help-10

[covert palm]

Idk how to solve this

May someone explain it for me step by step?

Mb pinged before 15 mins but it passed 8 and I saw other ppl getting help before 🙃

[brisk matrix]

cardinal sin

[covert palm]

What

Oh

😭

Ok I guess I’ll wait for the 15 mins to ping

<@&286206848099549185>

Now I can ping

.close

[obtuse pebble]

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[obtuse pebble]

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[solemn creek]

Understanding the Clues:
We have an 8x8 grid.
The pentominoes should be F-shaped.
The cells on the outside along the x and y axes represent the sums that each column and row must add up to.
When you enlarge, the value of each grid and the size of the entire grid is doubled.
Pentominoes can be rotated, reflected, and enlarged.
Place a ‘1’ inside each white cell of an unenlarged pentomino.
For an enlarged pentomino (factor N > 1), place an N in each cell.
The sum of numbers within each rectangular region must be the same.
The number outside the grid represents the sum of all numbered cells in the corresponding row or column once the grid is complete.

Image nr. 2 is an example of enlargement.

[obtuse pebble]

@solemn creek Has your question been resolved?

[solemn creek]

<@&286206848099549185>

[obtuse pebble]

@solemn creek Has your question been resolved?

[compact jay]

Hi

[obtuse pebble]

@solemn creek Has your question been resolved?

[obtuse pebble]

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[violet timber]

Did I show my work right?

I know I got it right but I wanna know if how I showed my work is valid (ignor crossing out on non practice questions I will erase

[obtuse pebble]

@violet timber Has your question been resolved?

[last pilot]

yo

what tbis

*this

@violet timber

[obtuse pebble]

@violet timber Has your question been resolved?

[violet timber]

wdym

[timid silo]

They found the answer

Can you close your ticket?

[obtuse pebble]

@violet timber Has your question been resolved?

[violet timber]

how is saying "yo what this" an answer

i dont understand

im wondering if the work i did is correctnly shown

and "yo what this" didnt explain that it is or it isnt

<@&286206848099549185>

[strong plinth]

maybe he couldn't read it lmao

it sounds like your calculation is correct, but if you're asking if your teacher would be happy with it if you showed him that, they might have some issues

1st recommendation is to always explain what you're doing, you can use words, that helps a lot to understand what the equations mean

2nd recommendation is that you should not combine calculations like you did there

like, if somebody told me "tom wants to build a 20m fence, 4m costs 5€, how much will the fence cost" you shouldn't write something like $20m/4m = 5 * 5€ = 25€$ because, generally, when you write equation strings like that, you want the very first thing in the equation to be equal to the very last thing in the equation

[warm shale]

Lartomato

[strong plinth]

and that's not true if you combine calculations

so an ideal presentation of your solution would be:
"we first convert the 39ft into meters:
[first equation]
so we know that he needs at least 11.8 meters. since he cannot buy part of a meter, he needs to buy 12 meters. the price for this is
[second equation]"

@violet timber does that help you?

also i don't wanna be a dick but if you want someone to help you with your math, it really pays off to write it down nicely

b/c if i'm scrolling through the #help servers looking for someone to help, and i need like 2 minutes to decipher what the screenshot says, i might just be like "eh maybe i'll let someone else do this"

[strong plinth]

this is sooo important btw, i've graded university students who just did not explain what they were doing, and you just end up having to guess your way through their equations

you're gonna get so much better at math if you learn how to give every equation you write down a description

@violet timber good luck & much love fam

[violet timber]

Yeah sorry my writing is just bad

[violet timber]

I have never been told I need to do that

And never got marks removed from it

[violet timber]

Thanks so much, have a wonderful day 😊

[strong plinth]

yeah the expectations might not be so high in school

and teachers are generally happy when you just get the right answer

but they're even happier if you can explain what you're even doing

[desert cobalt]

چرا سیگما نیستی

[obtuse pebble]

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[violet timber]

"why aren't you sigma"

[obtuse pebble]

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[zinc sky]

Hello i don't know how to solve the number 10 can someone help ? https://www.zupimages.net/up/22/01/kv10.jpg

[queen briar]

je ne sais pas

sorry

[zinc sky]

mince

[obtuse pebble]

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@zinc sky Has your question been resolved?

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[timid silo]

How do i solve for x?

[pine barn]

add them all up to 180

[timid silo]

so 90° + (2x+1)° + (5x+5)° = 180°?

then i evaluate?

[pine barn]

yep

[timid silo]

alr

solved

.clsoe

.close

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[spring frost]

so im trying to simplify an equation. here is it and this is the answer

[spring frost]

I tried to simplify it like so, what exactly did I do wrong?

by the way the way I simplified it outputs 3.92 instead of 0.39

<@&286206848099549185> pls help. idk what im doing wrong. thankyou.

[obtuse pebble]

@spring frost Has your question been resolved?

[tawny pivot]

i think yoou multiply the ones wi paranthesis first

[spring frost]

?

[tawny pivot]

but in this case start with the ones that dont have a

[spring frost]

ok so

slight rounding error

[tawny pivot]

oop

alr

[spring frost]

ok

[tawny pivot]

i think youd divide after multypling both idk

lowkey confused

[spring frost]

alr well

idk, I think I figured it out though

thanks for help

.close

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[deep ravine]

Why does it say an alternative method would be to find the area bounded by the curve for 0<=(theta)<=pi, instead of 0<=(theta)<=2pi?

