help-27

[lost laurel]

(u,v(

[faint gorge]

ok sure

[lost laurel]

so I have , uh

Okay, yeah

makes sense

thanks

I can do it from here

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[cosmic yacht]

Determine all complex numbers (picture).
Provide the solutions in the form (picture).

[devout snow]

Please don't occupy multiple help channels.

[cosmic yacht]

yes closed the other one

I dont know how to start. I dont know anything.

[faint gorge]

Du kannst 1/i mit i/i multiplizieren

und dann es in Exponentialform bringen

[cosmic yacht]

wieso kann ich das? :3

[faint gorge]

Weil i/i = 1 ist

Du veränderst nicht den Wert sondern nur die Form

Dann hast du z³ = -i

Jetzt versuch -i in Exponentialform zu bringen

[cosmic yacht]

mhmm bin gerade bei sqrt(1i/-1)

[faint gorge]

wie

hä

[faint gorge]

ja

z³ = -i

Im Prinzip kannst du auch eine Lösung erraten
i² = -1
i³ = -i
Also ist z = i eine Lösung

[cosmic yacht]

achso die 1 hat mich verwirrt vor dem i... da steht ja i / -1, was dann -i ist logisch

[faint gorge]

ja

[cosmic yacht]

denkfehler

okay dann probiere ich jetzt mit -i weiterzumachen und der expotentialform

[faint gorge]

Du kannst auch die Lösung dir graphisch ausdenken

[cosmic yacht]

okay die Exponentialform lautet so:

r scheint der betrag der komplexen zahl zu sein

[faint gorge]

ja

Aber den hast du geschenkt, was ist den der Betrag von z = -i?

[cosmic yacht]

-sqrt(-1)

[faint gorge]

|z| = sqrt(x²+y²)

sqrt(0²+(-1)²)

[faint gorge]

ja

[cosmic yacht]

Achso ja Z = 1

Ohne Wurzel

[faint gorge]

betrag z = 1

der winkel ist offensichtlich bei -i

[cosmic yacht]

Hab’s so jetzt aufgeschrieben

[faint gorge]

arbeite lieber mit exp form

[woven radish]

anti-algebraist 𝔸dωn𝓲²s

[cosmic yacht]

ich verstehe leider absolut nichts gerade

☠️

[faint gorge]

weißt du wo -i ist graphisch

der grund warum du mit exp arbeitest ist weil du leicht potenzgesetze anwenden kannst

Alle Lösungen von z^n = w sind gleichmäßig verteilt, wie als ob du gleichgroße Kuchenstücke schneidest

Das erste Stück liegt bei pi/2 das nächste bei +2pi/3 und das nächste bei nochmal +2pi/3

[cosmic yacht]

-i ist unterhalb der x achse

Hab’s jz so weit geschafft

[faint gorge]

ja genau

jetzt machst du auf beiden Seiten hoch 3

[faint gorge]

Schau dir das nochmal an

[cosmic yacht]

Okay, aber bei mir ist das I noch da

[faint gorge]

das i muss da sein

[woven radish]

anti-algebraist 𝔸dωn𝓲²s

[faint gorge]

du hast außerdem das minus vergessen

[woven radish]

anti-algebraist 𝔸dωn𝓲²s

[faint gorge]

-3pi/2 ist kongruent zu pi/2, da sind die gleichen winkel sozusagen wenn du 2pi addierst

sind beides auf der oberen y-achse

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[hasty flame]

I'm fairly sure everything is correct, but iffy on t

can the exponential growth function be "complete" if t is undefined?

[supple knot]

why do you think t is undefined

[hasty flame]

well its not actually "undefined"

t is years

but since I'm not given an amount of years to calculate for

[supple knot]

t is a variable so your answer should be in terms of t

[tropic skiff]

(better to use unknown in the future)

[slender mirage]

You're supposed to write Y(t)

[hasty flame]

oh shoot yes

then when I want to set the value of t I just plug that in?

or should I make one up now for the sake of finishing the function

[winter patrol]

wdym for the sake of finishing?
you're already done

no need to make up values for t

you could simplify the 1+0.12 to 1.12 though

[hasty flame]

I was wrong, I thought that the function couldn't be considered complete without knowing every variable other than Y

obviously that is "impossible" with the given info since it's a general formula not for a specific number of years

whoops

[winter patrol]

you encounter functions with unknowns all the time

[hasty flame]

yeah, I've only done problems like these with known values for all the variables before, so I guess that's why I was confused. Not exactly a smart mistake

thanks everyone

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[hasty flame]

.reopen

[devout snow]

✅

[hasty flame]

.close

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[green vortex]

hi, im lacking lots of foundational maths which makes a lot of things harder for me, could someone people help me understand where the 2 infront of the C comes from in the rearrange step, i understand divide both sides by 1/2 to get rid of it on the left and turn 1/3 into 2/3 on the right but how come there is a 2 in front of the C.

thhank you!!

[supple knot]

they multiplied by 2 on both sides in the top equation

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[scarlet falcon]

hi!!! could someone just do for me I'm incredibly bad at graphs

[acoustic leaf]

what is the question?

