help-17

[quick rain]

it’s 3x3 though

[median crane]

This is the determinant for the 3x3

[vast shale]

yeah, started with 3x3 then you did cofactor expansion to get these smaller 2x2 matrices

[quick rain]

yes

so i just use that and get those numbers

[vast shale]

yeah

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

[livid horizon]

What's the latex for matrix symbol

[vast shale]

$\begin{pmatrix} %
a & b \
c & d \
\end{pmatrix}$

[twin meteor]

$\begin{pmatrix} % 
      a & b \\
      c & d \\
   \end{pmatrix}$

[quick rain]

aha got it

thank you so much

[median crane]

\begin{vmatrix}
a&b\\
c&d\\
\end{vmatrix}

[vast shale]

[twin meteor]

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

[vast shale]

too late

sniped

[quick rain]

got -59

can y’all help with another one by any chance?

[median crane]

I’m on the tube

[vast shale]

it even got an error smh

[quick rain]

this one is solving a matrix but i have to find the bottom corner first

[median crane]

Same thing

[quick rain]

let me take a pic of it

[vast shale]

,w determinant ((2,3,1), (-1,5,-10), (3,-1,5))

[twin meteor]

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[median crane]

[quick rain]

that’s the whole problem

[quick rain]

i have to have pro to get the step by step

[vast shale]

so you want it as an upper triangular matrix

[quick rain]

i just need it like this

[vast shale]

are you comfortable with row operations?

[vast shale]

step by step

dont you understand the technique at least

[quick rain]

1234
0234
0034

with the zeros on the bottom corner

[quick rain]

yes i do

just troubling with these specific numbers

[vast shale]

yeah we call those matrices upper triangular

[clear oriole]

enzwi pro

[vast shale]

zan

[quick rain]

then yes

[vast shale]

lmfao my LA teacher is watching me

[quick rain]

are you able to help with this problem?

[vast shale]

do row operations then

[vast shale]

on your matrix

[quick rain]

ok

so multiply the first equation by 3 then subtract that by the third equation multiplying by 2?

[vast shale]

$\begin{pmatrix} %
2 & 3 & -1 & -1 \
-1 & 5 & 3 & -10\
3 & -1 & -6 & 5
\end{pmatrix}$

[twin meteor]

$\begin{pmatrix} % 
      2 & 3 & -1 & -1 \\
      -1 & 5 & 3 & -10\\
     3 & -1 & -6 & 5
   \end{pmatrix}$

[vast shale]

whichever way that you'll get you the form
$\begin{pmatrix} %

• & * & * & * \
  0 & * & * & * \
  0 & 0 & * & * \
  \end{pmatrix}$

[quick rain]

^yes or no

[vast shale]

rip tex

[vast shale]

what do you get after that operation?

i cannot like walk you through step by step since there are may ways you can go about this

[quick rain]

i got

2,3,-1,-1
-1,5,3,-10,
0,11,9,13

[vast shale]

try texing that

copy the template code

[quick rain]

i can’t for some reason

$\begin{pmatrix}%

[vast shale]

tex fixed
$\begin{pmatrix} %

• & * & * & * \
  0 & * & * & * \
  0 & 0 & * & * \
  \end{pmatrix}$

[quick rain]

like that?

[twin meteor]

tex fixed
$\begin{pmatrix} % 
* & * & * & * \\
0 & * & * & * \\
0 & 0 & * & * \\
   \end{pmatrix}$

[vast shale]

this is what i meant

[vast shale]

just copy the black box and replace your entries

[vast shale]

but in general just row reduce to get the zeros on the lower left triangle

[quick rain]

tex fixed
$\begin{pmatrix} % 
2* 3 -1* -1* \\
-1* 5 * 3*-10 * \\
0& 11 & 9 & 13 \\
   \end{pmatrix}$

[twin meteor]

GhostPepper

[quick rain]

did it wrong

[vast shale]

& &

[quick rain]

can we just use that

i need to get this done and i am pretty sure i got that first multiplication right

[vast shale]

$\begin{pmatrix} %
2 & 3 &-1 & -1 \
-1 & 5 & 3 &-10 \
0& 11 & 9 & 13 \
\end{pmatrix}$

[quick rain]

yes

[quick rain]

so now what 0 do i get next

i could get rid of the -1

[clear oriole]

-13 btw for the bottom right entry

[quick rain]

multiply 2 equation by 0then subtract that from multiply the 3 equation by -1

haha that’s crazy

i just get the same thing on the bottom

[vast shale]

wait I didnt check your row operation, can u use the format

R3->2R2 + ?R3 ...etc

[quick rain]

wdym?

[vast shale]

eh nvm we'll go with words then

,w 3(2,3,-1,-1) -2(3,-1,-6,5)

[twin meteor]

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[vast shale]

is this what you meant

[quick rain]

yes i get that

[vast shale]

.

