help-13

[tawny drum]

from nzmo

[opal basin]

oh, to practice?

bounding means, getting bounds on some numbers, you know that w + r + b <= 100, so w + r can't be greater then 100

[tawny drum]

oh

so

5r=7w

[opal basin]

Yeah

notice that r had to be even btw

[tawny drum]

wait so what do i do now

[opal basin]

what would r being even mean?

in terms of variables

[tawny drum]

um idk

sorry im new to very logical maths so im not very efficient

[opal basin]

r = 2k, for some integer k

[tawny drum]

whats k

oh

[opal basin]

every even number can be expressed as 2 times another

so now you just need to solve for k

[tawny drum]

so 46k=7b

10k=7w

[opal basin]

yeah

[tawny drum]

k=7w/10

[opal basin]

that's where the number theory comes in. You have 10k = 7w, look at divisors

[tawny drum]

can you elaborate further pleas

[opal basin]

7 divides 7w, right?

[tawny drum]

yeah

[opal basin]

so 7 divides 10k

[tawny drum]

then what do u do

[opal basin]

that would mean 7 divides k

and the other way around, 10 divides w

[tawny drum]

ohh

[opal basin]

What happens if k = 14 and w = 20?

[tawny drum]

they are equal?

[opal basin]

yeah, but for the value of b?

[tawny drum]

92

[opal basin]

Yeah, so there would be too many balls

k and w have to be smaller

[tawny drum]

so k=7 w=10

[opal basin]

yeah

[tawny drum]

b=46

w=10

r=14

46+10+1470

=70

is that the answer

[opal basin]

Yeah, k and w can't be bigger

I have to go now though, if you want help with other questions open a new channel

[tawny drum]

i have to go too

thanks for the help

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[torn talon]

why do they multiply the two functions?

what does it achieve

[livid hound]

fancy way of applying ivt
if the product is negative,
one value is negative and the other positive

[torn talon]

yeah i get that

[livid hound]

0 is between some positive value and some negative value

[torn talon]

yeah i get that as well

but why did they multiply

and after multiplying how do they solve for a and b?

[cedar kiln]

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[cedar kiln]

@torn talon Has your question been resolved?

[torn talon]

<@&286206848099549185>

[spice kraken]

to use intermediate value theorem, we want one of f(1) and f(-1) be positive and another be negative

that's the same as saying they have different sign

which means f(1)*f(-1) is negative

[torn talon]

yeah i get ivt

[spice kraken]

thus f(1)*f(-1)<0

[torn talon]

okay

so how did they get 6 and -6? trial and error?

[spice kraken]

they calculated f(1)*f(-1) and plugged in the a,b values

[torn talon]

ahh ok

[spice kraken]

you can solve inequality f(1)*f(-1)<0 if you want exact solutions

[torn talon]

so its kinda trying each option?

[spice kraken]

yes

[torn talon]

okay ty

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[spice kraken]

no problem

[cedar kiln]

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[echo oyster]

i wish to show the validity of the following argument form. this argument form basically says that (for propositional variables p and q) if (p -> ~q) leads to a contradiction, then p->q is true. did i prove the argument form correctly?

[echo oyster]

because it seems like 'proof by contradiction' but it's not it, since p -> ~q isn't the negation of p -> q

[cedar kiln]

@echo oyster Has your question been resolved?

[echo oyster]

<@&286206848099549185>

here is the argument form btw

[ebon pulsar]

?

whats the question

[echo oyster]

here

[ebon pulsar]

what u mean by ~

its diffrenet signs lol

[echo oyster]

negation

[ebon pulsar]

ah k

lemme see

[echo oyster]

mb

[ebon pulsar]

you need to prove that :
(p -> ~q )-> c
using truth table

if i understand

right?

[echo oyster]

i gotta show that if this is true, then p->q is true

[ebon pulsar]

sec

u still didnt finish the table?

[echo oyster]

if it is false, then the conclusion doesn't matter

yeah cause the hypothesis is false

[ebon pulsar]

k?

[echo oyster]

so the truth of the conclusion doesn't matter

[ebon pulsar]

whats ur question then

[echo oyster]

did i do any mistake on my truth table or anything?

i just feel like it is wrong for some reason

[spice kraken]

isn't the contradiction just same as negation?

[ebon pulsar]

how did u determine that all F for c?

[echo oyster]

c is a contradiction statement, so it's always false

[ebon pulsar]

i thought its a varible lol

[echo oyster]

not really, contradiction is an always false statement i believe

[echo oyster]

mb

[ebon pulsar]

if so then u shoudl have 2^3 lines

in the table

not 4

[spice kraken]

if a statement leads to contradiction, then the statement have to be false

[ebon pulsar]

yup

[echo oyster]

yeap that's if we're using the contradiction rule, but this is another argument form

i just wanna prove that

this is the contradiction rule

[spice kraken]

ok

[spice kraken]

your argument is valid

[echo oyster]

cool

just seems kinda weird at first

since i've seen people wanting to prove by contradiction and using that argument form instead of the contradiction rule form, but i suppose it's correct too then

thanks

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[spice kraken]

actually you can prove a stronger result, namely p and q is true

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[vernal kite]

If this is in vertex form, how come in my answer sheet both 1/2 and 3/4 are positive

[vernal kite]

is it always positive?

