help-10

[upbeat island]

yo no worries haha

[candid canopy]

so,

lemme do 3 realquick

reflecting the porabla

[versed cave]

My bad, I didn't notice you were doing the exercise 50

[candid canopy]

all g

what do i do for 4?

half the shape intercepts with the line

[upbeat island]

so you could think of it as going zero steps away

when it touches

but the same rule applies

[candid canopy]

this?

i drew it a lik too far

but somn like this yea?

[candid canopy]

^^

[upbeat island]

hmm

close

the bottom left of the og rectangle

should be to the left of the reflection line

[candid canopy]

oh

wait why to the left?

?

[upbeat island]

well it should be on the opposite side

ah noo

lol

[candid canopy]

whaaat

i dont get whatchu mean 😭😭

[upbeat island]

do you have an actual mirror?

[candid canopy]

no

[upbeat island]

or a piece of glass like a smartphone?

[candid canopy]

yes

[upbeat island]

you can put that against the line and maybe visualize what the other side should look like

to build some intuition

[candid canopy]

i get what u mean but i dont get it

[upbeat island]

it's ok

let's do it numerically then

take the bottom left of the og rectangle

you see how it's down one from the line?

[candid canopy]

yea

[candid canopy]

.

[upbeat island]

think of down turning into left and up turning into right

this is so hard to explain in text lol

better done visually

[candid canopy]

yea

i dontrly get it

[nocturne minnow]

So I could be thinking of this wrong since I'm not doing it on paper, but here's what you could try. Put your pencil, on a grid point, that is on the shape, move your pencil to dotted line and count how many squares it takes, then go perpendicular, on the other side(empty side of the line), that many spaces

[candid canopy]

?

[nocturne minnow]

Almost, because you need to do the bottom part too

[candid canopy]

wdym

[nocturne minnow]

The bottom part isn't reflected properly

[upbeat island]

ohh this might help

turn your paper so that the dotted line is straight up and down

then use your finger for the original and match it with your pencil for the reflection that you're drawing

[candid canopy]

hmm

[nocturne minnow]

Same idea, put your pencil on a grid square, say the bottom left corner of the rectangle, move up to the dotted line, counting how many squares it took, then go perpendicular, to the left, that many squares

[candid canopy]

idek why this is so complicated for somn so easy

i wouldve been able to do in seconds

last year

[upbeat island]

it's more like we can't just point to the same sheet of paper over the internet lol

[nocturne minnow]

I'm telling you what you should do. It's not complicated

[candid canopy]

idk i understand/dont

[candid canopy]

if ur on phone u can like screenshot and edit it and draw on it

[nocturne minnow]

Put your pencil, on the bottom left corner of the rectangle

[candid canopy]

done

[nocturne minnow]

Move your pencil up to the dotted line, counting the number of grid spaces it took

Then go perpendicular, to the left that many spaces

[candid canopy]

ahh

i got it

ty

[nocturne minnow]

What do you have now?

[candid canopy]

thats all

ive finished it

[candid canopy]

ty

.close

[obtuse pebble]

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[proven zephyr]

is there a particular way that you're taught on how to do this?

[dire agate]

This was in a test so no

[proven zephyr]

i have a manual way of doing it but idk how to prove it

[dire agate]

Can I see your manual way?

[proven zephyr]

171 * 6 = 1026

= 9

9* 111 = 999

[brave bramble]

Because it was on a test, I would think you were given a way to do it haha

[proven zephyr]

yeah exactly

[dire agate]

Well more so a competition

[brave bramble]

Fair I see

Then yeah whatever tools are good

[upbeat island]

can you factor out a six

then manually do some of it and find the pattern when you put the six back in

[brave bramble]

Taking mod 9, gives the sum of digits, mod 9.

But that's obviously not enough by itself

[proven zephyr]

wait.. do you add the sum again if it has more than 1 digit?

[dire agate]

I'm honestly not sure how to do this question

[proven zephyr]

no right?

[dire agate]

I don't think so

[proven zephyr]

oh ok

[dire agate]

I got 999 but I'm not sure it's right

[proven zephyr]

that's what i got too

[upbeat island]

so 1026 * 111...1 with variable numbers of ones looks like 113999...999886 with variable numbers of nines

[proven zephyr]

999 is the answer no?

[dire agate]

???

I don't know the answer

[upbeat island]

i think 111 ones will have 108 nines

[proven zephyr]

care to explain?

[dire agate]

It has to be lower than 1000 due to the only option being a 3 digit number

[upbeat island]

i crunched it out for different amounts of ones

i also get 999 then

[proven zephyr]

yeah, i think that's the answer

[dire agate]

I used a computer to solve it, it said 999 aswell, thank you :)

[upbeat island]

you can do it by hand

[proven zephyr]

i just.... multiply 171 by 6 to get 1026
add the numbers together
1 + 0 + 2 + 6 = 9
and since there is 111 sixes
111 * 9 = 999

this is my work

[dire agate]

I did the same way originally

[upbeat island]

can you do the digit sum out of order like that?

