help-10

[glossy basalt]

you did it

[hexed ravine]

ah awesome, thank you!

but wait if I try to get the ground speed using cos law I get 468.32

or do I obtain it a different way

[glossy basalt]

hmm, lemme try

,w sqrt(100^2+500^2-2100500*cos((180-41-25-10.527596656) deg)

[warm shale]

Results provided by WolframAlpha

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

Requested by OldBiscuit#6122

[glossy basalt]

there we go

[glossy basalt]

are you doing

[glossy basalt]

this just now?

[hexed ravine]

I only did cos of 10.52 etc..

[glossy basalt]

oh...

[hexed ravine]

I know it's wrong xd

[glossy basalt]

do you know why it is 180-41-25-10.527596656

[hexed ravine]

because that's the angle between w and v?

[glossy basalt]

correct

[hexed ravine]

ah okay

that makes more sense than whatever I did lol

thank you so much for your help!

.close

[obtuse pebble]

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[glossy basalt]

you're welcome!

[obtuse pebble]

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[twilit ivy]

in this problem i dont see where the singularity is
so i just used Cauchy integral theorem

but apparently the correct answer is non zero?
is that correct or is my answer correct

[twilit ivy]

sin(z) never explodes
e² is a constant (even if i assume that it was a misprint and it was supposed to be e^z - even then it doesnt explode in the given contour)
and z² doesnt go to 0 in that contour

[tacit basin]

Are you sure about that last statement?

[twilit ivy]

let me check

[tacit basin]

Oh wait, no I think you're right

I mean check it anyway

[twilit ivy]

yeh bcz it doesnt include z=0

[tacit basin]

Yeah, that's right. I have no idea then, sorry

[twilit ivy]

i see

btw what if the contour was |z-1|=1

will i say that z=0 is included?

[tacit basin]

I want to say that you need the function to be defined on the curve itself? But my complex analysis is really rusty

[obtuse pebble]

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[clear glacier]

what is a conditional distribution

[tardy epoch]

https://en.wikipedia.org/wiki/Conditional_probability_distribution

[obtuse pebble]

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[static shoal]

where should I start from?

[polar fossil]

convert the floor into an inequality

[static shoal]

i mean so many x are there

[polar fossil]

start with the outer one

so 10 <= x^(whatever) < 11

tbh it might not be possible to solve this analytically

[kind hawk]

I would actually start at the inside. can ceil(x)=1? or 2? or 3? or higher?

then you know what ceil(x) is so you can replace it which simplifies stuff

then the next ceil. which values can it be. and so on

[static shoal]

i actually confused

[polar fossil]

can x be 20, for example?

[static shoal]

no

[obtuse pebble]

@static shoal Has your question been resolved?

[polar fossil]

can x be 10? try to find some bounds for x

[obtuse pebble]

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[bitter canyon]

Is this the idiot triangle of Sin cos tan?

[bitter canyon]

or does sin cos tan needs to be on top? This is right light that right?

[fathom flicker]

What???

[bitter canyon]

Idiotic Triangle is like the thing that you dont have, you put your finger on it

and you can calculate it easy

[royal basin]

can you show some other idiot triangles

also what are G and H

[bitter canyon]

Gegenkathete and Hypotenuse

Im german

;D

[royal basin]

and what do those letters (U, R, I) stand for?

[royal basin]

what does the gegen part mean

[polar fossil]

gegen means "on top of" that's all I know

shoutouts to Rammstein

I would just memorize the relationships for sin, cos, and tan

in english we can say SOH CAH TOA but i'm not sure how it works in german

,tex .sohcahtoa

[warm shale]

Hayley

[royal basin]

presumably if you find yourself in need of an "idiot triangle" these are beyond you

[royal basin]

this triangle as written here is only good for sin, if that.

[bitter canyon]

i know, you just need to put cos and change the other things with it.

I just wasnt sure if it sin goes on top or lower left

[royal basin]

i think making these triangles is unhelpful tbh

because for example the opposite leg goes on the bottom for the sine but on the top for tangent

and you are definitely going to confuse yourself if you try to memorize it the idiot way

[spice thicket]

Why it is called an idiot triangle?

[bitter canyon]

Its simple to use.

[bitter canyon]

you just put it in there

its even easyer

in germany its common

even etachers, professors use it

[spice thicket]

Wow I didn't know about stupid triangle

[royal basin]

this is better i think

don't try to memorize what sin, cos and tan do in isolation from the right triangle

but do it with the right triangle

so that you have a mental picture

[obtuse pebble]

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[pseudo swift]

alright

got any ideas or no ?

[rain pendant]

i've never done this before

its an extra credit question

we've been doimng these

which are fairly erasy

im ust trying to see how it relates

[pseudo swift]

yeah it's just the reverse problem pretty much

but they don't give you a graph here

[rain pendant]

maybe its

9sin(8pit)

[pseudo swift]

you could try drawing one though if you want

using the info written

[rain pendant]

eh

[pseudo swift]

at least you're trying the numbers on the pic ^^

yeah let's try and interpret what they've written

[rain pendant]

Okay :]

[pseudo swift]

difference between highest and lowest point is 8cm

[rain pendant]

mhm

[pseudo swift]

here's a random sinus graph I plotted w/ geogebra

are you able to draw where the 8cm are ?

