help-13

[south tundra]

Wdym

[dense wing]

write it w/o abs value

then you have 2 cases

[crimson sedge]

First case it’s positive

Second case it’s negative

Ok

[dense wing]

yes those are the cases.

[jaunty mural]

You had me for a sec there.

[crimson sedge]

This question is easy

1, 21, 19

Are the solutions

[jaunty mural]

Utilise this definition properly and proceed.

[dense wing]

They were already there.

[livid hound]

1, 21, 19
you're missing a solution

[crimson sedge]

Hmm

Ah and -1

But I have to test that they work

1 doesn’t work

Nor does 19

21 and -1 work tho

Right?

[livid hound]

what's your issue with 1 and 19

why do you think that they don't work

[crimson sedge]

They equal -20

[livid hound]

wdym by "they"

[fallen heath]

the absolute value of which is?

[crimson sedge]

20

Ah

I guess?

[dense wing]

You guess?

What?

[crimson sedge]

You used passive voice as well

[dense wing]

That matters... why?

[crimson sedge]

Why the absolute on -20

[dense wing]

???

[crimson sedge]

I thought the absolute valuenn but is on the lhs

[dense wing]

bruh

where did -1 1 19 and 21 all come from then for you?

Thin air?

[livid hound]

wdym by "they"

[dense wing]

plug all 4 of those values into the LHS and you get 20

[tropic oxide]

where is the passive voice

[livid hound]

what exactly are you saying is -20 when plugging in x=1

[tropic oxide]

the answer is obviously "whatever the right answer is to that question that'll get you off my back because i hate being interrogated"

[crimson sedge]

Yes

[crimson sedge]

No

[dense wing]

Did you even read what Ramonov said?

[livid hound]

that wasn't a yes/no question

[crimson sedge]

Some have 2 gave -20 and 2 gave 20

[livid hound]

tell me exactly what you did that led to -20

when plugging x=1 into the lhs

(of the original equation)

[crimson sedge]

1-20-1=-20

[dense wing]

.

[jaunty mural]

Find all the real solutions to $$\abs{x^2-20x-1} = 20$$

[wraith dagger]

Shuri2060

[livid hound]

the left side of the original equation is
$$|x^2-20x-1|$$
and for some reason it seems that you're completely ignoring the existence of those absolute value bars

[wraith dagger]

ℝamonov

[crimson sedge]

Ah right

Nvm

They all work

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

.close

[cedar kiln]

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[ancient zealot]

@velvet mason you can go ahead my question is not that urgent

[glad kestrel]

it's already your channel

[ancient zealot]

ok

[glad kestrel]

what is your question

[ancient zealot]

Question is how do I find the normal Line

[velvet mason]

No it’s fine I’m in channel 1 now

[ancient zealot]

Ok thanks

[dense wing]

You found it already tho

[glad kestrel]

lol

[ancient zealot]

I guessed it what is the step by step

step by step process to solving it

[dense wing]

well why did you guess 3-x/3

instead of say 2-4x?

[glad kestrel]

pretty good guess tbh

[ancient zealot]

I saw a similar answer in the example problem

[dense wing]

so... follow the soln in the example problem?

[ancient zealot]

lol They don't show the step by step solution

[dense wing]

I doubt that

[ancient zealot]

They gave this instead

[dense wing]

if you were able to follow it enough, without steps, to get to the correct answer

[ancient zealot]

Use the point-slope form
y − f(a) = f '(a)(x − a)
to find the equations of the tangent line and the normal line.

I used this to find the first part. But then guessed the second part

[dense wing]

Ok but:

You followed a solution to get the normal line

So... there is sufficient information in the solution you read to get to the answer.

[ancient zealot]

I guessed the solution to the normal line. What is the step by step process?

[dense wing]

find the slope and a point.

[ancient zealot]

slope is 3

[dense wing]

slope of the normal line

[ancient zealot]

then what

[dense wing]

not slope of the tangent.

normals and tangents are perpendicular.

[ancient zealot]

It looks like slope of tangent line - inverse slope

[dense wing]

normal and tangent are perpendicular

there is a known relationship b/w their slopes then.

[ancient zealot]

what does the 3 in the normal line formula represent?

[dense wing]

....

[ancient zealot]

then the -x/3

[dense wing]

You've worked with linear equations.... right?

[ancient zealot]

yeah

I mean how do I calculate off of the tangent line

[dense wing]

So I shouldn't have to answer what the +3 and -1/3 mean in the normal like equation

[dense wing]

Find the slope of the line:
Find a point on the line:
Find the equation of the line:

I said this already

[ancient zealot]

Oh now I get

took me while lol

Thanks

@sick viper This helps a lot thanks.

.close

[cedar kiln]

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[glad kestrel]

mosh jr.

[cedar kiln]

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[sudden lance]

im not getting the same answer

[dire geode]

share your work

[sudden lance]

i just took the integral from a to y of Y with dy differential

@dire geode

but i got something completeley different

[dire geode]

show your actual work

[cedar kiln]

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[sudden lance]

i got it

u dont take an integral lol

[cedar kiln]

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[twilit crane]

Hello, I have problem with this one. To be honest I was struggling trying the methods our teacher showed but rip can't get it right (I'm writing on teams, I used to earase all of it before I found this server :') Can you just show me please how to get it right from the start with every step?

[formal nest]

Hi

So basically

Youll need to use the trigonometric circle

As you can see for the cosx=-1/2

If you use soh cah toa you’ll realise that the cos of an angle will give you a distance on the x axis

Whereas sin of an angle will give you a distance on the y axis

Do you understand @twilit crane ?