[dire agate]

But I just needed to double check

[dire agate]

I think so

Anyhow I got the answer thank you

.close

[obtuse pebble]

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[ivory pilot]

with f(x) = (2^x)(cos x) and the interval 0 ≤ 𝑥 ≤ π, find the critical points and state the nature of the critical point

[ivory pilot]

So I found f’(x) which is

2^x[(ln2)(cos(x)) - sin(x)]

I set it equal to zero, cancel out the 2^x but then I’m stuck with (ln2)(cos(x)) = sin(x)

Anyone got any idea how to solve this ?

[upbeat island]

make tangent?

[ivory pilot]

OH

OMG

SORRY

[upbeat island]

cosine is never zero in the domain that's the only thing to be careful of

[ivory pilot]

IDK HOW I DIDNT SEE THAT

OKOK

[upbeat island]

lool nw that took me a minute also

[brave bramble]

Still not really possible to get x by hand tho

[ivory pilot]

And wait

How would I know the nature of the critical point

Without second derivatives?

[brave bramble]

But x = arctan(ln(2))

[ivory pilot]

I think we can use calculators

It’s a practice question for an assignment we are getting

[brave bramble]

Why can't you second derivative?

[ivory pilot]

We technically didn’t learn it yet

But I kinda sped up and learnt most of the course on my own

But I’m not sure how she expects us to do it

Without using concacvity

And second derivatives

[brave bramble]

Well, okay. You can always check points near your point, to see if the curve is increasing or decreasing

f(arctan(ln(2)) is not really doable by hand though haha

Then check something like f(0) to see if that's a minimum or maximum

[ivory pilot]

Right right

Okok thanks

Wait sorry

Another QUESTUIN

Question*

So I got 0.61 rads right

But when I graph it on desmos

WAITTTTTTTT

I’m an idiot nvm

Hold up mb

Actual question now

So I got x = 0.61 rads right when making f’(x) = 0

[obtuse pebble]

@ivory pilot Has your question been resolved?

[obtuse pebble]

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[marsh sedge]

Should I try to rewrite the left side using the Taylor series of cos(x)?

[forest sinew]

yea that sounds reasonable i think

youre gonna have to deal with that powers of 2

[marsh sedge]

-(cos(x)-1) times the infinite series of [(-1)^n * 2^(2n+2)] = 1/2

Does the divergence of the the series of powers of 2 mean anything for solving for x?

[obtuse pebble]

@marsh sedge Has your question been resolved?

[marsh sedge]

$1/2 = \ -{(cos(x) -1)} * \sum_{k=0}^{+infinity} \ (-1)^n 2^{2n+2}$

[warm shale]

theworstbesttourist_20

[marsh sedge]

How do I solve for x on interval [0,pi] given this equation?

<@&286206848099549185>

[tardy epoch]

Your series doesn't even converge

[tardy epoch]

Oh this is the question

[tardy epoch]

This isn't a helpful expansion. Try to get something that converges

[marsh sedge]

I'm lost, the sum looks like an alternating series

ln(1+x), sin(x), cos(x)

Maybe arctan(x)

Wait got it!

Thanks @tardy epoch

.close

[obtuse pebble]

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[echo steppe]

he

[echo steppe]

Simplifying equation

cos a cos (b - a) - sin a sin (b-a)

I just need help figuring out how to simplify it

[timid silo]

what rules do you know

[echo steppe]

Not many my brains been off due to me being extremely sick and I’m just trying to get little bits done

[timid silo]

do you have a picture or note?

[echo steppe]

Hold on I’ll get the picture

C for reference

[echo steppe]

i did a and b in schoo and went home

[timid silo]

sorry i mean, do you have any trig rules that you know

that you can apply

[echo steppe]

I’ll check my folder

Like this?

[echo steppe]

not sure what i can use here

[timid silo]

look at your note for $\cos(\vartheta+\phi)$

[warm shale]

jswatj

[echo steppe]

Ohh thankyou!

.close

[obtuse pebble]

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[static shoal]

Help

[static shoal]

How should I continue from here?

[obtuse pebble]

@static shoal Has your question been resolved?

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[obtuse pebble]

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[fickle swan]

could i get some help on this question

[dapper shale]

1/(xy)=(5-x^3)/(4x^3)

[fickle swan]

mmmhm

ive done till that part

subbing the value of y in terms of x

but what do i do next

[dapper shale]

So try to write this expression as a function of X = 1/x^3

[fickle swan]

oook

[dapper shale]

You have to get an affine function and then youll be able to find the value of k and h

[fickle swan]

ahh kk

thanks

.close

[obtuse pebble]

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[upbeat yacht]

[Linear Algebra, Similar Matrices]
I don't understand the point of similar matrices in an exercise.
If I have
1] the linear operator T:V->V
2] The base B1 for both space vectors
3] an associated Matrix A1 with B1 as starting and ending point
4] a base B2 for both space vectors
And I want to find a matrix A2 why would I use the concept of similar matrices?
Can't I just find A2 through the use of B2 and T?

[obtuse pebble]

@upbeat yacht Has your question been resolved?

[upbeat yacht]

<@&286206848099549185> I need your help por favor

[obtuse pebble]

@upbeat yacht Has your question been resolved?

[kind hawk]

you will find that A2 is similar to A1 if you do that

in particular the basis change matrix from B1 to B2 will be the relevant invertible matrix

so it might be easier to compute that basis change matrix and then calculating A2 from A1 using it instead of starting new at T

[upbeat yacht]

Yeah it might

But I'm not so sure about that

By that I mean that calculating from T could actually be wrong.

