math-help

[cyan night]

Let the polynomial f(x) have integer coefficients. Does there exist five distinct integers a, b, c, d, e that satisfy f(a) = b, f(b) = c, f(c) = d, f(d) = e and f(e) = a ?

i used the property that f(x) - f(y) : x - y when f(x) and f(y) got integer coefficients

[ember kernel]

?

[cyan night]

uh

: means divisible by

im lazy to write divisible by

[ember kernel]

Oooh

Mb, I thought it just meant /

[cyan night]

ok

so uh

it actually doesnt exist

[ember kernel]

Let me check everything first

[cyan night]

but im struggling to find a way to subtract equation by equation in order to get this

uh

u still can help?

@ember kernel

[ember kernel]

Yup

[peak gale]

have you already studied the theory of groups?

[cyan night]

What is it about

[ember kernel]

(The Greek letter theta seems used as an exponent, but it's merely multiplied, I has forgotten to write it)

[cyan night]

Oh

What’s the final conclusion

[ember kernel]

No matter what integer values you get for alpha, beta, gamma, lambda and theta, such that their multiplication = 1, you will always get a system of 5 unknowns (a,b,c,d,e) for 5 dependant equations

That means an infinity of solutions

[cyan night]

Uh

So u aren’t subtracting equation by equation

[ember kernel]

And thus that there exist integers a,b,c,d,e such that f(a) = b, f(b) = c,...

[cyan night]

Wait let me show u this

[ember kernel]

(If I'm not wrong of course. I was helped by programmation language Python, but that's not the kind of questions I'm used answering to)

[cyan night]

this was how my teacher did it for 3 variables

and then subtract it like that

ket hop just means combine

@ember kernel

set as k

and then is invalid

so k =0 and => a = b =c which is invalid

so there is no possible solutions

[ember kernel]

Oh

Oh no

[cyan night]

but this was with 3 variables

[ember kernel]

I'm all wrong

[cyan night]

i think 5 u do the same

[ember kernel]

I just suck lol

Today's the day I retire from maths

[cyan night]

uh

but

the thing is

[cyan night]

i dont know how to subtract

to get it equal

[ember kernel]

I'm sorry

I'll just give up for this one. My brain's square, and I've just proved to myself that this is not the math I'm familiar with 😅

[cyan night]

oh

okay

[left knoll]

1. In ax’ +bx+c=0,00, delta =b² - 4ac is called discriminant. If theroots are equal, then

2. If the roots having opposite signs, then

3. If both the roots are negative, then

4. if both roots are positive then

a. delta = 0
b. -b/a greater than 0
c. c/a less than 0
d. -b/a less than 0

​Ayo can anyone help me in this thing pls

[tawdry dagger]

THANKS MAN ❤️

[spring crater]

just checked this channel and it reminded me about the fact that because of covid, online school and our new teacher i missed a whole year of maths, i am kinda fu... yea

[fading basalt]

Help me with this

I am gonna take a walk in the meantime

[cyan night]

what have u thought of yet

[thorny flicker]

this is my method by idk if it is a good method @fading basalt
1/16 x 100% = 6.25%
1/12 x 100% = 25/3 or 8.33 % (keeping in fraction for accuracy)

Let's say the work has to reach 100% to be completed.

Starting with A, the first day has 6.25% completed.
Then B on the second day, so 6.25 + 25/3 = 175/12
Then third day i add in A again, so 175/12 + 6.25 = 125/6

i keep on adding up the percentage until it reaches 100%, which will be at day 14.
Pretty sure there is an easier way xd.

[fading basalt]

I am back

I was taking a little walk

[cyan night]

keep it with fractions

1/16 and 1/12

3/48 and 4/48

[fading basalt]

I did it like that

[fading basalt]

Answer is not 14 days

[cyan night]

3+4 = 7

[thorny flicker]

ah ok im wrong

[cyan night]

48/7 = 6.85?

bet is wrong

[thorny flicker]

can i have the ans?

[fading basalt]

6.85 what?

[fading basalt]

Not even close

[cyan night]

then 13

[fading basalt]

Nope

[thorny flicker]

On the first day A completes 1/16 of the work. On the second day B completes 1/12 of the work. Therefore, in 2 days they complete 1/16 +1/12 = 7/48 of the work.