[iron kindle]

!noans

[devout snow]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[iron kindle]

!original

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[scarlet falcon]

Find the mean of all months ; and find the number sold in april

[scarlet falcon]

I tried doing it the normal way and I got 92 for April and 94 all together

[iron kindle]

that sounds about right

[scarlet falcon]

Thanks !!

[iron kindle]

wait actually

i read it wrong

april should be 94

[scarlet falcon]

.. oh 😭 okay

Is the mean correct ?

[iron kindle]

i dont wanna check all of those values tbh...

reading off this graph is annoying

[scarlet falcon]

That was my problem too 😭 I kept getting confused

[shell scarab]

If that's the issue you're not bad at reading graphs, the graph is just poorly made

[scarlet falcon]

😭😭

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[wet robin]

i cant get this question right i simply dont get what im doing wrong

however this is what im doing

i found the equation of the line

which is

-0.5x+8

i integrate between 16 and 2 of the expression -0.5x+8-(32/x^2+3x-8)

pls anyone

<@&286206848099549185>

[lavish raft]

why 16

pj

oh nevermind

oaky

i think you need to find the intersection of C and l

and integrate c from 2 to the point of intersection

and then integrate l from the point of intersection to 16

[wet robin]

ty

.solved

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[formal venture]

sending my work

i skipped a bunch of steps

[thin inlet]

you may need to put your exact answer

[formal venture]

oh that 14.... is the exact answer

[thin inlet]

,calc 525/37

[woven radish]

Result:

14.189189189189

[faint gorge]

like fractional

[formal venture]

oh wait

[thin inlet]

its a repeating decimal

so you'll need an exact answer

[formal venture]

not on my calculaotr it seems

but hold on

thank you

it works

i was going INSANE

[thin inlet]

👍

[formal venture]

.close

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[obsidian kindle]

hi i need help

[devout snow]

Please don't occupy multiple help channels.

[supple knot]

just wait in #help-49

[obsidian kindle]

whoops

.close

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[junior cradle]

how do i do 64 and 66?

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[hasty flame]

how do I find csc20 and it's cofunction?

[hasty flame]

no other info is given btw- no triangle

[wise bronze]

.open

how do i find a video on Finding the constant term of a linear equation.

[hasty flame]

if you are looking to open a channel for you question you just have to go over to the (available) section and type it out, there's no open command

[sacred hound]

you can start from sin(20). Hint: 60 = 3*20

I assume you might be familar with double angle formula and sine of sum of angle formula

[hasty flame]

slightly

[hasty flame]

the cofunction of csc would be

sec?

yes

[sacred hound]

yes but that is easy

[hasty flame]

do we not want easy?

[sacred hound]

what is hard is to find csc(20), agree?

because once you know that value, you can essentially derive everything out

[hasty flame]

agree

[sacred hound]

so you essentially need some kind of sin(3x) formula. which, if you know you can just plug in. if you don't then that would be fun

Hint: sin(3x) = sin(2x+x)

because I also assume you know what is sin(60)

[hasty flame]

I feel like we are speaking different languages

I should preface- complete trig noob

[sacred hound]

do you need to solve this, because this is not beginner trigonometry question

[hasty flame]

need- no, would like to- yes

but if i need to- i will give up for now and try to start with something easier

[sacred hound]

I can start step by step then:

1. you need csc(20), this csc(20) is essentially 1/sin(20) [csc(x) = 1/sin(x)]
2. we know sin(60) = sqrt(3)/2
3. then if we know sin(3x) formula, we might be able to solve for sin(x) [3x = 60, x = 20]

To get sin(3x)

1. Recall sin(3x) = sin(2x + x)
2. Also recall sin(a+b) = sin(a)cos(b) + cos(a)sin(b), then this means sin(2x+x) = sin(2x)cos(x)+cos(2x)+sin(x)
3. Recall sin(2x) = 2sin(x)cos(x), cos(2x) = cos^2(x) - sin^2(x), substitute that in and you should have sin(3x) formula.
4. Get everything in terms of sin(x)
5. Solve for sin(x)

you see why it is not beginner?

[hasty flame]

unfortunately

if csc 20 could be punched in to a calculator would that make things easier? just focusing on the cofunction part?

[sacred hound]

yes

way easier

because cofunction implies that say co f and f is just shifted slightly by some phase.

if you were to look at the graph, it is just the same, but it is like cofunction is slide a bit like this

red one is a bit to the right, but it is essentially the same

hence, if we know what csc(20) is, we can easily just add some phase (here is 90 degree). That is: csc(20) = sec(90-20)

[hasty flame]

ok this i can understand

"add some phase"

what do you mean by that?

[sacred hound]

say trigonometry function is a periodic function, do you agree?

if you can agree that trigonometric function is a periodic function, then I can say csc(x) is the same as sec(x), but it is moved by 90 degree. This is what I meant by adding phase. Essentially they are the same but their period is different by 90 degree

[hasty flame]

why 90 degrees?

will that be the case for every cofunction conversion?