I tried the classic method and what I got was a different value compared to the one with similar matrices

The exercise in question: (without the values needed to craft the similar matrix]
T: V -> V, B2 is a base for both Vs
T([x y z]) = [x + y - z, y + z, 2x]
B2 = { [1, 1, 0], [-1, 0, 1], [1, 1, 1] }

Actually I'll create a new help for this

.close

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[final plaza]

how to simplify 4a(x+y)+2b(x+y)

[thick oracle]

$4a(x+y)+2b(x+y)$

[warm shale]

Master Oogway

[final plaza]

ok

could i expand first then simplify?

[thick oracle]

yes

you just expand

[final plaza]

oh

[thick oracle]

you cant simplify it further

[final plaza]

oh

but the qn

ask me to simplify

[thick oracle]

yes

[final plaza]

no wait

factorise

[thick oracle]

ok

in that case

[final plaza]

do i still exaand?

[thick oracle]

no

[final plaza]

oh

[thick oracle]

notice

something is written 2 times

[final plaza]

x+y

[thick oracle]

what is it?

good

so that's a common facotr

[final plaza]

ok

[thick oracle]

so how do we write it?

[final plaza]

just x+y?

[thick oracle]

?

[final plaza]

2(x+y)?

[thick oracle]

no

[final plaza]

oh

[thick oracle]

4x+2

howdo we write this

[final plaza]

2(2x+2)

1

i mean

[thick oracle]

good

[final plaza]

ok

[thick oracle]

do the same

[final plaza]

i still dont understan

haha

[thick oracle]

)

[final plaza]

can i do

2x+2y

[thick oracle]

4a * x + 2b *x

do you know this?

[final plaza]

what is the a*

that symbol

[thick oracle]

times

multiply

[final plaza]

so

4a times x +2b times x?

[thick oracle]

yes

[final plaza]

lemme thinl

2x(2a+b)

?

[thick oracle]

no

expand it

you'll see it's not the same

[final plaza]

oh

i dont understand haha havent revised in a while

[thick oracle]

ok

i'll tell you

[final plaza]

oh

[thick oracle]

$4a\cdot x + 2b\cdot x = x(4a+2b)$

[warm shale]

Master Oogway

[final plaza]

oh

wait is the ans

2(x+y)1(2a+b)

[thick oracle]

no

[final plaza]

oh

[thick oracle]

instead of x

just write x+y

[final plaza]

yea?

oh

wait

what do you mean

i wrote x+y

[thick oracle]

goo

d

[final plaza]

oh

[thick oracle]

i got confused by that 1

in the middle

[final plaza]

ah

but the 1 what does it

acctualy represent

[thick oracle]

nothing

[final plaza]

oh

[thick oracle]

get rid of it

[final plaza]

ah

alright then ig im doing a revision ws without revising

so i might have some probelms

but thanks for the help:)

[thick oracle]

np

[final plaza]

do i close or

[thick oracle]

if you're done yes]

[final plaza]

i might have other qns not rlly sure tbh

[thick oracle]

it's gonna close automatically if noone types

for a while

[final plaza]

ah

alright then if i have qn ima open a new one tks and cya

.c

[thick oracle]

bye

[final plaza]

.close

[obtuse pebble]

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[boreal oasis]

is f(0) f'(0) and f''(0)

all 0?

because thats what i got

[random ocean]

did u find f'(0) by lebinitz rule? im getting f'0=1

[boreal oasis]

how did you get f'0 = 1?

i didnt learn lebinitz rule

[random ocean]

cos0 / 0^2 + 1

[boreal oasis]

what about f'(0) and f''(0)

[random ocean]

f`x = (cosx)/(x^2 +1)

by differentiating you'll get f''x

[boreal oasis]

thats weird

i got0 for all of them

[random ocean]

how did u get 0?

[boreal oasis]

with this

the thing is its x 0

[random ocean]

cos0 is 1 isnt it?

[boreal oasis]

integral 0 x *

integral is always 0 / 0

when you plug it in for x

anyone ?

[random ocean]

what did u meant by integral is always 0/0

[boreal oasis]

its x to 0

and x = 0

i just dont get how its

1

[obtuse pebble]

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[river plank]

this is a question from an Oxford course( available on the website i am just doing this to improve my understanding)

[river plank]

I am not sure about what the question is asking
what i understand so far is that it's asking me to divide the given polynomial by the given quadratic and give the quotient and remainder
now i tried a particular example and i can obtain the coefficients by matching coefficient on both sides it gives me a triangular system in that case
but what does it mean by deriving a method that uses nested multiplication
i know that it has something to do with Horner's method but that's all

Also this is numerical analysis

[obtuse pebble]

@river plank Has your question been resolved?

[river plank]

@jaunty stream

[obtuse pebble]

@river plank Has your question been resolved?

[obtuse pebble]

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[timid silo]

got to here but not sure on what to do after

[reef grotto]

well just get a common denominator

this channel is occupied already

[timid silo]

could you give me an example

[reef grotto]

well

if you have

$\frac{1}{4x+3}+\frac{2}{2x+1}$

[warm shale]

Asagao 朝顔

[reef grotto]

you can simplify them

so to get them on a common denominator you can do

$\frac{2x+1}{(4x+3)(2x+1)}+\frac{2(4x+3)}{(2x+1)(4x+3)}$

[warm shale]

Asagao 朝顔

[reef grotto]

and they have the same denominator

[reef grotto]

though notice here

you have (1-x)^3 and (1-x) as the denominator

[timid silo]

so like this rule? (a/b) +- (c/d) -> (ad+-bc)/(bd) ?

