral linden]

yep, so f' is decreasing

so f'' < 0

[coral lagoon]

so c would be negative, e would be positive

what about d?

it looks like 0 but that isnt correct

[floral linden]

f' is negative to the left of D and +ve to the right of it

so what is happening to f'?

[coral lagoon]

i thin i udnerstand it

its becoming less and less negative, so becoming more and more positive, then it goes positive

so itd have to be positive?

[floral linden]

yep

[coral lagoon]

aight than u

i have one more really quick quesiton if u have time ot answer

.close

[marsh citrus]

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[marsh citrus]

Please don't occupy multiple help channels.

[long sage]

<@&286206848099549185>

[native frigate]

you ignored like two rules at once

you already have a channel, and you have to wait to ping helpers

[long sage]

my bad

[native frigate]

make the ages of the women xyz, and the priests age p. what eqns can you set up for the first two statements

[long sage]

xyz= 2450

and x+y+z= 2p

[native frigate]

sure, now one more for the last statement

oh wait i misread that part

[long sage]

oh

what is the las

statement though?

[native frigate]

Im pretty sure you have to make some more assumptions here

"you are the oldest person here"

[long sage]

I dont get that part

how can I use that info

to find their ages

[native frigate]

i think the bishop has to be like 26 or something

[long sage]

hmmmmm

k

[native frigate]

wait this is such a silly problem

ohh i get it

women 7, 14, 25, priest 23, bishop is somewhere between 26 and 35 (inclusive)

its just like a little logic problem - you have to note the prime factors of 2450 are 2, 5, 5, 7, 7

[long sage]

but how did you find priest age?

[native frigate]

add the womens age, then divide by 2

[long sage]

but I got 4 possible ages for the women

[native frigate]

what other possibility do you have that isnt mine?

[uneven pier]

Why can't they be 5,10 and 49 or 7, 7 and 50?

[long sage]

10, 35 and 49

14, 35 and 5

10, 35 and 7

[native frigate]

the "you are the oldest person in the room" implies that you need the maximum age of the women to be minimized

otherwise it would surely be impossible for him to know the womens age?

oh im stupid, this is not the intended solution path

he knows the priests age

ok my b lemme get the right answer

[uneven pier]

Now I'm interested in this logic puzzle.

[native frigate]

the fact that the priest still doesnt know after being told the sum is twice his age limits it a LOT

which means two sets of womens ages add to the same thing

[long sage]

oh

[native frigate]

seems impossible? but im not thinking well rn

[uneven pier]

I think there is no one true solution with so little information.

[long sage]

???

[marsh citrus]

@long sage Has your question been resolved?

[long sage]

.closed

.close

[marsh citrus]

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[bronze valley]

Dude what kind of question is this ive never seen smth like it

[long sage]

IK

it took me 1 hour to even understand what the question is understanding

but thats the problem

<@&286206848099549185>

[ocean ruin]

Let me see

[still temple]

hmm

[long sage]

this is the 3rd time asking this question in server and no one could really solve it

[ocean ruin]

Not enough info

[long sage]

thats the same thing I told my teacher but she said there is enough

and thats what I don't get

[still temple]

well as far as i know

the priest HAS to be at least 40-70 years of age

in order for this to work

but as far as the ages go

i’m not so sure

[long sage]

.close

[marsh citrus]

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[mental kayak]

How to prove that any non square number put in a radical is irrational?

[mental kayak]

*positive real numbers

[marsh citrus]

@mental kayak Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[muted cove]

got a test back with a percentage and with the amount of points lost, how would i calculate total points in exam

[fleet basalt]

points lost / (1-percent)

[muted cove]

i got a decimal when doing so

is that normal?