[twilit crane]

To be honest not really, the teacher just showed us a video of these, can I dm you link to this video to show what I mean?

[formal nest]

its ok ill explain everything you need to know

Just tell me what you understand about trigonometry

do you understand soh cah toa ?

or sin=o/h …

cos=a/h

Tan=o/a

[twilit crane]

We had the basics as these, I understand it ↓

[formal nest]

https://www.mathwarehouse.com/trigonometry/images/sohcohtoa/sohcatoa-all-v2.webp

Do you understand this ?

[twilit crane]

Yup yup

[formal nest]

alr then

[formal nest]

Sketch a circle like this one

The radius is 1

unit

Thats why there is no angle that can give you a cos of a higer value than 1

[junior pond]

yeah it's the quadratic circle

[formal nest]

ok since you understand that the cos of an angle is the x value and sin the y value

You agree that when it says cosx=-1/2

it means that the x value is -1/2 ?

Right ?

On the circle ?

U here @twilit crane ?

[twilit crane]

Yes, watching everything

[formal nest]

Did you draw a circle on your notepad ?

[twilit crane]

Sorry I'm little translating, it's not my first language, never tried write about math before

Yes I got it

[formal nest]

Do you speak french?

[twilit crane]

Unfortunetly no, my first language is polish

[formal nest]

oh ok

anyways did you draw a circle ?

[twilit crane]

Yes

[formal nest]

Ok and what you’re trying to find is an angle that will give you a distance of -1/2 right ?

[formal nest]

The problem is that there are 2 angles that will give you an x of -1/2

The calculator will give you one of them, however you might want to know the other.

Thats why you will have to adjust the answer that the calculator gave you in order to find the other

The trigonometric circle will help you visualize the other one

[twilit crane]

Okay

I'm sorry I feel stressed can't clear my mind and think about it.. I don't want to take your time and bother you, I'm so sorry and thank you so much that you wanted to help me but It can be really hard to solve this in this situation..

It's a shame but can't lie about it

[formal nest]

its okay i have time

[twilit crane]

Don't you mind drawing and showing this on image? Maybe I can understand it better

[formal nest]

ok ill draw another circle

Do you see the 2 angles that both will give an answer of cosx=-1/2 ?

[twilit crane]

Yes now it's more clear to see

[formal nest]

the angles also create a triangle

thats why their answer is a distance

Lenght

Or height

Ok now if you write arccos(-1/2) or cos^-1(-1/2) in your calculator

It will give you as an answer one of the 2 angles

you can then find the other one by doing some simple arithmetic

[cedar kiln]

@twilit crane Has your question been resolved?

[hot garden]

my calculator gives both bcz I got CAS 🤣

If you have it, you can try getting the other solution like that

Otherwise it's a matter of adding PI

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

I’m a bit stuck here

trying to get to the arctan form

[fallen heath]

Hey Cenni, looks like a struggle

[crimson sedge]

Right?

[fallen heath]

a bit rusty on algebraic manipulation?

[crimson sedge]

I guess I am

this isn’t supposed to be a hard exercise

[fallen heath]

Yep. I can see that

[crimson sedge]

is there an easy way i’m missing?

[fallen heath]

$\int \frac{15}{4(x-\frac{1}{2})^2 +25} \dd{x}$

[wraith dagger]

Ansh

[crimson sedge]

I think I can proceed from here

[fallen heath]

No wait

of course you can but I'm kind of concerned seeing your approach from earlier

[crimson sedge]

Why?

I a process to complete the square

[fallen heath]

mhmm. yk, whenever I'm exposed such a dilemma, I write the two "to be similar" forms side by side and compare each step

like here you want your expression into an arctan form

[crimson sedge]

yes, I did this also

[fallen heath]

$\int \frac{15}{4(x-\frac{1}{2})^2 +25} \dd{x}$ vs $\int \frac{1}{y^2 +1} \dd{y}$

[wraith dagger]

Ansh

[crimson sedge]

to eliminate the x2

but it’s not helpful

[fallen heath]

no no. I mean

[crimson sedge]

dy?

[fallen heath]

here.. notice what stuff fits where

[crimson sedge]

can I consider y to be 2(x-1/2)?

[fallen heath]

yep but the remaining thingy in the denominator isn't 1

but 25

[crimson sedge]

I know that

that’s the whole issue

[fallen heath]

A good practice would be changing from 25 to 1

divide the whole thing with 25

[crimson sedge]

yeah that works

[fallen heath]

$\frac{15}{25} \int \frac{1}{\frac{4}{25}(x-\frac{1}{2})^2 +1} \dd{x}$ vs $\int \frac{1}{y^2 +1} \dd{y}$

[wraith dagger]

Ansh

[fallen heath]

Ah.. now they kinda look like twins :o

[crimson sedge]

yeah that’s it

[fallen heath]

the integral I mean

[crimson sedge]

ye i’m not a baby

[fallen heath]

XD

[fallen heath]

what's your "y" going to be then?

[crimson sedge]

2/25(x-1/2)

[fallen heath]

🙅

[crimson sedge]

egh I meant 2/5

[fallen heath]

mhmm

[crimson sedge]

now to take the derivative

[fallen heath]

also, don't forget to adjust your "dx" to "dy" accordingly

[crimson sedge]

I am not sure I see why we changed to dy

[fallen heath]

👀 wdym "why" ... we were comparing the two no?