In 12 days they will complete 6*7/48 = 42/48 of the work. The remaining work is 6/48.

On the 13th A completes 3/48 of the work. The remaining work is 3/48. Now, B can complete this work in (3/48)/(1/12) = 3/4 day.

Hence, the whole work is completed in 13 and 3/4 days.

[cyan night]

what have u thought of?

[thorny flicker]

oh i did it again but with fraction

[fading basalt]

It’s 13 3/4

[cyan night]

yep

this is for grade 5 right

[fading basalt]

Nice

[fading basalt]

Ye

[fading basalt]

I thought of is total work as 1 and A’s one day and B’s one work to be 1/16 and 1/12 respectively

And assuming the total no. of days to be x

[cyan night]

thats what dusk said

[fading basalt]

Ye

[cyan night]

1/16 = 3/48 1/12 = 4/48

2 days = 7/48

12 days = 42/48

and its A's turn

[fading basalt]

Ye

We gotta find it for the last day separately

That’s what i was missing

Thanks

[thorny flicker]

https://tenor.com/view/hololive-shirakami-fubuki-2d-animation-head-banging-gif-16938082

[fading basalt]

Before i got 13 6/7 days

[cyan night]

,flip

,flip

,rotate

.rotate

.flip

,help

!help

!rotate

[cyan night]

uh

[fading basalt]

What’s this

[cyan night]

its my math solution

how do i rotate it to submit to the teacher

bots!

!bots

[fading basalt]

You don’t need bot for that

You can retake the pic

[cyan night]

how do i flip the image

my camera wont fit horizontally

[fading basalt]

Takes a sec

K wait a sec

Here

Not that hard

@cyan night

[cyan night]

its kinda blurry

[fading basalt]

I used this

[cyan night]

uh

nvm then ill send it horizontally

lol

[fading basalt]

…..

[coarse temple]

Hello, i just wanna ask when we apply newtons second law and we project on the y or x axis and then if a vector is angled we multiply it by cos or sin but the thing is idk which angle to use and why we actually do this
Plz if some one can give me a quick answer or link me to a document i would really appreciate it

Im litteraly abput to be a swcond uni year student anf i keep forgeting how this works

Well i think i figured it out

I think yeah i did

I just thought logically how to do it

Its just when the x and y axis are angled confused me

[stuck pewter]

Pls help me with this, I'm gonna get depression coz of it

[proper sail]

Can we apply D.I method here?

[bleak tree]

The answer is D) none ]-1;0]

[stuck pewter]

But my teacher said B

[ember kernel]

⬆️ Answer above, in MathBot

For the first equality (1st line), I used f' = e^u and g = sin(u)

For the second equality (2nd line), I integrated a second time with f' = e^u and g = cos(u)

For the equivalence of the 3rd line, just add the integral of e^u sin(u) to both sides of the equality

[fierce vector]

Option B

[stuck pewter]

Sir I worship u.

[fierce vector]

As ome is not included in the number line you have to include it in the remaining set D

[fierce vector]

xD

[stuck pewter]

How did u do that

Modulus is torture

[stuck pewter]

Ok thank you

[fierce vector]

Did u get it??

[stuck pewter]

I mean yea but I still have like a lil problem with modulus

Like it sucks.

[proper sail]

@ember kernel OK thanks

[cyan night]

yo can anyone help me
im wondering how i can prove vf^2 = vi^2 + 2ad
i learnt the formula in my kinematic unit at school, but my teacher didnt prove it to us as she said it required calculus

[paper bone]

You do know the first 2 equations of motion right?

[cyan night]

disstance formula?

or vf = vi + at

[paper bone]

Yeah those two

[cyan night]

d = vi*t + at^2/2

yeah

[paper bone]

Yeah I'm not sure how your teacher derived them

But my textbook derived it from a graph

[cyan night]

oh

[cyan night]

i cant really see it well

[paper bone]

Here v is the final velocity, u is the initial velocity and x is the distance

[paper bone]

It seems fine to me?

