help-23

[tame zephyr]

ahhhhhhhhhh

i just realized

now

i only have one day to study for my real analysis 2nd midterm

😱

what did i do

the lin alg quiz ive got isnt even worth that much

holyyyyyyyyyyyy

ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

[junior smelt]

https://tenor.com/view/shiet-angry-gif-13515034

[tame zephyr]

fckkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

[junior smelt]

Pain

[tame zephyr]

okok

so plan is its already too late anyways rn im gonna sleep soon

so

ill finsih up lin alg, skim through the next chapter💀

take quiz

and then cram

ahhhhh what was I doing the few days before???

i dont think i was completely wasting my time??

i cant remember

[junior smelt]

Same I really should've rested sooner

[junior smelt]

Good question you did do a bit I think

[tame zephyr]

😭 yea... prob😭

is the equivalence of dimension of the two basis enough to show that its isomorphic

sweet

[junior smelt]

Yep pretty much in any case it's easy to create an actual isomorphism between them (e.g. send the diagonal entries to a corresponding vector)

[tame zephyr]

I cant just skim through the next chapter..

its too much🥲

[tame zephyr]

no

just

no

[junior smelt]

what's that referring to?

[tame zephyr]

alright Im gonna go to sleep

yea

good night

[junior smelt]

Goodnight sleep well

[tame zephyr]

.close

[safe radish]

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[versed wave]

In a 3D coordinate system (Oxyz) given the sphere (S)

[versed wave]

<@&286206848099549185> reasking cuz i was not answered

if there is anything you need me to provide im more than welcome

[safe radish]

@versed wave Has your question been resolved?

[versed wave]

<@&286206848099549185>

[vagrant night]

Hello Ill try my best to help.

What exactly is the question

[versed wave]

this

[vagrant night]

Is that the whole question

[versed wave]

yes

[vagrant night]

So what are u trying to solve

[versed wave]

oh i forgot

[versed wave]

find thr radius of this circle

[vagrant night]

alr Ill attempt the question now

[glass flame]

!show

[safe radish]

Show your work, and if possible, explain where you are stuck.

[glass flame]

i apologize for that

[vagrant night]

So I've done the working on a different program to ensure that u can read and understand it better bc my handwrighting is terrible.

[vagrant night]

So Im assuming that the first 3 photos were parts of the question and then u wanted me to find the radius of the sphere.

I got a different answer but I hope that helped

@versed wave U there ?

Hello

[safe radish]

@versed wave Has your question been resolved?

[versed wave]

you didnt do this

this is chatgpt

[versed wave]

it was a test

but my work was questionablr

so thats why im asking

[vagrant night]

Yes I used Chat Gpt. That was the different prgogram. I didnt quite get it so I thought it would help u.

[versed wave]

!nogpt

[safe radish]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[versed wave]

i appreciate your spirit, but i suggest you should know what you are truly doing

[vagrant night]

I apologise, but did it atleast help u in the slightest. Its fine if I get banned or kicked for using ChatGpt. I just felt bad when I saw that nobody answered ur questioned so I thought GPT wil help. Once again I do apologise for everything but did ChatGPT atleast help u in some way. PLs tell me it helped in some way.

Like was Gpt correct

TBH bro Im a y12 Student I got no clue abt this. I just felt bad when there was nobody answered ur question and knowing I ceratinly can't I used Ai.

[versed wave]

uh no not really since i dont trust chatgpt in a slightest

[left gyro]

you forgot to say that M is on the sphere

[versed wave]

chatgpt cant even do a 5 term sum correctly, i aint trusting it

[vagrant night]

ye it's not a reliable resource. ISTG it even makes up numbers putta knowhere

[versed wave]

oh yeah, what the question meant was the locus of all point M such that...

it doesnt necessarily belong to the sphere i believe

[left gyro]

also is the * in there dot product of the vectors MB and MC or multiplication of the lengths MB and MC

[versed wave]

dot product

so essentially i label M(a,b,c), after expanding iirc i got something along the line of

$a^2+b^2+c^2 +... = 4$

[flat frigate]

FungusDesu MSC2020 34A05

[versed wave]

it was extremely similar to a sphere equation

but i cant figure out what to do with it

[left gyro]

Ive graphed the possible coordinates for M, assuming M does not need to be on S, and the locus of M is ||the surface of a sphere||

[left gyro]

you need to say more than just that

if it had the form of a^2 + b^2 + c^2 + ...ab + ...ac + ...bc + ... = 4, where ... are coefficients, then it is a sphere equation and you just didnt notice

[versed wave]

thats what im confused at, the question did not specify M must belong on the sphere, yet the locus is a sphere

[left gyro]

no

[versed wave]

either way, would just finding its radius suffice?

[left gyro]

the locus is a different sphere

it cant even be the same sphere

the locus if M must belong on a sphere is a circle, because a circle is the intersection of the surfaces of 2 spheres

[versed wave]

hmm

so lets say we want to find the radius of the intersection circle of the two spheres

how to do so

i might redo this problem when i get home

[left gyro]

typo earlier
a^2 + b^2 + c^2 + #a + #b + #c + G = 4 where # can be any coefficient, and G is sufficiently negative, is guaranteed to be a sphere

you should try this problem again now that you know that the locus of M (ignoring that M should be on S) is a separate sphere whose equation you can find

you should get that the equation of the locus of M (ignoring that M should be on S) is (x - ||1||)^2 + (y - ||1||)^2 + (z - ||0||)^2 = ||3||^2

seeing both the equation equation and the original S equation (x - 3)^2 + (y - 3)^2 + (z - 2)^2 = 3^2 should give you an idea on what their intersection must geometrically satisfy

remember that you only need the radius of the circle, you do not need an equation of the circle in 3D space

[versed wave]

alright brb, i just got home and gonna redo this problen

[safe radish]

@versed wave Has your question been resolved?

