math-help

[ember kernel]

Anyway

My math Bible for beginners

YouTube
(This is not a joke)

Not so great to learn core things (formulas, applications,...) but very convenient for visual introductions and getting a concept taught to you from another perspective when you struggle with the way your teacher explains things

^

[woven orchid]

eyo anyoene good at physics

cus uhmmmmmmm... me fucked

[ember kernel]

#science-help

[woven orchid]

thank you sensai

[terse niche]

Hi, can anyone help me with question 5.b

[hard swan]

not exactly sure what the meaning of relation is (haven't done this topic in detail), but I believe you might just have to shade the region of points satisfied by x >= y

on an x, y graph

[unique solstice]

Anyone good with Calc 3? I was wondering what are some good websites or YouTube videos that will help me understand the topic.

[topaz hamlet]

maybe khan academy has something related to the stuff you're learning in calc 3? (i'm not in calc but I hope that helps)

or maybe there's some online tutoring websites

[terse niche]

i've figured out, the condition applies to all pairs below the line, so i just shaded it

and yeah you right, thank you

[thick bluff]

Um hi, so what is the function from y of y=x^2?

[lost rune]

@unique solstice ProfRobBob and Professor Leonard. both on youtube

[cyan night]

can you elaborate?

wdym

[lost rune]

if you mean what is the function in terms of y, then it would look like this

[thick bluff]

Thank uuu

[ember kernel]

I'm not American. What's Calc 3 about ? I might have learnt it under a different name

[unique solstice]

It’s calculus 3 which we learn functions of two variables which is described in three dimensional coordinate system. Idk if that made sense but yeah

[ember kernel]

Oh yeah I've already seen that

What do you struggle with ?

@unique solstice

[unique solstice]

Well right now we’re doing double integrals and I’m just having a hard time grasping how to do that.

[ember kernel]

Ckassic 1D integrals, curve/surface integrals, or Divergence/Stokes theorem (if you know about them) ?

@unique solstice

[unique solstice]

Double integrals over rectangles, general regions. I’ll post an example of a question if I stumble upon one later.

[ember kernel]

Don't hesitate to dm me (no friend request needed), I'm quite familiar with the concept

[ember kernel]

Which question ?

[proven orchid]

2 and 3

just need to crosscheck answers

any steps are appreciated

[ember kernel]

3. Curves intersect at x=0 and x=4 (that's your integration domain)

Integral of the first curve = 3x² - x³/3
Integral of the second curve = x²

(Curve 1 > Curve 2 in your integration domain)
Total area of Curve1 - Curve2 in your integration domain :
[(3x² - x³/3) - (x²)] from 0 to 4
= (48 - 0) - (64/3 - 0) - (16-0)
= 32 - 64/3 = (96-64)/3 = 32/3

[ember kernel]

^

[proven orchid]

yeah i got the same overall answer for 3

[ember kernel]

2. follow the same steps :
   -Find intersections
   -Substract the curves (the bigger minus the smaller over the integration domain)
   -Integrate the substraction between the intersections

Except here curve 2 is y = 0

[ember kernel]

^

[echo temple]

anyone know how?

i got 15 as an answer but it was wrong so im a bit confused

[proven orchid]

thank you so very much

[left knoll]

help pls

[civic notch]

Thats blurred oof

? Maybe? I haven't checked it yet tho, so my numbers might be wrong

But im pretty sure the method is right

[left knoll]

ah, thanks!!

rly neat working out btw lol

[cyan night]

what type of quadrilateral is abcd?

[cyan night]

hm im not sure, but there are some triangles that are equal in area such as apd = bpc, bcd = acd, abc = abd

[left knoll]

How do i find the y int of number 5 pls help

@ember kernel

@west halo

[marble canopy]

they ez

[left knoll]

Wahts the y

[left knoll]

Oi whats the y int

[heady egret]

I cant too much hard lol

[marble canopy]

cba

we did that not too long ago

[left knoll]

@left knoll wait i am doing it

Ok

Thanks

[left knoll]

Cap

[ember kernel]

1. Stay patient
2. Y-intercept = the value of y when x = 0, try substituting value 0

[marble canopy]

How bout, you learn in class eh?

[left knoll]

What about x

[oblique current]

Use this app ”photomath”

[regal oxide]

its the opposite

[left knoll]

Wasnt paying attention

[marble canopy]

when y = 0

[left knoll]

simple quadratic put y=0 then calculate for x

[ember kernel]

x-intercept = root of y = the value of x when y = 0

[marble canopy]

factorise it first tho

or that's what we did at least

[teal tree]

anyone can help me with one question about indefinite integral? (my english isn't good)

[marble canopy]

ah yes we did that, we just didn't use the word polynomials

[ember kernel]

Sure go on

[teal tree]

i have f(x) = 1/x^2, and i don't understand why the answer is -1/x + k

[marble canopy]

Because you did it wrong.

@left knoll how old are ya

[teal tree]

what i saw this in the list

[left knoll]

?

[marble canopy]

H o w o l d a r e y o u ?

[ember kernel]

Ok so use the power rule : integral of x^p = (p+1) * x^(p+1)

Here, you have p = -2

(Because 1/x^2 = x^(-2) )

Therefore, integral of x^(-2) = (-2+1) * x^(-2+1) = -1 * x^(-1) = -1/x

[left knoll]

oh you did it too...

[teal tree]

nicee, i didn't put x/t on top

thanks

[vernal scaffold]

What’s meant by a locus

[ember kernel]

Is there context ?

[vernal scaffold]

Circle

[spice basin]

Does anybody know why is this sequence always bigger than zero? I mean technically it’s right but it seems that it always equals 2 so I don’t know why would my teacher use 0 as the limit of this sequence

*is equal or higher

[ember kernel]

Is it for sequence convergence ?

@spice basin

[spice basin]

I don’t think so, judging by my calculations it’s always going to be 2 if we start from n=1 and use only natural numbers

[ember kernel]

What I mean is : are you analysing the sequence to determine whether it converges/diverges ?

[spice basin]

Yes

Sorry I misread

[ember kernel]

Then, it doesn't really matter. You're trying to prove the sequence has a lower bound to prove it can't diverge towards the negative values

You can use 0 or 2 as the lower bound, the main point is that the sequence has a lower bound

[spice basin]

Ah okay, i though that i have to be 100% precise, thanks

[ember kernel]

Depends on the context of course, but that doesn't seem to be the case here

[flat badge]

can someone please help me with the 1.20 and 1.21 T-T

thank you

[left knoll]

Hi guys

Can someone explain to me why this ellipse has tangent points (-0.5,0.5) and (-0.5,-0.5) with respect to the Triangle

Sorry for my english

P is the center of the ellipse

[left knoll]

W H Y?

[marble canopy]

@left knoll

[left knoll]

?

I mean
I need to know how to find that tangent points

[marble canopy]

I was deadass going to help you

My parents woke up at 02:30, I thought it was 05:30, now I’m head back to sleep

[left knoll]

What does this mean?