Again, if you can't really understand it

I'd suggest you leave it

[cyan night]

i have to zoom in its pretty blurry

no i understand

[paper bone]

Alright yay

[cyan night]

but i cant really see this

oh wwait

did it say

x = vi + 1/2 at^3

[paper bone]

That's the distance formula

[cyan night]

can u type it out for me i think its not the standard one

[paper bone]

it's x = vi*t + 1/2at^2

[cyan night]

oh

ok

whats the next one

[paper bone]

vi is just written as vo
And vf is written as v

[paper bone]

I'm not really sure why the quality is that compressed

Is there any other way I could send it?

[cyan night]

this is what i got

oh wait its fine

wait whats the line above the letter?

[cyan night]

ayy thanks

[paper bone]

Is this better?

[cyan night]

yep a lot better

[paper bone]

Alright ^^

[cyan night]

wait i dont reall yunderstand whts the line above v

average?

ah ok

thnkas

[paper bone]

Yup

[nimble tendon]

hi, is there a specific college major that invole in Economics, Finance and Maths?

[left knoll]

@ember kernel I'm starting with integration. Can you suggest me how and where to start

[covert anchor]

How am I sup to do this

[cyan night]

Econometrics?

[heady tree]

If you separate a condensed fraction (left) into individual fractions that you’re adding together then they of course have to have the same denominator (3+x)(4x-^2)

Its like they meshed addition of fractions for the numerator with multiplication for the denominator lmao

[ember kernel]

Multivariable integration I guess ?

I've got a whole 8-or-so-page summary on it if you wish

[ember kernel]

I'm not available right now for the whole explanation of what to learn in what order, but dm me and I'll answer when I can 😄

[south token]

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

https://tenor.com/view/spongebob-squarepants-dance-happy-dance-歡快-gif-5084836

[drowsy fog]

Anyone here ?

[left knoll]

I'm bad at transposing what to do ?

[drowsy fog]

I need help with answering this, its supposed to be easy ???

[peak gale]

maybe use : x-y= the sum of fractions, then transform the denominator into only one variable thanks to x-y=1, and by playing withs the denominators, you will obtain a big polynomial of the 4th degree, then pay attention to the common factors to reduce the polynomial and find roots

[cyan night]

cant u just substitute x = y+1 and solve it

x = y+1 => 4y/3(y+1) - 10(y+1)/3y = 1

=> 4y/3y+3 - 10y+10/3y = 1

4y/3(y+1) - 10(y+1)/3y = 1

[drowsy fog]

I think i got it

[cyan night]

then reduce to the same denominator and solve for 1 variable

yes its easy

do it my way its very simple

good luck and have fun with studying fuji ^^

[drowsy fog]

I got,

y = -2 , x = -1

Or

y = - 5/9 , x = 4/9

[cyan night]

hmm let me try

[cloud night]

can someone help me with this q pls

[cyan night]

try drawing it out

[fierce vector]

is the answer 16 and 8 kmph??

[cloud night]

nvm I solved it

thanks

[fierce vector]

ok

[warm coral]

Heyo I have a quick question about surface integrals, I am trying to solve this question without using Gauß/Divergence but just by using surface integrals. Is my working here correct? How would I do the surface integrals for the faces z=0, x=0 and x=3?

When I do the 1st surface integral, my answer is already bigger than what I get when I do Gauß/Divergence so I am currently a bit confused.

[celest hamlet]

I'm not getting the right ans, pls can anyone send the sol of this Q

[left knoll]

anyone

[hearty tapir]

Hi guys.There's something i didn't understand completely.Why n.root of x^n is equal to x when n is odd, and is equal to |x| when n is even?Can you explain this with persuasive ways please?

[loud igloo]

still need help ??

[celest hamlet]

no i got it

thanks for replying tho

[loud igloo]

ok no problem i'm here if you need any help

[celest hamlet]

actually i need

i posted a phy Q on #science-help

if you can explain, it'll be a great help

[loud igloo]

i didn't study that yet

maybe i'll explain it to you next year

[celest hamlet]

oh k lol but i can't wait that much tho

[peak gale]

because when n is odd , (-x)^n = (x^n) * (-1)^n = -x^n; the n-th power of x can be negative when n is odd, and not when n is even. So the only solution for x^n is the former case is a positive solution

[hearty tapir]

Hmmm.I think i found something else about this:

When n is odd, ⁿ√x is a real number in every case because nth power of a real number can be negative and positive when n is odd.But when n is even, ⁿ√x is not a real number in every case, sometimes it can be a complex number.x must be nonnegative if we want ⁿ√x to represent a real number, because nth of a real number is nonnegative when n is even.Keep this in mind.
nth root function is inverse of the nth power.If you take nth power of a number and then take nth root of that number, nothing will change.Keep this in mind too.