[safe radish]

@versed wave Has your question been resolved?

[versed wave]

solved! sqrt6 was indeed the correct amswer

.close

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[versed wave]

@left gyro thank you

[safe radish]

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[tough ravine]

<@&286206848099549185>

[tough ravine]

hi any idea how to solve this

[rose plume]

dya know surface of revolution formula?

[tough ravine]

no

[uncut skiff]

do you even need that though

rotating the circumference just gives a spherical shell

[tough ravine]

its a semi circle

so we have to use pi

[uncut skiff]

basically it just wants the surface area of the sphere that you get after rotating

[tough ravine]

ye

[mental dew]

I can help if needed

[tough ravine]

sure

[mental dew]

are you trying to do this question using calculus

[tough ravine]

ye

i think it need integral

[uncut skiff]

is it really not just a spherical shell?

[tough ravine]

from -2 to 2 integral and subtract eachotehr

thats my guess

[mental dew]

do you know the disk/washer method for calculating area

[tough ravine]

no its a semi circle

[tough ravine]

idk

[uncut skiff]

but the semi circle rotated becomes a sphere

[rose plume]

ryan you're correct

[tough ravine]

its a 2d circle

[rose plume]

it's just the AREAf iaw90-kj f0-a wvk0-otk 0aw-powst of the sphere

[uncut skiff]

try visualising the rotation

[tough ravine]

what do you suggest?

[tough ravine]

im not allowed to use a calculator

so i have to do it pure on numbers

[magic owl]

it would still be a 2d semi circle if it is rotated about z axis

[uncut skiff]

that isn’t even the calculator doing the rotation

that’s just the plain circle with some extra lines to indicate that it’s 3D

[mental dew]

the equation tells you the hight at every moment, so if you create a circle at ever dx and add the together then you get the surface area.

you can do this by using integrals

[tough ravine]

yeah thats what i suggested. but since its circle do i need to multiply by pi?

[magic owl]

why can't we just find the area of the entire circle and just divide it by 2?

[uncut skiff]

If you really wanted to evaluate it with integrals it would be integrating pi times (f(x))^2

[mental dew]

correct surface area = (pi) * integral( the hight eqution )

[tough ravine]

wait pls

[uncut skiff]

but like, why would you evaluate it with integrals if you could just pull out the formula for surface area of a sphere

[mental dew]

not every equation has a standardied equation

[uncut skiff]

that’s just intentionally making your life harder

[magic owl]

can someone please answer this

[uncut skiff]

generality is good but we don’t have to derive the same thing every time lol

[tough ravine]

guys

[uncut skiff]

the answer is just 16pi

[tough ravine]

[rose plume]

know what better idea

1. is it mandatory to use surface of revolution formula?

[tough ravine]

no

[rose plume]

ok then ryan's idea is better

you basically have half perimeter of a circle rotated 360 degrees

that's a sphere

[tough ravine]

its not a sphier

[rose plume]

why not

[tough ravine]

BECUASE ITS NOT

its a semi circle

circle is diffrent then sphier

[rose plume]

it is asking you to rotate the semi circle around the x-axis

[rose plume]

like here

[tough ravine]

rotate x axis

not z

[rose plume]

wat

ok that's the x axis?

[tough ravine]

oh fuck im an idiot

[rose plume]

so it is indeed a sphere

[tough ravine]

wait

no

[tough ravine]

its a circle

a sphier will move it y axsis

[rose plume]

wat

[tough ravine]

oh yeah i just checked you might be correct, since the document i found said the amount of area that been rotated

[tough ravine]

yeah your correct

im sorry for making this more complicated

[tough ravine]

friend?

[rose plume]

?

[tough ravine]

so um how do you solve it now we found out its a sphier?😅

[rose plume]

surface area of sphere

[tough ravine]

ok

[rose plume]

V = 4pir^2

[tough ravine]

so what about the x^2+y^2=4?

[rose plume]

that's half of a circle

for y >= 0

[tough ravine]

oh okok

[rose plume]

upper semi circle

you rotate the upper semi circle 360 degrees

thats a sphere

[tough ravine]

ok thanks

[safe radish]

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[oak violet]

does my logic look sound?

[safe radish]

@oak violet Has your question been resolved?

[oak violet]

.close

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[winged bolt]

Texit broke my post so here it is in all its awfully-formatted glory again

[winged bolt]

The bottom was saying "even though this definitely represents the capacitance at a given area covered by a dielectric, I feel like I'm supposed to integrate this, but what would the resulting sum mean?"

Anyway I just feel like I'm supposed to integrate this because of the way it comes together, but in the end I'm actually supposed to be finding the current $I=C \frac{dV}{dt}$, so I'm probably not actually supposed to

[flat frigate]

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

[winged bolt]

texit please

I guess the integral of velocity isn't "the sum of velocities", but rather the displacement, so I guess I'm just confused on the whole "sum" aspect of integrals
Now that I know that taking the integral isn't correct, though, I can search for my actual question instead of asking it here

.close

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[cinder cipher]

Hi can someone help me with the euler method for differential equations?

[cinder cipher]

i dont understand what to do to calculate h when i have de exact solution of the problem

i endeed up with this, but dont know how to continue

should i replace x(t) for the exact solution or what should i do?

[flat frigate]

svc20060505

svc20060505

svc20060505

svc20060505

[cinder cipher]

Thanks for translating!

