help-10

[hidden kraken]

Very true actually

Yes I was going to do a similar thing to what we did for x again for y but this is nicer

[south oracle]

If the function was sin(x²) instead of 1/x we can't use this approach

But since this fn is very simple and has a nice restriction that p and q can't be 0

[hidden kraken]

Yeah that’s true

In terms of the whole m and b being constant stuff btw

I’m pretty sure that means that the locus of points is for all the possible values of the one that isn’t constant

That should be manageable once you find an expression for the x ordinate and the y ordinate

[south oracle]

So anyways as pq is -1/m, we can find that the midpoint y-coordinate should be 1/2 (-b/m)/(-1/m) = b/2

So the coordinates of the midpoint that we have found is (-b/m, b/2)

so i) when m is constant

Let x' be -b/m and y'=b/2

We here find that -mx' =2y'

Thus y'= -m/2 x'
We can write it like this as m is a constant other than 0 here

ii) when b is constant, y' is the half of b, which should be constant as well, regardless of what values x' has due to the variable m

So the equation of the locus becomes y'= b/2 except for when x=0

Why?

we stated here that x'= -b/m

When b is a constant other than 0, x' cannot be 0 no matter what m is

[obtuse pebble]

@latent knoll Has your question been resolved?

[south oracle]

Basically, a constant means a fixed value, like 3

To find the answer of this question

You have to find the x and y coordinates of the midpoints of A and B where A and B are the two points that the graph t=1/x and y= mx +b meet at

Note that since the question said "the midpoint of A and B", A and B are two different points on the graph

[latent knoll]

wait to find the x coordinates for A and B, dont u have to find the values for m and b?

[south oracle]

Ultimately yes

But the question is asking how can we find the general case when either m or b is a variable

That is, when either m or b is changing just like how we can change tje value of x in y=f(x) freely; and the other is a fixed value

[south oracle]

We will start here by setting A has the x coordinate of p, and B has the x coordinate of q

Since A and B are on the graph of the function y=1/x, A is the same point as (p, 1/p) and B is the same point as (q, 1/q)

@latent knoll understood?

[obtuse pebble]

@latent knoll Has your question been resolved?

[obtuse pebble]

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[timid silo]

hi

[timid silo]

i need help

[burnt cipher]

i was typing here lol

oh well gl

[timid silo]

sorry

hi i need help

[tardy helm]

Um, what are you trying to solve @timid silo

[timid silo]

express the letters in the brackets as subject of formula

[tardy helm]

Ah, so you are solving for u?

[timid silo]

for u means?

oh i understand now

[tardy helm]

Well u is in the brackets so I assume you are trying to find what u =

[timid silo]

yes

[tardy helm]

Ok. Well. To do that we need to get u by itself correct?

[timid silo]

yea

correct

[tardy helm]

What could we do first to attempt this?

$5q = \frac{3}{u} - 5$

[warm shale]

jpsz

[timid silo]

eh?

[tardy helm]

This is the problem above

[timid silo]

yes

[tardy helm]

So, (u) being on the bottom of a fraction is a problem right?

[timid silo]

can u type (u) pls

[timid silo]

yes

we need to move to the other side

[tardy helm]

Yea, lets move the 5 to the other side first

$5q + 5 = \frac{3}{u} - 5 + 5$

[warm shale]

jpsz

[tardy helm]

$5q + 5 = \frac{3}{u}$

[warm shale]

jpsz

[tardy helm]

So now how do we get (u) to not be on the bottom of a fraction?

[timid silo]

sorry pls repeat

[tardy helm]

hm, all we have done so far is added five to both sides

make sense?

[timid silo]

so u move the single 5 to the other side

[tardy helm]

Yup

[timid silo]

ok makes sense

[tardy helm]

Now we can multiply both sides by (u)

[timid silo]

now u move the 5q+5 and replace with u?

[tardy helm]

do you mean swap the 5q+5 with the u?

[timid silo]

yes

[tardy helm]

Yup

$u = \frac{3}{5q+5}$

[warm shale]

jpsz

[tardy helm]

There we go. You solved it

[timid silo]

ok i understand now

ty

[tardy helm]

Nice job!

[timid silo]

alrigth thats it thanks

should i close the server?

[tardy helm]

If you are done, then yes!

[timid silo]

.close

[obtuse pebble]

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[calm carbon]

I don’t get this method

[calm carbon]

The way I did it was different - but this seems more efficient

What does it mean sub x = sqrt y and rearrange

Doesn’t that mean the poly is in terms of y ?

[north acorn]

can i see the question

[calm carbon]

here’s the q

<@&286206848099549185>

[tardy helm]

So, what are the roots of the equation $2x^3-8x^2+3x+5 = 0$?

[warm shale]

jpsz

[tardy helm]

Did you find those?

[obtuse pebble]

@calm carbon Has your question been resolved?

[obtuse pebble]

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[silver plover]

for this question, does this diagram satistfy what the question is talking about?

[wise talon]

ye looks right

[silver plover]

and at time t = 0, will the y position be 0?

[pseudo swift]

nope it ain't right

[silver plover]

yh my answer is wrong

[pseudo swift]

the projectile is fired over a cliff

you're starting on top of the cliff

then going down

[wise talon]

errr

[silver plover]

surely it would say instead then 'it is fired off a cliff'

[wise talon]

that diagram is a projectile being fired over a cliff....