[tired shell]

the wacky thing about polar coordinates is that negative r sometimes looks like positive r with 180 plus the angle, here when you trace out r=sin(7theta), the loop doubles over itself before completing 2pi

you can see it if you like trace your pencil over the curve and see when it completes the loop or just plug in points

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[whole tiger]

solve this question

[whole tiger]

i have solved the equation of a ladder by doing the slope and 2 point form to find the equation of the ladder

toget 4/3x + y - 4 = 0

what about the rest of the question

[eager niche]

The 5’9 part is weird, get that in meters, and do you know how to find a perpendicular line?

[whole tiger]

no thats the problem

so in m that is 1.75

<@&286206848099549185>

[runic zodiac]

yes

[eager niche]

https://www.khanacademy.org/math/geometry/hs-geo-analytic-geometry/hs-geo-parallel-perpendicular-eq/v/perpendicular-lines

[obtuse pebble]

@whole tiger Has your question been resolved?

[whole tiger]

but where are the lines perpendicular

<@&286206848099549185>

.close

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[ember frost]

Given the following nuclear reaction:
$\newline^{23}{11}Na + ^1_1H \rightarrow ^4_2He +^{20}{10}Ne\newline$
Determine whether the reaction is exoergic or endoergic. Calculate the amount of energy (MeV) it releases/absorbs

[warm shale]

Fungus 34A05

[ember frost]

so i found that the reaction is exoergic, but i have trouble calculating the amount of energy it releases

my answer is 4,37805

[brisk matrix]

cardinal sin

[ember frost]

[brave bramble]

There's a couple of methods to do this. Did your class make it clear which method to use?

[ember frost]

the formula i used is

$W = (m_t -m_s)c^2$

[warm shale]

Fungus 34A05

[ember frost]

m_t is the total rest mass befote reaction

m_s is after

because m_t-m_s > 0, the reaction is exoergic

[ember frost]

but for some reason i get this, which is incorrect

[brave bramble]

I forget, is that the amount of energy given one mol of both reactants are introduced?

[ember frost]

not sure what you mean by that

its the energy that the reaction itself produces

or absorbs

[brave bramble]

How did you find the masses of the inputs and outputs?

[ember frost]

the rest mass?

for each element i apply this formula

$m=Z(m_p) + N(m_n)$

[warm shale]

Fungus 34A05

[ember frost]

Z is the number of proton, mp is proton mass, N is number of neutron

this is the rest mass, the atomic mass thr problem also gave

the nuclear mass of Na, H, He and Ne, are 22.9837, 1, 4.0015, 19.9869 respectively

all in dalton

[deep ravine]

How can i detect this when looking at an equation in polar form?

[obtuse pebble]

@ember frost Has your question been resolved?

[ember frost]

solved

i miscalculated

.close

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[verbal niche]

when would it not be normal?

[obtuse pebble]

@verbal niche Has your question been resolved?

[unreal musk]

Considering the population was normally distributed to begin with, you're forced to have the sample mean as normally distributed in that case

[obtuse pebble]

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[bold zinc]

the linear equation of the first line

would have a c value of 0 right?

in y=mx + c

because

no y-intercept

right?

[cunning burrow]

no

[bold zinc]

oh?

how would i find what 'c' is then

[cunning burrow]

So the first line segment goes from x = -3 to x = -1

It does not intersect the y-axis because the function is piecewise

[bold zinc]

ok

i need to mention this function

how do i find 'c'

i know i include domain

but like

that doesnt make 'c' disappear in the actual function

if that makes sense

[modest ingot]

bro

[bold zinc]

oh

hold on

💀

i figured it out

that was so much easier than i thought

function of first line would just be

y = x + 3

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[cyan oxide]

Hello fellas, this isnt more of a math help thing as it is a life help thing but i suppose its related. I want to be in a field where math is dominant as i feel like its the only subject i can really excel at, now ive been seeing stuff on social media like oh engineering is mostly math and all that but the problem is here in my country we have "strands", essentially what you take in senior highschool which is grade 11 and grade 12, ABM, STEM, HUMSS, etc. STEM is mostly science based and has like the calculus and stuff which i want to take when im in like college or whatever (call me crazy but i just like math), but the problem is that strand focuses heavily on science, physics, biology, and chemistry and they only have pre calc and basic calc, and trust me when i tell yall i am extremely dumb when it comes to science, idk what it is but i just am, on the other hand I want to take ABM and become an accountant but they only have business math over there, i guess im just looking for an external opinion to help me out, ill take this down if need be, i know i troll alot but id appreciate it if i could have a serious discussion with people here

[cyan oxide]

note that engineering is bound to STEM, so if i wanna take it, I have to take STEM

[fathom flicker]

Do you think this might be better in #math-discussion or #advanced-lounge or #discussion ?

[cyan oxide]

it might be

mb

.close

[obtuse pebble]

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[surreal forge]

(i wonder what that 2nd channel is)

[cyan oxide]

also what is that no access channel

[fathom flicker]

advanced lounge

[robust sleet]

For smarties

[cyan oxide]

damn, so we just dumb out here

[robust sleet]

Dw there's also a genius lounge

[viscid gull]

i thought there was a self assignable role for adv lounge

[unreal musk]

The undergrad one I think should give you it

[cyan oxide]

you guys have a role that does nothing?