[rose plume]

it just says bound* in Spanish. I guess it would be an upper bound(?) Sorry, I don't know about ODE's

[cinder cipher]

hahah no problem

thanks for the translation

its okey if i stay here till i found help or what would you recommend?

[rose plume]

in 3 minutes you can ping helpers

[cinder cipher]

great! How can i do that?

sorry for my english i dont use it pretty often

[flat frigate]

svc20060505

[cinder cipher]

its upper, sorry i didnt understand you before

i want to calculate the maximum possible error

how can i ping?

[rose plume]

<@&286206848099549185>

[cinder cipher]

thanks

<@&286206848099549185>

[rose plume]

no se de esto pero depronto puedo

te dieron el tamaño del paso q tienes q tomar?

[cinder cipher]

nop, lo tengo que calcular sacando la cota creo, porque de ahi puedo despejar h que me da la cantidad de pasos indirectamente

hablas español? jajajaj

[rose plume]

sí

espera

q significa h?

[cinder cipher]

jajaja y yo hablando en ingles como un boludo

espero

ahi te doy la definicion

[rose plume]

ok

[cinder cipher]

me descoloca un poco aca porque no tengo un tiempo final

asi que supongo que primero tengo que calcular el error

que es este

calcule la derivada segunda

pero despues estoy perdido con el tema de que acotar o como reemplazar

[cinder cipher]

.

[rose plume]

si efectivcamente esto esta en chino me rindo

[cinder cipher]

jajaj

no pasa nada

se agradece la intencion

sabes si hay alguien por aca que sepa del tema para mandarle?

<@&286206848099549185>

sigo spameando esto hasta que llegue alguien jajaj

[rose plume]

si pero para q lleguen

[cinder cipher]

jajaj a quien apuramos? ah

[rose plume]

creo q ya entendi

$|x(t)-\text{approx}| \leq |x''(t)|\frac{h^2}{2}$ ?

[cinder cipher]

claro

una cosa asi

[flat frigate]

svc20060505

[cinder cipher]

onda la diferencia esa vendria a ser mi error

que es igual a lo que viene despues de la inequidad (no se como se dice)

onda el segundo termino vos queres que sea menor o igual a ponele en este caso 10^-2

tengo calculada la derivada segunda pero ahi me quedo porque no se bien que datos tengo que usar para reemlazar x(t)

[rose plume]

puedo ver como obtuviste esa cosa rara de 4t2x(t)-2x(t) o algo asi

[cinder cipher]

ahi te muestro

[rose plume]

creo q el t desaparece porque es t = 1 $|x(1)-\text{approx}| \leq |x''(1)|\frac{h^2}{2}$

[flat frigate]

svc20060505

[cinder cipher]

import numpy as np
import matplotlib.pyplot as plt
import sympy as sp

x, t, x_sym= sp.symbols ("x t x")
x_func= sp.Function("x")

funcion= 2x_func(t)-5sp.sin(t)

def euler_error_truncamiento(funcion):
f_subs= funcion.subs(x_func(t),x_sym) #Me permite derivar mas facil
ft= sp.diff(f_subs, t) #Calculo derivadas parciales
fx= sp.diff(f_subs, x_sym)
df_2= ft + fx*funcion
df_2= df_2.subs(x_sym, x_func(t))
return df_2

euler_error_truncamiento(funcion)

siii

el t si medio que se como sacarlo

pero x(t) me mata

[rose plume]

x(1) sería e^-(1)^2 ?

[cinder cipher]

yo estaba pensando en eso

pero no siento que tenga mucho sentido

que use esa funcion para aproximaer

[rose plume]

nose jajjaj

[cinder cipher]

jajaja

yo tampoco

onda trate de resolverlo asi

y siento que tiene algo de sentido

pero no me termina de cerrar

bah no se en realidad

para vos tiene sentido?

[rose plume]

pues siguiendo la definicion seria asi

[cinder cipher]

si no?

tiene sentido

supong

[rose plume]

ah no lei mal

[cinder cipher]

si alguien mas lo piensa jajaja

F

[rose plume]

seria t0 + nh

[cinder cipher]

si

[rose plume]

sabes q ignora esto

cierralo y vuelvelo a abrir para q quede todo mas ordenado haces la pregunta desde 0

con .close y .reopen

[cinder cipher]

dale muchas gracias

puedo copiar lo que hcisite como latex?

[rose plume]

ya te lo paso

[cinder cipher]

gracias!

[rose plume]

Consider $x'(t) = -2tx(t)$, $x(0) = 1$, $t \geq 0$, where the exact solution is given by $x(t) = e^{-t^2}$

a) Using Euler's method to approximate x(1), find an upper bound in terms of the step h for the error incurred in the approximation.

b) What should be the value of $h$ to ensure |error| < $10^{-2}$?

c) Compute the solution at $t=1$ using the value of $h$ from (b), and compare the values of the approximation and the exact solution.

[cinder cipher]

Gracias !!!

[rose plume]

para q te atiendan mas rapido has esto

[rose plume]

manda esta foto

[cinder cipher]

que?

[rose plume]

luego esta

y luego el latex

y mejor cierra este canal y abrelo en otro distinto

[cinder cipher]

jajaj bueno

hago eso entonces

muchas gracias!

[rose plume]

dnd

[cinder cipher]

.close

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[crystal summit]

is this done correctly? Evaluating limits with more than 1 variable and trying to prove they dont exist Thanks!

[crystal summit]

i wrote y^2 at one point should be y^4

nvm

noticed what i did wrong -> plugged in 1 instead of 0

.close

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[still patrol]

Can someone help me with 13-17? I wasn’t here when we learned this and I got an 11/45 on the test, and it’s on the quarterly tomorrow. I think I can do the rest after getting help with the first couple

[fallen grotto]

@still patrol do you have anything so far for those problems. Otherwise I might just have to type it out

also knowing the quadratic formula would be helpful on some of these

[still patrol]

no, im not sure where to start

[still patrol]

ik that

[fossil forum]

Am I just tired or is the answer for 13 not listed?