[silver plover]

btw aPlatypus is correct i just dont get why

[wise talon]

is it not?

strange

[silver plover]

yh im baffled as well

[wild swallow]

"horizontally away from the cliff edge"

[silver plover]

yh

my i is away from the cliff edge no?

[wild swallow]

wait thats about the i and j

[pseudo swift]

yeah

:/

[silver plover]

surely it would say instead then 'it is fired **off **a cliff'

which is baffling me

[pseudo swift]

yeah I can understand it

[silver plover]

but as i said aPlatypus is correct

[pseudo swift]

but you wouldn't have any starting position then

[wise talon]

maybe its time to ask the prof lol

[silver plover]

yh i prob should

[wise talon]

wording is so weird

[pseudo swift]

yeah

[silver plover]

i wanna see my working anyway?

[wise talon]

did you take analysis 2 btw

[silver plover]

still considering it

gotta make my choices soon

[wise talon]

lmao

[silver plover]

icl it looks very calculus based

@pseudo swift if we go via ur logic, then would the starting position in the y direction be 25j?

at t=0

[pseudo swift]

that's how I understood it at first

[silver plover]

cuz the reason my answer is wrong is cuz i said at t = 0 y displacement is 0

but in fact it should be 25

ight kl thanks for the help lads

[pseudo swift]

with my interpretation of the question you'd get this

cliff being 25 units high

[silver plover]

yh thats it

ur correct

just didnt get why

but i guess i gotta kinda accept the wording now

[pseudo swift]

yeah the wording is a bit ambiguous indeed

[silver plover]

ight kl thanks mate

.close

[obtuse pebble]

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[rigid nacelle]

Common factoring, quadratics

[rigid nacelle]

What is this and how do I do it

[high cypress]

quadratics are a way to express an expression/equation with 2 unlike variables

such as in this form

$ax^2+bx+c$

[warm shale]

gekka

[high cypress]

we cannot combine that ax powered and that bx because ax powered and bx are 2 different terms

[rigid nacelle]

So what do we do to solve when doing common factoring

[high cypress]

factoring is a method that break a quadratic into two expressions, and that when both expressions are being multiplies, it results to the same quadratic as before

[rigid nacelle]

Ah okay

Sweet, that's all I had to ask

Thanks man, g'day!

.close

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[upbeat valley]

how did they get $A^{-1}$?

[warm shale]

TheWhiteShadow

[royal basin]

$A^3 - 4A^2 - A - 6I = 0$

[warm shale]

Ann (glomed)

[royal basin]

from this, get $A^3 - 4A^2 - A = 6I$, thus $(A^2 - 4A - I)A = 6I$

[warm shale]

Ann (glomed)

[upbeat valley]

ahhh i see thanks!

how would you get $A^{-2}$ then?

[warm shale]

TheWhiteShadow

[royal basin]

by a variety of methods

probably just multiply the expression for A^-1 by itself and then replace A^3 and A^4 with their epxressions

[upbeat valley]

right

so

[royal basin]

gonna be mildly painful

[upbeat valley]

$A^{-2} = \frac{1}{6}A^3-\frac{2}{3}A^2-\frac{1}{6}A$

and then sub A^3

[warm shale]

TheWhiteShadow

[royal basin]

eh?

where did you get this from

(-1/6 A^2 + 2/3 A - 1/6 I)^2 is def not that lol

[upbeat valley]

oh wait messed up

$A^{-2} = \frac{1}{6}A-\frac{2}{3}I_3-\frac{1}{6}A^{-1}$

then sub for A^-1

[warm shale]

TheWhiteShadow

[royal basin]

oh yeah you could do that too

might be less painful that way

[upbeat valley]

sick

tqtq

.close

[obtuse pebble]

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[mellow forge]

Why can we use the taylor series to find the error margin in eulers method?

[obtuse pebble]

@mellow forge Has your question been resolved?

[timid silo]

elaborate?

[mellow forge]

so euelrs method

yn+1=yn+hf(xn,yn)

[timid silo]

ok idk

[mellow forge]

you know anyone who knows?

[timid silo]

ping helpers

[mellow forge]

<@&286206848099549185>

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

test each case

it'll be really obvious if you test them

[viral orchid]

wdym

[lean trail]

substitute the options into the given equation

[viral orchid]

oh

yeah got it

ty

.close

[obtuse pebble]

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[silver plover]

where do they get v hat from?

and where does v hat and v^2 come from?

[pseudo swift]

v hat is v/|v|

therefore v=|v| * vhat

Replace in the quadratic drag equation and you're done

@silver plover

[silver plover]

ok but then what about the v^2

where does that come from

cuz u get f = -gamma vhat

and u know gamma is given as 1/2 C p A

but when u sub that in, where does the v^2 come from?

@pseudo swift

[pseudo swift]

i think in the solution they write |v| as v

(not boldface should mean it's a scalar value)

@silver plover

[silver plover]

so what should it be instead of v^2?

[pseudo swift]

what I mean is that when they write v^2 here, they mean |v|^2

it's a common enough convention

v (boldface) is the vector

v (not boldface) is its norm

[silver plover]

ohhh

ok

makes sense

gotcha

thanks pal

wow

a super react

bro u used it

😭

cant un do it

once its used its used

[pseudo swift]

I didn't know there were super reactions lol

was surprised

[silver plover]

u get one free

i think youve used it now

rip

[pseudo swift]

yeah rip

you got another question or can we close ?

.close

[obtuse pebble]

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[coarse laurel]

Hi i got 57% percent for my cycle test (math) any advice?

