help-0

[gaunt magnet]

hey can u check my work real quick

[astral dagger]

neat

[glass lichen]

Yep

[willow wadi]

is that for me

[glass lichen]

Oh someone else felt the need to post the answer

@astral dagger refrain from posting the answer

[willow wadi]

wouldnt that be not right cause those are vectors and it wants us to do in an algebric method

[glass lichen]

...

it's algebra of vectors

[willow wadi]

oh so it is right

[astral dagger]

sorry

[gaunt magnet]

@astral daggerthanks

for doing that

[glass lichen]

$\vec{OB}=\vec{OA}+0.5\vec{AC} \ \vec{OB}=\vec{OC}-0.5\vec{AC}$ then add them together

[ocean seal]

Mosh

[astral dagger]

https://tenor.com/view/you-got-that-right-pointing-you-will-smith-gif-12658812

[gaunt magnet]

@glass lichen Moshill may u check my work please

☹️

@glass lichen

Resolve a force of 200N in the direction S 22 degree E into its regular componets
i corrected this and got
x = 200cos(-68), y = 200sin(-68)

@glass lichen

[stable mirage]

Hello! It would be kindly appreciated if someone helped
How the do you ∫√(Cot[x]) dx
(Integrate the square root of Cot[x])
I have no clue on how to do it

[gaunt magnet]

it would be kindly appreciated if you read #❓how-to-get-help and use proper channels

im using this channel

[astral dagger]

that's a really difficult one

,w integrate sqrt(cot(x)) dx

[ocean seal]

Results provided by WolframAlpha

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

Requested by SubGui#6716

[stable mirage]

ahhh damn I see

[willow wadi]

@astral dagger can u check this question for me to see if i doing it right please

[rich basin]

the answer says CAD is an iscocles triangle and i wonder why is it

[glass lichen]

2 radii

[gaunt magnet]

hey

can u check

[astral dagger]

but if you want a real good step by step, try integral-calculator.com

[stable mirage]

alright thank you so much, thats exactly what I need!

[gaunt magnet]

.___.

[astral dagger]

what is the name of the segment that connects the center and a point of a circumference?

[rich basin]

@astral dagger that weould be the radius

[alpine sable]

Is this channel being used rn?

[glass lichen]

you were using 4....

[willow wadi]

i had this channel

[alpine sable]

Ya but you sent me a how to get help thing? Idrk anymore x-x

[glass lichen]

yeah.. cause you pinged helpers right away..

[alpine sable]

Oops sorry

I deleted it

[forest drum]

Help pls

[gaunt magnet]

#❓how-to-get-help

delete before

@glass lichen yells at u

[glacial hedge]

r u doing this in bed?

[gaunt magnet]

lol wtf

he deleted it right after u said that

[willow wadi]

A mass of 60kg is suspended from the ceiling by ropes.The angles between the ceiling and the ropes are 31 degree and 59 degree.Determine the tensions in the ropes include a diagram.

The acceleration due to gravity is 9.8m/s^2
Include a diagram

Sorry if asking this question again 🥺
I got

T1 = 645.1
T2 = 1073.6

just tell me if im right or wrong

[astral dagger]

and what about the angle opposite to the sides of equal lengths on a triangle?

[violet iris]

nvm

[wary stream]

Did not get those values

[willow wadi]

heres my diagaram

[wary stream]

One of your equations is set up incorrectly, watch the signs

[willow wadi]

which one

[wary stream]

X driection

[willow wadi]

do u get $T_1=-\frac{\frac{60}{\sin(59)(1-\tan(31))}\cos(59)}{\cos(31)}$

[ocean seal]

yungdagger

[wary stream]

No

[willow wadi]

Ok 🤔

so whats wrong about my equation?

[wary stream]

A lot

[wary stream]

Show steps/work on how you got this

[willow wadi]

got that by just looking at the ques, like i guessed

[wary stream]

You can't just mentally do it, you'll make mistakes, write it out

[wary stream]

You label mass as grams here, which is incorrect too, units don't match with original problem you gave

And tension is a force, so units are wrong too

[willow wadi]

ok so my T_1 cos31 _ T_2 cos59=0 is wrong?

[wary stream]

Yes, the x direction is incorrect

[willow wadi]

whats wrong in it, like what should i change in it

[wary stream]

Also, stop reply pinging

Some signs at wrong

[willow wadi]

oh sorry

so how should it look like?

[wary stream]

Figure out the proper coordinate system to use

[willow wadi]

how

[wary stream]

That's up to you to determine

[alpine sable]

Say you have a function y=x^2 why whenever you subtract from x the function shifts to the right

[wary stream]

https://youtu.be/F5oqJ5t-pa4?t=238
Watch this, it'll help, starts at that time I linked, ends around 12:30

[fossil valve]

https://math.stackexchange.com/questions/4214784/assuming-convergence-of-the-power-series-of-a-complex-number-prove-values-of-co

[wary stream]

Don't post in multiple channels

[fossil valve]

oh ok sorry

[valid crypt]

can someone explain this better:

also an example would help too

[grim shadow]

this is something like a method to solve "symmetrical" polynomials to find their zeros

dont learn the formula

mb remember by an example

@valid crypt

[tawny totem]

https://math.stackexchange.com/questions/4216229/how-to-you-prove-that-the-heat-kernel-of-the-sub-laplacian-is-given-by-this-form

Help

[valid crypt]

I don't understand the steps, can you do an example and post a pic so I can follow it

[grim shadow]

ok

lets take x^2+x+1

its symmetrical

[valid crypt]

yes

[grim shadow]

so we have to find the zeros

divide it by x

so it'll become f(x)/x = x+(1/x) + 1

ok??

[valid crypt]

yes

[grim shadow]

now put x+(1/x)=t , a new variable right

[valid crypt]

so then it's t+1=0?

[grim shadow]

and remember one thing that x cannot be zero or it'll become not defined

so t=-1

and then substitute x

and solve the quadratic

but u must be thinking that it dumb to solve a quadratic for a quadratic, right??