[still patrol]

I didnt know it would help but I know the quadratic formula

[still patrol]

I couldnt tell ya what the answer for 13 is
How would I use quadratic for 13 if there is no a b or c

[fallen grotto]

get everything on the left first

[still patrol]

8x^3-8x

[fallen grotto]

when they ask you to solve for x an equation, they're usually asking for what value(s) of x does the equation equal to 0.

worst case, you plug the values given if you don't know what to od

for this, factor out an 8x first

[still patrol]

rn im at 8x(x-1)

so +1 would be an answer I think

maybe 0 cuz of the 8x

[fallen grotto]

exactly

so B

moving on

[still patrol]

Wwait where did 8 come from

and where did 0 go

oh wait nvm

I thought C said B

[fossil forum]

You get 8x(x^2-1) which is difference of 2 squares, so 8x(x+1)(x-1) so the answers are 0, -1, +1

I think they typo'd A and meant to write 0 not 8

[fallen grotto]

thanks for the catch, i thought they factored properly

@still patrol for 14, let x^2 = y and substitue y in. Solve for y using the quadratic formula, then solve for x

[still patrol]

im so lost

[fallen grotto]

15: factor out the s, use quadratic and consider the extra s you factored out

ok $a^4 -17a^2 + 16$

[flat frigate]

Fuzzy_Alpaca

[fallen grotto]

you just "see" that it factors into

$(a^2-16)(a^2-1)$ or let $x = a^2$, so you have $x^2-17x+16$

[flat frigate]

Fuzzy_Alpaca

[still patrol]

i'll try to ask my teacher about that one tomorrow because I am still confused

[fallen grotto]

and factor $x^2-17x+16$ via the quadratic formula to get the zeros: x=16,x=1. This gives you $a^2 = 16$ and $a^2 = 1$, so $a = -4,4,-1,1$

[flat frigate]

Fuzzy_Alpaca

[fallen grotto]

ok

[still patrol]

im going to try 15 and I will send a picture of what I have

[fallen grotto]

@still patrol 15 I already mapped out. 16 is straightforward. 17 is a toughy. $a^3+b^3 = (a+b)(a^2-ab+b^2)$ and $a^3-b^3 = (a-b)(a^2+ab+b^2)$. Use the first property, and it's good to know off the top of your head with the second property. Either memorize it or figure it out on the spot.

[flat frigate]

Fuzzy_Alpaca

[still patrol]

whats first and second property?

[fallen grotto]

the two aforementioned equations

[safe radish]

@still patrol Has your question been resolved?

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[crystal badger]

im trying to find the eigenspace of this matrix and this is what i have so far. what do i do next?

[crystal badger]

and when im doing the nul space am i supposed to bring the matrix to REF or RREF?

[light shoal]

you can row reduce if you want to, but you only have one nonzero row so you can just work with it directly

2x - 4y = 0

where x,y are the components of an eigenvector

you just need a nonzero solution

[crystal badger]

so i just need 1 row of only 0's in which i dont have to continue the nul space?

and what do i do with the C1 and C2 thing?

[light shoal]

the row of only 0's tells you nothing, it's just 0=0

well using your C1/C2 notation, it's:
2C1 - 4C2 = 0 ,right?

any C1/C2 that satisfy this (as long as they're not both zero) is a valid eigenvector

so just pick a number for C2 and find the corresponding C1

[crystal badger]

but i thought you're supposed to use numbers from the last matrix
like x -2y = 0

[light shoal]

that's the same thing

as 2x - 4y = 0

one equation is just 2 times the other equation

they have the same solution set

[crystal badger]

so i can use numbers from the 1st or 2nd matrix in any case?

[light shoal]

yes

either way, there are infinitely many valid solutions (any nonzero scalar multiple of an eigenvector is also an eigenvector)

that is reflected in the fact that you have two unknowns but only one equation

so you have one degree of freedom

[crystal badger]

ohhh okay

thank you

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[prime mason]

can someone tell me how this is wrong?

I flipped the answers and its still wrong

like I wrote it as -0.89 and 1.33 and that showed up wrong too

.close

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[frigid locust]

maybe try fractions

the answer above is in fractions

[prime mason]

I figured out that it doesn't want us to approximate it

.close

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[solar gust]

help in this

[restive niche]

What have you tried

[solar gust]

tried by using L'Hopital rule but doing some error

my limit is not existing

[restive niche]

What did you get after using lhopital

[solar gust]

w8 sending

[restive niche]

How did you get 1/(24x^2+3)?

Thats wrong

[solar gust]

ya I also thought that

[flat frigate]

smidgin

smidgin

[solar gust]

yes how to calculate it

[restive niche]

What is the defining property of an inverse function?

[solar gust]

the function should be bijective to have inverse

[restive niche]

Yes

[flat frigate]

smidgin

[solar gust]

yes

[restive niche]

Let's make use of this fact to find (f^{-1}(x))'

Just differentiate both sides

What do you get?

[flat frigate]

smidgin

[solar gust]

yes

[flat frigate]

smidgin

[solar gust]

yes

[flat frigate]

smidgin

[restive niche]

Which is what we wanted

[solar gust]

then how to find derivative of f^-1(8x)

[restive niche]

Chain rule

[solar gust]

can you give me a solution for this question

[restive niche]

There are solutions on the internet, i was trying to help you get to it on your own

https://math.stackexchange.com/questions/1326952/if-fx-8x33x-x-in-mathbbr-how-do-i-find-lim-x-to-infty-frac-f

[solar gust]

thanks I got my mistake now

[safe radish]

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[woeful agate]

could i get help with lagrange error bounding

[woeful agate]

pleaser

[safe radish]

@woeful agate Has your question been resolved?