[vast shale]

go on

[quick rain]

i got rid of it

my answer is the same as the vottom

but the 11 is -

0,-11,9,13

bottom is 0,11,9,-13

[vast shale]

multiply Row 2 by 0?

[quick rain]

yeah

is that wrong?

[vast shale]

can you list for me the elementary row operations

[quick rain]

just gonna ask my teacher for this help, too confusing and i’ve learned differently

it shouldn’t be this complicated to solve a matrix

thanks for the help bro

[vast shale]

Row Swap, multiply a row by a constant and add a multiple of one row to another row.

right

[vast shale]

you'll get the hang of it

[vast shale]

so basically multiplying by a row by zero then adding is the same as doing nothing

R2-> R2 + 0R3

[quick rain]

yes

that’s what i did to get the 0

[vast shale]

@clear oriole wanna halp i think the first operation was wrong

[quick rain]

did the inverse of the number to cancel out the number to get a 0

[clear oriole]

can't help seriously rn, third row is fine, you shouldn't multiply a row with 0 though

just continue eliminating

[vast shale]

that does not change the row

think of another way

[quick rain]

ok

multiple first then subtract it by multiple second

[vast shale]

we want row 2 to have a zero at the first entry right

[clear oriole]

(or add it, yes)

[quick rain]

yes

[vast shale]

what can we multiply by row 1's 2 to get 0

-1 + x(2) = 0

lmk if thats not clear

[quick rain]

first equation would be multiplied by -1

then subtracted from the second equation multiplied by 2

[vast shale]

you can do this compactly by just multiplying one row by a constant

[vast shale]

what is x

[quick rain]

1/2

[vast shale]

okay and what happens when u multiply row 1 by that

then adding it to row 2

[quick rain]

ok let me try

[vast shale]

im just gonna just make u learn my way

[clear oriole]

||LU decomposition with partial pivoting||

[vast shale]

shoo

[quick rain]

new row is

0,7.5,2.5, -10.5

[vast shale]

lemme verify

,w 1/2(2,3,-1,-1) + (-1,5,3,-10)

[twin meteor]

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[vast shale]

okay

[quick rain]

is 7.5 right?

[vast shale]

you shld get 6.5

[quick rain]

it’s 6.5

ok

all i need is the 11 a 0

and i’m done

[vast shale]

$\begin{pmatrix} %
2 & 3 &-1 & -1 \
0 & 6.5 & 2.5 & -10.5 \
0& 11 & 9 & -13 \
\end{pmatrix}$

[quick rain]

what’s next?

[vast shale]

yup

any ideas on getting a zero there

[quick rain]

dividing them both?

[vast shale]

dividing by what

[quick rain]

wait, it would have to be 11

[vast shale]

we just want a zero on the 11, we could divide by 11 to get a one there, then subtract from row 2

[quick rain]

yes

wait

a 1?

oh right

[vast shale]

$\begin{pmatrix} %
2 & 3 &-1 & -1 \
0 & 6.5 & 2.5 & -10.5 \
0& 1 & 9/11 & -13/11 \
\end{pmatrix}$

[vast shale]

what next

[quick rain]

subtract by row 2

[vast shale]

11/11

[quick rain]

haha

forgot

[vast shale]

subtract from what and by what constant

[quick rain]

6.5

[quick rain]

i honestly don’t know

[vast shale]

try to write it for me in this way

[quick rain]

not familiar with that

leave for class in 10 mins…

🤣

[vast shale]

basically means 1/2 was multiplied to row 1 and we added it to row 2

[quick rain]

oh ok

[vast shale]

R2 -> ?R3 + R2

fill in

[quick rain]

so R2-> 6.5R3+R2

wait

it would be 6.5R3-R2

that’s the only way to get 0

[vast shale]

leave the row you're applying it to what can u do to row 3

-6.5 right

[quick rain]

yes

[vast shale]

so
R2-> -6.5R3+R2

[quick rain]

so it would give us a 1 and a 0

1 in the second row

0 in the third

[vast shale]

yeah and remembering all your row operations

[vast shale]

what can you then do to get it in this form

[quick rain]

would you be able to do the rest of the math in that formula thing so i can see the full operation

[vast shale]

.

[quick rain]

multiply third row by 6.5 then subtract it from 2

[vast shale]

.

[quick rain]

second row

[vast shale]

im saying after you've done that, and you've got this form

0 @ @ @
0 0 @ @
0 @ @ @

[quick rain]

i need to get going, i’ll ask my teacher the rest.

once again, thank you

[quick rain]

you would finish the math for the @s

[vast shale]

you would then do row swapping then you're done

[quick rain]

ok

[vast shale]

terrible helping from me, but thats it

[quick rain]

not at all

super helpful

[vast shale]

if we're done i can close this

[quick rain]

yes

[vast shale]

alr good luck

.close

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[maiden river]

I've filled in the blanks as far as I could. I've spent the past hour trying to figure out what x, y, and z (angles) would be with no luck.