[shy vigil]

prattice

prattice makes perfect

[coarse laurel]

thats the problem i practiced my heart out

[tame narwhal]

practice more

[coarse laurel]

any easy methods to help me understand it better

[tame narwhal]

what is "it"

[coarse laurel]

well, everything, but specifically on fractions and decimals

[tame narwhal]

I don't know, maybe Khan Academy

[coarse laurel]

what is that

[tame narwhal]

an educational website. google it

[shy vigil]

have u tried all questions of your textbook

[coarse laurel]

what do you mean by that

[shy vigil]

the questions in your textbook

[coarse laurel]

yeah, i did some revision on stuff in there if thats what you mean

[shy vigil]

you should try all that you can

if you barge into any trouble with any question you can always ask here or even dm me personally 🙂

[coarse laurel]

ok thanks

I know this is a math channel on math help, but any idea how to explain to my parents that i got 57%

[shy vigil]

i dont know man i suck at that. I'd just tell them what I got and tell true reasons why its not what I expected. Also assure that I'll prepare harder for the next tests.

But in any case, remember that you're trying to improve, and that's what matters. I'm proud of you, and I know you'll improve in your next tests 😄

[coarse laurel]

thanks

can you dm me quick

[shy vigil]

Of course! Sorry, I forgot to open my dms for this server

[tame narwhal]

just rip the band-aid off, so to speak

[obtuse pebble]

@coarse laurel Has your question been resolved?

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[rain pendant]

unsure how to go about this

[rain pendant]

Nvm, got it

.close

[obtuse pebble]

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[warm canopy]

<@&268886789983436800>

[odd ermine]

👍

[rain pendant]

jesus that was crazy

[odd ermine]

damn I got snipped metal actually got the ban

this is so sad

[warm canopy]

Battle of the mods

[oblique glacier]

Yes

im on fone too

Get good

[obtuse pebble]

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[noble kindle]

Not sure how to relate |a+b| to |Axb|

[timid silo]

use the cos rule

along with this fact:

[noble kindle]

what is the cos rule?

[timid silo]

we want to find theta which is the angle between a and b

[noble kindle]

so the dot product $a \cdot b = |a||b| \cos \theta$?

[noble kindle]

right.

[timid silo]

yeah

[noble kindle]

but we dont have |a x b|, instead we are given |a+b|

[timid silo]

expand |a+b|^2 in terms of the dot product

[noble kindle]

$|a+b|^2 = (a+b) \cdot (a+b) = |a|^2 + |b|^2 + 2(a \cdot b)$

[warm shale]

KN

[runic void]

Isn’t this Law of Cosines?

[noble kindle]

So then $a \cdot b = \frac{-1}{2}$

[warm shale]

KN

[timid silo]

yeah, it is

[noble kindle]

so $\frac{-1}{2} = 2(1) \cos \theta$

[warm shale]

KN

[noble kindle]

But how would you sove $\cos = \frac{-1}{4}$

[warm shale]

KN

[timid silo]

you dont need to

draw a triangle and use pythagoras to find sin theta

[timid silo]

are you sure its sqrt 3?

,calc 4^2 - 1^2

[warm shale]

Result:

15

[noble kindle]

yeah, youre right..

i got it. ty

.close

[obtuse pebble]

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[empty salmon]

How do I do the 18th question

[rich plume]

separately write values of x,y and z

[empty salmon]

done that

[rich plume]

now write x-y, y-z,z-x using the AP values

[empty salmon]

aight but like how'd you come to that step

i mean how'd you figure out that we've to substract the given terms

[rich plume]

that's intuition and practice

Also this seems to simplify

[empty salmon]

fair

[rich plume]

I guess you need to check the options now

[empty salmon]

done

[empty salmon]

yea aight

[rich plume]

Like the B option seems that it can be simplified

by setting RHS=1

[empty salmon]

wait but I didn't touch the GP values of x y z at all

only used the AP values and left the gp ones unused

[rich plume]

I guess that will come in use while checking for options

[empty salmon]

true

aight i got it now

.close

[obtuse pebble]

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[silent cave]

is my solution correct? or am i doing something wrong

[tardy helm]

yup, that is correct

[silent cave]

oh its you again

[silent cave]

thanks for previous reply btw

.close

[obtuse pebble]

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[silent cave]

.reopen

[obtuse pebble]

✅

[silent cave]

ok finish 😂

[tardy helm]

LOL, glad to help :😂

[silent cave]

might ask more as a sanity check, dont want to clog the help channels so much

so ill solve alil

[tardy helm]

Nice, sounds good!

[silent cave]

yeye

ok see u in the next help channel 😂

.close

[obtuse pebble]

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[timid silo]

how would i approch d

[kind hawk]

what did you find in b?

[pseudo swift]

@timid silo

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

i got

this sorry

its a bit messy

i got basis to be 1 0 0 1

0 1 -1 0

and dimension to be 2

@pseudo swift

[pseudo swift]

sounds alright

ok so we know the A^n live in C_A (thanks to part c), a space of dimension 2

if we look at, say {A^0, A^1, A^2}, can this set be linearly independent ?

[obtuse pebble]

@timid silo Has your question been resolved?

[pseudo swift]

@timid silo

[obtuse pebble]

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[timid silo]

Can somebody with a graphing calculator please help me? I don't have one on hand!

[trail musk]

Just use excel

[timid silo]

any tutorials available?