[woeful agate]

Shibababarr

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@woeful agate Has your question been resolved?

[verbal drift]

what's the problem?

[woeful agate]

I just need the general way of solving problems doing it, I do not have any examples on hand but can find some

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[hollow elbow]

hey guys just want some clarification on this: (the range of g is [0, sqrt(3/2)]
γ and δ would equal to the range of g right? and to find α and β it must equal the codomain? pls :')

[safe radish]

@hollow elbow Has your question been resolved?

[hollow elbow]

<@&286206848099549185> sorry for the ping

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@hollow elbow Has your question been resolved?

[quasi gull]

you are correct

[hollow elbow]

alright thank you!

[lean otter]

how to ask help i'm new in discord , i want help to calcul this intégral

[hollow elbow]

you’d have to post that on an available math help channel

.close

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[stark jackal]

help pls

how do u figure out p, q, and r without a protractor

[past birch]

if the green angle is a right angle, p+36 is also a right angle

adjacent angles on straight line

[weary oracle]

q=36, p = 90-36. r = 180-36

[past birch]

i think youre supposed to say how to do it

[weary oracle]

it's so obvious

[past birch]

instead of straight up giving answers

!nosols

[safe radish]

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.

[stark jackal]

why do u have to do 90-36

[weary oracle]

Do you understand why q=36?

[past birch]

do you agree p+36 = 90

[stark jackal]

ohhh

what about these ones then

these are so hard

and ive got a exam tmr

[lean otter]

A line is 180 degrees, on that line we have 3 angles
180 = 90 (given) + p + 36
We can subtract 90 to simply get
90 = p + 36

[past birch]

for J look for isosceles triangle

for C use vertically opposite angles

for E use sum of interior angles of parallel lines = 180

for F use adjacent angles on straight line

[stark jackal]

ohhh

[stark jackal]

wait what wdym

[past birch]

x+y=180

[stark jackal]

wait so what would be the equation onn my paper

[stark jackal]

im kinda confused on how u figured it out

[past birch]

u mean the proof?

[stark jackal]

yeah its so confusing

[stark jackal]

i meant for e

[past birch]

E+C=180

[stark jackal]

how did u find e

[past birch]

E=180-C

[upper arrow]

p = 54, q = 36 and r = 144

[stark jackal]

wait but howd u know it was 180

[past birch]

because the lines containing C and E are parallel

[upper arrow]

what is e here?

[past birch]

therefore the sum of the angles inside it are 180

[past birch]

i'm proving that the sum of x and y are 180

[stark jackal]

wait so parallel lines are 180

[past birch]

here x+y=180

just use this but in your case

x is E and y is C

[upper arrow]

you can just draw an imaginary line somewhere east of x and y to turn this into a quadrilateral. and by it's angle sum property we know that all angels of a quadilateral add upto 360. The line that you drew will be 90+90 thats 180 so the left is 180

[past birch]

my proof also works soo

sure i guess

[upper arrow]

yea i mean i will just giving an alternate approach since V seemed to have trouble understanding

[stark jackal]

bro my 7th grade brain cant handle this

is there like a method

[upper arrow]

7th grade? how old are you?

[stark jackal]

13

[upper arrow]

i see

[past birch]

I learnt this in 7th grade too

[upper arrow]

basically consider the four angles of a square right? what do the four angles in a square add upto?

[stark jackal]

360

[upper arrow]

that's very good

now i will give you a diagram

[stark jackal]

ok

[upper arrow]

@stark jackal look at this

do you know what a quadrilateral is

?

[stark jackal]

yes

[upper arrow]

what do all the angles in a quadilateral add upto ?

[stark jackal]

360

or a square

[upper arrow]

very good

now as you can see the angles in the quadilateral ABCD in the diagram i just showed also add upto 360

two angles are 90. C and D

so that's 90+90+x+y = 360

this means that x+y = 180

do you understand it now?

[stark jackal]

ohh yes

[upper arrow]

🙏

if you do you can close the thread now

[stark jackal]

wait but im still confused on like how you got e

[stark jackal]

like for E

[upper arrow]

what is E?

[stark jackal]

this one

[upper arrow]

oh

ok let me look at it

[upper arrow]

https://byjus.com/maths/alternate-interior-angles/

consider reading this

this explains it pretty nicely

[stark jackal]

think i got it on just E it would be 37 because 180-143

[stark jackal]

but how would you find G

[upper arrow]

to find G you would have to first find H which is 96. Then you will find F which is 180-37-85. Then you will subtract F and H from 180 to get G

[upper arrow]

that's not it entirely but i guess this is something you will learn with more practice

[stark jackal]

can u teacch me how to do g

[upper arrow]

@stark jackal this right here is how you do G

[stark jackal]

how did u find h

[upper arrow]

it is extremely obvious that H is 96 because of the 48 and 48 being added together

[stark jackal]

how did u know to add 48 and 48 tho

[upper arrow]

as you can see that the lenght of the two sides of that triange are equal. An unnecessary detail but since we know that the total angle of a triangle is 180, we are left with J being 180-48-48. Now H and J lie on the same like which means that H+J = 180, implying that H = 180-J. But we already know that J = 180-48-48. So H = 180 - (180-48-48) = 180 - 180 + 48 + 48 = 96.

you should go back and learn basic properties of a triangle

[safe radish]

@stark jackal Has your question been resolved?