[valid crypt]

x+1/x=-1

x+1/x+1=0

how does that help?

[grim shadow]

so its helpful in higher powers

if it was cubic or biquadratic

lets take cubic

[valid crypt]

wait what's the polynomial from the 1st example?

*x?

so x^2+x+1?

[grim shadow]

x^2+x+1

wait

[valid crypt]

lol that's the same as the start

[grim shadow]

lets take f(x)= x^4+x^3+x^2+x+1

divide it by x^2

so it'll become

[valid crypt]

x^2+x+1+1/x+1/x^2

[grim shadow]

now substitute x+1/x=t

but

[grim shadow]

(x+1/x)^2= x^2 +1/x^2 + 2

[valid crypt]

this chat is occupied

[grim shadow]

so x^2+1/x^2 = (x+1/x)^2 - 2

so x^2 + 1/x^2 = t^2 - 2

[valid crypt]

t^2+t-1=0

[grim shadow]

yep

you reduced it to quad

now u can easily solve it

but u'll get 2 values of t, right??

[valid crypt]

yeah my brain isn't working lol can't brain solve the quad

[grim shadow]

use the formula

lets not focus on finding the answer

just understand

[valid crypt]

oh it is +-sqrt2

so then x+1/x=sqrt2

[grim shadow]

wait

t= (-1+sqrt(5))/2 and its conjugate

did u apply the correct formula

[valid crypt]

yeah woops

I desmos-ed it lol

[grim shadow]

lmao

[valid crypt]

and saw 1.something

it's been a long day...

[grim shadow]

ok so u'll get 2 values of t and when u resubstitute it as t= x+1/x

u'll get 2 quads in x

[valid crypt]

anyway so then I end up with x^2-tx+1=0

[grim shadow]

whose roots are the roots of f(x)

so in total u'll get 4 roots

[valid crypt]

and x^2+tx+1=0

and then solve those and I'm done

right?

[grim shadow]

bro u'll have to substitute the numerical value of t

after solving

[valid crypt]

yeah

(duh)

[grim shadow]

and u'll get 2 values of t

[valid crypt]

oh sorry they aren't always +-

yeah

[grim shadow]

yep

[valid crypt]

I get it

[grim shadow]

and u'll get 2 quads in x

[valid crypt]

yeah

I see how this works, thank you

[grim shadow]

this can be similar for x-1/x

[valid crypt]

I'll do some practice with it tomorrow because I obviously cannot do math right now lol

[grim shadow]

what time is it there??

[alpine sable]

Yo do I @ helpers since my question wasnt answered

[valid crypt]

not late just kinda brain dead in the evenings

it's almost 7 pm

[grim shadow]

hmmm its morning here

like 7:30 am

[valid crypt]

so we're on opposite sides of the world then

[grim shadow]

lol

[valid crypt]

I'm is washington state where are you?

[grim shadow]

mumbai, india

[valid crypt]

interesting

well thx for the help

[grim shadow]

u got it right??

[valid crypt]

yeah

[grim shadow]

and dont forget to practice it later

[valid crypt]

def

[grim shadow]

it became a headache for me coz i procrastinated it in past

[valid crypt]

nah it's not homework, I'm learning it myself

[grim shadow]

hmmm how old r u

[valid crypt]

for math competitions when school starts again

16

[grim shadow]

good at ur age

well im 18

[valid crypt]

aight I gtg

[grim shadow]

see ya then

[alpine sable]

Yeah @grim shadow can you help me pls

[grim shadow]

yep

[alpine sable]

I didnt want to @ helpers yet

So my question, say you have a function y=x^2 why whenever you subtract from x the function shifts to the left

Like im trying to understand it intuitively

[wanton spoke]

You can @ helpers when it's been 15 minutes since you posted your question and it's still unanswered

[grim shadow]

hmm intutively

[alpine sable]

Im aware but i at least wanted to ask someone who was alr present helping and had finishdd

[wanton spoke]

i see. go on.

[alpine sable]

Alr did

[grim shadow]

have u studied parabola??

[alpine sable]

Uhh wym

[wary stream]

Second equation has incorrect trig function

[grim shadow]

mb u can understand from that

look for a parabola

[wary stream]

But once you fix that error, that's the correct system of equations to use

Then use substitution to find one of the unknowns

[grim shadow]

actually u dont require conics

so on which side r u subtracting from

[willow wadi]

ok so one i fix it thats the correct answer

[alpine sable]

So like you graph the parent function x^2=y but when you also graph the function (x-2)^2=y it appears to have moved right 2 from only subtracting 2

Im wondering why is this the case

[willow wadi]

here

[alpine sable]

Like is there an intuitive way to think abt this

[wary stream]

It's the correct system of equations now

[grim shadow]

guys can u pls use some other channel

[grim shadow]

its basically shifting of origin and shifting of curve

lets say Y=X^2 is the curve for now

[alpine sable]

Wdym by its the curve

[grim shadow]

look

Y and X are linear in y and x resp

so origin will lie where X=0 and Y=0

right??

[alpine sable]

I think i need some other knowledge here

[grim shadow]

this is an entire topic in conics

but u dont need that deep knowledge to understand

[alpine sable]

I thought transformations of functions was alg 2

Is conics apart of alg 2

[grim shadow]

bro im not from US so idk

[alpine sable]

K lol holup

[grim shadow]

whats interesting is that later u'll learn to rotate the curve

it gets hectic there

so look

when curve was y=x^2 vertex was at x=0 and y=0, basically origin

[alpine sable]

Whenever u say curve are u just saying another word for function

[grim shadow]

ya

[alpine sable]

They dont go over conics in alg 2 so ill try

But yeah i get what ur saying so far

[grim shadow]

ok listen carefully now

now lets say our new curve is y=(x-2)^2

[alpine sable]

Yep

[grim shadow]

so the vertex will lie at y=0 and x-2=0

[steel palm]

can someone check channel 3 once they are done pls

[grim shadow]

ok

[steel palm]

thx

[grim shadow]

so vertex will lie at y=0 and x=2

[alpine sable]

