help-28

[hallow walrus]

I got $R_{net}=\frac{3}{2}R,C_{net}=2C$ so the time constant, $\tau=3RC$

[glossy valve]

kheerii

[hallow walrus]

so we will get $i=i_0e^{-\frac{t}{3RC}}$ where $i_0=\frac{2E}{3R}$

[glossy valve]

kheerii

[hallow walrus]

here I used i_0 = E/R_net

am I right so far?

for the first part of the equation, finding the current through the battery as a function of time

I don't really get what the key K closing and the switch S opening at t=3RC signifies

doesn't that just short circuit the whole thing

<@&286206848099549185>

[untold birch]

Hello

Oof

This is hard for me

[hallow walrus]

did u figure it out?

[thick hedge]

try asking on electrical engineeing stack exchnage maybe

[hallow walrus]

hmm

[thick hedge]

https://electronics.stackexchange.com

here

[hallow walrus]

hmm

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[novel gazelle]

I have a math oral coming up soon and this is a theorem for the formula of the distance between a point and a surface (sorry its written in french on the document), and i dont understand how we go from HP • n to || HP ¦¦•¦¦n¦¦• cos(alpha)

[static cobalt]

im just guessing but could it be dot product

[sharp hemlock]

ça me semble faux

les vecteurs HP et n sont colinéaires si je lis bien ton schéma

[novel gazelle]

oui

[sharp hemlock]

alpha c'est quel angle ?

[novel gazelle]

je sais pas justement, c'est un théorème que j'ai copié en classe

[sharp hemlock]

ah, d'accord

[novel gazelle]

ca c'est ce quej'ai ajouté de plus à mon cahier, au crayon

[sharp hemlock]

aaaah

une définition du produit scalaire entre 2 vecteurs u et v c'est u•v=||u||*||v||*cos(theta)

[mild charm]

thats not the definition but something that is true for the standard scalar product of two vectors

[sharp hemlock]

(trahi par le markdown)

[sharp hemlock]

I'm aware, a scalar product is bilinear symmetric etc

it doesn't seem relevant here though

[sharp hemlock]

où theta c'est l'angle entre les vecteurs u et v

[novel gazelle]

ahh okay, pcq y a un autre théorème qui montre ca aussi mais je croyais que ca fonctionnait pas pour tous les vecteurs peu importe leurs directions

[mild charm]

yes but even the standard scalar product is defined as $\sum_i x_i y_i$

[glossy valve]

Maya

[novel gazelle]

woah

i didn't know there was a bot that could do that

[sharp hemlock]

mais ici l'angle entre n et HP c'est 0, donc le cos vaut 1

[novel gazelle]

ooook d'accord

[sharp hemlock]

donc c'est juste norme de n * norme de HP

[novel gazelle]

merci bien
j'ai d'autres questions donc je ferme ce salon et en recrée un ?

[sharp hemlock]

tu peux continuer je pense

[novel gazelle]

Jsp si c'est trop flou l'image

si c'est flou je me connecte sur discord sur mon ipad

[sharp hemlock]

ça va, je te demanderai si j'arrive pas à lire une ligne en particulier

[novel gazelle]

mais en gros j'aimerais savoir comment ou quelles étapes tu dirais pour expliquer la définition de la limite

[sharp hemlock]

comment l'expliquer ?

c'est pour ton grand oral ?

[novel gazelle]

j'avais fait un oral blanc et je suis tombée sur ca, le prof a pas vraiment aimé comment je le démontrais

oui

par exemple pour le graphique je sais pas vraiment comment l'expliquer

[sharp hemlock]

il y a deux choses sur le screen que tu m'as envoyé, la définition de la limite et le théorème d'unicité de la limite

[novel gazelle]

alors juste la définition de la limite avec le tableau

[sharp hemlock]

ok

en gros l'idée de f(x) -> L quand x -> a c'est la suivante

tu peux choisir une tolérance quelconque autour de la limite

ça c'est ton epsilon

et c'est ce qui est représenté sur l'axe des ordonnées par L + eps et L - eps

en gros tu veux contraindre f(x) à valoir L, à epsilon près

et si x s'approche suffisamment près de a

dans un intervalle ]a - delta, a + delta[

alors tu sais que f(x) vaudra L à epsilon près

plus tu choisis un epsilon petit, plus le delta correspondant va être petit

mais il y en aura toujours un

[novel gazelle]

okok

[sharp hemlock]

ça te semble parlant ?

[novel gazelle]

euhh du coup si on peut pas trouver de x dans cet espace c'est que la limite existe pas ?

[sharp hemlock]

je comprends pas la question

[sharp hemlock]

ah

en fait c'est plutôt si tu peux pas trouver de delta

[novel gazelle]

la fonction sur le deuxième tableau a un trou ou je sais pas trop comment ca s'appelle, ca veut dire qu'elle existe pas la limite ?

ah

[sharp hemlock]

dans ton deuxième dessin, tu peux resserrer delta aussi petit que tu veux

tu n'arriveras jamais à restreindre f entre L+eps et L-eps

(au fait, ça s'appelle une discontinuité ou un point de discontinuité)

et en effet, la limite n'existe pas

[novel gazelle]

mais concrètement, c'est quoi la limite ? j'crois que j'ai toujours pas compris

[sharp hemlock]

concrètement, c'est la valeur dont la fonction s'approche

quand tu regardes la fonction autout de a, les valeurs qu'elle prend sont proches de L, sa limite

et elles sont d'autant plus proches que tu regardes près de a

[novel gazelle]

donc quand x tend vers a, sa limite c'est L

mais si x tendait vers b par exemple, L sera à un autre endroit de l'axe y ?