I want a role that does nothing

[viscid gull]

it makes ur name grey

[cyan oxide]

i have no roles lol

[viscid gull]

thats self assignable too

[unreal musk]

No that’s not very ppl

[viscid gull]

Oh

[robust sleet]

You've the helpers rule blud

I want it

[unreal musk]

Role that does nothing is self assignable tho, channels and rules haha

[cyan oxide]

right i forgot to assign some roles for myself

[viscid gull]

hmm take the STEM strand

better than business math

phys is p mathy too

[cyan oxide]

ahhh

chat gpt

my friend

my dumb smart friend

[viscid gull]

Chem isnt too bad either

[cyan oxide]

yeah ig drugs are alright too

[viscid gull]

only bio is p perpendicular to math at that lvl

[fathom flicker]

!redir

[obtuse pebble]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

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[timid silo]

help-?

[zenith raft]

what is this cropping

[timid silo]

i work with what i got

[robust sleet]

Casually cropping out the point marks

😝

[timid silo]

there..happy? lmao

[zenith raft]

yes because i was about to suggest using a calculator

[timid silo]

lmfao

[ember frost]

this is way better than your abominable cropping

[fathom flicker]

Do not suggest this, their doubt says "without using a calculator"

[zenith raft]

anyway what is giving you trouble?

[timid silo]

HEY MANNN DONT JUDGE MY BEAUTIFUL CROPPING..IT WAS PERFECT LMAO

[zenith raft]

i was a fan of the cropping, it was funny

[timid silo]

the whole damn thing💀

[safe haven]

-pi is around -3.14, point e is like -2.8 or 2.9

aint no way -pi is on point e

[timid silo]

bish idfk

lmao

[ember frost]

this question is kinda stupid, unless theres an error range you cant really tell the exact value of the points

[timid silo]

LIKE BRO..ITS 2:30 AM.. AND IM DOING MATH BC I CANT SLEEP

[ember frost]

calm down

[timid silo]

sorry

[zenith raft]

i think the points were there on the line already

and if it's 3 only options you can tell

[ember frost]

oh my bad, i thought you writr value with given points

[timid silo]

they were.

[zenith raft]

it's just matching

[timid silo]

basically what i need to find out is just an estimate of where the dots could be

[ember frost]

like shioshi said, d and e seems to be wrong

[safe haven]

also sqrt(8) is definitely less than 3, because its less than sqrt(9)

[ember frost]

the middle point is -2,8, while the left point is around -3.1

[safe haven]

and greater than 2 because its bigger than sqrt(4)

[timid silo]

okokok...can u guys just help me with this?

[brisk matrix]

too many cooks guys, one at a time

[zenith raft]

10/10 cropping

[timid silo]

especially that ONE straight line

[zenith raft]

lasso tool moment

[ember frost]

actually im kinda liking this crop, im using this

[timid silo]

lmfaoo

i need this

[zenith raft]

well do you know what pi approximately is?

in decimals

i think that one is a little easier if you know it

[timid silo]

3.14

wait

im not sure

[zenith raft]

yea

[timid silo]

ok lol

[zenith raft]

so -3.14 should be to the left of -3

[timid silo]

ok so i got it right lol

[zenith raft]

[timid silo]

ty guys.. im adding u shay

[safe haven]

you put -pi on e

e is on the right of -3

[zenith raft]

[brisk matrix]

based

[timid silo]

ty guyss

[zenith raft]

mount rushmore or something

[timid silo]

YEP >:D

Ok ty guyss

.close

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[stuck lantern]

A person rolls a dice 3 times in a row. What is the probability that the value of the third roll is in between the values of the first and second rolls?

I'm not sure where to even begin with this question

[surreal forge]

u could count the number of ways this can happen

[gloomy vector]

i dont know if this is correct or not, but could you find the number of diffrrent numbered rolls then divide by 3 cause of permutation since from 3 diffrent numbers there is 2 diffrent ways to arrange it so that the third value is between the first and second, and that would be the possibilities that it would be between the first and second, then just divide by 6^3

[stuck lantern]

im not quite sure i understand

[gloomy vector]

yeah ok nvm

[stuck lantern]

so would i just list all the possibiilties

huh

ok

[stuck lantern]

i think i get what ur saying now

so its like

the first roll doesnt matter

the second roll has a 5/6 chance of being different

and the third has a 4/6 chance of being different from both

so then the third has a 1/3 chance of being in the middle?

[gloomy vector]

i think so? since if you imagine it being 3 diffrent numbers a, b, c where a>b>c then the only possible ways to arrange yhe third number to be in the biddle is abc and cba

[stuck lantern]

i see

so that gives 5/27

i get it now

thanks for the help

.close

[obtuse pebble]

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[gloomy vector]

though i dont know if its correct

[obtuse pebble]

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[bold zinc]

can someone give me a hint of what form i would use

so i got the coordinates

(2, -6) <-- vertex

(6, 0) <-- one of the x-intercepts

well ngl its not even the form

idk how i'd go about solving this

oh wait

i got it nvm

[dense imp]

well a parabola is symmetric over its principal axis

[bold zinc]

yeah

so it'll look smthn like this

$\frac{a+6}{2}=2$

[warm shale]

pixel

[bold zinc]

where 'a' is the 2nd x-intercept

[dense imp]

right

[bold zinc]

i think the answer is

$y=2(x-1)^2-2$

[warm shale]

pixel

[bold zinc]

can someone check please?

having the two equations dilate by a factor of 2

[brisk matrix]

looks good

[bold zinc]

would make them the same shape

woo

$2=a(3-1)^2-2$

[warm shale]

pixel

[bold zinc]

correct?

solve for 'a' then put back into vertex form

a = 1

answer should be

$y=(x-1)^2-2$

[warm shale]

pixel

[brisk matrix]

yup looks like you have the right idea

[obtuse pebble]

@bold zinc Has your question been resolved?