[trail musk]

Yeah for sure

Go on YouTube and search up how to do a linear regression on excel

Really useful skill to learn too, fast quick and easy

[timid silo]

thanks

.close

[obtuse pebble]

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[timid silo]

.reopen

[obtuse pebble]

✅

[timid silo]

reposting for new people

anybody have a graphing calculator and know how to find correlation coefficient r?

please help

[dusk pasture]

you could use excel and do linear regression

[timid silo]

😭

please show some tutorial if you're gonna say that

[dusk pasture]

no i won’t show any tutorial

[timid silo]

???

[dusk pasture]

there’s plenty online

[timid silo]

where man

they're all different versions of exce;

or modified

[dusk pasture]

it’s all the same man barely any differences

[timid silo]

@tame narwhal would YOU mind providing a link seeing as how you reacted

also this is one outlier problem

i dont deal with these problems

so i dont want to do all that for one problem

[tame narwhal]

well there are no shortcuts here. use Excel, a graphing calculator w/ regression function, R, Python, Matlab, SPSS, whichever

[timid silo]

ok i pulled up desmos atleast tell me how to configure it

[tame narwhal]

I've never used desmos for that. try googling

[timid silo]

bruh...

[candid oracle]

Do you own any graphing calculators?

[timid silo]

<@&286206848099549185> scroll up and help please somebody i don't deal with these problems this is like the first and only time and i don't have a graphing calculator

[timid silo]

yes but they are kept at my school

[spring elk]

Yes

What do i need to do

[timid silo]

and these people are just telling me to google it like bro

why would i ask if i could

[timid silo]

help me

can you plug this in i don't have any graphing calculator on hand

[candid oracle]

Okay. The calculator basically does the problem. Without a calculator, you're more or less asking us to do the problem FOR you, which is why people keep giving you indirect answers (I imagine).

To the person who DOES own a graphing calculator - what model? TI-83 or 84?

[timid silo]

i dont use excel cause i usually dont deal with theese problems

[candid oracle]

NT lucid, I'll see if I can get you a link to a regression site. One sec.

[spring elk]

I have a casio cg50

But no clue how to do that question on it

[timid silo]

would you mind showing some sort of tutorial to plug this in please

i swear im not asking for answer

just how to do it

[candid oracle]

Hm. I'm unfamiliar with casio. I might have to refer you to the same web site I sent lucid to.

[timid silo]

?

[candid oracle]

Okay. Shortened steps:

1. Find regression calculation tool (calculator, web site, etc). For now, use: https://www.graphpad.com/quickcalcs/linear1/
2. Type the YEARS into the "x" column, and the corresponding TICKETS SOLD into the y column.
   2.5 CHECK FOR TYPOS. This is the big killer for questions that use regression tech. Typos, typos, typos.
3. For this website, press "calculate now". For other tools, there will be different buttons to press to basically "Get the result"

[timid silo]

what about this label button sir

[candid oracle]

Think it through. "Label" means exactly what it says - label "x" and "y". So, you can type "year" as the label for x (if you want to), and "tickets sold" for the label of y.

[timid silo]

oh thank you dude

[candid oracle]

Labelling is not mandatory. But it can be useful just so you're not left asking "wait, what was x again?"

real quick before I go - you need the correlation coefficient, right?

[timid silo]

yeah makes sense

[timid silo]

yes r

[candid oracle]

Okay. I'm not a fan of thow this particular site shows you that answer, so this'll take some explaining.

In stats, there are two closely-related number: r and r^2

In the "goodness of fit" output, they give you r^2

take the square root of that to get "r"

BUT

when you square root the number, it just spits out a positive answer

in general cases, the r may not be positive. It has the same sign as the SLOPE does

[timid silo]

ok

[candid oracle]

since your slope IS positive, the r will also be positive

[timid silo]

thank you man

[candid oracle]

but in future cases, watch out. It may be negative

You got it. Best of luck!

[timid silo]

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[timid silo]

I wrote something in my notes: 'The sum of lines is equal to the sum of lines, and the product lines are equal to the sum of products.' This should have reminded me how we arrived at this result on the right side

[worn yoke]

,rotate

[timid silo]

how we got this on the right

where can i see that definition when we have sum

and why we got -1?

[tame narwhal]

,tex .exp rules

[worn yoke]

exponent rules:
$a^{b+c} = a^b \cdot a^c$

[warm shale]

cloud

cwatson

[worn yoke]

then we factor out $e^{x_0}$

[warm shale]

cloud

[timid silo]

.close

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[timid silo]

translate

[fathom flicker]

Are you telling us to translate for you?

This looks like a test, (Testul = Test?)

<@&268886789983436800>

[teal turret]

Bro what how u gonna tell us to translate

[tardy epoch]

This is a language server, no?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[ocean nexus]

hi, i kinda forgot how to find the line of a rotated line. What is the new line of the line y = 2x, rotated 90 degrees around the origin? is it just y = -2x?

[ocean nexus]

is there an equation?

so i just flip the slope?

ok thanks, what if it was like 45 degrees or 33 degrees?

oh ok

thanks for the help

.close

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[terse sparrow]

I have a value, for example 8. I want to convert it so that 8 means 1 and 16 means 2 and 4 means 0.5 etc. I dont know what this is called in math, it's probably a concept? I need to program to set a multiplier based on deviance from the "1" value, which in this case is 8.

[tame narwhal]

just divide by 8?