Wait how did u get the vertex from the equarion

[grim shadow]

a points properties will never change

the point earlier was at Y=0 and X=0

and now it'll remain at the same place

tho the origin can shift left or right

so now our vertex is at (2,0) and earlier it was at (0,0)

and tho it shifted right

if u take y=(x+2)^2 then it'll shift towards left

[alpine sable]

K wait lets rewind a lil bit

[grim shadow]

ya

[alpine sable]

You said this

now lets say our new curve is y=(x-2)^2

[grim shadow]

np

[alpine sable]

so the vertex will lie at y=0 and x-2=0

[grim shadow]

yep

[alpine sable]

How do you get the vertex from just the equation

[grim shadow]

do u know the graph of y=x^2

vertex lies at (0,0)

desmos it

[alpine sable]

Yeah im aware

So we got the vertex from graphing it

Is what ur saying

[grim shadow]

wait

the characteristic of vertex is that y=0 and x=0

so when u replace either of the y and x, with a "linear" in y and x

u can just equate that linear in y or x to 0 and find the vertex

for eg, (y+4)= (x-9)^2

vertex will lie at y+4=0 and x-9=0

so its y=-4 and x=9

or (9,-4)

if u dont trust me then u can desmos it

[alpine sable]

Wdym by a linear

Like a constant

[grim shadow]

conditions for linear are that coefficient of x/y remain same and addition or subtaction of a constant

well what i said that the characteristics of point never change (thats fundamental) but the graph may become rotated or with a different slope pattern

[alpine sable]

Hmm I see

So wdym when you reference linear?

[grim shadow]

when u replace x with (2x-5), the graph will not exactly be the same

tho u can find the vertex with my method

[grim shadow]

well i just told u what i mean by linear

conditions for linear are that coefficient of x/y remain same and addition or subtaction of a constant

[alpine sable]

I dont exaclty understand what that means thrn

Then*

Are you saying the conditions of a linear equation are that the coefficients will remain the same

[grim shadow]

nah

it basically means that replaced term should be similar to the original

well just graph y=x^2, y=(x-2)^2 and y=(2x-5)^2

desmos it

and u'll see

[alpine sable]

Aight

[grim shadow]

i know its vexing

u can see that graph 1 and graph 2 are exactly the same but just shifted

but graph 3 is entirely different in its properties

@alpine sable u there??

[alpine sable]

Wait

Im doing it rn

[grim shadow]

ok

[alpine sable]

Yeah i see that

[grim shadow]

so thats what i meant by linear

its just a term

[alpine sable]

So linear means the 2 graphs will be the same when u subtract/add to x

But once u multiply something by x its diff

[grim shadow]

basically yes

and u can imagine difficult rotating of curve might be

@alpine sable got it???

[alpine sable]

So if you only have subtraction or addition to the x and y you set their expressions equal to 0 to find the vertex of the parabola

[grim shadow]

yep

personally i would suggest u to study conics first, so u can understand how graphs work better

[grim shadow]

and u dont have to memorize this statement

[vital dagger]

heyy

could someone help me w circle theorems

[dull oak]

Hey guys

i have this question

how many ways can 6 and 1 add up to 20?

[rigid smelt]

infinitely many ways?

supposed you can add multiple of 6's and 1's

hmm oh wait, only addition are allowed

nvm then

basically you have got a linear equation there

[covert berry]

A square $(n \times n) (n \ge 2)$, filled with numbers from $1 \to n^2$ randomly. Prove that there exists two squares sharing a side whose difference is at least n

[ocean seal]

moono

[primal fern]

help pls

this too

[alpine sable]

@grim shadow how abt this way of looking at it, "when you subtract from x you are taking out of its equilibrium and so you need a value simply greater enough to cancel the subtraction"

[grim shadow]

well yes

thats why we have to make it back at its equilibrium and thus we equated it to 0 before

[alpine sable]

Dude this actually is vexing like you said desmos literally shows 2 graphs that look so similar

[grim shadow]

haha i know

basically i taught u an entire chapter

[grim shadow]

whatever we discussed above is valid for any damn curve in the real plane

[manic glade]

is this correct phrasing?*

[grim shadow]

sorry bro im not good with symbols so mb i cant help u

[primal fern]

can you guys answer my question

[austere bridge]

the correct phrasing for what?

[manic glade]

for what im trying to say

asking about the commas and arrows

and just organization of it

[austere bridge]

what are you trying to say, though

in words

[manic glade]

the right half side of it

that |(0,10^-10)|=א

[karmic raft]

i

[velvet pelican]

shouldn't it be Aleph 1 tho

[manic glade]

we just use aleph

[karmic raft]

I have a matter, can somebody help me？ thanks

my brother give me this matter ，let me finish when he come back？ who can help me？

[crisp grove]

$|a+b| \leq |a|+|b|$

[ocean seal]

Ryuzaki

[celest anvil]

alright u substitution

it's given that u = ln(x)

du/dx=1/x so du=1/xdx

can someone tell me what i did wrong?

[tough hatch]

(1/x)dx

[rigid smelt]

that box is supposed to be the answer of the integral

else there would be an integral sign in front

[celest anvil]

no they want me to write the integral in terms of u and du

wait ill try to put the integral sign in front

[rigid smelt]

then there need to be an integral sign infront of the u^12 du

[celest anvil]

yes mb

well shit

dont work

[rigid smelt]

then just write the answer to the integration?

its not very hard

[celest anvil]

ok ill try that

[rigid smelt]

and im pretty sure they expect an answer to the integration

because they clearly mentioned that you should use C for the arbitrary constant

[tough hatch]

well they do want you to ...(Then) evaluate.

[rigid smelt]

🤷‍♂️

[celest anvil]

yuh

[obsidian marsh]

hey yall

is the gain matrix G of a square matrix A defined by: G=A'*A?