[azure horizon]

they aren't always positive.

basically, the vertex is when the squared bit is 0. so (x - 0.5)^2 = 0, so that's when x = 0.5

if it were (x + 0.5)^2, it would be when x = -0.5, to make the bracket 0

[spice kraken]

um

y=(x-0.5)^2+3/4 though

[azure horizon]

yeah so it's at (0.5, 0.75)

[spice kraken]

so y is always positive

[vernal kite]

ohhhhh

so what if like (x-1/2)^2 - 3/4, would 3/4 be negative then

[spice kraken]

then both 1/2 and 3/4 are negative

[vernal kite]

so if 'k' here were to be negative, h is negative then right? and vice versa

[spice kraken]

?

h is the x coordinate of vertex and k is y coordinate of vertex

they can be any value

[vernal kite]

ohhh i get it now

[azure horizon]

yes

the vertex would be at (0.5, -0.75)

[vernal kite]

and then if it was (x+1/2)^2 - 3/4, then both are negative?

[azure horizon]

yes

(-0.5, -0.75)

[vernal kite]

alrlr thank u

so for (x-2)^2 -3 = y, then 2 is positive and 3 is negative?

@azure horizon

[azure horizon]

yep

you got it

[vernal kite]

so for this quesiton: Find the maximum or minimum value of y=x^2 - 4x + 7, and the value of x for which it occurs by completing the square

[vernal kite]

which links to this

the answer states that it's 3 at x=2

i get the x=2 part but how is it a positive 3

[azure horizon]

so what do you get when you complete the square

[vernal kite]

(x-2)^2 -3 =y

[azure horizon]

oh

that's not right

when you complete the square it should be (x-2)^2 + 3

check your working maybe

[vernal kite]

so its x^2 - 4x = y-7 right

[azure horizon]

no

wait yes

that's right

[vernal kite]

then x^2 - 4x + 4 = y - 7 + 4

[azure horizon]

yep

[vernal kite]

OHHH

i put 3 instead of -3

[azure horizon]

yeah probably just a minor arithmetic mistake

yeah

[vernal kite]

yepyep

ty for that

so this is a minimum value correct?

minimum value of 3?

[median hazel]

#help Find the limit as x approaches 0 of (sin(x))/x.

[vernal kite]

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[small slate]

Hey guys I can you help me out how do i define this function in point(0,0)

i mean i will do it with lim aproaching the values but I cant wrap my head around how to do it

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[sterile jasper]

I am looking for resources on how to perform rotations on a vector3 using Euler angles.

I am in the process of implementing a linear algebra library in C++, but I do have many weaknesses in math. (though I am trying to fix this and learn!) I am looking for resources that break down how to perform rotations on vec3 using Euler angles. (Ideally something geared towards beginners!)

ALSO, side question. Would it make sense mathematically to allow rotations to be performed on vec4?

(CONTEXT: This is for a linear algebra library thats primary focus is game development)

[sterile jasper]

Any insights on the matter would be greatly appreciated!

Also ideally any resources provided that takes computational complexity in mind would be extremely preferred as I am trying to implement performant code. Though it is not a requirement!

[cedar kiln]

@sterile jasper Has your question been resolved?

[floral arrow]

A quick google search yielded this:
https://gamemath.com/book/orient.html

As for your side question, of course it makes sense to perform rotations on 4D vectors, but I can't tell you whether you want to or not

4D space has 10 degrees of freedom: 4 for position, 6 for orientation

As opposed to the 6 degrees of freedom of 3D space, 3 for position and 3 for orientation

So in 4D you'd need 6 "Euler angles" to represent an orientation

[sterile jasper]

I guess the question is if it makes sense in the context of a game setting. I've personally never had to perform an operation as such as quaternions generally handle anything like that, but I'm also wondering if others may desire that functionality.

[floral arrow]

Probably not unless they're making a game that deals with 4D in some way

[sterile jasper]

True, it does feel to me as an issue for high dimensions and not so much 3d, but I'll leave this open for a bit longer to hear other inputs. ^^

[sterile jasper]

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[winter lynx]

Hello, what’s the derivation of a number^2?