Expand (x+y)(m-n) how can I do this?

[cyan night]

(a+b)(c+d) = ac+ac+bc+bd

you distribute each term on the first parentheses to each term on the second one

[left knoll]

So it will be xm-ny?

[cyan night]

no

you distribute x with both m and n

[left knoll]

ok let me do it

[cyan night]

(x+y)(m-n) = x(m-n)+y(m-n)

= xm - xn + ym - yn

[left knoll]

I think I got it

So you did a(c+d)+b(c+d)

[cyan night]

yes

[left knoll]

Yes! I got it thanks

So is the answer of (y-3)(y-8)=y^2-8y-3y+24 =y^2-11y+24

I am so poor in math....I don't know when I will be an expert in math

😔

[cyan night]

yes this is correct

[tender breach]

hello

had a trig question

[left knoll]

Anyone can help me with (x+3)^2=4

Anyone?

Please someone help me

[cyan night]

the the square root of both sides

=> x+3 = +-2

+- implies either +2 or -2

so x = -1 or -5

[left knoll]

what will be it in steps?

[cyan night]

what

[left knoll]

(x+3)(x+3)=4?

[cyan night]

no

take the square root of both sides

[left knoll]

(x+3)^2=4^2

[cyan night]

2^2*

its either 2^2 or (-2)^2

[left knoll]

I am still confused..

[cyan night]

ok so

(x+3)^2 = 4

[left knoll]

yes

[cyan night]

4 can be written as 2^2 or (-2)^2

right?

[left knoll]

yes

[cyan night]

so there are 2 cases

4 = 2^2 or (-2)^2

case 1 is 2^2

so (x+3)^2 = 2^2

or

x+3 = 2

hence x = -1

case 2 is (-2)^2

you solve it

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

[left knoll]

x+2=-2^2

[cyan night]

uh

uhh

[left knoll]

x+3 =-2^2

[cyan night]

(x+3)^2

[left knoll]

sorry

[cyan night]

also remember to use parentheses

[left knoll]

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

[cyan night]

(-2)^2 = 4 because a negative times a negative is a positive

yes

[left knoll]

so x can be positive 2 as well as negative 2

x= -2 or 2

[cyan night]

no

x+3 = 2 or -2

not x

[left knoll]

ohh

[cyan night]

yes

[left knoll]

Thank you very much

[cyan night]

np

[left knoll]

(a-4)^2=25 => (a-4)^2=5x5 => a-4=5

Am I right?

@cyan night

[alpine mulch]

yea thats right

square root both sides to get a-4=5

[cyan night]

you missed one case

25 = 5x5 or (-5)x(-5)

[left knoll]

ohhh yeah

[left knoll]

Anyways why is you name in blue color can I also get colored text.

[cyan night]

go to #roles

[left knoll]

I did I completed all informations.

[left knoll]

could someone help me please

You have to write 3,2 in to equation

I believe it is y=mx+b

okay

So it will be..

waite

dw i got it

thanks

cool

[cyan night]

no

[cyan night]

rewrite it under the vertex form y= a*(x-h)^2+k

(h,k) is the vertex

[left knoll]

Can you explain factorising 6p^4-2q^2

can you sub it for me

[cyan night]

2(3p^4-q^2)

[left knoll]

I get it lol it was easy

[cyan night]

f(x) = a*(x-3)+2

wait

maybe im mistaken the question

f(x-5) implies a shift

shift right 5 units

vertex is (8,2)

please correct me

if im mistaken again

[left knoll]

why x-2

won't it be 6(p^2+2q)(p^2-2q)

when it's -5

[cyan night]

hm?

x-5 = shift right 5 units

[left knoll]

ah nvm

[cyan night]

ok.

[left knoll]

Shouldn't the answer be 6p^4-2q^2

[cyan night]

what

[left knoll]

I means the answer should be 6(p^2+2q)(p^2-2q)

[cyan night]

(p^2+2q)(p^2-2q) = p^4-4q^2

no

6(p^2+2q)(p^2-2q) = 6p^4-24q^2

[left knoll]

yeah it is factorising

[cyan night]

but your question was 6p^4-2q^2?

[left knoll]

factorsie 6p^4-2q^2

[cyan night]

yes?

[left knoll]

so it is right

[cyan night]

no

[cyan night]

.

[left knoll]

yeah so the answer is 6(p^2+2q)(p^2-2q)

[cyan night]

NO

the coefficient is 2 not 24

[left knoll]

sorry sorry i messed up

it is 24^q

[cyan night]

...

ok.

correct then

[left knoll]

ok

Thank you

[cyan night]

np

[left knoll]

https://tenor.com/view/thank-you-thanks-to-you-thanking-you-grateful-thanks-gif-17361424

Is it fine to write the answer directly

if not so I have to write in reverse

@cyan night

[marble canopy]

Hegarty?

[cyan night]

this isnt even a good way to do it

[lunar atlas]

Can anyone help with this precalc?

[supple island]

can you writ it or post a pic where you can see the text better?

[marble canopy]

bro just rotate your pc

[pliant shuttle]

not done maths in abt 2 years and i have to prep for SATs lol. doing A levels, applied 2 UK unis and i wanna do SATs for US apps
what r the most important topics for those?

[fringe timber]

The question is about the function f(x) = ax3 + bx2 ; a is not 0 , b is not 0. This function has 2 extreme points between these you can draw a straight line. Put a=4 and b=3 and decide both of the extreme points. B) Decide the equation for the straight line you can draw between the extreme points. C) Now decide the equation between the straight you can draw between the extreme points in the general case- partly for the arbitrary(possible) values for a and b. Please help, i don't get C.

[marble canopy]

ain't SATs primary stuff?

SATs, GCSE, A levels

[pliant shuttle]

USA SATs :)

[marble canopy]

aight

[pliant shuttle]

ya i need to take SATs or ACT for US applications, i plan on taking SATs as they r apparently better for humanities type ppl like me

[left knoll]

can someone help me with some basics of linear functions? i don't really understand how to solve it

[cyan night]

the inverse of function is just swapping x and y but then isolating y again

y = 2x/x+100

x = 2y/y+100

x = 2+y/50

y = 50(x-2)

[pulsar scaffold]

could someone help me with this?

i found perimeters of this triangles, did system of equations, found out if BC=x then CD=x-6, if AB=y then AD=y-6 and AC=54-x-y

this is all my progress, no clue what to do next

[cinder ibex]

Hi, can anyone please help me simplify this expression?

Under the square root it says 16+36*t^4

[oblique siren]

Hi, can anyone help me find the characteristic polynomial of this matrix pleasee? I can't find the same eigenvalues as in the correction.