[rain pendant]

it would be at the very bottom

of the middle scoop

[pseudo swift]

so you mean this ?

[rain pendant]

i have an idea

time period = 1/2 second right

so 2pi/w = 1/2

w= 4pi right

so amplititude would be

4

[pseudo swift]

wdym 1/2 second

[rain pendant]

OH

It says EVERY Second

[pseudo swift]

there's 9 cycles per second

[pseudo swift]

and that has nothing to do with amplitude

[rain pendant]

i thought i was onto something

t bh

[pseudo swift]

I mean your formula is fine

it's for the frequency tho, not the amplitude

9 cycles per second

[rain pendant]

mh

[pseudo swift]

so our time period is 1/9 second

then 2pi/w = 1/9

w = 2pi/9

[rain pendant]

so is it

8sin(2pi/9 t)

[pseudo swift]

almost

[rain pendant]

whys he making this so hard

LMAO

[pseudo swift]

we still haven't finished with the amplitude

[pseudo swift]

do you agree with that pic?

[rain pendant]

the amplitude is usually given thats why

[rain pendant]

no

wait

lemme c

yes

it would be that middle line

[pseudo swift]

well you shouldn't

[rain pendant]

how, theyre both 2 points away

or is it completely 4

from the lowest to highest point?

[pseudo swift]

highest point vs lowest point

[rain pendant]

so its the tip to the bottom of the graph

[pseudo swift]

I mean yeah the 0 is not the higest point of our sin

[pseudo swift]

yes

[rain pendant]

which is 4 points

[pseudo swift]

so amplitude is 8/2

[rain pendant]

on that graph

[rain pendant]

which is 4

W

W

W
W

so its 4sin(2pi/9t)

amplitude has been FOUND

[pseudo swift]

and now we won

profit!!!

[rain pendant]

that right

[pseudo swift]

yea

[rain pendant]

LMFOOOOOOO

its wrong

[pseudo swift]

wat

[rain pendant]

hoowwwww is this wrong

[pseudo swift]

wait i screwed up somewhere

[pseudo swift]

^

this is alright

but then we shouldn't get w=2pi/9

it was just to see if you followed

[rain pendant]

wait

O_O

im confused

F

[pseudo swift]

I screwed up the calculation

2pi/w = 1/9 is the correct equation

but I told you complete shit as for the solution

you can find the correct solution on your own I suppose

[rain pendant]

so ur saying its

4sin(1/9t)

[pseudo swift]

no

[pseudo swift]

solve this equation correctly and you'll find the correct w

unlike me

[rain pendant]

w is 18pi if u solve that

[pseudo swift]

yea

should be 18pi

[rain pendant]

got it

tysm

[pseudo swift]

.close

[obtuse pebble]

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[zinc stump]

Hopefully the English here is good enough, any help will be appreciated. I know I'm suppose to somehow use Zorn's lemma but got no clue how.

[obtuse pebble]

@zinc stump Has your question been resolved?

[zinc stump]

I've been thinking maybe I could show that for all R, (A, R) is partially ordered set, and then I could use Hausdorff maximal principle in order to claim that there is such max?

[obtuse pebble]

@zinc stump Has your question been resolved?

[timid silo]

Zamn

[obtuse pebble]

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[fossil plume]

Can anyone tell me why ln() returns undefined for any negative number but ln(-x) returns this graph? ln(-3) should return -1.1 when considering this graph right?

[tardy epoch]

what's the sign of -x when x < 0 ?

[fossil plume]

well from my understanding -1

[tranquil sonnet]

negative times negative equals positive

[normal oracle]

-1 is not a sign

[coarse ibex]

ln(-x) when x = -3 is ln(-(-3)) which is ln(3)

Which of those two steps was giving you trouble?

[fossil plume]

I need 2 minutes i think I am about to understand it

[tardy epoch]

-(-1) = ?

[tired sparrow]

because ln(-(-x)) = lnx

[fossil plume]

Alright I think I got it, its kinda tough to explain but I do get the same y values for ln(-x) and ln(x) if I actually use this ln(-(-x)) I didnt know that. thank you all!

[obtuse pebble]

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[frozen kindle]

if there are 6 children holding hands in a circle how many ways can they hold hands

[fathom nebula]

do you have the result?

but i think is 6! (factorial)

[lone dirge]

6! is the number of arrangments on a line

on a circle it's different because many line orders are the same

for instance

1 2 3 4 5 6, and 2 3 4 5 6 1

on a circle are the same

[fathom nebula]

so the result is 720?

[lone dirge]

you actually count each arrangment 6 times

,calc 5!

[warm shale]

Result:

120

[fathom nebula]

aaahhh i got it now

because 6 times will be for 7 children

[lone dirge]

for instance, (1 2 3 4 5 6), (2 3 4 5 6 1), ( 3 4 5 6 1 2), (4 5 6 1 2 3), (5 6 1 2 3 4), (6 1 2 3 4 5) are all the same

So you need to divide by 6

[fathom nebula]

yeah that true the position will be repeated

[lone dirge]

6! / 6 = 5!

In general (n-1)! for circular arrangment

of n people

But, I'm not sure what the question means, they might hold hands in odd ways 😉

[fathom nebula]

yeah btw interesting

[obtuse pebble]

@frozen kindle Has your question been resolved?