A'=A transposed

also

can the gain matrix also be retrieved from a redundant matrix (rectangular) of coefficients?

also

suppose i have a non-symmetric matrix A

if i find the gain matrix G of A

and LU factorize the gain matrix G

will the L matrix be equal to R' (transpose of R) from the QR factorization?

i heard this is a theorem

but i don't recall it's name or it's proof

[woeful pulsar]

no idea what is "gain matrix"

[willow wadi]

hi

[rose crypt]

can someone solve these for me?

[ionic jewel]

solve them?

[rose crypt]

mhm

[ionic jewel]

if you happen to mean simplify, you will want to find the common factors to each term, and pull them out

[rose crypt]

no, I mean factorise

I just learnt them tdy so yea

[ionic jewel]

okay so for the first one

what are the common terms?

[rose crypt]

x,y and 5

because factor of 10 is 5 and 2

[ionic jewel]

okay

so what's the simplified form?

when you pull that out?

[rose crypt]

is the answer for 1?
5x power of 5 y (x power of 4-2)

[ionic jewel]

^ is the exponent symbol

[rose crypt]

wdym?

[ionic jewel]

like "x to the power 2" is x^2

[rose crypt]

OH

so its basically for example x power 2 is x^2?

[ionic jewel]

yes

[rose crypt]

ah thanks

[ionic jewel]

$x^2$

[ocean seal]

bunny

[rose crypt]

not used to online lol

[ionic jewel]

anyways for the first one, you said the common elements are $5,x,y$

[ocean seal]

bunny

[ionic jewel]

so we want to pull a $5xy$ out of both the first and second term

[ocean seal]

bunny

[rose crypt]

mhm

[ionic jewel]

so what do you get when you pull $5xy$ out of $5x^5y$?

[ocean seal]

bunny

[rose crypt]

5xy (x^4-2)

[ionic jewel]

where is the -2 coming from?

ohh

yeah that's the full answer

okay do you know how to do the other ones from here?

[rose crypt]

ya, thanks

its summer holidays now

and I have work, so yea. Im learning this when I'm going into yr 8.

kinda new for me

[ionic jewel]

ah nice

always good to learn

[rose crypt]

ya

[obsidian marsh]

ok sorry

[wanton spoke]

why is it only the odd-numbered problems that have an answer?

might not be math related but

[ionic jewel]

half the work, but you still cover all the relevent topics

also often it's to match/not match the answer keys in the back of the book

[wanton spoke]

oh, i see

is it common to math textbooks?

[ionic jewel]

oh wait i thought you asked why teachers assigned work like that

[wanton spoke]

oh, no lol

[ionic jewel]

for books it's pretty standard to do that, yes

probably for ease of professors/teachers for the reasons i stated as well

also less work for them

[wanton spoke]

i see. thank you

[manic glade]

is this ok?

[distant violet]

Uhh

[quick surge]

can anyone teach me this

[rigid smelt]

have you heard of intercept theorem aka thales' theorem

or proportionality theorem, i have heard that being called sometimes

[quick surge]

yes

[rigid smelt]

then apply it

[quick surge]

ok

[rigid smelt]

you have parallel lines and triangle

you can apply that here

[quick surge]

PR/QL = HR/RL = HP/HK

like this?

[rigid smelt]

no your ratios are all mixed up

[quick surge]

what

isnt it like this

[rigid smelt]

if you choose PR/QL, that means you are looking at triangle HQL

hence HK is not relevant here

secondly

PR is the small side, so PR/QL equals RL/HR

because small correspong to small,

big correspond to big

[quick surge]

hm

[rigid smelt]

and thirdly, PR/QL does not help because we dont have any of the two length

why not try the two pairs HR-RL and HP-PQ

[quick surge]

oh

HR/RL = HP/PQ
l

like this?

[rigid smelt]

yeah

[quick surge]

oh ok

40/10 = HP/12

[rigid smelt]

mhm

[quick surge]

HP = 4 x 12

HP = 48

like this?

[rigid smelt]

yeah

[quick surge]

oh

how do i find the line QR

[rigid smelt]

do the same thing

but different triangle

same thing as in same theorem

not the same ratio

[quick surge]

wait no

i

i dont know how

[rigid smelt]

look at HKL, it must have something to do with the pair QR-KL

[quick surge]

i still

dont understand

[rigid smelt]

its the same theorem

try to find a ratio equal to QR/KL

ok i gtg, just look at how QR divides HL

[quick surge]

oh ok

bye bye

[vast lance]

If sec(x)=3 and 3pi/2<x<=2pi, is cot(x)=1/2√2?

[obtuse cove]

how would i solve this for x

[jagged imp]

wrong sign

[obtuse cove]

oh and also the rhs is 0.6

[jagged imp]

3pi/2<x<2pi means cosx is positive and sinx is negative, so cotx=cosx/sinx is negative

[vast lance]

Ah

True

Thanks!

So -1/2root 2 right

[alpine sable]

@obtuse cove what have you tried?

[glad aspen]

Hey guys I need some help
im sure im doing something really dumb
but here goes
We've got a rhombus
Sides ABCD
A (1,3), B (7,5), C (9, 11), D (3, 9)
A circle is placed inside the rhombus
that circle has a radius of sqrt(10) - sqrt(2)
and it is tangent to the AD and AB lines
I have to find the coordinate of the center of that circle
What I went for was to use the equation to get the distance between a point and a line
and rearrange it to get the vector between point A and the Center of the circle
from there since I know the coordinates for A, i'd be able to tell the coordinates for the center
does the reasoning make sense so far?

[glad aspen]

I found my issue

I was using the equation for 3d space

not 2d space

[jagged imp]

sorry for the late response but yes if you're still wandering

[alpine sable]

Can anyone give me a clue?