[weary arrow]

Hi, remember than the square of a fraction is the square of the fraction components, meaning you can get rid of the square root in your equation. Hope this helps :)

[cinder ibex]

yeah i did that but it just gives me a really weird result that dnsnt look like the solutions maybe i just need to simplify my result but idk how, anyway thanks for trying to help : )

[peak eagle]

that's how I learned it.
Tell me if you need help to understand how I wrote it.

It's not a solution, just a little hint

[oblique siren]

don't worry I understood what you wrote, thanks!!!

[weary arrow]

this is what a computer simplified it as. Does that make sense to you?

[cinder ibex]

ah not really, this was my solution (with t instead of x)

this is the solution tho

so basically the solution is equal to what computer simplified it to

so its definitely me who is doing something wrong simplifying

i will try again, thanks!

[left knoll]

Can anyone find the next number in the series :
10, 16, 39, 140, 675, ?
Do tell the logic behind it. Thanks in advance 🙂

[peak eagle]

try it here https://oeis.org/

[lost rune]

sorry my handwriting sucks, but hopefully this helps @cinder ibex

[cinder ibex]

well thank you but i didnt understand how u got from 2 step to the 3 nor from the 5 to the last, im sorry @lost rune

actually i just understood the 3 step

alright i understood the last as well, i though it was a 6 but it was a t ahah

thank you very much!

[wicked olive]

How do I get rid of the "A" to solve for "h"?

[cinder yacht]

a good first thing to try is to sum all the columns

and see if it gives u a multiple of the column [1 1 1]

because here ur matrix (that I'll call A) verifies :
A * [1 1 1] = 2 * [1 1 1]

so u can already know that 2 is an eigenvalue

so u know that u can factorize the characteristic polynomial by X-2

another person here explained to u the method of finding the characteristic polynomial and eigenvalues, I just gave u a tip that is sometimes pretty useful to speed things up (cause now u know that u can do the euclidien division of ur characteristic polynomial by X-2 to look for the 2 remaining eigenvalues, and it's easier to factor a 2nd degree polynomial than a 3rd degree one)

it's 2+5x

look at how many "leaps" u have to take to go from 2 to 37

not how many cells there are in between

three are 5 leaps, so x = 7

it's like how between ur thumb and ur pinkie, u have only 3 fingers, but u have 4 empty spaces

here u have 4 cells (fingers) , but the important thing is that u have 5 "empty spaces"

[willow basin]

oh shit

my half asleep ass just did that

[cinder yacht]

happens dw XD

[willow basin]

thanks man

[cinder yacht]

yw

[cinder yacht]

it's a/h = a/2 - 1/2?
in that case, our goal is to switch the two sides of the equations to their reciprocals.
Before doing that, since the reciprocal of the right side would be ugly, we should change the right side a bit first : by writing it a/h = (a-1)/2

that way the reciprocals' equation is : h/a = 2/(a-1)

so now u just gotta get the left side "a" over to the other side, which gives :
h = 2 * a /(a-1)

PS: normally we should pay attention that we don't compute the reciprocal of 0. it's not the case here, i would explain why but i think ur course is not at a point where u need to worry abt that yet

[wicked olive]

ohh i see- thank you I get it now!

That's all I needed

[cinder yacht]

yw ^^

[cinder yacht]

so

to go from 10 to 16

u multiply by 2 and subtract by 2*2

from 16 to 39 : multiply by 3 and subtract by 3*3

from 39 to 140 : multiply by 4 and subtract by 4*4

140 to 675 : same with 5

675 to ???? : u multiply by 6 and subtract by 6*6

2664 I think

I'm too lazy to Google (another way to do it would be : subtract by 2 THEN multiply by 2. same with 3, etc)

[cinder yacht]

oh wow I didn't know computing determinants was taught like that in other countrie

s

in France u "develop" according to one column or one line of ur choice, idk how to explain it with words XD

[peak eagle]

yes that's also a way if you have a reaaaally big matrix with hopefully many zeros 🙂
In the country I am studying (Germany) it's called Entwicklungssatz nach Laplace (in english it should be the Laplace expansion)

What I did was the rule of Sarrus. If it's just a 2 x 2 till maybe a 4 x 4 it can be quite nice. But if it's a bigger matrix n x n with n>4 it's just really easy to make mistakes.

[left knoll]

@cinder yacht @peak eagle Thanks a ton. 🙂

[oblique siren]

It's a really great tip, thank you!

[wraith dock]

morning y'all! Could someone help me with this one? I'll write the question and after that, the alternatives

If A = {-1, 0, 1} and B = {-2, -1, 0, 1, 2}, check the option where a given law defines a function (f) of A with values ​​in B:

a) f(x)=2x+1
b) f(x)=-x+4
c) f(x)=1-2x³
d) f(x)=3x
e) f(x)=2x

I tried the whole night and now i gave up on trying by myself, so . . . if someone could help me i'd be very grateful!

[left knoll]

In mathematics, a function[note 1] is a binary relation between two sets that associates each element of the first set to exactly one element of the second set.

a) is not possible: 2*(1)+1=3 and 3 is not in B so the answer is not valid

same for d) and b)

idk if u wrote correctly c)

but the answer should be e) if I'm not wrong

@wraith dock

you basically have to put each element of A in the equation, and every one should result like an element in B

[wraith dock]

@left knoll Sorry, i've wrote something wrote. Can I correct it very fast???

I saw now my error up there . . .

It's 2x³ and not 2x * 3

[left knoll]

ah c) is wrong as well

try it yourself

[wraith dock]

yeah, i know. i tried all the answers

but no one was correct

[left knoll]

e) is correct

[wraith dock]

yeah, got that

[left knoll]

look at the image, is ok if some elements of B aren't taken

[wraith dock]

Jesus, i didn't saw on my own notebook that the B number set had 2 and -2 . . . .

Bruh, forgive me, bro

but thanks again for the support!!!

[left knoll]

no problem 🙂

[small dove]

ay guys

would anybody help me out

can't figure one simple thing out

[amber bluff]

(n+1)! - 2(n!) / (n-2)! + (n-1)! =25

what is ''n'' for this question? i need help with this question

[cinder ibex]

what is ur question?

[amber bluff]

another factorial question i need help: (4n+3)! = (5n-3)!

[small dove]

do i have to derive the equation so that xguns is before the equal sign?

[amber bluff]

how do we solve this kind of questions?

[small dove]

if so, is the answer xguns=xbanjos^1/2 -u

[cinder ibex]

dnt know how to help sorry, u should put directly the question tho so that other people can see and hopefully be able to help

[small dove]

gotcha

[amber bluff]

thank you

[hallow arch]

welcome :))

[sharp ferry]

i think the formula for average rate of change is just (f(b)-f(a)) / (b-a)

[fast schooner]

i got it alr

thanks tho

i did use that one

[sharp ferry]

oo nice 👍

[cinder yacht]

the two equal angles don't look equal on ur drawing, u should change it because it makes everything look weird

when u do it properly u'll see that X is a symmetry axis, and AB=AD, CB=CD

wtf I didn't mean to send this lol

anyway the answer is tan(alpha)

[sand python]

It doesn't look the same in its original form either, I'll pay attention next time.