    /////////

    When n is _odd_ ,
    x can be positive or negative, so xⁿ can be positive _and_ negative.That way, ⁿ√xⁿ can be equal to a positive or negative number like x.This means, it's okay for ⁿ√xⁿ to be x since x can be positive or negative.
    Also, we can write this expression as _(xⁿ)^1/n_ .xⁿ can be positive/negative as we said, 1/n. power of a positive/negative number can be equal to a positive/negative number.So, there's no problem when ⁿ√xⁿ is equal to _x_ and n is odd.

    _ⁿ√xⁿ = x_ ✓ 

   /////////

   When n is _even_ , things change.
   x can be negative/positive again, but xⁿ is always nonnegative.We may think that when x is _negative_ , ⁿ√xⁿ can be equal to x because nth root function is inverse of the nth power function.However, things are different here.
   We know that ⁿ√xⁿ = (xⁿ)^1/n and xⁿ is always nonnegative.Because of that 1/n. of a _nonnegative_ must be _also nonnegative_ .This means the result of simplification of ⁿ√xⁿ must be equal to something _nonnegative_ and we can make the result nonnegative with taking its _absolute value_ .This might be a logical cause of the this equation:

     _ⁿ√xⁿ = |x|_ ✓

I'd be very grateful if you tell me if there's any mistake in this thought

[left knoll]

hey can anyone help me with a problem

[silk rivet]

So i found out that the answer to the question is the letter B. anyone here can help me to explain how did that happened?

[heady tree]

factorise both the top and bottom into (x+-number)(x+-number) and cancel

[silk rivet]

I tried to factor the bottom part but I couldn't factor the top, i maybe missing some rules in factoring binomials or maybe the question is wrong idk

[left knoll]

I guess it is x^2 - 25 so that numerator cancels with the denominator

X^2- 25/ (x- 5)^2 => (x+5)(x-5)/ (x-5)^2

[silk rivet]

so the simplification would be (x+5)/(x-5) right?

[left knoll]

yessir

[silk rivet]

I see thank you all, you guys are cool

[late thunder]

The 2nd one.

[left knoll]

faisal must be 10 years old

[left knoll]

Differentiate product equation and equate it to zero

[thorny pasture]

Can anyone help me with the question?

[cyan night]

do u guys know any websites that allow me to freely annotate on a unit circle?

[peak eagle]

look beneath this message: you need to solve the equation that's written there

[thorny pasture]

THANK YOU SM

[covert anchor]

hi

[ember kernel]

hey

[covert anchor]

how do i do a

[solid anvil]

what is the condition to have a determinant of a matrix?

i dont think there is any condition

so pls help

[kindred otter]

Since, the given solid is a triangular prism, ABEF must be a rectangle by definition.
Hence, BGE is a right angled triangle with BG as hypotenuse.
Now, EB = BG (cos60) = 50cm

[zenith quarry]

Determinant is special number that defined only for square matrices . Determinant is used to know whether the matrices can be inverted or not. It Is useful to slove simple linear equations, used in calculus to find area of traingle and more !!

[nocturne forge]

Yes

[sterile hare]

number 5 pls

[drowsy fog]

Hmm i have yet to learn that

[cyan night]

do u guys know any websites that allow u to freely annotate on a unit circle?

[cyan night]

Given isosceles triangle CAB (CA = CB). CM is a median line ( AM = BM, M lies on AB). AD is the bisector of angle CAB. Calculate the measurement of angle B known that AD = 2CM.

🤔

[frail geode]

anybody who needs help with math, ill be here for the next hour or so ^^

[frail nova]

Can any one explain the 3N+1 theory?

