help-10

[timid silo]

i just joined this server like half an hour ago 💀

i like helping in active servers

[fading tree]

how the flip do i solve that

[fading tree]

LOL

i joined 2nd day of school last week

[timid silo]

you just need to solve any two of the three equations

[fading tree]

ohhh

[timid silo]

cause all of them are equal anyways

[fading tree]

and then the third one

or just two

[timid silo]

if you are to find UX

you do 3x+1 = 4x-6

and you get the value of x

but this time its a bit different

as UX is the whole line

you add all of these together

and you will get 9x + 3

[fading tree]

i got

x=7

[timid silo]

yes that is right

[fading tree]

for

oh ok

and then

[timid silo]

now you do this

[fading tree]

for UX

[timid silo]

you add all segments together

[fading tree]

where is 9x+3 from

[timid silo]

cause UX is the whole line

from adding all of these segments together

[fading tree]

couldnt i just add 7 3 times

[timid silo]

all those 3 segments form UX

no you cant

7 is the value of x

but the segment isnt x

[fading tree]

so 9 times 7

• 3

[timid silo]

they are 3x+1 4x-6 2x+8 respectively

[timid silo]

nono

you add the three segments

and you will get 9x+3

now substitute the value of x into 9x+3

you will be able to get the value for the length of UX

[fading tree]

would it be

9(7+3)

[timid silo]

no

9(7) + 3

cause the x+3 is not bracketed

[fading tree]

so 66?

[timid silo]

yes

[fading tree]

so UX=66

[timid silo]

yes you are right

congrats on completing (part of) your assignment

[fading tree]

thank u man

this server has saved me a lot

[timid silo]

dw

[fading tree]

i appreciate the help bro

[timid silo]

my pleasure

[fading tree]

have a goodnight 💯

[timid silo]

its the morning for me lol

on aug 28

diff timezones

[fading tree]

oh goodmorning then 🤣

alr cya bro

[timid silo]

wonderful

feel free to dm me if you need more help

[fading tree]

oh fr

ima add u then

[timid silo]

of course you can also open tickets here

[timid silo]

sure why not

[fading tree]

thank u bro i appreicate u

[timid silo]

my pleasure

[fading tree]

ima head to bed

bye man

.close

[obtuse pebble]

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[unreal musk]

Maybe see if you can argue that f'(0) would be zero?

[heady turtle]

how

[unreal musk]

Try from the first principles definition: that $f'(0) = \lim_{x\to 0} \frac{f(x) - f(0)}{x}$, and then try and manipulate that limit to make use of the fact that $f$ is even, so $f(-x) = f(x)$

[warm shale]

@unreal musk

[heady turtle]

[ f’(0) = \lim_{{x \to 0}} \frac{{f(x) - f(0)}}{x} = \lim_{{x \to 0}} \frac{{f(-x) - f(0)}}{x} ]

[warm shale]

miyo

[heady turtle]

wait what does this say?

The sign of f’(x) is opposite on both sides of the origin?

[unreal musk]

Well, not quite yet-

Have you made substitutions in limits before? (have you dealt with limits like the one for derivatives much, for that matter?)

[heady turtle]

f'(0-)=-f'(0+)

[unreal musk]

Well, I mean, you can conclude that f'(0) = -f'(0), which is stronger-

[heady turtle]

this should be correct, why is this wrong

[unreal musk]

As for here, have you? e.g. have you done something like saying
[
\lim_{x\to 0} \frac{\sin(3x)}{3x} = \lim_{u \to 0} \frac{\sin(u)}u
]
before?

[warm shale]

@unreal musk

[heady turtle]

you assume f'(0-)=f'(0)=f'(0+) by default, why? is it bc of the continuity of f?

[unreal musk]

And you haven't yet gotten to a point where you can conclude that: the statement is right, but you have to justify it

[unreal musk]

I don't assume it: what happens...

[unreal musk]

...if you set u = -x in the limit on the right hand side here?

[heady turtle]

but you said f'(0) = -f'(0) after i said f'(0-)=-f'(0+)..?im confused sorry just want to make it clear

[unreal musk]

I said that you can conclude [i.e. you should show] that it's true, and that's what the aim is (and I'm hinting how you get there)

[unreal musk]

Well, see here

[heady turtle]

ooh

yes i get it now

I actually thought of that substitution early on, but my thought process was too scattered and not rigorous enough. Thank you.

[unreal musk]

sometimes you just gotta try your ideas out

[heady turtle]

.close

[obtuse pebble]

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[obtuse pebble]

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[main acorn]

the method on the left seems different from the one used on the right

they use a different matrice . one uses ♪I - [T]B . while the other uses [T]B - ♪I . yet they arrive to the same eigenspace

does the order not matter?

am i dumb and it is just the same concept as :
5-x = 0 => x = 5
x-5 = 0 => x = 5
?

[obtuse pebble]

@main acorn Has your question been resolved?

[main acorn]

<@&286206848099549185>

[obtuse pebble]

@main acorn Has your question been resolved?

[main acorn]

.close

[obtuse pebble]

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[obtuse pebble]

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[cold brook]

can i please get help with this

[kind hawk]

w=c1v1+...+cnvn for some scalars c1,...,cn

now compute <ck,w> using that w=c1v1+...+cnvn

[fossil crag]

<vk,w>?

[kind hawk]

whoops

[obtuse pebble]

@cold brook Has your question been resolved?