[winter lynx]

For x^2 it’s 2x

But for a normal number idk

[idle tusk]

derivative of constant is always 0

whether it's 9, 9^2 or whatever, the derivative is 0

[surreal cave]

$\frac{d}{dx}[c]=0$

[wraith dagger]

PajamaMamaLlama

[winter lynx]

Right right I forgot that

Thanks 👍

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[crimson sedge]

hi

[idle tusk]

[crimson sedge]

Please, read last two lines

All these four formulas are same ??
Please tell me the what are these formulas/symbol called?

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[heavy steeple]

Im reviewing a practice midterm for theroy of computation. b seems more correct than c since c is dependent on the number of states of the NFA. 2^n = 77. where n is the number of of states in the NFA.

Both seem correct but b is not dependent on anything where as c is. This is where im confused.

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[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

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[manic pumice]

is this isomorphic?

[south tundra]

What do you think?

[floral arrow]

"this" is pretty vague

But, after making some assumptions about what "this" is, I believe your question translates to "can you pair up every u with a v such that the two figures are identical"

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[drowsy egret]

you have: sin²x+sqrt(2) cosx + cos(pi/3)
prove that it's equal to -(cos x - sqrt(2)/2)²+2

[drowsy egret]

also x [0 ; pi] if that matters

[long arrow]

so

in other words prove that

[wraith dagger]

Modus

[long arrow]

yea?

[drowsy egret]

ye

[long arrow]

what have you tried?

[drowsy egret]

like

simplifying the square

but at the end i can't manage to get the cos pi/3

without a +1 or smth

[long arrow]

cos(pi/3) = 1/2

[drowsy egret]

yeah ik

[long arrow]

so it gives

-(cos^2x - sqrt(2)cosx + 1/2) + 2

= -cos^2x + sqrt(2)cosx - 1/2 + 2

= (1-cos^2x) + sqrt(2)cosx - 1/2 + 1

= sin^2x + sqrt(2)cosx + 1/2

= sin^2x + sqrt(2)cosx + cos(pi/3)

[drowsy egret]

and the +1 from here

[long arrow]

-1/2 + 1 = 1/2

[drowsy egret]

oh

thanks

lol

forgor that it's NEGATIVE 1/2

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[dire geode]

is that supposed to be math

if so just ask

[queen ice]

Just ask the question you opened this channel for

No need to wait for someone who is familiar with the topic of the question

They’ll stumble in here eventually

[vast pike]

you are fair

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[eager lance]

I need help

[eager lance]

with physics

Its like math right

[tropic oxide]

!da2a

[cedar kiln]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[eager lance]

oh sorry

but is anyone able to help me with this

idk if im doing it right

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[verbal compass]

help

<@&286206848099549185>

[dire geode]

!15m

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[pulsar turret]

I need help with complex numbers

[pulsar turret]

It doesnt look like it is in cartesian form like i usually see it

what should i do?

any help would be great thanks

[vast pike]

in general $|z-w|=d(z,w)$ where $d$ is the classic euclidian distance

[wraith dagger]

everg

[pulsar turret]

so in this case

the x-axis

refers to z?

which is 1?

[vast pike]

in fact $|z|=|a+ib|=\sqrt{a^2+b^2}=d(z,0)$

[wraith dagger]

everg

[vast pike]

sorry i must go

[pulsar turret]

ok thanks

i will try to understand

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[uneven osprey]

why isn't this just ln(e^x^5) or x^5?

[austere merlin]

Chain rule

[uneven osprey]

so how would i get the answer here

[austere merlin]

F'(x) = f(x), but F'(g(x)) = g'(x)F'(g(x))

ill let u figure it out

[uneven osprey]

i might need a bit more help than that

let me try to work thru here

is it just e^x(ln(e^x^5))

[cerulean sail]

Yes, but this can be simplified (also for the exponents)

[cerulean sail]

[Note ln( (e^x)^5 ) and ln( e^(x^5) ) aren't the same]

[uneven osprey]

x^5e^x(ln(e)) which could just be simplified to x^(5)e^(x)?

[cerulean sail]

Not quite either

(Think that you're putting u= e^x in the place of ln(u^5) - maybe think about the rules of exponents/logs a bit)

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[runic prawn]

how would u start this if x^2+2x+10 isn't factorable

[crimson sedge]

Partial fraction decomposition

[runic prawn]

yea but I don't need to factor it furgher?

further

[echo sedge]

It can be written as (1/(x+3)(x^2+2x+10)) * 2x^2+x+11

[crimson sedge]

Would you use integration by parts?

For that

[runic prawn]

that looks messy tho

and the ws is specifically partial fractions decomp

[crimson sedge]

Like what I would do is distribute x+3 into x^2+2x+10 and use long division

[runic prawn]

ohhhhh

ok

like distribute

[crimson sedge]

yeah

[runic prawn]

wait

I'm dumb

how does that work

[crimson sedge]

Now use long divison

once you find that

[runic prawn]

but the numerator has a bigger degree

smaller degree**

[crimson sedge]

it'll be easier to integrate

[runic prawn]

could I do partial fractions like this

[crimson sedge]

my apologies for the incorrect methods I gave you before, I have just figured out how to do it after grabbing a piece of paper and working on the problem myself

yes you do that

just perform partial fractions like you would using that setup

[runic prawn]

okok

[crimson sedge]

tell me what you got for A, B, and C and ill check if I got those as well

[runic prawn]

would the x value just be -3

[crimson sedge]

wdym x value?