[maiden river]

My bad

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[random prism]

.close

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[fallen agate]

Need some help proving using structural induction

[fallen agate]

this is the proof i have written

am i missing anything

or doing something incorrect?

[vocal sleet]

@fallen agate Has your question been resolved?

[fallen agate]

<@&286206848099549185>

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@fallen agate Has your question been resolved?

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@fallen agate Has your question been resolved?

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@fallen agate Has your question been resolved?

[fallen agate]

<@&286206848099549185>

[elfin laurel]

It looks good to me in terms of - all the claims you make are indeed true,
but the claim that "Any sequence created with rule 1 or 2 will always have at most 1 R at the beginning" might need some proof in addition because it feels non-trivial as compared to most of the other claims made in this proof

[fallen agate]

Yea i had the same though but I am not sure how to prove it. Do you have any directions?

[elfin laurel]

you can do another structural induction proof for it

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[steep merlin]

do i plug in 4x into the t position here?

[steep merlin]

little confused on what to start doing here

[maiden iron]

hmm, best you can do is let u = 4x maybe

[steep merlin]

use u substitution?

[maiden iron]

yea

[steep merlin]

let me see

isnt this FTOC tho?

[maiden iron]

it is

[steep merlin]

honestly no idea

got lost in my work

.close

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[kind light]

how do people get d/dx 2^x= 2^x ln2?

[kind light]

like whats the method or is there like some sort of rule

[oblique gazelle]

Its a "rule" but there is a proof

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[wanton stone]

Hey y’all can anyone help me with this please

[livid horizon]

,rotate

[wanton stone]

i've proven the fraction but have no idea how to solve the lim

sorry that was really badly worded

this is the question im stuck on

not the lim of summation

[gaunt sparrow]

Factor n^2 out top and bottom and then discard terms that go to 0.

[wanton stone]

expand first?

then factor out?

[gaunt sparrow]

Yeah you can do

Or you could factor n out of every factor

ALternatively if you don't want to expand

Comes out the same

[wanton stone]

ok thanks done it

.close

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[young hollow]

I need help with number 3 it says solve k for every X

[paper depot]

more precisely, it says:

find k such that the inequality is true for all x

[livid horizon]

What have you tried?

[young hollow]

Im not sure but i mightve solved it

I got two solutions cause of the absolute value

[livid horizon]

Ye

-2 < x²+Kx+.....<2

[young hollow]

ive also noticed the bottom part is always positive which is very helpful

[livid horizon]

Yes

[young hollow]

is the final solution the intersect or the union of the two solutions?

[vast shale]

do the integral firtst

let t = 5tan theta

[young hollow]

i think thats already been solved

[vast shale]

yo i cannot read that language

what does that eman

[young hollow]

i said bellow

find k such that the inequality is true for all x

[vast shale]

k

basically function is equal to or more than 0 less than 2

[vocal sleet]

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[young pawn]

A grazing paddock is to be fenced on all sides. It is rectangular in shape, with a length of 244 m and a
breadth of 178 m. If fencing costs $20 per metre, find the cost of fencing required. (2 marks)

[young pawn]

?

[elder scaffold]

perimeter?

[young pawn]

umm no its asking in dollers

well yes you need to do area

[elder scaffold]

when you fence do you fill the entire area, or just the perimeter?

[young pawn]

perimeter

[elder scaffold]

right, so start with that

[young pawn]

so 244 + 178

so 442

[elder scaffold]

not quite, a rectangle has 4 sides

[young pawn]

442x2

[elder scaffold]

right

[young pawn]

so 884

884 dived by 20

so $44.2

[elder scaffold]

think about it, its $20 per metre of fencing, that means that each metre of fencing costs $20.

we have 884m of fencing and each metre is worth $20

[young pawn]

44.2

20 x 44.2

[elder scaffold]

no that just gets us back to the start

we dont want to work with 44.2

we are looking for 884 x 20

we can think of it as, if each metre of fencing costs $1 then 884m would be $884 dollars, but $20 is 20 times larger than $1 so we have to multiply it by 20

[young pawn]

884 x 20

?

?

[elder scaffold]

say you wanted to buy 3 apples and each apple costs $3 how much would you have to pay?