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[robust acorn]

its reflection right, if so how would i know which one to choose

[robust acorn]

my only resource supporting rn

[violet sentinel]

take your points on one side of the supposed reflection axis and their reflected counter points and see how they differ.

For example:
F = (1, 1), Q = (1, -1).
G = (4, 5), R = (4, -5)

do you see the pattern?

you are correct it's reflection, but which one?

[robust acorn]

E= (-3,1) F=(1,1) G=(4,5)

P=(-3,-1) Q=(1,-1) R=(4,-5)

only pattern im seeing is that all the Y’s are negative in the reflection

[violet sentinel]

yep exactly

which is this one

[robust acorn]

which would be the second answer

thank you for the help

im thinking a 90 degree rotation about the point (-3,-3)

but being its not named im not sure if thats the answer they’re looking for

[obtuse pebble]

@robust acorn Has your question been resolved?

[robust acorn]

<@&286206848099549185>

[polar fossil]

if it's rigid the side lengths will be the same

[robust acorn]

then it would be a no?

[candid yarrow]

check whether the side lengths are all the same

[robust acorn]

they aren’t

[sacred wadi]

yeah, so it would not be a rigid motion

[robust acorn]

alright thanks

.close

[obtuse pebble]

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[vestal breach]

for this problem am i using 1400Sin50 and 1400Cos50?

[vestal breach]

Sin for the vertical and Sin for horizontal??

[candid yarrow]

sin for vertical yes

[vestal breach]

no for horizontal then? @candid yarrow

[candid yarrow]

not sin for horizontal

[vestal breach]

so i would use tangent

?

[nocturne minnow]

No

[nocturne minnow]

They were saying that sin is not used for the horizontal component

[vestal breach]

oh okay

so for sin i got 1072.5

for the vertical

<@&286206848099549185>

[obtuse pebble]

@vestal breach Has your question been resolved?

[obtuse pebble]

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[prisma glacier]

Can anyone on here help with physics?

[native inlet]

There's a physics server in #old-network :))

[quaint sonnet]

You can find many servers related to physics on the public server list

discoverable servers

You can ask questions related to physics if you need help with the mathematics part of the physics question

[fierce lagoon]

🗑️ 🔥

[teal turret]

Post ur question and see if someone can do it anyways

May or may not be worth it

[tardy epoch]

What MrFancy said

.close

[obtuse pebble]

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[glossy nest]

Wondering if I did this correctly

My thought is that I may need to get c alone before I can find dC

but solving for C would give me a positive and negative sqrt for a^2 + b^2

<@&286206848099549185>

[polar fossil]

You've got derivatives on here but it's really just asking for a start-to-end difference

[glossy nest]

well wouldn't dC represent the change from start-to-end?

[polar fossil]

well... ΔC would represent that

it doesn't change at a constant rate so using the first derivative at the start point won't give you the right answer

[glossy nest]

what would give me the difference?

[polar fossil]

just doing the problem the simple way lol

just figure out the hyp length with the new values

[glossy nest]

damn that makes sense fr

that just seems suspiciously easy for this calc 3 assignment 🤨

word

.close

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[oblique sandal]

why is tan(50)-cot(50) = 0

[oblique sandal]

idk why that workds

i know they are complementry angles

[high lily]

it doesn't

tan(50) - cos(50) isn't 0

[oblique sandal]

you mean cot

[golden oriole]

Still not 0

[oblique sandal]

then what is it

[golden oriole]

3.4

Or 0.35

In degree

[oblique sandal]

right

wait nvm its 0

[golden oriole]

What are you doing?
How are you getting 0

[oblique sandal]

tan(90-40) - cot(40)

cot(40)-cot(40) = 0

because they are complementry angles

@golden oriole

[obtuse pebble]

@oblique sandal Has your question been resolved?

[golden oriole]

Yes but that isn't what you originally said

[golden oriole]

This would mean tan(50) - tan(40) = 0

Which isn't true

[oblique sandal]

yea but tan(50) - tan(40) arnt comlemntry angles

like tan and cot

[golden oriole]

50 and 40 are complementary

[oblique sandal]

they are because triangles are 180 degrees

[golden oriole]

Im confused on what your trying to show with this ?

Because in no way does this support your original claim

[royal basin]

what is the question at hand

[golden oriole]

.

[royal basin]

,w tan(50°) - cot(50°)

[warm shale]

Results provided by WolframAlpha

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

Requested by ann_dec#0

[royal basin]

@oblique sandal does this look like zero to you?

[polar fossil]

,w tan(50 radians) - cot(50 radians)

[warm shale]

Results provided by WolframAlpha

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

Requested by kitten.in.a.teacup#0

[golden oriole]

Why did it turn cot into tan an tan into cot ?
For no reason

[oblique sandal]

simplyfing the expression give you 0

[royal basin]

no, it does not.

if you insist it does, then show us exactly how you are simplifying it to give 0.

[golden oriole]

He broke math 😔

[golden oriole]

That isn't what you originally said

You said tan50 - cot50

[royal basin]

^

[golden oriole]

Here you do tan50 - cot40

[oblique sandal]

my bad i mean simplify

[royal basin]

do you understand that you gave us a different expression than you put into photomath?