I wanna understand it

[frail nova]

I think i knew how to solve it but im too stupid in english to explain it

Sorry

[weary arrow]

Bro no one can solve that 😂

[frail nova]

Ik

But need a little closer look on

I just want to see how far i can go with solving even tho i wont

I have doubts on it

[weary arrow]

Well, veritasium or Smth did a YouTube video on it

Maybe that'll help you

[stable ridge]

I need help with these issues

[cyan night]

Hi

[left knoll]

hi yea

so obv u can learn them, but the way i do it is

[opal rock]

What topic is this in math I can help

[left knoll]

theta is an acute angle- so 180-theta is in the second quadrant where sin and cosec are postiive

[left knoll]

trigonometry

[cyan night]

yeah

[opal rock]

OK I learn't that already I can help

[left knoll]

similarly 360-theta will lie in the fourth quad where cos and sec are positive

if it is a mutilple of 180 then dont change anything

but if it is a multiple of 90 then u change it, like: sin(90-theta)=cos theta, sec(270+theta)= -cosec theta

[cyan night]

yeah the astc

[left knoll]

the 90 ones u gotta learn

[opal rock]

SO like in trigonometry you should always remember there is sine, cosine and tangent

[cyan night]

and the 270?

[left knoll]

there u can just apply ur understanding na

like hm

[cyan night]

i remember doing a problem on khan today and i got wrong

the evaluate sin 240

[opal rock]

All I can tell you trigonometry is about using a calculator and its all about triangles

[cyan night]

wait

[left knoll]

270 is a multiple of 90 so u gotta change the ratio

[cyan night]

what was it again

[left knoll]

then the positive or negative is acc to the quad

uh give me questions, sahil help

[cyan night]

uh

so

there was the exercise

evaluate sin 240

so i thought i take 270 - 240 =30 and find sin 30 but i was wrong

[red spruce]

SIN (90 X 2 + 60 ) 90 EVEN MULTIPLY SO NO CHANGE sing 60 now for sign 240 third qaud and sign is negative so -sin60

[left knoll]

why are u hindi

[cyan night]

what

[left knoll]

can blueberry understand hindi

[red spruce]

oop

[cyan night]

no i cannot

[left knoll]

yea

ok let me attempt to explain

[red spruce]

nvm i edited it

[left knoll]

ok so

[cyan night]

i know it third quad but normally i do it that way for quadrant II and IV

[left knoll]

sin(240)= sin(270-30)

[cyan night]

yes

[left knoll]

sin(270-theta) = -cos theta

[opal rock]

What are we trying to find

[left knoll]

since 270 is a mutilple of 90, we change sin to cos

[opal rock]

Ok Im outta here Im useless

[cyan night]

isnt 90 and 180 and 360 also a multiple

[red spruce]

all school to college --- this is the trick , all positive in first quad
school - i e sin functions un second
to - ie tan function in third
college ie cos function in 4th
are positive

[left knoll]

and since 270-theta, (where theta is an acute angle) lies in the third quadrant where Sin is negative we put the - sin

[left knoll]

sorry im an idiot i mean odd multiples of 90

[cyan night]

u can just say odd quadrants

[left knoll]

hm

[cyan night]

oh

[left knoll]

what if u get sin 750

[cyan night]

then go in circles

[red spruce]

HMPH it will be fun when u start solving in complex equation u wont remember odd quads and eveything

[cyan night]

750-720 = 30

[red spruce]

keep it simple

[left knoll]

multiple ways to understand

[left knoll]

nono, 750 isnt a multiple of 90, u will have to do 720+30

[red spruce]

sin 90 x 8 + 30

[cyan night]

then in that case i have to brute force memorize identities

[cyan night]

its in the first quadrant so i can just take sin 30or can i

[left knoll]

so here u got -cos30 which is -root3/2

[red spruce]

at the end pf day u have to lmao there no shortcut for that

[left knoll]

yea it is, so the answer is 1/2

[cyan night]

ya i thought it was -sin 30 = -1/2

[red spruce]

or - sin 60 huhuhu

[left knoll]

brother chill out

lmfaoo i think blue is mroe confused by us

i feel so bad

[red spruce]

hmph see the meme u sent i made another one

i told him to watch youtube lectures

[left knoll]

yea wait

[cyan night]

if i have to evaluate cos 330 then is it just cos 30 because cos is positive in the 4th quad

[left knoll]

yes

better to go step by step

cos(360-30) = cos 30

[cyan night]

if its sin 330 then i do -cos 30?