[reef grotto]

so there's an easy way to get them on a common denominator

(1-x) times what gives you (1-x)^3

[timid silo]

(1-x)^2

[reef grotto]

yeah

so multiply top and bottom of the fraction by (1-x)^2

$\frac{x^3}{(1-x)^3}-\frac{4x(1-x)^2}{(1-x)(1-x)^2}+1$

[warm shale]

Asagao 朝顔

[reef grotto]

like so

and now you can just combine like terms after you expand

also you can do the same thing with that 1

at the end

you can write 1 with the common denominator as well

since

anything divided by itself is 1

[timid silo]

ok

for the (1-x)^2 i can just apply the power and get 1 - x^2?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

.close

[obtuse pebble]

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[frosty stump]

This is the question. I tried to get an equation with just one variable and got z^4 - 14z^2 - 96z + 301 = 0. I can't factorize this so I am not sure what to do.

[frosty stump]

Please could anyone give a clue as to how to approach this?

[royal shard]

are you familiar with matrices?

[frosty stump]

nope

oh wait

yh i will look that up

would that be enough to solve the problem?

[royal shard]

nvm i read it wrong

give me a moment

[frosty stump]

k

[tardy epoch]

3 equations, 3 unknowns. the tedious part is to solve for one variable in one equation, then plug it into the second and repeat for 3rd

e.g. solve for y in the first, plug into 2nd, solve for z and plug into 3rd

[frosty stump]

I have tried that, but i have ended up with z^4 - 14z^2 - 96z + 301 = 0, and i am not sure how to factorise it

[royal shard]

btw, i have put the functions into a program, and it looks like to me that there are no such x,y,z

[tardy epoch]

,w solve z^4 - 14z^2 - 96z + 301 = 0

[warm shale]

Results provided by WolframAlpha

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

Requested by riemann#2529

[royal shard]

huh

[tardy epoch]

pretty

[frosty stump]

lol

i am unable to use a calculator/any other sites for this as it is from an olympiad paper

[tardy epoch]

yea it doesn't look like two of those parabola sheets intersect

which 2 are they martin?

oh the axes are labeled i'm blind

[frosty stump]

so there are no solutions?

[royal shard]

the one with x^2 and the one with y^2

[tardy epoch]

yea if you plot these two, they don't intersect

[frosty stump]

oh ok cool

thank you so much!

.close

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[tardy epoch]

wait does that make sense

[frosty stump]

kinda

there are no solutions right?

cause no value of x corresponds to any value of y to make the equations true

[obtuse pebble]

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[tardy epoch]

@royal shard can you double check the graph? why would two sheets be in the same plane ?

[tardy epoch]

can you plug one of these z values in for y and x and see if they satisfy your equations approximately?

[frosty stump]

.reopen

[tardy epoch]

the second one should by 6z

[frosty stump]

i just realised

x = 1, y = 2, z = 3 seems to work

[royal shard]

oooohhh

[frosty stump]

trial and error

[royal shard]

oops haha

[warm canopy]

Sorry what's the question here

[tardy epoch]

question: ||your butt||
https://cdn.discordapp.com/attachments/903430332324405288/1007381015683145768/Screen_Shot_2022-08-11_at_9.10.50_PM.png

[warm canopy]

Have u tried bbc bitesize

[tardy epoch]

lol

[royal shard]

somewhere here 🙂

[frosty stump]

is there just 1 solution?

cause x = 1, y = 2, z= 3 works

[tardy epoch]

z^4 - 14z^2 - 96z + 301 = 0
probably messed up some algebra there then

[frosty stump]

yh

thx for your help 🙂

[tardy epoch]

fun graphs, ty martin

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[royal shard]

@tardy epoch
i use mathcha.io for those
it is really easy to use supports latex is free and looks cool

[warm canopy]

Aha was just about to ask

I always use geogebra 3d but it's so janky

[tardy epoch]

i don't think you realize how lazy i am

[timid silo]

mathcha op for tikz

[tardy epoch]

like if i can't do it from here or python console, i just won't

[royal shard]

for anyone interested, in matchcha you press "alt" and "enter" and then type something in, like mathbb{R} and it shows you what you could have meant

[warm canopy]

Lol you even run python here

[royal shard]

so it is usfull if you dont know all the latex names

[tardy epoch]

what

,python print('hello world')

f

that would be too amazing

i meant i just alt tab to terminal

[timid silo]

I tried putting sage in my preamble, but they said no

[warm canopy]

I swear you did it the other day

Oh ig you just copied from your terminal

You counted the number of words in something i sent

[obtuse pebble]

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[timid silo]

.close

[warm canopy]

WHATS YOUR QUESTION?

[timid silo]

lol

[warm canopy]

[timid silo]

my phraseexpander is misbehaving

[obtuse pebble]

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[gray aspen]

can you only refer to perimeter to two-dimensional shapes?