[crimson sedge]

well I just simplified it

[fallen heath]

if your "y" is 2/5 (x - 1/2), there's gonna be something going on between "dx" vs "dy" as well no?

[crimson sedge]

you want me to say dx=D(y) dy?

[fallen heath]

ahem kinda 👀

dy = 2/5 dx in this case

so you need to adjust that factor 2/5 x in as well

[crimson sedge]

Just add 5/2 outside of the integral

[fallen heath]

💠 ✅

[crimson sedge]

pretty nice

thanks for the help

[fallen heath]

btw

I'd suggest trying the integral $\int \frac{1}{x^2 + t^2} \dd{x}$

[wraith dagger]

Ansh

[fallen heath]

as it'll help simplify your future problems a bit ._.

[crimson sedge]

whatever’s that?

I think I saw something similar in the past

[fallen heath]

you just did XD

[crimson sedge]

Oh

yeah you’re right

[fallen heath]

$\int \frac{15}{(2x-1)^2 +25} \dd{x}$

[wraith dagger]

Ansh

[fallen heath]

looks similar :o

[crimson sedge]

should I treat it like a special case? I mean it’s just some manipulation

[fallen heath]

it's just some manipulation

treat it like a generalized case

[crimson sedge]

Sure I will

[fallen heath]

Have fun integrating

[crimson sedge]

Yessir ^^

.close

[cedar kiln]

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[sour glacier]

Can someone please help me? I have (c1)v + (c2)w + (c3)u = 0 only and don’t know where to go

[upper abyss]

That's backwards. You want to end with that, you don't want to start with it

Take everything in the "suppose". Write them as equations.

Or wait, are you doing a contrapositive? That could work too

[sour glacier]

Thank you, so (c1)u + c2(v) + (c3)w = 0. I think I’m supposed to take the dot product of something but not sure what

I’m not sure what a contrapositive is so probably not

What if I just dot v w and u by each of the respective 3 parts of my equation

Can I do this?

[calm sierra]

this is nonsense

[sour glacier]

Hahaha, sorry math is not my strong suit

As you can tell

[dire geode]

write down the definition of them being orthogonal to each other. that's the only information you're given.

[sour glacier]

Right, so v•w = 0 etc

I get that part but I’m not sure how else to use it if I don’t dot a vector by a vector

Thanks tho

[calm sierra]

What if I just dot v w and u by each of the respective 3 parts of my equation
this isnt a valid move

[sour glacier]

I thought linear algebra was creative :/

Lol

Please help me understand it 😅

[calm sierra]

its like having 1+2=3

but randomly multiplying 1 by 4 to get 4+2=3

[sour glacier]

Oh yes sorry that really is a little weird

A friend said to dot through by v1, is that correct?

[calm sierra]

but taking dot products IS the right idea

idk what v1 is

we're just given u,v,w

[sour glacier]

Sorry just v*

[calm sierra]

yeah so take (c1)u + c2(v) + (c3)w = 0

dot each side with v

[sour glacier]

Ok dot products for sure, but do you think I can dot it all by v?

Thank you!!

I’ll do that and show you one second

[calm sierra]

use dot product properties & ur assumptions

[sour glacier]

Actually this doesn’t look like it’s working out

Pretty sure my brackets are wrong too. Dot both sides? So even the 0?

[calm sierra]

yes

whatever u do to one side u do to the other

[sour glacier]

Thanks. So = 0•v on the other side. But then what? v•v is not zero

[calm sierra]

also dots are needed for each dot product

[sour glacier]

It’s ||v||2

[calm sierra]

$v\cdot($stuff )

[sour glacier]

Length of 2

[wraith dagger]

RokabeJintaro

[calm sierra]

now recall ur assumptions

[sour glacier]

Ok so like this

Yes my assumptions I see

But the problem is v•v isn’t 0

Length of v squared?

[calm sierra]

thats not a problem, in fact it helps us

recall we're showing linear independence

so we must show c1=c2=c3=0

exactly what assumptions am i talking about?

[sour glacier]

Thank you I appreciate it

[calm sierra]

do u know what 0.v is?

[sour glacier]

Assumptions about orthogonal vectors? The ones mentioned in the question

0

Right? Unless it’s incorrect and I’m missing something but I hope it’s 0

OH

Hahaha I’m dumb

[calm sierra]

what do u get now?

[sour glacier]

can I just stop at c2 (length of v)^2 = 0? C2 must be zero

[calm sierra]

$c_2(v\cdot v)=0$

[sour glacier]

I think that’s what I get.

I got c2 (v • v) = 0

[wraith dagger]

RokabeJintaro

[calm sierra]

typo

but v.v!=0

[sour glacier]

C2 must be zero, right? There’s no way another scalar can get us 0

[calm sierra]

we can divide by nonzero scalars

so we can divide by v.v

[sour glacier]

No way I didn’t know we can divide

That’s cool

I think this is finally correct

Thank you soooo much!

I really appreciate it this was driving me nuts

[calm sierra]

its high school algebra

$2x=2$

[wraith dagger]

RokabeJintaro

[calm sierra]

$2\ne0$ so u can divide by 2

[wraith dagger]

RokabeJintaro

[calm sierra]

$x=1$

[wraith dagger]

RokabeJintaro

[sour glacier]

I thought uni linear algebra was a unique concept with no division

So it’s not right? I have to keep going or is this an example? Bc I can’t register this

[calm sierra]

scalars are fancy name for numbers

[sour glacier]

Oh thank god this is an example

[calm sierra]

all the highschool algebra laws apply to scalars

[sour glacier]

Ohhhh

Thank you that makes sense

Do you think I’ve sufficiently proved c2 = 0 or is there another hidden step? Because I’m not sure of anything with this course

[calm sierra]

thats it

u now need to also show c1=c3=0

[sour glacier]

Oh god

No way there’s more

[calm sierra]

u only showed c2=0

[sour glacier]

That’s true I see ur point

Ok I’ll do the same thing except with vectors w and u

Thank you you’re a lifesaver

[calm sierra]

exactly, and np

[sour glacier]

Have a good one

[cedar kiln]

@sour glacier Has your question been resolved?