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[obtuse pebble]

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[hot urchin]

i tried getting a common factor and put it as x(X^2+25X)

it didnt help me at all

[obtuse pebble]

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[hot urchin]

nvm i got it

[half mortar]

,w integrate 1/tan(x)

[warm shale]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by integrateseverything

[half mortar]

ok thank you channel owner

[hot urchin]

thank you a lot man

i appreciate your answer and your time

[half mortar]

yea its the answer everyone is lookin for

the billion dollar equation

[hot urchin]

btw

i didnt learn how to work with integrals yet

i solved it another way

much more simple

[half mortar]

how can you solve integral of cot(x without learning integration

[hot urchin]

do you want me to show you what i did?

its simple math

[half mortar]

no im good

[hot urchin]

nothing complicated

alrighty

thank you anyway man

[half mortar]

can you close the channel

[hot urchin]

.close

um

how?

.done

[half mortar]

.getout

[hot urchin]

.getout

?

what

[half mortar]

.close

[hot urchin]

i was nothing but nice to you

[half mortar]

try this

[half mortar]

im not saying that to you blud

alright i dlted it

fine

[hot urchin]

.close

.resolved

[half mortar]

now type .close

what are you doing

this should work

[hot urchin]

.close

[half mortar]

this isnt your channel

[hot urchin]

this is lol

[half mortar]

its someone elses

no it isnt

[hot urchin]

it is?

i typed my question in the available catagory

and it got me to here

[half mortar]

whatver bro just dont text in this channel

they will automatically close it

[obtuse pebble]

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[unkempt vortex]

guys help

[unkempt vortex]

only need 2 and 3

teach me pls

[gloomy vector]

find the ratio for each term, since you got 1 im sure you know what to do

[fathom light]

If you know the formula, if you know the ratio I think you will be good

[obtuse pebble]

@unkempt vortex Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[near saddle]

I forget how to factor the x^3 :/

[near saddle]

Do I need to add +ax^2 -ax^2 +bx -bx ? If so how do I decide what a and b are?

[fathom light]

a³+b³ = (a+b) (a²-ab+b²)

Use this formula

[near saddle]

how do i memorize all of these formulas

[fathom light]

😅few formulas you need to remember

Like (a+b)² (a-b)² , cube etc etc

[hidden compass]

With practice

And if you only remember one of the first two bits you can always find the other by expanding it out

[near saddle]

I'm taking async calc and it's hard to know what I will be allowed to use on the tests 😓

Using notes, books, cheat sheets, apps, other devices, or any other aids, real or virtual, including websites like Chegg is not permitted. You may not interact with others, in person or virtually, to give or receive help during the test. If you are being proctored online, be sure to keep your eyes focused on your computer screen and your written work and always remain in view of the camera.

A TI-83/84 graphing calculator is permitted on all tests.

This is so strict only a TI calculator :/

[rancid zenith]

A TI calculator will allow you to do anything almost

[rancid zenith]

you could compute this by just graphing it or letting x values being really close to -3

but its always better to know how to solve it analytically

[near saddle]

I prefer graphing on desmos than ti

[rancid zenith]

obviously

[near saddle]

😅

[rancid zenith]

desmos has a much better UI

but TI is still very useful in exams

[near saddle]

I need to continue reviewing the algebra practice before my test to review the factoring etc

Thank you all

.close

[obtuse pebble]

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[obtuse pebble]

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[final prism]

so im working on a mathematics problem where I attempted to use the same method where we divide the circle into concentric circles and approximate the circles as rectangles to reach A=pir^2i for ellipses. since ellipse circumference is weird and uses complete elliptic integral of second kind (4aE(k)) I used agm to make a method to compute E(k).so at this point I knew I wasn't gonna be able to simplify my result to piab but I decided to try and explore if I can reach an equivalent or at least approximate results. I have attached the current formula I have reached and I want to know how I can compare this formula with the regular formula and determine whether its equivalent or approximate or completely wrong, btw w represents the width of a concentric elliptical ring and A represents the semi-major axis

[haughty stump]

approximate it looks like, itll be more accurate as w gets smaller

[final prism]

yes of course

that is evident

but thats not my question

i am asking how i can prove the equivalency of this function and A = piab

what mathematically rigorous method can i use to realise whether or not this is equivalent to piab

[haughty stump]

alright sorry this is a bit out of my league lol

[final prism]

i dont understand it either man

but it seems to work

its ok

[haughty stump]

well its going to be hard to prove

but it is inherently limited and cannot perfectly represent the curvature of the ellipse, cause its approximating the ellipse as a series of concentric rectangles

so cause of that I would say it probably isnt equivalent, but I cant be sure about anything else

[final prism]

concentric ellipses actually, which when stretched out approximate to rectangles...

bit convulated

[final prism]

hmm

okay

[haughty stump]

¯_(ツ)_/¯

[obtuse pebble]

@final prism Has your question been resolved?

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[high glacier]

hello! i need help with a problem i cannot determine properly

[high glacier]

i do not know how to solve this

[grizzled shore]

just check points

what points are impossible

then check that you need the product of 2 things to have the same sign to be positive

[wild swallow]

its a good idea to multiply through by the denominator

[high glacier]

thank you frosst

[high glacier]

wait hwat do you mean by this?