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[bold zinc]

$y=a\left(x+\frac{1}{a}\right)^2-\frac{1-a^2}{a}$

[warm shale]

pixel

[bold zinc]

is my answer correct

[brisk matrix]

,w complete the square ax^2 + 2x + a

[warm shale]

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[brisk matrix]

looks fine

[bold zinc]

$\left(\frac{1}{a}, -\frac{1-a^2}{a}\right)$

[warm shale]

pixel

[grizzled shore]

I feel like that does not like on the parabola

[bold zinc]

my answer to (a) or (b)?

[bold zinc]

also im thinking the 'x' vertex coordinate would be negative

[dense imp]

the x coordinate of the vertex on a parabola will be at -b/2a if the equation is ax^2 + bx + c, i think you just forgot the negative sign: -2/2a = -1/a

[bold zinc]

wait

let me double check my formula

isnt the formula to perfect the square: $y=a(x+\frac{b}{2a})^2-\frac{b^2-4ac}{4a}$

[warm shale]

pixel

[dense imp]

i dunno, im not really into memorizing formulas especially something like that but i know the -b/2a for finding the vertex because it also makes sense in terms of calculus: the derivative will be 2ax + b, you'd set that equal to 0 to find a horizontal tangent and you get x = -b/2a

[grizzled shore]

I think it’s y = a(x-h)² + k has vertex (h, k)

[bold zinc]

yeah

[grizzled shore]

But your think is not on the parabola

[bold zinc]

which is why i said this

because s

[grizzled shore]

You need to swap the - signs

[bold zinc]

since its a +

that means its -h which would be $-\frac{1}{a}$

[warm shale]

pixel

[bold zinc]

really?

[dense imp]

$y=a(x-(-\frac{b}{2a}))^2-\frac{b^2-4ac}{4a}$

[warm shale]

Soosh

[dense imp]

think of it like this, your h = -b/2a

[bold zinc]

yeah

[bold zinc]

yeah see that just simplifies down to a(x+b/2a)^2

[bold zinc]

is my equation wrong tho

do i need to swap the signs like he said

$\left(-\frac{1}{a}, -\frac{1-a^2}{a}\right)$

[dense imp]

but memorizing a formula like this is a bit useless, just remember to find the x vertex -b/2a and then plug the x value into the equation to find the y

[warm shale]

pixel

[bold zinc]

yeah ig but like I've already memorised it ages ago

its become like the quadratic formula for me, it just comes to me

[dense imp]

i've noticed from various channels you've opened that you're very into formulas and thinking things out analytically and technically so i guess yeah learn however works for you, what works for me may not work for you : )

[bold zinc]

yeah haha thank you

[bold zinc]

is my answer incorrect for (b)

[twilit meteor]

its correct

[dense imp]

seems right, but maybe a more simpliied version for the y value is (a^2-1)/a

but yes, correct

@bold zinc oh also another reason why the -b/2a for the vertex is easy to remember: its built into the quadratic formula, to find the roots because of symmetry is basically like youre starting at the vertex x = -b/2a and you go to the left or to the right (the plus or minus) by a certain amount, so you can see how the roots are symmetric to either side of x = -b/2a

[obtuse pebble]

@bold zinc Has your question been resolved?

[obtuse pebble]

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[stiff remnant]

I am a bit confused with this

[paper dome]

do you know exterior angle property

[obtuse pebble]

@stiff remnant Has your question been resolved?

[stiff remnant]

@paper dome yes

[paper dome]

and also that each angle of a equilater is 60

so basically in the triangle with 40 degree uh, we know two angles which are 60 and 40 so the exterior angle is 100

now in the triangle with x

we also know that one angle is 60, 100 and the third is x

can you solve it now

[stiff remnant]

Yup

Thanks for your help

[paper dome]

!done

[obtuse pebble]

If you are done with this channel, please mark your problem as solved by typing .close

[paper dome]

npnp

[obtuse pebble]

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[stiff remnant]

.close

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[valid ivy]

Can anyone help me trying to explain how to use PQ method to solve x2−8x+17=2 ?

[timid silo]

what's PQ method

[valid ivy]

pq

ehh

[slim lake]

Factorization

What have u tried?

[valid ivy]

i tried using it and got the answer wrong, so i must not understand how to use it

[slim lake]

Find a p and q such that pq = the constant
p+q=coefficient of x

[last pilot]

^^

[slim lake]

Show ur working

[timid silo]

oh alr

[last pilot]

remember that $(x+a)(x+b) = x^2 + (a+b)x + ab$

[warm shale]

ren

[valid ivy]

i Got this:

[brisk matrix]

impopster

[last pilot]

so here, ab = 15 and a + b = -8

[valid ivy]

x2 = -8/2 -+ Root (8/2)2 -15 = 0

[slim lake]

Lmao

[valid ivy]

But like, its just stuck in my brain i cant get anything out of it

Im having trouble deciphering the formula

[crude solar]

First of all try equal the equation to 0

[valid ivy]

So -2 on both sides

[crude solar]

Yes

[valid ivy]

that i did

[crude solar]

What did you get

[valid ivy]

2x-8+15=0

[crude solar]

Yes

[valid ivy]

8x*

[crude solar]

Just use this now

[valid ivy]

P is 8

q is 15

[crude solar]

Wait, it’s supposed to be x² not 2x

[valid ivy]

x2, ye i got that aswell

how do u type that ^2

[crude solar]

I just have it in shortcut

[valid ivy]

So im at x2 -8x+15 = 0

p is 8 and q is 15

[crude solar]

Find a numbers p and q such that pq = multiplatication of coefficient of x² and the constant

And p + q = the coefficient of middle term, which is x here

[valid ivy]

Ouff damn

I need to find a swedish tutor

its too much doing this in other language aswell

im not getting this at all

[crude solar]

Yk the coefficient of x² and you know the constant right?