[sharp hemlock]

oui

ça serait une autre valeur de L

[novel gazelle]

ca marche
d'ailleurs je peux t'ajouter en ami vu que tu parles francais ? (mais vraiment pas obligé c'est juste plus simple pcq j'ai plusieurs questions)

[sharp hemlock]

oui, tu peux ^^

[novel gazelle]

ah, je peux pas t'ajouter apparemment

[sharp hemlock]

je l'ai fait

[novel gazelle]

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[novel gazelle]

ah

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[vagrant tangle]

I am stuck on c) for this question. The answer is supposed to be 14.9, I can't see what I am doing wrong or get any other answer. Basically, I solved for y(t) = 20, then solved for x(t), then subtracted 15.

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@vagrant tangle Has your question been resolved?

[brisk obsidian]

The time you calculated is incorrect. Just use the Quadratic Formula to solve for t.

Also, when you Completed the Square, you solved for when (t - 5sqrt(3)/8))^2 = 0, but it doesn't equal 0, it equals 75/64 - 1.

[vagrant tangle]

Did I complete the square incorrectly?

[brisk obsidian]

No. The problem is that you solved for the RHS equal to zero, but on the second to last line you can see that it is equal to 75/64 - 1.

[vagrant tangle]

Oh wait I see

Yeah ok

Thank you

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[viscid thunder]

how would i do part a? i did dy/dx=0 which got me x=-2k and then i subbed in x=-2k into the original equation to get y=4k-8k^2+1 but apparently its 1-4k^2

[naive vapor]

complete the square

[viscid thunder]

(x-4k)^2 - 16k^2 + 1?

[naive vapor]

remember you have to divide b by 2

[polar valve]

you did this wrong: then i subbed in x=-2k into the original equation to get y=4k-8k^2+1

[viscid thunder]

oh ye for some reason i was thinking this and did something different, so (x-2k)^2 - 4k^2 + 1?

is the inside - or +?

[viscid thunder]

why wouldn't the differentiation method work?

[polar valve]

it does.

you made a mistake in subbing

[viscid thunder]

oh ye, 4k^2 - 8k^ + 1 is also -4k^2 + 1

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[pallid sequoia]

hey! i am really struggling with this one and could use some help. Heres the question: Assuming items a and b, show that there is no rational number α which is the solution for "cos(απ)= 1/3"

[final juniper]

hmmm

[pallid sequoia]

the questions from the previous items (a and b) are: a) Let β ∈ R , tan(β) = 2√2. Suppose that for every positive integer n, there are An

and Bn integers, with bn ̸= 0, such that

tan(nβ) = An√2 / Bn.

Find formulas for an+1 and bn+1 in terms of an and bn.

(b) Show that, for every positive integer n, an ̸= 0 and bn ̸= 0. Hint: It may be useful to calculate an

and bn for small values ​​of n.

if it helps, i got the answers for a and b

[final juniper]

show that there is no rational number α which is the solution for "cos(απ)= 1/3"

is this ur quesiton

question

[pallid sequoia]

yess

[final juniper]

well

are you familiar with Chebyshev polynomials

[pallid sequoia]

kind of

[final juniper]

?

[final juniper]

you are?

[pallid sequoia]

kind of

[final juniper]

well

[pallid sequoia]

yep

[final juniper]

there is already a proof online

which uses them to prove

[pallid sequoia]

is it?

[final juniper]

alpha is irrational

ill link it here

https://math.stackexchange.com/questions/398687/how-do-i-prove-that-frac1-pi-arccos1-3-is-irrational

[pallid sequoia]

Thank you! Really helped out a lot!

[final juniper]

np

[pallid sequoia]

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[orchid shell]

how do i solve this volume question?: find the volume of the solid generated byb the triangle with vertices at (1,1), (4,1) and (4,3), rotated about the x axis

[orchid shell]

im not too confident on whatevers better, shell or washer

and idk what to do since it gives vertices rather than a function

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[orchid shell]

<@&286206848099549185> srry for ping, i appreciate any input

[orchid shell]

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[unborn gulch]

@orchid shell did you get the solution?

[orchid shell]

not really, i just tried the chatgpt explanation which idk is correct

it says that since its constant at y=1 and x=4, you can plug those into washer method

after trying that i got 35 pi

[unborn gulch]

Well you will get a cone when u rotate about the x-axis

the 3 points on the graph look like this

now rotating this will cause rotation about vertex (4,3) making a cone with radiue 3 units and height 2 units

now you can solve it

@orchid shell

[orchid shell]

uhhhh

[unborn gulch]

its answer should be 6 pi

[orchid shell]

using washer method?