[warm canopy]

Yes

The 3d equivalent would be surface area

[gray aspen]

ahhh okay so to find the distance around a 3d shape, i would have to calculate the surface area right

[warm canopy]

Sure yeah, not sure distance around is the right word though

[obtuse pebble]

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[gray aspen]

thank you!

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[exotic ledge]

hi

[exotic ledge]

can someone help me with this

the first thing i tried was studing the sign for the numerator

making x-2 = -1/3 gives the solution x=3/2

for which x<3/2 make the inequality x-2>-1/3x not true and x>3/2 prove it

idk if i'm correct until here

[obtuse pebble]

@exotic ledge Has your question been resolved?

[exotic ledge]

<@&286206848099549185>

[bleak maple]

i suggest splitting this into cases since we have the absolute value

consider x >= -1/2 and x < -1/2 separately

[obtuse pebble]

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[crude forum]

Is the answer sqrt(50,000) km?

[timid silo]

,calc 200^2+100^2

[warm shale]

Result:

50000

[timid silo]

yes

[crude forum]

alright I just wanted to be sure It should be a triangle and stuff

thx

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[fiery raptor]

how is this true?

[timid silo]

im assuming multiplicative inverse?

$3\cdot 2=6$ and $6\equiv 1\mod(5)$

[warm shale]

jswatj

[obtuse pebble]

@fiery raptor Has your question been resolved?

[warm canopy]

I would bet that the "previous equation" mentioned is more or less this

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[devout grove]

Hello, is this correct??

[timid silo]

hmm

[wise token]

no

[timid silo]

no

x=0 is an exception

[wise token]

divding by 3 x gives x=15

[devout grove]

So, this is incorrect

[timid silo]

ig he is trying to solve an equation thats

$45x=3x^2$

[warm shale]

Normal Zeta

[devout grove]

?

[obtuse pebble]

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[crude forum]

I have no idea what to do, I think 200 is the hypotenuse and that 100 a side?

[proven zephyr]

yes, that's pretty logical, and that is the only way of doing the question

[crude forum]

I just don't know how to find the angle

[fierce lagoon]

That's also assuming that the ground is flat and parallel to the horizon

[high lily]

start by drawing a diagram

[crude forum]

ok

now?

[proven zephyr]

no, next year

[crude forum]

I need the angle where the string and horizontal meet right?

[crude forum]

ok, that was uncalled for

I meant what do i do now

[high lily]

show is your diagram

[crude forum]

ok

[high lily]

ideally sufficiently labelled

[crude forum]

what counts as sufficent?

[high lily]

well I would want to see a triangle
the given info
and indication of the angle you're trying to find

[crude forum]

ok

[high lily]

and labels for the position of the kite and person flying it

[high lily]

no, that isn't the angle you want

do you know what horizontal means

[crude forum]

I'm stupid

I remember now

so, how do I find the angle?

I know the horizontal is sqrt30,000

[high lily]

trig

and/or special triangles

[crude forum]

what is a special triangle

[high lily]

30-60-90,
45-45-90

[crude forum]

how does that help here

how can I know if one of the angles is 30 60 or 45?

[high lily]

look up ratios of special triangles

[crude forum]

ok, and?

I don't see how those are helpful

[high lily]

what exactly are you looking at

[high lily]

note that here the ratio of a leg to hyp is 100/200 = 1/2,
and if you're looking at the correct stuff you should see something related to that

[crude forum]

but that isn't the case for this triangle

[high lily]

wdym

what isn't the case

[crude forum]

the ratio of that leg isn't 1/2 h

the other one is

[high lily]

huh?

rotate/reflect your triangle if that helps

[crude forum]

what

[high lily]

I don't see what your issue is

[crude forum]

check the pin

[high lily]

?

[crude forum]

I said what the issue is in the pin

[high lily]

i know what the problem is

what's your issue with applying the info about special triangles to your problem

[crude forum]

it doesn't follow either of those ratios

[high lily]

note that here the ratio of a leg to hyp is 100/200 = 1/2,
and if you're looking at the correct stuff you should see something related to that

[crude forum]

only one of the legs match 1/2 h but its the wrong one

[high lily]

wdym by wrong one?

[crude forum]

on my triangle the vertical is 1/2 h

[high lily]

so?

[crude forum]

on the special triangle the bottom one is

[high lily]

so?

consider the relative position of the angles

[crude forum]

oh it would just mean the other one is the angle

oh

[high lily]

instead of insisting the must have the same orientation

[crude forum]

ok

I think I get it

[high lily]

rotate or reflect the image if you have to

[crude forum]

on a side note, how could i solve it with trig

[high lily]

consider what your me trying to find, and the relative positions of the given information

and use the appropriate trig function

[crude forum]

its the inverse of the cosine ?

[high lily]

what's it

[crude forum]

the ?

angle

like the angle I labelled as a question mark ?

[high lily]

inverse cos of what

[crude forum]

theta ig

[high lily]

what's theta represent

[crude forum]

a question mark

[high lily]

$\theta = \cos^{-1}\theta$?

[warm shale]

ℝamonov

[crude forum]

yes

[high lily]

no

that doesn't make any sense

[crude forum]

?

[high lily]

and I gtg someone else would need to take over

[crude forum]

what do I do to get someone else?

ping helpers?

