help-23

[drowsy jolt]

example: x^2 is continuous, {x = 0: 2, x != 0: x^2} is not

its alg 2

[lean otter]

Oh then I'm screwed...

[drowsy jolt]

like i mean literally draw it

[lean otter]

Ooh, continous means no holes

Got it

.close

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[rough matrix]

Hello

[rough matrix]

Help plz

I will send question

I have to find height and how many triangles are possible to be formed

[frank parrot]

Is 31 and 6 like inches?

or cm

if so then its one possible triangle

because they would need to connet to one point

[rough matrix]

Can u draw a diagaram plz

Im so confused

31 and 6 like cm or inch

Either

[frank parrot]

ok bad example

i just made this on paint a few seconds ago

but basically what i'm trying to say is

if you already have an angle

+dimensions wouldnt you just connect them

like that

and wouldnt that just give you the answer

and if we do it like that

couldn't we cut it into a right angle triangle

[rough matrix]

How is that right?

[frank parrot]

i think im confused on the question

[rough matrix]

Isnt it this?

[frank parrot]

my bad i didnt label it

[rough matrix]

Am i right?

[frank parrot]

yea

but thats an isoceles

i think this is more accurate

[frank parrot]

it should only be one angle formed

[rough matrix]

Wdym

How do i find height

Also

Thas first step

[frank parrot]

oh ok then

ok after looking at it more

we can do it like this

there's a formula that allows for you to find the height by getting h and b then doing it so

ab/c

to find height

but to find base we should do the c^2 -a ^2 = b^2

30.41

the height is 5.885

roughly

[rough matrix]

It says uts 6.4

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[rough matrix]

What does <ABC mean?

[shadow tapir]

it reads as angle ABC, where point B is the vertex of the angle

[rough matrix]

Can C be on left side or right?

Does it matter?

[split ether]

<ABC and <CBA should denote the same angle, yes

[rough matrix]

Also they all add to 34 degrees

U dont understand that

[split ether]

Show context

[rough matrix]

Draw a diagram of this triangle: ST: 18, TV: 14, <SVT: 36 degrees

[devout shale]

always be clearin

[rough matrix]

?

[zinc crown]

@devout shale what to do?

[devout shale]

ABCD - always be clearin doubts

ABC = always be clearin

[rough matrix]

What

Wdym

[devout shale]

I believe that is what it means

[zinc crown]

🗿

🔠

[rough matrix]

Bro help

[zinc crown]

Does it want you to just sketch a visual or to find the last side?

[devout shale]

What's your doubt? My sir ultimate chad will clear it

[rough matrix]

Draw it

[zinc crown]

Can you send the full question description

Do you have like a ruler / compass too?

[rough matrix]

It just says draw sketch and see how many triangles can be made

No compass or ruler

[zinc crown]

Ah

Have you tried drawing the first two sides

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[eager mica]

hey

[stoic dune]

It's far less clean

Try putting the mods on one side, everything non-mod on the other, and squaring.

You'll need to square again after

[eager mica]

what do you mean by this?

let’s say
| 3x+5 | +5x -7 = | 6x-6 | + 7x + 5
bring mod to one side, rest to the other

| 3x+5 | - | 6x-6 | = 2x + 12
square both sides
( 3x+5 - (6x-6) )^2 = (2x+12)^2
= (11-3x)^2 = (2x+12)^2
= 121 - 33x - 33x + 9x^2 = 4x^2 + 24x + 24x + 144

then i just solve for x through factorisation right?

this’d only require squaring once tho

[stoic dune]

(|a| + |b|)² ≠ (a + b)²

[eager mica]

Oh

how then

[safe radish]

@eager mica Has your question been resolved?

[hollow crest]

To solve x

U need to take constraints

For X>1 you can remove the mod

And solve the equation

If u get X as above 1, then the X is a solution for your equation

Next condition is X> -5/3 and X<1

Do the same and if the value of X u get is between -5/3 and 1, then this value is a solution for equation too

@eager mica

[stoic dune]

The idea I had in mind is that, by expanding:
(|a| + |b|)² = a² + b² + 2|ab|

Which is one less mod than before. We can follow that up as if it were a one-mod equation

But this is really, really not ideal

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[opal moat]

im so lost

[opal moat]

what do they want

i tried (-e^u + C) and (-e^u) aswell was alsoo wrong

[indigo wave]

Is that the full question

[opal moat]

OH sorry LOL

Got everything else but idk what it wants

I tried 1/3,1 but that didnt work either

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[opal moat]

.close

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[solar briar]

another stupid question but

[solar briar]

how do i visualize a plane?

like how do i plot this in a 3d graph?

if i plug in x i still have 2 variables to solve

am i overthinking something? it’s like 1am and my brain is hardly functioning

[grim plover]

probably open something like desmos and geogebra if u need help to see it

visualising in ur mind...that can only come with practice

[solar briar]

so i can’t graph it without a software

[grim plover]

u can graph it in ur mind if u have seen enough 3D graphs

[solar briar]

i can’t like plug values in i meant

actually wait

can’t i plug x=0 and say y=0 and z=20

y=-20 and z=0

[grim plover]

that would be saying that

at 0,0

the plane is 20 units above the x and y axis

[grim plover]

sure

[solar briar]

so i can kinda visualize a plane then if i get uh

3 points?