[wild swallow]

,, \f {(x + 7)(x - 3)} {(x - 5)^2} \cdot (x - 5)^2 > 0 \cdot (x - 5)^2 \textqq{as long as} x \ne 5

[high glacier]

wait

[grizzled shore]

well i was just gonna go with as long as x is not 5, the bottom will be positive because you square it

[grizzled shore]

that was the point of this

[wild swallow]

but now you get to throw it away

out of sight out of mind

[high glacier]

give me a moment

[grizzled shore]

this doesn't flip the inequality only because the thing is always positive

[high glacier]

ahhh wait

i get it

[wild swallow]

which is why i suggested multiplying it through

[grizzled shore]

if you try this trick on other quotients it might get weird

[high glacier]

thank you

[wild swallow]

always kill squared things

[high glacier]

.close

[obtuse pebble]

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[obtuse pebble]

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[plucky trail]

Hello, I need help to understand this again

I am clearly doing something wrong/illegal

I was told there was supposed to be a K^2 in the denominator

I don't understand where it comes from

[rich plume]

what have you assumed x and y to be

wait

is that even a part of your doubt

[plucky trail]

sorry x = K^2

and y=-1

z=2

[rich plume]

I don't get why you would excatly do such a substitution when you end goal is just to simplify

[plucky trail]

Yes how would I go about simplifying this

[plucky trail]

I am clearly doing something wrong

So I am going about it the wrong way

[rich plume]

you are very right up until third step

[plucky trail]

The third step meaning where I expanded the parenthesis?

[rich plume]

line 3

[plucky trail]

Or attempted to

Line 3

Okay

Line 3 is correct/wrong? Sorry language barrier about "up until"

[rich plume]

notice both the parenthesis have their individual 'k' in denominator

[rich plume]

line 3 is correct

[plucky trail]

Yes they do have an individual K

Okay okay

Then I am with you

[rich plume]

so while expanding you must multiply them out

[plucky trail]

1/K?

break it out I mean

[rich plume]

not necessarily

[plucky trail]

no okay

am I supposed to square it?

like (1/K)^2 = 1/K^2

[rich plume]

$\left(\frac{a}{b}\right)\cdot\left(\frac{c}{d}\right)=\frac{ac}{bd}$

[warm shale]

B-eard

[plucky trail]

b = k

d = k

[rich plume]

product of numerator divided by product of denominator of individual parenthesis

[plucky trail]

bd = k^2

[rich plume]

in your case, yeah

I'm pretty sure you are confusing this with addition

$\frac{a}{b}+\frac{c}{b}=\frac{a+c}{b}$

[warm shale]

B-eard

[plucky trail]

I think my confusion is deeper than I could describe

I don't think I am confusing it with that: Truth be told I am not familiar with even basic concepts like that. I just lack understanding in the most fundamental things.

[rich plume]

tbf it's a really simple one, you're just having a not so brain moment

[plucky trail]

I hope so

I started Uni three days ago

Have had 3 math lessons

Everyone is so far beyond my level

Not impostor syndrome

Rare case where it is true

[rich plume]

well all I can do is hope for your best

you can backup using some youtube lectures

[plucky trail]

Yeah, I am thinking about switching to an easier program

[rich plume]

well that's another option

[plucky trail]

This is civil engineering

If I am struggling with this I have a problem

L education system moment

[rich plume]

you could bring this issue out to your teacher

[plucky trail]

I haven't given up totally but honestly I think the mature thing would be to switch

[rich plume]

maybe that could help

[plucky trail]

I will talk to them for sure

This is painful

Thanks for your input and help with the original question

I still like math and physics* but I am not there yet

.close

[obtuse pebble]

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[rich plume]

welcome

[rich plume]

I get it, you have an abstract interest in those fields, but you hate it when it comes down to those formulas

[obtuse pebble]

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[brisk arrow]

I’m so depressed

[brisk arrow]

Can someone explain how it works

Like what the numerator is V_R + V_A instead of just V_R

[ancient drift]

Resistance is equal to the sum of the voltages over the current

[brisk arrow]

"the current", do you mean the current that passes R2

[ancient drift]

Yeah ig

[brisk arrow]

what do you mean the sum of the voltages

The voltages produced by the A stuff and by the resistor?

[haughty stump]

V_R is the voltage drop across the resistor R.

V_A is the voltage drop across the ammeter A.

yeah sorry corrected myself lol

I_R is the current flowing through the resistor R.

[brisk arrow]

Why the numerator is the sum, of the voltage rather than just V_R itself?

[haughty stump]

because this is a series

[brisk arrow]

that confuses me

[haughty stump]

and in a series the total voltage drop across the circuit is equal to the sum of the voltage drops across each individual component

[brisk arrow]

In my opinion, if you are trying to get the resistance of a electric unit, then you should just put it into the formula R=V/I, where R is the resistance of the unit, V is the voltage drop of the unit, and etc...

So, I cannot stand that the numerator is the sum of voltage.

[haughty stump]

well thats the exact equation we're using here

its just that we need the voltage drop across resistor R and ammeter A and sum them up because it is a series

V in the equation R=V/I is the same as the sum of the voltage drops across resistor R and ammeter A

aka V_R + V_A

[brisk arrow]

why we need the voltage across R and A.

we need only the R.

[haughty stump]

This question is giving you the voltage drop BECAUSE OF the resistor, not the resistance itself

hence why it is V_R & V_A and not just R and A

[brisk arrow]

but we are trying to the the value of R_2, no?

therefore, the numerator should be just V_R

R_2=the voltage drop of the resistor/ the current passes through the resistor

that's it.

[haughty stump]

well you still need to include the voltage drop of the ammeter

no matter how negligible it is

[brisk arrow]

why I have to include it, that's something I cannot figure out and justify

[haughty stump]

Because the voltage still drops there, so you have to include it in a series calculation

[haughty stump]

well if you really want to you could drop the V_A altogether, because the value is likely very close to 0, usually the resistance of an ammeter isnt actually included in the calculation since its such a small and negligible number. But I guess this teacher just wanted to be extra exact

[obtuse pebble]

@brisk arrow Has your question been resolved?