<@&286206848099549185>

[oblique sage]

Hi

What’s the question

[crude forum]

hi

its in the pin

but now its about specifically using trig

how would I solve this with trig

and not special triangles

[oblique sage]

uhhh soo I think you should solve for the two other angles, then subtract them from 180

Wait now

I was looking at wrong spot

[crude forum]

yeah you were

[oblique sage]

It’s just using sin

[crude forum]

inverse of sin or just sin

[oblique sage]

Since angle between horizontal and string has opposite and hypotenuse

Sin is when you have angle, , sin(angle) is just opp/hyp

[crude forum]

I have angle not side

I mean side not ngle

angle

[oblique sage]

Arcsin of opp/hyp is angle

[crude forum]

yeah

so I just need that and thats the answer, right?

[oblique sage]

yes arcsin(op/hyp)

[crude forum]

k

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[obtuse pebble]

Please don't occupy multiple help channels.

[lone vault]

when i completed the square

i got (5,-1) for the vertex

but idk how they gpt thus

[mint tendon]

notice that b is 5/2 and c is -1/2

exactly half of your coordinates

the reason it is half is because of the leading term of x^2 is 2

[lone vault]

ok so now i simplied it

i got vertex (2.5,-2.25)

aight all good

[obtuse pebble]

@lone vault Has your question been resolved?

[slim cove]

(I'm assuming this question was solved)

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[eternal cloud]

I don't understand why |x| = -x here when x < 0

[eternal cloud]

shouldn't the | | make it always positive?

[hot sonnet]

Hello, when x<0, x is negative

[eternal cloud]

but the absolute value signs turn it positive, no?

[slim cove]

suppose you have x = -3

then |x| = -(-3) = 3

for example

when x is negative, -x is positive

[eternal cloud]

ok, i forgot -x is not the input

i thought they said the result of |x| = -x

which is incorrect right

[slim cove]

yeah x is the input, and -x is the output when x < 0

so |x| = -x is true

when x < 0

this is just saying, if x is positive, then just return it unchanged

if x is negative, then flip it to positive

[eternal cloud]

yeah i get it now

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[eternal cloud]

thanks

[slim cove]

awesome! no problem

[obtuse pebble]

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[eager portal]

How do I do this

[slim cove]

first what is the prime factorization of 240

[eager portal]

3 x 5 x 2^4

[compact shadow]

Chinese remainder theorem

[austere folio]

fr

[obtuse pebble]

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[warm thunder]

ideas on this?

tried writing out the first few terms but, since the index variable has to be zero kinda stuck

[compact shadow]

$c_{n}=(\frac{1}{4})^{[\frac{n}{2}]}(\frac{1+(-1)^{n}}{2}c_{0}+\frac{1-(-1)^{n}}{2}c_{1})$ so $\sum_{n \geq 0} (\frac{1}{4})^{[\frac{n}{2}]}(\frac{1+(-1)^{n}}{2}c_{0}+\frac{1-(-1)^{n}}{2}c_{1})x^{n}$

[warm shale]

Cogwheels of the mind

[obtuse pebble]

@warm thunder Has your question been resolved?

[warm thunder]

how did you get this?/

[compact shadow]

From your recurrence relation

[timid silo]

well you can quite literally skip the whole Ansatz of the infinite sum since you're given the relation, all you need to do is plug in values of $n$

[warm shale]

Aude

Cogwheels of the mind

[compact shadow]

More generally, suppose you are given $a_{km+i}=a_{i,k}$, then let $w=\cos(\frac{2π}{n})+\sin(\frac{2π}{n})i$, we have $a_{n}=\sum_{i} \frac{\prod_{j \neq i}(w^{n}-w^{j})}{\prod_{j \neq i}(w^{i}-w^{j})}a_{i,[\frac{n}{m}]}$

[warm shale]

Cogwheels of the mind

[obtuse pebble]

@warm thunder Has your question been resolved?

[obtuse pebble]

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[upbeat yacht]

Question, in regards of space vectors and eigenspaces.
Given the linear operator T:V->V, can V have multiple eigenspaces each one different form the other?

[kind hawk]

sure

one with respect to the eigenvalue 2, one with respect to the eigenvalue 17, whatever

[upbeat yacht]

Oh I see, thanks ehehe

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[icy acorn]

"If a population grows at a constant rate of 2.8% per year, then what percent will it grow over the next 10 years?"

[icy acorn]

How do I solve for that

[obtuse pebble]

@icy acorn Has your question been resolved?

[rugged acorn]

"grows at a constant rate of 2.8%" actually means each year is 2.8% larger than the year before
So for example if in 2022 it was let say p, in 2023 what will be the population ?

If we generalize the process for any year, we get something recursive

[icy acorn]

So it's just basic multiplication?

[rugged acorn]

Yes

[icy acorn]

2.8 x 10 is all I need to get the answer?

[rugged acorn]

Mhh no

[icy acorn]

Or is it like working with exponential bases

[rugged acorn]

when we say that the population increased by 2.8%, it is the population + 2.8% of the population, so if p is the population it's p + 2.8% of p, or p(1+0.028)
And we do the same thing each year, we multiplie the population by 1+0.028

[icy acorn]

Is there a way I can convert that to exponential form

[rugged acorn]

no need number e, just power

Population after 10 years is then
((p x 1.028) x 1.028) x 1.028 .... 10 times

[icy acorn]

I got 28.672 as a final answer

28.6%?