[fierce lagoon]

That's legit just some number, n, divided by 8

Think about it

if n = 8, n/8 = 1

[terse sparrow]

yes of course, im sorry, I asked a weird question and I found the solution. thanks for the help!

How do I indicate succces?

[fierce lagoon]

Wdym

Like close the channel?

[terse sparrow]

yes exactly, sorry, im just occupying it for no reason now

/close

[fierce lagoon]

.close

[obtuse pebble]

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[terse sparrow]

.close

ty

/.close

.close

[fierce lagoon]

I already closed it...

[terse sparrow]

ups I messed up

can I close it again then?

.close

[obtuse pebble]

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[opaque galleon]

guys i need help

[opaque galleon]

am I doing this correctly?

my answer is the last line

[obtuse pebble]

@opaque galleon Has your question been resolved?

[obtuse pebble]

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[opaque galleon]

my solution:

am I wrong?

plssss

plsssssss

🙏

[chrome horizon]

circles are called parabaloids now ??

[obtuse pebble]

@opaque galleon Has your question been resolved?

[brisk matrix]

x^2 + y^2 = z does not make a circle

[chrome horizon]

is z a constant

[brisk matrix]

no

[chrome horizon]

ah ok I assumed it was radius squared

[obtuse pebble]

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[light hull]

How does this work?

[brisk matrix]

they factored a 5

[median skiff]

both terms on the numerator are being multiplied by 5, so they have a common factor of 5, meaning you can just pull it out

^

[light hull]

oh

understood

.close

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[flint ivy]

i dont understand this formula they used

am i just supposed to use the general formula

like memorise it

and just put some numbers in it

or can somebody explain to me how it works

[obtuse pebble]

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[spare belfry]

I need help

[fallow cove]

ok so what you got?

[spare belfry]

Nothing

Idk how to do it

[fallow cove]

you learnt similar triangles right?

[spare belfry]

Ye

not triangles looking like that tho

[fallow cove]

ok so I need you to recall the identities of angles on parallel lines

[spare belfry]

Huh

I already know the answer

[fallow cove]

ok

[spare belfry]

I just wanna see steps on how to do it

Answer is 260

all i wanna see is how that happened

<@&286206848099549185>

[obtuse pebble]

@spare belfry Has your question been resolved?

[obtuse pebble]

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[upbeat valley]

Im trying to proof $\langle f, g \rangle = \int_a^b f(t) \overline{g(t)} , \mathrm{d}t$ is a inner product of $\mathcal{C}[a, b]$

[warm shale]

TheWhiteShadow

[upbeat valley]

for axiom 4

$\langle f, f \rangle = \int_a^b f(t) \overline{f(t)}$

[warm shale]

TheWhiteShadow

[upbeat valley]

what does this equal?

Is it $\int_a^b \mathfrak{I}(f(t)^2) , \mathrm{d}t$

[warm shale]

TheWhiteShadow

[brisk matrix]

what is that

[brisk matrix]

you should recognize f(f*) to be >= 0

[obtuse pebble]

@upbeat valley Has your question been resolved?

[upbeat valley]

how so?

[brisk matrix]

do you agree that z(z*) >= 0 for all complex values of z?

[upbeat valley]

yea cuz its just Re(z)^2 + Im(z)^2

[brisk matrix]

then notice that f(x)(f*(x)) is also just a value in C

and hence positive

[upbeat valley]

C as in complex?

[brisk matrix]

let me rephrase that

f(x) is in C, the set of complex numbers

so f(x)(f(x)*) >= 0

does that make sense to you

[upbeat valley]

am I not trying to proof that tho?

[brisk matrix]

i don’t see how you are

you’re showing <f,f> >= 0

but what i said above is a fundamental fact of complex numbers

[upbeat valley]

wait is f(x)(f(x)*) = Re(f(x))^2+Im(f(x)^2

[brisk matrix]

yes

set f(x) = z

[obtuse pebble]

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[drifting sluice]

When I rotate this around y-axis, why can't I calculate the volume as V = 2π*A?

[royal basin]

by A do you mean the area of the triangle with these points as vertices

[drifting sluice]

Yeah, the closed area

[royal basin]

why would you be able to calculate it that way? moving it further away from the y axis wouldn't change A, but it would make the volume greater.

[drifting sluice]

So I suppose to calculate trapezoidal area * 2π and subtract square * 2π?

Nvm it's still π

Do I have to use formula for margins or am I expected to find volume solely through area of triange?

Here I still find it π via integrals

What am I missing

[obtuse pebble]

@drifting sluice Has your question been resolved?

[obtuse pebble]

@drifting sluice Has your question been resolved?

[obtuse pebble]

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[thin bough]

why in this question BOD=360-140

i got 140, here is my working

but can some explain why is BOD=360-140

[unkempt perch]

AB0 is a right angle

and so is AD0

90+90+40=220

[thin bough]

oh ok but the reason i have this confusion is

due this problem

like here i got the correct answer via the method i tried

here have a look

[unkempt perch]

u got it right?

[thin bough]

yeh

[unkempt perch]

and u dont know how to do the other

[thin bough]

but why in this case we don't do 90+90+50 for 130 like u said for the first problem

[unkempt perch]

you do tho

then 130/2 is 65

like the answer

90+90+50=230

360-230=130

130/2=65

[thin bough]

ah got it

thanks

[unkempt perch]

np

[thin bough]

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[ruby elm]

Open.

[ruby elm]

Just studying limits, came across this question.