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[crimson sedge]

how does one find area of turbine blades

[dense wing]

what shape do they trace out?

[crimson sedge]

Ah a circle

[dense wing]

and the radius is?

[crimson sedge]

7.5 because of the blades

[dense wing]

yep

[crimson sedge]

ty

.close

[cedar kiln]

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[dense wing]

[cedar kiln]

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[cedar kiln]

@crimson sedge Has your question been resolved?

[open forge]

I need help guys been studying for my exams and we were given a practice sheet(not graded) and I just cant find the answer
The length of a rectangle is 12 m more than thrice its width. The perimeter of the rectangle is 200 m. Find the length of the rectangle.
its been almost an hour

[dire geode]

Can you do it if p=5?

Try it and then come back

[cedar kiln]

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[abstract lion]

can anyone quickly explain to me how this happens

[glad kestrel]

rewrite the radical as an exponent and use exponent rules

[abstract lion]

oh right

thanks

.close

[cedar kiln]

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[crimson sedge]

why is it that ds = sqrt(x^2+y^2)dt

[cedar kiln]

@crimson sedge Has your question been resolved?

[dire geode]

read this
https://tutorial.math.lamar.edu/classes/calciii/lineintegralspti.aspx

[cedar kiln]

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[cedar kiln]

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[hexed palm]

LaTeX: \frac{1}{2}\frac{ }{ }\times\frac{2}{3} =

[hexed palm]

help meeeeeeeee

[hollow osprey]

$\frac{1}{2}\frac{ }{ }\times\frac{2}{3} = $

[hexed palm]

sorry wrong one

[hollow osprey]

latex is buggin

[hexed palm]

1

Elizabeth had {1}{4} meters of green ribbon. She used {1}{2} of it for her project in Science. How many meters of ribbon did she use?

[obsidian coral]

I think because of the space at the end before the $

[hollow osprey]

When I didn’t it made an emoji lmao

😒

This one

[cedar kiln]

@hexed palm Has your question been resolved?

[cedar kiln]

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[worn rapids]

I can't figure out the number of pairs as asked

I've tried putting values as 1 and 2 for x and y but that won't be a good solution

Can anybody help?

[gentle lintel]

this looks like a contest 🤔

there should be a suggested solution

[worn rapids]

No

Sample paper

[gentle lintel]

of..?

[worn rapids]

Yeah like that's the issue

I don't know the source

My friend sent me

@gentle lintel can you help solve it?

[worn rapids]

<@&286206848099549185>

[fallen heath]

Well one obvious attempt is $x^3 +y^3 +n^3 -3nxy =m +n^3$

[wraith dagger]

Ansh

[worn rapids]

Yes

[fallen heath]

and x+y+n =0 gives you an infinite pair (x, y) of solutions, given m + n^3 = 0

[worn rapids]

Right

So it's like m = -n³

And we have to figure out the values 2014 can take by considering 1 as n or anything else?

[fallen heath]

the point is... u need to find "all possible" m, n such that (x, y) have infinite integer solutions

[worn rapids]

Yes

[fallen heath]

not just x+y+n=0

[worn rapids]

Is that even possible?

[worn rapids]

Oh

So what do we do next?

@fallen heath

[cedar kiln]

@worn rapids Has your question been resolved?

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Ansh#3502

[worn rapids]

Oh

[fallen heath]

,w plot x^3 + y^3 = 1-3xy

[wraith dagger]

Results provided by WolframAlpha

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

Requested by Ansh#3502

[worn rapids]

Yes that's the equation

But that's when m and n are 1,-1 each

[fallen heath]

1 and -1 you mean

[worn rapids]

Yes

But I didn't understand how will that help in finding possible integers 😅

[fallen heath]

maybe if you could show x + y + n = 0, is the only solution :o

[worn rapids]

Oh

Then the possible ordered pairs would be technically all the numbers from -2014 to 2014

Right?

[fallen heath]

not really

[worn rapids]

Cause n can take all values from -2014 to 2014 and if m is -n³ then it would definitely be within the range

[worn rapids]

Oh

[fallen heath]

you also need the condition: $n^3 + m = 0$ to be fulfilled

[wraith dagger]

Ansh

[worn rapids]

Yeah like it will be within the range

Oh no we can't go irrational sorry

[fallen heath]

so basically, $-2014 < -n^3 < 2014$ should do the trick

[worn rapids]

I was thinking of underrots

[wraith dagger]

Ansh

[worn rapids]

Yes

[fallen heath]

$-12\leq n \leq 12$

[wraith dagger]

Ansh

[fallen heath]

smh

[worn rapids]

Yeah in integers definitely

So 25 cases possible cases

[fallen heath]

$f(x, y) = (x^3 + y^3 + n^3 - 3xyn)$

[wraith dagger]

Ansh

[worn rapids]

Yes

[fallen heath]

is independent of "x, y"

[worn rapids]

Yes

[fallen heath]

so must be 0?