[grim plover]

well getting a few more points might be better

but i mean

in this example its obvious u get a linear plane

[solar briar]

i’m only working with linear planes atm

[grim plover]

yeah sorta like that

[grim plover]

well once 3d-graphs get more complicated u will need additional points

[solar briar]

i hope i don’t see any of that in this class 😬

i think that’s for upper calculus classes

alright thanks

[grim plover]

heres an example

[grim plover]

probably

[grim plover]

np

u can close ur channel if u dont any further help

[solar briar]

.close

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[leaden ember]

i need help

[leaden ember]

sorry i dint see there was anothrt back side

<@&286206848099549185>

[pale river]

question?

[leaden ember]

4,76 x 10⁷

4123 x 3⁸

[split ether]

Those are not questions

[leaden ember]

what

[versed wave]

!xy

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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.

[split ether]

Do you have a question?

[leaden ember]

ye its writed in pen

[split ether]

Show it

[leaden ember]

k

[split ether]

Those are not questions

[leaden ember]

what is it

?

[pale river]

what do you wish to find out

what do you need help with

[split ether]

Some numbers

[leaden ember]

the awnsers

why is it writed

im so anoyed

let me ask my teacher

brb

i quit im to dumb for math

bye

[pale river]

nooo

never give up

[leaden ember]

nah im sick i dont fell goo dmy lungs burn

ty for the help

.close

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[sick cipher]

The prime number of them are 17 and 41 so 41-17=24

[lean otter]

cant i make it like
17-41

[sick cipher]

Difference is usually positive

[lean otter]

why not negative

[sick cipher]

It's the math

[lean otter]

no

[sick cipher]

Why can't you just add when they give a '-' sign

[lean otter]

sh

it can have a negative diff too

[sick cipher]

May be

[stuck crest]

u can think of difference as like the distance between them in a number line

if ur numbers are 5 and 10 for example starting from 5 and going to 10 is the same distance as starting from 10 and going to 5

[lean otter]

5 units in both ways

[stuck crest]

yes

thats the difference

[lean otter]

why is distance always positive btw like

????

[stuck crest]

cuz u cant travel a negative distance

[lean otter]

-5 eould mean going to the left direcrion to 5 units

[sick cipher]

You are adding direction to it making it a vector or displacement

[stuck crest]

distance doesnt take direction into account

[stuck crest]

yes but the distance is still 5

because u still travel 5 units

[sick cipher]

Distance means without the direction

[stuck crest]

make sense? or confusing still

[lean otter]

displacement is distance + direction right

so therefore distance can be negative

and so

the difference can be 17-41

[lean otter]

look

@helpees

<@&286206848099549185>

@stuck crest

comebac

[lean otter]

why cant this be reffered to displacement

[stuck crest]

nooo

distance is how far you traveled

displacement is the difference from the final position to your origin in a straight line

take this number line right

if you start at 0

[lean otter]

ye

[stuck crest]

and walk 5 units forward and 5 units back

[lean otter]

mhm

[stuck crest]

your distance is 10 units

[lean otter]

i got it

now

t

[stuck crest]

np

[lean otter]

okbye

.close

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[loud ember]

What is least area of line BP when area of triangle inside of trapezoid is less 32 and side lengths are 7 and 5 but the 2 points of triangles are on edge of trapezoid ABCD

[safe radish]

@loud ember Has your question been resolved?

[safe radish]

@loud ember Has your question been resolved?

[loud ember]

Also line AB=7

DC=5

While P is between B and C and line BC= 12

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@loud ember Has your question been resolved?

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[warm moss]

I don't know how to use this server, first time

anyway, the limit is either sqrt2 or sqrt3 ( it's a multiple choice question, and I've reduced the others)

[safe radish]

@warm moss Has your question been resolved?

[brave wolf]

Interesting question

[warm moss]

if you put it in desmos and calculate it to a high value you do get sqrt 3, however i don't know how you get there

it's not as simple as a polynomial over another polynomial, since you get infinity to infinity over infinity to infinity

which is an indeterminate form

[brave wolf]

Ann just proposed a quick solution, if you know stirling approximations, you can just rewrite it in terms of double factorials and then apply those approximations

[warm moss]

hmmm, i've heard about stirling aproximations, however i've stood away from them since they generally really complicate problems from the book i'm working on

i wondered if i could somehow reduce it to a riemann sum

[brave wolf]

that sounds like an interesting thing to do

by applying logarithm?