Find limit (x tends 0) of (e - (1+x)^(1/x))/tan x .

The one in black.

[unkempt perch]

have u tried lhopital?

differentiating numerator and denominator

[hybrid gyro]

I think you should guide them to realise the question satisfies the criteria of L'Hopital and hence use L'Hopital, since you can't always use L'Hopital

[hybrid gyro]

so do you know anything familiar to (1+x)^(1/x) ?

[ruby elm]

I was thinking the same thing but not sure if l hopital can be used here.

[unkempt perch]

yeah true

[hybrid gyro]

well L'Hopital can be used here (I believe), but we need to first show it satisfies the criteria of L'Hopital with a bit of twist. So here's my strategy towards this question (with a bit of reverse engineering ish):

1. can I use L'Hopital? If I can, then it should satisfies 0/0 or +-infinity/inifinity. And obv tanx = 0 as x tends to 0.
2. so then next question is, can I show that the numerator = 0? And this is my hint

[obtuse pebble]

@ruby elm Has your question been resolved?

[ruby elm]

Still could not get through. Need something more.

[hybrid gyro]

ok, so do you understand what I was saying before?

[ruby elm]

Yes.

I just found something. Limit x tends 0 of (1+x) ^ 1/x = e

So the numerator too tends 0

[obtuse pebble]

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[zinc stump]

Looking to disprove the following:
let $\alpha, \beta, \gamma$ infinite ordinals.
\ 1. $\alpha < \beta \Rightarrow \alpha^{\gamma} < \beta^{\gamma}$ \
2. $\alpha < \beta \Rightarrow \gamma^{\alpha} < \gamma^{\beta} $

[warm shale]

meitar5674

[obtuse pebble]

@zinc stump Has your question been resolved?

[ruby elm]

What does infinite ordinals mean?

[zinc stump]

like $\aleph / \aleph_0$

[warm shale]

meitar5674

[zinc stump]

set theory

[obtuse pebble]

@zinc stump Has your question been resolved?

[obtuse pebble]

@zinc stump Has your question been resolved?

[zinc stump]

never mind guys 🙂

solved it

[royal karma]

yameleh

[obtuse pebble]

@zinc stump Has your question been resolved?

[obtuse pebble]

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[fathom sigil]

Hey, I wonder how is the "t" variable just gone in the third row?

[fathom sigil]

and it's only sin(n*4pi) that is left

so the "t" is there in the first and second row, but not in the third. I dont understand how/why it can just dissapear

[latent walrus]

they applied the integrals limits, which are the lines with the numbers at the top and bottom

so t gets replaced by those

[obtuse pebble]

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[thorny canyon]

help in french please

[obtuse pebble]

Please don't occupy multiple help channels.

[thorny canyon]

je ne comprend pas la deuxième question

[alpine raven]

Qu'est ce que tu as essayé ?

[royal basin]

attendez un instant

[thorny canyon]

j'ai fais la premiere question mais la deuxième je ne comprend pas

[royal basin]

n'avons-nous pas fait cette même question, les deux parties, il y a quelques jours ?

[thorny canyon]

ahhh je me souvient de toii

oui c'etais pour un controle merci bcp franchement ça m'as grave aidé j'ai pu avoir une bonne note

[royal basin]

#help-20 message

[alpine raven]

Sinon théorème de complétion de base

[thorny canyon]

sauf que je ne comprend pas comment on avait fais pour réondre

quel est le theoreme ?

[royal basin]

réondre ?

[thorny canyon]

*répondre

[royal basin]

ah

[thorny canyon]

faute de frappes sorry

[alpine raven]

théorème de la base incomplète

[thorny canyon]

et que dit le théoreme precisemment ?

[alpine raven]

Elle dit que toute famille libre d'un espace vectoriel E peut être complétée par des vecteurs de E pour former une base

[thorny canyon]

mais avec quel genre de vexteurs on complete ?

*vecteurs

[royal basin]

attends je vais écrire en détails

[alpine raven]

J'ai pas fait énormément d'exercices à propos de ce théorème mais je crois qu'il faut rechercher une base canonique de vecteurs de E

[thorny canyon]

d'accord merci bcp !

enft les coordonnées d'une base canonique sont : (1,0,0) (0.1.0) (0.0.1)

[alpine raven]

Ca c'est dans R^3

[thorny canyon]

mais dans cette exercice on a utilisé les coordonnées de X²

bah justelent dans cette exercice on est dans R3 nn ?

*justement

[alpine raven]

une base canonique de R_2[X] c'est (1, X, X^2) si je dis pas de bêtises

[alpine raven]

Non

L'exercice le dit bien qu'on est dans un espace de polynomes de dégré inférieur ou egal à 2

[thorny canyon]

bah la dim de R_2 [X] c'est R3 normalement

[alpine raven]

Oula

[thorny canyon]

c'est dans la correcxtion -_-

j'invente pas

[alpine raven]

tf ?

[royal basin]

1. la dimension de R_2[X] est 3.
2. pour qu'une famille de vecteurs soit une base, il faut 2 choses:
   2.1. la famille doit être libre (c'est la première partie du problème)
   2.2. elle doit contenir exactement autant de vecteurs que la dimension de son espace
3. notre famille (P_1, P_2) contient maintenant 2 vecteurs.
4. il en doit y avoir 3 pour une base, donc il nous faut 1 de plus.
5. ce 3ème vecteur peut être presque n'importe quoi, avec une seule condition: la famille étendue doit être aussi libre.
6. pour que la famille étendue soit libre, il faut que le troisième vecteur ne soit pas une combinaison linéaire des deux précédents.
7. un exemple de candidat pour le troisième vecteur est X^2.