[worn rapids]

I didn't understood here

😅

Oh yeah got it

Like x³ + y³ - 3xyn you replaced it with -n³

So it's like the zero of the equation

And independent of x and y

Right?

[fallen heath]

i'm quite lost here lol. the intent should be showing x + y + n =0 is the only possible place you could get infinite soln.
but uhh, the solution isn't very convincing

you should check the official soln. or something

[worn rapids]

I tried Google but

I can't find the problem

I don't have the source

😅

[fallen heath]

$(x+y+n)$ divides $m + n^3$ for all x, y

[wraith dagger]

Ansh

[fallen heath]

Yeah that's it I think

[worn rapids]

Oh

But that's like 0 divides m + n³ for all x and y

Oh yeah that's why it's infinite solutions

[worn rapids]

So now the answer should be 25?

[fallen heath]

[worn rapids]

-12 ≤ n ≤ 12

[fallen heath]

Yep

[worn rapids]

Oh great

:)

[fallen heath]

I can't find any other. Again, it's only a hunch lol.. An official soln. would truly help

or ask your friend for one

[worn rapids]

Sure

Thanks for the help :)

.close

[cedar kiln]

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[thorn grail]

Is this channel available

[fallen heath]

👀

[thorn grail]

Nice

[fallen heath]

Hey salty

[cedar kiln]

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[thorn grail]

This is official solutions and part b is confusing me

[thorn grail]

@fallen heath

[fallen heath]

why ping me ;-;

i'm not even a helper lmao

[thorn grail]

Ah sorry

[fallen heath]

what's confusing about the second part though? 👀

[thorn grail]

I don't get how drawing the graphs helps with saying that Right hand riemann is too large

like how it proves the point

[fallen heath]

aaaah

well, for starters they broke the graph down into bits.. I hope you get why?

[thorn grail]

I kind of get why they broke down the graph (riemann rectangles) but the rest is all confusion to me

[fallen heath]

"kind of" is not good enough 👀 do you, for sure, get why they broke down the graph?

[thorn grail]

Ok yeah I don't get why

[fallen heath]

$f(x) = x^2 \sin (2x)$ right?

[wraith dagger]

Ansh

[thorn grail]

yes

[fallen heath]

from 0 to pi/2 this graph is positive, lying above the x-axis right?

[thorn grail]

yes

[fallen heath]

I mean, x² is obviously always positive so the only reason for the minus sign comes from the non-grateful sin (2x) term

okie

so .. what happens after pi/2 then?

the graph goes under the x-axis for a while

and the integral: the area under the curve changes it's sign

[thorn grail]

yeah this is the part I don't get

when I looked at the graph of sinx and cosx

I thought that it'd result in 0

[fallen heath]

why sine and cos x? 🤦‍♂️

isn't it combined sin(2x) for you now?

[thorn grail]

right...

very confused why they drew sinx and cosx separately

[fallen heath]

ah because :O there exists people who tend to not understand sin(2x) and wish to break it down to sine and cosine x

well, anyways 👀 if you understand the sign changes in sin(2x), that's good enough

your next reasoning is: from 0 to pi/2, the graph is +ve, i.e., above the x-axis and so, the area of this region, say Area A is +ve

However, this curve is governed by the sin(2x) factor which is always between 0 and 1 in this interval, and so, the max the area can go is the same as the curve x² (which is just hypothetical)

For the next part however, the sign has changed but your interval starts from (pi/2) this time upto (2pi).... the curve y = x² this time would be vast compared to the previous one and the factor sin(2x) being periodic.. would assist you in claiming that the area of this region > area of the previous region

.
.
Put together with the fact that this area actually came along with a minus sign, plus the negative area is greater than the previous positive one, you'd also have the conclusion that: integral of f(x) from 0 to pi is actually -ve

similarly for the next two intervals and you can conclude that the net area is actually -ve and far from 0 at that

[thorn grail]

I need time to digest this brb

[cedar kiln]

@thorn grail Has your question been resolved?

[thorn grail]

So this kind of proof is only valid for periodic functions

[cedar kiln]

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[smoky ridge]

let's say I know that the population mean score for some test is 100, and I know that the standard deviation for the population is 15. let's say I have a guy take this test on two different occasions. let's say the first time he scored 85 and the second time he scored 115. I want to try to figure out if this was a significant leap or not.

my first guess based on google searching is to try to set up some sort of a t-test where each score is one of the means, and the sample size of each would be one. but most of the formulas I'm seeing want something like the standard deviation for each sample. the problem is that with a sample size of one, there won't be any deviation. so this doesn't work.

so what can I do? I can tell obviously that the two scores are two standard deviations apart. can I use that to say that the two scores are significantly different from each other and likely not due to chance?

this isn't for any sort of class. I'm looking more for an answer about how to solve this problem. I haven't taken a math class since 2008, so my math is really rusty.

[smoky ridge]

if it helps, I also know that the population score distribution is normal

[cedar kiln]

@smoky ridge Has your question been resolved?

[smoky ridge]

<@&286206848099549185> I would settle for a link that teaches me about this sort of problem. I'm not versed enough in stats to know what all my options might be :(

[wraith crypt]

look up hypothesis testing

[smoky ridge]

if I'm not mistaken, I thought hypothesis testing was for comparing a score to a (theoretical) mean. so I could take one of the guy's scores and compare it to the population mean of 100 and determine if he significantly deviates from that. correct?

would hypothesis testing help me determine if his first score is significantly different from his second score?