[golden oriole]

?
There is no way to simplify that to what you sent

[obtuse pebble]

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[royal basin]

refusal to admit own mistake huh

[obtuse pebble]

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[ripe sleet]

can someone help me solve this

ive been moving numbers around for like 10 minutes but dont know what to do

[obtuse pebble]

@ripe sleet Has your question been resolved?

[ripe sleet]

<@&286206848099549185>

[obtuse pebble]

@ripe sleet Has your question been resolved?

[obtuse pebble]

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[sand otter]

.reopen

[sand otter]

Someone helps me problem h) pls

<@&286206848099549185>

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[sand otter]

Sorry

[timber island]

.

[sand otter]

Step 1 actually

[timber island]

Have u done similar problems before?

[sand otter]

Nah

[timber island]

...

[sand otter]

You have any ideas

Well

I think that i have some right ideas

[timber island]

Well let's do f

Have been taught u identities

Like (a-b)^2 and a^2-b^2

[sand otter]

Yeah

What's more

[timber island]

Well notice that in the first equation on the left side, u can simplify the square root using (a-b)^2

[sand otter]

Wait

[timber island]

Wait oh u can't 💀

I thought it's +1 not +2

[sand otter]

Problem h) bruhhh

[timber island]

Well then ur only option is to subtract 1 from both sides, square and isolate either variable

[timber island]

Oh

[timber island]

Have u tried substituting the value of xy (u can find xy from the second equation)

[sand otter]

I don't think so

I have tried but nothing happens

Bruhhh

I find the solution

Tks for helping me

[timber island]

Np

[obtuse pebble]

@sand otter Has your question been resolved?

[obtuse pebble]

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[karmic hedge]

https://i.imgur.com/KIJlNA9.png

[karmic hedge]

I dont get how its -infinity

I got Positive number/0

When I sub in -4.99

[gilded needle]

,w plot (x^2 - 5x + 6)/(x^2 + 7x +10) from x = -6 to x = -4

[warm shale]

Results provided by WolframAlpha

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

Requested by bungo.the.original#0

[gilded needle]

looks like -infinity to me

[karmic hedge]

What did I do wrong

[high lily]

consider whether the denominator is approaching 0 from above or below

[gilded needle]

can you show your calculation?

[karmic hedge]

I factorized both to get (x-2)(x-3)/(x+2)(x+5)
subbed in -4.99, greater number than -5 because its a right limit
Got Positive number/0 which means its positive infinity

[high lily]

0 itself is meaningless

by itself gives no info

[gilded needle]

did you sub -4.99 into both the num and denom?

[high lily]

you can't assert that 0 is positive

[karmic hedge]

It shows its going to infinity no? Cause its not defined

[high lily]

undefined is not the same as inf

[gilded needle]

could go to -infinity, you need to check what's happening as x->-5 from the right

[karmic hedge]

But the positive numerator shows its positive?

[pseudo swift]

look at 1/x when x goes to 0, on one side you get -inf, on the other +inf

[high lily]

1/0 is undefined

[gilded needle]

consider f(x) = 1/x and f(x) = 1/(-x), what do these do when x->0 from the right?

[high lily]

consider a simpler case 1/x as x approaches 0

[gilded needle]

oops, @pseudo swift beat me to it 😁

[karmic hedge]

Positive infinity both

[gilded needle]

1/(-x) = -1/x, so clearly that can't be true

[karmic hedge]

Hmmm ok

I guess I learned how to do it wrong haha

I factorized both to get (x-2)(x-3)/(x+2)(x+5)
subbed in -4.99, greater number than -5 because its a right limit
Got Positive number/0 , what do I do next?

[obtuse pebble]

@karmic hedge Has your question been resolved?

[pseudo swift]

well as with 1/x, the sign of the denominator is important

if x approaches 0 from the positive side, you'd get a positive number out (positive/positive)

[karmic hedge]

its negative

so the overall fraction is negative

aaaah

[pseudo swift]

if x approaches 0 from the negative side, you'd get a negative no. out (positive/negative)

[pseudo swift]

yea

[karmic hedge]

I think I get why its negative infinity now

I just forgot about the denominators sign

Yeye I get it now

Thank you!!!

❤️

.close

[obtuse pebble]

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[obtuse pebble]

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[karmic hedge]

https://i.imgur.com/fAkyiIG.png

[karmic hedge]

Just checking so that im sure, if it was x -> positive infinity instead it would still be the same answer, right?

!status

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[karmic hedge]

4

[sage geode]

Yes

[karmic hedge]

Oke awesome

Thanks!!

❤️

.close

[obtuse pebble]

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[obtuse pebble]

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[sinful eagle]

|x-3|<7

[sinful eagle]

What I Did was

|x|+3<7

-7 both sides

|x|-4<0

+4 both sides

|x|<4

[willow ravine]

Do this: |x-3| < 7 -> -7 < x-3 < 7

Does that help any?

[sinful eagle]

What

No

[willow ravine]

When you have an absolute value inequality you can take the absolute away and work from both sides of the right hand side.

Because if x-3 is greater than -7 its absolute value will be less than 7

[royal basin]

|x|+3<7
incorrect

[sinful eagle]

Wait i dont understand

[royal basin]

|x-3| ≠ |x|+3 generally

[willow ravine]

for instance x=-2 |-2 - 3| = |-5| = 5 <7

[sinful eagle]

Which is true

But what made you put -7

[warm shale]

adamchebil33

[sinful eagle]

Wait so

|x|>1

[inland zenith]

-x < 1 < x probably?

