help-0

[surreal meadow]

do you see the sequence

[nocturne iris]

yes

[surreal meadow]

do you see how we pieced it together

[nocturne iris]

yes so the epsilon here is 1/n?

[surreal meadow]

no

what we have done is constructed a sequence

[nocturne iris]

for epsilon = 1/n >0, ...?

[surreal meadow]

no

what we have done is constructed a sequence

[nocturne iris]

ok

[surreal meadow]

{a_n}, satisfying, d(a_n,x) < 1/n

is this a convergent sequence

[nocturne iris]

well it should be of course but im not sure why

uhmmm

[surreal meadow]

what we have is a sequence where the nth element is at most 1/n units away from x

as n gets very very very large

what happens to the distance

[nocturne iris]

it becomes 0

[surreal meadow]

it gets close to 0 yes

you should make this rigorous

so what we did is constructed a sequence

and now you need to show it converges to x like we want it to

[nocturne iris]

yes

soo

[surreal meadow]

showing this converges to x is almost the same argument as showing 1/n converges to 0

[nocturne iris]

why is that?

[surreal meadow]

because d(a_n,x) < 1/n

distance is monotonically decreasing to 0

i implore you to try and prove this sequence goes to x

[nocturne iris]

i mean if the distance converges to 0 then it's all intuitive but if i need to get something similar to the limit definition with epsilon and stuff im stuck

[surreal meadow]

well try it out

you've only had this sequence for like 5 minutes

[nocturne iris]

ok

[surreal meadow]

the epsilon definition should follow almost immediately

the distance converges to 0
more importantly, it converges to 0 monotonically

so the distance always gets smaller

[nocturne iris]

ok so to show that a_n converges to x, let's take epsilon > 0, so to satisfy the definition i would need a n0: for all epsilon > 0, there exists n0 in N such that for all n>= n0, d(x, a_n) < epsilon; i already know that d(x, a_n) < 1/n for all n0, so we know that whatver N is, for all n>= n0, d(x, a_n) < 1/n0

then now i dont know, here we have 1/n0 but not epsilon,

[surreal meadow]

ok so to show that a_n converges to x
you're done

oh

i read that as "now that"

let me read again

im not quite sure what you're trying to say

what we want is that

given some eps > 0

the distance is eventually less than epsilon

[nocturne iris]

yes, that's where im stuck

[surreal meadow]

can you construct a proof that a_n = 1/n converges to 0

[nocturne iris]

lim 1/n when n approaches infinity

[surreal meadow]

no

an epsilon-N proof

[nocturne iris]

ok

for all eps > 0, exists n0 = ??, such that n >= n0 --> d(x - 1/n) < epsilon. I dont know what n0 could be

[surreal meadow]

that's what im asking you to solve for

it is the exact same argument for the other sequence

first of all

if a_n = 1/n, and we are converging to 0

what is x

[pseudo ice]

[you want that last bit to be d(x, a_n) < 1/n presumably] nvm just noticed

[nocturne iris]

the kimit

[surreal meadow]

but what are we trying to prove convergence to

what is the proposed limit value

[nocturne iris]

ah ok 0

[surreal meadow]

yeah

so now we can rewrite the final statement
d(x,1/n) < eps
as
d(0,1/n) < eps

what does d(0,1/n) simplify to

[nocturne iris]

|1/n| < eps

[surreal meadow]

have you heard of the archimedean principle

[nocturne iris]

no

[surreal meadow]

oh

[nocturne iris]

but what is it

[surreal meadow]

well it is pretty fundamental

[nocturne iris]

maybe i dont remember

[surreal meadow]

it says that for any eps > 0, there exists a natural number N such that 1/N < eps

[nocturne iris]

ok

[surreal meadow]

so we have some epsilon > 0

what does the archimedean principle give us

[nocturne iris]

you mean in what context? you want a link with |1/n|?

[surreal meadow]

i was asking for you to restate the second part of the principle

there exists a natural number N such that 1/N < eps

maybe i was unclear, but we have that now

so if we are given some arbitrary epsilon > 0
the archimedean principle tells us there is some natural number N so that 1/N < eps

now remember we defined a_n = 1/n

so a_N = 1/N, and d(0,a_N) = d(0,1/N) = 1/N < eps

[nocturne iris]

yes so the limit is 0

[surreal meadow]

no

we've only shown d(0,a_N) < eps

not d(0,a_n) for all n >= N

[nocturne iris]

mmmhh yes

[surreal meadow]

how would we do that

[nocturne iris]

we just check if the inequality still works if we take n bigger than N

[surreal meadow]

yes perfect

now why does the inequality still work

if n >= N, then d(0,a_n) = 1/n < ...

[nocturne iris]

if n>= N, 1/n <= 1/N < eps

[surreal meadow]

flip the first inequality

but yes

perfect

and now we've shown d(0,a_n) < eps for all n >= some N

and in fact

we've shown it for arbitrary epsilon

so this indeed converges to 0

now everything we just did

translates to the other sequence

[nocturne iris]

ohh ok i see

so i first i create the sequence a_n and then i show it converges to x and i just replace evrything with x and a_n

[surreal meadow]

sort of yes

we created the sequence

and it satisfies d(x,a_n) < 1/n

if you apply the same arguments we just applied, you should be able to show that for any epsilon

d(x,a_n) < eps for all n >= some N

which is exactly convergence to x

[nocturne iris]

ok thank you 😭

[nimble fern]

Sorry for stalking, but i have been following through the progress and found this very inspiring

i remember back in my year one analysis, I failed to construct a sequence similar to this situation, and I bet I would never know what's happening if no one told me the answer

[nocturne iris]

i feel you XD

[nimble fern]

I am glad to see maximo is guiding through patiently and precisely to reach the goal

[surreal meadow]

that is generally the feeling i see from people doing this sort of exercise

[nimble fern]

and I'm also glad to see lilisworld. is getting closer to closer to understand it all!