[sacred hound]

yes

but I need to be careful because they are not exactly the same. they are essentially the same, but not exactly

like, this is reciprocal of sine and cosine, we can't divide by zero

also, I need to be careful because it is true in general for trigonometry function kind of cofunction

and why 90? that is because unit circle and sin^2(x)+cos^2(x) = 1

in unit circle, we can denote x as cos(t) and y as sin(t)

[hasty flame]

ohhhh

I think I get it

so-

sin47 for example

the cofunction would be

cos43

[sacred hound]

yes

[hasty flame]

sine / cosine and 43+47=90

ez ez

[hasty flame]

i'll figure out this stuff later I guess

[sacred hound]

yeah trigonometry is quite complex so take it easy

[hasty flame]

thanks for the help

.clos3

.close

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[restive river]

Help

[restive river]

I can do 18-20 there was a backpage i did but i forgot how to do the triangle (10-17) since i fall asleep alot in class

.close

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[restive river]

.close

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[devout snow]

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[chilly magnet]

Hi Could someone help me with this idk if I’m going on the right track

[wicked turtle]

look ok so far

[faint zinc]

You might want to leave this as sqrt(2) though

Easier to work with

[chilly magnet]

Wouldn’t that be wrong tho

Oh u mean for the increase?

Instead of 0.83

[wicked turtle]

yea

[chilly magnet]

Alr bet ty

[wicked turtle]

the answer will work out nicely if you leave it as sqrt(2)

[chilly magnet]

Ya that makes sense

Ty!

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[kindred juniper]

How do I find the diameter of the pool if building it costs $4240?

[kindred juniper]

I know that every time you go up a segment, price increases by $200 and its length goes up by 1 meter

[vital hinge]

is this for a circular pool?

[kindred juniper]

Doesn’t specify in the problem

I can send it but it’s in French

[vital hinge]

no i forgot all my french lol havent used it since grade 10

[kindred juniper]

Lmao

[vital hinge]

i am not sure how you'd do this if it wasnt circular (or square i suppose)

let me think about this for a second

[kindred juniper]

✅

[vital hinge]

what do you mean by up a segment?

you're referring to the graph correct

[kindred juniper]

Yes

Like every time

It goes up by 200$

The segment’s length goes up by 1m

I don’t need the exact diameter I THINK it’s relevant to the second part of the question tho

If you can understand

First page

[vital hinge]

i find it weird that the price is not a multiple of 200 despite the price increasing by 200 every time

[kindred juniper]

Yeahh

[vital hinge]

after second thought I do not think I am certain enough about how to solve this problem to be able to properly help you

best of luck

[kindred juniper]

It’s all good

Ty

[kindred juniper]

Omg.

I’m such an idiot

I’m an actual idiot

LMAOO

IT’S 400$ FOR EXCAVATING RHE GROUND

AND LIKE

I DIDNT READ THE TITLE PROPERLY

[strange arch]

My french scope doesn't go beyond Je parle francais juste un peu but doesn't the task say she has 400 for excavation and 3840 for the pool

And with 400 excavation cost she can get a pool with diameter between 4m and 6m?

[kindred juniper]

YES IM

IM AN IDIOT

[strange arch]

and for 3840 you just need to evaluate f(x) twice for the enterprises

[kindred juniper]

OK ILL TRY IT RN

IM ACTUALLY DUMB

MB

[strange arch]

no need to apologize

[vital hinge]

i am very happy for you both

[pure bay]

Je ne sais pas comment aider

Mon français n'est pas meilleur non plus mdr

[kindred juniper]

So for entreprise 2

If x = [4, 6[

Do I have to use

Both of these

@vital hinge What do you think

[strange arch]

Ah yeah

[kindred juniper]

So

[strange arch]

yip yip

[kindred juniper]

150x + 1100
150(4) + 1100 =1,700

195x + 2000
195(6) + 2000=3,170

Does that seem right…

[strange arch]

yes

[kindred juniper]

f(x) = [1700, 3170[

?

Ait

Wait

Nvm

[strange arch]

it's not a single interval since f(x) jumps from 1700 to a higher value right after x=4

But just to be clear, the question asks for the greatest diameter given that she spends all of it?

[kindred juniper]

Uhhh

Doesn’t mention it unless I’m blind

[strange arch]

Bc I have no idea what the texts state

"laquelle des deux enterprises Dimitri etc Anna devraient ils choisir?"

[kindred juniper]

It says they have to excavate and compact the ground, price depends on the diameter

[kindred juniper]

Yes

[strange arch]

which enterprise should they choose I suppose

[kindred juniper]

Yes

Which one costs less to install the pool basically

They wanna spend $400 for excavating and compacting the ground

Same as their friends

[strange arch]

yea and 3840 for the installation

But do they have to spend 3840

[kindred juniper]

No

[strange arch]

like is the goal to get the largest pool

[kindred juniper]

I don’t think so

Doesn’t specify :/

[strange arch]

It's unclear to me how they pick an enterprise

[kindred juniper]

😭

[strange arch]

it must say somewhere what the goal is

largest pool under given budget?

lowest cost for given size?