[sinful eagle]

Wait i need think

[warm shale]

adamchebil33

[sinful eagle]

Wait before this

[sinful eagle]

.

So if |x|>1

[willow ravine]

No, you apply the opposite side to the side that does not have the | |

[fierce vale]

wait i thought this was the original equation

[willow ravine]

So, |x|>1 -> -1 > x > 1

[sinful eagle]

It is hold on

Is it because x can be -1

[warm shale]

adamchebil33

[sinful eagle]

So it means x can also be x<1

[willow ravine]

Because |x| < 1 is true for any number that isn’t between -1 and 1

You have to include everything less than -1 because it’s absolute value is greater than 1

[sinful eagle]

Wait i think you wrong

[willow ravine]

No.

|0.5| is not greater than 1

|-0.6| is not greater than 1

[sinful eagle]

No wait ill do a sketch

[high lily]

is wrong

[sinful eagle]

Yeah he is wrong

[willow ravine]

How?

[sinful eagle]

Its -1<x>1

[willow ravine]

|-7| = 7 > 1

[high lily]

there are no values that are simultaneously less than -1 and greater the 1

[sinful eagle]

It has to be less than -1 to be greater than 1

[fierce vale]

|x| > 1 => x<-1 or x>1

[willow ravine]

If you were to graph

Those are the values of x

Which is greater than 1 and less than -1

[high lily]

you did not properly represent the solution

[willow ravine]

You’re not solving for a specific x, you’re finding a range of x such that it fits the equation

[high lily]

there are no values of x that satisfy the inequality you wrote

[willow ravine]

Oh what you want (-inf, -1) U (1, inf)?

[sinful eagle]

How Did you guys do |x-3|<7

[willow ravine]

|x| > 1 then there is no solution and that Desmos graphing it is just coded wrong?

[high lily]

the compound inequality you wrote represents
x < -1 AND x > 1 for which

there are no values that are simultaneously less than -1 and greater the 1

you want OR

and you can't combine the inequalities like you attempted

[fierce vale]

yeah that's the correct solution for |x| > 1

[high lily]

keep them separated

[sinful eagle]

How Did you guys think here

[willow ravine]

He was providing an example and I was showing him how he can find x values that solve |x| > 1

So he can better understand absolute value and inequalities

[sinful eagle]

Ops sorry i called

[willow ravine]

He doesn’t understand why you switch signs and work from both directions removing the absolute value.

[sinful eagle]

No hold on How Did you think here

[willow ravine]

I’ll let Adam and Ramanov take over. Good luck!

No point in arguing about how to do something multiple ways, only do it the way they teach you.

[sinful eagle]

Yeah

[fierce vale]

this a the general form i just applied it there

[sinful eagle]

I understand that you put -a on side

Okay so I think its easier if we see a lenght

[fierce vale]

lenght?

[sinful eagle]

I think im confusing myself

Why was it inaccurate the way i Did it was to remove 3

[fierce vale]

where?

[sinful eagle]

|x-3|<7

|x|<4

And then

[high lily]

that's invalid

[sinful eagle]

Why

[high lily]

there arent any mathematical operations that justifiy that

[fierce vale]

u can't seperate addition in absolute value

[sinful eagle]

Why is that?

Like What’s the issue

Since |7+3| =|10|

=10

Or do you think |x-3| is one number?

[fierce vale]

it might be equal in some cases but generally it's not

when u're solving equations with absoulate values u need to solve for all the possible cases

[sinful eagle]

And is absolute value a number of it’s own?

Oh 😐

Okay so what made you think cus we have <7 that we have <-7 on the other side

[sinful eagle]

Is this always the case

[fierce vale]

of course it is

you can understand it by intuition if you don't want to blindly apply the formula

[sinful eagle]

That’s the thing though it don’t make sense to me

[warm shale]

adamchebil33

[sinful eagle]

What is this

[fierce vale]

just so u know that a is real number

do u agree that if |x|<a then x<a ?

[sinful eagle]

Okay this is my mindset

I think of it like a lenght

That the number in x is less than a therefore x is <-a

[sinful eagle]

Yes and x<-a

Im gonna try another example

|x-2|<5

Because number x-2 can be positive or negative

[barren trellis]

Yup thats a facr

Fact*

[sinful eagle]

And that 2 is less than 5

[barren trellis]

Question is |x-3| is less than 7?

[sinful eagle]

That’s not the question

Its a equation

|x-3| is less than 7

[barren trellis]

We wanna find x?