[alpine raven]

la dimension de R_2 [X] est 3

mais pas R^3 comme tu dis

[royal basin]

la dimension d'un espace est un NOMBRE et non soi-même un espace

[alpine raven]

^

[royal basin]

un nombre naturel, pour être plus exacte

[thorny canyon]

oui c'est ce que je disais la dim ici est bien 3

[alpine raven]

Tu as dit R^3, j'ai buggé

[royal basin]

lis ceci et dis-moi le numéro de la première ligne qui te confond

[thorny canyon]

ahhh mince dsl c'est de la faute

je me suis mal exprimé

dsl @alpine raven

[alpine raven]

npnp

[royal basin]

fais BONNE attention parce que, si tu dis que c'est par exemple 4. je vais présupposer que tu comprends les lignes 1, 2, 2.1, 2.2 et 3

[thorny canyon]

au liue d'utiliser X^2 je peux utiliser quoi d'autre ?

*lieu

[royal basin]

bah y en a un million

2X^2

-X^2

1

X-1

420X^2 + 69X

[alpine raven]

💀

[thorny canyon]

je peux prendre 3x+1 ?

par exemple

[royal basin]

bien sûr

[thorny canyon]

mais je peux vraiment prendre tout et nimporte quoi ?

[royal basin]

pas tout

[thorny canyon]

justement qu'est ce que je ne peux pas prendre

[alpine raven]

pas tout, faut que la famille reste libre et genératrice

[royal basin]

seulement ceux qui ne sont pas exprimables en combinaison linéaire de X^2+1 et X+1

moi je parle de la préservation de sa liberté

[thorny canyon]

qu'est ce que ça veut dire ?

[royal basin]

et son égalité (en cardinal à la dimension)

[royal basin]

qu'elle reste libre ...

[alpine raven]

que la famille reste libre

[thorny canyon]

ahhh mdrrr ok dsl

[royal basin]

je sais pas comment faire un jeu de mots avec fraternité

[thorny canyon]

par exemple j'ai un autre exercice ou j'ai P1= X^3-X, P2= 2X^3+X²+2X, P3= 3X^3+X²+2X

il faut que je dise si c'est une base ou pas sinonn faut que je complete

la du coup je choisit les coefficient a, b et C

aP1+bP2+cP3=0

c'est bon poiur l'instant ?

[royal basin]

eh

c'est pas très productif

de cette façon tu vérifieras la liberté

[thorny canyon]

ah oui c'est vrai

faut que je montre que c gene aussi

[royal basin]

mais ça se passe dans R_3[X] non ?

dim = 4

[thorny canyon]

oui c'est ça

[royal basin]

donc cette famille de trois n'est pas une base

[thorny canyon]

du coup faut que je complete absolument

ouiii

[royal basin]

il faut 1 de plus

[thorny canyon]

yess donc la je dois choisir un vecteur

[royal basin]

ici se trouve facilement le polynôme constant 1

ou 2 ou 3 ou n'importe quoi sauf zéro

[thorny canyon]

ok je comprend c'est juste c encore flou dans ma tete, je n'ai pas trop compris pourquoi on ne peut pas choisir ce qu'on veut l'histoire de combinaison lineaire et tout

[royal basin]

si l'on choisit un vecteur exprimable en combinaison linéaire des précédents, on perd la liberté

la liberté signifie qu'il n'y a pas de relations linéaires entre les membres de la famille

l'ajout d'un nouvel élément avec elles défenestre la liberté

[thorny canyon]

ah en gros lorsque je fais le determinant pour prouver si c'est libre ou pas faut que je trouve u'importe quoi sauf 0 c'est ça ?

[royal basin]

eh

[thorny canyon]

eh = oui ?

[royal basin]

tu peux faire le déterminant s'il n'y a pas de candidat évident

choisis n'importe quoi et vérifie si ça marche

[thorny canyon]

ok je dois verifier que c'est different de 0

purée c'est dingue je comprend tout avec toi devient ma prof particulère 😂😂

d'ailleurs pour mon controle tu m'as sauvé l'autre fois ! J'ai carrement fais un cours a mes camarades de classe

[royal basin]

bonne chance

[alpine raven]

L'algèbre linéaire c'est fun quand on comprend tout

[thorny canyon]

je suis trop nul en algebre lineaire

les applications lineaire je ne comprend rien

avec le noyau et tout

notre prof fais que du cours magistral

[alpine raven]

Bruh

Après, j'ai réussi à tout comprendre en 2ème année de prépa

[thorny canyon]

jsuis en L1

L1 PC précisemment

[alpine raven]

ah oé cool

la fysik

[royal basin]

⚠️ ANGLICISME ALERT ⚠️

[thorny canyon]

non c'est trop dur !

[alpine raven]

[thorny canyon]

c'est quoi anglicisme hahaha

[royal basin]

fun

[thorny canyon]

yaaa okkk

[royal basin]

bon je ne suis pas l'académie

[thorny canyon]

je pense que cette année je vais redoubler encore mon année..

[alpine raven]

we can say "fun" in french too 😭

[thorny canyon]

hier j'ai vider toutes les larmes de mon corps

[royal basin]

[alpine raven]

merde...

Courage

[thorny canyon]

mercii.. jai rater un partiel hier

vous penser c'est rattrapable ?