[kindred juniper]

Doesn’t say idk 😭 😭 😭

Ok but entreprise 1

I know f(4) = 2300

And then I have to find the equation right

[strange arch]

"Dimitri and Anna want to choose a company that will allow them to pay the same total amount as their friends for the pool" do I get that right

if so then that'd be the goal

then they have to pay 4240

[kindred juniper]

No they want to pay the same price for the excavation as their friend

$400

And then they wanna find the best offer

Economically

For the installation of the pool

WHICH IS WEIRD

[strange arch]

Ok and they can spend 3840?

And they want to get the largest pool for 3840?

[kindred juniper]

Doesn’t say anything about how much money they want to spend after

They just want the best offer for the diameter given

Ahhh

I can’t do this

What if the teacher made a mistake and I’ve been

Working my brain for no reason

[strange arch]

"Dimitri and Anna do not know which offer is best for their pool installation, but they absolutely want to pay the same price as their friends for excavation and soil compaction"

is the translation of the last segment

[kindred juniper]

Yes

[strange arch]

Which doesn't state that they want to pay 400 for excavation

they just want to pay exactly 4240 in total

and now you need to find which enterprise provides the larger pool

if they spend exactly 4240

so they could for instance spend 600 on excavation and 3640 on the pool

if that leads to the largest pool for some enterprise

[kindred juniper]

Wdym

[strange arch]

They want to pay the same total price

[kindred juniper]

Their friends spent 400 for excavating and compacting

[kindred juniper]

Wait what

[strange arch]

ah nvm soil compaction is part of the excavation

[kindred juniper]

Yeahh

[strange arch]

Ok so they have to pay 400 for exc and seek the best pool for 3840

[kindred juniper]

But then

If they’re paying 3840

THE DIAMETER ISNT GONNA BE [4, 6[

😭 😭

Which is what it should be if theyre paying 400 for excavating and

Compacting

[strange arch]

hm yeah I don't see where it's mentioned how much they want to spend on the pool itself

[kindred juniper]

Yeahh

Maybe the teacher did smth wrong man…

If the second function was [4, 7]

Then it’d be easy

But like

I guess not

[strange arch]

Ok well we know that their pool size is in [4,6[

so I guess you can then compare within that interval for which prices which enterprise is better

[kindred juniper]

Yeahh

If x = 4 then enterprise 2 is better

But if 4 < x < 6

Entreprise 1 is better??

[strange arch]

Yeah

Sounds correct

[kindred juniper]

Alrightt

[strange arch]

I suppose that is the answer

[kindred juniper]

How would I compare intervals tho

Like

[strange arch]

Still sounds odd though

[kindred juniper]

Yea

[kindred juniper]

Caus

[strange arch]

You just need to show f(x) > g(x) for x e ]4,6[ for the functions f,g for the enterprises 2 and 1

Enterprise 2 has 195x+2000 in that interval

and Enterprise 1 has 350x+900 in that interval

[kindred juniper]

So 2300 > 1700 for x = 4

[strange arch]

Not for specific values, but for all values in the interval ]4,6[

[kindred juniper]

How did you get that 😭

I’m confused

[kindred juniper]

Oh ok

[strange arch]

Since you want to show it's better for any size they pick within the interval

[kindred juniper]

Wait if I tell you all the values could you show me

Cause I wanna make sure

I do it right

[strange arch]

Enterprise 1 has a slope from (4,2300) to (7,3700)

difference in Y is 1400 for a difference of X of 3

[kindred juniper]

1400/3 =466.667

[strange arch]

so the slope is 1400/3 = 466. Yeah

Oops I wrote 350

ys

[kindred juniper]

2300 = 466.667(4) + b

?

Then u find the equation

Or what

[strange arch]

yes

[kindred juniper]

Okk

And then witht hat u get like

b = 433.33

[strange arch]

yup exactly

now you can compare them

[kindred juniper]

So for f(6)

U have f(x) = 466.667(6) + 433.33

f(6) =3233.333

I think

[strange arch]

yes

[kindred juniper]

So now for entreprise 2

f(4) =1700

How would I write the interval

For x = [4,6[

[strange arch]

wdym

[kindred juniper]

f(x) = ]2780, 3170[ for x = ]4,6[

?

Wait

No

[strange arch]

since you used [4,6[ it'd be [2780, 3170[ otherwise yes

[kindred juniper]

Yeahh

[strange arch]

yes now correct

[kindred juniper]

I changed it

[strange arch]

yop

ok so we can see that the performance differs

[kindred juniper]

Could you show me how to write this

😭

[strange arch]

sure a sec

[kindred juniper]

Ty

[strange arch]

k so starting with the functions

we know that Enterprise 1 is better at first, but then Enterprise 2 becomes better

in the interval ]4,6[

[kindred juniper]

Yup

[strange arch]

since f(x) and g(x) are linear functions

we can look at where they meet

[kindred juniper]

Holy moly we have NEVER done this yet

[strange arch]

and we know they must meet in ]4,6[ since it changes which enterprise is better within the interval

so let's look at f(x) = g(x)

because that's exactly where they meet

[kindred juniper]

195x + 2000 = 1400/3x + 1300/3?

x = 5.766

[strange arch]

5.76687116564 yoppa

ok neat

now we know where they meet

so we have for x=4, Enterprise 2 is better

for x e ]4,5.766...[, Enterprise 1 is better

for x = 5.766..., both are the same

for x e ]5.766...,6[, Enterprise 2 is better

[kindred juniper]

Bro this is for a friend and I know she’s not gonna understand this 😭 😭

[kindred juniper]

I’ll still

Explain

[strange arch]

Is this the desired answer? I don't know 😄

[strange arch]

kk

[kindred juniper]

PORBABLY NOT

SHES IN LIKE REGULAR MATHS AND IM IN STRONG MATHS AND I HAVENT SEEN THIS AT ALL

[strange arch]

probably better to ask for clarification

yeah just ask for clarification

[kindred juniper]

Alrightt

Ty for the help

[strange arch]

np!