C has a value of -3 yes

I assume that’s what you meant

[runic prawn]

uhhhh

like to find b or c

I got 2 for a

[crimson sedge]

That’s right

To find C, you have to use the coefficient of the constants

Remember constants are multiplied by x^0

[runic prawn]

what ab b

[crimson sedge]

did you solve the system of equations

because those depend on the coefficients

of the x^2, x, and constant terms

[runic prawn]

uhhh idk how to find b

[crimson sedge]

A+B = 2

10A + 3C = 11

2A + 3B + C = 1

A = 2-B

C = 1 - 2A - 3B

Plug those in to 10A + 3C = 11

[cedar kiln]

@runic prawn Has your question been resolved?

[runic prawn]

c = 1

b = 0

a = 2?

[crimson sedge]

your b and a values are right, but your c value is wrong

[runic prawn]

oh

1 = 3b+c

b = 0

1 = c

[crimson sedge]

what happened to A

[runic prawn]

OH

mbmb

i forgot the 2ax

[crimson sedge]

np

lmk what you got for your integral

[runic prawn]

c = -3

[crimson sedge]

yep

[crimson sedge]

\int \frac{2}{x+3}dx+\int \frac{-3}{x^2+2x+10}dx

[crimson sedge]

yeah

you have to complete the square for the 2nd integral

for the denominator

[runic prawn]

but its not factorable

[crimson sedge]

yeah thats why you complete the square

x^2+2x +1 = -10 + 1

(x+1)^2= -9

(x+1)^2 + 9 = 0

I assume you've learned how to do trig integration ?

[runic prawn]

yeah but don't rly remember 😭

[crimson sedge]

np ill help you

ok so

,,\int \frac{-3}{x^2+2x+10}dx

[wraith dagger]

Paradox

[crimson sedge]

,,-3 \int \frac{1}{(x+1)^2+9}dx

[wraith dagger]

Paradox

[crimson sedge]

,,u = x+1, du = dx

[wraith dagger]

Paradox

[crimson sedge]

,,-3 \int \frac{1}{(u^2)+9}du

[wraith dagger]

Paradox

[crimson sedge]

okay, are you seeing what's happening here

you have to make it equal to the derivative of arctan

for this problem specifically

but for other problems, depending on the terms and values, you have to make it equal to the other inverse trig functions

[runic prawn]

ohhh

yes

[crimson sedge]

,,\frac{d}{dx}\left(\arctan(x)\right) = \frac{1}{x^2+1}

[wraith dagger]

Paradox

[crimson sedge]

can you do the rest on your own

[runic prawn]

uhhhh

[crimson sedge]

or do you still need help

[runic prawn]

wait so where does the 9 go

[crimson sedge]

you have to do another sub

v = u/3

dv = 1/3 du

or you could factor it out

and get u^2/9 + 1

then set v = 3u

it doesnt matter

[runic prawn]

sorry gimee a sec

[crimson sedge]

np

[runic prawn]

what is after this step?

,rotate

[crimson sedge]

do another substitution

[runic prawn]

u can do that?

[crimson sedge]

Yeah

[runic prawn]

o didn't know thats why was confused

[crimson sedge]

Np, now you know so that’s good

[runic prawn]

and where u get u/3 from?

[crimson sedge]

You can factor 9 out from u^2 + 9 to get 9(u^2/9 + 1) and then you can find that u^2/9 = (u/3)^2

so you set v = u/3

then find that u=3v

so you plug that in

for u

and du = 3dv

[runic prawn]

ok uhh

what would i then plug in du for

or is it just 9 intgral(1/(9(v+1)))

[crimson sedge]

3/9 = 1/3

1/3integral(1/v^2 +1)dv

-3(1/3)arctanv

-arctanu/3

-arctan(x+1/3)

[crimson sedge]

du = 3dv

[runic prawn]

but how is it 3/9 if it's 3dv

then 1/3du

then outside integral is 3 no?

[crimson sedge]

because you’re replacing the du with 3dv

[runic prawn]

,rotate

[runic prawn]

sp it would be dv for the last integral?

so

[crimson sedge]

it's supposed to be v^2 = u^2/9

v = u/3

[crimson sedge]

yes it would

3dv

[runic prawn]

so v^2+1?

and then the number in front of integral is 9

[night matrix]

do u know this?

hey!

[cedar kiln]

@runic prawn Has your question been resolved?