[warm moss]

yeah

[brave wolf]

imma try that too

(n+1)/n is just 1 + 1/n, so it might work

wait

[warm moss]

however, the major problem with that is that it does exactly opposite of what you want

[brave wolf]

but then it will be 1 + 1/(n+2) etc

[warm moss]

+2, but yeah. . .

[brave wolf]

yes, right

[warm moss]

the problem as a whole is a gross oversimplification of the bigger problem

since i've got an exam in a couple of days, and idc if it doesn't work in general, i need to solve it on a multiple choice basis

ill put the bigger problem here as well, i've got a solution on that, but it's honestly way to long to do it in an exam, so i'm guessing it could be easier

[brave wolf]

whats the solution?

what method did they use

[warm moss]

too much to explain in a message, can i just send the link?

[brave wolf]

sure ig

also you are pre-uni and this will be in exam?

[warm moss]

not this exactly

but yeah pre uni

the exam to get into said uni

but it's known to be one of the hardest math exams in the country at a pre university level

[brave wolf]

ig i can see why

[warm moss]

https://www.youtube.com/watch?v=aB7xDuV2uyc&t=5s

[brave wolf]

ah so loging it

i mean applying logarithm

quite an insane solution though

[warm moss]

well yeah, i understand that solution

however you get 3 hours for the whole exam

and you get 30 questions, 7 of which are at this difficulty

so i'm guessing there's a shorter, more elegant solution

since you really don't have time for all of that

[brave wolf]

that might be through stirling approximations

[warm moss]

hmmmmmm

maybe, i'll give it a try

[brave wolf]

the top and bottom can be easily rewriten through double factorials, and then some stuff should cancel nicely

(after applying stirling ofc)

[warm moss]

yeah, i see what you mean, idk how stirling aproximation works though, but i guess i'll read up on it

[brave wolf]

it approximates it for large values of n

as n gets larger and larger, the approximation is more and more accurate

so as n tends to infinity, you can just replace it with the approximation

[warm moss]

ok, ill try this, thanks a lot

[brave wolf]

np, good luck with that exam

[warm moss]

.close

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[native sluice]

<@&286206848099549185>

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[native sluice]

<@&286206848099549185>

Bro

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[versed thistle]

Im not sure how to interpret this

this is my work :3

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[versed thistle]

<@&286206848099549185>

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@versed thistle Has your question been resolved?

[versed thistle]

:(

.close

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[junior smelt]

. you may potentially have better luck in the #probability-statistics channel maybe
[sorry I'm not sure myself for these ]

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[versed thistle]

ty for the suggestion ^-^

[versed thistle]

oopsie

.close

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[proud sinew]

hello, im struggling with this ODE question i dont even know where i should start on this. it would be great if someone can show me how to tackle a question like this, tks!!!

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[proud sinew]

@raven vessel

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[thorn flare]

Shouldn’t 4- be 0

41 be 0

Why is it 1/4

[safe radish]

Show your work, and if possible, explain where you are stuck.

[high gate]

use log rules to group the ln

and then you can apply lhospital

[tardy mango]

(Alternatively, recognise this is the derivative of ln x at x=4)

[eternal carbon]

better way of thinking abt it ye

[thorn flare]

Oh u don’t plug in 4 as a constant initially?

[tardy mango]

?

wdym

[thorn flare]

Like u don’t straight up do dx ln4?

[tardy mango]

no...?

[thorn flare]

For 28 what am I doing wrong

[tardy mango]

idk what you're doing tbh

you should use logarithmic differentiation

[thorn flare]

I’m using the exponential detiartive rule

[eternal carbon]

in x^(2lnx), x is not a constant

so that rule does not apply

[thorn flare]

Oh that’s only for constants

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@thorn flare Has your question been resolved?

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[ionic oxide]

Essentially, the question asks to find the number of possible paths a ball dropped in the leftmost top track can take to the bottom of this sample Plinko board. I was able to use Pascal's triangle to deduce the number of possible paths the ball can take when dropped from the center track, however I'm having trouble finding a way to calculate the number of paths from the top left.

[ionic oxide]

This is how far I got, however I'm not very confident that this is right 😅

[steel stag]

that's all right 👍

[ionic oxide]

oh? wouldn't have thought lol

my professor told me that some modifications were needed beyond just using pascal's triangle/method but I'm confused as to what he meant by that

[steel stag]

yea it's called pascal's method it's legit

[ionic oxide]

because this feels like a similar thing

[steel stag]

stuff like this

[ionic oxide]

ah I see

just confused by how he worded the question then or what he meant by "modifications to Pascal's method"

anyways, thank you so much for the input! 🙏

.close

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[somber cape]

is this enough to prove the symmetric property?