[young pawn]

9

3x3

[elder scaffold]

right

same situation 884m of fencing and we are buying it at $20 per metre

[young pawn]

so we have to find how much tht would be

right

[elder scaffold]

right

[young pawn]

so wouldnt it by 884 x 20

than or what

im not sure

[elder scaffold]

yes it would be

[young pawn]

16880

$16880

right

[elder scaffold]

yep

wait

slightly off

[young pawn]

what do u mean

[elder scaffold]

its $17680

800x2 is 1600, 80x2 is 160 4x2 is 8, add them all together 1600+160+8=1768 then times that by 10 = 17680

[young pawn]

ohhhh okkk thanksss

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[young pawn]

Each face of a 4-sided die is triangular, with a base of 2.1 cm and a height of 1.8 cm. Find the total area of
all 4 faces of the die

[young pawn]

so wait is it 1/2 x b x h x 4

???

anyone

[urban edge]

calm down its been like three minutes

and yes

[young pawn]

ok thanks

sorry another queestionn

Number B

i get how we have to times and all

but is it 3 x 2.1cm?

and is the area of a square?

[young pawn]

2 min

[mild flower]

,tex .plane geom

[twin meteor]

hayley

[mild flower]

yeah parallelogram is base times height (top right)

[young pawn]

ok so 3 x 2.1

kkk thankssss

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[young pawn]

omggg sorryrrrr

again

hehheehhee

[mild flower]

split this up into smaller shapes, maybe one big shape minus a cutout of a smaller shape

[young pawn]

so it could be 20 x 14 ( square) and and 1/2x14x20( triangle)

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[still dune]

guys

is my teacher trippin

or am i trippin

how is this equal to positive sqrt(6)/4

[urban edge]

whered da negative go

[still dune]

isn't it negative

exactly

ok thanks

imma go sue my teacher now

[young pawn]

umm i was doing a qustion

why is it like this

huh

RIGHT HERE

[urban edge]

it looks like this was part of a sin sum of angles, you can check using a calculators

[young pawn]

!!!

where can i complain?

[still dune]

well thanks for confirming @urban edge

[young pawn]

.close

[still dune]

ig my teacher posted the wrong answer key again

😭

[urban edge]

oof, i hate wrong answer keys

[young pawn]

..

HELLO

oops capital

.close

[elder scaffold]

oh apparently i own this i think

.close

nvm

theres another galaxy

of course

[nimble elm]

it seems their question was answered as well

.close

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[shadow oak]

If E is the intersection of the extension line of diameter AB in circle O, then line DO = line DE, and arc BD = 6, what is the area of ​​arc AC and line AB passes through the center O, and there is B between lines OE.

[vocal sleet]

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@shadow oak Has your question been resolved?

[patent nymph]

Have you drawn a picture?

[patent nymph]

Are you given any information about the length of the radius?

[patent nymph]

can you send the original question?

[shadow oak]

There's nothing about area of radius

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[rough pine]

im trying to find solutions to this separable ODE

i have made some notes from this in a lecture but im struggling to understand whats going on after a certain point in my notes

would someone be able to work with me to understand what is happening plz

im up to here at the minute

i dont rly understand the 'two regions' part and below

[somber portal]

that has to do with the absolute value of a number (in this case, of the big ass fraction)

if the number would be negative, the absolute value changes the sign

so where the fraction is positive (either both numerator and denominator are positive, or both negative), the fraction is as given
where the fraction is negative (each has a different sign) then you're operating with the opposite fraction

[rough pine]

but then in the diagram

its positive when x < sqrt -3

[somber portal]

because that inequality is not correctly solved

$\frac{x-\sqrt{3}}{x+\sqrt{3}}>0$

[twin meteor]

LordFelix

[rough pine]

how would i solve this

[somber portal]

that inequality has two cases

[rough pine]

my lecturer said to treat the numrator and denomitor separetelty

[somber portal]

case 1: both numerator AND denominator are positive

[rough pine]

so numeratr > 0

and denoitr > 0

[somber portal]

they are both positive when x is larger than sqrt3

case 2: both numerator AND denominator are negative

they are both negative when x is lower than minus sqrt3

so the fraction is positive IF you are in the red marked area

not including the actual points

[rough pine]

so my lecturer is wrong

💀

[somber portal]

the fraction is negative if you are in the purple marked area

[somber portal]

it's not "wrong". It's just very badly explained

[rough pine]

but x >sqrt3 and x > -sqrt 3

is defo wrong

for it being >0

the fracrion

[somber portal]

no. that's the "very badly explained" part

those inequalities are, respectively, "numerator is positive" and "denominator is positive"

[rough pine]

i see

[somber portal]

at the very least i ask those two inequalities to be identified by my students

now, remember that i mentioned the purple and red areas, and explicity said that sqrt3 and minus sqrt3 specifically were not included in either

[rough pine]

yes

[somber portal]

on sqrt3, numerator is 0

so the fraction is equal to zero

so you cannot take the logarithm of the fraction

on minus sqrt3, the denominator is 0. So you cant compute the fraction, and thus cant compute the logarithm either

is it clear now?

[rough pine]

sorry i was reading ur comments

makes A LOT more sense now thanks alot

also

how does the red circle also mean ---> modulus x > sqrt3 ?