[unborn gulch]

I dont exactly know about that i just visualised it

[orchid shell]

can you explain ur process pls

wouldnt it be a cone kinda with a hole bored through the middle

[unborn gulch]

i am spinning the line (1,1) to (4,1) circularly

[unborn gulch]

yes exactly what i am saying

[orchid shell]

what was your equation

[unborn gulch]

i didnt get any equation i just visualised rotating the line parallel to x axis which will become the radius and the other perpendicular line as height

@orchid shell before continuing i guess you should reopen this problem

or the help channel will go to the available criteria

[orchid shell]

.open

sry how do i open it

[unborn gulch]

.reopen

[orchid shell]

.reopen

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✅

[orchid shell]

awesome ty

[unborn gulch]

yeah done

[unborn gulch]

...

[orchid shell]

i found a problem in the textbook thats identical, im gonna try to see what the textbook says

[unborn gulch]

okk

[orchid shell]

im having a hard time visualizing this srry, im still kinda new to volumes

i appreciate your help though

[unborn gulch]

well you are rotating along the x axis

so the line parallel to x axis is your base

and line perpendicular to x-axis which remains unchanged will become your height

@orchid shell something like this

[orchid shell]

but what are you plugging these values into

[unborn gulch]

here your BC is line parallel to x-axis

[unborn gulch]

its just an example

i am showing how i am visualising

[orchid shell]

ohh

[unborn gulch]

like this

so ur radius becomes the length of the parallell line

and height remains const

this actually doesnt require you to make equations

[unborn gulch]

this is a better example i guess

[orchid shell]

wait so youre finding the area of the triangle and multiplying it by 2pi?

base 3 * 2 height / 2 * 2pi

6pi

[unborn gulch]

1/3 * pi * r^2 * h

r will become the line parallel to x axis

h will become the line perpendicular to it

because the perpendicular line has no changes while rotating and stays const

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[orchid shell]

hey so i talked with my prof

[unborn gulch]

okk

[orchid shell]

sorry 😭

[unborn gulch]

???

[orchid shell]

he said that the slope of the triangle is found with the point/slope formula

and you subtract that by the inner radius

the correct value is 10 pi

ill send my work in a sec leme get a pic

[unborn gulch]

how is there a inner radius in a single sheet cone?

[orchid shell]

.

[orchid shell]

the outer radius is the slant and the inner radius is y=1

[unborn gulch]

oh i see so you are rotating up and down

i was rotating left and right

i was visualising along the wrong axis mb

[orchid shell]

ohhh alright all good

ty

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[fleet tundra]

why the answer is 1/sqrt(x²+1) and no -1/sqrt(x²+1)

[civic bay]

notice that arctan gives you an angle corresponding to a particular tangent value

so we can say arctan(x) = theta

then tan(arctan(x)) = tan(theta) so x = tan(theta)

this means that there is a right triangle with an angle theta so that the opposite is x and the adjacent is 1

since the tangent is x and tangent is opposite over adjacent

[civic bay]

by the pythagorean theorem, the hypotenuse would be sqrt(x^2 + 1)

and since cosine is adjacent / hypotenuse, you'd get 1/sqrt(x^2 + 1)

[fleet tundra]

look here

here how i can know

@civic bay

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[smoky jackal]

What is a Euclidean triangle?

[smoky jackal]

Lol I should ask google

Sorry guys

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[vale cipher]

How do you do b)

Ik the first two terms are 0, 1

but I dont understand the actual formula itself

[velvet sedge]

if you know the first terms are 0 and 1, this tells you t1=0 and t2=1 yes?

[vale cipher]

yes

[velvet sedge]

so, what happens when you plug in 3

in for n

[vale cipher]

well how do i plug it in....

[velvet sedge]

think of it like a function

[vale cipher]

is it just 2(3-1) + (3-2)

im confused why the n-2 and n-2 r like small at the bottom..

[velvet sedge]

t(n)=2t(n-1)+t(n-2)

this is equivalent

its just notation for sequences

so you know t(1)=0, and t(2)=1

so plugging in n=3

what happens :3

[vale cipher]

RAHHHH IGET 5 IDK

[velvet sedge]

t(3)=

what happens

[vale cipher]

ummmmmmmmmm

[velvet sedge]

when you plug in n=3 into the other side

just make n=3

[vale cipher]

2(3-1)+(3-2)

[velvet sedge]

f(x)=x^2

what is f(3)

[devout valley]

(they're small because they represent the (n - 1)th and (n - 2)th term of the sequence, respectively)

[vale cipher]

9

[velvet sedge]

yes

f(x+1), if I plug in 3, what is it

[vale cipher]

16

[velvet sedge]

mhm

now we know t(1)=0 and t(2)=1

and we have an expression for the function t

the sequence

t(n)=2t(n-1)+t(n-2)

what happens if n is 3

just replace n with 3

in this

[vale cipher]

t(3)=2t(3-1)+t(3-2)

[velvet sedge]

YES

and what happens to the things inside of t()

you can simplify those

t(3)=

[vale cipher]

t(3)=2t(2)+t(1)

[velvet sedge]

and you KNOW t(2) and t(1)

[vale cipher]

.

oh

how do i show that though

[velvet sedge]

its just as simple I wrote it

but instead of parens you can do subscript

[vale cipher]

can u show me

[velvet sedge]

so $t_3=2t_{3-1}+t_{3-2}=2t_{2}+t_{1}$

[glossy valve]

🫎 Chmoosey 🫎

[vale cipher]

ok

[velvet sedge]

just instead of () its subscript

do you understand?