[fair fossil]

bro i have a dout

[versed wave]

! occupied

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[faint seal]

What you need to prove for symmetry is $$x^2\equiv y^2 (mod 4)\iff y^2\equiv x^2 (mod 4)$$

[flat frigate]

kheerii

[faint seal]

You can prove any one side of this

Prove that $$x^2\equiv y^2 (mod 4)\implies y^2\equiv x^2 (mod 4)$$

[flat frigate]

kheerii

[somber cape]

Oh ok

[faint seal]

We can say that $x^2=y^2+4k$ for some integer k, which makes $y^2=x^2-4k=x^2+4(-k)$

[flat frigate]

kheerii

[somber cape]

Oh ok I see

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[weary osprey]

can you check if this steps are correct?

i'm getting $\frac{n}{p_n / \ln p_n} = \frac{n \ln p_n}{p_n}$

[flat frigate]

Dubs

[weary osprey]

but they have written the reciprocal of this

this is the intial page

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[solar hazel]

yea it's ok, if a sequence tends to 1 so does its reciprocal

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[umbral swan]

F7 represents a finite field of 7 elements right

[umbral swan]

and F7 doesn't have to be a subset of N I'd presume

[mortal sandal]

Yeah

[umbral swan]

kk thank thees :]

[mortal sandal]

And yeah it's kind of abstract though if you were to pick a specific set it would probably be residue classes mod 7

[umbral swan]

kk

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[zinc bluff]

need help calculating the limit of xarctan(1/x) when x goes to infinity I know its 1 but i cant get past the indetermination of inf0

[pine horizon]

convert it to zero by zero form

then apply l'hopital

[zinc bluff]

how do i do that

i cant find a way to correct it

[pine horizon]

xarctan(1/x) -> arctan(1/x)/(1/x)

the x in multiplication can be written as 1/(1/x)

[zinc bluff]

ooooooooooooh yeaaaaaah TY

forgot it could be as simple

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[smoky trellis]

Hello I am 9th standard i want to know why does an irrational number never end means why doesnt it stops anywhere.

[manic path]

it doesn't stop anywhere because you can infinitely calculate its remainders

do me a favour and try to divide 2 by 3 by any method you want

[smoky trellis]

yes i have tried that and it just keep going

[manic path]

hence thats why its called irrational

it just goes on and on

[civic loom]

or 22/7

[smoky trellis]

oh thanks for your help

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[thin bridge]

2/3 isn't irrational though

incomplete definition

irrational numbers goes on without repetition

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[hoary wind]

-4ac

a=e, c = -2

-4 * e * -2

Because $e^{2x+1} = e\cdot e^{2x}$

[flat frigate]

TayBee

[pliant egret]

there is u^2*e

so a = e

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[bitter steppe]

why's this not e^2x/2x

[velvet flax]

d/dx e^(2x)/2x = (2e^(2x)*2x - 2e^(2x)/(4x^2)

doesn't look like e^(2x-1) to me

[bitter steppe]

wym

[velvet flax]

try differentiating $\frac{e^{2x}}{2x}$

[flat frigate]

Ham

[bitter steppe]

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[rugged harness]

can someone help me with this exercise?
Task:
Consider the linear transformation (f: \mathbb{R}^4 \rightarrow \mathbb{R}^2). The general solution of the resulting inhomogeneous linear system of equations (LSE) is:

[x = \begin{pmatrix} 2 \ 4 \ 1 \ 1 \end{pmatrix} + \lambda \begin{pmatrix} -2 \ 0 \ 4 \ 3 \end{pmatrix} + \mu \begin{pmatrix} 8 \ -3 \ 1 \ 2 \end{pmatrix} + \nu \begin{pmatrix} 0 \ 2 \ 0 \ 1 \end{pmatrix}]

where (\lambda, \mu, \nu \in \mathbb{R}).

Evaluate whether this solution could indeed be a solution, and provide a detailed explanation.

[flat frigate]

skenox

[rugged harness]

I don't really know how to solve it, maybe I need to prove whether the vectors are linear independent or not to say whether it is a general solution or not?

[whole forge]

I don't really understand why there would be a resulting inhomogeneous LSE, where does it come from? is there anything that is missing from the question?

[rugged harness]

this was 4a) we also have 4b):

Task:
Consider the linear transformation (f: \mathbb{R}^2 \rightarrow \mathbb{R}^4). The general solution of the resulting linear system of equations (LSE) is:

[x = \begin{pmatrix} 2 \ 4 \end{pmatrix} + \lambda \begin{pmatrix} -1 \ 3 \end{pmatrix} + \mu \begin{pmatrix} 5 \ -3 \end{pmatrix}]

where (\lambda, \mu \in \mathbb{R}).

Evaluate whether this solution could indeed be a solution, and provide a detailed explanation.

[flat frigate]

skenox

[rugged harness]

but it's pretty similar hm

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[somber ridge]

Hi all - a bit of a puzzle question for you. Mainly looking for ideas of next steps. I have an image here which I'll attach. I'm trying to generate additional GPS coordinates that I can plot on a map that should lead to a solution location.