[runic prawn]

hhh

what does it say after the +

[runic prawn]

.close

[cedar kiln]

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[hasty skiff]

anyone know a faster way of factoring polynomials? The videos online dont help and neither does the lesson. I know how to find the gcf and factor with that, as well as finding with using the provided formula Ax^2 + bx = c, but this just really twists my mind and I need a easier way to achieve the answer.

[plucky owl]

Those are just big numbers, there's not going to be a very fast way with large numbers

[hasty skiff]

Alr, how can I solve this

[plucky owl]

Can you use a calculator?

[hasty skiff]

i can but I prefer paper

[plucky owl]

It's large numbers so a calculator would benefit a lot

[median holly]

If you notice this the discriminant will always give 1

[hasty skiff]

this is how they show me

[median holly]

Wait it will add nvm

[median holly]

You wont be able to do that

For bigger number

Easily

[hasty skiff]

😦

[plucky owl]

Because with large numbers, I would use quadratic formula tbh to make it faster

[median holly]

Notice this

Its 33 34 35

You k ow the quadratic formula?

You can write 33 x 35 as 34²-1

[hasty skiff]

nope

[median holly]

Hmm

Then you factor 33 x 35

[hasty skiff]

I know that when we multiply it doesnt matter where the numbers are in the equation

[median holly]

3 x 11 x 5 x7

And try to combine this

In such a way

That difference

Is 34

[hasty skiff]

correct

[median holly]

Of the two terms

Can you spot it

?

Faster spotting will only come by practice

[hasty skiff]

A faster method I can think for me is to factor the 33r^2 and the -35, which u just did there

alr lemme re organize my brain rq

alr

@median holly ty lol

its funny how its so easy after u get it

and so impossible and mind twisting before

but yea, I got (3r-5)(11r+7)

Basically once you factor the Ax^2 and C, they will have the same results as the middle nukmber

ty

close sesame

wait its not a -5, its a -7

[cedar kiln]

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[ionic wolf]

What am I even reading😭

Wat does "every point A" mean

[kindred storm]

It means you can pick any point on the line and name it "A".

So, you point somewhere at the line, and that's the point A you're talking about.

So, basically what it's saying is that every point on the line maps to itself.

[ionic wolf]

Uhhh is it rotation

Idek

It says circle

[kindred storm]

Well, what you want to do is draw a line.

Just get some paper and draw a line.

Then, draw a point that's not on the line.

Then try to see what P' would be for that point.

For your guess at P', draw a line between P and P'.

Halfway along the line, it has to hit the line l you started with.

[ionic wolf]

Uhhh where's the transformation fr

[kindred storm]

What do you mean?

[ionic wolf]

Uhhh

Where is the transformation

Idk what else to say

Wat is transforming

[kindred storm]

Well, it takes a point and gives you another point.

That's what the transformation does.

It's like a function.

[ionic wolf]

Uhhh

Is it reflection

[kindred storm]

OK, why do you think it's reflection?

[ionic wolf]

Cuz like

P reflected across line L

Idk

😭

[kindred storm]

Well, let's say you draw the line l and a point that's not on it.

[ionic wolf]

Oh I guess it prob wouldn't be perpendicular

Most likely idk

[kindred storm]

So, you have a line l and a point P.

So, P and P' are on a circle together.

And the center of that circle is on the line l.

[ionic wolf]

Idk what else

[ionic wolf]

I thought of it as a slanted line

[kindred storm]

Does that make sense so far?

[ionic wolf]

It also say line p p' is perpendicular frfr

[kindred storm]

Right.

[ionic wolf]

But I guess rotation idk

[kindred storm]

Well, let's see.

Keep working with the rules.

P' is on a circle with P.

The line from P' to P is a diameter.

So, it has to go through the center of the circle.

So, P' can't just be anywhere on the circle.

It has to be directly across from P.

[ionic wolf]

Idk it looks like it could be any 😭

[kindred storm]

So, that's the circle part handled.

Does that make sense so far?

[ionic wolf]

I guess

But wat if the line was like slanted fr

Idk

[kindred storm]

Well, just rotate my picture a bit.

[ionic wolf]

okkk but couldn't u just translate it

To the left

[kindred storm]

Yes, it works for any line.

I just made it vertical.

You could move it around on the x-y plane or rotate the line.

So, it can be slanted if you want.

But what they're trying to do here is to give you a new mathematical ability.

They don't just tell you what kind of transformation it is.

They give you the conditions it has to meet.

Then you have to think through how those conditions affect how it has to work.

Then you have to figure out what kind of transformation it is.

To think through how those conditions affect it, you need to go through them one by one and figure out how the first one limits you, how the second one limits you further, and so on.

[ionic wolf]

Is it rotation

Or reflection

Idk

[kindred storm]

Well, that's the thing.

[ionic wolf]

I FEEL LIKE IT COULD BE ANY

[kindred storm]

You're skipping the lesson they're teaching.