[somber portal]

look at the red area

the red area is x either smaller than minus sqrt3, or larger than plus sqrt3

which means that the red area is the absolute value of x larger than sqrt3

[rough pine]

ofc

that makes so much sense

ur a legend

thank u so much

amazing teacher

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[midnight walrus]

I just filled in the table sodoku style but how do I verify that this is a group?

[paper depot]

hold on

was the task to fill in the table SUCH THAT it defines a group, or what

[midnight walrus]

yes I think in the end it has to be a group

[paper depot]

hm

alright so it looks like x is the identity

verifying assoc is gonna be kind of painful

[midnight walrus]

can you maybe explain how to do that

I haven been a bit puzzled on how to do that

[paper depot]

go through all 216 trios of elements and compute the two products that assoc says have to be equal

it is long and painful

also i have noticed something sus

according to this table, yy = b

(yy)y = by = a but y(yy) = yb = c

so assoc is already violated lmo

lmao*

[midnight walrus]

oh ye I see haha

[paper depot]

your table's wrong and you have to redo it i think

[midnight walrus]

ye thank you for help lol

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[stoic mica]

hello, i'm confused on what's done here. can somebody please explain?

[stoic mica]

why is 5/12 pi = sqrt(6) - sqrt(2) / 4 ?

[raven owl]

Just is. Guess they treated it as a commonly known value. If u wanna calculate it do cos(pi/6+pi/4) and use addition identities

[vocal sleet]

@stoic mica Has your question been resolved?

[stoic mica]

Yeah, using the addition identities it gives me the same result although I wonder if that's something I should know by heart or if the professor had already calculated it beforehand, because 5/12 isn't one of the trig known values and it's not something I can easily calculate in my head

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[rustic chasm]

Can I get your help on this

[heavy yoke]

do you know u-substitution?

[vocal sleet]

@rustic chasm Has your question been resolved?

[rustic chasm]

Yeah

My bad

I sent the wrong one

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[slim monolith]

please help me

[flat whale]

What did you try

[lucid bane]

!status

[vocal sleet]

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

[lucid bane]

!showwork

[vocal sleet]

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

[slim monolith]

I dont know how to solve the Q

[urban laurel]

i recommend making in terms of sinx only

Hint : ||cos² x = 1-sin² x||

[vocal sleet]

@slim monolith Has your question been resolved?

[slim monolith]

How...i dont get it

[urban laurel]

by using this identity cos² x = 1-sin² x

$\int x\sin x \cos ^2 x \ \dd x = \int x \sin x \left( 1-\sin ^2 x \right) \ \dd x \ = \int x\sin x -x\sin ^2x \ \dd x$

[twin meteor]

Adam Chebil

[urban laurel]

then split the integral

[slim monolith]

Ok, i try

Thanks a lot

[twilit locust]

On the stream we solved it by IBP with u=x and dv=sinx cos^2(x)
Would be a good approach to the problem!

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[tame pasture]

sooo...

[tame pasture]

I have many multidimensional scalar fields.

p_1, p_2, ... p_m

each a function R^n -> R

(all with the same n)

All functions are continuous, defined and differentiable everywhere along any axis

And for every dimension, there is a maximum v_{max,n}

Can I always find a m*n 'matrix' of variables v_{1,1} ... v_{m,n}

so that

the sum of all variables along a column (v_{1,n} + ... + v{m,n} for a given n) equals v_{max,n}

and for every m, p_m(v_{1,k}, ..., v_{n,k}) is never higher for k ≠ m than for k = m

[vocal sleet]

@tame pasture Has your question been resolved?

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@tame pasture Has your question been resolved?

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@tame pasture Has your question been resolved?

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[midnight walrus]

So G is a group and I have to fill the table but I dont really know how to continue. So far I just added the obvious ones what are the steps to fill such a table?

[lucid bane]

are you given the group is commutative/abelian?

or is all you are given the table

[half imp]

Look at b*y

It's like sudoku

[lucid bane]

yes, tables behave like sudoku, each column/row has exactly one of each element

[half imp]

Not commutative, look at bc vs cb

[lucid bane]

ah right

[midnight walrus]

it says G {a,b,c,x,y,z} and * on G such that it is a group

[lucid bane]

alright

[half imp]

But, if two elements multiply to the identity, then you can reverse the order

[midnight walrus]

does that mean there is more than one right solution here?

[half imp]

no

[lucid bane]

and you've already deduced x is identity by process of elimination

[half imp]

Well, because x*x=x

[midnight walrus]

would this be correct then?

i already attempted it but I wanted to ask for some advice

[half imp]

probably

[midnight walrus]

is it not important for the table to show assoc?