[sand python]

I don't understand how they are equal :(

[cinder yacht]

oh lol they're tricking u

-but ehm basically, n(ACB) = n(ACD)
-and also, n(CBA) = 90° = n(CDA) [cuz each of those 2 triangle has 2 points of the diameter of the circle, and a 3rd point in the circle (B or D) ]

• which means that n(CAB) = n(CAD) [cuz for each of the triangles ABC and ADC, the sum of the three angles is 180°. Since 2 angles of the first triangle are identical to 2 angles of the second triangle, their respective 3rd angles are identical too]
  so basically ACB and ADC are "identical" (i forgot the technical term), and u can see in a proper drawing that they're symmetrical (on opposite sides of the X axis)
  so AB = AD and CD = CB

but I could explain it in a faster way I guess, maybe this is too complex for ur current course, so here is a 2nd version :

n(BCA) = n(ACD)
so sin(BCA) = sin(ACD)
so AB/AC = AD/AC
so AB = AD
so AB/CD = AD/CD
but we know that AD/CD = tan(alpha)
so AB/CD = tan(alpha)

[sand python]

got it thank you so much

https://tenor.com/view/makise-kurise-steins-gate-anime-like-gif-22035300

[unreal lion]

hi guys I need some help for this one

I know I should be using bays theorem, but how could you find the PMF without knowing the P(Cj), P(Ck) in the first place?

[cinder yacht]

huh

i can be 1, 2, or 3

P(C2) = P(C3) = P(C1) = 1/2

the first thing to ask urself (even tho it's obvious, I think it's important to set things clearly) : what values can X be equal to? well : 0, 1, 2, or 3
then the endgoal is to compute all the following probabilities but one :
-P(X=3)
-P(X=2)
-P(X=1)
-P(X=0)

P(X=3) = P(C1 and C2 and C3) btw

I'd suggest to compute first of all P(Ci and Cj), then P(Ci and Cj and Ck) [where i j and k are all different one from another]

you can do so by using the conditional probabilities

so once u do that :
P ( X=3) = P(C1 and C2 and C3), u got that one. U still need to find the rest

-to get X=2, u need either (C1 and C2 and not C3), or (C1 and C3 and not C2), or (C2 and C3 and not C1)
so gotta compute P(Ci and Cj and not Ck) [where i, j, k are distinct]

well, P(Ci and Cj and not Ck) = P(Ci and Cj) - P(Ci and Cj and Ck), and u have those 2 latter probability values at this point, so hg

gg*

P(X=2) is computed then

hmm X=1 or X=0 seem a bit more complicated, I'm not sure how to do it without a pen and paper to write out things

knowing that P(Ci and not Cj) = P(Ci) - P(Ci and Cj), we can compute that probability.
But we know that :
P(Ci and not Cj and not Ck) = P(Ci) - [P(Ci and Cj and not Ck) + P(Ci and Ck and not Ck) + P(Ci and Cj and Ck)]
to convince urself of this formula, use a Venn diagram
so we can compute P(Ci and not Cj and not Ck) for every distinct i, j, k
but P(X=1) = P(C1 and not C2 and not C3) + P(C2 and not C1 and not C3) + P(C3 and not C1 and not C2)
so we can compute P(X=1)
so we can compute P(X=0) = 1 - P(X=1) - P(X=2) - P(X=3)

I think the route I took was maybe longer than needed... I'm tired and i don't have a paper in front of me to find an easier route, sorry

the answer is P(X=n) = 1/4 for every n lol

much easier with a Venn diagram

[cyan night]

Given angle xOy. A ∈ Ox, B ∈ Oy. Find the set values for M such that it satisfies S_AMO = S_BMO

can someone help me plz

[ember kernel]

S means surface right ?

[cyan night]

area actually

[ember kernel]

That's the same

[cyan night]

Ok my bad

[ember kernel]

Looks pretty much like M = (x,x)

[cyan night]

so what ive thought of is dividing into 2 scenarios: when M is a point on segment AB and when M isnt

when M is on AB then M must be the midpoint of AB

[cinder yacht]

i think there's a tiny difference, surface meaning the points inside the shape and area meaning the number of units within the surface or smth lol

doesn't matter tho

[cinder yacht]

i don't think so

is xOy = 90°?

[cyan night]

no

[ember kernel]

?

[cyan night]

xOy is any arbitrary angle

i was replying to coly

[cyan night]

why so?

[cinder yacht]

im trying to understand why i don't feel like it is

maybe im wrong

[cyan night]

i think when M is on AB then the triangles have the same height right

i think thats the case

the next scneario is M is not on AB

i was planning to divide this into 2 more cases (when OA=OB or OA # OB) but idk if i shoul

[ember kernel]

Ooooh, I thought it was the cartesian coordinates system, given the notation O, x ,y

[cyan night]

oh

this is just angle xOy

[ember kernel]

M = the set of points such that dist(M,Ox) = dist(M,Oy)

[cinder yacht]

oh yea ur right

i didnt study geometry since middle school almost i sukk =(

[cyan night]

okk should i divide into these cases

[cinder yacht]

no

[ember kernel]

Which means the set of points located on the bisector

[cinder yacht]

i think that's pointless

[cyan night]

because i think that

if oa = ob

[cinder yacht]

in the case iof OA = OB it's an obvious answer

but they want the hard answer

[ember kernel]

Oh no I'd just misread the question since the beginning

[cyan night]

but it said all sets of values

so we need several scnearios

[cinder yacht]

you need a general rule that works regardless of where A and B are

[cyan night]

i think that if oa = ob then m just needs to be any point on the midperpendicular of ab since then o is equidistant to a and b

[cinder yacht]

u can separate the special case OA = OB if you want but that doesnt solve the problem

yeah

[cyan night]

maybe i can generalize it without needing oa = ob tho

[cinder yacht]

if u want to treat every special case, u can consider : if O = A then it's any point on the Y axis

(cuz the surface is 0)

[cyan night]

maybe the general formula can treat every special case

i want to do some addition subtraction among the heights and bases of the two triangles to make them equal in area

[cinder yacht]

oh i think i got it

it's any point of the midperpendicular of ab

lemme try to prove it properly

[cyan night]

what if oa # ob?

[cinder yacht]

works

lemme try to prove

[cyan night]

heights are the same?

[cinder yacht]

nvm

i was wrong

ahaha

i mean

[cyan night]

a

[cinder yacht]

i got confused : it's the lines (OI) where I is the middle of [AB]

line*

lemme make sure

[cyan night]

this is want i wanan do

im not sure if it works at all

[cinder yacht]

S(OIB) = S(OIA)

and

S(AIM) = S(BIM)

so S(AMO) = S(BMO)

[cyan night]

wait whats i

[cinder yacht]

I is the middle of the segment [AB]

in red

[cyan night]

yes

ok

OH SO

OHH

M IS A POINT SUCH THAT

O,I,M are collinear

[cinder yacht]

yeah

[cyan night]

ty!