[wraith crypt]

you'll need to define a theoretical distribution of this guy's first score

say, you expect his mean to be 85 and so and so

then you can do a hypo testing

[smoky ridge]

do I need to also assume what this guy's standard deviation would be? or can I use the population standard deviation?

thank you for the help by the way.

[wraith crypt]

you need to assume it

[smoky ridge]

hmm. okay.

thank you for your help. i think this means that there just isn't a good way to scientifically determine if the two scores are significantly different from each other with the given information.

[wraith crypt]

the problem when you're talking about a single guy's ability/test scores, is that the population statistics have zero impact on how well he's expected to do

yeah

a priori he has a native ability level which doesn't depend on the population

so when you're trying to determine if he improved or something, you have to compare with that native level, not with the population

[smoky ridge]

yeah. that makes sense. it's disappointing, but if that's just how it is then it can't be helped.

thank you.

.close

[cedar kiln]

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[toxic garden]

lesson is rationalizing denominators

[reef lance]

how about u learn how to rationalize?

[toxic garden]

wdym

[reef lance]

holup

take the minus common out from the denominator and put it in the numerator

[toxic garden]

wdym

[reef lance]

making it:
-5 / (4 - sqrt5)

now rationalize

[toxic garden]

why would the negative go on top tho

[reef lance]

u see the new minus in the deno?

[livid sluice]

kermit can i ask how to get help and create my own chat

[toxic garden]

yes

[toxic garden]

#❓how-to-get-help

[wet rover]

just multiply by conjugate i'm not sure why the other person wants you to factor out the negative lol

[toxic garden]

I see

so the conjugate would look likee

5/-4 + root 5

times

[reef lance]

to make it "look" easy

[toxic garden]

5(-4-root 5)/ -4 - root 5?

[wet rover]

the conjugate would just be the reversed sign of whatever's in the denominator:

[wet rover]

it really doesn't make it look any easier(?) hmm

$\frac{4+\sqrt{5}}{4+\sqrt{5}}$

[wraith dagger]

Yone

[toxic garden]

ohh

so the negative in the 4 t urns into positive

so it would be times 5 (4 +root 5)/ 4 +root 5?

[reef lance]

kids don't like to see the deno as
$\frac{\sqrt{5}+4$

[wraith dagger]

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

[toxic garden]

what

[reef lance]

as sqrt5 - 4 whose conjugate then becomes sqrt5 +4

[wet rover]

no the point of rationalizing the denominator is so that you don't have to deal with surds (in the denominator obv)

[reef lance]

yes there is no dealing with surds

[toxic garden]

sooo how wld it look like

[wet rover]

numerator is fine but denominator is not fine; it should rather be (-4+root(5)) * (4+root(5)) which should ofc result in an integer value

the end product (well not really the end) but the coherent followup of multiply w the conjugate would be something like this:

$\frac{5\left(4+\sqrt{5}\right)}{\left(-4+\sqrt{5}\right)\left(4+\sqrt{5}\right)}$

[wraith dagger]

Yone

[toxic garden]

hmm

[wet rover]

expand the denominator and you'll see that you indeed end up with an integer value

[toxic garden]

meaning

[wet rover]

try expanding the denominator? you'll see what i mean

[reef lance]

yes; but do u see the problem with @toxic garden is to realize that the conjugate of -4+root(5) is 4+root(5)

[toxic garden]

how do I do that

[wet rover]

the conjugate of x-y is x+y so if you simply switched the signs there is no relevance of factoring out the negative other than adding an extra step(?)

[wet rover]

i mean like.. multiply (-4+root(5)) (4+root(5))

[reef lance]

people don't see -4+root(5) as [root(5) - 4]

[reef lance]

whose conjugate, easily, then is root(5) + 4

[toxic garden]

is it 4 + 5?

[reef lance]

yes U ARE CORRECT. but @toxic garden doesn't know this u see lol

[reef lance]

yep.

[wet rover]

no

[toxic garden]

what

[wet rover]

a shortcut to multiplication w the conjugate would just be squaring both the terms (-4)^2 + (root(5))^2

[reef lance]

with the sqrt ofc

[wet rover]

anyway if you manually multiplied you'd get -16 + 5 = -11 not 4+5

[toxic garden]

ahh

oh

mb

[wet rover]

okay... lol but the addition of an extra step doesn't change the fact that you still need to multiply but sure...

[toxic garden]

so thats 5(4 + root 5) / -11?

or is it not

[reef lance]

exactly. lol

[wet rover]

sure

[toxic garden]

is it not

[wet rover]

sure as in yeah it's fine

[toxic garden]

what does that mean

[wet rover]

lol...

i'm just saying what you wrote is fine for an answer

[toxic garden]

ahhhh

sooo whats the next step

[wet rover]

that's it

[toxic garden]

hmm

but answer key says its

[wet rover]

if you want to expand you can do that but otherwise leaving it this way is also fine

[toxic garden]

how does the negative go to 5

[wet rover]

it's equivalent to what you have

notice you have a negative on the denominator (-11)

[reef lance]

I TOLD YOU RIGHT LOL

[toxic garden]

yes

[wet rover]

which is equivalent to having a negative sign on the numerator

[toxic garden]

ohhh

is that how it works

[wet rover]

you might have to brush up on your fundamentals lol

[toxic garden]

yea

[wet rover]

but yes i guess

[toxic garden]

so either way its the same answer?

[wet rover]

yess

[toxic garden]

ahhhh alrr

[wet rover]

1/(-3) is the same as -1/3

same logic here

[toxic garden]

ahhh alr

so it applies to all stuff like that?