[kindred juniper]

How do I close

.close

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[worthy heath]

find $S_n = \frac{1}{4}+\frac{1.3}{4.6}+\frac{1.3.5}{4.6.8}... n$ terms

[woven radish]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[worthy heath]

i wrote the general term as $T_n= \prod_{r=1}^{n}\frac{2r-1}{2r+2}$

[woven radish]

∮Ē.dĀ = Qₑₙ꜀/ε₀

[worthy heath]

but i cant figure what to do next

apparently i should get the general term as a difference of 2 things which make it telescoping

but

how

my first thought was i can take 2 common from the denominaor but it didnt really help

[slender mirage]

Write (S_n)/2 for starters, and multiply numerator with (2n + 2) - (2n + 1)

Should get the telescope you're looking for

[worthy heath]

for that wouldnt n have to be even??

[slender mirage]

(S_n)/2

[worthy heath]

oh

yeah i got this

now i take the difference of this and the original expression?

[slender mirage]

Huh? no

Did you do what I said?

[worthy heath]

yes i got this

this

sry

[slender mirage]

General term here is: $$T_n = \frac{1\cdot 3 \cdots (2n - 1)[(2n + 2) - (2n + 1)]}{2 \cdots (2n) \cdot (2n + 2)}$$

[worthy heath]

shouldnt the 3rd term have 2.4.6.8 ?

[slender mirage]

Is this fine now?

General term in (S_n)/2 ?

[worthy heath]

wait

[worthy heath]

hold on

im tryna figure why u did the 2n+2 -(2n+1)

ik it works but

[slender mirage]

Cause you want a telescope

[worthy heath]

hm

okay i kinda get it

[slender mirage]

$T_n = \frac{1\cdots (2n - 1)}{2\cdot 4 \cdots (2n)} - \frac{1\cdots (2n + 1)}{2\cdots (2n + 2)}$

[worthy heath]

so now i split the fraction?

yeah

[worthy heath]

i got the same

now its telescoping ig

yeah the 1st and the last term remain if im not wrong

[slender mirage]

now, plugging ns in this for (S_n)/2 = T_1 + ... + T_n, should give you the telescope

[worthy heath]

got it

thanks

have a good day

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[slender mirage]

.@worthy heath, Just a quick mention, the given sum is $$\sum_{i=1}^n \binom{2i}{i} \frac{1}{4^i (i+1)!}$$

[woven radish]

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

[worthy heath]

got it thanks

[worthy heath]

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[worthy heath]

??

[slender mirage]

Well, the point was, it could've also been computed without telescoping

[worthy heath]

oh

[slender mirage]

If you're up for differentiation that is

[worthy heath]

i generally prefer not using integration and differentiation for sums

[slender mirage]

Got you

[worthy heath]

also the guy on yt said to do it by telescoping

thank you, imma try this again later

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[unborn quail]

For question 73 I am trying to figure out what I did wrong

[unborn quail]

this is what i got

sorry it’s sideways 😟

this is what the textbook has as the answer

oh wait

nvm i just found it 🫠

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[mighty pendant]

Isn't this more correct?

[mighty pendant]

d/dx f(g(x)) = d/dx f(g(x)) d/dx g(x)

this is a circular definition

[mighty pendant]

im just not sure if this is the correct syntax

[acoustic leaf]

usually for functions we use d(g(x)) just to make extra clear what the derivative is doing

although honestly it's probably easier to write the right side using the lagrange notation

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[arctic field]

<@&268886789983436800> 998482640476852224

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[shadow dirge]

can someone explain how this was concluded?

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<@&286206848099549185>

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[shadow dirge]

can someone explain how this was concluded

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[sand dove]

u2,...,uk are all in LS({u1,...,uk}) and LS(w_1,u2,...,uk)}

so all we really need to show is u1 is in LS(w_1,u2,...,uk)} and w_1 is in LS({u1,...,uk})

that way we will have double inclusion

well

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[sand dove]

w1 is in LS({u1,...,uk})

[sand dove]

and u1 is in LS({w_1,u2,...,uk})

so done

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[shadow dirge]

oh right, got it

thank you

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[woven radish]

938c2cc0dcc05f2b68c4287040cfcf71

[spring oasis]

(v+w) = (1,0,0)+3(-1,1,0)-(-1,-1,1)
(v+w)=(1,0,0)+(-3,3,0)+(1,1,-1)

(v+w) = (1-3+1,3+1,-1)

[slender mirage]

Lin Alg is my woe

[spring oasis]

is just definition of coordinates wrt to a basis, means that the coefficients of the linear combination of the basis vectors from B are (1,3,-1)

(v+w)=(2-3,4,-1)

(v+w)=(-1,4,-1)

[radiant hare]

Ig just transform the equations to the standard basis

[spring oasis]

(v-w) = -2(1,0,0) + 3(-1,1,0) + 5(-1,-1,1)
(v-w) = (-2,0,0) + (-3,3,0) + (-5,-5,5)

[slender mirage]

So basically, (v, w) = [(-5.5, 1, 2), (4.5, 3, -3)]

[spring oasis]

yeah and now sum them I think

we get 2v

2v = (-1,4,-1) + (-10,-2,5)

v = (-11/2,1,2)

wait, we get a fraction in first coordinate of v?