[gusty ravine]

$\gcd(2^a - 1, 2^b - 1) \= \gcd(2^a - 1 - (2^b - 1), 2^b - 1) \= \gcd(2^b(2^{a-b} - 1), 2^a - 1) \= \gcd(2^{a-b} - 1, 2^b - 1)$

[ocean seal]

soαρ

[gusty ravine]

assuming a >= b

now this should spark some ideas @alpine sable

do u know how to read the rules

#❓how-to-get-help

[hearty nest]

[alpine sable]

If you define vector addition and scalar multiplication, all mathematical objects can be vectors?

[sage summit]

yes you can put whatever you want as elements of a vector space

[undone lantern]

hi !

I need help

For a proof

Can you help me pls ?

[alpine sable]

Don't ask to ask.

Just ask.

[undone lantern]

Let E a set (not empty) and n=card(E)
We define $Pk(E) = \left { x \in P(E) ; card(x) = k \right }$

Show that exists $x \in P{n}(E), x'' \in P{n-1}(E), ... , x^{(n+1)} \in P{0}(E)$ such that $P_0(E) ⊆ P_1(E) ⊆ ... ⊆ P_n(E)$

[ocean seal]

じさつ しょうねん
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[alpine sable]

card(x) represents the cardinality of the set X?

[cold cedar]

Anyone available for question?

[velvet condor]

that is?

[cold cedar]

Wanna ask how this is done

Gotta figure out variable, so both vectors are 90 degress

Got a few more questions like these, so just need help once 😅

[velvet condor]

what abt vectors:))

oh

just draw

[cold cedar]

It's just general vector questions, trying to read up on it

[velvet condor]

ig....express the vector?

since it's 90 degrees

[cold cedar]

Like i figured out this, but I'm not sure if it's the right approach

I express both vectors, and equal them to 0

[velvet condor]

lemme recall some vector stuffs :)))

havent touched that for a long time

[cold cedar]

Fair enough 😄

[glass lichen]

what can you say about their dot if they're orthogonal?

[velvet condor]

i mean you can just use trigo

[cold cedar]

it's 0

[glass lichen]

right.. so you get an equation for t

[cold cedar]

yeye, is that right tho?

just wanna make sure

[glass lichen]

yes..?

[cold cedar]

Oh lol

[glass lichen]

$a\cdot b =0 \iff a\perp b$

[ocean seal]

Mosh

[velvet condor]

y not -1

[glass lichen]

what?

[cold cedar]

then it would be under/over 90

[glass lichen]

what do you mean....?

[velvet condor]

how to derive this

[glass lichen]

geometric definition of dot product.. or vector projections

$a \cdot b =\norm{a}\norm{b}\cos(\theta) \ \theta =\frac{\pi}{2}\iff a\cdot b = 0$

[velvet condor]

ok...........

[ocean seal]

Mosh

[velvet condor]

yes got that part

[glass lichen]

yeah, QED

[woven oriole]

help me in #help-3 pls

tq

[cold cedar]

so what if i wanna find out variable, if they need to be over/under 90 degrees?

what would i equal the vectors to?

if 0 is for 90 degrees

[glass lichen]

then you'd do an inequality

[clever locust]

That equivalence arrow is a bit sketchy, as a * b = 0 does not necessarily imply theta = pi/2

[glass lichen]

for an aside proof it's fine

[cold cedar]

so how would i find for 45 degrees

i'm not sure i understand

[glass lichen]

find what?

[clever locust]

just use an implication

[cold cedar]

like variable so vectors make a 45 degree angle

[glass lichen]

$\norm{a}\norm{b}\cos(\frac{\pi}{4})=a_1b_1+a_2b_2$ is also an equation in just t...

[ocean seal]

Mosh

[cold cedar]

i'm lost

[glass lichen]

$\sqrt{65}\sqrt{(t+1)^2+1}\frac{1}{\sqrt{2}}=-4(t+1)+7$

[ocean seal]

Mosh

[velvet condor]

can we just use trigo instead:)))

i think thats also possible

[cold cedar]

ooh, i wasn't looking at the task nr. 6 lol

now it makes a bit more sense

[glass lichen]

ok.. well you failed to provide actual details about the question you had so I went with what I had

[velvet condor]

wait lemme absorb this equation

[cold cedar]

i'm such a dumbass, mb

[thin wolf]

hey, can somebody help me with Q3 (i) 🙏🏻 i know i have to somehow cancel out (1+cosx), but i’m not sure how to get a (1+cosx) in the numerator

[glass lichen]

cant read it

[thin wolf]

ah dammit, was clearer on my phone.

Prove that (cosec x + cot x)^2 = (1+cosx) / (1-cosx)

ill take another picture of my working hold up

[glass lichen]

ok.. so $(\csc(x)+\cot(x))^2=\left(\frac{1+\cos(x)}{\sin(x)}\right)^2$

[ocean seal]

Mosh

[glass lichen]

yeah no, dont expand it

[thin wolf]

oh

ah damm

aight hold on lemme try again

[glass lichen]

ok

[thin wolf]

aight i got it, thanks!!

[unreal valley]

C.P of 100 cards = $850 (example)
S.P of 100 cards is $1360
Therefore, Profit = $510
Percent of profit =
Profit or Loss/C.P × 100
= 510/850 × 100
= 60%

Is it correct?

[thin wolf]

yep its correct

[unreal valley]

Thanks

I an just doing homework so yea

[thin wolf]

not sure if 100% reliable, but my method of "checking" is taking C.P. multiplied by the profits (e.g. $850 x 160%)

[unreal valley]

Ya

[thin wolf]

you should get the S.P. if its correct

[unreal valley]

The question is about to give profit%

[thin wolf]

ah i see

[unreal valley]

So its 60% right

[thin wolf]

yep

[unreal valley]

Alr thanks

can i use this as therefore?
∴

[thin wolf]

yea u can

[unreal valley]

Ok

[burnt pond]

hey people

can anybody explain how this works?

I don't know where the n comes from??

[left knot]

when you solve for that equation, you are essentially finding these interceptions

n pi, because as you can see the answer repeats for every pi

does anyone here know how to derive an equation for arg(a + bi)?

[tall ermine]

Hey guys I'm doing a decision tree, may I get some help?