[vale cipher]

yeah

[velvet sedge]

do you see how you would get t_4 and t_5?

[vale cipher]

yea

[velvet sedge]

awesome :3

[vale cipher]

yay

tysm

https://tenor.com/view/happy-cat-cat-kedy-mutlukedy-gif-5853214187617629569

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[desert cypress]

when x is large and negative how did they get a negative answer bec i got positive

oh

wait

hruh

negatives cancel

omg

😭

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[smoky jackal]

Y’all is my answer correct? It’s the only one that makes sense to me but it’s still weird because logically, just because you are wearing a scarf that doesn’t mean it is snowing!!!!

This is geometry class 🥲

[heavy breach]

not quite

if conditional (given) is true, only the contrapositive is true
inverse and converse are not necessarily true

[smoky jackal]

Oh right yes

[smoky jackal]

Would it be “if I’m not wearing a scarf then it is not snowing”?

[heavy breach]

👍

[smoky jackal]

Ty

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[upper silo]

ik this question you need to use composite functions to answer it.... but idk how to start

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[upper silo]

<@&286206848099549185>

any starting tips?

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[young scaffold]

help

[civic bay]

what do you know about corresponding and adjacent angles and stuff

for parallel lines intersected by a transversal

[young scaffold]

idrk

im js tryna whip thru these lessons but this q has me stuck

[civic bay]

notice that if you look at the four angles produced by the intersection of the diagonal line with L, each of the four angles kind of "matches up" with the four angles made by the intersection of the diagonal line with M

for example, the angle labelled 6x + 3 degrees matches up with the top right angle for L

another useful fact here is that opposite angles are congruent, and adjacent angles are supplementary

that's actually everything you need to solve the problem, then this just turns into middle school algebra

[young scaffold]

so whats the equation??

this is the only q i need everything else is done for that quiz

[civic bay]

i understand you're in a rush but maybe when you do problems with the only motivation of getting them done, you don't learn much in the process

the point of the server is to get help, not answers

[young scaffold]

alr

[civic bay]

notice that if that angle is 6x + 3, the angle opposite it is also 6x + 3 since opposite angles are congruent

[young scaffold]

yea

[civic bay]

that angle corresponds to an angle associated with the line L

what equation does that give you?

[young scaffold]

6x+3=8x-31

??

[civic bay]

yes

you can solve that for x to figure out the exact measure of that angle

and notice that that angle and the angle (5y + 35) are supplementary, so they add to 180 degrees

[young scaffold]

x=17

[civic bay]

there's another equation which you can solve for y

looks right so far

so what's the angle

[young scaffold]

how do i find the equation for solving y

[civic bay]

find the angle measure first

of either 6x + 3 or 8x - 31

they're the same angle

and you've found x

[young scaffold]

105

[civic bay]

okay

now notice how that angle is adjacent to the (5y + 35) angle

together the two angles make up 180 degrees

[young scaffold]

14

??

[civic bay]

no

[young scaffold]

hmm

[civic bay]

what's the equation you get

[young scaffold]

5y+35=105??

idk

[civic bay]

the angle on the left is 105, you found that

and notice that the two angles, 105 and (5y + 35), are adjacent on this line

the angles add up to make that big angle, 180 degrees

we say the two angles are "supplementary"

[young scaffold]

8??

[civic bay]

yes

[young scaffold]

so the answer is 8

[civic bay]

yes

[young scaffold]

??

thanks man

appreciate it

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[woeful basalt]

i need help solving the derivative of this

[woeful basalt]

these are examples my professor gave me

we’re solving it the short way with the equations

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[modest rain]

By using the linearizations at the other critical points we find that (2,−1) is a saddle point
and (−2,−1), (−2, −1) is an unstable node. Furthermore you have to use the properties of the other two critical points as found above. The following picture shows the direction fields on squares of length 2 centered around the four critical points.

[modest rain]

how i can i read these graphs

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[cerulean verge]

I have no idea what I did wrong with this question

When i plug the values into my calculator

I cant find my lambda because the answer I got was with decimals

Which is not the answer

[cerulean verge]

wait why is it wrong?

[ashen crescent]

-(3-x)(4) ≠ -12-4x

[cerulean verge]

its cause of the infront

[(-3)(-4) - (3-x)(4)}

(-3)(-4) = 12

(3-x)(4) = 12 - 4x

12 - 12 - 4x

am i wrong?

im stupid im so sorry hahaha which part

[ashen crescent]

It would be 12 - (12-4x)

[cerulean verge]

but i still have to subtract it no?

[ashen crescent]

Because (3-x)(4) are a product

[cerulean verge]

12 - 12

[ashen crescent]

12 - (12-4x) = 12 - 12 + 4x

[cerulean verge]

yeah i know this

so its 0?

[ashen crescent]

No, its 4x, but you have it down as -4x

[cerulean verge]

OH

THERE IT IS

okayokay thank u man

i failed my final because of this question

now i can resit peacefully

thanks man i really love u

no homo

[ashen crescent]

Be careful with signs, long as fuck when you mess up a question because you miss a single sign

[cerulean verge]

yeah

ty

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[zealous rose]

have I done question 4 right?