[cinder yacht]

thats what i meant by "M is on the line (OI)"

[cyan night]

tysm

[cinder yacht]

yw

[cyan night]

👍

[cinder yacht]

ehm there's one special case : if O is the middle of [AB], then I = O, which would mean that (OI) is not even a line so the answer doesn't rly stand. In that case, the answer is just : "any point in the plane" i think

[cyan night]

i think thats the case i demonstrated earlier

that m lies on ab

[cinder yacht]

i dont think so

[cyan night]

O?

if O is the middle of ab then isnt the angle xOy a 180?

[cinder yacht]

yeah it is

that means O lies on ab

not M lies on AB

[cyan night]

i think then

m playss the role of o

in the case when m lies on ab

[cinder yacht]

here is what i meant

^^

[cyan night]

whats the condition of m

[cinder yacht]

nothing

M can be anything

and it still works

i took a random M and it worked

[cyan night]

but o isnt the midpoint of AB?

[cinder yacht]

so the final answer is :

• if O is the midpoint of [AB] : M can be anywhere
• if it's not : M is in the line (OI) such that I is the midpoint of [AB]

[cinder yacht]

it is

[cyan night]

why

A and B are arbitrary points on Ox and Oy

[cinder yacht]

yeah but they didn't tell you "we konw that O isn't the midpoint of [AB]"

that means that "O is the midpoint of [AB]" is a possible case, which means it needs to be covered by your answer

[cyan night]

wait let me ask the teacher if i need to evaluate the case if xOy = 180

[cinder yacht]

if you answer just "M is in the line (OI) such that I is the midpoint of [AB]" then that makes no sense when O = I = the midpoint of [AB]

[cyan night]

if O is not the midpoint of AB then whever M lies doesnt matter because u cant make them equal

[cinder yacht]

you're mixing up O and I

xD

btw when O = I, (OI) is not a line

[cyan night]

nono

[cinder yacht]

that's my point

[cyan night]

yes

but

im considering 2 cases is when O is on AB and when O isnt, when O is on AB i dont include I under any circumstances

when O isnt on AB then i include i as the midpoint of AB

[cinder yacht]

huh

[cyan night]

since then O and I couldnt converge

[cinder yacht]

why

[cinder yacht]

what's your issue with the answer that i gave here?

if O is the midpoint of [AB], it looks like this :

https://cdn.discordapp.com/attachments/838057776584327198/904662988085153792/unknown.png

if O is not the midpoint of [AB], it looks like this :

https://cdn.discordapp.com/attachments/838057776584327198/904661627452932176/unknown.png

[cyan night]

because i think i am able to exclude the case that OAB = 180

[cinder yacht]

the case "AOB = 180 and O isn't the midpoint of AB" enters in the case "if O is not the midpoint of AB" and it's the same answer, look :

[cyan night]

ahh i c

[cinder yacht]

here "M can be any point in the line OI" holds

[cyan night]

ty!

[cinder yacht]

yw

[cyan night]

wait, can you help me with just one more thing

How do I draw hexagon ABCDEF that satisfies AB // DE, BC // EF, CD // FA

[cinder yacht]

draw 6 lines such that each pair of opposing lines are parallel

so the 6 vertices are necessarily the intersection points of each pair of adjacent lines

maybe i'm phrasing stuff in a way that's too complicated for middle school course XD

if 2 lines are of the same color, they're parallel

[cyan night]

i got it thx

[oblique inlet]

hi, can someone help me find the limit (not using the L'Hôpital's rule)

[thorny swift]

Ahh Limits😶

[ember kernel]

Why the constraint of not using L'Hospital specifically ?

[oblique inlet]

At my university, we have not studied this topic yet. So, the limit can be found without using L'Hôpital's rule

[cinder yacht]

what about Taylor expansions?

4^(3x^3) - 1 = exp ( ln ( 4^ (3x^3))) - 1
= exp ( 3x^3 * ln(4)) - 1
which is equivalent to 3x^3 * ln(4) when x->0 [if u have studied that]

as to the numerator : [1+(t * g^3 * 5) x] ^ (1/5) - 1 is equivalent to (t * g^3 * 5)x/5 = t * g^3 * x

so overall the fraction is equivalent to
(t * g^3) / (3 * x^2 * ln(4))
so if t * g^3 is <0, the limit is - infinity
if it's >0, the limit is + infinity
if it's = 0, the limit is 0

[oblique inlet]

i understood this

but in the numerator tg means tangent :), sry

in Russia we write tg instead of tan

so it will be (tan(5x))^3

[cinder yacht]

oh

that's fine, the numerator is equivalent to tg^3(5x)/5

but tg(5x) is equivalent to 5x

so tg^3(5x) is equivalent to 125x^3

so the numerator is equivalent to 25x^3

so the fraction is equivalent to 25/(3 * ln(4)) if i didnt make any mistake

so its* limit is 25/(3 * ln(4))

[oblique inlet]

yes that's right, thank you so much

[late axle]

need a quick check

B ≡ -1789π / 3 [2π] <=> B ≡ 3π/2 [2π] ?

[cinder yacht]

nop

the denominator doesn't change

[late axle]

oh

i see my error

[cinder yacht]

1789 = 1788 + 1 and 1788 is a multiple of 6

so 1789 = (1788+6)-5
so - 1789 = - (1794) + 5

and - 1794 is a multiple of 6, so -1794/3 is a multiple of 2, so -1794pi/3 = 0 [2pi]
so - 1789 pi / 3 = 5pi/3 [2pi]

[late axle]

thank you!

[left knoll]

how would i factorise x^2 + px + qx + pq

[grizzled seal]

(x+p)(x+q)

[left knoll]

5(xy)^2/15xy^3 help meeeeee

Help me do the algebraic fraction someone

[left knoll]

take 5xy common

ok

[left knoll]

yes

explain me please

ans is x/3y ?

waite

is it x/3y?

How can I do this what is the command?

@left knoll cancel the 2 (5xy)
then cancel y

[left knoll]

#how-to-use-math-bot

/frac {x}{3y}

[left knoll]

u need to put
=tex

[left knoll]

Thanks

😀

[left knoll]

right?

Is this right?

@left knollIs this right?

[left knoll]

y^3

[left knoll]

Is this right?

[next lark]

help plz

[cyan night]

inverse of y = 2+11x/5x-13 is x = 2+11y/5y-13 and then solve for y u get the inverse

[left knoll]

is this correct

[cyan night]

yh

[left knoll]

ok

@cyan night Are you a math professor?

[cyan night]

im a middle schooler

[left knoll]

😮 Your so good at math! How can I get good at math?

[cyan night]

im not that good

i just

learn on khan academy

and some books

also its you're

i suggest u trying khan academy go from algebra I

[left knoll]

I will.