[surreal meadow]

until all the definitions click and become super natural, it's tough to see it all piece together

[nimble fern]

it's a long and difficult process, but both of you nailed it!

me glad

[nocturne iris]

bruuhh thanks to maximo

[nimble fern]

and to you too! lilisworld.

i might have given up and .close this if i were you

[nocturne iris]

i struggle a lot tho but thanks ❤️

ok bye guys and thank you again!!!

.close

[lone heart]

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[lone heart]

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[vapid shuttle]

@surreal meadow

[surreal meadow]

please dont ping specific helpers

[vapid shuttle]

I'm working on this and assuming that gamma has piecewise C1 boundary (which I think is necessary here) I can apply Green's theorem to equate it to

[ocean seal]

Austin

[vapid shuttle]

and then I'm stuck about how to go about maximizing this

[surreal meadow]

bro \iint please

[vapid shuttle]

I didn't know that was a thing

[surreal meadow]

i think there is an argument here that the center of mass should be at the origin

because otherwise you are wasting magnitude from the integral

[vapid shuttle]

center of mass of our region?

[surreal meadow]

like

the average location for a point

[vapid shuttle]

ok

[surreal meadow]

if you take some region S centered at 0, and shift it in the x and/or y direction, then the integral gets smaller

[vapid shuttle]

Yeah

Still stuck even knowing that

[surreal meadow]

if you rewrite this as

[3\iint_S,\dd A - 3\iint_S(x^2+y^2),\dd A = 3A(S) - 3\iint_S(x^2+y^2),\dd A]

[ocean seal]

maximo

[surreal meadow]

you can see what i was saying about maximizing area while minimizing the distance from the origin

[vapid shuttle]

I do see what you mean by that

[surreal meadow]

so if you humor this idea

[vapid shuttle]

humoring

[surreal meadow]

and fix the area, we want a curve that minimizes perimeter for a given area

and alternatively

if you fix the perimeter, you want a curve that maximizes area

[vapid shuttle]

circle?

[surreal meadow]

yes

and per our other argument, centered at 0

[vapid shuttle]

o

so we go polar on they ass

0 to 2pi, 0 to r

dr dtheta

and integrate it out

and then maximize that bih

[surreal meadow]

yes for some radius r_0

i think that works but the argument is shaky

[vapid shuttle]

I'm going to be upset with you if it is the unit circle

[surreal meadow]

it should most definitely not be the unit circle

ok i take it back

im not that certain, but i doubt it is

ah shit

it probably is

[vapid shuttle]

https://tenor.com/view/monkey-funny-door-stuck-open-gif-17183333

[surreal meadow]

i made a grave error

i subbed x^2 + y^2 for a fixed radius of 1

but x^2 + y^2 would just be r^2 for r from 0 to 1

it probably is 1

[vapid shuttle]

https://tenor.com/view/dubya-gif-23030610

[surreal meadow]

and after 1 it probably goes down

i give broski the proof and he gets mad that i shot down his guess

smh

[vapid shuttle]

broskis proof was

uhhh

it is probably a circle

just think about it bro

and if its a circle then

uhh

[surreal meadow]

better than "prob want to be the unit circle"

"if i had to guess"

[vapid shuttle]

nearly the same

XD

jk

(not jk)

[surreal meadow]

i said "fix the area"

we are not the same

[vapid shuttle]

so I did the integral

we want to maximize

6pi(r_0)(1-(r_0)^2/3)

if r_0 > 1 this is negative

[surreal meadow]

im gonna scream

[vapid shuttle]

if r_0 is 0 this is 0

[surreal meadow]

[6\pi r_0(1-\frac{r_0^2}{3})]

[vapid shuttle]

hahahaahjaa

hahahaa

[surreal meadow]

is it 2r/3

[vapid shuttle]

no you buffoon

[surreal meadow]

or

[vapid shuttle]

r_0^2

/3

[surreal meadow]

r^2)/3

[vapid shuttle]

that ne

one*

[surreal meadow]

ight

[ocean seal]

maximo

[surreal meadow]

we good

[vapid shuttle]

lmao

yes that

the answer is somewhere in (0,1]

[surreal meadow]

distribute and derive and set = 0 and so on

[vapid shuttle]

bros telling me min max theorem

https://tenor.com/view/alpha-wolf-sigma-male-gif-26562640

[surreal meadow]

me when the curve is not piecewise C1

https://tenor.com/view/noooo-no-no-emote-emote-nooo-emote-gif-23071619

[vapid shuttle]

@surreal meadow it is the unit circle

u bot

u r a bot

r_0 =1

wow ur sucha bot

[surreal meadow]

fuck it we ball

[vapid shuttle]

me when I remember

this question had 2 part

https://images-ext-2.discordapp.net/external/nT0HLfMKyrkZCrdwjuPr56jvWB-q9SS0YUI0Duq8kWE/https/media.tenor.com/Yv2ty8bP3kgAAAPo/noooo-no.mp4

wtf

.close

[lone heart]

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[west cipher]

You have one length and an angle

that is enough information to find the other side

you can use soh cah toa to determine which one

you dont know soh cah toa?

ok

soh means

sin(angle)=opposite (o)/hypoteneuse (h)

cah means cos(angle)= adjacent/hypoteneuse

its a way of remembering

what sin is

toa means tan(angle)=opposite/adjacent

In the first image, there is an angle and two lengths

how are the two lengths related to the angle?