[hazy lance]

your answer is correct but it seems like you just guessed 16 and checked it

which is totally valid as long as you know how you would solve it if you couldnt guess that

[zealous rose]

ahhh okayyy

i have another question

[hazy lance]

what is it

[zealous rose]

do you need a close up?

[hazy lance]

what was your answer

and what is answer choice E

[hazy lance]

im not following your work here tbh

[zealous rose]

none of the above

[hazy lance]

and what did you get for your answer

[zealous rose]

I didn't work out an answer

[hazy lance]

ok so ill give you a hint

first observe that you have two congruent triangles here

[frosty kettle]

Does this server have a VC?

[hazy lance]

you can subtract these areas from the total area

pretty easy right

[frosty kettle]

A and R are paralell

[hazy lance]

now that youre left with the little slice down the middle, observe that you can drag the bottom part over to the top to form a rectangle

[frosty kettle]

Could you explain me how to do number 4

[hazy lance]

you can set up an equation this way in terms of x, the length of a side

[hazy lance]

get your own help channel

[zealous rose]

omfg

I get it now

I love ya
helped my ass to get some sleep

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[hazy lance]

[frosty kettle]

How? i am new here can you assist me a bit?

[hazy lance]

read #❓how-to-get-help

[zealous rose]

#❓how-to-get-help , after reading go pick an open channel

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[devout spear]

if i throw a ball from 3.05 m and it lands 18m ahead how do i find the angle it falls under just inverse of tan(3.05/18)?

[devout spear]

but some1 told me I have to to the inverse tan of vy/vx

and idk which one is right

they both seem logical

but u get 2 diff answers

it is

its pararel

to the ground

the throw of the ball

no it was on a test today and im calculating my points lmfao

but it just asked the angle

and the throw is directly straight

[buoyant wigeon]

the throw doesnt really matter the ball is on the ground so it's a triangle and what you said works 👍🏼

[devout spear]

yh thats what I did by my nerds said i need to do the inverse tan of final vy/vx

ik lmfao but wait ill sketch it rq

final velocities

the bottom triangle is the final velocities

[spice orchid]

yes you need to use final velocity, if i throw a ball up really high, it'll likely come down nearly straight, so you're expecting an angle of nearly 90. but this isnt the same angle as in the triangle like you've drawn

[devout spear]

damnn

just depressing

ggz then

ending it all

tysm

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[frail karma]

How can I prove that lim x->0 (sen(x) ÷ x) = 0

[fossil river]

do you mean lim x-> 0 sin(x)/x

[proven mango]

it’s not 0 tho

[grave elm]

yeah, sen is spanish / portugal

[frail karma]

Sorry, that is 1

How can I prove it?

[frozen coral]

L'hopital's rule?

[proven mango]

that’s not proving really

i’d recommend searching it up online

[frozen coral]

i mean if he wants to prove L'hopital's rule

[proven mango]

it’s kinda hard to explain it all out here

the reason why lhopital is bad for this is because this limit is required in finding the derivative of sin(x)

[grave elm]

Well this limit is used for finding the derivative of sin(x)

[proven mango]

using lhopital’s rule for this kinda creates a loophole

[grave elm]

Anyway, the proof usually relies on squeeze theorem

[twin wolf]

🥪

[grave elm]

https://www.youtube.com/watch?v=Ve99biD1KtA

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[placid sluice]

(Ignore my choice)

I’m not really sure how this works.

[worldly tapir]

The opposite angles of a cyclic quadrilateral add up to 180°

So you can use angle k and l to find their opposite angles

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[hot wadi]

Isn't this straight up 1/2 x base x height

[glossy valve]

Pro_Hecker

[torn jolt]

Idk

How do I solve the surd thingy

What foil

[bleak adder]

dont we rationalise it

[torn jolt]

I dont

Know

I did the 1/2 x b x h thing

But I don't get how to multiply those

[storm karma]

GCSE Maths goes crazzzyyyy

[bleak adder]

Rationalise

[storm karma]

Do you guys still need help here?

[torn jolt]

YES

Pls

[torn jolt]

How

[glossy valve]

faiyrose

[torn jolt]

uhuh

[storm karma]

Do you know how to expand (x+2)(x-3)? Its the same concept as the above

[bleak adder]

Solve that and rationalize youll get

Like

X+Y 3

X and y = a and b

[torn jolt]

im sorry

im a bit dumb

i dont get how to multiply surds

[bleak adder]

-1 + 9 underrot 3 is answer and

A = - 1 and B = 9

[full forum]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[glossy valve]

~! vil

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As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[bleak adder]

sorry ok

[torn jolt]

thank

you

but

what is the foil method

[bleak adder]

Multiplication method

[torn jolt]

ah

ok

thank u

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[bleak adder]

1/1+log a^bc + 1/1+log b^ac + 1/1+log c ^ ab

Eval.

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[ancient heath]

Hii sorry if this might sound dumb english isn’t my first language, I was studying for my GAT and I encountered this problem,
It tell us to compare between 2 values:
1 - the area of a rhombus with the diameters of 8m & 10m
2- the area of a square that is equal to 25m^2
I’ll attach a picture to how I solve it, but I’m honestly really bad when it comes to geometric shapes. Please help. 🥲

[ancient heath]

I guessed since the diameters are 10,8 that the other side would equal to 6, since (10,6,8) is a famous triangle, so I just multiplied it by 4 to get the area

Wouldn’t that still make the area of the rhombus < 24m^2 ?