[obtuse pebble]

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Remember:
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[buoyant lance]

|z+1|^2 = |z-3i|^2

Is it true that:

The solution to z, must fufill the following:

3*(Im(Z)) - Re(Z) = 4

[buoyant lance]

Because I dont know what program can be used to check my answers.

[modern idol]

Are you using |z+1| to refer to the magnitude of z+1?

[buoyant lance]

I am refering it as a distance between z(a random complex number) and the real number (-1)

because:

|z + 1 | = | z - (-1) |

[modern idol]

Oh duh okay

[buoyant lance]

ye

[tardy epoch]

do you have any work proving this

[buoyant lance]

I have it on a piece of paper yeah

[tardy epoch]

show

[buoyant lance]

ok

[buoyant lance]

there

oops

lemme flip this upside

[tardy epoch]

,rotate

[buoyant lance]

ah

[buoyant lance]

I am uncertain if you can take the square root of |x|^2 and just get the absolute value of x.

But I think you can, yes?

Exponent laws and such.

[tardy epoch]

you took the square root and then you squared again?

this line follows right after the first one

[buoyant lance]

Yeah I squared both sides of the expression here.

But the other picture, I am uncertain

[tardy epoch]

i only see one picture

[buoyant lance]

uhhh... Really?

wait

So basically

[tardy epoch]

only one i see

[buoyant lance]

Oh

Uh...

Like those steps make sense

Because you need to isolate x and y.

Without those pesky square roots in the way

[tardy epoch]

i didn't say it didn't make sense

[buoyant lance]

oh

[tardy epoch]

i just meant it was unnecessary

[buoyant lance]

ah... Well its just my brain going brr

but fair

[buoyant lance]

Otherwise, am I incorrect about my solution?

3y - x = 4

^ Where y is the imaginary part, x is the real part.

[tardy epoch]

i can't read this

[buoyant lance]

6y-2x = 8

Turns into:

3y - x = 4

after division by 2

[tardy epoch]

,w solve |(x+iy)+1|^2 = |(x+iy)-3i|^2

[warm shale]

Results provided by WolframAlpha

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

Requested by rie.mann

[buoyant lance]

huh

[tardy epoch]

looks like you have a sign error

[buoyant lance]

oh

I am confused about where it is

xD

Because:

(x-1)^2 + y^2 = x^2 + (y-3)^2

Does not equal to x+3y = 4.

I checked using desmos as well.

[tardy epoch]

then your mistake is earlier

[buoyant lance]

yes

I cant find it though, thats the frustrating part.

[tardy epoch]

keep plotting at every step before

[buoyant lance]

ok, sure

[buoyant lance]

Ok, so this doesnt match

meaning I at least simplified correctly

but idk about the previous steps though. As I am 90% sure I did them right too...

[buoyant lance]

wolfram gives me a different solution too

Its likely the |z+1| part that has screwed it up.

But I know, that |z+1| = | z -(-1)|

So I am not sure why it is (Re(Z) +1)^2 instead of the (Re(Z) -1)^2

because the real part is not 1, it is minus 1. For |z+1|

[tardy epoch]

wot

$|a + ib|^2 = a^2 + b^2$

[warm shale]

riemann

[tardy epoch]

$a = x + 1$

[warm shale]

riemann

[buoyant lance]

Sorry I mean, if we are trying to express |z+1| as a distance then we would say:

between z and -1. At least according to my teacher lmfao.

My professor, I mean, I study at uni lmao

[tardy epoch]

yea your professor is right?

[buoyant lance]

...ye?

Ofc

[tardy epoch]

that doesn't contradict what i said

[buoyant lance]

I was contridicting what wolfram alpha said actually

not u

[tardy epoch]

yea this is wrong

[tardy epoch]

use this with the a i said below it

[buoyant lance]

hold on im confused

uhh

What part is wrong here? Is it assuming that you can express |z+1| as a distance between z and -1.

Is that wrong.. @tardy epoch

unless if I am making two random claims at the same time, idk if I did

[buoyant lance]

Hold on. Did I accidently modify the grade of the polynomial?

Because, two absolute values multiplied with eachother. Yeah, they become one single product. Same with the right hand side.

But I took the square root, both sides. That means, I might lose roots?

[tardy epoch]

i don't understand what you're saying

[tardy epoch]

|z+1| is the distance from z+1 to the origin

[tardy epoch]

plug a=x+1 into this expression

[buoyant lance]

a = x+1

ok

| a + bi |^2 = a^2 + b^2

| (x+1) + bi | = (x+1)^2 + b^2

[buoyant lance]

There

[buoyant lance]

Yes. Thats fair.

But my professor said you cannot use distance with addition, you need to express it in terms of subtraction. For these two terms at least.

[tardy epoch]

right. now use this in your problem

[tardy epoch]

to answer your question

[buoyant lance]

Wait a moment.

[buoyant lance]

Am I supposed to write this like:

|x+b|^2 =| x^2 + b^2|

And not take the square root of the absolute value squares?

of LHS and RHS at the start of the equation?

Because I think I know where the error lies, it lies in the fact I have taken the square root of two polynomials on LHS and RHS. Which is something you should never do, because it modifies the maximum amount of solutions an equation can have

...Fucking damn it, now that I realize it. I am dumb dumb.

[buoyant lance]

Also, isn't it: |a^2 + 2ab + b^2| xD?