[somber ridge]

The starting location is the Statue of Liberty (40.689247,−74.044502) and a Lucas spiral and series are shown. I'm unsure what the " + 3.75 degrees" part indicates, nor the symbol above that

But I imagine there is a formula I can use to generate additional coordinates

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@somber ridge Has your question been resolved?

[somber ridge]

Ok so it seems the phi symbol represents the golden ratio. Still digging into what the (δ + 3.75°) might be

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@somber ridge Has your question been resolved?

[meager merlin]

@somber ridge

What is this for?

Generally, u can try convert Cartesian Coordinates to Polar Coord

[somber ridge]

It's for a treasure hunt I am following. I think I came up with some code to help me generate additional coordinates along the spiral but hard for me to know if they are correct

In particular I'm just not sure about the 3.75 degrees note at the bottom

[meager merlin]

Not sure if Im wrong but the line doesnt line up the golden ratio

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[scarlet comet]

What are all the elements of Z[x]/(3, x^2+1)?

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[granite timber]

I'm having trouble with trig

[granite timber]

The question is sinxtanx=-sinx Solve over the interval [0,2pi)

[pine horizon]

make rhs zero

then sinx common

you'll get 2 factors which can be zero

[granite timber]

how do i know which ones equal zero

[pine horizon]

sinx(tanx+1) = 0

solve for sinx = 0 and tanx+1 = 0

[granite timber]

ok i got it i just had trouble finding the tan

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[surreal idol]

Am I understanding transition matrices correctly?

[tranquil ore]

I don’t know what transition matrices are

But the problem says b loses 30% to a and c loses 20% to a

So the first row should be 0.8 0.3 and 0.2

[surreal idol]

ohhh myy godddd thanks

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[river niche]

hi how do I convert this into a table?

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[river niche]

anyone?

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[umbral cedar]

just take x and y values

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[surreal idol]

The top equation is a typo(not formatted correctly)

So basically, I need to find the maximum value of the second derivative which is pretty difficult. I don't know where to start

[surreal idol]

Differenting that, finding stationary point is too ceebs

I looked onto desmos, the maximum lies within 1 and 1/10

Is there an easier way?

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@surreal idol Has your question been resolved?

[lean otter]

U want to find the max of d^2 f/dx^2 at [1/10, 1]

that should be reasonable enough to do no?

its the exponential function so

[surreal idol]

with chain rule and x terms in the denominator it gets clunky

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[pine horizon]

what is v

[lean otter]

the angle

between a and b

are you trying to prove the area of the parallelogram being the magnitude of cross product

[lean otter]

that it is equal to the deter determinant of a and b or lengt of a and b * sin (v)

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.close

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[restive epoch]

hello i was watching a video on resistivity and conductivity and there was this problem, and i do not understand how to do articulate the final number? trying it on a calculator gives me syntax error or i might just be stupid

[lean otter]

articulate? as in put it in a nicer form?

[restive epoch]

understand how to find the number if say the lenght of the copper wire is 42m

[pliant egret]

do you mean in scientific notation?

[restive epoch]

nah

decimal noitation

[thin bridge]

still not sure what your issue is
and how your calc is giving you an error

[restive epoch]

im probably writing it wrong

[thin bridge]

can you take a pic of your calculator input

[restive epoch]

i should write it in mixed fractions right

[thin bridge]

no

[restive epoch]

then?

[thin bridge]

this is a product

$\rho \frac LA = \rho \times \frac LA$

[flat frigate]

ℝαμΩℕωⅤ

[restive epoch]

like that

oye

i got it

ty

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[balmy sky]

should the answer be -1 or -4

if i take k = -1, i get x < 2
if i take k = -4,i get x < 3

[raven hawk]

then its -4

[balmy sky]

why

@raven hawk

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[balmy sky]

<@&286206848099549185>

[willow herald]

and according to the 2nd ineqality 4x-1<7, x<2

[balmy sky]

yupp

[willow herald]

-1 should be right

is it

[balmy sky]

i dont know the correct answer

the answers aren't given

[willow herald]

-1 is correct

[safe radish]

@balmy sky Has your question been resolved?

[balmy sky]

why

[willow herald]

okay now on a second look

x<2 right

lets say x = 1.9999999999999999....999

which is approximately 2 but not =2

for this x,
-2-(approx 6) < k < 5- approx(6)

value of k lies between -7.999..99 to -0.99..99

now -1 and -4 left huh

smaller of these is -4

[balmy sky]

i guess your explanation is not quite right

if we choose k = -4, we get x < 3 and we get x < 2 if we choose k = -1

now the set of solutions for x < 3 is (-inf, 3) and since we know that x < 2 which means the solution set is (-inf, 2), then (-inf, 2) is a subset of (-inf, 3) so the answer should be -4

but what if the options weren't given

can't we push k to -infinity

there won't exist any least possible k if the options were not given

<@&286206848099549185>

<@&286206848099549185>

[wheat scarab]

Write the first inequality only with x in the middle

[balmy sky]

(-2 - k)/3 < x < (5-k)/3

[wheat scarab]

Yes

Now compare it with the inequality you get from the second equation

[balmy sky]

x < 2

so (5-k)/3 >= 2

?