[wet rover]

i mean yeah if it happens to follow this exact scenario

[toxic garden]

I see I see

alr thxx

ima close the channel ig

.close

[cedar kiln]

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[thin rover]

how do i solve

[north copper]

4x - 84 = x

3x - 84 = 0

therefore 3x = 84

[quartz sage]

x=28

[thin rover]

.close

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[eager moss]

Literally have no clue what the hell to do here....

Like I drew the two circles....and that's it

I don't know how to even approach this problem

Graph on desmos

The answer key says this is the answer... but some of my classmates were saying it was a typo and it was supposed to be = 1

But I tried graphing that and it doesn't make sense...

[reef lance]

did u graph the supposed to be one graph and did it make sense

[eager moss]

No it didn't

[tropic oxide]

may i suggest plotting the centers of both circles on your desmos graph

theres something geometric that you can derive here

[eager moss]

are those supposed to be like foci of my answer...?

[tropic oxide]

dunno

[reef lance]

locus simply means a relation between x and y. in short terms, the question is asking you for the "Equation of a circle" whose tangent.....

[eager moss]

to the centers...

i'm so confused is the answer supposed to be a circle or an ellipse

[reef lance]

circle

this is clearly mentioned in the question

[eager moss]

ok so my math teacher must've been like on something when writing the answers then lmao

but apparently other students are getting similar answers to him....?

idk how

[tropic oxide]

okay so

i think i got it

[reef lance]

ask the 'other students'?! hehehe

[tropic oxide]

let O1 be the center of the big circle

and O2 be the center of the small circle

and also let r_1 and r_2 be their radii

and let P be the center of a circle tangent internally to the big circle and externally to the small

and let r be its radius

then we have $O_1P + r = r_1$ and $O_2P - r = r_2$

[wraith dagger]

Ann

[tropic oxide]

thus $O_1P + O_2P = r_1 + r_2$

[wraith dagger]

Ann

[tropic oxide]

okay so yeah it'll be an ellipse with foci the centers of your two circles

[eager moss]

I am still so lost lmao

[reef lance]

ohkay so i misread the question here; nowhere is mentioned it WILL be a circle

[eager moss]

wait so how do you know the collection of points will be an ellipse

huh

[eager moss]

what does this even look like...

[eager moss]

wait what

ok i think i'm misunderstanding external vs internal tangent

wait what....it's supposed to be externally/internally tangent to eh centers though

[cedar kiln]

@eager moss Has your question been resolved?

[cedar kiln]

@eager moss Has your question been resolved?

[cedar kiln]

@eager moss Has your question been resolved?

[eager moss]

…sure. Why not.

I give up.

[cedar kiln]

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[drifting plover]

How do I prove the divergence of this?

[drifting plover]

can't figure out a comparison test

[fallen heath]

$y = e^{-x}, , x > 0$

[wraith dagger]

Ansh

[fallen heath]

⁉️

[drifting plover]

doesn't that tend to 0?

[fallen heath]

Does! but really so bad as an example

[wraith crypt]

I gotta ask: divergence?

[drifting plover]

or convergence

but i honestly don't think this converges idk

[fallen heath]

it's diverging tho I believe :o cause the periodic hitting the 0 🤦‍♂️

[drifting plover]

don't think I really understand this sentence 😛

[drifting plover]

divergent integrals...?

[wraith crypt]

...this is absolutely convergent yknow

[drifting plover]

huh?

🤔

so uh

how do i prove its convergence?

🤣

i get here and i'm like "oh shit why did i spend the whole lesson playing video games"

this is probably my fault

[wraith crypt]

Well, you could always integrate by parts, but an easier way is to look at \int |5e^(-t)sin2t|

[fallen heath]

hmmm, yeah saw it

[wraith crypt]

Sorry typing with one hand is pain

[drifting plover]

hm

\int |5e^(-t)sin2t|

so basically i should look at this part and...?

oh-

got it now

thanks

[fallen heath]

and video gameee timeee 🙊

[drifting plover]

.close

[cedar kiln]

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[fallen heath]

$y = e^{-x}, , x > 0$

[wraith dagger]

Ansh

[fallen heath]

$y = sin 2x, , x > 0$

[wraith dagger]

Ansh

[fallen heath]

one is always < 1 and the other is at max 1

[drifting plover]

so if both converges, the big one converges?

[fallen heath]

can easily find a function to compare: e^{-x} here

[drifting plover]

kk thanks

[cedar kiln]

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[minor delta]

help

[solar grail]

a)

ok so what is AG in the first place

bro you there

[leaden snow]

no

[solid quarry]

AG is obv the vector from A to G

[solar grail]

no no i mean-

[minor delta]

i think its talking about points A and G on the cube

[solar grail]

what is it a diagonal of

[minor delta]

idk tho

[solar grail]

i am trying to make a question that will lead you to the answer

💀

[minor delta]

bruh

[solar grail]

we aren't supposed to directly give you an answer

so anyway

[minor delta]

yeah idk

[solar grail]

what is GA in terms of the cube

you dont know?

[minor delta]

naw i dont

[solar grail]

ok so its ze diagonal

[minor delta]

ok

[solar grail]

now what if we add GH and AH

ok since yall are doing vectors

what is AH = to?

[minor delta]

bruh honestly idk im probably gonna drop this class loll, i missed 3 lessons already im hella lost😭

[solid quarry]

do you know how to add vectors?

[solar grail]

u proly not in the right server bro- and wait wdym drop-

[minor delta]

nope...