Ultimately, it would be better to settle before trial right?

No it's homework

@alpine sable oh lol, you think I could get some help bro?

[tidal mica]

I think the picture is inconsistent with the words. The picture says there's a 40% chance they'll give 2500 but the words say it's 30%

[tall ermine]

I believe the text meant 40%

[tidal mica]

I don't have much math experience but I multiplied 0.6 by 1200 cause I assume you have a guaranteed ability to settle for 1200 before the jury which came out to 720 so I think going to trial and then settling if you lose would be best

[tall ermine]

@tidal mica wait, so I'm wrong?

[tidal mica]

I think so

[tall ermine]

So than what does the -1680 and -1600 represent?

[tidal mica]

I don't really understand your work in the second line but how I think it is is that you would always choose to settle before the jury if you lose because it adds up to -1600 which is more loss than -1200 so whatever would happen if the jury gave their verdict wouldn't matter

[willow wadi]

Hi

$I got \sqrt{25^2+8^2-2258*cos(60)}$

[ocean seal]

yungdagger

[mortal marlin]

Guys, what's the biggest math discovery that has been done in this server if one has been done?

[glacial hedge]

tan(x) is a periodic funciton... meaning the cycle repeats every pi thats why there is an n

[undone lantern]

yes

can someone help me ?

<@&286206848099549185>

[vague coral]

with ?

this ? https://cdn.discordapp.com/attachments/490557019623915520/872448634078253096/623962367138725919.png

there is something wrong with the notation...

what is x

[willow wadi]

<@&286206848099549185>

[vague coral]

did you convert the velocity in m/s ?

[willow wadi]

i added them up using cosine theorm

[tardy comet]

where does x dy/dx come from

[willow wadi]

#❓how-to-get-help

[tardy comet]

._.

[vague coral]

can you show your process ?

[willow wadi]

ye

[ionic jewel]

the 1y+x(dy/dx) is product (then chain) rule from d(xy)/dx

[tardy comet]

oh wait

ya

thanks

[willow wadi]

just added them using the cosine theorem
$\sqrt{25^2+8^2-2258*cos(60)}$

[ocean seal]

yungdagger

[willow wadi]

and thats my diagram btw

thats it

[undone lantern]

Let E a set (not empty) and n=card(E).
We define Pk(E) =\left { $ A\in P(E) ; card(A) = k } \right . $

Show that exists $x \in P_{n}(E), x'' \in P_{n-1}(E), ... , x^{(n+1)} \in P_{0}(E)$ such that $P_0(E) ⊆ P_1(E) ⊆ ... ⊆ P_n(E)$

[ocean seal]

じさつ しょうねん
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[undone lantern]

Corrected

Look at here

[vague coral]

I think it's fine, but I don't know how you got 2258 there, shouldnt it be 2x25x8=400 ? or maybe I'm too rusty for geometry

[willow wadi]

oh this is vectors

[glass lichen]

unless I drew it wrong.. it's just pythagorean theorem for the magnitude of the resultant

[gaunt magnet]

who u talking too

[undone lantern]

@vague coral m

*?

[willow wadi]

nvm i was doing it wrong

this should be the correct way

206.16 km/h East, 35.96 degrees north

[wary stream]

All it looks like is sum of forces in the x direction and y direction

[willow wadi]

ye

[outer steppe]

Could someone send how to solve 81^x = 1/3? Thanks

[ionic jewel]

log base 81 of both sides

x = log base 81 of 1/3

because of the property $\log_a(a^b) = b$

[ocean seal]

bunny

[gray isle]

express 81 as a power of 3

[outer steppe]

this?

[ionic jewel]

that's the higher effort solution but perhaps intended?

[outer steppe]

333*3

oop

[gray isle]

requires minimal to no calculator use

[ionic jewel]

this works, left side reduces to x, right side is a constant, but in retrospect ramonovs solution is cleaner with no calculutor

[outer steppe]

okay thanks

[ionic jewel]

if you are allowed a calculutor my solution is faster, if not ramonovs is easier

[outer steppe]

Ramonov what do I do

[outer steppe]

I'm just learning on khan academy

[ionic jewel]

so you get $(3^4)^x = 3^{-1}$

[ocean seal]

bunny

[ionic jewel]

you see where this came from?

81 = 3^4 and 1/3 = 3^(-1)

[outer steppe]

Oh yes

[ionic jewel]

yeah

then using exponent rules $3^{4x} = 3^{-1}$

[ocean seal]

bunny

[outer steppe]

my potato brain forgot I could write (3^4)^x

[ionic jewel]

which means $4x = -1$, and I'm sure you can finish

[ocean seal]

bunny

[outer steppe]

Yes, thanks a lot

[gray oxide]

Hello

I need help with A and B

I haven't understood the concept of the limit of infinity

But I get thst the closer we get to infinity the closer we get to 0 right?

So the first A) is 4 as answer but the second one is tricky

Srr 4

Nvm i figured it out lmao thank you

[ionic jewel]

second one goes to 1

oh u got it

[alpine sable]

I am working on a ray box intersection and I do not fully understand the maths for it. After doing a couple diagrams I was able to determine that a ray will hit the boundaries when the intersection of [t0x, t1x] and [t0y, t1y] is not empty. The problem is I am having trouble of determining that.

[warped phoenix]

Hey guys, does anyone know any good resources to help me review the entirety of the Algebra 2 course? I was thinking of taking the Mastery Course on it in Khan Academy and then strengthening whatever I didn't do well on it via his videos, but I was wondering if anyone knew any other resources that might help?

[austere bridge]

Use the binomial coefficients

[void estuary]

This should be a really easy thing but my head is fried...
Say we have a circuit that has an 1 in 9 chance of giving an output when triggered...
If i stack three of these circuits together, what's the chance of an output being given out in the end?

Button-Ca-Cb-Cc-LAMP
C = Circuit```

[shell widget]

@void estuary (1/9)^3

[void estuary]

Is that really so lol

[shell widget]

Yes

[void estuary]

dang

alright

ty!