[cyan night]

what grade are you in

[left knoll]

7

[cyan night]

ah ok

[glacial robin]

мужики че тут делать

[hushed jacinth]

howw

[random flame]

Have you learned the laws of indices?

[hushed jacinth]

oh is it just that?

[random flame]

Yeap

[hushed jacinth]

damn i thought it was more complex ok ty LOL

[random flame]

XDD

Np man

Anyways umm anyone out there who can help me with part b-? Ping if you do thankiew

[cyan night]

just recall fractional exponents and negative exponents 😉

[hushed jacinth]

yessirr

[cyan night]

the rational zero theorem still requires recheck for extraneous solutions right?

[sharp ferry]

are you allowed to use inverse trigonometric functions?

[random flame]

Yeap

But nvm, I solved it ady

Thanks tho!!

[sharp ferry]

welcome 👍

[weary arrow]

The expression is a sum, and i is to denote the parts to be summed. Its hard to explain in English but basically, i=1,2,3... Will give Collateral1 (nr 1), collateral2 (nr 2) in the next addition etc.

[slow cobalt]

pls help

[ember kernel]

Two ways of solving :

• Rigorous : h'(x) = 0
  (if you've already learned about derivatives)
• Intuitive : -4sin(x) is maximised when sin(x) is at its lowest, which means sin(x) = -1

[slow cobalt]

ahh that makes sense so it would be 7

thank you

[ember kernel]

That's for h yeah

[slow cobalt]

im not sure if differentiating sin is any different from normal

is it?

[ember kernel]

Don't forget to indicate the x-value for which h = 7 tho

[slow cobalt]

ok

kool thnx agane

[ember kernel]

(sin(kx))' = k cos(kx)
(cos(kx))' = -k sin(kx)

If it helps

[slow cobalt]

ooh i do not think i have done this but it kinda makes sense

i will go with the second way anyways

much simpler

[ember kernel]

Just forget about it and use the intuitive way then ^^

[slow cobalt]

yep

[left knoll]

can you help me?

[thorny stump]

Please help

[lime junco]

I don't even know how to start doing the problem

[toxic bobcat]

Hello, can anyone help me with my math, please? It's nothing complicated.

[strong stump]

I can help with anything below Geometry or Geometry

[weary arrow]

just post your question here, and someone will help you 🙂

[toxic bobcat]

My handwriting isn't anything special, sorry about that.

[strong stump]

oh ok

thats ok

sont worry bout it

dont*

[toxic bobcat]

and what would be if the angle was 45 or 30 or 60

[strong stump]

ok what is it asking for

[toxic bobcat]

x

[strong stump]

oh

[toxic bobcat]

c

[cinder yacht]

lets call the angle alpha
tan(alpha) = a/b
so b = a / tan(alpha)
so b² = a² / tan²(alpha)
let's use pythagore theorem : c² = a² + b²
so x² = (4 * sqrt(3) )² + (4 * sqrt(3) / tan(alpha) )²
so x² = 48 + 48/tan²(alpha)
so x = sqrt ( 48 + 48/tan²(alpha) )

and u can replace alpha with 45 30 or 60

sqrt(blabla) means : square root of (blabla)

[strong stump]

oh dang

yea im not that good sorry

lol

[toxic bobcat]

yeah man i haven't gotten to trigonometry

[cinder yacht]

oh

wot

[toxic bobcat]

no no, please try

[cinder yacht]

idk how to do it then

without trigonometry

[strong stump]

does it give any other angle?

[cinder yacht]

the angle BCA is 90°, and the angle ABC = 90°-BAC

(so in this case ABC = 30 if BAC = 60)

[toxic bobcat]

no, they want me to find out what is CA abd AB

[strong stump]

oh may ga

[toxic bobcat]

or if you like it this way b or c

[cinder yacht]

i don't think u can do it without trigonometry

[toxic bobcat]

ill send you a picture from my test

[strong stump]

oh ok

[toxic bobcat]

handwriting was even worse i didnt have time

[strong stump]

To find angle C, we simply plug into the formula above and solve for C. A + B + C = 180. C = 180 - A - B C = 180 - 40 - 60. C = 80.

[cinder yacht]

huh

[toxic bobcat]

how do i find x and x1

[cinder yacht]

sorry i might sound rude, but this is wrong xD

C = 90°, A = 60°
A+B+C = 180° so B = 180°-A-C so B = 180°-60°-90° = 30°

[strong stump]

oh sorry lol

i got it off of the internet

im not a professional

[toxic bobcat]

how do i find c and b guys

[strong stump]

Are you taking Algebra 1 or 2

I dont know how to solve this i am so sorry

[toxic bobcat]

i think its 2

i am in 8th grade

europe

[strong stump]

https://schools.shmoop.com/home/

go there and somebody will help you a lot better than i probably can

[toxic bobcat]

english isnt my native language, so we dont use the same terms

[strong stump]

ohhhhhh

maybe thats why its so hard for me to do

im sorry dude idk what to tell you

[toxic bobcat]

its ok, thank you regardless for trying to help me

[strong stump]

yw 🙂

[slow cobalt]

r=1.06 here

so why th is it the bottom that

im so confused

the modulus of r<1

how does that change this

nm the working out is wrong

ignore it

[left knoll]

Can anyone please help me with this maths

[ember kernel]

Integration by parts rely on 3 important theorems/rules :

[candid moth]

An internal error occurred.

If this keeps happening, you should contact the developers on the official mathbot server: https://discord.gg/JbJbRZS

[ember kernel]

To get the integration by parts formula, just develop this equation :

[ember kernel]

and you get :

=tex \int (f(x)g(x))' dx \= \int (f'g+fg')dx \ = \int f'g dx + \int fg' dx \ Leftrightarrow \int f g' dx = \int (fg)' dx - \int f'g dx = fg - \int f'*g dx

[candid moth]

An internal error occurred.

If this keeps happening, you should contact the developers on the official mathbot server: https://discord.gg/JbJbRZS

[ember kernel]

... and you get :

[ember kernel]

Regarding the integral with h(-1)(x), use the same formula, but with f(x) = x and g(x) = h(-1)(x)

[ember kernel]

⬆️

[strong stump]

anybody need Algebra or Geometry help?

[left knoll]

Can anyone please see this

This is not that easy

[cyan night]

if the special 30 60 90 triangle isnt trig then its defo doable in a trivial manner

[frozen meadow]

im starting calculus in my college currently, is there any advice anyone can give, and anything i should give more importance or look out for when studying it

[ember kernel]

On wolfram you asked for P(x), not the antiderivative of P(x) (read the question again)

But yeah the antiderivative of P(x) can't be expressed with classic functions such as sin, cos, polynomials, exp, ...

[left knoll]

So how should we view it sir

@ember kernel

[cinder yacht]

did you study

exponentials of complex numbers?

like e(i * t)

[left knoll]

no

u?