[tardy epoch]

i don't know what you're asking anymore. you're jumping all over the place

[buoyant lance]

I am asking if the issue comes from the fact I did this

[tardy epoch]

this was your main issue. fix this and redo the problem

[timid silo]

graph it

lol

[buoyant lance]

what

I cant graph complex numbers dud xD

[timid silo]

sad

[buoyant lance]

unless if you want me to treat Z as a regular variable

[timid silo]

lol

[timid silo]

treat it as a vectorl ol

[buoyant lance]

xD

uh lemme see

[buoyant lance]

Yeah uh, I am not getting anywhere

Also, it appears the "not taking the square root of LHS and RHS" only applies for inequalities

[timid silo]

uh

yes

but still avoid square roots

[buoyant lance]

Well yeah I guess I will avoid using square roots for now. If I see variables on two sides an expression

I will only use them only if theres like an isolated variable or a bracket and a value

[timid silo]

sadly its none of the standard loci in complex plane which i know

[buoyant lance]

On the bright side, you can actually express:

|a+b|^2 as = |a+b|*|a+b| which is actually |(x+b)^2)|

[timid silo]

if |z+1| = |z-3i|

Its bascially equation of the bisector

[buoyant lance]

lets say

Im(z) = 1
Re(z) = -1

That becomes 4.

[timid silo]

And |z+1| + |z-3i| = 0 is not possible

[buoyant lance]

| (-1+i) | = ? = | (-1 + i) - 3i |

| -1+i | =?= | -1 + i -3i |

| -1 + i | = ? = | -1 - 2i|

[timid silo]

so it represents the bisector

[buoyant lance]

I wonder if these

actually become the same

if we square these distances

lemme see on a cord plane

[buoyant lance]

So, apperently it is not true that the absolute values are the same

therefore my thing does not hold

[timid silo]

bro

there is nothing such as absolute value

[buoyant lance]

yes there is

[timid silo]

in complex numbers

[buoyant lance]

xd

[timid silo]

ok

[buoyant lance]

What else are we supposed to call |x|. We name it abs value of x.

[timid silo]

we call it the modulus

[buoyant lance]

wtf

[timid silo]

uh

ok

whatever

[buoyant lance]

but regardless, they dont give the same distances

so... im back to being confused about what I did wrong

| (-1+i) + 1 | = ? = | (-1 + i) - 3i |

^ Ah I see now

I forgot the +1

lol

| i | = ? = | -1 -2i|

[timid silo]

Well dude this is basically perpendicular bisector of line segment joining (-1,0) and (0,3i)

so all complex numbers lying on this line will be the solutions

so write the equation

[buoyant lance]

| z + 1 |^2 =| z-3i| ^2

oops

there

[buoyant lance]

So I took the square root of both sides. At the start of the equation
But, a lot of shit can go wrong there so... Technically you should not

and I suppose that was my mistake

[timid silo]

ok

dude

[buoyant lance]

yes

[timid silo]

literally most of your calculation was stupid

lol

[buoyant lance]

lol w h y

[timid silo]

just get the equation of curve

And plot it

Why the fuss bro

[buoyant lance]

Because you cannot use complex numbers here. Its desmos

Like... The program does not accept z as a complex variable

[timid silo]

Bro when did i say use desmos

[buoyant lance]

well what am I supposed to use

le tell me

[timid silo]

Do it yourself

on a paper

[buoyant lance]

I literally calculated the result to:

3Im(Z) - Re(Z) = 4

What is the stupid part of my calculation

[timid silo]

Then why crying

[buoyant lance]

Because I cant verify it

[timid silo]

You already got the answer

Verify it with whay

What

[buoyant lance]

Wolfram alpha says it is wrong.
But in my calculations, I dont see the error

[timid silo]

Do you want me to get pen and paper and try it myself

Lol

[buoyant lance]

Like...

I have:

Im(z)^2 + (Re(Z)-1)^2 = Re(Z)^2 + (Im(Z) - 3)^2

[timid silo]

Cut the crap bro ill do it my way

Wait

[buoyant lance]

ok

[tardy epoch]

why do you keep doing (Re(z)-1)^2

[tardy epoch]

.

[timid silo]

Its wrong

[buoyant lance]

Because the problem is.

It is no longer making sense for me to do:

|a + bi | ^2 = | a^2 + (bi)^2 |

It makes more sense to do:

| a + bi | ^2 = |(a+bi)^2|

[timid silo]

The result is 3(Im(z) + Re(z) =4

[buoyant lance]

yes, which is something I got

xD

[timid silo]

no

[buoyant lance]

yes

I got 3y - x

oh wait

[timid silo]

.

[buoyant lance]

nvm

lol

[buoyant lance]

Yeah, and I cant find the sign error issue

[timid silo]

can i send what i did ?@rie.mann if its allowed

@tardy epoch

[buoyant lance]

i mean go ahead

[tardy epoch]

sure

[buoyant lance]

doesnt bother me

[timid silo]

do this

[buoyant lance]

Uhhh the solution does not make sense

[timid silo]

why ?

[buoyant lance]

I dont even know what ur doing

[timid silo]

uh

[buoyant lance]

like idk if its hand writing

xd

[timid silo]

then i should teach you

if z = x + iy

then its conjugate ž = x - iy

[buoyant lance]

yes it forms a real number

I know

[timid silo]

so (z)(ž) = |z|^2

[buoyant lance]

that misses its imaginary part

[timid silo]

i used this

[buoyant lance]

wait a moment

So that is an identity that my uni professor hasnt gone through yet

or maybe he has, but skimmed through it

[timid silo]

who cares now you know it

use it

my teacher didnt even teach me anything he just finished complex number in a week

stupid

i had to learn most of it on my owb

own

[buoyant lance]

maybe because thats why I am so stupid about complex numbers rn

xd

uni professors tend to be very fast with going through shit

[timid silo]

just use it bro

[timid silo]

i am into in uni yet so no idea

[buoyant lance]

okay, this would have been very useful to know

[timid silo]

But did you get what i did

yeah use them all

[buoyant lance]

Yeah you exploited the property z * z conjugate = |z|

[timid silo]

you know what there is one more cool way

[buoyant lance]

And you could use that to calculate shit

[timid silo]

Of doing this problem

Wanna see ?