[gray oxide]

So I have This problem of limits I've tried 3 methods

1. Just putting the value in but I get 0/0

I tried factoring there is nothing to factor the whole thing

3. I tried rearranging things becuase it is a division but that doesn't work either becuase not all of the numbers there have an x

Here's the problem btw

And here are my calculations

[shell widget]

x^2 - 2x - 8 = x^2 - 4x + 2x - 8 = x(x-4) + 2(x-4) = (x-4)(x+2)

4x-x^2 = -x(x-4)

[gray oxide]

Where did the 4x come from and why did you change the minus sign at

X^2 — 2x - 8?

[shell widget]

-2x = -4x + 2x

Yes?

[gray oxide]

Yrs

Yes*

[shell widget]

That's all we needed.

then its simple factoring

[whole charm]

(x-4) factors out

[gray oxide]

I see didn't know we could do that lmao I guess you really learn stuff every day

I got the bottom part for the fraction

But the upper part where the 8 was disturbed me deeply

[shell widget]

do u know the rational root theorem

[whole charm]

x²-2x-8 = (x-4)(x+2)
4x-x² = x(4-x)
Divide em', we're now left with -(x+2)/x → -6/4 = -3/2

[gray oxide]

@glacial hedge another room pls

[whole charm]

@gray oxide does this make sense?

[gray oxide]

No sorry lemme try to re read it

You're saying that the upper part

Is a conjuction?

[whole charm]

Well, the upper quadratic has 2 roots, 4 and -2
That's how we factorize it

Similarly, the lower quadratic also has 2 roots, 0 and 4

[gray oxide]

I mean these things btw

[whole charm]

These are expansions of perfect squares, neither the numerator nor denominator are perfect squares in the Q

[whole charm]

So this is a different story

[gray oxide]

I see but how am I supposed to know which number it is?

What method do I use?

[whole charm]

Are you aware of how to find the roots of a quadratic expression?

[shell widget]

https://www.youtube.com/watch?v=YtN9_tCaRQc

[whole charm]

Thanks mate, have a look over this
You'll understand :)

[gray oxide]

That problem that women is showing if it equals 0 I would know how to solve it

But it doesn't so unfortunately no

So I'll watch

[shell widget]

same thing

[whole charm]

Well, in the end, this is what it boils down to, so maybe have a look over this again at the end

[alpine sable]

Question 5 bit (ii)

Please somebody help. I have spent hours on this.

[gray oxide]

Is there any other way to solve this

Becuase I don't think we're supposed to use this

And this would mean I would need to memorize a new method

But if you can't think of anything I will just use this

Would this work as an answer?

[wary stream]

What exactly are you trying to do? Find the roots?

[fleet glacier]

It's kind of tough

I got a_n^2+2a_n * a_(n-1)

and that's really similar to question 4 part 4

[random pelican]

hi friends! that app told me that, but its kinda confusing, if b = 2/6 why the last one is 1/3? why did they changed? is the app wrong or im just confused?

[slender forge]

2/6 simplified is 1/3

[random pelican]

oh, so both forms are correct right? ty

[sturdy glen]

so

i have this problem (a) and the solution shows the basis they found

however, i have absolutely no clue how they got from the RREF of A-I to the actual basis

can someone please help explain this? thank you!

[wary stream]

All they did was A - I then row reduced it

[sturdy glen]

yeah but

im talking about the basis

the [1 0 0], [0 1 0]

[sturdy glen]

<@&286206848099549185>

[lament hazel]

Let 0<a<pi/2 and -pi/2<b<0. Find the value of tan(a+b) when sina=1/3 and cosb=1/3

help

<@&286206848099549185>

[sturdy glen]

😐

[sly mantle]

@lament hazel as a helper you should be familiar with the rules, in particular don't post in an occupied channel and wait 15min before pinging helpers

[sleek elbow]

[sly mantle]

@sturdy glen we seek a basis of ker(A-I)

[sturdy glen]

what's ker()

[sly mantle]

kernel aka nullspace

[sturdy glen]

ahh ok

so how come we need to find the nullspace of A-I?

[sly mantle]

recall 1 is an eigenvalue of A. the related eigenvectors are the nonzero solutions to (A-I)x=0. in other words the 1-eigenspace is ker(A-I)

[sturdy glen]

ahhh ok i see, that makes so much sense looking at it now

thank you !

[sly mantle]

no prob

[alpine sable]

alright i need help with two relatively basic problems

1.What is the length of the diagonal of a 13″ square?

2.A rectangular prism has a surface area of 72 square inches, its width is 3 inches, and its height is 6 inches. the length of hte internal diagonal is what

[wary stream]

You have to wait at least 15 minutes before pinging helpers

[alpine sable]

i would appreciate any help i can get @ helpers

[wary stream]

And deleting that doesn't remove a ping

[alpine sable]

i know i meant to write it like that^ and then edit it after

[wary stream]

13" square? All the sides are 13"?

[alpine sable]

are you sure thats what 13 inch means?

i think it means the area

[wary stream]

Also not a question, this is a statement

[alpine sable]

of the square

my bad, fixed it

but anywasy

if its area it would be the square root of 13^2+the square root of 13 squared=c^2

so 13+13=c^2

[wary stream]

So let's assume that 13 is the area. How do you find the area of a square?

[wary stream]

This is correct

[alpine sable]

but the square root of 13 plus the square root of 13 is not an option

neither is the square root of 26

and you cant factor the square root

[wary stream]

Then, I'm not 100% sure. If it has choices, post the entire question

[alpine sable]

ok

What is the length of the diagonal of a 13″ square?
Select one:

a.26 in

b.13*(the square root of 3) in

c.13 in

d.13(square root of 2) in

[queen nova]

How do you find the domain of a function given the range?