Yess

[glossy valve]

faiyrose

faiyrose

[ancient heath]

The original answer from my tutor stated that the area of the rhombus is greater than 24m^2

Ohh I see, thanks!

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[ancient heath]

Yess I just now noticed that you mentioned it, I’m getting brainrots from over studying haha

[worldly tapir]

Wouldn’t it be +?

[radiant trail]

🧌

[glossy valve]

faiyrose

[ancient heath]

Thanks a lot!💖

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[sharp flame]

So regarding (b)

I've done some thinking

Basically the algebraic numbers of "order" 1 require linear equations

order 2 require quadratics

order 3 cubics

and so on

sqrt(3) + sqrt(2) would be order 4, it needs a quartic

So basically any algebraic number can be associated with a certain index, that being a natural number, and thus A_n is countable?

(c) is easy, algebraics are countable, reals aren't, therefore transcendentals aren't

[grave elm]

is that index unique for each algebraic number tho?

(By index you mean the least order of a polynomial such that it's the root?)

[sharp flame]

yeah

[grave elm]

The reals could be countable too then

if i just map them to their integer parts

you need to ensure that the mapping is injective

multiple algebraic numbers cant be mapped to the same natural number

that doesnt show that it's countable

e.g. both sqrt(2) and sqrt(3) are associated with the number 2

this is wrong

[sharp flame]

no its not

[grave elm]

to be able to count the algebraics, you need each natural number to correspond to at most one algebraic number

[sharp flame]

but atleast for the single term ones

they're all gonna be naturals

so you have a union of countable sets

[sharp flame]

yeah sorry i was away for a bit i didnt elaborate

[grave elm]

countable union of countable sets, thats important

but yeah, this explains it

[sharp flame]

If I have a union of infinite countable sets thats not fine?

[sharp flame]

I got it then?

Kind of unsure how I'd write this down formally though

[grave elm]

Yeah

[grave elm]

Form set for each order

there will be countably many of the sets and each will have countably many elements

[sharp flame]

Right sounds simple enough

[grave elm]

mhm

[sharp flame]

I think I can do it now, thanks man!

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[grave elm]

np

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[ebon ibex]

how do i quickly calulate the eigenvalues of this matrix

[ebon ibex]

i need them for to check if the function is definite or indefinite

i tried going - lambda on the diagonal and use the rule of sarrus

but the solution of that was 3 lambda -8

no idea what to do there further

[vast fossil]

Show your work

[ebon ibex]

(1-λ) x (1-λ) x (4-λ) +2 +2 - [(1-λ) x (1-λ) x 4(4-λ)]

this is what i get from using sarrus

how do i get the eigenvalues out of this

Like calculating that is 3λ-8

I want to get the 3 eigenvalues

[vast fossil]

Are you sure you are applying the rule correctly? That's not the characteristic polynomial

[ebon ibex]

oh sec i made a mistake in the second part

they should be +

(1-λ) x (1-λ) x (4-λ) +2 +2 - [(1-λ) + (1-λ) + 4(4-λ)]

like this

hmm

result is still complicated how do i get eigenvalues from that

[split swift]

expand it into a cubic then try factor/remainder theorem to determine linear factors

if it's an exam question o.e. that'll generally work, as long as you're not given non-integer eigenvalues

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[compact mulch]

if m and n are positive integers ,prove that m root n is either a positive integer or irrational.

[compact mulch]

i’m sort of confused, because if m=2 and n=8, then we have 2 root 8 (the square root of eight), which is neither a positive integer nor an irrational number.

[minor crater]

why is it not irrational?

is it rational?

[compact mulch]

i’m pretty sure

i could be wrong

[minor crater]

can you write it as a fraction?

[compact mulch]

i’ll check

oh it is irrational

that’s my bad

[minor crater]

there you go

[compact mulch]

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[edgy yarrow]

I am trying to prove this formula for even powers of cosine. Clearly I am going wrong somewhere, I suspect the issue is where I take the real parts of the equation, but I’m not sure exactly how.

[fiery fern]

Answer?

[viral jasper]

<@&268886789983436800>

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[edgy yarrow]

<@&286206848099549185> please help

[viral jasper]

You have it right so far. You need to keep going

[viral jasper]

Try n=2 and n=4 to see what you need to do next

[edgy yarrow]

Okay, will do. I assumed it was wrong because the LHS is the same but the RHS isn’t

[viral jasper]

Nah, so far you been right

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.reopen

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✅

[edgy yarrow]

I've tried substituting in n=2,3,4,5 and all of them work. So now I'm wondering why I can't find my version of the formula online anywhere, especially because it's so simple compared to the one I was trying to prove. And the one I was trying to prove is for even n, but my equation seems to work for all n.

Surely I'm missing something

[edgy yarrow]

Have I invented a formula here? <@&286206848099549185>

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[round plank]

can sm1 explain what func g(x) i use for the fixed point method
im kind've lost
i tried g(x) = 25/x^3
not sure waht else i can do can any1 help?

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@round plank Has your question been resolved?