[stark jackal]

@alpine wigeon i need help with r, how do u figure out this

[safe radish]

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[potent seal]

.close

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[light shoal]

<@&268886789983436800> google translate says that was very naughty..

[royal lantern]

mere moments before I got the mute...

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[upper arrow]

$f(x) =
\begin{cases}
\frac{e^{3x}-e^{-5x}}{x} & \text{if x =/= 0}
k & \text{if x = 0}
\end{cases} \newline
\text{is continuous at x = 0 for the value of k as?}$

[desert pasture]

this channel is now closed, open a new one

[upper arrow]

ok this must work

[desert pasture]

see this

[upper arrow]

oh i totally missed that message i am so sorry

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[urban vale]

I was watching this physics video and I swear they did the math wrong. This isnt right is it?

[uncut skiff]

feels right to me

What exactly are you confused about

[urban vale]

tan^-1 (8.0/-6.0) is -53 degrees

right?

[warm sinew]

sure dont seem right

[uncut skiff]

I guess they just missed the 180 there

that’s more of a principle value thing

[urban vale]

oh they rotated it clockwise

ryancant u are a genuis

[uncut skiff]

no it’s just that tan(theta)=tan(theta+180) so you have this really stupid thing where arctan only spits out -90 to 90

[urban vale]

yeah I get it now

[shadow glade]

it's not stupid, it's so that it's a function, it can only have 1 output

:p

[urban vale]

thanks

.close

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[uncut skiff]

it’s really annoying to account for sometimes lol

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[uncut skiff]

cut the area

[little scaffold]

I'm working right now, at the moment I can give you a hint but cut it like this and use the formulas for the area of the trapezoid and a rectangle (yeh urs is better)

k peace

[uncut skiff]

yeah

You know how to compute the area of those right

the height of the triangle is not 3 cm

What’s the length of the black line

Vertical one

the one that aligns with the triangle’s height

use the other vertical lengths to find it

and you can consider it with the 3 cm above it too

close

on the left you see that the 5 cm+3 cm would be the same as the 3 cm and the black line

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[cedar tiger]

What would be a good mathematical statistical analyze to get an idea to go buy or short on a stock or forex market on that practicular day ? based on past data

[lean otter]

lagrange polynomials are generally good for predicting stuff based on past data

[crisp bluff]

I think I remember seeing the Black Scholes/Merton equation

Apparently it was used to tell whether a stock was worth investing in

[cedar tiger]

Cool thanks alot i need to research that terms never heard off 💪

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[onyx turret]

I have been stuck on a few questions like this for ages and i am following the videos and still cant get it

[safe radish]

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[wicked hull]

Why is the absolute value dropped for y at the end?

[wicked hull]

<@&286206848099549185>

[snow goblet]

think about it this way, c times the absolute value of y means the same thing as c * y & c*-y right?

[crisp bluff]

Isn't absolute value not supposed to have negative

[snow goblet]

?

[wicked hull]

is it because it doesnt matter since is +/-C

[crisp bluff]

The only way I can think of it going away is if the y being positive is implicit

It's occupied

[snow goblet]

how you tryna answer a question abt calc and then post a 5th grade problem 😭

[crisp bluff]

Hehehe

Oof

[snow goblet]

kidna

kinda

the sign doesnt matter bc the sign is a constant

at least afaik

wait actually one second

[crisp bluff]

How come the integral of 1/x does not have a constant

But 1/y does

[snow goblet]

i think im mistaken

[crisp bluff]

With what

[snow goblet]

you can kinda get rid of it

by skipping a step

subtract c2 from both sides

c2 being x’s constant

[crisp bluff]

Yes

[snow goblet]

constant minus a constant is js a constant

[crisp bluff]

Ohhhh

Okay

[snow goblet]

ok

[crisp bluff]

So c1 is c2-the constant for 1/y

[snow goblet]

i thought about the absolute value thing and i think im still right

[crisp bluff]

Okay

Tell the original poster

[wicked hull]

my professor said you dont need it because its still a constant anyway

[snow goblet]

your professor is wrong then????

[crisp bluff]

What is still a constant anyway?

[wicked hull]

you just combine them c1 and c2 and it become c3

[snow goblet]

your js skipping a step

[snow goblet]

oh dear 😭

[crisp bluff]

Wait

No

I did not want it like that

[snow goblet]

are you in calc 2

[crisp bluff]

I meant what he was referring to

Like it sounded like he said that integral of 1/x is a constant in of itself

Which sounds wrong

[snow goblet]

i can near promise you he didnt say that

[crisp bluff]

Probably

But it sort of sounded weird

[snow goblet]

unless idk what im talking about

[crisp bluff]

Also I am not in Calc 2

[snow goblet]

i just finished calc 2

[snow goblet]

what????

[crisp bluff]

Cool

What

[snow goblet]

what calc is this

[wicked hull]

2

[snow goblet]

omg

[crisp bluff]

What

[snow goblet]

im mixing people up

[crisp bluff]

Mixing who up

[snow goblet]

i thought you were op

[crisp bluff]

Nope

[wicked hull]

youre propably right though bc it makes sense

Would I drop the absolute value in this case bc of the constant or bc it’s a root? @snow goblet

[lethal path]

Which is the problem ?

[lethal path]

This is the problem correct?

[wicked hull]

yeah. the the answer is Csqrt(x^(2)+32)

i just want to make sure my understand and reasoning is correct

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@wicked hull Has your question been resolved?

[wicked hull]

<@&286206848099549185>

[wicked hull]

.close

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[loud osprey]

let f be a function given by f(x)=lnx. What is the approximation for ln(3/2) found by using the line tangent to the graph of f at x=1?