[valid ivy]

x should be coeffcient

and constat the 2?

[crude solar]

No

Constant means the number which has no variables, it just a number, which is 15 here

Coefficient means what is x² multiplied by

[valid ivy]

yea

ok

[crude solar]

So you know which ones they are now?

[valid ivy]

yes

[crude solar]

Ok

[valid ivy]

So 15 is the constant and the x is the coeffecient

[crude solar]

No

[valid ivy]

Then no

[crude solar]

Coefficient means the number in front of the variable

Since there is nothing in front of x², it’s coefficient is 1

[valid ivy]

alright

[crude solar]

Ok, so their multiplication is 1x15, which is 15

So we need p and q such that pq=15

And p+q= the coefficient of the middle term, which is -8x here

[valid ivy]

x2-8x+15=0

[crude solar]

The middle term is -8x, what is the coefficient of it?

[valid ivy]

-8

[crude solar]

Yes

So now find factors of pq

[crude solar]

Also write it as x2, 2x means something completely different

[valid ivy]

-8/2

[crude solar]

No

You know what factors mean?

[valid ivy]

its multiplication

so 8x15 ? i dont understand this at all bro

I dont think im gonna get this

Sorry

[crude solar]

Factors means numbers whose multiplication is equal to 15

[valid ivy]

3x5 but

i dont know why we got there

[crude solar]

Yes 3x5

[crude solar]

This is the pq method

You have find two numbers such that their multiplication is equal to 1 and 15, but their sum should equal to -8, in this question

[valid ivy]

so -3x-5

[crude solar]

Yes -3x-5x

Also if you are still confused, there is a video that explains it really well

[valid ivy]

Im very much still confused but thats why im gonna sit with this for a while now

But very much thanks for the help

[crude solar]

K

[valid ivy]

.close

[obtuse pebble]

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[ruby forge]

um hello

[ruby forge]

i wanted to ask where i went wrong on this middle term splitting question

OH WAIT NVM 4x9 not equals to 27 i read thwt

Wrong

As 3x9

my bad

you can close this

[warm canopy]

You can write .close

[ruby forge]

.close

[obtuse pebble]

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[graceful marten]

Is this true?

-ln urMom=

ln 1/urMom

?

Bc thats crazy

Ive never seen that before

Sry im a bit annoyed

I shouldnt be typing like this

[timid silo]

that just log properties

like the integers at the start go as powers

and anything to the power negative 1

is a reciprocal

[graceful marten]

like ik ln 4 - ln 6 = ln 4/6

[timid silo]

ya

[graceful marten]

right!!

Ill take a note

thank you for that

: >

[timid silo]

welcome

[graceful marten]

: )

.close

[obtuse pebble]

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[graceful marten]

yeah i remember now

Ty

[obtuse pebble]

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[steady bear]

What is a Markov Chain and what is the application of matrices in modelling and analysing stochastic processes like Markov chains?

[viscid gull]

do u have a question that is either smaller in scope or not googleable

[steady bear]

I dont really understand what Markov Chain's are, and google didnt really help me

[obtuse pebble]

@steady bear Has your question been resolved?

[obtuse pebble]

@steady bear Has your question been resolved?

[obtuse pebble]

@steady bear Has your question been resolved?

[obtuse pebble]

@steady bear Has your question been resolved?

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[fading anvil]

hi

[obtuse pebble]

Please don't occupy multiple help channels.

[fading anvil]

sorry

can someone please give me a brief explanation about this platform

please

pleae

please

please

[kind hawk]

well you can ask math questions here and hope that someone volunteers to help you

[obtuse pebble]

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[tight zealot]

this is a more general question, why is analytic continuity a thing which is considered acceptable to do

[tight zealot]

doesnt it literally expand the graph into domain where its undefined just cz it wants to

[kind hawk]

well yes thats the point

[zenith raft]

wdym “acceptable to do”

[obtuse pebble]

@tight zealot Has your question been resolved?

[tight zealot]

like why is it even a thing

[kind hawk]

why shouldnt it

[tight zealot]

it just seems like something someone made up

[kind hawk]

(dont ignore the bot)

[tight zealot]

cz they couldnt finish their question

lol

[kind hawk]

expanding the domain of stuff is something you are very used to

[tight zealot]

is it

[kind hawk]

in elementary school you learned addition only for natural numbers. and then you expanded it to integers, rationals, reals, complex numbers

[tight zealot]

yes but in those cases we learn it is defined on integers, rationals etc

in this case we are expanding the graph to areas where it's litearlly undefined?