using soh cah toa

there is two

y and the number

y and 12cm

is the angle directly across from y?

ie is the the side y adjacent to the angle?

if it is adjacent, then would you use sin? (opposite/hypoteneuse)

so y is opposite of the angle then?

not adjacent?

since sin(50)=opposite/hypoteneuse

ok

[small ice]

can i call somone i need help badly

[west cipher]

why do you say y is opposite

its not

yes

cos(50)=adjacent/hyp = y/12

soh cah toa

are you in radians mode

make sure youre in degree mode

ok

looks reasonable

idk

looks reasonable tho

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[alpine sable]

.reopen

[alpine sable]

I need help with # 2

I've spent this whole weeking

trying to solve this

but I'm just slow

[pine shard]

couldn't you say that angle AFG is congruent to angle GFE because of supplementary angles

oh

[alpine sable]

My teacher

wants me to solve it

in proofs

with steps

[pine shard]

I just realized its the triangles

not the angles

[alpine sable]

Yeah

[pine shard]

aight

[alpine sable]

I've spent all weekend

but I keep messing up

[narrow kelp]

think of all the ways to prove a triangle congruent

[alpine sable]

alright

I mean

if the sides are congruent

[pine shard]

you can say that GF is congruent because they share the same side

[alpine sable]

two sides equal?

[pine shard]

and then use AAS congruency

with the supplemental angles

[pine shard]

AFG and GFE

[alpine sable]

this is what I did

for the first

proof

[alpine sable]

3 steps?

[pine shard]

and that GHF and GIF are congruent because its given

would that be possible?

[alpine sable]

I'm lost

[pine shard]

aight full proof I'll try

gimme a sec to make a table

[alpine sable]

thanks bro

😭

[pine shard]

lemme know if I'm missing anything

[alpine sable]

WAIT

BRO COOKED UP

@pine shard

TY BRO

YOU SAVED ME

[pine shard]

EYYYY

NICE

[alpine sable]

now I have to write a essay

about

why triangles

are triangles

n shi

😐

[pine shard]

lmao

but at least you have the proof now

[alpine sable]

last question, it might sound dumb

but

Introduce triangle congruence and at least 3 ways it can be applied in real life

[pine shard]

is that a question they ask?

[alpine sable]

yes

like I know triangles

in real life

but how can they be applied

it doesnt make sense

[pine shard]

It can be used for making triangular patterns

[proven leaf]

triangulation and how to find yourself on the globe as well as architecture/construction and engineering on top of that, geometry is by far the most applicable branch of mathematics

[pine shard]

ye thats a good answer

[alpine sable]

thank you

[alpine sable]

thank you

[pine shard]

no proble

[lone heart]

@alpine sable Has your question been resolved?

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[gray crescent]

Hello

[stark crater]

Write the polynomial in factored form and combine

[gray crescent]

Wdym

Could u explain a bit further thanks

[stark crater]

You are given the roots so it can be written as (x-alpha)(x-alpha bar). You can rewrite alpha as a+bi

[gray crescent]

agh ok i see

thankyou

[lone heart]

@gray crescent Has your question been resolved?

[lone heart]

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[twin lynx]

How do I use implicit differentiation here to find dy/dx in terms of x and y

[mighty crest]

differentiate each side to start off

[lone heart]

@twin lynx Has your question been resolved?

[twin lynx]

So 2(2y^3 + x^3) times dy/dx ( 2y^3 + x^3)

Which is equal to 1

[mighty crest]

like this

chain rule

[tacit arch]

Not dy/dx (2y^3+x^3)

d/dx (2y^3+x^3)

[twin lynx]

Figured it out

.close

[lone heart]

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[fast swift]

So I've got the problem ((n+1)!+(n-1)!)/n! = 11

[fast swift]

sorry, i hit send before finishing typing accidentally, I'll put the rest of my work in soon

I simplified it by doing ((n+1)n!+(n-1)!)/n! = 11

then n + 1 + (n-1)! = 11

then got to n + (n-1)! = 10

and from here, I'm stuck

[inner bough]

(n-1)!/n! should be 1/n, so you should get n+1/n=10

[fast swift]

sorry, I formatted it badly, the entire left side of the equation is divided by n!

I'm trying to figure out how to use the bot to format it better

so here is the original equation: $\frac{(n+1)!+(n-1)!}{n!} = 11$

[ocean seal]

kneeecaps

[fast swift]

And here are my working steps so far:
$\frac{(n+1)n!+(n-1)!}{n!}=11\
n + 1 + (n-1)! = 11\
n + n)n-1)! = 10$

and now i've stuffed the bot again :/

sorry, im not good with this thing

[inner bough]

when you divide (n-1)!/n!, you get 1/n because (n-1)!/(n*(n-1)!), the (n-1)!s cancel out

[ocean seal]

kneeecaps

[fast swift]

oh ok, I completely forgot about that. So would I want to turn it into two fractions then to solve it properly? Like this:

$\frac{(n+1)n!}{n!} + \frac{(n-1)!}{n!} = 11$

[inner bough]

yes

[ocean seal]

kneeecaps

[fast swift]

ok, thank you so much, that helps a lot

so far I got
$\frac{(n+1)n!}{n!} + \frac{(n-1)!}{n!} = 11\
n + 1 + \frac{1}{n} = 11\
n + \frac{1}{n} = 10$

is that on the right track or have i messed something up?

[ocean seal]

kneeecaps

[inner bough]

thats right

[fast swift]

ok, give me a second to try and figure out the rest of it

[lone heart]

@fast swift Has your question been resolved?