[plush tundra]

Your g would be 25/x^2 since x = 25/x^2 is equivalent to x^3 = 25 not your g. However the derivative of this new g is likely too big to converge nicely. Can you think of other functions such that x = g(x)?

Hint: perhaps you can introduce another factor/fixed point

[round plank]

not really

[plush tundra]

It will work if you add a factor of (x-4) so you are solving (x^3-25)(x-4)=0 between 2 and 4

[round plank]

25/x - x^2? idk

[plush tundra]

Does this have the same solution and bring g’ below 1 in your area of interest?

[round plank]

what?

how r u getting x-4

[plush tundra]

I worked backwards

Start by adding a linear factor x - a and write out the expression for g

Then pick your starting point (3 is good), and try some values to get g’ as low as possible

In this case it is clear a=4 ensures |g’(3)=0| which is as good as you can get

[plush tundra]

Starting with a guess of a linear factor makes sence since it is the simplest modification to f you can make

[round plank]

this makes no sense

to me

[round plank]

what are u saying

the way i get g was just solving for x

whats the easiest way to do it

[plush tundra]

I’m trying to paste a screenshot

The first 3 steps transform a function that has 25^(1/3) as a solution into x = g(x) form

Then g has a fixed point ar the solution by definition of fixed points

And you can iteratively apply g to some close starting value (in this case 3) to get your approximation

3 steps should be enough for your desired accuracy

I chose x-4 by first doing the first 3 steps with x-a and writing the formula for g’

Then a=4 minimizes |g’(3)| so it will converge fastest

And since the derivative is 0 which is less than 1 it will not diverge

So your idea to solve for x is correct but you need to solve a function that yields a g’ with low absolute value to ensure convergence

Introducing another factor is a good way to do this

[round plank]

why r u randomly adding x-4

i dont understand this

is there not an easier way to do this

[plush tundra]

It’s the same method as yours. Take a function f with root 25^(1/3) and solve for x to get g

[round plank]

no

[plush tundra]

However your f(x) = 25 - x^3 does not work

Because the resulting g has too large a derivative to convergw

[round plank]

this is how i do it

idk where ur getting (x-4) from

[round plank]

i see

but where r u getting (x-4) from

[plush tundra]

We need to modify f to keep the root but add some other root(s) to minimize |g’|

I guessed that adding a linear factor (1 root) should do the trick

So I solved for g with an extra x-a factor

Then solved for the a that minimizes |g’| which 4 does

There are plenty of other modifications that work, x-4 is just the one I chose

A good hint that 4 as a root will work is that we can tell the value is between 2 and 3 and hence can use 3 as the starting value. Then 4 is further from the starting value than the solution so the fixed point we find will be the desired solution and not the fixed we added at 4

[round plank]

or i can just use newtons method as an equation

[plush tundra]

That should work

[round plank]

x - f(x)/f`(x)

[plush tundra]

Newtons method is just a special case of fixed point iteration

[round plank]

what is f(x)

that i use

x=25/x^2?

[plush tundra]

f(x)=x^3-25

This is the function that has the solution that you are after

[round plank]

ok so

x - x^3-25/3x^2

is what i get

is that correct?

[round plank]

i was confused how sm1 got 1/2(x+25/x^2)

[plush tundra]

Assuming you meant to place parenthesis around the numerator then yes

[round plank]

how did this person get 1/2(x+25/x^2)

then

doesnt amtter

wtv

[round plank]

how would i determine the interveral

to tell if it converged i just take derviative of f(x) and plug in something to see if it is less than 1 right?

[plush tundra]

Yes

[round plank]

so if it coverges g(0)

and g(1)

i could say interval 0,1?

[plush tundra]

Depends on g. If g’ is 2-x^2 for instance, that reasoning won’t work

It is better to calculate the derivative for g and check when it is less than 1 in magnitude

[round plank]

how

[plush tundra]

Differentiate g and find an interval by looking at the inequalities

Solving g’(x) < 1 can be done by finding the zeros of g’(x) - 1

[round plank]

bro

can u explain it better

like not complicated

i dont understand what your trying to say

and also

[round plank]

for this someone did this isntead is this correct?

[plush tundra]

They are equivalent

They just did some algebra to the same answer (and have a misplaced equality between f and g at the beginning)

[plush tundra]

Are you familiar with derivatives? If so, find the derivate of your g(x)

Are you familiar with solving inequalities?

[round plank]

i did

[plush tundra]

Ok, then solve |g’(x)| < 1

[round plank]

u cant i dont think

[plush tundra]

You just need an interval that satisfies it

Also, what is your g’?

[round plank]

(2x-e^x)/3

[plush tundra]

This one works in [-1, 1] for instance. You just have to look at g’ - 1 and see when it crosses 0

[round plank]

how do i solve

for this

-3<2x-e^x < 3

[plush tundra]

Think about the shape of g’, what does that tell you?

[round plank]

i cant shape this out tho 💀

no idea how it looks

[plush tundra]

Well, when x is negative, the linear factor dominates

[round plank]

how do i do this

without like graphing it or smth

[plush tundra]

When x is positive e^x dominates

[round plank]

yea

[plush tundra]

My wifi is not really working rn, messages take many minutes to send. I suggest you put it into an online algebra calculator and look at the steps. It’s just an inequality

[round plank]

doing that

yeah i dont get any results

,wolgram

,wolfgram

[glossy valve]

Results provided by WolframAlpha

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

Requested by ramb00z

[round plank]

,wolf \left|\frac{2x-e^x}{3}\right|<:1

[glossy valve]

Results provided by WolframAlpha

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

Requested by ramb00z

[round plank]

this is so confusing lmao

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@round plank Has your question been resolved?