[alpine raven]

aïe

Ca dépend du prof je pense

[thorny canyon]

je veux compenser enfaite

[alpine raven]

C'est en controle continu ?

[thorny canyon]

en gros au premier partiel javais eu 17 dans cette matiere

et maintenant jaurai max 5

[alpine raven]

Ça va, essaie de te débrouiller pour au moins 14

[thorny canyon]

ensuite jai une note de TD de 14

jespere purée

[alpine raven]

si tu as 14 tu es clean

[thorny canyon]

tu me rassures vrm

bon je vous laisse je vais aller bosser, si jmais j'ai uen question j'aimerai bien que un de vous me répond svp 🥰

[alpine raven]

si vous avez encore une note après ça, faudra juste essayer de garder la moyenne autour de 12

[thorny canyon]

non ya plus de note

[alpine raven]

Bon courage, dis toi que actu la thermodynamique veut ma peau, je peux doubler aussi

[thorny canyon]

bah justement c'etait en thermo hahaha

mais en soi c grv simple

[alpine raven]

Fck thermodynamics

[thorny canyon]

c juste pendant le partiel je sais pas ce qui m'as pris

c'est quoi que tu ne comprend pas ?

c'est sur quoi ?

[alpine raven]

c'était au premier semestre, c'est juste les bails de premier principe etc.

Mais c'est de la thermodynamique pour ingenieur

[thorny canyon]

c les changement de phase et tout ?

[alpine raven]

donc c'est pas les mêmes réalités comme en prépa

changement de phase yep

[thorny canyon]

liquefaction et vaprorisation

je vais pas te mentir je comprend pas trop non plus a partir de la

[alpine raven]

On utilise tellement d'abaques et de table thermodynamique

[thorny canyon]

oui d'un coup c nCliq d'un coup c la chaleur latente

d'un coup c mvap jppp

[alpine raven]

Je hais cette matière

je veux pas faire carrière en énergétique

[thorny canyon]

purée courage a nous !

[alpine raven]

je m'en sors qu'en électricité et mécanique mdr

go faire de la robotique

[thorny canyon]

oula la meca loin de moi

[alpine raven]

Courage à nous effectivement 😭

[alpine raven]

c'est fun pourtant mdr

[thorny canyon]

kiss ❤️

[thorny canyon]

non merci garde le pour toi

[alpine raven]

allez, j'ai du SQL a faire

cya

ferme le channel

[obtuse pebble]

@thorny canyon Has your question been resolved?

[obtuse pebble]

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[full tusk]

3 π/4 < x < π
__ _ _ __ __ __
And √1+2.sinx.cosx

What is the simplest form of the expression?

[full tusk]

|sinx + cosx| coming from it I know but

On unit circle it’s on 2nd Area so

Cos is (-) and Sin is (+)

It shouldn’t be sinx > cosx here?

@tight thunder

[tight thunder]

Oh hi

[full tusk]

Hey 👋

[tight thunder]

I don't get what ur asking

[full tusk]

It is asking simple form

So YouTuber teacher saying cosx > sinx after 135 degree on unit circle

But my knowledge saying that sinx > cosx always on 2nd area

And he saying on 2nd area of unit circle after 135 degree, if the value of degree gonna higher, cosx value gonna higher but hell nah it should be gonna lower

For this question he saying sinx is + and cosx is -

But

[tight thunder]

he means |cos x| > |sin x|

because cos is horizontal

and sin is vertical

[full tusk]

But here different

[tight thunder]

because that is less than 135

so |sin x| > |cos x|

[full tusk]

What’s the point of 135?

[tight thunder]

135 is where |sin x| = |cos x|

[full tusk]

I know

So I only remember that 0<x<90 , 90<x<180 , 180<x<270 and 270<x<360

Only I remember that this points changing that sin and cos is positive or negative

So asked why 135

[tight thunder]

u don't rly have to remember if u can draw the circle

positive or negative is different from which is bigger

[full tusk]

Can you explain a bit about 135 degree stuffs?

Really I didn’t get it still

After 135 degree till 270 cosx always negative right?

[tight thunder]

sorry

I meant this

[tight thunder]

cos is negative 90 to 270

[full tusk]

And 90 to 180 when the degree increasing

Cos value is decreasing

Right?

[tight thunder]

yes

cos is decreasing but |cos| is increasing

[full tusk]

So why the hell this YouTube man saying that after 135 cos increasing when degree increasing?

Not talking about |cosx|

Only he said cosx

And tried to get him

[tight thunder]

he's not being very specific ig

[full tusk]

So on this type questions only I should care |cosx| > or < |sinx|

Right?

[tight thunder]

I suppose yes

basically you have to think if sin x + cos x is positive or negative

[full tusk]

So Idc 135 degree stuffs then hahaha

[tight thunder]

no u need to

[full tusk]

Bruh

[tight thunder]

because you need to see if |cos x| > |sin x|

[full tusk]

Idc cause I can draw circle and x-y coordinate system and can check values of sin and cos

[tight thunder]

yes

what im saying is

you don't have to memorise it

but just draw the circle

[full tusk]

So like that can see |sinx| and |cosx| and so I don’t need to care 135 degree

[tight thunder]

ye

[full tusk]

I hate YouTubers

Hahahaha

Ty bro btw

[tight thunder]

Yw

[obtuse pebble]

@full tusk Has your question been resolved?

[obtuse pebble]

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