[fast swift]

alright, so after too much time trying to format this so it is readable, I've got the rest of my working out:

$n + \frac{1}{n} = 10\
\frac{n^2+1}{n} = 10\
n^2 + 1 = 10n\
n^2 - 10n + 1 = 0\
\frac{-(-10)\pm \sqrt{-10^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}\
\frac{10\pm \sqrt{100 - 4}}{2}\
\frac{10\pm \sqrt{96}}{2}\
\frac{10\pm \sqrt{16 \cdot 6}}{2}\
\frac{10\pm 4 \cdot \sqrt{6}}{2}\
5 \pm 2 \cdot \sqrt{6}$

[ocean seal]

kneeecaps

[fast swift]

is that the same answer that you got or did i stuff something up?

[inner bough]

yup, thats right

[fast swift]

ok, thank you so much for your help

[inner bough]

for sure

[fast swift]

.close

[lone heart]

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[smoky condor]

In this why ilate rule is invalid?

I mean cos2x should be the first one

[tacit arch]

Who says it is

[smoky condor]

But here it is the 2nd one

[smoky condor]

If we apply we get a different answer

[tacit arch]

Show

[hearty quartz]

Why don't you use an integral calculator. ILATE is just a rule thumb. Give it a couple years and you end up forgetting it. Normally I just use experience when using integration by parts

[smoky condor]

@tacit arch

[mighty crest]

,rotate

[mighty crest]

huh wait

[lone heart]

@smoky condor Has your question been resolved?

[smoky condor]

No

[lone heart]

@smoky condor Has your question been resolved?

[smoky condor]

No

[lone heart]

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[void summit]

Hi, math newbie here. How can I find the (20x1) + (19x2) + (18x3) ... (1x20) ? any formula?

[vale wigeon]

there is but it's more complicated than just doing it the dumb way

[void summit]

• The radius of the spheres is 2
•Maximum spheres will be placed in the primary cylinder, the number of spheres will be successively increased by one in subsequent cylinders.
•The weight of the spheres placed in the primary cylinder will be 1 gram, and the weights of the spheres placed in the other cylinders will be increased sequentially.

so I need the total weight of spheres

also radius of cylinder is 2

[vale wigeon]

http://xyproblem.info/

... can we see the entire problem please?

screenshot the whole thing

asking because i am kind of suspicious now

[void summit]

I translated to question from my book

[vale wigeon]

show the original

just in case

[void summit]

You sure? it's not english

:( sorry

[vale wigeon]

i know it's not in english, you told me that by saying you translated it.

post it anyway.

[void summit]

here you go

[vale wigeon]

ok so it's in turkish and i don't speak turkish

[void summit]

yeah

right

[vale wigeon]

however that formula at the top is important!

and i didn't know until now that they gave it to you.

[void summit]

Right... my bad

[vale wigeon]

• The radius of the spheres is 2
•Maximum spheres will be placed in the primary cylinder, the number of spheres will be successively increased by one in subsequent cylinders.
•The weight of the spheres placed in the primary cylinder will be 1 gram, and the weights of the spheres placed in the other cylinders will be increased sequentially.

did you mean that the number of spheres is DECREASED by 1 in subsequent cylinders?

[void summit]

yeah

[vale wigeon]

and maybe "first" instead of primary

[void summit]

and weight increased

[vale wigeon]

anyway ok

do you know sigma notation?

[void summit]

no

[vale wigeon]

ok

do you know how to add 1+2+...+n?

[void summit]

n*n+1/2

[vale wigeon]

you are missing some parentheses

(fatally)

[void summit]

((n)*(n+1))/(2)

[vale wigeon]

now there are too many parentheses

which is not incorrect but it is just clutter

n(n+1)/2

anyway ok

[void summit]

:0

[vale wigeon]

so you have expressed the total mass of the spheres as 20 * 1 + 19 * 2 + 18 * 3 + ... + 1 * 20

[void summit]

Yes

[vale wigeon]

in other words,

the k'th bin contains k(21-k) grams in spheres

[void summit]

YES

[vale wigeon]

or in other words that's 21k - k^2

and you add that up for k from 1 to 20

so if you group the 21k terms together that's 21(1+2+...+20)

and if you group the squares together that's -(1^2+2^2+...+20^2), which you can calculate with that formula at the top

[void summit]

(20*(21)/2)x(21)?

[vale wigeon]

do not use the letter x as a multiplication symbol

also a lot of those parentheses are redundant again, but yes

[void summit]

discord changes the font when I use 2 *

[vale wigeon]

\*

or put spaces directly around the asterisk

[void summit]

aight

[void summit]

(20*21)/2*(21)-(20* 21*41)/6

=1540

[void summit]

so option D) is answer @vale wigeon Thanks a lot!!

I appreciate the help

as a last question do you have any advice for me? I want to study math in university

[lone heart]

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[exotic canopy]

solve what? what's going on?

what's F_n?

fibonacci number?

also what does this say?

[lone heart]

@alpine sable Has your question been resolved?

[devout crest]

what is the question?

[lone heart]

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[ocean seal]

murphey

murphey

murphey

murphey

murphey

murphey

murphey

murphey

murphey

murphey

murphey

[vale crag]

so what are you trying to do ?

finding the generating function of the fibonacci numbers ?

@alpine sable

what's your F_0 and F_1 here ? @alpine sable

if you have F_0 = 0 and F_1 = 1 there's no error there

[lone heart]

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[rigid nova]

Pre Calc Help

[lethal belfry]

[rigid nova]

Composite Function: [G(x)=x^3+2] [F(x)=3x-1] evaluate

[alpine sable]

isnt that a nice attitude to have... <@&268886789983436800>

[rigid nova]

Assume G(F(1))

hello?