[slender mirage]

a(-5.5, 1, 2) + b(4.5, 3, -3) intersection S is required

[radiant hare]

Is <v,w> the span

?

[spring oasis]

yeah write as a generic point and make it satisfy equation of S

[spring oasis]

yes

[radiant hare]

Ah ok

[slender mirage]

-2.5a + 4.5b = 0

5a = 9b

a = 9k, b = 5k, should give intersection

[spring oasis]

(-5.5a + 4.5b, a + 3b, 2a -3b) = (x1,x2,x3)

(-5.5(9k) + 4.5(5k), 9k + 3(5k), 2(9k)-3(5k)) = (-11(9k) + 9(5k), 18k + 30k, 4(9k)-30k) = k(-99 + 45, 18 + 30, 36-30)

k(-54, 48, 6) = k(-27, 24, 3)

k(-9, 8,1)

S ∩ <v, w> = <(-9,8,1)>

anyways ty

.solved

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[frigid igloo]

I need a little bit of help with this- basically I have to prove this with implications and I’m not sure if I’m doing it right or more likely I’m quite lost

[frigid igloo]

<@&286206848099549185>

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[unkempt crow]

The translation is: Find the polynomial with real coefficients in two variables such that its zero locus is this. I didn't even know hot to tackle such a problem, here is what i tried to do: first i found out that P(0,0) Is different than 0, therefore there is a constant at the end. Then i figured out that you can change x with -x and the graph will remain the same, therefore p(x,0) can only have x raised to even powers.. same goes for y. I'm not really finding any way to continue. I can't even find their degree because considering for example P(x,1) There seem to be 6 solutions but there could be more just imaginary. This is an admission problem for a university

[pastel pasture]

Is this the actual test to admission or is it a mock

[unkempt crow]

admission to the normale (pisa) one of the most prestigious universitiez

technically not a university but still

[pastel pasture]

I meant is it the actual test that will give you the admission

[winter torrent]

the main question is are you allowed to get our help on this

[pastel pasture]

Because we aren't allowed to help with a graded assignment

[feral agate]

i doubt you'd be able to access the internet during such a test

[unkempt crow]

no it isn't

2023

sorry didn't understand the question

[narrow bloom]

if it's a previous year's exam that's fine

[unkempt crow]

it is

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[unkempt crow]

is anyone trying to solve it?

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[narrow bloom]

sorry I was trying to solve it with other helpers, and someone got to something but idk how to explain it properly

the idea is, we'll want p(x,y) to have terms with y^2, (x^2-1) and (x^2-4) since p(x,0) and p(0,y) will have some specific roots

so one might try something like p(x,y)=y^2+(x^2-1)(x^2-4). But that doesn't work (here I'm not 100% sure how to explain why), however raising (x^2-1) to 2 does (again, not entirely sure why. something to do with local branches when you try to parametrize the curve p(x,y)=0)

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[runic prawn]

the sum can be negative

if you just bound the sum between 2 values, it doesn't imply that it converges

so ur saying $$ -\sum_n |a_n| \le \sum_n a_n \le \sum_n |a_n|$$ so $ \sum_n a_n$ converges?

[woven radish]

G-torsor

[tardy edge]

only if $\sum (a_n^+ - a_n^-) = \sum a_n$

[woven radish]

alex <3

[tardy edge]

where $a_n^+ = max(a_n,0)$ and $a_n^- = max(-a_n, 0)$

[woven radish]

alex <3

[runic prawn]

isn't a_n^+ - a_n^- just a_n then

wdym last bit

[tardy edge]

i meant if both of them converge

their difference would be $\sum a_n$

[woven radish]

alex <3

[tardy edge]

and it would also converge

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[polar ivy]

error on khan academy or something i did wrong?

[winter patrol]

you shouldn't be multiplying the powers

[polar ivy]

ohhh

ok,thanks.

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[mystic quail]

helppp

[mystic quail]

im not sure what i did wrong but i got close to answer 😭

[restive river]

hi

[mystic quail]

Hi

[devout snow]

@mystic quail Has your question been resolved?

[mystic quail]

<@&286206848099549185>

[mystic quail]

plsss 😢

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[mystic quail]

[delicate fossil]

in ur last step

$-\tan\left(180^\circ-\alpha\right)=-\left(-\frac{3}{4}\right)=\frac{3}{4}$

[woven radish]

@delicate fossil

[delicate fossil]

granted i didnt check the working in the first pic

[mystic quail]

omg yeah

makes sense now tysmmm

i looked over so many times i didn’t even see that

[delicate fossil]

lmao happens

[mystic quail]

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[still leaf]

Hello, I'm trying to learn about linear manifolds for my college exam and I need help with some exercises.