[wheat scarab]

Yes

[balmy sky]

or should it be just (5-k)/3 = 2

because i think that it could be greater

the solution set would just be a superset of x<2

[wheat scarab]

Yeah that's not what they're asking for here

But that is the most appropriate inequality

[balmy sky]

i just wanted to ask that what if the options were not there

then would there exist any least possible k

[wheat scarab]

Yeah

[balmy sky]

how

[wheat scarab]

You are still getting the least value of k here

[balmy sky]

this would result in k <= -1

there is not least value in this case

it would approach to infinity

we won't get a definite value right?

[wheat scarab]

Yeah that's why you can't consider supersets here

[wheat scarab]

You're right, my mental calculation went wrong

[balmy sky]

what do you mean

lets say i take k = -1

[wheat scarab]

You'll have to consider this

[balmy sky]

but what if we take k = -4

it would give us x < 3

and it is in the options

so would it be correct to choose k = -4, considering the given options?

[wheat scarab]

It's not satisfying the second inequality

[wheat scarab]

No, you need to consider the intersection of those inequalities

[balmy sky]

oh so here comes the point about "supersets"

they don't want the new inequality to be the superset of x<2

they want it to be exactly equal?

[wheat scarab]

Yes

[willow herald]

could be

[balmy sky]

is the language of the question confusing or am i just too dumb

[willow herald]

it is

i also thought the same

its just not framed quite right

[wheat scarab]

I didn't read the whole thing
I just saw the inequalities

[balmy sky]

hmmm

[balmy sky]

so what do you think now

its still k = -1?

[wheat scarab]

Still seems pretty clear to me😅

[wheat scarab]

Yes

[balmy sky]

guess i thought too much about it

😭

thanks to both of you

[wheat scarab]

Yeah that's probably it

No problem

[balmy sky]

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[balmy sky]

@wheat scarab you had smth to say?

nvm

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[cloud rock]

How do I solve for the exponent x?

[burnt goblet]

Try using logarithms

[cloud rock]

is there a way without using log because my teacher hasnt taught us it yet

he said thats gr 12 stuff

[burnt goblet]

You can factorize 76

75

[pliant egret]

in that case you wanna have same base on both sides

note that 75=3*25

[cloud rock]

yeah thats what i was taught

what do i do for the exponents then

[pliant egret]

so you have 3* 5^(x^2-x)=3* 25

[cloud rock]

because i need to solve for x

[pliant egret]

do you see how to proceed?

[cloud rock]

do i factor 25?

[pliant egret]

you can divide both sides by 3 first

[cloud rock]

ohhhhhhh

[pliant egret]

and 25=5^2

[cloud rock]

thank you i get it now :D

[pliant egret]

np

glad i helped

[cloud rock]

wait for the exponent i do x^2 - x = 2?

i was going to square root it but idk if its right

or maybe i do quadratic formula

[pliant egret]

quadratic

and obviously take everything on one side first

[cloud rock]

yea

alr

thanks again lol

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[dim ravine]

i cant find the angle to convert to polar how is this supposed to be simplified

been stumped on this question all day this is my last hope

supposed to be solvable without calculators

help would really be appreciated 🙏🙏

[dull sequoia]

Perhaps try let the question = a + ib

And then square both sides and compare real and imaginary parts to solve for a and b

[dim ravine]

ohh havent thought of that

ima try it thanks

[dull sequoia]

That’s usually the trick for these “what’s this complex number equal to?”

[dim ravine]

no way it worked 😭

ur a life saver thank u so much 🙏🙏🙏

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[strange portal]

how do i do this (either with mean value theorem or taylor's theorem)

[obtuse jackal]

I suspect a typo, as the LHS should be n y^n-1
Otherwise it fails when x - y -> 0

[strange portal]

but 0 < y < x so x-y wont tend to 0 right

[obtuse jackal]

it can
you're just saying x - y > 0
I can take values of x (with fixed y) such that x-y -> 0

[strange portal]

oh okay yeah mb

[obtuse jackal]

this is clearly supposed to be symetric, as then it states that the rate of change is bounded between the extremal derivatives, which amounts to convexity of a clearly convex function

[strange portal]

this makes sense

even if it was just ny^n-1 how would i prove it

[obtuse jackal]

did you study anything involving convexity?

but honestly here you don't need it, the MVT is enough since you easily prove the derivative is increasing

[strange portal]

all i know is the derivative is increasing for convex functions

[obtuse jackal]

you have all you need
Put it together

[strange portal]

okay i got it

thank you sm

.close

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[olive acorn]