[indigo solar]

So you need G(F(1))
?

[rigid nova]

Sorry Let me show u

F(1)= 3(1)-1

    =3-1
   = 2

Then...

G(F(1)) --> g(2)

g(x) = x^3 + 2

G(2) = 2^3 + 2

G(2) = 18

[indigo solar]

2^3 isn’t 16

[rigid nova]

Oh shit!!my bad small mistake messed me up

G(2) = 10

so my probem is that i need to make a composite function from this in a simplified equation

how do i do this?

[indigo solar]

So just do G(F(x))?

[brittle ember]

you can substitute x inside G(x) to F(x)

As G(F(x)) means calculating G(x) with setting value of x to F(x)

[lone heart]

@rigid nova Has your question been resolved?

[lone heart]

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[buoyant thunder]

I forgot how to do part b

[gray isle]

doing part a)
gives you slope-intercept form

the gradient and y-intercept will be visible in that form

[buoyant thunder]

i got y = (12-x)/3

ohhh wait

there was a formula right

@gray isle

[gray isle]

are you familiar with slope-intercept form?

[buoyant thunder]

i did it a while back

but ik

3 is the y intercept

[gray isle]

3 isn't the y-intercept

[buoyant thunder]

can you remind me of the formula pls

[gray isle]

do a quick seach on slope intercept form

then split this fraction

(12-x)/3

[buoyant thunder]

y = (gradient)x + y intercept

right

@gray isle im confused on how i find the gradient and interceptt

[gray isle]

then split this fraction
(12-x)/3

[buoyant thunder]

how

[gray isle]

$\frac{p-q}{r} = \frac pr - \frac qr$

[ocean seal]

ℝαμΩℕωⅤ

[buoyant thunder]

4 - x/3?

[gray isle]

yes

[buoyant thunder]

So gradient is 4 and intercept is x/3

[gray isle]

no

[buoyant thunder]

-x/3

[gray isle]

y = (gradient)x + y intercept
note that the gradient the the amount being multiplied to x, (the coefficient of x, this should not contain x)
and the y-intercept is the constant being added

the gradient will be the coefficient of x in

-x/3

[buoyant thunder]

Ok

@gray isle can it be rearranged as y = -x/3 + 4

[gray isle]

yes

[buoyant thunder]

@gray isle

hey

why is my graph

different

its not -x/3

That’s -2/3

Wait no

It 4/12

So 1/3

But why does it look like a -2/3 on the graph

also

should i write it as -x/3

or -1/3

[lone heart]

@buoyant thunder Has your question been resolved?

[lone heart]

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[coarse vale]

`

[coarse vale]

dunno how to approach.

Not even able to solve the integral

[lone heart]

@coarse vale Has your question been resolved?

[lone heart]

@coarse vale Has your question been resolved?

[coarse vale]

<@&286206848099549185>

[buoyant thunder]

😨 what da hel

[coarse vale]

help..?

[buoyant thunder]

i cannot help you bro💯

[coarse vale]

This is the given solution btw (still can't understand)

[buoyant thunder]

😨

[coarse vale]

With smol writing mistake sry

[buoyant thunder]

that looks more like computer code than math at this point

[buoyant thunder]

but can you help me

[buoyant thunder]

dont worry

u just need a warmup question

the square in between the square

is -2/3

[coarse vale]

What do u even need to find out

Question?

[buoyant thunder]

i found the slope which was 4/12 = -1/3x

but in the graph

it doesnt look like a -1/3x

but more like a -2/3x

[coarse vale]

Have u learned differentiation?

[buoyant thunder]

no

i dont know nothin

im in 9th grade

[coarse vale]

Just show me the question—

[buoyant thunder]

i did the question bru

im asking

why it looks like -2/3

[coarse vale]

How will I know—

[buoyant thunder]

look

the gap is a -2/3

but its actually a -1/3

[coarse vale]

Look bro. I'm not google lense.
I can't just tell if a function is of what slope just by looking at it—

I need to equation u trying to plot

[buoyant thunder]

i finished the thing man

its a visual question

rise/run

=

4/12

=1/3

so why it loookin like 2/3

[coarse vale]

like... sir, I haven't used that formula in my entire life.

I just

different

the equation

And that gives the slope to me.

No one rises no one runs.

[buoyant thunder]

NOOO

ITS RISE

DIVIDED

BY THE RUNNNN

[coarse vale]

sir... This is my channel rn.
And I was asking a question.
But u now intrude and demand an answer...
Without the dang question...

[buoyant thunder]

ok wait

i will get the question

[coarse vale]

I'm not sum dang robot that looks at a graph and goes "hm-m, that sure is tan inverse of 2/3 right there"

[buoyant thunder]

bruh

cmon

u shud know

[coarse vale]

Bruh... I've by hearted angles 15,75,18,72,36,54,53,37.
And none of em has 2 and 3 as non hypotenuse sides.

[buoyant thunder]

why

[coarse vale]

Cause no one wants those. They useless.

[buoyant thunder]

so

did you get the 2/3

or the 1/3

[coarse vale]

I don't get nothing.

[buoyant thunder]

fr

the question too difficult

[coarse vale]

Show me 🗿

[buoyant thunder]

ok

@coarse vale

[coarse vale]

<@&286206848099549185>

[coarse vale]

1

[buoyant thunder]

ok

u wrong

its fine

we all mess up

[coarse vale]

Now be on ur way kid

[buoyant thunder]

dont u need help

[coarse vale]

And u not being any help.