The exercises I need help with are something like:
Find the parametric and nonparametric equations of the manifold W with the lowest possible dimension in R^5 that contains the vectors x=(1, 0, 1, 1) and y=(2, 1, 0, 1) .

1. What is the dimension of the manifold W?
2. Write some vector different from x and y that belongs to the manifold W. Justify why so.
3. Write some two different vectors that do not belong to the manifold W . Give a reason why this is so.

Or questions like: Write the parametric and non parametric equations of linear manifold in Z^4 modulus 5 that contains the vectors x=(0,1,1,1) and y=(1,0,0,1)
Write the parametric equation of this manifold so it has the highest dimension possible and contains the vectors x and y.
Write the non parametric equation of this manifold so it has the lowest dimension possible and contains the vectors x and y.

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[zealous vault]

Yo I'm studying vectors and I have a little trouble when doing vectorial calculus to see the result bc I am sometimes kind of loft of which vector should I use, which point should I add, etc

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[upper fossil]

Is it true that ¬P∨(¬Q∨R) ≡ (¬P∨¬Q∨R) by associativity ?

[frozen aurora]

yes

[upper fossil]

we can just remove the brackets like that?

[frozen aurora]

yes

[upper fossil]

ty

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[quasi reef]

if a question ask to use the factor theorem to show that (x+2) is a factor of x^3+5x^2-2x-24, how can i solve it?

[quasi reef]

is it by x=-2 then fill it if its 0 then its a factor

[stone stump]

yes

[quasi reef]

or like divide it and if there is no remainder then it is a factor

[stone stump]

that also works but is more effort

and not quite the factor theorem statement

[quasi reef]

ahh kk thank youu

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[quasi reef]

the question is to find a

[quasi reef]

im confused on the way to do it

[distant helm]

the simplest way is just to do what they tell you to do

• perform polynomial long division for both case

[slender mirage]

f(a) = f(-2a)

[quasi reef]

ohh okk

[slender mirage]

Use remainder theorem

6a² + a + 7 = 24a² - 2a + 7

And it's given a ≠ 0, so the other value should be ur ans

[quasi reef]

i tried this onee but im stuck in this part

[slender mirage]

HuH? What's in there to get stuck on?

It's a quadratic equation.. solve it

[quasi reef]

ohh right

tyy

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[native stone]

so i got this exercise

[native stone]

and i did this

thought ,, oh, c) is correct,,

its not

and i checked for m=1 and i see how its not correct

but then how do i solve it

this shouldnt be a hard exercise

[devout snow]

@native stone Has your question been resolved?

[slender mirage]

So m = 1 gives no solution right?

What about m = 3?

@native stone

[native stone]

yo

[native stone]

prolly no solutions either

[slender mirage]

Check and tell me

[native stone]

yea

same thing

-1/2

[slender mirage]

Cool ._.

Now do you know what a parabola like like?

[native stone]

ya

[slender mirage]

Draw a parabola with it's mouth upwards and two x intercepts both on opposite sides of 0

[native stone]

how do you know its concavity

wasnt that determined by the second derivative?

[slender mirage]

... Wow, is this a "I won't do it unless I know why I'm doing it"?

[native stone]

ill do it

sure

ill trust u

[slender mirage]

.

[native stone]

ight so, what does mouth upwards mean?

concave or convex?

[slender mirage]

Concave up

[native stone]

right

oki

ight

[slender mirage]

Now, how many positive roots does y = 0 have?

[native stone]

hmmm

it should just be 1 right?

[slender mirage]

Good

[native stone]

the other should be negative

otherwise there multiple real roots, which we dont want

[slender mirage]

Now, draw another parabola, mouth downwards, two x intercepts both opposite side of 0

[native stone]

alright

done

[slender mirage]

How many positive roots y = 0 has?

[native stone]

1

[slender mirage]

Great you're learning

[native stone]

so ure saying that it doesnt matter whether m-2 is positive or negative?

[slender mirage]

So then, in both cases. Suppose I replace x with 2^x, how many values of x will satisfy y = 0?

[native stone]

1

[slender mirage]

Great :) Now you know how to solve the question

[native stone]

i already knew this tho

thats why i checked for both m=1 and m=3

[slender mirage]

So you checked (m - 2)f(0) < 0 ?

Is what you're saying

f(x) = (m - 2)x² + 2(2m - 3)x + 5m - 6

That should've given you your answer then

(5m - 6)(m - 2) < 0

=> m in (6/5, 2)

Mark option b) next. ^^"

[native stone]

hmm

why should it be smaller than 0

[native stone]

whats the logic here

[slender mirage]

Idk, you tell me

You "knew it" apparently

[native stone]

knew that a1 * a2 <0

so that the theres one positive root and one negative

[slender mirage]

(⁠ヘ⁠･⁠_⁠･⁠)⁠ヘ⁠┳⁠━⁠┳

[native stone]

so that theres only 1 real root

[slender mirage]

Alr, from start

When is f(x) concave up ?