[solid quarry]

theres ur problem

[minor delta]

like leave the class

[solid quarry]

its not hard tho

[minor delta]

ok do u mind explaining it?

[solid quarry]

sure :)

[minor delta]

🙏

[solid quarry]

see the little arrows? that denotes a vector

[minor delta]

yeah

[solid quarry]

vectors can be thought of arrows with a length and a direction in space

[minor delta]

ohhh ok

[solid quarry]

and they add like so ^, from head to tail

[fallen heath]

head to tail

[minor delta]

so would i do AG like this?

[solid quarry]

yeah

you should put the head/tip of the arrow at G

[minor delta]

okok so then how would i get an answer?

[fallen heath]

It's like connecting two positions together using the path defined :o

Say I were to visit you.. and I can only reach you via Detox.. so I first go to Detox and then Detox takes me to you

[minor delta]

ohhh ok that makes sense

so i wouldnt have to do any calculations for this question then just draw the lines?

[fallen heath]

🐖 similarly,in your question, you move from A to B, then B to C, then C to G

[solid quarry]

so try to figure out how you would add vectors a, b and c to get to the desired vector

since its a cube, a vector from B to C can also be used to get from A to D, E to H, F to G

[minor delta]

alright so
a) AG = a + b + c?
b) CH = -a + c
c) DF = c - b + a

would that be the answer?

[solid quarry]

yup!

[minor delta]

ok thank you so much🙏

[solid quarry]

good job

[minor delta]

how do i end this loll

[solid quarry]

.close

[minor delta]

.close

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[formal nest]

Cut the areas into small shapes

[cedar kiln]

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[pallid cobalt]

hi, i just want to ask for some advice with learning math for undergraduate physics as well as relevant physics olympiads

[pallid cobalt]

my friend recommend me to read math methods

but i find that to be quite dense and it doesn't seem to give an intuitive understanding of the math concepts presented

so is there a better way to learn math for my purpose ?

[wraith crypt]

This might be better asked on the physics server tbh

Math for math and math for physics can be quite different, esp in competitions and uni

[kindred ridge]

Hey I need help with a question please

[wraith crypt]

See #❓how-to-get-help

[cedar kiln]

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[last night]

Hello, what is the equation of deriving k from a period?

[last night]

so far I have...

$k = 2pi/p$

[wraith dagger]

Samm

[dense wing]

looks right ignoring the bad TeX

[last night]

sorry....I am not that experienced

How do I put the pi symbol?

[cedar kiln]

@last night Has your question been resolved?

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[cedar charm]

So I have to calculate tan(2𝛼 + 𝛽) where tan 𝛼 = −1,5 and tan 𝛽 = −2, I'm really confused as it's been a while.

[cedar charm]

There's formulas, I know, but that's exactly why I'm confused.

This is what confuses me a bit.

[proven fable]

Here ull get tan(2alpha)

So just put the value of tan(2aplha )in the second eqn

[cedar charm]

So just replace it with the tan(alpha)?

[proven fable]

Yeah i

[cedar charm]

Man this stuff is so easy that it messes with my head sometimes.

[proven fable]

I mean just think as the alpha beta in the brackets are variables

[cedar charm]

Yeahh I got it now:)

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[cedar kiln]

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[cedar charm]

How do I find the quadrant of this?

[graceful karma]

Impossible

You can limit it down to two quadrants but

[cedar charm]

It has to be given right?

Alright so if the quadrant of sin 𝛼 = 3/5 is given and I calculate the cos 𝛼, so 4/5, how would I find the quadrant of cos?

[graceful karma]

It'd be the same as the sin

Since the angle is the same

[cedar charm]

oh

Thanks!

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[cedar kiln]

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[cedar kiln]

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[crimson sedge]

hey i need help

[graceful karma]

@ Chromium#2633

Help with what

[crimson sedge]

1 sec im seding

any 1?

[glad kestrel]

what have you tried

[graceful karma]

What are you having trouble with

[crimson sedge]

well idk how to get perimeter

in any

[graceful karma]

Do you know what perimeter is

[crimson sedge]

yeah

[graceful karma]

What is it in your words

To get an idea of what you know

[crimson sedge]

like a boundry of the shape and u have to get the lenth of it

[graceful karma]

So how would you do the very bottom left one for example

[crimson sedge]

bottom left wouldnt that be adding them all

[graceful karma]

Yes

[crimson sedge]

oh

thanks

but what about the curved ones

[graceful karma]

Use your circumference formula

pi * diameter

Or 2 pi r

But keep in mind some are only half of a circles circumference

[crimson sedge]

so pi r + 2r

for semi circle circumfrence

[graceful karma]

No just pi r

Because a whole circle is 2 pi r

So half of that is pi r

[crimson sedge]

ok

[graceful karma]

Or wait

Ohh

I see what you mean

If it was just a semi circle yeah exactly

[crimson sedge]

ok so i take the circumfrence of the circle and add it to the rest of the stuff right

[graceful karma]

Yes

Just keep in mind that if the bottom part of the semi circle isn't part of the perimeter just use pi r

[crimson sedge]

could u do q2 part 2 so i can just see if im doing it right

[graceful karma]

Have you done it already?

[crimson sedge]

yeah

is it 46

[graceful karma]

That's not quite what I got

What was your work?

[crimson sedge]

oh wait

i see

is it 55.98

[graceful karma]

Yeah about

[crimson sedge]

yesss

thanks a lot

[graceful karma]

Np

[crimson sedge]

.close

[cedar kiln]

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[worthy ruin]

A =3
B = 2/3
C = -13/3