[buoyant thunder]

ok ill help now

show

question

[buoyant thunder]

why are there letters

other than one letter letters

[coarse vale]

<@&286206848099549185>

[nimble fern]

Hello ` ! Is your function
f(x)=(-c²+(b-1)c-2)x + int_0^x sin²t+cos⁴t dt ?

[coarse vale]

This is the function, yes.

[nimble fern]

since it's a decreasing function for all x in R and c in R, we will have
f'(x)<0 for all x in R and x in R

[coarse vale]

okey

[nimble fern]

looks like your solution the first 2 lines, f(x) should be f'(x) instead

[nimble fern]

and this should be non-increasing function instead of decreasing function

anyways

d(f(x))/dx
= d((-c²+(b-1)c-2)x + int_0^x sin²t+cos⁴t dt)/dx
=(-c²+(b-1)c-2) + sin²x+cos⁴x

all good till here?

[coarse vale]

Yes

[nimble fern]

now,
f'(x)≤0
we have
(-c²+(b-1)c-2) + sin²x+cos⁴x ≤ 0

[coarse vale]

Wait... The limit on the integration can just vanish?

[nimble fern]

Do you know the fundamental Theorem of Calculus?

[coarse vale]

I don't remember it by the name I'm afraid.

[nimble fern]

https://granitegeek.concordmonitor.com/wp-content/uploads/2016/08/fundamental-theorem-of-calculus.jpg

[coarse vale]

wbt "a"

[nimble fern]

we are using the bottom one

a is just any constant

[coarse vale]

no... where'd f(a) part go in 2nd equation

oh..

Okey. Pls continue with the question explanation.

[nimble fern]

d(constant)/dx = 0

ok

(-c²+(b-1)c-2) + sin²x+cos⁴x ≤ 0
-c²+(b-1)c-2 + sin²x+(1-sin²x)² ≤ 0
-c²+(b-1)c-2+sin²x+1-2sin²x+sin⁴x ≤ 0

[coarse vale]

okey

[nimble fern]

sin⁴x-sin²x-c²+(b-1)c-1≤0

by completing square, we have
(sin²x-1/2)²-1/4-c²+(b-1)c-1≤0

which arrived
(sin²x-1/2)²-c²+(b-1)c-5/4≤0

now that's where i start to be puzzled, why would your teacher/prof choose the max value of (sin²x-1/2)²

which is 1/4

[coarse vale]

It's actually a question from a mock tests and the app didn't alow screen shots.
So I just jot it all down.

[nimble fern]

note that:
(sin²x-1/2)²
= ((2sin²x-1)/2)²
= (-cos(2x)/2)²
which is within [0, 1/4]

if i were to do this question, i would have different solution starting from here

[coarse vale]

Kool. Proceed.

I've kept a page empty after that solution for better explanation to myself.

[nimble fern]

so, we have
c²-(b-1)c+5/4-(sin²x-1/2)²≥0

sorry, typo

since this quadratic of c is always ≥0, we have the discriminant ≤ 0

[coarse vale]

How?

[nimble fern]

is this ok?

[coarse vale]

Still don't get how.
What if b>4 and c<4 while a=1?

[nimble fern]

then there will be 2 distinct root

[coarse vale]

If D<0, doesn't that mean complex root?

[nimble fern]

for ax²+bx+c≥0 for a>0
it will either just touch the 0 line or not touch at all, therefore it's a repeated root or no real root
therefore the discriminant ≤ 0

[nimble fern]

though we are not concerned about the roots of this c²-(b-1)c+5/4-(sin²x-1/2)² in this question

[nimble fern]

see if you understand this first

[coarse vale]

okey, so... f(x)≥0 means y≥0, thus the graph

Now I got the answer...

Thank you for the help.

[nimble fern]

well

i haven't finished

[coarse vale]

U can, if u want to.

[nimble fern]

(b-1)²-4(5/4)(sin²x-1/2)²≤0

so we have
(b-1)²≤5(sin²x-1/2)²

oh wait

sorry

[coarse vale]

How this happened—

[nimble fern]

my bad

lemme do it again

(b-1)²-4((5/4)-(sin²x-1/2)²)≤0

there we go

b²-2b+1-5+4(sin²x-1/2)²≤0

[coarse vale]

Wait, still confused on how ths happened—

[nimble fern]

discriminant ≤0

[coarse vale]

Where that 4 come from and where c go

Oh. 4 mystery solved.

Oh

[nimble fern]

c is from the quadratic expression c²-(b-1)c+5/4-(sin²x-1/2)²

[coarse vale]

Taking c as variable.

But x?

[nimble fern]

yes

[nimble fern]

that's the main point

which i think it's the hardest part of the question

[nimble fern]

since we have this

let's fix 2 cases for the
b²-2b+1-5+4(sin²x-1/2)²

it is when (sin²x-1/2)² is max or min

when it's max,
the quadratic of b becomes
b²-2b-3≤0
when it's min,
we have
b²-2b-4≤0

lemme check things a bit

sorry for taking your time

[coarse vale]

No worries.

[nimble fern]

so for both situations, we have
b²-2b+1≤4
or
b²-2b+1≤5

since it's an 'or' relationship, we have to choose the largest suitable range for both situations

which is b in [-1,3]U[1-√5,1+√5]

which finally we have
b in [1-√5,1+√5]

which is slightly larger than the range in the final answer of the marking scheme

that's all i think

[coarse vale]

It's given in the options, but the [-1, 3] is marked as correct—

[nimble fern]

interesting

anyways thanks for your time, I've had a great time doing this question!