[native stone]

wasnt it when m-2 >0?

idk for sure

[slender mirage]

Yeah

So, for (m - 2) > 0

What's the condition for f(x) to have zeroes on opposite side of x = 0

[native stone]

idk

[slender mirage]

You're on your own for this one

For a upward facing parabola, that is, given coefficient of x² is greater than 0

What is the necessary and sufficient condition for it to have two roots both opposite of x = 0

[native stone]

its that 2(2m-3) should be smaller than delta

oh

so after doing the calculations we get that a*c< 0

so exactly what you said

ive seen this with you, you dont prove anything anymore, you just have so much experience that you skip like 5 steps in your explinations

[slender mirage]

?

I believe what I'm trying to do here is guide you step by step. And I don't think I've left you half hanging on any solutions

But anyways

[native stone]

theres just some stuff that you'd consider easy bud

[slender mirage]

Here's a hint: Look at the graph of the parabola

[native stone]

i also got that 2(2m-3)(5m-6)<0

[slender mirage]

You got that? Can you explain how?

Cause that seems to be inaccurate

[native stone]

a1,2 = -b+- sqrt delta / 2a

2a > 0

[slender mirage]

Yeah

[native stone]

therefore b < sqrt delta so that we get 2 diff signed solutions

so i raised it to the second power

and got that 4ac<0

therefore ac<0

thats the steps i was saying u skipped

[slender mirage]

Smart, but that gives (m - 2)(5m - 6) < 0

[native stone]

yea

isnt that what we wanted?

[slender mirage]

.

[native stone]

oh

yea i typed that wrong

my bad

[native stone]

so ure saying that this doesnt help?

[slender mirage]

Anyways, instead of looking at the quadratic formula

What you should've done was looking at the graph I had you make before I left

One upward facing, one downwards facing parabola

Both have x - intercepts on opposite sides of x = 0

f(0) < 0 for a > 0
f(0) > 0 for a < 0

Is how I wrote (m - 2)f(0) < 0 is your desired equation

[slender mirage]

Since that is necessary and sufficient for f(x) = 0 to have only one positive root, this suggests replacing x with 2^x will also have only one real soln.

Hence you're done :D

[native stone]

wow

[native stone]

i completely forgot about these

dayum

thx dude

wait

doesnt f(0)=10m-14?

so then it would be (5m-7)(m-2)<0

[slender mirage]

I bet a goldfish has better attention span

[native stone]

jesus fucking christ

[slender mirage]

.

[native stone]

dont bother

[slender mirage]

You replaced 2^x with a and now you want f(a) = ... (your quadratic), to have exactly one zero positive

[native stone]

the first condition i put when doing the substitution

that 2^x = a

is that a>0

[slender mirage]

Your choice of variable is so wrong, you'll confuse yourself within two steps of your solution.

(m - 2)y² + 2(2m - 3)y + (5m - 6) = 0
Suppose f(y) = (m - 2)y² + 2(2m - 3)y + (5m - 6).
For given equation to have exactly one real solution, we must have exactly one positive zero of f(y).
Case 1) ∆ = 0, gives no solution.
Case 2) Two zeroes both opposite side of x = 0 :
=> coeff[y²] • f(0) < 0
=> (m - 2)(5m - 6) < 0
=> m \in (6/5, 2)```

Let me know if there's any step in there that I didn't walk you through, or if you want to revisit

[native stone]

y=0 would imply x = - infinity

[slender mirage]

And?

[native stone]

yea its fine

i get it

[slender mirage]

Since for every m in that interval , f(y) has a positive zero y_0, 2^x = y_0 is all I'm interested in

[native stone]

yup

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[copper dagger]

I got 48% of people do not read any of these newspapers by drawing a venn diagram

but the answer is apparently 52%?

I got that 52 was the union of I, II and III but I may have made a mistake but I used a venn diagram for this

This was my own working

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[copper dagger]

.reopen

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✅

[copper dagger]

.close

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[sour shard]

I need help

[rough copper]

with?

[sour shard]

I need to determine the domain and range for these answer but I don’t even know what the choices mean on the right

[rough copper]

just asking do you know what domain and range are?

[sour shard]

No

[rough copper]

ah ok

so simply putting in for the specific example, your domain will be the spread of values in which x can be defined

and your range will be the spread of values for where y can be defined

so for the figure in question 1

how high up and low down can the function go in terms of y

so there will be a minimum and maximum value for y and thats where the inequalities come in

does this make sense?

or do you want me to give some more detail?

[sour shard]

What are the inequalities

[rough copper]

these things -5<y<5

they represent what values y can be

in the case -5≤y≤5

y can only be on this interval

y cant be -6 or 6 for example

[sour shard]

what does - 7 < y < 7 mean

Does the greater sign have a different meaning in these subjects

Or is it the same

[rough copper]

so this symbol <. means strictly less than

≤, this you can say is less than or equal too

so for example

a < b

a is always less than b and cannot be equal to it

a ≤ b

a = b is a possibility too

so for your example

-7 ≤ y ≤ 7

y= -7 y =-6 .... y = 6, y = 7, are all possible values for y in this case