[buoyant lance]

yes

is it graphically

[timid silo]

Yeah lemme draw it on my paper

[buoyant lance]

but yes, I get it now.

[timid silo]

Cringe

I am not getting the ans geometrically

Oh wait

I am

Got it

@buoyant lance u can turn this into a problem of analytic geometry if you can interpret the equation well

learn to graph things on the argand plane

[obtuse pebble]

@buoyant lance Has your question been resolved?

[buoyant lance]

Oh I have 0 knowledge about analytic geometry

But that sounds cool, I guess?

[timid silo]

well you probably know about lines and slopes

[buoyant lance]

yes

[timid silo]

atleast

thats enough for learning other stuff

[buoyant lance]

Yeah but I have to wait with that

I want to be in sync with the course

But otherwise, cool stuff

[buoyant lance]

I also remember four new identities as a result

1. Re(Z) = (1/2) * (Z + Z(conjugate))
2. Im(Z) = (1/2) * (Z - Z(conjugate))
3. |Z_1 * Z_2 | = | Z_1| * | Z_2|
4. | Z_1 + Z_2| <= |Z_1| + |Z_2|

The third one is just exploiting the understanding of modulus

And four is just... Triangle inequality stuff

[timid silo]

yes memorize triangles inqualities in both forms

[buoyant lance]

Thankfully, the lecturer does not require us to know this yet.

The other form

the first one is enough yes?

for now that is

[timid silo]

idk dude i find both of them as important results

[buoyant lance]

oh ok, well regardless. I know how to solve absolute value equations (that are squared) with complex numbers now.

thats important

[timid silo]

ok

good

[buoyant lance]

Now I can extrapolate upon that to apply it to harder things or unfamiliar things

so yes, mission success

[obtuse pebble]

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[timid silo]

Ok bro shut up

Just do it

Dont say

[obtuse pebble]

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[austere pagoda]

so when not provided the function and you're just provided with the information about what it looks like how do you solve this again?

[dire carbon]

There is a formula for average rate of change, do you recall what it is?

[austere pagoda]

f(b) - f(a) / b - a?

[dire carbon]

Yep!

So firstly, what are our a and b values?

[austere pagoda]

b is 1 a is -3

[dire carbon]

Yep!

So, you can analyze the graph and find the f(a) and f(b) values

[austere pagoda]

yeah im just trying to figure out what function graphs like this

[dire carbon]

Ohh ok

[austere pagoda]

i know its negative

im just tryna figure out what it is

is it just in the form y = a(x - h)^2 + k?

[solemn basin]

U will need to know the roots for that

[solemn basin]

Kinda yea

[austere pagoda]

dont you need the function to find what the roots are though

[dire carbon]

Not really, you could use the graph and come to an approximation

[austere pagoda]

enlighten me

[solemn basin]

U have the roots in the graph

[solemn basin]

the points at which the graph crosses x axis are the roots

[austere pagoda]

well it crosses at points that you cannot define

4. something

[dire carbon]

The graph looks like it crosses through -4.5

[austere pagoda]

2. something

[austere pagoda]

true but unless I could prove that wouldn't be making this assumption wrong?

besides the -2 root is not clear

[solemn basin]

-4*

[dire carbon]

The assumption would just be an assumption

[austere pagoda]

okay so a is 1, h is -4.5 x we can use -3 and where x is -3 y = 1

[solemn basin]

Well u can get the graph in someways i think. First of all u know the minima
Second we know how much of the graph is shifted from origin 3 units
And how much it's shifted above -1 unit
So we can get the constant and the x coeff by taking derivative and finding maxima

[austere pagoda]

2 = 1(-3 - (-4.5)) + k

k = -0.25?

y = 1(-3 - (-4.5)) - 0.25?

[solemn basin]

Idk abt the calculations
I told u the method seems
correct to me tho

[austere pagoda]

okay

so now I swap that -3

for 1

and then subtract the result of 1

from the result of -3

and then divide it by 1-3 yeah?

[solemn basin]

Yes

To obtain average rate of change

Rather it should be 1-(-3)
x2 -x1

[austere pagoda]

oh yeah you right

and we already knew y for the other question cause we used it to find k

so its 30-2/1-(-3)

rate of change is 7?

[obtuse pebble]

@austere pagoda Has your question been resolved?

[obtuse pebble]

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[mellow slate]

how do I get turn pythagorean theorom into simplest radical form?

[native inlet]

what exactly do you mean?

[silk quail]

the pythagorean theorem is already in its simplest form.

[mellow slate]

the hypotenuse is supposed to be in simplest radical form

i dont knwo radicals 😭

oh mu god i got it right

is radical form just organizing the square roots while keeping only integer numbers?

[kind hawk]

well radical means root

"simplest" radical form means what you mean I guess

[mellow slate]

so if I'm trying to say find the simplest radical form of √90 what steps should I be taking?

[kind hawk]

write 90 as a product square*nonsquare number. then use sqrt(ab)=sqrt(a)sqrt(b)

for the first step you can also find the prime factors of 90 and then pair them up

[mellow slate]

tyvm

thanks for the help!