[coarse vale]

I'll jot down all this question and ask my teacher tomorrow.

[coarse vale]

Thanks to u too for helping with this.

[nimble fern]

sure, I am thinking i might have something wrong, dunno 😛

Cheers!

[coarse vale]

Cheers

[lone heart]

@coarse vale Has your question been resolved?

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[fair vigil]

I have asked this question before, but maybe someone has a shorter and less complex way to solve this?

[fair vigil]

oh lovely, people typing. Took me hours to get an answer last time

[quasi vector]

try writing $T_r=\frac1{4(r+3)\cdot 4(r+4) \cdot 4(r+5)}=\frac1{64(r+3)(r+4)(r+5)}$

ah..

[ocean seal]

kheerii

[quasi vector]

and then

make this into a difference of two consecutive terms in some sequence

[fair vigil]

wait how did you get that

oh its like the pattern

what is that 64 tho

[lethal belfry]

$4^3$

[ocean seal]

Why am. I here

[fair vigil]

why is that

[lethal belfry]

wdym

why is 4^3=64?

[fair vigil]

nah, where did that 4^3 come from

[lethal belfry]

444 is 64

• 4 into 4 into 4

[fair vigil]

No I know how that works

lemme word the question better

at what point did we come to the conclusion that we should write 64 there

[lethal belfry]

did you understand how the general term was obtained?

[fair vigil]

thats what im asking

[quasi vector]

well

there are three consecutive multiples in the denominator of each term

[fair vigil]

yeah

for example in the first one its 16 20 24

[quasi vector]

yes

say that is T_1

[fair vigil]

yep

[quasi vector]

16=4(1+3), 20=4(1+4), 24=4(1+5)

[fair vigil]

oh I see and if we put 1 instead of r we get 16 20 and 24

and 4 is common so we take it out

[quasi vector]

yep

[fair vigil]

so its a sequence

[quasi vector]

correct

[fair vigil]

so I could calculate the sum

[quasi vector]

so we have defined the series

[fair vigil]

S_n = n/2 * (a1 + an)

[quasi vector]

now we need a way to add it up

[quasi vector]

no, that only applies for arithmetic series

which this is not.

[fair vigil]

its geometric then

[quasi vector]

why?

there exist many sequences which are not arithmetic or geometric

infinitely many, in fact.

an Arithmetic sequence is one in which the difference of two consecutive terms is constant

a Geometric sequenec is one in which the ratio of two consecutive terms is constant

none of those two conditions work here

[fair vigil]

yeah ur right

okay my bad its neither of those

[quasi vector]

yep

adding these terms up will be hard without some sort of cancellation

we need to find some way to cancel these terms

[fair vigil]

every neighbor fractions share two common numbers

[quasi vector]

yeah

[fair vigil]

we have 20 24 in the first two

we have 24 28 and the next two and so on

soo

[quasi vector]

yes, that is a key observation here

basically whenever you have a term like 1/r(r+1)(r+2)(r+3).... something like this

it can turn into a telescoping series

meaning you can manipulate that term in such a way that the intermediate terms all cancel out and you only have to calculate something at the ends of the summation

[fair vigil]

oh dang I remember using this for a less complex problem

I represented 1/(4*5) as a subtraction of two fractions or something and then cancelled common fractions

[quasi vector]

yes, you need to use a similar concept here

[fair vigil]

let me try it on my own first. Thanks

[fair vigil]

So I tried rewriting the first fraction as a subtraction of fraction

and did the same for the second one

so I get 6 smaller fractions like 1/4, 1/5, 1/6 and some of them would cancel each other but they are multiplied by some random stuff

like -14, -23 and so on which prevents the cancellation

[quasi vector]

Try writing $T_r=\frac1{64(r+3)(r+4)(r+5)} = \frac{(r+5)-(r+3)}{2\cdot 64(r+3)(r+4)(r+5)}$

[ocean seal]

kheerii

[lone heart]

@fair vigil Has your question been resolved?

[fair vigil]

and then?

I am not sure I see what to do next

[quasi vector]

we can say $T_r = \frac1{128} [\frac1{(r+3)(r+4)}-\frac1{(r+4)(r+5)}]$

[ocean seal]

kheerii

[quasi vector]

you see how?

[fair vigil]

yeah I get this

and then should I divide these smaller fractions into even smaller ones?

or nah nvm we should be fine

[quasi vector]

then we can define $v_r = \frac1{(r+3)(r+4)}$, which gives us $T_r=\frac1{128}(v_r-v_{r+1})$

[ocean seal]

kheerii

[quasi vector]

that's the crucial step

this v_r - v_r+1 gives us that telescoping effect

[fair vigil]

so the next fraction will be 1/128(v_(r+1) - v_(r+2))

and that v_(r+1) will get cancelled

[quasi vector]

yep

[fair vigil]

okay lemme calculate the rest, brb

@quasi vector I got the answer. Thanks a lot!

.close

[lone heart]

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[cold mist]

still don't get it. What happens there?

[nimble fern]

i think those mean "using double angle formula"

[azure frigate]

sin(2x)=2sin(x)cos(x)

[lone heart]

@cold mist Has your question been resolved?

[ionic oak]

What's stopping you from using double angle formula?

[cold mist]

I got it 👍
thanks everyone!

.close

[lone heart]

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[frigid mirage]

after a reduction by p% the wardrobe costs s zlotys, a)what will be the price of this wardrobe after the increase of p%? b)Determine the price of the wardrobe before the reduction

zlotys is like dollars you know guys. I need help with it, im not sure my equation is correct.