[kind hawk]

well is it defined for rationals, or did we give a new definition which agrees with the old one for integers but also now works for rationals

[zenith raft]

domain of a function is the choice of the person declaring the function

[tight zealot]

but in many cases doesnt analytic continuation result in absurdity

like the famous

1^2 + 2^2 ... = -1/12

[kind hawk]

I know that the comparison isnt perfect. but the rough idea is still there. we can give a new definition of the function which agrees with the old definition for the old domain

well if you tried to interpret it that literally, yes

think about how exponentiation originally was repeated multiplication

but we generalized that and at some point got to e^(i pi)=-1

how can that in any way still be "repeated multiplication"

you shouldnt still actually try to interpret it that way

[tight zealot]

damn how'd they manage that without using the identity 💀

[upper kindle]

(1+2+3+4...=-1/12*) ^^

[tight zealot]

was it harmonic series

wait how is it harmonic series

oh wait it prob is

[zenith raft]

zeta is defined by zeta(s) = sum of 1/n^s for s with real part greater than 1, and defined differently for all other complex numbers

[tight zealot]

yes

but doesnt analytic continuation find convergence of the zeta function for when the real part is less than 1 too

[zenith raft]

“find convergence” is a weird thing to say

[tight zealot]

doesnt analytic continuation say it converges then

[zenith raft]

you don’t evaluate zeta(s) with re(s) <= 1 by plugging s into the sum from earlier

[tight zealot]

for real parts less than 1

at certain values

[tight zealot]

ye my english is just bad lol

[kind hawk]

analytic continuation gives you a new thing which agrees with the old thing in the old domain, but is something else in the new domain

you dont have the series anymore

you have something else

[zenith raft]

think like a piecewise defined function. if s has re(s) > 1, zeta(s) is defined by the sum. otherwise it’s something else

[upper kindle]

idk what u guys are talking but I'm enjoying

[tight zealot]

riemann hypothesis stuff

lol

[tight zealot]

ah this explanation makes sense

but then again why cant you just make like

any type of graph with analytic continuation

what's stopping you from just making some random scribble

[zenith raft]

you can

[tight zealot]

and saying this is the extension

[zenith raft]

it just might not be useful

well

analytic continuation has certain requirements

[tight zealot]

ye it's if the function has a derivative everywhere or smth right

[zenith raft]

but yes there are different ways you can extend functions

[kind hawk]

if you have a normal graph which ended somewhere and you wanted to continue it but with it staying "smooth", then also intuitively you know that you cant just draw some random scribble. and now if you are working over C, this restriction is actually even more restricting

[tight zealot]

sounds like a bit of thigns i should pick for my next yaer modules

lool

would this fall under complex analysis

[kind hawk]

yes

[tight zealot]

at least i know im def going down the pure route

lol

[tight zealot]

just curious did u ever take fluid mechanics/ dynamics

[kind hawk]

no

[obtuse pebble]

@tight zealot Has your question been resolved?

[obtuse pebble]

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[timid silo]

solving this integral

I integrated sec^5x and am trying to see real quick does uh

sec(arctan(y)) simplify to anything im not seeing?

because it feels like the second integral would be a pain in the ass if I cant simplify that

[obtuse pebble]

@timid silo Has your question been resolved?

[alpine verge]

i think you should try swapping the order of integration first, since i don’t think you can even integrate what you have there at the end

[timid silo]

I just realized that 😭

ALL THAT WRITING FOR NOTHING AHFUHDJFKH

the bounds I think would be like

tan(x) <= y <= π/4?

[alpine verge]

not exactly

it should be from 0 to tan x

[timid silo]

[alpine verge]

the top function is tan x and the bottom is 0

yeah doing it this way it’s literally just a simple u sub at the end

[timid silo]

just curious, how do you figure out what should be the top and bottom?

[alpine verge]

but it’s quite impressive that you got the integral of sec^5(x)

[alpine verge]

i usually just draw out all the bounds on a graph

so you can visually see what’s happening

[timid silo]

gotcha

so my bounds then are 0<=y<=tanx, 0<=x<=1

[alpine verge]

yep

oh wait no

it’s 0 to pi/4 for the x one

[timid silo]

oh wait yeah mb thats what i mean

[alpine verge]

alright

[timid silo]

okay time to redo this rip my 30 min

also cute nahida pfp!

tysm for helpign!

[alpine verge]

ywyw

[timid silo]

.close

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[heady widget]

I need help with another one:
A 10.0 kg piece of brass at 71°C is placed in 20.0 kg of water that is 10.0°C. What is
the final temperature of the brass and water (cbrass = 380J/k°C)?

[silent thorn]

what have you tried so far?

[normal imp]

this is more of a physics problem, no?

[silent thorn]

yeah this is a heat problem

they should ideally ask this in a physics server

[normal imp]

its fine i can help

@heady widget have you learned Q = mcΔt?

[heady widget]

yes

[heady widget]

im not sure where to start

[normal imp]

ok, so the temperature change for brass and water will be opposite because one gets hotter and one gets colder, right?

[heady widget]

yes

[normal imp]

so Q1 = -Q2

and m1c1Δt = -m2c2Δt

for Δt, we can replace it with initial temperature minus final temperature: (Ti - Tf)

since Tf is the same but Ti is different, we have m1c1(Ti1 - Tf) = -m2c2(Ti2 - Tf)

we need to find Tf, so rearranging we have

m1c1Ti1 + m2c2Ti2 = Tf(m1c1 + m2c2)

then

Tf = (m1c1Ti1 + m2c2Ti2) / (m1c1 + m2c2)

we plug in our values and assume 1 is for water, 2 is for brass

m1 = 20 kg
c1 = 4180 J/k°C
Ti1 = 10°C

m2 = 10 kg
c2 = 380J/k°C
Ti2 = 71°C

so

Tf = (20*4180*10 + 10*380*71) / (20*4180 + 10*380)

Tf = 12.6527 °C

= 13°C if you account for significant figures

if you have any questions feel free to ask

[heady widget]

i dont totally get this part sorry, how did you get everything in the same equation?

[normal imp]

do you understand the process of how we get to Tf = (m1c1Ti1 + m2c2Ti2) / (m1c1 + m2c2)?