[sinful eagle]

Yeah

[barren trellis]

So mod opens about 3

Break it to 2 eqs

[sinful eagle]

Wait hold on

We about |x-2|<5 rn

So what i Said earlier

Is i think it works like this

Ah shit i forgot now

[barren trellis]

When x is less than 3

Than eq is 3-x<7
X>-4
When c is greater tha 3

X-3<7

X<10

X lies between -4 and 10

[sinful eagle]

We about x-2<5

[sinful eagle]

not this

[barren trellis]

What is the question

😭

[sinful eagle]

|x-2|<5

[barren trellis]

Ok

So

Mod open about 2

[sinful eagle]

Wait

Idk what you mean mod or open

[barren trellis]

Ok

So when x is below 2

Than value inside mod is negetive

But because of mod it becaomes +ve

So to remove the mod we put -ve sign

And when its grrater than 0 than value inside mod is +ve so we just open with +ve sign

[sinful eagle]

What is ve

[barren trellis]

+ve= positive

-ve=negetive

[fierce vale]

https://www.chilimath.com/lessons/intermediate-algebra/absolute-value-inequalities/

[sinful eagle]

Yes it means we have 2 x

A area of 2 x

[barren trellis]

So back to question when c is below 2

2-x<5

X>-3

When x is more than 2 than

X-2<5

X<7

Hence x lies between -3 and 7

[sinful eagle]

No your wrong here

[barren trellis]

Dude

U know how to remove a mod?

Opening the mod means to remove mod

[sinful eagle]

What is mod

[barren trellis]

| |

This is modulus function

[sinful eagle]

Hold on i made a sketch

[barren trellis]

I had learnt this question in 10th grade in my school

[barren trellis]

Yup

Opening mod is a fancier way to do it

[sinful eagle]

What you mean

[barren trellis]

This way also correct

[sinful eagle]

My way is better

[barren trellis]

Nvm its hard to explain

I use my way because it saves time

Especially for competitive exams like jee

[fierce vale]

you want to understand why this is true ?

[sinful eagle]

This means that the distance between x to 0 is less than a

[fierce vale]

you agree that |x| represents the distance from 0 to x right??

[sinful eagle]

You mean x to 0?

[fierce vale]

same thing3

it's a distance

|x| is the distance between x and 0

[sinful eagle]

Ah Yeah

I can see that

[fierce vale]

now if that distance is less than a where should x be ?

[sinful eagle]

This means that the distance between x to 0 is less than a

Where x is. X<a or x<-a

[fierce vale]

if -a<x<0 does that satisfy the equation |x|<a

[sinful eagle]

Yeah

[fierce vale]

correct

[sinful eagle]

No why you put 0

[fierce vale]

but we need to find all the solutions

[fierce vale]

i'm just splitting the solution

to understand better

[sinful eagle]

Oh

🤔

Ok thx i Will work more about this

.close

[obtuse pebble]

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[fierce vale]

try to gaph it u'll understand better

[obtuse pebble]

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[celest helm]

how to integrate without u-substitution

[celest helm]

I'm aware its easier with u-substitution and aware on how to do it

but if I weren't relying on u-technique, just purely anti-deriv

how do i solve it

[upbeat gazelle]

You can try integration by parts

[desert sinew]

u sub with u = denominator, i.e. 1 + t^2

[celest helm]

i said without u sub

[thick fog]

The derivative of the denominator is a constant multiple of the numerator so the integral is a constant multiplied by the log of the denominator

[desert sinew]

I dont see why you dont want to use it if it is obviously the most easy way to go 🙂

I mean you can also do trig sub with t = tan theta, but why waste your energy on easy question…

[celest helm]

cause i have test on anti-derivs and we arent allowed to use u sub

antidifferentials

[high lily]

without explictly doing u-sub,
you're pretty much still applying the same concept of chain rule

[desert sinew]

If you cheat by having u-sub to get the final answer, then you can work backward and pretend you get the integral by reversed-differentiation. 😂

[high lily]

the integral identity they supposedly want you to apply comes from u-sub and chain rule
so no idea why u-sub isn't expliclty going to allowed

[obtuse pebble]

@celest helm Has your question been resolved?

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[molten lance]

hello

[molten lance]

is this always true and if so how do we got from lhs to rhs?
$$
\sum_{i=a}^{i=b}x^{i+1}=\sum_{i=a+1}^{i=b+1}x^{i}
$$

[warm shale]

NotAlexander

[tame narwhal]

looks right. you just plug in the index and see that the sum on the LHS is x^(a+1) + ... + x^(b + 1)

[molten lance]

i know but how do i generalise it?

i'm working on a proof that requires this manupilation

i just want more understanding

[tame narwhal]

is it not already generalized?

[molten lance]

i dont think so

let me show you part of the proof

$$
=\sum_{i=0}^{i=t}(-1)^{i+2}x^{t-i+1}(\text{lists of length i})+\sum_{i=0}^{i=t}-(-1)^{i+2}x^{t-i}(\text{lists of length i})\cdot z
$$

[warm shale]

NotAlexander

[molten lance]

$\text{i want the } t-i+1 \text{to be reduced to t-i in the first sum}$

[warm shale]

NotAlexander

[tame narwhal]

I don't know, but I'd write out some of the terms to see if a pattern emerges

[molten lance]

.close

[obtuse pebble]

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[split lintel]

It throws me off when the function is just one number like here...so is h(-2)=3?

[frosty wagon]

You could write it as $0x+3$ though.

[warm shale]

Tsumi

[high lily]

Yes

overthinking/complicating

[frosty wagon]

Basically, at that range, the function doesn't depends on the x value.

[split lintel]

So let me try to absorb

ooo and is that latex in discord? that's so dope.

anyway...umm...

So it's 0, because there is no X coordinate?

er i mean 3

[native inlet]

it's just a straight line for any input

for any input between-5 and 3

[split lintel]

I understand that concept....but format wise...