.close

[obtuse pebble]

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[umbral flume]

i have smth like
cos(x) = 5y
im trying to get x alone, how would i get the cos off of it?

[umbral flume]

i feel like its something really simple but i just dont remember

[worn yoke]

you apply the inverse function

[umbral flume]

yea i forgot what that is lol

what would happen to 5y after applying the inverse function

[marsh geyser]

Do you know the name of that function?

[umbral flume]

co sine

[marsh geyser]

!occupied

[obtuse pebble]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[umbral flume]

man im still here 😭

[marsh geyser]

I meant the name of the inverse

<@&268886789983436800>

[umbral flume]

secant right

im tryna get help with my math hw man let me do this in peace 😭

[umbral flume]

i forgot 😔

[marsh geyser]

I recommend you to review your notes

It is arccosine

[umbral flume]

we went over cos sin etc last year i just forgot lol

[marsh geyser]

No problem, just look for information about that and read it

[umbral flume]

alr i sorta got it

its would be cos^-1(5y) or arccos(5y) right

alr so i got
a(x) = arccos(-16x^2+4)
b(x) = arccos(16x^2+16x+4)

but its wrong

i setup the inequality like
-x<=g(x) <= x

ohh wait

i was supposed to divide by 4 not multiply 😔

hollon

🤦

still got it wrong 😔

damn im dumb ik why

i added instead of subtracted 2

its been too long

i got it resolved

so uhh

!resolved

idk

.close

[obtuse pebble]

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[obtuse pebble]

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[marble epoch]

In discrete, Is there a general rule for something like ¬J ∨ C if the conclusion is ¬J? since C doesn't really matter to the equation?

[obtuse pebble]

@marble epoch Has your question been resolved?

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[timid silo]

C is the midpoint of AB, D is the midpoint of AC, E is the midpoint of AD, F is the midpoint of ED, G is the midpoint of EF, and H is the midpoint of DB. If DC = 16, what is GH?

[timid silo]

did you try to draw a line and mark down the different points and distances

if you did i think it would have been very sf

i did

i'll send a pic hold on

kk

wait also i have another question, when I describe a line for example, a line with points a b c, could I say line ab?

[thick timber]

ye better to tell the initial point and the final point and the other points lie on them too

[timid silo]

wdym

is line ab okay or do i have to put line ac

[thick timber]

nah as they say AB are the first and the last points

and the other points lie between them

that diagram is better

[timid silo]

so in general when i need to name lines i just put the first and last points?

[thick timber]

ye

other points are also there lie on the same line

@vale ember

[obtuse pebble]

@vale ember Has your question been resolved?

[thick timber]

@vale ember

[obtuse pebble]

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[versed eagle]

Help with #34

[versed eagle]

since im not given a point im not sure what to do

[rancid zenith]

u find the point

what is the derivative of the given quadratic?

[versed eagle]

4x-2

so y = -2

[rancid zenith]

we then have 4x-2 = 2 right

[versed eagle]

oh right

🤦‍♀️

always forgetting the damn basics

[rancid zenith]

its okay

so then u find the x value and then plug it into the original function

to find the y value

and u have ur points

[versed eagle]

ty

[rancid zenith]

yw

[versed eagle]

.close

[obtuse pebble]

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[rancid zenith]

do u remember the unit circle

[cosmic pecan]

No

I know that it is equal to 300 degrees. However you can't match that to a right triangle of course.

Here it is presented that I must map it out on the unit cirlce and determine it is pi over 3

To find a corresponding right angle

My question is, is there a way where I can configure 300 degrees to an corresponding angle that forms a right traingle without mapping it out to find it

In which case, 300 degrees would correspond to 60 degrees, and I can craft a right angle triangle out of that and solve

But how would I be able to do this systematically? Is there a forumla for this?

Or would I actually have to remember the angles and their corresponding angles

[obtuse pebble]

@cosmic pecan Has your question been resolved?

[obtuse pebble]

@cosmic pecan Has your question been resolved?

[versed eagle]

I’m not a mathematician or anybody who works to solve problems, but can I ask do you know how to find absolute value?

Do you know about 30,60,90 right triangles and 45,45,90 right triangles?

[cosmic pecan]

yes

[versed eagle]

Memorize this

All you need to remember is 60, 45, and 30.

60 is pi/3, 45 is pi/4, and 30 is pi/6. If you know those and your 2 right triangles then that’s all you need

If you want to find 5pi/3, just add pi/3+pi/3+pi/3+pi/3+pi/3

[versed eagle]

The only thing necessary to memorize is this information

@cosmic pecan

[cosmic pecan]

https://tenor.com/view/orange-justice-super-fast-hyper-energetic-gif-16361975

ok thx you made me realize i just had to know the denominator to backtrack to 60, 45, 30 on the unit circle

.close

[obtuse pebble]

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[cedar ginkgo]

How to solve this question:
Vikings always tell the truth, and Saxons always lie. Given the following information, use a truth table to determine what type each person is or if their status cannot be determined. Be sure to provide a conclusion based on your work. Person X says: “Y is a Viking if and only if I am a Saxon" Person Y says: “If Z is a Viking, then X is not a Viking" Person Z says: “Both X and I are Saxons".

Did like this:

Let p be "X is a Viking"
Let q be "Y is a Viking"
Let r be "Z is a Viking"

[unkempt vault]

X says

Y is a Viking if and only if I am a Saxon
this tells us that
p <-> (r <-> ~p)

because the statement is "r <-> ~p", and it's true iff p is true

evaluate that stsatement cross off all the lines where it's false

and so on for the other two statements

[cedar ginkgo]

Wait why is there a r <-> p

I'm kind of confused a little bit?