[heady widget]

no

[normal imp]

is there a step specifically you dont understand or the entire thing

[obtuse pebble]

@heady widget Has your question been resolved?

[obtuse pebble]

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[fluid snow]

#1195800467880558672 message

[fluid snow]

.close

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[signal dagger]

Hello if someone could help me in this exercise i just need to understand the steps

[signal dagger]

https://media.discordapp.net/attachments/903430333888884736/1219745303289401394/image.png?ex=660c6b56&is=65f9f656&hm=b5624cd91acb71022dd99e54438126ae025e7b50023d6b83c51fcc4e1a8111c8&=&format=webp&quality=lossless

so lets say for vertical its impossible because it needs to be infinite and there is no 0

[obtuse pebble]

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[modern palm]

And heres my work so far:

[warm shale]

hamr

[brisk matrix]

looks fine

[modern palm]

What is phase angle?

Is phase angle just zero?

Period is 2pi

And amplitude is 1?

Or wait

cos(theta +2pi) = sin(theta)

So the phase angle is 2pi? hmmm

[modern palm]

Is the phase angle 2pi? or 0?

[obtuse pebble]

@modern palm Has your question been resolved?

[obtuse pebble]

@modern palm Has your question been resolved?

[obtuse pebble]

@modern palm Has your question been resolved?

[obtuse pebble]

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[limpid spindle]

how many turning points are on this graph and how many points of inflection does it have?

[obtuse pebble]

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[unkempt rover]

HI ! Simple math problem here, my issue is that I can't seem to wrap my head around removing the square roots. here is the math problem I need to resolve:
sqr(x) - sqr(x-9) = 1
I've attempted to put everything to the power of 2 but since it ends up being x-x-9 =1... well it kinds of fail, hahaha.

Thanks in advance ! 😄

[polar fossil]

it does not end up being x-x-9

[unkempt rover]

Could you explain how come ?

[polar fossil]

if you're going to square both sides, you need to square the entire side

$\sqrt x - \sqrt{x-9} = 1$ \
$(\sqrt x - \sqrt{x-9})² = 1²$ \
$(\sqrt x - \sqrt{x-9})(\sqrt x - \sqrt{x-9}) = 1$

[warm shale]

hayley!

[polar fossil]

and then do FOIL

[unkempt rover]

would multiplying sqrt(x) by sqrt(x-9) result in x-9x ?

[polar fossil]

no

it would result in $\sqrt{x(x-9)}$

[warm shale]

hayley!

[unkempt rover]

therfore I would need to square both sides again afterward ?

[unkempt rover]

to remove this sqrt as well

[polar fossil]

yes... but isolate the square root first

[unkempt rover]

okay, I'll try things out, it already clears up a lot of things. Thank you very much ! Have a great rest of day !

.close

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[gritty geode]

need help with this. I have tried to write an equation in arithmetic and geometric form but neither work.

[knotty pewter]

Ok

Where are you stuck

[gritty geode]

I guess at the very beginning. Not sure how to do this problem. Like which explicit formula i should use

[normal imp]

try to look for patterns

[gritty geode]

I set them all to the same denominator to make things easier and I could not find a pattern

[knotty pewter]

Do you see an arthemetic progression on the numerator?

[gritty geode]

yeah

[lyric basalt]

the r value is the common ratio

[knotty pewter]

And a geometric series in the denominator

[lyric basalt]

which you can find by dividing the second term by the first term

[gritty geode]

yeah

For my ratio I got 5/6

[lyric basalt]

that's good

next, what is a (the first term)

[gritty geode]

the final geometric sequence I got was 5^n-1/4

but that didnt work when I tested it

[lyric basalt]

(3/2)((5/6)^(n-1))

[knotty pewter]

Because it is not a geometric progression

[gritty geode]

that was my question

Im not sure how to approach the problem

[knotty pewter]

You know the nth term of the numerator

[lyric basalt]

oh shoot mb

so sorry

[gritty geode]

no worries, stumped me too

[knotty pewter]

Correct?

[lyric basalt]

look at the numerator and denominator separately

numerator: 3, 5, 7, 9
denominator: 2, 4, 8, 16

can you find the equation for the arithmetic series for both the numerator and denominator? (put in terms of n)

numerator: 2n - 1
denominator:

[gritty geode]

numerator would be 6n-3 I believe

no

[knotty pewter]

How

3+(n-1)×2

[gritty geode]

6n-6

[knotty pewter]

That's 2n-1

How do you get 6(n-1)

[gritty geode]

2n+4

[knotty pewter]

Nth term of an ap is a +(n-1)d where a is the first term, d is the common difference

[lyric basalt]

okay numerator is 2n-1 right?

numerator: 2n - 1
denominator:

does that make sense?

[gritty geode]

yes

I couldnt do 6th grade math for a minute

idk why

[knotty pewter]

Nth term of a gp is?

[lyric basalt]

numerator: 2n - 1
denominator:

what is denominator

[gritty geode]

denominator would be 4^n-1?

[knotty pewter]

No

[lyric basalt]

try again

[gritty geode]

first number is 2

[knotty pewter]

It's ar^(n-1)

[lyric basalt]

remember the form a(r)^(n-1(

[gritty geode]

rate is 2

[knotty pewter]

Not (ar)^(n-1)

[gritty geode]

2(2^n-1)

[knotty pewter]

Yes

Z^m*Z^n=Z^(m+n)

[gritty geode]

I have not seen Z I dont think

what does that represent

or m

[knotty pewter]

Yes I was trying to help you reach the final answer

[lyric basalt]

numerator: 2n - 1
denominator: (2)(2^(n-1))

[knotty pewter]

You have (2^1)*2^(n-1)

[lyric basalt]

now simply put it in a fraction

[gritty geode]

I see. So it would be 2n-1/2(2^n-1)?