[cinder yacht]

yea

i guess you studied integration by parts right?

let u(t) = sin( pi * t² / 2 ) and v'(t) = 1, do integration by parts

and you'll find the answer (which is x * P(x) - sin(pi * x² / 2 ) / pi i think)

[left knoll]

Like this?

@cinder yacht

💯

[cinder yacht]

lol

you have to actually prove it though

not just give the end result

did you study partial integration or not?

[left knoll]

I studied but can't catch it up could u please support

[ember kernel]

Read your first line and the question again

The first line should be

[ember kernel]

Instead you have as your first line (which is not what you were asked)

[ember kernel]

Forget about the value of P(x) for now, we'll just ignore it

[ember kernel]

Thus, if we refer to the formula of integration by parts, we have f(x) = x and g(x) = P(x) (no matter its value), right?

Then, just sub x, P(x), 1 and P'(x) in the integration by parts formula

Hint : knowing the equality hereinbelow could help you simplify the expression you end up with AT THE VERY END OF YOUR ANSWER (do not pay attention to it unless you're done answering)

[ember kernel]

@left knoll

[cyan night]

Hey guys I need help with clarifying: So with my knowledge so far, I am capable of identifying the total number of complex roots/zeroes of an nth degree polynomial, which is basically n complex roots (according to The Fundamental Theorem of Algebra). And the number of turning points in a graph is always 1 less than the degree of the polynomial, implying that the number of turning points is also 1 less than the total number of complex roots in that polynomial. But since it refers to complex roots (which comprises of both real and imaginary roots), the x-intercepts of the graph do not wholly represent all the roots of a polynomial (another concern right here, if only x-intercepts (real roots) are visual to us, how can we determine the degree of the polynomial (because it exists imaginary roots that aren't visible as well))? Now back to the turning points, in the book I'm using, it said the number of turning points is always one less than the degree of the polynomial. And the turning points also hence determine the total number of complex roots (right?). So is using the number of turning points more reliable than using the x-intercepts of the graph to determine the degree of the polynomial? The x-intercepts would definitely be more useful since it also gives us the idea of the entire polynomial.

[cinder yacht]

(it's not n complex roots, it can be less, but the sum of multiplicities of roots is n. For example : P = (X-1)^4 has one root whose multiplicity is 4)

and i'm not sure how you define turning points. It's local maxima and minima?

[cyan night]

yes

[cyan night]

lets say we assume multiplicity here, the number roots are 1,1,1,1

[cinder yacht]

and by graph I assume you mean the graph (x, P(x)) where x are real numbers and P is the polynomial function? Because we could define a graph (z, P(z)) where z are complex numbers, but you wouldn't be able to REPRESENT that graph in 2D. You would need to represent it in 4D ( real part of z, imaginary part of z, real part of P(z), imaginary part of P(z) )

[cinder yacht]

sure

[cyan night]

yes

the complex roots are not representable

[cinder yacht]

also if P's coefficients are not real then the graph (x,P(x)) can't be drawn in 2D either

even if x is real

[cyan night]

wait why

oh

yeah

i misread saying not rational

ok so the number of turning points is reliable on determining the degree?

but not the x-intercepts

the book that im using says use x intercepts to generate the function

[cinder yacht]

I don't even think so

[cyan night]

why so?

[cinder yacht]

I mean

what is a turning point

[cyan night]

when the graph changes direction

i just take it visual intuitively

they directly relate because graphically, you need to turn in order to "reach" the next zero

including complex zeroes

are you there

[cyan night]

a

[cyan night]

please

[lilac flint]

Where the gradient is zero

dydx=0

[ripe hamlet]

help help pleaseeeee

pleaseeee

[lilac flint]

b

b^2-4ac is smaller than 0

roots are non real

[cyan night]

yes this is correct but keep in mind if this are a test it is prohibited to answer 😉

[lilac flint]

haha fail and become part of the normal distribution

kekw darwinism

[high rock]

can someone help me solve this problem.

[cyan night]

not me, i cannot.

[green cliff]

I fail to comprehend basic math

It's truly unfortunate

But where the hecc did that -2x come from exactly

[high rock]

@green cliff this -2x?

[green cliff]

Yes

[high rock]

so

on

was used this formula

you see?

[green cliff]

Oh

Oh I entirely forgot how that works didn't I

[high rock]

haha

[green cliff]

It just foils (x-)(x-1) doesn't it

I am but a simple fool dear lord

[high rock]

take these may help you in the future ❤️

[green cliff]

I love you

[green barn]

Guys I need guidance.
Currently my situation is as follows: I haven't done math for 2 years, and to transfer to a different 4 year college by next year, I need to have completed 4 math classes NEXT SEMESTER. I have to do the following:
Linear Algebra
Discrete Mathematics
Physics with Calculus
Calculus 2
How the hell do I prepare for this so I actually have chances of getting a transfer to my dream school?

[green bolt]

Hello! Isn't the answer 0? Can someone explain? Thank you so much!

Or is it -2?

[sharp ferry]

because this was the graph of f'

not f

[ember kernel]

• Ok, first of all, when is the deadline exactly ? (European fella here, so I'm not sure of when next semester is for you)
• Secondly, by "completing", does it mean learning and understanding, or passing actual, official exams ?
  The amount of work you'll have to put in will depend a lot on these two questions
• Regarding Linear Algebra, there's an excellent, complete, free course given by G. STRANG himself on YouTube (course name : 18.06 Linear algebra). Strang is an excellent MIT lecturer, and author of the book "Introduction to linear algebra", which I recommend as a textbook if your study type requires one.
• If your Discrete Mathematics course is the same as mine, it'll be heavy in definitions and theory. But the best way to get to remember them all is by practising.
• Calculus requires a good amount of visualisation, and its textbooks usually have lots of formulas, spreading everywhere on every sheet of the book. I experienced it and it was really discouraging being face to face to so many very generic formulas, that you need to know by heart. Never hesitate to take a break and switch to YouTube to have them explained to you with visual examples (what textbooks rarely do). To get good at calculus exercises, you need to deeply understand the formulas (where they come from, what value they give you, what they're used for) in the first place. While reading the textbook, take some time to pause for a minute, and focus on understanding what you've just been reading (was it important stuff, or just some example of what a previously-seen formula could be used for ?). Calculus will most likely be the hardest thing to learn out of the four classes, but DO NOT GIVE UP, even when you're feeling lost. Just take more time to understand what you've already read, without going further. Most of the time, it only takes a little realisation for your brain to make links between all the concepts, and then everything will suddenly seem clear.

[sharp ferry]

if this was the graph of f(x) then x= 0 would be correct

but rn you have to check where f'(x) changes its sign so it should be 2

-ve to +ve

the first derivative test remember ? for local min if f has local min at x=a then f'(a-) <0 and f'(a+)>0 and f'(a)=0

[green bolt]

Ahhh, alright. Got it, thank you!