[noble sinew]

depends on your function

[gaunt magnet]

hi

🙂

[noble sinew]

but then 1st robot will still be at an advantage, because now the first robot to pull above 0.75 will win (is the idea)

which the 1st robot is most likely to do because 1st robot starting

not really sure how to find it but doesn't seem possible to do it with simulation when they want it to 7 significant figures

[pliant ridge]

hello, i tried half angle formulas but no avail, wat do sirs, give hinterino

tyty

[lusty plaza]

plug in each and see if it's true

[vague coral]

$\int (\sin(2x) - \cos(2x)) dx = \int \sin(2x) dx - \int \cos(2x) dx$
and you can do a u-sub (u = 2x)

[ocean seal]

Herels

[pliant ridge]

oh my god im stupid LOOL

[pliant ridge]

yh that would work but not going towards a trick rn i guess

[pliant ridge]

tyty ❤️

[vague coral]

np

[desert basin]

I'm so nervous. I'm going for my CFAT test soon and math was never my strong suit. Anyone know of any good math sites or anything I can use to study a bit?

[vague coral]

brilliant

[red scarab]

is this channel free?

[vague coral]

kinda

[desert basin]

thanks for your help! 🙂

[slender forge]

U can use YouTube and khan

[desert basin]

ty

[elfin snow]

wolframalpha can help you answer questions and verify results @desert basin

[red scarab]

so i solved this cubic equation with the cardano formula: x^3 - 15x + 4, the solution is third root of -2+11i - third root of 2+11i (complex numbers obviously), which is (2+i) - (-2+i), however the solution should be 4 then, right?

in my ,,solutions book'' it says -4

[desert basin]

awesome! I'm just nervous cause I only really understand the most basic of math lol. Like, theres one question on the practice sheet I have here and its 0.5 divided by 0.05 apparently, that = 10?

[raven cipher]

Hey sorry for this bothering but i cant wrap my head around this equation 5a2− 10b + 9b (the 2 is a exponent)

this might be simple for yall but the negative got me fucked up

[elfin snow]

think of 0.5 as 1/2 and 0.05 as 1/20 and divide 1/2 by 1/20, and when you divide fractions you multiply by the inverse, so 1/2*20/1=20/2=10 @desert basin

if that makes sense

@raven cipher that's not an equation, that's an expression

and you can use ^2 for the exponent

[desert basin]

thank you so much!

[raven cipher]

oh

a= 3 and b=2

[elfin snow]

well you need it to equal to something

[red scarab]

???? just calculate it then

[raven cipher]

the negative like i said keeps messing me up i get 4

when i break it down i get it down to a 15 - 20 and i feel like this is where i mess up

[elfin snow]

so you have $5a^2-10b+9b$ but what is it equal to?

[ocean seal]

gmod

[elfin snow]

zero?

[raven cipher]

is it actually?

[lusty plaza]

he's asking you

[raven cipher]

ohh

no

[lusty plaza]

you didn't post an equation

you need to include what that expression is equal to

$5a^2-10b+9b = ?$

[ocean seal]

CreamyBoy

[raven cipher]

yea

[lusty plaza]

So...

[raven cipher]

ye

[elfin snow]

so then plug the a value and b value into the equation

[raven cipher]

oh

[elfin snow]

5(3)^2-10(2)+9(2)

what is that equal to?

[raven cipher]

yeah sorry

i got 4

[elfin snow]

not quite

what's 3^2?

[raven cipher]

9

[elfin snow]

what's 5*9

[red scarab]

Wow

2

[elfin snow]

?

5*9-10*2+9*2 is equal to what? @raven cipher

[red scarab]

I think he just has to calculate the term

[elfin snow]

yeah hes evaluating the expression

[raven cipher]

158?

[elfin snow]

order of operations

[raven cipher]

Shi

[elfin snow]

multiplication before addition and subtraction

[raven cipher]

Give me 2 mins

[elfin snow]

ok lol

[raven cipher]

43?

[red scarab]

Correct

[raven cipher]

Okay ty guys

[elfin snow]

gg

yw

[red scarab]

ggs

[random pelican]

hi, is my answer correct?

[willow wadi]

how will i find t1 and t2

,w (5-x)^2+9=(x+1)^2

[ocean seal]

Results provided by WolframAlpha

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

Requested by yungdagger#3426

[elfin snow]

guys please one person per channel

@willow wadi try to solve it by finding T1 and then substituting that result in

think of like a system of equations

[willow wadi]

how can it be substituted because

[elfin snow]

from algebra class

[willow wadi]

the cosines arent equal

[elfin snow]

hmm

lemme think

I feel like the numbers are chosen for a purpose, since 31=30+1 and 59=60-1

OH

ARUSLLY

[random pelican]

[elfin snow]

you can turn the cosines into sines

[random pelican]

it was my question

oh

its there

[elfin snow]

the definition of cos(x) is sin(90-x) @willow wadi

[willow wadi]

oh 🤔

[random pelican]

ty

[elfin snow]

so this can be turned into:
$\ T_1 \sin 59 - T_2 \sin 31 =0
\ T_1 \sin 31 + T_2 \sin 59 = mg$

[ocean seal]

gmod

[elfin snow]

@willow wadi does that help?

now you can manipulate these and solve them like a proper system

[willow wadi]

ohh

[glacial hedge]

In matrices is ABCD = (AB)(CD)??

[oak chasm]

Yes, matrix multiplication is associative.

[alpine sable]

Can someone verify that this proof is correct?

I have lim(w->2)$\sqrt{2w+5}=3$, $\delta=\min(1,3\epsilon/2)$

[ocean seal]

jswatj

[alpine sable]

and here is my proof of the limit statement

[ocean seal]

jswatj

[alpine sable]

i did |2w+5|<16 since i knew it was gonna get rooted

<@&286206848099549185>

$\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\boxed {\LaTeX}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}$

[ocean seal]

clockwurk

[hardy mirage]

What does the question mean?

What does it mean "5 units from (5,5) and 2 units from (2,2)?"

ermm

no

So it is square root of (5-2)^2 - (5-2)^2?

Oh ok

And what does plug mean

Is it subsitute?