[loud warren]

Can you show what you tried?

[opal ingot]

It's a common misconception that raising the price of something 10% (for example) and then lowering the price by 10% will get you back to the original price, but it won't. Is that the source of your confusion?

[loud warren]

@frigid mirage we can't help you if you don't answer...

[frigid mirage]

omg sorry

i was doing other ones

[loud warren]

don't worry

[frigid mirage]

yeah wait

let me find the piece of paper

[frigid mirage]

not really

[frigid mirage]

let me edit the photo a little bit for u cuz my datas are in Polsih

Polish

ugh wait ill just write the translation write its too annyoing

1. wardrobe cost
2. wardrope cost after a raise
3. wardrope cost before the reduction

@loud warren

[opal ingot]

,rotate

[opal ingot]

Does that say s+2p%?

[frigid mirage]

where? in a)?

yes it does

[opal ingot]

OK, how did you get that?

[frigid mirage]

Well, i was thinking like that: if the wardrobe's cost is now s zlotys (like dollars) . so i multiplied s * p% (To get it to the original price). so now if it has raised by another p% i did that.

im very sorry English is not my native

[opal ingot]

Let me re-read the original problem

[frigid mirage]

ok 😭

[opal ingot]

OK. So, if something starts at 150 and you reduce it by 20% how would you compute that?

[frigid mirage]

150 * 0,80, yes?

[opal ingot]

Exactly. OK, let's say the original price was x. If you start at price x and reduce by p% what do you get?

[frigid mirage]

i don't know.. i can't think rn

[opal ingot]

That will make it difficult. Do you want to continue or come back when you're ready?

[frigid mirage]

i can continue sure

ur choice tbh

[opal ingot]

OK, let's say the original price was x and you reduced it 20%. What would the new price be?

[frigid mirage]

0,80x

[opal ingot]

Right! What if that was 30%?

[frigid mirage]

0,70x

[opal ingot]

Now, can you do it in general for p% ?

[loud warren]

100% = 1 (maybe this helps you)

[frigid mirage]

wait

am i thinking it good? x * p/100 ?

[opal ingot]

Very close. However, if you have a 20% reduction, do you multiply by 20/100?

[frigid mirage]

no, i multiply it by 0,80, yes?

[opal ingot]

Correct. So, can you change your formula to show thaT?

[frigid mirage]

this forumla, eh?

[opal ingot]

Yes

[frigid mirage]

x * 100/100? can we count p as 100?

[opal ingot]

Well, x*100/100 is just the same thing as x

[frigid mirage]

yeah

i can't think of any other things...

[opal ingot]

OK, when I said 20% reduction, you said multiply by 0.8. How did you get from the 20% to the 0.8?

[frigid mirage]

1 - 0,2 is 0,8

[opal ingot]

Right! So, if it's p%, what do you multiply by?

[frigid mirage]

by x

and x is equal to 1, right?

[opal ingot]

Well, right, but what do you multiply x by? When it's 20% you multiply by 0.8. When it's 30% you multiply by 0.7. When it's p % (any number for p), you multiply by what?

[frigid mirage]

so if x is ex. 1 and lets say p is equal to 0,50 then:
1x * 0,50p, right? sorry if ur losing patience for me. i am in 7th grade.. and my brain does not process it so fast..

[opal ingot]

No worries 🙂 Let me think if I can explain it another way

[frigid mirage]

aight

[opal ingot]

So, when I give you 30% off, you divide it by 100 and subtract it from 1 to get 1-30/100 = 0.7 right?

[frigid mirage]

yeah

[opal ingot]

So, if you had p in place of 30, you'd get 1-p/100. Do you see how we got that?

[frigid mirage]

yea i can see that

[opal ingot]

OK, so if the price of something starts at x and we reduce it by p%, what's the reduced price?

[frigid mirage]

wouldn't it be just x - p% since we don't know what p is equal to?

[opal ingot]

Well, when it's 30% off, you divide by 100, subtract from 1 and then do what to get the new price? What do you do with the 0.7 you get?

[frigid mirage]

what do u mean what do i do with 0.7? its the equation to the formula.

[opal ingot]

Right, but you multiply it with the original price, right?

[frigid mirage]

u mean 0,1p% by that? 😭

[frigid mirage]

yeah

[opal ingot]

So, the new price after p% reduction would be (100-p)/100 * x. Do you see how we got that?

[frigid mirage]

hmm

100-p is like percent - percent?

[opal ingot]

Sort of, yes

You have to divide by 100 to get back to a number that's not a percent

[frigid mirage]

okay...

[opal ingot]

So, do you see how to proceed from there?

[frigid mirage]

sorry I don't... and after all the explanations u gave me i feel embarassed. don't take me as a bad student please i just don't understand it.

[opal ingot]

No, that's OK. You're not a bad student, but I'm not sure I can explain any better, so feel free to call in other helpers

[frigid mirage]

ok.. but 100-p is x yes? cause we dont know what the equation is

[opal ingot]

x is the original price

[frigid mirage]

alright ty.. ill try to do my best but wait can you please check if chatgpt is doing the equation good? maybe it will explain it somewhat better?

[opal ingot]

I'll wait, but ChatGPT isn't always good at math

[frigid mirage]

yeah i know

[opal ingot]

(reading)

The text after (a) is correct, but the calculation is not. That's now how you increase something by p%

[frigid mirage]

oh okay..

alright well. I guess ill leave it now. But just in case can u give the equation to this? maybe i will understand it better or at school with friends.

[opal ingot]

I'll let you work it out for yourself, sorry

[frigid mirage]

alright, ty for all these explanations. have a fantastic day!