[sharp ferry]

man you're so inspiring

[ember kernel]

Regarding the organisation of your study schedule, it all depends on you. Only you know what your average week looks like, and when/in what conditions you'll be able to study these additional classes.

I don't want to lie to you : it's gonna be tough (even more if you're already following other classes). But with good organisation and as little schedule overloading as possible (schedule overloading, which results from bad organisation, is the #1 reason of why people give up on new habits/goals)

Anyway, for every class you need to follow, find one good, complete course/textbook, and stick to it. Having all the important information centralised in one single place is crucial, and will save you a lot of time (the time you won't spend on finding this information)

[green barn]

Beginning of semester starts on January 5, and by completing I probably need at least a B in all of those classes so I could get accepted to my transfer schools

Is there some textbook you can reccomend to re study fundamentals of math? I haven't done it in so long so I forget even basic algebra here and there

[ember kernel]

Also, don't underestimate the importance of doing exercises, even when you struggle with them, MOSTLY when you struggle with them ; you'll have to be able to solve these anyway, so don't just avoid them

Tho it looks tempting to do so (my experience speaking here)

[ember kernel]

Unfortunately, no, I'm sorry. I've always used my high school's printed notes when learning maths fundamentals :(

[sharp ferry]

i have a pdf that i use in highschool rn will that help?

[green barn]

Anything helps tbh. Just trying to get my resources together so I can figure out what I need to learn

[sharp ferry]

i has basic algebra trig log graphs and stuff

but my system here is a bit different will it still help

[green barn]

I believe so

[sharp ferry]

okay then imma dm you the pdf you can go through it if it helps

[ember kernel]

Maybe you could try to find maths-related entrance exam preparation courses. I followed one, and at the beginning of each lecture we were given folders with all the important formulas and theorems, followed by exercises to apply them. Maybe you could find similar folders somewhere. Ask your college/old highschool about it

[sharp ferry]

sent

[left knoll]

@ember kernel Thanks man I solved it finally by your help

[ebon lintel]

If there's a Quadratic Equation, how to find out whether its a "Symmetric Expression or not"?

[cinder yacht]

idk how ur teacher defines symmetric expression

oh I just googled

I think the answer is there lol

u need to try to switch every possible pair

for example if u have x y and z in ur expression, then u need to try to switch :
-x and y
-y and z
-x and z

if the expression stays the same whenever u switch any pair, then it's a symmetric polynomial

[buoyant stag]

Hey does anyone know how to help with this it is higher gcse maths

I’m really stuck and idk the method

Or the formulas needed

[maiden prism]

Hi guys, I need help with question a. we know that to be linear inde then c1v1 + c2v2 + c3v3 + ... = 0 as c are coefficients and v are vectors. then rearrage them I have w1(c1-c3) + w2(c2-c1) + w3(c3-c2) = 0. So how this tell abt linear independent? Tks

[cyan night]

why do i feel like you would just straight up calculate investment A and B then compare

investment A is (150x12)x1.025

B is 3500*1.03^2

i got 1868.15

[ebon lintel]

Thanks, that helped!

[willow basin]

Zac buys a watch for $y including sales tax at 15%. Show that the sales tax paid by Zac is $3y/23.

huh

someone translate this into english please.

so... if i do $yx1.15 = $3y/23...

why do i still not get it.

...

[left knoll]

how would i start this

y = a * b^x btw

[weary arrow]

Make two equations from the formula and the two points

12=ab^1 and 36= ab^2

[left knoll]

thanks

i got it now

[cobalt mural]

i don’t know if it’s totally right but like,, i did: y*0.15=3y/23

then solved for y

oh shoot wait it doesn’t work either

[cobalt mural]

wait do you even have to solve for y?

[willow basin]

No. You have to show that the sales tax paid by Zac is $3y/23. ALGEBRAICALLY said the teacher. But still. I don't get it.

[hushed jacinth]

how would u solve this without a calculator?

[tender scaffold]

log(10^(-1/4))

[hushed jacinth]

tahnk

you

[cobalt mural]

sorry can’t be much of help but i found this online tho: https://www.accountingcoach.com/blog/calculate-sales-tax

it said to divide i think

by decimal of percent

[cinder ibex]

Hi, can someone pls help me derivate this, i have been trying for a while but i can never get to the solutions result

i searched online for the derivation and got this, i do have the first expression in this picture but idk why i cant simplify it to that result on the right

[polar heron]

Take (x^2+1) * (-12x) common from the numerator.

[cinder ibex]

just tried but idk i cant get to that expression thanks tho

[smoky berry]

I almost always just use negative exponents with product rule

[cinder ibex]

oh wow i just spent more than a page trying to simplyficate it and u made it wayy easier thankss : )

[smoky berry]

tried to write out the simplifications to make it easier to follow

[cinder ibex]

yeah i got it i was using the (f/g)' = (f?g-fg')(g^2 ) but it got way harder

thank you

[smoky berry]

np, when you do a lot of them they get easier

[cinder ibex]

yeah thats the spirit : )

i mean looking back im a bit confused, did u use the division or multiplication rule? and how u got rid of the negative expoents from the first step to the second?

ah u used the multiplication rule the - comes from the expoent

[smoky berry]

yeah. let me know if it still needs clarification

[cinder ibex]

ah i didnt understand what happened there, sorryy

i mean what happened to those in the second step

[smoky berry]

so there I imagine that those could be on the denominator, and to simplify the expression we need a common denominator, so for that to happen the left term needs to be multiplied by (x^2 + 1)^-1

[cinder ibex]

yes

but woudnt it disappear from the numerator then?

[smoky berry]

well if you multiply the bottom, you need to also multiply the top right?

so that's where the -12x(x^2+1) comes from in the third step

[cinder ibex]

i did itt, thank you so much

[smoky berry]

that's great

[cinder ibex]

: )

i had forgotten to include 2x from derivating (x^2+1)^-2

[smoky berry]

I am stuck trying to find the recurrence relation for this ODE. On the right there's a term x^(n+1) but on the left it's x^n how can I cancel them?

[bright haven]

hi does anyone know how to find the nature of roots for a quartic function?

[smoky berry]

nevermind, I figured mine out

[slow cobalt]

yo need some help quick

[patent rain]

3a: Equilibrium means net force = 0, tension makes up the only horizontal force, T1cos35deg = T2cos35deg —> T1=T2
3b: Tension = weight
3c: vertical component of tension force

[livid atlas]

Can anyone help me with exponent and logarithm equality?

[left knoll]

could someone answer this?

[patent rain]

f(x) = f(-x)

substitute x with -x and see if its the same function

for example, y = x^2 is even because (-x)^2 = x^2

[fringe stratus]

Hi !
Anyone has an idea of what that can mean ? Maybe that i is inferior than j in the domain of N ?
(It's in a discrete maths class)

[cinder yacht]

maybe "i is smaller than j considering the natural order relation of N" (as opposed to some weirdass order relation we could invent) but not sure i never saw it