[knotty pewter]

Yes

Simplify

[knotty pewter]

Simplify this

[lyric basalt]

I don't think it's necessary?

[knotty pewter]

You get 2^n

Which looks more pleasing to my eyes

[gritty geode]

Ok I get it

Thank you guys

[knotty pewter]

2n-1/2^n

[lyric basalt]

oh I see

[gritty geode]

.close

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[timid silo]

what would the reference angle be for -660

[timid silo]

is it 60 because 720-660=60

[brisk matrix]

it is indeed 60

[timid silo]

thanks lol

i am having trouble with ref angles

appreacite it

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[surreal skiff]

Question 8 pls 🙏

[knotty pewter]

You want someone to check your answer?

[surreal skiff]

No I got it wrong, I need help with how to do it

[tepid yoke]

Where did 2.17 come from

[lapis charm]

Where did the 8 go

Real talk

[tepid yoke]

$8(1)^{1/2}+c=5$ just divide though by 8 to find c

[warm shale]

Luca M

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[untold harness]

Need help in #precalculus

[untold harness]

Please

Dude I gotta go to bed 💀

[obtuse pebble]

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[glass pilot]

Please explain how to do this with shell method

[chrome crypt]

Sure, I'll help.

So with the shell method, we need to ensure we choose a rectangular strp for our riemann sum that is parallel to the axis of revolution.

[obtuse pebble]

@glass pilot Has your question been resolved?

[chrome crypt]

@glass pilot There you go, let me know if you have any questions.

[obtuse pebble]

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[past walrus]

Can someone tell me how they managed to get u=x^6?

I understand the concept of integrals but prblms like these throw me off when they get a u-sub like this

[zenith raft]

you want an x^5 to show up when you differentiate u

[past walrus]

to like cancel out?

[zenith raft]

so, u = x^6 will do the trick

yea

[past walrus]

is this always the case?

Cuz a lot of integrals is just plug and check

[zenith raft]

depends what ‘always’ means ig

[ember frost]

right, but as you do more integrals, you develop an intuition of knowing what to plug

[glass pilot]

Thank you so much!!!

[past walrus]

I see, so if I am doing an integral which I cant find its u-sub easily for instance, I just plug something in to cancel out huh

[ember frost]

your top goal is to plug something so that the function becomes more elementary

[past walrus]

Ah ok

@ember frost Quick question

if ur going at an integral prblm and ur havin a hard time making it look more elementary what would u do?

[obtuse pebble]

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[zinc stump]

Hey guys, I need some help reviewing my work 🙂
I need to find the degree and minimal polynomial of $\alpha = \sqrt{3+\sqrt{2}}$ in \mathbb Q
this is what I've done, is this correct?

[warm shale]

mtr123
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[zinc stump]

and actually the right way to do it

I think I got something wrong there in the end cuz (a^2-7)(a^2+1) isn't the right polynomial but I need to use the quadratic formula and be fine

this should be better

[brisk matrix]

a^2 - 7 is certainly not the minimal polynomial, since a^2 = 3 + sqrt2

i'd say you want to start with
a^2 = 3 + sqrt(2) -> a^2 - 3 = sqrt(2) -> (a^2 - 3)^2 = 2
so the minimal polynomial would be
a^4 - 6a^2 + 7, which was what you found, though your factorization was sketchy (-7 * 1 =/= 7)

[zinc stump]

yeah I wrote it is wrong 🙂
so it is indeed $a^4-6a^2+7$? and is it the correct way to find it?

[warm shale]

mtr123

[brisk matrix]

i wouldn't raise to the fourth power right away, but i would do what i showed there

you should somehow argue that this is indeed a minimal polynomial though

i just said it was without reason

you could probably say this has at most 2 solutions, which must be of the form x = a^2, then showing those arent in Q. im not entirely sure though

[zinc stump]

so as I did here the solutions are $\alpha$ and 3-$\sqrt2$ which is indeed not in Q

[warm shale]

mtr123

[zinc stump]

not sure if that's what you meant though

[brisk matrix]

well you solved that x^2 = a^2 or x^2 = 3 - sqrt(2), which sounds right

so id say with a few words this should be sufficient

[zinc stump]

thank you! 🙂

@brisk matrix

[zinc stump]

.close

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[graceful marten]

I have just started this chapter, and it has the worst explination possible

[last pilot]

what's the question...?

and also *explanation

[graceful marten]

Integration by subsitution

How would I do it for Part B

I had to watch a ytb video for this stuff

explination was so bad

[obtuse pebble]

@graceful marten Has your question been resolved?

[graceful marten]

Bro integration by subsititution is so cringe

.close

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[sage dagger]

jus u wait til u get to partial fractions

[graceful marten]

.reopen

[obtuse pebble]

✅

[graceful marten]

LOL

Also

If you dont mind

To take u = 1+ sinx

Is it generally easier / better to always do

u^2 = 1 + sinx

Or does it not really matter?

--
Anyone can answer this, first come first serve

[brisk matrix]

i would not do u^2 = 1 + sinx

[graceful marten]

okok

Bc i saw a question where it took u = 1+ sin x

and then u^2 = 1+ sin x

And I was wondering if its a better choice

to do when you have it rooted

okokok tyty

.close