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@round plank Has your question been resolved?

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[dusky drift]

can someone help with this?

I have the answer key for it but I am not exactly too sure on how they got their answers

like why did they do half cirlces etc

[sick karma]

because integrals are just essentially measuring the area under a curve

in (1) for example, you are integrating from 0 to 2, so you are finding the exact area that is shaded between the function and the x axis from x = 0 to x = 2

[dusky drift]

right but how is a half circles below x axis -pi/2?

[sick karma]

because remember, any area under the x axis is considered negative

[dusky drift]

no yea I know that but the pi/2 specifically

like did they use a unit circle here?

[sick karma]

what is the radius of the semicircle?

the one from x = 0 to 2

[dusky drift]

I dont remember that rlly but If I had to guess its prolly pi/2 huh

[sick karma]

are you sure? the radius?

well think about it this way

it's a half circle, and its centerline is 2 units long because it ranges from x = 0 to x = 2

[dusky drift]

ok

[sick karma]

what is another name for this centerline?

[dusky drift]

Not sure to be completely honest

[sick karma]

ring any bells?

[dusky drift]

diameter?

[sick karma]

yes

so the centerline of the entire semicircle is the diameter, and the diameter has length 2 units

how do you find radius from diameter?

[dusky drift]

d=2r?

[sick karma]

yes; so do you multiply or divide diameter by 2 to get radius?

[dusky drift]

multiplying

[sick karma]

no

[dusky drift]

oh I see cuz d=2r is finding the diameter so we do d/2=r to find r right?

[sick karma]

yes

because remember

radius is defined as the length from the center to the circumference of a circle

[dusky drift]

yea

[sick karma]

so radius will always be less than diameter

so anyway, your diameter is 2, what would your radius be for the semicircle?

[dusky drift]

pi?

[sick karma]

?

where did you get that from

[dusky drift]

just a wild guess to be honest

[sick karma]

well, refer to this equation you just wrote out yourself

your diameter is 2

what is your radius?

[dusky drift]

d/2?

[sick karma]

what is the value??

[dusky drift]

Idk Im lost

[sick karma]

alright, what does the letter d represent in that equation?

[dusky drift]

diameter

[sick karma]

and we just specified what about the diameter?

[dusky drift]

that it is 2

[sick karma]

so, d = ?

[dusky drift]

2

[sick karma]

now, r = d / 2 = ?

[dusky drift]

are u asking what r is or what d/2 will be?

[sick karma]

we just said that r is equal to d/2 no?

[dusky drift]

yea

[sick karma]

then using this information, what is r equal?

[dusky drift]

2/2 or 1?

[sick karma]

yes

r = 1

[dusky drift]

ok

[sick karma]

the radius is equal to 1 unit because it is half of the diameter which is equal to 2 units

the radius of the semicircle is equal to 1 unit

[dusky drift]

ok

[sick karma]

now that you have the radius of the semicircle, what is the formula for the area of a circle?

[dusky drift]

A=(pi)r^2?

[sick karma]

yes

however, the area is a semicircle, not a circle

a semicircle is a circle that is cut in half

therefore, what is the formula for the area of a semicircle?

[dusky drift]

A=(pi)r^2/2

[sick karma]

yes

what is r?

[dusky drift]

1

[sick karma]

so plug it in

the area of your semicircle is?

[dusky drift]

(pi)*1^2/2= pi/2

[sick karma]

there you go

[dusky drift]

A=pi/2

but its A=-pi/2 since its under x

[sick karma]

yes

[dusky drift]

ok that makes more sense

sorry for your trouble

Imma try and do the other ones and see if I get em right @sick karma

[sick karma]

alright

[full forum]

@dusky drift Has your question been resolved?

[dusky drift]

this is what I got but its wrong

@sick karma

Oh wait I figured it out

[dusky drift]

it matches the same answer as the answer key but lmk if its wrong @sick karma

[full forum]

@dusky drift Has your question been resolved?

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• Show your work, and if possible, explain where you are stuck.
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[echo prism]

Can someone explain how to do this??

[hoary ember]

it wants you to find which is the graph right?

[echo prism]

Yeah

[hoary ember]

notice that the constant infront of x^2 is positive, so the graph will be some sort of u shape

[echo prism]

Ok

[hoary ember]

and you can try to plug in some random value x

like take x=3, tmif you plug it in you get y=3

you can easily see which one is correct

[echo prism]

???

I dont understand

Hold lemme show u what I have

[brittle steeple]

So if x=3, then y=3

[echo prism]

??

nvm thanks anyways

.close

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[glossy valve]

kisnar

kisnar

[granite torrent]

can't simplify like this

yes

degree of numerator is greater than or equal to degree of denominator

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[junior geode]

I’m confused about row 3 of E

You subtract row 1 from row 3 and then you subtract row 2 from row 3

i was thinking it should be [ -l 31 , -l 32, 0 ] for row 3 of E