[rugged acorn]

Mhh

[icy acorn]

This next problem is similar to it

[rugged acorn]

Have you ever seen the geometric progression ?

[icy acorn]

"The half-life of a radioactive material is the amount of time it takes for 50% of it's radioactivity to decrease. If a particular material has a half-life of 35 years, then what percent will remain radioactive after 100 years?"

[icy acorn]

What's that

[rugged acorn]

If you understand this kind of problem you understand geometric progression, it's just the name for something general for all this kind of problem

[rugged acorn]

We can translate this in terms of power

[icy acorn]

My problem is mostly setting the problem up

[rugged acorn]

A summary

if we want " what percent will it grow over the next 10 years?", we want the ratio in percentage between the 10th and the initial year so p(n)/p(0) * 100

[obtuse pebble]

@icy acorn Has your question been resolved?

[icy acorn]

Alright I finished it

Thanks

[obtuse pebble]

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[fiery raptor]

How do I tell if it’s y coordinate is 160 or if it’s X coordinate is 160?

[warm canopy]

cos is always x coordinate

sin is always y coordinate

On the unit circle

It's how cos and sin are defined

[fiery raptor]

Okay thank u

[obtuse pebble]

@fiery raptor Has your question been resolved?

[blazing sentinel]

this is true because if you take the cos of the angle the ratio would be that adjacent side over 1 so it would just be the length of the side

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[slim cove]

just to clarify, the problem is that you have $$f(x) = \ln\left(\sqrt[3]{1+2x^4}\right)$$ and you want to figure out what $f'(x)$ is?

[warm shale]

Eric Tao (he/him)

[timid silo]

g(x)=lnf(x)
g'(x)=f'(x)/f(x)

[slim cove]

okay got it

so basically you have to apply the chain rule 3 times here

your notation is not quite correct, but I think you are doing it right

[harsh remnant]

I think you haven't applied chain rule here

[slim cove]

they do it on the next line

[harsh remnant]

Ohh

[slim cove]

I think they just wrote the wrong symbol

It confused me too lol

okay

well first can you tell me what the chain rule says

yup!

or rather

the derivative of f(g(x)) with respect to x is f'(g(x)) g'(x)

so in this example let's choose f(x) = ln x and g(x) = cube root of (1 + 2x^4)

then, $$\f\dd{\dd x} \ln\left(\sqrt[3]{1+2x^4}\right) = \f1{\sqrt[3]{1+2x^4}} \f\dd{\dd x}\sqrt[3]{1+2x^4}$$

[warm shale]

Eric Tao (he/him)

[slim cove]

yep

so applying the chain rule again, you will get

then, $$\begin{aligned}\f\dd{\dd x} \ln\left(\sqrt[3]{1+2x^4}\right) &= \f1{\sqrt[3]{1+2x^4}} \f\dd{\dd x}\sqrt[3]{1+2x^4}\&= \f1{\sqrt[3]{1+2x^4}} \f83x^3(1+2x^4)^{-\f23}\end{aligned}$$

[warm shale]

Eric Tao (he/him)

[slim cove]

don't forget the factor of 1/3 :)

and then you can simplify that a little

yep!

and then simplify a little

[kindred oasis]

Rewrite the expression to the power of -2/3 as a fraction with a cube root at the denominator

Then you'll be able to multiply it by the other fraction to simplify

No it's an exponent

It's negative, so the expression is going to go in the denominator

Let me write it

$$\frac{1}{³\sqrt{(1+2x^4)^2}}$$

[warm shale]

Andrea276

[kindred oasis]

Can you see how you can rewrite it that way?

$(1+2x^4)^{-\frac{2}{3}}$

You have this

[warm shale]

Andrea276

[kindred oasis]

It becomes:
$\frac{1}{(1+2x^4)^{\frac{2}{3}}}$

[warm shale]

Andrea276

[kindred oasis]

Right?

[kindred oasis]

Then this

Then you can multiply that with the other fraction

Yes

8x³/3

$$\frac{1}{³\sqrt{(1+2x^4)^2}}\times\frac{1}{³\sqrt{(1+2x^4)}}=\frac{1}{1+2x^4}$$

[warm shale]

Andrea276

Andrea276

[kindred oasis]

Yep that's right, that's what you should get

[kindred oasis]

You've got this multiplied by 8x³/3, which is the same as your answer

[kindred oasis]

That's basically $\frac{1}{a^{\frac{2}{3}}}\times \frac{1}{ a^{\frac{1}{3}}}=\frac{1}{a^{\frac{2}{3}+\frac{1}{3}}}=\frac{1}{a}$

[warm shale]

Andrea276

[kindred oasis]

The thing I did there

[kindred oasis]

this was a pain to write

correct, good job

yw

.close

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[smoky panther]

double checking, is this answer right?

[mint tendon]

how does this follow?

[smoky panther]

oh

it should be 5 shouldnt it

[mint tendon]

yeah i think so

[smoky panther]

instead of 6

that means there are 20 ways no?

because it leads to the 11 becoming 10

then 10 + 7 + 2 + 1 is 20

[mint tendon]

the 7 should become a 6

[smoky panther]

ah yes

19 ways

.close

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[timid silo]

Hi there, can someone help me with this problem? I got the answer wrong although I don't know where I went wrong. After checking everything I realized that for the Volume of the cone I squared the height instead of the radius, but even after correcting this mistake my answer was still wrong. Any help would be greatly appreciated :)

(Also sorry about the vertical image idk how to fix it)

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

I haven’t gone through the whole problem, so this might not be the only mistake. But the volume of your cone is incorrect since the radius should be squared not the height.