[native inlet]

the output is 3, no matter the input

[split lintel]

I wonder what it's asking me to put in the box

3?

[native inlet]

^

[high lily]

h(x) = whatever for x satisfying the specified condition

and for that second piece, that value is 3

that is all

[split lintel]

So I put 3 in the box?

[native inlet]

why would you put anything else? :))

[frosty wagon]

You could try that on desmo, or any kind of graphing calculator. Even Microsoft Encarta is useable. And you'll see a straight line perpendicular to the y-axis.

[high lily]

consider a simpler example, stripping away the piecewise component

[split lintel]

Because I'm 37, haven't done math since high school (and barely passed then), and am almost in tears just trying to survive this college algebra class because it's my last math class

😉

thank you

[high lily]

consider
y = f(x) = 7
f(99) = ?

[split lintel]

7

[high lily]

exactly the same idea applies here

[split lintel]

i suppose it was more of a wtf formatting question

.close

[obtuse pebble]

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[timber island]

question, why dont we divided by the sum of all weights when we calculate weighted generlized mean?

[timber island]

this is the formula I found

this is the weighted AM

[timber island]

which should follow from this formula, but it doesnt

we r missing the the summation at the bottom

[normal oracle]

because the weights already accounted for in the sum

[timber island]

no I am asking why dont i get smthing equal to weight AM when I put p=1

this is what wikipedia says

[normal oracle]

you want to normalize the weights so they sum 1?

[timber island]

eh not necessarily

[timber island]

is the summation at the bottom for normalizing the weights?

i am very new to this stuff, so pls be a little patient with me 😅

[normal oracle]

I think you're talking about weighted arithmetic mean formula

[timber island]

Okay lemme rephrase my question

[timber island]

When I use this formula, when I set p=1

I should get the formula for the weighted AM

[timber island]

but this is different than what I get

[normal oracle]

if you want the weighted generalized mean formula to reduce to the weighted arithmetic mean formula when p=1, you would need to include that

[timber island]

So doesnt that mean its wrong? and wait include what?

[normal oracle]

this one is the one you're looking for right?

[timber island]

Mhm

this is the arithmetic one

But I want to generalize it

So say if I wanted to calc WGM, WHM, and so on and so forth

I could find those formulas

oh wait-

am dumb

i ignored this 😐

but this confirms my suscipision

that formula i found online is missing the summation at the bottom

[normal oracle]

that's for weighted power mean, u said generalized

and that's this one

[timber island]

is there a difference?

[normal oracle]

for WGM u substitute p with 0

WHM p = -1

aritmethic p = 1

[timber island]

wait lemme think abt this real quick

the weighted power mean can give me the weight GM, or is it the weighted generlized mean

oof i cant put p=0

i have to use limits

[normal oracle]

yes, for geometric u can't put directly p=0

but u use the limit

[timber island]

can u explain this?

this is the proof to obtain GeoMean from GeneralMean

the last step i dont get

If the bottom weights sum up to 1

why dont the upper ones do so

[normal oracle]

this assumes that the weights sum to 1

[timber island]

its the sum of w_i, not w_j

i dont get these 3 steps at all

definitely not the last step

[normal oracle]

send me the link

[timber island]

https://en.wikipedia.org/wiki/Generalized_mean

[timber island]

also why th r we involving weights? isnt obtaining Geometric mean from generlized mean something that should be done without involving these weights?

I am way tooooo confused

ohh

i get why we involve the weights

nvm that question

[timber island]

just explain this

[normal oracle]

need more time to explain this, i will leave in 2 minutes

[timber island]

😭

can u share some resource

[normal oracle]

it's ok, a lot of ppl will know this here for sure

[timber island]

well i have to close in a few minutes as well..i amma most likely ask this tmrw again

[timber island]

welll lets hope someone comes to help

if someones looking for the questions:

1. https://en.wikipedia.org/wiki/Generalized_mean
   proof of how one can derive geometric mean using genelized mean formula (the last 3 steps of the limit)
2. Weighted Power Mean Vs Weighted Generlized Mean
3. Do i use WPM or WGM for deriving weighted geometric mean, weighted harmonic mean, etc.

oof i gotta close

.close

[obtuse pebble]

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[scenic fjord]

Can I know how to change from the first matrix to the second one, which is 0?

[tardy epoch]

is there anymore to this problem

[scenic fjord]

It’s only asking us to show laplacian in spherical Polar coordinates

And I’m referring to this website

[scenic fjord]

http://gmjacksonphysics.blogspot.com/2019/10/how-to-derive-laplace-operator.html

[scenic fjord]

However, I referring to one of the book, it’s saying that it’s easier to solve with vector operators in orthogonal curvilinear coordinates but I don’t know how after reading that so I’m referring to the website

Also I would like to know how to see this part since the website only give this directly

[tardy epoch]

Are you looking for a brute force proof?