Hey guys, I'm supposed to do the following:
(Determining the real and imaginary parts) Determine all complex numbers z = x + iy with −x + 4iy + 3ix − 2y = −6i + 4

[olive acorn]

Honestly I'm not even sure how and where to start. Am I supposed to just rearrange the equation to fit z? are there more than 1 complex numbers

Sorry if the wording is not the best, I'm translating

[inner parrot]

group the real and imaginary parts together

and set those parts equal to the real and imaginary parts that we are given

wait are you trying to find what x and y are or is it something else

[olive acorn]

honestly, no idea

the wording is odd even in the original langauge

[inner parrot]

damn

let’s just assume it wants what x and y are

and we’ll just plug it back into z ig

[olive acorn]

alright, so by grouping you mean I should first rearange so I have -x + 3ix + 4iy -2y = 4 + 6i?

this way I'd have the real parts with x and the imaginary party with y and i seperated

[inner parrot]

like rearrange the left hand side so that the i terms are together

then we can try to match the coefficients

and solve a system of equations

[olive acorn]

so for instance I'd do +6i to get rid of the -6i on the left?

and get the i to the right=

and then solve for x?

I'm like a fish out of water here

[flat frigate]

y0shi

[inner parrot]

now we can set the real and imaginary parts equal

so like -x-2y=4 and 4y+3x=-6

[magic tendon]

3x+4y = - 6
-x-2y = 4

Multiply the one below by 2

Then add them together

You get x = 2

And find y

[olive acorn]

so we can just drop the i?

[magic tendon]

In complex numbers, the real part and the imaginary part are like water and oil

So
3ix + 4iy has to be equal to -6i

[olive acorn]

Im not getting it. I think I'll call it a night and try again tomorrow. Getting late.
Thanks for the help anyways guys

.close

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[tender dawn]

hello

[tender dawn]

how do you derivate 2xy in terms of x

so d(2xy) / dx

y is variable not constant

[inner parrot]

is y a function of x?

[tender dawn]

yes

[inner parrot]

we don’t know what that function would be, so we would have to use something called implicit differentiation

essentially we have to use the product rule here

since we have a product of two functions, x and y

[solar hazel]

eh…

[tender dawn]

oh yeah right., but the product rule only works because y is in terms of x right?

[dull sequoia]

it'll work even if y is constant wrt x

the dy/dx terms will just become 0

[tender dawn]

alright

thanks

@dull sequoia btw midnight is the best album

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[dull sequoia]

D:

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[lean otter]

Is this asking where y is decreasing exponentially (slope) and x is positive

I mean

Wait I worded that wrong

Ok whatever

Yea how do I do this

Can someone explain

I’m dumb sorry

Wait is it C

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@lean otter Has your question been resolved?

[lean otter]

<@&286206848099549185>

[leaden light]

C is decreasing, but it's rate of decreasing is slower than D

Graph D is decreasing at an increasing rate. So "over time" it's decreasing rate increases

[lean otter]

Okay I see

Thanks

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[orchid lake]

need help with finding the taylor series inequality (b)

[orchid lake]

tried to do t4 of x with x = 1.2 but it is wrong

Am i supposed to do something with the x value range?

(i already know how to do (a), this is just a 2nd try)

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@orchid lake Has your question been resolved?

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[simple oak]

The term with x^5

What is 2a raised to

If x is raised to 5 then 2a is raised to 8-5

[warm rain]

Idk what ur talking about it

[simple oak]

Binomial theorem

[warm rain]

How do I apply it In this context

<@&286206848099549185>

[rustic goblet]

if you know the relationship between the coefficients of the binomial expansion and Pascal's triangle, you can use that here

[simple oak]

This is eighth row and fifth term

Coefficient 8C5

Of x^5

Multiply by (2a)^3

Equal to 56

Solve for a

[warm rain]

Where did they get the 8 and 3 from the solutions

<@&286206848099549185>

For question 3

[simple oak]

Terms from pascal triangle are nCr

[warm rain]

That’s isn’t my question

[warm rain]

^^

[simple oak]

8

Is the n

Or the row

The exponent of the thing being expanded

[warm rain]

Ok then how did they get 3

[simple oak]

Because if we know one term is x raised to 5

Other term must be raised to 3

Or the third term of the eighth row

[warm rain]

How

Do u know that

[warm rain]

This

[simple oak]

That’s how the binomial expansion work tho

It has to add up to n

Like

[warm rain]

Can u explain how u know it’s 3

[simple oak]

If (x+2a)^8

And we know one of them is x^5

Then the other term must be raised a degree such that it will add to 8 (the total)

[warm rain]

Does this apply to every binomial question

[simple oak]

Yh

[warm rain]

?

[simple oak]

Yes

[warm rain]

Let me think

I don’t get it

I still don’t get how u know the other term must be raised to 3

Maybe I should,d watch some YouTube videos

[safe radish]

@warm rain Has your question been resolved?