This is supposed to be somewhat difficult.

But for a good purpose.

A lot of more advanced math stuff is like this.

[ionic wolf]

I think it is rotation cuz it says circle

[kindred storm]

Rotation about what point?

[ionic wolf]

Umm idk the middle of the circle centered on l

[kindred storm]

OK, so which point on l?

Rotation on a 2D plane is where you pick a point and spin everything around it.

This method of figuring it out won't work, though.

[ionic wolf]

Oh

[kindred storm]

You can't just skim over the things, see a few words that stick out, like circle, and then guess.

You have to really try to understand how the first condition restricts you, how the second condition restricts you further, and so on.

[ionic wolf]

I think reflection

[kindred storm]

Why?

[ionic wolf]

Ummm

NVM

Idk how to make it perpendicular

[kindred storm]

Well, look at my prior drawing: https://cdn.discordapp.com/attachments/903430334681608232/1169465221115543552/F40D9C6D-572E-4590-B1D8-4B76A0E9F54A.jpg?ex=6555805c&is=65430b5c&hm=30efdc23f597c316d41c00e2d723b553527fbd3243997f0d5278e669818fef75&

That's not perpendicular, right?

[ionic wolf]

Oh wait doesn't it have to be a reflection cuz of the first thing

[kindred storm]

How does the first thing make it a reflection?

[ionic wolf]

Uhh every point A on line l map to itself

So likee

It will map to itself

[kindred storm]

Right.

[ionic wolf]

If it's the line of reflection then it would just be itself

After

[kindred storm]

Yes, that's true.

Can the other two transformations meet that requirement?

Like does translation allow for that?

[ionic wolf]

I don't think soo

[kindred storm]

Why not?

[ionic wolf]

It would move the line somewhere else

And it would not map to itself

Anymor

[kindred storm]

Well, it would have to move a point on the line somewhere else.

Right.

What about a rotation?

[ionic wolf]

Uhhh

I think that would move it around too

[kindred storm]

Why?

[ionic wolf]

Cuz like

Rotations like

Move thing around

[kindred storm]

Yeah, rotations will move every point except the center of the rotation.

[ionic wolf]

Fr

So tis reflection right

Its

[kindred storm]

Yes.

[ionic wolf]

Fr

So cool

[kindred storm]

When you add in perpendicular, P' is right across from P.

So, if you were to make a bunch of Ps that spelled out Hello, it would do a mirror image on the other side of the line.

[ionic wolf]

Fr

.close

[cedar kiln]

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[bronze briar]

i don't really know what i am doing. So the change in f(x,y) is the vector <2x, 2y>.
Am i just taking the dot product of this and the unit tangent vector?

[cedar kiln]

@bronze briar Has your question been resolved?

[bronze briar]

<@&286206848099549185>

[cedar kiln]

@bronze briar Has your question been resolved?

[cedar kiln]

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[grim yacht]

https://gyazo.com/c2bcc7beb031db6b0d042056dfa53bb6'

[grim yacht]

can anyone help me on how i should start this problem?

https://gyazo.com/c2bcc7beb031db6b0d042056dfa53bb6

[runic garnet]

Draw a triangle

Right triangle

[cedar kiln]

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[cedar kiln]

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[random kelp]

ive found that x=1 at some point

but im not sure of how to write the intervals

[tropic oxide]

can you show your work...?

[random kelp]

oh wait

i think its just equal to 0

[steep minnow]

.close

[random kelp]

...

[tropic oxide]

this isn't your channel. you can only close your own channel.

@random kelp show your work!

[random kelp]

i will

[steep minnow]

oh

[tropic oxide]

what is this equals sign doing here?

[random kelp]

i just wrote it

it helps me

but ignore it, they arent equal to each other

[tropic oxide]

you should not misuse the equals sign this way.

[random kelp]

okay

but ignore it

[tropic oxide]

you applied the ratio test to this series.

do you agree or disagree that that is what you did?

[random kelp]

yes

that is what I did

[tropic oxide]

ok

can you tell me what the ratio test says?

i.e. how to properly interpret the value you get from it

[random kelp]

if the result is less than 1

it is convergent

if it is greater than 1 it is divergent

[tropic oxide]

right.

[random kelp]

if it is equal to 1 it is inconclusive

[tropic oxide]

not happy about your overuse of "it", but whatever.

you got that the result is zero for all x.

what does this mean for the region of convergence of your series?

[random kelp]

It converges to the value of 0?

[tropic oxide]

wrong

[random kelp]

damn

[tropic oxide]

nobody said anything about the value of the series,

and it is NOT under consideration AT ALL here.

[random kelp]

then I dont know

[tropic oxide]

for which values of x does the series converge?

keeping in mind that the ratio test returns 0 across the board.