[timid silo]

I think the rest of it is right it’s just a calculation error

[timid silo]

yes I know, I said I realised this - after substituting 96 pi instead of 128 pi it was still wrong
if it's a calculation error can you double check? becuase I've checked numerous times and I get the sme answer

which is wrong

the actual answer is 159/2 pi

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

I'm not sure where you got (0,8)

this is what I did and it looks right. I revolved it from (0,4)

[obtuse pebble]

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[sand crystal]

how do computers factor math. Like what is the exact formula they use to factor
or logic i guess
i want to understand factoring and simplifying completely. Because ive always been terrible at it

[obtuse pebble]

@sand crystal Has your question been resolved?

[timid silo]

what is "factor math"

are you talking about sympy? https://www.sympy.org/en/index.html

[sand crystal]

i said "how do computers factor math" i dont recall saying "factor math"

sorry my connection is bad

i just want to know the exact, logic/steps taken by a computer to factor a mathmatical expression

so i know exactly how to factor

i said "how do computers factor math" i dont recall saying "factor math"

sorry bad connection

[sand crystal]

i didnt mean to repeat that message

[timid silo]

by mathematical expression, do you mean polynomials?

[sand crystal]

polynomials

or trinomials

or i guess

in general

[timid silo]

https://en.wikipedia.org/wiki/Factorization_of_polynomials

it also depends on the cas ig

if its over C, then its equivalent to root finding

[sand crystal]

i just want a deterministic ruleset for factoring or simplifiying

that i can follow

instead of using my intuition

[timid silo]

give an example

[sand crystal]

ok gimme a sec

it feels like factoring techniques are very specific, and sort of rely on the persons judgement to figure out how a math expression is factored or simplified. like this for example: the tutorial I was watching says that if you have an expression with an x^2 and a x^3 you factor out the x^2 along with the number next to it. But if it was 5x -15 i would be factoring out the 5 instead. It just feels like there are specific rules for almost every eqaution i need to memorize or detect using my intuition. I wish there was some unifying rules that different forms of factoring all had in common

[timid silo]

ah, I totally misread your situation lmao

I'm sure a cas factoring algorithm is deterministic, but I doubt most people (me included) would understand how its done

[sand crystal]

i would like to know how its done

[timid silo]

look it up then

sage is open source

sympy is also open source

but humans usually factor using what you call "intuition"

[sand crystal]

yeah im not into that whole intuition thing

[timid silo]

theres a set of stuff to do, such as first finding the gcd of all the terms, then finding linear factors

but any more than linear factors is usually guessing

https://www.wikihow.com/Factor-Algebraic-Equations

[timid silo]

its intuition, guessing or educated guessing most of the time unfortunately :/

if its a quadratic, then we have the quadratic formula

and there are special forms of polys like difference of two cubes that are easy to remember

[sand crystal]

i dont get why there isnt a branch of mathematics dedicated to deterministically factoring/simplifying equations

[timid silo]

there probably is

and factoring is deterministic

but its not feasible for a human to do by hand most of the time

I'll look up the sage source code for you

[sand crystal]

what is sage

[timid silo]

its a cas

computer algebra system

looks like they pass it on to pari

lol

idk how to look up source for pari, so would have to research it yourself @sand crystal

[sand crystal]

alright

thanks anyway though

[timid silo]

https://d1wqtxts1xzle7.cloudfront.net/50513159/s0747-7171_2885_2980012-220161123-19129-1s6g24z-with-cover-page-v2.pdf?Expires=1660353683&Signature=dZLq1psx~3mQprmtfNIaufuPyb605fhKSYlwAPf9jZk9kO8zlI6hkQQYLZUghKN70vO~8wjqGMREfvG4U9tD5I4wtJyTqdCeM518yPc4fVbFDT1wHhwhsZQwI-OTriWSHSDLyO9oKa01R6l2L~P36QAod8kz3m474vL59AjfDF-E0FW4jONwD4Spe3JS8f7Z5MkPLwnYdoFvRbPcQlzzZtsw8dOP8RZtU7wVJDfe9erOa8FpooespWvunTQn0d42XimU2-Lnj0lsw8p7y4~RROK9offfUl8bFeUMMdg-XYYnMfM1LR38prbrZmEF0VmU9hJZMmrw3NfTA3Un5gRPhA__&Key-Pair-Id=APKAJLOHF5GGSLRBV4ZA

[sand crystal]

interesting

ill definitely do some research into it

[obtuse pebble]

@sand crystal Has your question been resolved?

[obtuse pebble]

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[main abyss]

Have a question around sets/subsets

[main abyss]

A={u,v,w,x,y,z}

then what does the following mean?

{X = P(A) | u=X, v doesnt= X}

where the = are the weird C= thing

.close

[obtuse pebble]

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[azure lark]

Hi, i have a question and im a bit confused.

A person competing in an orienteering course travelled 5.0 km due East, then 4.0 km North East and then finally 6.0 km due South. 

a. what distance has the person travelled?