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[glossy cypress]

so i found this question from the gcse online and my brain is not functioning

[glossy cypress]

i have

zero clue how to solve it rn

I was trying to calculate the area of either circle outside of B, but it's impossible because it first needs me to calculate the intersection area

[quiet plume]

You can consider just the area between two of the circles. Then the shaded area is the area of the middle circle - twice the overlapping circles.

For computing that area, you can consider that, and then notice that you can compute the area as one of the yellow portion I've highlighted. Then, considering just one of the circles, we can see that the area of the red sector - the area of the green triangle gives the desired area.

Units are wrong here though, the hypotenuse should be 4 and the bottom leg 2

[glossy cypress]

pardon my stupidity, but how exactly is it obvious that the circles intersect at exactly half the radius?

wait nevermind i am dumb as fuck

okay so

my problem here is that computing those triangles still includes 4 of those little wedges

[inland monolith]

you could also just do the area of the 120* sector

it's the same thing but very slightly less work

[glossy cypress]

no i see that but i still dont know how you'd go about calculating these

[quiet plume]

The yellow part contains this

[inland monolith]

you don't

[quiet plume]

If you add it up for both circles the in between is in there

[inland monolith]

you need the yellow area

[slender sable]

Isn't this section's angle 90°?

[inland monolith]

then your total shaded area will be one circle area - 4(yellow)

[inland monolith]

the entire red sector is 120°

[slender sable]

About the angle of the triangle, i mean*

[inland monolith]

yes it is a right triangle

[slender sable]

why is it 60° then

[inland monolith]

A different angle is 60°

[glossy cypress]

so the area of the left half of the kite thing is equal to the base of one of the right triangles (2) times the height. tan60 = sqrt3. sqrt 3 times 2 = sqrt12. so we have the area of the left half of the kite is equal to sqrt48 (i think?)

[lean otter]

You don’t need to do that, actually

[glossy cypress]

if my math is right, i can subtract this from the 120 degree sector which has an area of 16pi/3. so i get the area of all the intersections is equal to 4(16pi/3 - sqrt(48))

actually nope

thats

wrong

it's negative

wait

no its not

[inland monolith]

16π-4(16π/3-4√3)

[inland monolith]

that looks right

[glossy cypress]

oh wait im not wrong? lets fucking go?

[inland monolith]

lol

,w simplify 16π-4(16π/3-4√3)

[flat frigate]

Results provided by WolframAlpha

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Requested by praneel.g

[glossy cypress]

48pi/3 - 64pi/3 + 16sqrt(3)
16(sqrt(3) - pi/3)

?

[slender sable]

after drawing the triangle i'd assume if i doubled it's size it'd be a square since triangle's angles are 90,45,45. So half a slope of this square is 2, the slope itself of this square is 4.

[inland monolith]

yeah that works I think

[glossy cypress]

see the problem with this is

i can absolutely do the math

but i cant actually piece together the parts

its like

cool. they interesct and have the same radius

now what

[inland monolith]

no, it's a 30-60-90 triangle

[glossy cypress]

like

[inland monolith]

yeah I've been doing these kinds of problems for years since I do some competitive math

[slender sable]

my monke brain can't comprehend that

[glossy cypress]

forming an equilateral triangle by connecting the radii is something i would probably never wouldve thought of

this feels like one of those questions that you specifically need to plan for

[inland monolith]

well I'd just do this and then try to find the area of one of the slices

which is sector-triangle

[glossy cypress]

i tried watching a bprp video on it and he somehow solved for one of the goofy little wedges

but i had no clue how he got any of it so i just came here

circle hurt my brain

[inland monolith]

✨ practice ✨

these kinds of questions are pretty common in comp math so I'm used to them

[slender sable]

can you explain me how it's a 30-60-90? I don't get it

[safe radish]

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[wary widget]

Anyone wanna give me an intro to Calculus lol

[sleek sentinel]

https://youtube.com/playlist?list=PLZHQObOWTQDMsr9K-rj53DwVRMYO3t5Yr

[wary widget]

Thanks homie

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[night oxide]

given 2 vectors a and b, which of the following statements is/are correct?
the dot product of these vectors is 0 if they are orthogonal
the cross product of these vectors is 0 if they are collinear
the magnitude of the resultant sum of these vectors is maximum if they are anti-parallel
the magnitude of the resultant sum of these vectors is least if they are parallel

[dusk nova]

what do you think?

[night oxide]

i know option 1 is right, option 2 is probably right since the cross product is 0 if collinear or parallel and 3 and 4 im not too sure

[dusk nova]

i agree with you

the 1st and 2nd ones are correct

anyway for the 3rd one, do you know what anti-parallel means?

it's just "parallel" but the vectors point in opposite directions

so hmm why don't you consider two vectors that are parallel and antiparallel

and do a comparison in their magnitude