[half imp]

thr main diagonal doenst look right

[half imp]

It is important

[midnight walrus]

what do you mean

[half imp]

The only nonabelian group of order 6 is D_3

D_3 has 3 elements which are squares: e, r, r^2

your table has 4

[midnight walrus]

sry I am new to algebra

what is 4 in my table in plain words

[half imp]

a^2 = y

so a and y should commute

(a*a)*a = y*a = z a*(a*a) = a*y = c

Not associative

[midnight walrus]

ah ok i understand

so it is not just straight sudoku it is also important to look at associativity since that is definition of a group

if I get the table like that it is correct I assume?

[half imp]

yes

Probably you should not be filling in and hoping it works

[midnight walrus]

is there a system to solve or just trial and error?

[half imp]

Just logical deduction

What does the table look like right now?

[midnight walrus]

what do you mean right now? the first picture I send was where i am stuck

[half imp]

I gave you a few suggestions though

[half imp]

.

[half imp]

.

[midnight walrus]

ye I am gonna try thank you for your help

[half imp]

np

[midnight walrus]

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[naive garnet]

I cannot solve d please save me

[vocal sleet]

@naive garnet Has your question been resolved?

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@naive garnet Has your question been resolved?

[naive garnet]

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[weak hornet]

what is the formulae for calculating all roots of an (n)th rooth

[weak hornet]

like for example J^3 = 1 has 3 possible roots, J^n has n possible roots

[orchid pier]

you take it to the polar form

first

[vocal sleet]

@weak hornet Has your question been resolved?

[weak hornet]

how do you convert to polar form again?

as in how do you calculate r and theta

[orchid pier]

r = sqrt(re^2 + im^2)

[weak hornet]

what's re

and im?

[vocal sleet]

@weak hornet Has your question been resolved?

[orchid pier]

the real part and the imaginary part

[weak hornet]

ah

so that's r

how do you calculate theta

[orchid pier]

arctan(im/re)

[weak hornet]

ah that's not good

thank you for your help

wait

it's 0

awesome i can calculate this for once

thankyou

so I have it in polar form, how do i calculate roots?

[weak hornet]

<@&286206848099549185> i got told that conveting my number to polar form would help me calculate the nth root, but im not sure how go from there

[vocal sleet]

@weak hornet Has your question been resolved?

[vocal sleet]

@weak hornet Has your question been resolved?

[flat whale]

https://brilliant.org/wiki/roots-of-unity/

[weak hornet]

thank you so much this helps alot

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[vast shale]

How to figure out if something is a homogeneous function or not

[vast shale]

For example

[paper depot]

check it against the definition of a homogeneous function

[vast shale]

I dont understand it

[paper depot]

do you have the definition in front of you?

[vast shale]

Yes

[paper depot]

uh huh

show us the definition and the function whose homogeneity you're interested in

[vast shale]

Okay

How to know which formula is to be used where

[paper depot]

none of these 3 pictures show a definition of "homogeneous function" do they

[tidal dock]

these are not definitions, these are corollaries

[paper depot]

but also i might have to stay out of this

[vast shale]

I understand this, i guess i font understand the corollaries

[tidal dock]

recall $(a+b)^2 = a^2 + 2ab + b^2$

[twin meteor]

artemetra

[tidal dock]

but also it seems like the third corollary is a generalization of the first one

[vast shale]

Here why did they say note that it is not a function

And then they worked it out without the cosec inverae

And then it is a function

I read this

But then here

What am i missing

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

<@&286206848099549185>

.close

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[paper depot]

cropping

does this question actually have 4 answer options that you didn't show us?

or is it free response

ok

do you know in general how to calculate euler's totient function?

you didn't answer my question

my bad, i was trying to ask if you knew of any formulas by which phi can be computed.

or if you just threw it into a calculator

[vocal sleet]

@rapid bane Has your question been resolved?

[vocal sleet]

@rapid bane Has your question been resolved?

[lyric fossil]

@rapid bane please don’t call ann “bro”

then don’t expect to get help

[hard atlas]

if you want to get help its usually a better idea not to piss people off

[urban edge]

*she

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[stone mauve]

Hello can anyone help me with this question

checking in with WolframAlpha it says that lim x -> -inf of this expressions is 1 / (1+sqrt2)
but the answer I got is the answer of lim x -> inf of this expression not lim x-> -inf

can anyone point out what and where goes wrong
thanks
(sorry for the bad hand writing)

[vast shale]

i dont think 1/x when x is -oo will be zero

[graceful surge]

this is the mistake

$\sqrt{x^{2}}=\left|x\right|$

[twin meteor]

Combustion

[graceful surge]

and |x| when x<0 is -x

[stone mauve]

ah thank you

[graceful surge]

[stone mauve]

i simply forgot the fact that

sqrt (x^2) is just |x|

let me see if i can solve this

[stone mauve]

how do you say that |x| is -x because lim x -> -inf

my teacher requires rigorous explaination

[graceful surge]

do you know what |x| does?