[loud osprey]

How do i do this

[icy lance]

find the equation of the tangent

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[fleet axle]

Hi, I need some help with this equation (looking for a)

[fleet axle]

This is basically what I've gotten so far... Prob wrong tho I was sadly missing on the lesson we my teacher explained this

[scenic saffron]

that all looks correct to me I think?

you’ve correctly calculated the integral and found the area in terms of a

you’ve gotten a quadratic in a now, so you’ll need to solve that (eg try factorise - if that doesn’t work then use the quadratic formula)

then you’ll probably get two solutions, as we are told a>0 we will choose the positive one

[fleet axle]

I think I'm just to dumb to solve the last part

[restive sigil]

The question is a bit tricky imo. The integral gives the signed area, which means that any area below the x-axis will be negative.

[fleet axle]

Haha

[restive sigil]

Instead of setting it equal to 7, try setting it equal to -7

[fleet axle]

Yes true I think you have to use the absolute value (idk what the right term is in English)

[restive sigil]

Yeah you could do that too

thats probably more right as it works for any general case

[fleet axle]

Yeye I think my problem is just with simple algebraic manipulation (again idk the English term sry)

In the last part so the a/4-a^2/2=7

[scenic saffron]

ooh ur right my bad!

nice spot

[scenic saffron]

so their point was that it should be =-7 not 7, since the area is negative

so from here you can simplify to get a quadratic

so -2a^2 + a + 28=0 i think?

and from here you can apply the quadratic formula

[fleet axle]

Yeah I think you're right Dahm so my problem was that I didn't pay attention in 7th grade hahahaha

To dumb to simplify

Thy a lot for your help tho both of you ❤️

[scenic saffron]

ok so we have

a/4 -a^2/2 = -7

same as

-a^2/2 + a/4 = -7

adding 7 to both sides

-a^2/2+a/4+7=0

multiply by 4

-2a^2 + a + 28=0

[scenic saffron]

no worries :)

[fleet axle]

Thy for explaining!

2 am for me now, gotta catch some sleep for the exam

.close

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[late briar]

someone guide me through this problem?

[buoyant shadow]

it's not hard

there are 12×11×10 ways to buy three plants

you can divide by 6 but it's not even necessary

how many ways to buy 3 female plants? @late briar

[late briar]

56 ways?

[buoyant shadow]

you divided by 6

[late briar]

12 x 11 x 10 = 1320

[buoyant shadow]

then we should also divide the total, so there's 220 ways

220 total, 56 all female

find all male

subtract, and you get "both types"

[late briar]

okay

220 - 56 = 164

[buoyant shadow]

no

220 − 56 − 4 = 160

[late briar]

4 = all male plants?

[buoyant shadow]

yeah

[late briar]

so divide 160 by 220 and that is the answer?

[buoyant shadow]

no

divide 160 by 220 and that's the answer

[late briar]

yea i just changed it

my bad

160/220 = .727

[buoyant shadow]

very good

[late briar]

ok

thank u

.close

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[remote crag]

Need help finding the derivative. Idk what I did wrong

[torpid fable]

You wrote 4x instead of 4^x

[remote crag]

OMGOSH Ty 😭😭😭

.close

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[gritty umbra]

1. A point charge at the center of a
   sphere contains an electric field of 1.25 ✕106N/C
   from a distance of 0.150m. Calculate the charge’s magnitude passing through the
   surface of the sphere if it has a radius of 0.150 m

[daring crescent]

I am assuming this is a physics question?

[gritty umbra]

yes

[daring crescent]

You might find more help in #old-network

[gritty umbra]

I have tried solving it, just came here to check

[gritty umbra]

I actually find more help here because there's more active people in this server 😔

[safe radish]

@gritty umbra Has your question been resolved?

[safe radish]

@gritty umbra Has your question been resolved?

[gritty umbra]

no 😔

<@&286206848099549185>

[lean otter]

<siunitx>
so i guess [
1.25\210^6 ;\t{V/m} =\41{4\pi\eps_0} \4q{(0.150 ; \t m)^2}
]
and solve for $q$?

[gritty umbra]

is that the same as $K\frac{q}{r^2}$

[flat frigate]

drutus

[gritty umbra]

But I used gauss' law

[lean otter]

yes

[lean otter]

right. I don't think i quite understand how your question is written

can you show me like a picture of the question

[gritty umbra]

that is the exact question, I just pasted it

but wait

item 1

[lean otter]

right okay so i guess they want you to use Gauss's law

[gritty umbra]

yes

[lean otter]

so by "calculate the charge's magnitude passing through the sphere" are they asking you to calculate the flux?

the wording really seems hideous from my perspective

[gritty umbra]

yeah I don't get it too

[lean otter]

but like [
\oiint \vj E \vd \vc*n \dd A = \4{Q_{\t{enc.}}}{\eps_0}
]
The flux is independent of your surface's shape or size or whatever

[lean otter]

do the flux is going to be Q/epsilon

now if they want you to calculate the charge's magnitude

[gritty umbra]

wuduhel

I literally do not understand any of that

[lean otter]

oh what form of Gauss's law did they teach you

[gritty umbra]

the general formula that we use is electric flux = EA = Q_enc/Epsilon thing

[lean otter]

ok

that's fine

,, 4\pi r^2 E = \4{Q_{\t{enc.}}}{\eps_0}

[lean otter]

this is the same as this again because coulomb's law and Gauss's law are equivalent

[gritty umbra]

okay I get that

it's asking for the charge so I solved for Q

[lean otter]

right

so that should be correct

[lean otter]

this seems fine

[gritty umbra]

okay, thank you!

[safe radish]

@gritty umbra Has your question been resolved?