[tardy epoch]

This just looks like dot product

Author left out that notation

Google "gotohaggstrom laplacian"

[scenic fjord]

This is my question

[scenic fjord]

Oh, just know that it’s also called this

Btw, I looked through the pdf uploaded for spherical polar coordinates, it’s very long

[tardy epoch]

yea so that's why i'm asking what kind of proof you're looking for

[scenic fjord]

I’m actually looking for the most simplest method to prove the question

[scenic fjord]

Here

Like any method

Cause our lecturer ask us just go into the internet and search for it

[tardy epoch]

This one has spherical coordinates but using the cylindircal coordinates first
https://www.math.cmu.edu/~rcristof/pdf/Teaching/Spring2017/The Laplacian in polar and spherical coordinates(1).pdf

it's marginally shorter

[scenic fjord]

Omg, proving is hard

[tardy epoch]

indeed

especially if you're given such open instructions

probably gonna take you a couple hours doing it step-by-step

[scenic fjord]

Thanks

I think I’ll look into it

.close

[obtuse pebble]

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[lost quartz]

It is multiple option correct type question
I just want to know how A is not an answer as I transformed the given function to integration as int of (1+x/1+x²)dx with limits 0 to 1 for n=infinity then if k starts from 0 then the sum should be greater than that when n=infinity so why is A incorrect?

[lost quartz]

Thnx for explaining

[cunning breach]

Apologies, i had my stuff on the keyboard, and I accidently sent that.

[lost quartz]

No problem

[obtuse pebble]

@lost quartz Has your question been resolved?

[lost quartz]

<@&286206848099549185> help me plz

[obtuse pebble]

@lost quartz Has your question been resolved?

[obtuse pebble]

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[lilac oak]

Ignore my workings out on the side!

Need help like asap im so confused loll

[timid silo]

use the conditional probability formula

[lilac oak]

whtas that?

[timid silo]

this thing

[obtuse pebble]

@lilac oak Has your question been resolved?

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[crisp elk]

im stuck in this problem

[native inlet]

can you translate?

[dusk mantle]

Cauchy pfp, doesn't speak french

[native inlet]

[crisp elk]

i didn't find a way to prove (OA')//(BB')

[native inlet]

I eat pizza and pasta, doesn't mean I'm Italian :((

[crisp elk]

<@&286206848099549185> i didnt know how to prove that BOA'B' parallelogram or even (AO)//(BB')

[rapid mesa]

Hmmm

[drowsy coral]

The reactions keep growing but no help given 😔

[obtuse pebble]

@crisp elk Has your question been resolved?

[gray pulsar]

hi

[crisp elk]

no

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[crisp elk]

2

then ?

yeah

yes

yeahh u have too

[obtuse pebble]

@crisp elk Has your question been resolved?

[obtuse pebble]

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[long hedge]

so if the period is pi4

why doesen't one cycle end at pi4 why does it end at pi3

in this video he says this is one cycle

if thats one cycle then the period of the above graph must be 3Pi not 4Pi ????????

[rocky goblet]

...ok wait which graph are you referring to? since you've posted 12 graphs

[long hedge]

then ones i choose an answer for

[rocky goblet]

...and why do you think those have a period of 3pi?

[long hedge]

based on the screen shot in the video

[rocky goblet]

well the video doesn't say anything about 3pi

[long hedge]

wait its because the video is of sine and this graph is cosine

[rocky goblet]

...what do you think the "period" of a function actually is?

[long hedge]

what shrinks/stretches the graph

one full cycle of the graph happens within the period according to the video i screenshotted

NOW I SEEEEE

.close

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[long beacon]

I need help with infinite limits

[long beacon]

How do you solve for a function for x going towards positive infinity versus x going towards negative infinity? I am struggling with the algebraic difference

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[rancid leaf]

How can I know that I have to use a line integral and no a sufurce integral?

[warm canopy]

do you have a question in mind where it wasnt clear?

[rancid leaf]

For example this, why I know that I have to use surfuce integral an no line?, find the area of the cylinder x^2 + z^2 = 1 who is inside the cylinder x^2 + y^2 = 1

[mellow ether]

if you're finding a surface area you need a surface integral

surface integral is for 2d things, line integral is for 1d things

[rancid leaf]

Oooh

So, if the exercise mention a something like this is a surfuce integral?

[warm canopy]

it depends what it is on an object they are wanting you to calculate

a surface integral can calculate surface area

after all surface integrals are just "adding up" a bunch of things on a surface, those things can be areas

[mellow ether]

yeah it depends on the context

[rancid leaf]

ooh

[rancid leaf]

So, in this context the object that I want to calculate the area is 2d object, then is a surface integral?

[mellow ether]

yeah

whenever something says area of a surface they mean the surface integral of 1 over the surface

[rancid leaf]

I think that I get it!

thank you so much

[mellow ether]

nw

[rancid leaf]

a quick question

surfaces always are 2d?

[mellow ether]

yeah

[rancid leaf]

they can't be 3d? why?

[mellow ether]

it's just the definition

like surface just means 2d space

there isn't really a name for 3d spaces as far as i know

[rancid leaf]

surface doesn't have "stuffed"?

[mellow ether]

like filled in?

[rancid leaf]

yes

[mellow ether]

that wouldn't be a surface

it still exists you just wouldn't call it a surface

it would need to be at least 3 dimensional

[rancid leaf]

what do you mean?

[mellow ether]

surface = 2 dimensions

if a space has volume (like a filled in sphere) it has to be bigger than 2 dimensions

so a filled in sphere is 3d for example

[rancid leaf]

ooooh

I see, thank you so much, that concepts are a little difficult for me

[mellow ether]

no worries

it's hard to have good intuition for these kind of things when you first see them

but it will come with experience

[rancid leaf]

.close