[fleet basalt]

whats the percent and whats the amount of points lost

[muted cove]

74% and 47 lost

[fleet basalt]

~180.77 points

% give to you is probably rounded

[muted cove]

ohhhhhhhh

okay thank you so much for the help

[fleet basalt]

nw

[muted cove]

.close

[marsh citrus]

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[lethal bridge]

why does this graph face the other direction as if its being reflected across the y axis even tho theres no -x in the exponent

[iron meadow]

because 0.8 < 1, as you raise it to a larger and larger exponent it becomes smaller and smaller

[burnt abyss]

for big x values the y values get smaller and smaller

[iron meadow]

0.8 = 4/5 so it's essentially:

y = 10 (5/4)^(-x)

like a 5/4 base with a negative exponent if you think about flipping the 4/5

[lethal bridge]

is the 10(a)^x where a is changing the direction the graph is facing?

[iron meadow]

sorry, i don't understand the question?

[lethal bridge]

like for example

if we hav e the form

y = 10(a)^x

right

is a the quantity that makes the graph change direction

or is it 10

[still temple]

a

[iron meadow]

"change direction" is a weird way to describe it

all the 10 does is stretch it vertically

no matter what else happens

[lethal bridge]

so then

a is the one changing the graph

[burnt abyss]

no one is changing the graph

[lethal bridge]

a is

[iron meadow]

if a > 1 then it's gonna increse without bound if you keep multiplying it with itself

if a < 1 it'll get smaller and smaller

if you have -x exponent instead of x then all it does is a becomes the reciprocal of itself so it just changes whether it's greater or less than 1

[marsh citrus]

@lethal bridge Has your question been resolved?

[lethal bridge]

so um

if

0<a<1

It’s the same as y axis reflection

?

@iron meadow

[iron meadow]

reflection compared to what

y = a^x i guess will be a reflection of y = (1/a)^x over the y axis because you can rewrite y = (1/a)^x = a^(-x) if thats what you mean

so its the same as f(-x)

[lethal bridge]

If y = 10(a)^x then is 0<a<1 the same as f(-x)

??

[iron meadow]

I don't know how i can clarify this any more, that's about all i have to say

a < 1 and a > 0 just have different behaviors in this type of function, that's about all there is to it

and a= 1 is just a flat line

[lethal bridge]

I don’t think you understand what I am saying

I’m saying if a is a value between

0 and 1

Like 0.8

That has the same effect as f(-x)???

[floral linden]

if you have a function f(x) = 2^x, then the function g(x) = (1/2)^x is a reflection of f(x) about the y-axis

so g(x) = f(-x)

[floral linden]

it has the same effect as what f(-x) does for a>1 since that is literally what f(-x) would do

[marsh citrus]

@lethal bridge Has your question been resolved?

[lethal bridge]

so im incorrect in what i said?

[floral linden]

not really, just that the way you said it was confusing

[lethal bridge]

oh my

<@&268886789983436800>

[knotty shard]

I was so confused for a sec

[lethal bridge]

like about the image?

[marsh citrus]

@lethal bridge Has your question been resolved?

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[broken flicker]

Hi can somone please help me?

[marsh citrus]

@broken flicker Has your question been resolved?

[still temple]

@broken flicker which question?

[marsh citrus]

@broken flicker Has your question been resolved?

[broken flicker]

@still temple all of them my teacher just slapped me with this for homework and I don't understand a thing

.close

[marsh citrus]

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[still temple]

@broken flicker @broken flicker

[marsh citrus]

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[still temple]

.reopen

You should be patient

[still temple]

.solved

.close

[marsh citrus]

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[sour flame]

$a\times(a+2)+c(c-2)-2\times a\times c$

[elfin berry]

shotgun

[sour flame]

$a-c=7$

[elfin berry]

shotgun

[sour flame]

hi

i need help

i dont know how to do it

im stuck

i simplified the expression first and got

a² + 2a + c² - 2c - 2ac

i moved the -2c behind the 2a and took +2a-2c out and replaced it with 7

so now i have a² + 7 + c² - 2ac

and since a-c=7 then 7+a=c

it also works for exponents i believe

so i replaced a²+7 with c²

then i cot c²+c² - 2ac

add the c² together and 2c² -2ac

now i dont know what to do

[marsh citrus]

@sour flame Has your question been resolved?

[rocky scaffold]

What do you have to do with this

Oh find value

[rocky scaffold]

This step is wrong