[stone mauve]

yes

[graceful surge]

when x is negative, it multiplied x by -1 to make it positive again

so basically |x| when x<0 is -x

i don't know how you would rigorously explain it 🤷‍♂️

[stone mauve]

same

i did get the correct answer but i dont know how to word it
too bad i guess haha thanks combustion

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[earnest briar]

Hi! I have a calculus 2 question. I solved the question but I'm not sure if I showed all work.

[earnest briar]

"Determine the convergence or divergence of the sequence with the given nth term. If the sequence converges, find its limit. (If the quantity diverges, enter DIVERGES.)"

The issue is towards the end, I know that n is larger than ln(n), and as n increases, n becomes exponentially larger than ln(n), meaning eventually it'll leave ln(n) in the dust, making it effectively 0

...but how do I show work for that?

[graceful surge]

l'hopital?

[earnest briar]

how do I apply that here? I saw videos online saying to treat n as x, but is that allowed?

[tidal dock]

yes

since you already taking the root to infinity

[earnest briar]

so this then?

[graceful surge]

yep

[earnest briar]

alrighty!

Thank you guys!

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[vocal sleet]

Please don't occupy multiple help channels.

[river kettle]

you shoud really listen to @vocal sleet

[vocal sleet]

Please do not ping individual helpers unprompted.

[outer warren]

we want you to read the #rules and guidelines and #❓how-to-get-help

which makes it smoother to ask for and receive help

failure will only antagonise would be helpers

like being informed of the 15m rule moments ago yet deciding to ping before 15m anyway.

going to close, stay in your original channel and try to wait paitently

.close

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[tidal dock]

just ask

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[vast shale]

I got it wrong apparently

Idk how

Pls help

[astral needle]

,rotate

[vast shale]

Sorry lol

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@vast shale Has your question been resolved?

[vast shale]

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[gloomy plume]

For differential equations, given solutions y1 and y2, I am asked to show if they are linearly independent. Do I simply find if the wronskian is unequal to 0?

[tribal moss]

yes it is correct

[vocal sleet]

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[pure thorn]

how do I simplify this?

[white whale]

!status

[vocal sleet]

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

[pure thorn]

I could only convert bottom part to (√x + √y)²

[white whale]

There is also a common factor in the numerator part

[pure thorn]

(x^7/4 - y^5/4)^2?

i don't know what to do after that

[white whale]

No, that's a wrong factorization

[pure thorn]

oh right

wait why did I even do that

[pure thorn]

so it is actually x^2,5 * y ^2,5 (x-y)?

[white whale]

Yes

[pure thorn]

okay I think I solved it now, thank you 👍

.close

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[crystal drum]

can I get some help with this pls?

part b

[vocal sleet]

@crystal drum Has your question been resolved?

[crystal drum]

.close

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[long dagger]

(2^16/3^6) = 9 * (4/3)^n

[long dagger]

(2^16/3^6) = 9 * (4^n/3^n)

(2^16/3^6) = (3^2 * 4^n) / 3^n

and now I'm stuck

[civic drift]

$\frac{2^{16}}{3^6} = \frac{2^{2n}}{3^{n-2}}$

[long dagger]

yeah that's kinda what I did, I put 3^2 and 4^n together in the same fraction though

so (3^2 * 4^n) divided by 3^n

[twin meteor]

ColdTee

[civic drift]

There you go

[long dagger]

3^n - 3^2 = 3^(n-2)

makes sense

[civic drift]

$3^n \cdot 3^{-2} = 3^{n-2}$

[twin meteor]

ColdTee

[civic drift]

This is incorrect

[long dagger]

hm?

[civic drift]

$3^n - 3^2 = 3^{n-2}$

[twin meteor]

ColdTee

[civic drift]

This is what you wrote

[long dagger]

yeah

[civic drift]

Compare it to what I wrote

[long dagger]

I'm confused

[civic drift]

You are subtracting while I multiplied

[long dagger]

oh

[civic drift]

Fine lets

[long dagger]

wait I'm dumb

[civic drift]

Take an example

[civic drift]

Put n = 2 and find out if this is true

[long dagger]

(3^n) / (3^2) = 3^(n-2)

[civic drift]

Or any natural number

[long dagger]

^ is that correct now?

[civic drift]

Yes

[long dagger]

yeah I just realized haha, sorry

[civic drift]

It's fine

[long dagger]

ok I'm with you now, (2^16) / (3^6) = 2^(2*n) / ( 3^(n-2))

can't use calculators btw 🙁

[civic drift]

3^{n-2}

$\frac{2^{16}}{3^6} = \frac{2^{2n}}{3^{n-2}}$

[twin meteor]

ColdTee

[long dagger]

yes

[civic drift]

Compare powers on both sides of the equation

2n = 16 and n-2 = 6

[long dagger]

ooooooh

both sides/ fractions have the same bases

so you can just compare the exponents ?