[random kelp]

if it just returns 0

does that mean it doesnt converge at all for all values of x

[tropic oxide]

if the result is less than 1
it is convergent

[random kelp]

ok then

it converges

[tropic oxide]

"it converges EVERYWHERE" is what i wanted to hear from you.

[random kelp]

okay

so the bounds are -inf and inf

[tropic oxide]

yes of course

[random kelp]

can I ask about this then

I solved for the value of x

but I dont understand why its not including 0 and 14

[tropic oxide]

replace x with 0 in your series

what do you get?

[random kelp]

-1

[tropic oxide]

no, you do not simply get -1

do it properly

on paper right now

[random kelp]

-7^n/7n

[tropic oxide]

do it properly and on paper

or in your notes app or whatever

you are being sloppy right now and this is your downfall

[random kelp]

well I get something like 7^n/7^n+1

shiiiiit idk

[tropic oxide]

fucking hell.

[random kelp]

i do

[tropic oxide]

ok

then show me

SHOW me

a screenshot or pic

of what you're doing

[random kelp]

??

[tropic oxide]

ok first off you should really be crossing your sevens

but also where did the sigma symbol vanish to?

also (-7)^n not -(7^n).

[random kelp]

okay

[tropic oxide]

what you should have had is: $\sum_{n=1}^{\infty} \frac{(0-7)^n}{7^n} = \sum_{n=1}^{\infty} \frac{(-7)^n}{7^n} = \sum_{n=1}^{\infty} (-1)^n$.

[wraith dagger]

Ann

[tropic oxide]

do you understand this yes or no

[random kelp]

yes

[tropic oxide]

do you understand why this diverges yes or no

[random kelp]

shit

im dividing by the coefficients bruh

[random kelp]

well i know there is divergence when the answer is greater than 1

oh hold on

[tropic oxide]

fucksdgjhldjfg.

[random kelp]

I only get 2 ansqwers

so it doesnt converge

so it must diverge

i understand why it diverges

[tropic oxide]

ngl "i only get two answers" kinda reeks of bullshit.

the reason that the series $\sum_{n=1}^{\infty} (-1)^n$ diverges is simple: the terms fail to approach 0.

[wraith dagger]

Ann

[random kelp]

i see

ok i understand

[cedar kiln]

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[barren lodge]

@cedar kiln

[hollow cloak]

/?

[barren lodge]

Hi
Do you remember that formula
I mean the sigma function that we can calculate sin of an angle?
I try 150 with that but it was sooooo diffrent with calculator

[hollow cloak]

You mean sin of 150?

Did you change it to degrees?

Or maybe it's supposed to be used with only radians

[barren lodge]

How I should?

[hollow cloak]

The sum probably is only used for radians not degrees

[cedar kiln]

@barren lodge Has your question been resolved?

[cedar kiln]

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[vital sphinx]

Find the McLaurinseries of 1/(2-x) using the reference series 1/(1+x)

[vital sphinx]

How would i start with this? I have rewritten 1/(2-x) as 1/(1+(1-x))

do i Just substitute (1-x) in the reference series?

[ebon pulsar]

yes

exactly

u already have the series for 1/(1+x)

so ye just sub 1-x

[vital sphinx]

The answer should be the last line. Substituting doesn't seem to work

[ebon pulsar]

u should get something like:
1/(1+(1-x)) = 1 - (1-x) + (1-x)^2 - (1-x)^3 + (1-x)^4 - ...

its working

[vital sphinx]

So the series are equal even though they are written differently?

[ebon pulsar]

ye

[vital sphinx]

How would I find the convergence area after that?

[ebon pulsar]

apply the ratio test

use this:
L = lim (n -> ∞) |a_(n+1)/a_n|

[vital sphinx]

Is this correct? According to my answer sheet is should be ]-2, 2[

[cedar kiln]

@vital sphinx Has your question been resolved?

[barren lodge]

So what I should do

[cedar kiln]

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[supple pike]

Sue bought 2 raffle tickets for a fundraiser. 3 tickets are drawn out of 100 for 3 different prizes. What is the probability of Sue winning exactly 1 prize? I am lost

[tropic oxide]

how much combinatorics do you know?

[supple pike]

Uhm... none?

Idk what that is

[tropic oxide]

do you know what the notation nCk means

or maybe $\binom{n}{k}$

[supple pike]

no sorry

[wraith dagger]

Ann

[tropic oxide]

no?

[supple pike]

no I'm at a year 10 level

[tropic oxide]

"year 10" differs from place to place. i've known students younger than that but familiar with the notation.

so that doesn't tell me much

anyway, ok, let's try a different angle

do you know basic probability

[supple pike]

yeah

[tropic oxide]

ok

can you make a tree diagram?

[supple pike]

yeah

[tropic oxide]

ok, then make one.

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[somber prism]

How do i escalate after tan = 3/4 simple question

[cedar kiln]

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[ebon pulsar]

use inverse trigonometric functions.