Could you break down from the sentences how to get the logical statements

wouldn't this mean: p <-> (q <-> ~p) --> X is a Viking if and only if Y is a Viking if and only if X is a Saxon.

[unkempt vault]

yes

sorry, I meant to use q, not r

[cedar ginkgo]

Hm I don't understand where the p <-> comes from

[unkempt vault]

X says

Y is a Viking if and only if I am a Saxon

in other words, X asserts that (q <-> ~p) is true

this means that X is a Viking iff (q <-> ~p) is true

so p <-> (r <-> ~p) must be true

[cedar ginkgo]

Ohh

Gotcha

so

wait I'm still a little confused

[cedar ginkgo]

like for this

Because if X is a Viking how can the end that X is a Saxon make sense

[unkempt vault]

X makes a statement about himself being a saxon. If he said "I am a Saxon", that would be p <-> ~p, which would be a problem (same paradox as "this statement is false"), but p <-> (r <-> ~p) is fine

it's a logical statement, so evaluate it for the 2 different valules of p and the 2 different values of q

[cedar ginkgo]

Ok gotcha

[cedar ginkgo]

what do you mean by this?

[unkempt vault]

make it a column in the truth table

[cedar ginkgo]

ah yeah and then evaluate for different pqr values?

also for y's statement would it then be: q <-> (r -> -p)?

and for z's statement: r <-> (-p ^ -r)

wait would x's statement also not be : (p -> (q <-> -p)) and (-p -> -(q <-> -p))

could you verify?

[obtuse pebble]

@cedar ginkgo Has your question been resolved?

[obtuse pebble]

@cedar ginkgo Has your question been resolved?

[obtuse pebble]

@cedar ginkgo Has your question been resolved?

[obtuse pebble]

@cedar ginkgo Has your question been resolved?

[obtuse pebble]

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[safe kite]

can you even find the rate of change from knowing t=2 ?

i found dr/dt but idk what r would be when t=2

[opaque dome]

Take the volume and minus the rate of melting knowing every second it lost 2cm^3 and there is 2 second

Oh wait

[dense imp]

is that the complete problem?

[safe kite]

yeah

[dense imp]

yeah, kind of seems like they left some info out, is this from a book or just some worksheet your teacher is more likely to have made a mistake on?

[safe kite]

yeah its a worksheet

[dense imp]

you'd have to know the volume or radius at t = 0 or something

so you can get r at t = 2 like you said

[safe kite]

ok thanks

.close

[obtuse pebble]

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[obtuse pebble]

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[modern bison]

Find GB

.

I tried this much

Then got stuck

[obtuse pebble]

@modern bison Has your question been resolved?

[modern bison]

<@&286206848099549185>

<@&286206848099549185> please help

[subtle wasp]

what is the question ?

ok find GB

what theorem can u use here ?

@modern bison there are 2 ways to calculate GB

[modern bison]

Can you tell me

[modern bison]

Idk tried by trying to show triangles similar but got stuck after finding AC

I wanted to also fing AF but how?

Idk I might be off track

[subtle wasp]

ye is there any special restrictions to calculate this length ?

like "only use trigo" ...

[modern bison]

Trigonometry?

[subtle wasp]

yes

or theorem ?

is there restrictions ?

[modern bison]

No

[subtle wasp]

ok

so what are the 2 biggest theorems and well-known to apply in a triangle ?

[modern bison]

I'm sorry I am not aware about theorems yet. I did learn a lot of theorems but I forgot it all ( I am preparing for SAT)

[subtle wasp]

ok do u remind of thales ?

[modern bison]

I might recall if you can tell me

[modern bison]

Oh yeah I know that one I did it here too

[subtle wasp]

where did u apply it

[modern bison]

O find AC

To

[subtle wasp]

so u didn't use here this theorem u use the law for the similar triangles it's not the same

[modern bison]

Ohh really

My bad

[subtle wasp]

but it's right to calculate AC

u did it well

BUT

to calculate GB u didn't get all the conditions

[modern bison]

Yes

[subtle wasp]

so where can u apply this theorem in ur exercise to calculate GB

[modern bison]

I didn't know how to get AF

[subtle wasp]

is the question calculate AF before the one for GB ?

[modern bison]

No

[subtle wasp]

so that's bc we NEED GB to calculate AF

[modern bison]

Finding GB is the question

[subtle wasp]

so in other terms where can u see the same figure as in the theorem in the exercise

where can u use it

[modern bison]

In ∆AFC

[subtle wasp]

good

why

[modern bison]

Because we need to find GB which is in the same triangle as ∆ACF

It has 2 triangle in same triangle

[subtle wasp]

yes

but in fact i was asking about conditions of the theorem (DE and BC parallels in the picture i sent) but nvm

so what is the equality that claims the theorem in ur shape

[modern bison]

AG/AF = AB/AC = GB/FC

[subtle wasp]

exactly

which one interests us the most

[modern bison]

GB/FC

[subtle wasp]

no bc we can't calculate this ratio rn

we don't have GB

so we need to find another ratio that we can calculate

[modern bison]

AG/AF

[subtle wasp]

exact

[modern bison]

That's 1/2?

[subtle wasp]

why aren't u sure about it

ofc it's it

[modern bison]

Ahh

[subtle wasp]

bc AF = AG + GF

and bc AF=GF

[modern bison]

GB is 12

[subtle wasp]

We can tell that AF = AG + AG

YES