[civic drift]

Yeah exactly

Yes

[long dagger]

aah makes sense

2n=16
n=8

[civic drift]

Yep

[long dagger]

aah, thanks man

appreciate it

[civic drift]

Np

[long dagger]

.close

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[vast shale]

can i get some elaboration on this?

[vast shale]

im pretty confused on what they are trying to indicate

[median crane]

I think it just means you can choose a reference point when calculating GPE ,but for KE, x=0 is only when the spring is at the natural length.

[vast shale]

okay but again why

why this distinction

oh wait

F = kx

x = 0 implies F = 0

should we use something else as our origin

suddenly x = 0 should not be F = 0 anymore

wait so i guess to solve this we like

uh

F = k(x-x_0) with x_0 being where the spring is neutral?

[median crane]

[vast shale]

lets say we have x = 0 somewhere

and its neutral at x = 3

then k(3-3) = 0

yeah i guess that checks out

so for the work we now have to like

[
W = k\int_{x_1}^{x_2} x-x_0 \dd x
]

[median crane]

\i

[vast shale]

[
W = k\p{\f{x_2^2 - x_1^2}2 -x_0x_2 +x_0x_1}
]

[vast shale]

okay thats annoying

does this simplify like at all

wait maybe im wrong hol up

[
W = k\int_{x_1-x_0}^{x_2-x_0} x-x_0 \dd x
]

[vast shale]

would this be correct

[median crane]

Idk what youre trying to do

[vast shale]

im trying to find an expression for the work of a spring given a shifted origin

[median crane]

we need @cobalt crypt help

[vast shale]

oh is it plus

[
W = k\int_{x_1+x_0}^{x_2+x_0} x-x_0 \dd x
]

[vast shale]

wait is this it

snow in his im too lazy to write so i only react era

[vast shale]

ok yes the bounds make sense now

lets integrate

ok so

[\dfrac{k\cdot\left(x_2-x_1\right)\left(x_2+x_1\right)}{2}]

[vast shale]

does this make sense

processing

[full arch]

yes

[vast shale]

[full arch]

and thats fine

idk what the textbook is on about

[vast shale]

what is U_el in this case

how do u represent this as a negative change of U

oh wait

wait

hol up

hol up boys and gals

expanding gets u

no i mean

difference of squares

[
\f{kx_2^2}2 - \f{kx_1^2}2
]

[vast shale]

which is like

exactly the same expression

so wait

it just

doesnt matter?

well it just means that we are going to be measuring x_2 and x_1 from whatever new origin point we have

but otherwise, the expression remains consistent

so yeah wtf is up with the book

[full arch]

which is why apparently 'x must be zeROO'

[dull kayak]

[vast shale]

yeah odd distinction

[vast shale]

hlo airy

[magic wasp]

I'm pretty sure it's because that formula only applies within a range of deformation

[full arch]

maybe this also

[vast shale]

elaborate?

[magic wasp]

If you stretch a spring too much, it won't come back to how it was before

[dull kayak]

[magic wasp]

https://en.wikipedia.org/wiki/Elastic_energy#Elastic_potential_energy_in_mechanical_systems

Within a certain range of deformation, k remains constant

Implying it doesn't outside of that range

[vast shale]

so what should i takeaway from this

[cobalt crypt]

i was too lazy to type but yall are fucking this up so bad

@vast shale do you know about potentials

[vast shale]

like potential energy in general?

[cobalt crypt]

like

in calc 3

where you learn about scalar potentials

[vast shale]

oh you mean those

yeah i took it, but is it not possible to stray from like a highschooler explanation lmao

[cobalt crypt]

i mean its the same explanation

like

the gravitational and elastic forces are both conservative

so you can find a potential for them

F = grad U or something

in the gravitational case it happens to be U = mgy

in the elastic case it happens to be U = 1/2 kx^2

[vast shale]

yeah

i get that

[cobalt crypt]

the only reason you cant "shift" the zero point for elastic energy while you can for gravitational energy is because like

usually you model gravity as being spatially uniform near the surface of the earth

so F = g or something

probably -g

F = -grad U

[vast shale]

right right

[cobalt crypt]

whereas the equation F = -kx is definitely not spatially uniform

the amount of force you experience depends on x

if you do shit like F = -GmM/x^2

then you start having to worry about what your x is

[vast shale]

yeah because its varying with x

so i guess the point is that you need to have a "fixed" reference point when your thing isnt spatially uniform?

[cobalt crypt]

well you need to know what x is

like you can change your frame of reference

[cobalt crypt]

like you did here

but then you have to change where you integrate too

its just a substitution

[vast shale]

yeah

[vast shale]

wait im just being dumb but is this conclusion still fine then?

o

wait something doesnt feel right

its because like

that equation should not be the same for different frames of reference