In this case, you can use the arctan

or tan^-1

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[paper summit]

title: basic linear algebra steps of introductory qm.

[paper summit]

for the first one, all I know is that they are using A|a'> = a', since a',a'',... are the eigenkets of A. Not sure though how it simplifies.

for the second one, I just need help filling in 1.3.33. I get delta_a'a"" = <a'|a''> but even adding that the step is not obvious to me.

just some pinged hint would be great, thank you

other givens are that A is hermitian

<@&286206848099549185>

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.close

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[prisma nymph]

How?

[prisma nymph]

finding the adjacent

[crimson sedge]

AC or AB?

[prisma nymph]

AC

[crimson sedge]

Do you know the sine function?

For example what is sinC?

[prisma nymph]

Sin = opp/hyp ?

[crimson sedge]

Yes

So replace opp/hyp with the lines

Eg. AC, BC, AC

[prisma nymph]

I got sin = 9/hyp for examples or tan = 9/adj (AC)

how to solve to get 26?

I tried

[crimson sedge]

$sin(A) = frac {AC} {BC}$

[floral arrow]

@crimson sedge You may want to take a look at the figure again

[crimson sedge]

@floral arrow isn't this correct?

[floral arrow]

No

[prisma nymph]

I got it

tan 19=9/x
0.345x=9
x=9/0.345
x=26

thank you

[floral arrow]

Yes

[prisma nymph]

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[eager steppe]

Complex and Imaginary numbers, Exponential Function. I can translate from French to English. Any takers?

[dawn junco]

des idées ? @eager steppe

[eager steppe]

aucune.

[dawn junco]

ok

déjà calculer z^2019 ça devrait pas trop poser de problème

en notation exponentielle en tout cas

[eager steppe]

Mais I n'a pas de valeur?

[dawn junco]

?

[eager steppe]

i

[dawn junco]

ok et alors ?

[eager steppe]

L'imaginaire n'a pas de valeur veritable pour etre utilise comme exponentiel.

[dawn junco]

faut que t'ailles lire des trucs sur les exponentielles complexes alors

[eager steppe]

Z^2019= 5e^i*(pi/3)+2019

[eager steppe]

Ok je vais aller regarder quelques videos.

[dawn junco]

nope

ouais il y doit bien il y avoir des tonnes de trucs sur yt

[eager steppe]

comment nope?

[dawn junco]

c'est pas bon ton calcul

[eager steppe]

(a^2)^2 = a^4 ouais?

[dawn junco]

(a^2)^2 = a^(2*2) oui

alors pourquoi **+**2019 dans ton cas ?

[eager steppe]

?

une seconde

[dawn junco]

(a^m)^n = a^(m*n)

[eager steppe]

(a^2)^2 = a^(2+2) ????

Ah c multiplie

[dawn junco]

t'es juste tombé sur un cas un peu con parce que 2+2 = 2*2

dommage

[eager steppe]

bon, meme comme ca c...

[dawn junco]

et puis aussi $z^{2019} = (5 e^{i\frac{\pi}{3}})^{2019} = 5^{2019}e^{2019i\frac{\pi}{3}}$

[eager steppe]

z^2019= 5e^i(2019*pi/3)

[dawn junco]

faut pas oublier le 5 non plus

[wraith dagger]

aPlatypus

[eager steppe]

ouais mais

comment peut-on dire que c'est un reel???

peut-etre car i ici est un simple expo?

[dawn junco]

2019pi/3 c'est l'argument de ce nombre complexe

[eager steppe]

ok

[dawn junco]

si ça se trouve 2019pi/3 = 0 ou pi mod 2pi

[eager steppe]

Bon arretons

[dawn junco]

et dans ce cas là, le nombre est réel

[eager steppe]

il faut que je revois ma lecon

[dawn junco]

ouais

allez courage

[eager steppe]

Merci pour l'aide.

J'apprecie.

🫡

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[versed plover]

Can someone help me with this

[versed plover]

or like can someone help me revise it to make it a proper inverse proportion problem

4x/4 2/4
x = 1/2

1/2 is = to 30 mins ?

So it will take 30 mins for 4 farmers to plow the field ?

Hmm alryt alryt ty

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[high thicket]

Are there any textbooks that cover all of high school maths (from rudimentary arithmetic to calculus) in one volume?

[tropic oxide]

such a textbook would have to be massive, no?

[high thicket]

I’m an adult learner who dropped maths a long time ago who wants to have a solid foundation.

[high thicket]

I’m thinking of something like a modern version of Serge Lang’s “Basic Mathematics”

[primal socket]

Hm

Are you strictly limited to textbooks?

[cedar kiln]

@high thicket Has your question been resolved?

[high thicket]

Not necessarily, anything that can be gone through independently (without a teacher)

[tropic oxide]

http://khanacademy.org/