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[royal mulch]

what am i doing wrong here? the number comes out negative which is clearly wrong

[versed wave]

[versed wave]

,rccw

[sick jackal]

the number comes out positive?

45/7

[royal mulch]

no to the right

finding time was the first step

[versed wave]

he meant x

[sick jackal]

what's the next step

[royal mulch]

i need to find position travelled from the car stopping

which came out as 2025/7 - 2025/2

[sick jackal]

just a tip for your working write the unit at the end not while you're doing the calculations it makes it hard to read

[royal mulch]

it helps me keep track of the units im in, id rather get correct units

[sick jackal]

but you're calculating distance. the only unit you need is the m at the end

it comes out as positive for me

[royal mulch]

okay so what did i do wrong?

[sick jackal]

0.5 times -7 times (45/7)^2

equals -144.6428571

[royal mulch]

oh fuck forgot to square the 7 too

thank you

[sick jackal]

add that to 2025/7 and you get 2025/14

no problem

are you doing this by hand

[royal mulch]

everything but the 2025 x 7 ive done by hand

[sick jackal]

if you get a calculator in the test don't do it by hand - just substitute the values into the formula and you will always get the right answer

[royal mulch]

alright, he grades off of work shown so i like so show all the steps so if i did setup wrong i still get points. But ill def do that to check my work

.close

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[steep yacht]

if you have a number k, then find the closest number that is n^2 + n to k, why does sqrt(k) roughly equal n?

[icy lance]

closest integer?

[bleak delta]

what is n actually?

[steep yacht]

an integer

[bleak delta]

it can be zero then

[untold sky]

actually, does integer or real n matter here?

[bleak delta]

and k can be 1839271983

and sqrt of this isnt n

[icy lance]

so its n^2+n ~ k

[untold sky]

you find n after being given k

[steep yacht]

yeah

[bleak delta]

okay what if k = 9

then what will we do

because i dont understand it clearly

[untold sky]

n=2 or 3 i guess? both 3 away, i think the tiebreaker isn't exactly relevant

and if you allow real n then obviously just solve

[bleak delta]

so √k = n?

[steep yacht]

yes roughly

[bleak delta]

okay what is the question then

[steep yacht]

why is that the case?

[bleak delta]

because we defined it that way

was there a reason for this?

[steep yacht]

really how?

[bleak delta]

just want to define and it's defined

[untold sky]

well yes but that's not a given, they want to know why sqrt(k) ~= n

and the answer is probably just that n/n^2 = 1/n ~ 0 (insert limit argument here) so (n^2+n)/n ~ n = k/n so k ~ n^2

[bleak delta]

what is the wave thing like ≈

approx equal?

[steep yacht]

yeah

[bleak delta]

okay

[untold sky]

laziness for approximately equal, it's not an actual notation

[bleak delta]

but why 1/n ≈ 0

[lean otter]

complete the square

[untold sky]

1/n is asymptotically 0 which is why i said limit argument

[hasty wagon]

sorr to interrupt, i know we are helping tonnes of help in this channel, but heres my thought:
consider
n²+n=n(n+1)=(n+0.5-0.5)(n+0.5+0.5)=(n+0.5)²-0.25
if we take squareroot, we will have
sqrt((n+0.5)²-0.25)≈(n+0.5) when n is large

[steep yacht]

ohhhhh it's becuase it's pretty much asking what is the closest sqaure

[bleak delta]

why do we think this way?

[steep yacht]

so ofc it is near the sqrt

[bleak delta]

it is asymptotically ∞ either

[untold sky]

specifically here the important part is that n compared to n^2 is negligibly small for large n

i really should latex but im too tired today

[bleak delta]

okay but why did we choose large n

[steep yacht]

yeah makes sense, but will it get more inaccurate the larger the number?

or stay the same, because n will always be to small compared the n**2

[untold sky]

because k can be anything you want, and that argument shows that as n gets larger and larger the approximation only gets better (if n^2+n < 1.0000001n^2 then it's more accurate than a value of n where 1.0000001n^2 < n^2+n < 1.5n^2)

you can quantify the error better if you straight up solve the quadratic and look at the roots

[steep yacht]

well in my case k is a semiprime but i don't think that matters to much

[bleak delta]

whatever, i still dont understand what is the question and what they want from us and why do we think like this

[steep yacht]

well you guys basically answered my question so I'm happy

.close

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[urban eagle]

How do I do q6

1/2mu^2 = 1/2mv^2 + mgh
v^2 = u^2 - 2gh
T - mgcos(theta) = mv^2/r

That’s all I’ve really got

For h I have 0.3(1-cos(theta))

[safe radish]

@urban eagle Has your question been resolved?

[safe radish]

@urban eagle Has your question been resolved?

[urban eagle]

.close

[safe radish]

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[safe radish]

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lone arch]

Determine concretely the elements of the set [C = {x \in \mathbb Z \mid |z| < 35 \text{ and } \exists a, b \in \mathbb Z:x = a^3 - b^3}.]

[lone arch]

Does anyone have a hint for this?

I tried out a = 1, then b = 1, b = 2, ...

Then a = 2, then b =1, b=2, ...

And saw what I got

Maybe I just looked at too little

Maybe we can just list all the cubes up to 50 or so

Then look at their differences

$1, 8, 27, 64, 125, 216$

7, 26, 63, 124, 216

19, 56, 117, 208

37, 98, 189

61, 152

91

[runic pine]

How are the differences that?

[lone arch]

I looked at 1 first

difference from 1 to 8, difference from 1 to 27, difference from 1 to 64

Then at 8

etc.

[runic pine]

The differences of this ?

[lone arch]

No

Of this

[runic pine]

Ohk