[left knoll]

no

[cyan night]

oh uh

[left knoll]

ok how does ur reasoning go about it

why do u think it is -cos30

[cyan night]

wait let me retry

cos 30 = sqrt3/2

^2 = 3/4

[left knoll]

yea

why

[cyan night]

then 1- 3/4 = 1/4

take root = 1/2

[left knoll]

ah

youre calculating sin30?

[cyan night]

so sin 330 =

-sin 30

[left knoll]

yea

[cyan night]

yes i use pythagorean identity

[left knoll]

but i dont think ur method

sin330= sin (360-30)

[cyan night]

sin 330 = -sin 30 wait what it just works

then why doesnt it work that way for cosine

[left knoll]

sin(360-theta) is -sintheta, since 360 is even multiple of 90 and it lies in the fourth wuad where sin is -ve

[left knoll]

wdym

[cyan night]

when cosine in 3rd quadrant in the exercise aforementioned

[left knoll]

cosin in third quad is negative

[cyan night]

sine is negative in 4th quad

[left knoll]

are u calculating cos330

[left knoll]

yes

[cyan night]

but sin 330 i can just subtract 360-330 = -sin 30 why cant i do the same to cos 240

both are negative

[left knoll]

cos 270-30=cos240

answer should be -sin30

[cyan night]

ya why cant i take -cos 30

[left knoll]

because 270 is odd multiple of 90

[cyan night]

uh

if sin 240

then is it -cos 30

apparently it isnt

yoshi?

[left knoll]

yea wait

my net

[left knoll]

it is

[cyan night]

ok i realized i was in radian mode

[left knoll]

oh what

[cyan night]

yeah it is

[left knoll]

ive been talking degree the whole time

[cyan night]

yeah just on the calculator

i got it lol

so its just the 3rd quadrant

i take 270 and subtract whatever angle

and i get an acute

then i change ratio and symbol right

[left knoll]

yesyes

[cyan night]

not for 45 cuz its the same for both sine and cosine

[left knoll]

change symbols according to quadrant, change ratio according to odd mutliple of 90

[cyan night]

ok thank you

for the first quadrant i already memorized so no need xd

[left knoll]

i mean just do the whole method

[left knoll]

yea u are supposed to know that

ill suggest u solve a bit more, or ping me so that we can solve some together

just to u can fully understa

ofc if u prefer to learn then go ahead

[cyan night]

okay

i have to fix my notes

[left knoll]

i find that understanding is easier

let me see if there is a youtbe video

[cyan night]

ok

iss there

[opal tulip]

Can anyone help me in this

[cyan night]

the squares are equal right

whats is 16 and 24

[opal tulip]

Ohh sorry

I forgot to mention that the pink one is a square and the second one is a rectangle

@shut violet @cyan night

[cyan night]

whats 16 and 24

[opal tulip]

And the white dots on the sides mean that they are equal

[opal tulip]

when you sum 16 and 24 you get the length of the square side

[cyan night]

ah ok

wait ill do it for u later

i have class rn

[opal tulip]

thanks a lot

Good luck man

[cyan night]

you 2 have a good time

ok that was a weird statement

[opal tulip]

Ikr

[shut violet]

.-. I just helped you with the solution above

[cyan night]

i think theres an easier way ill figure it out

[opal tulip]

yea i sawww it sorry

[cyan night]

no need heron

[opal tulip]

Thank you too!

I used the hardiest way

[cyan night]

try exploring his way first

then ill make my way

better set a different variable than x

[shut violet]

x is the original variable

I didnt change it

[cyan night]

ah okay

@opal tulip what grade math is this

[opal tulip]

Lol its was a challenge between me and my brother

[cyan night]

oh alright u are cheating btw

xd

[opal tulip]

tbh I forfeited this challenge once he started it

[cyan night]

is he an older or younger bro

[opal tulip]

waaay older

[cyan night]

ah understandable

[opal tulip]

he just saw me solving some equations and wanted to ruin my day

[cyan night]

he bully

[opal tulip]

this is not his first time doing it

[cyan night]

yeah it was pretty obvious if he challenged u with this problem lol

so what was the bet

[opal tulip]

if u lose you are a looser 😭

[cyan night]

ah wait let me finish class

[opal tulip]

gl

[fossil wedge]

a? and F?

[peak gale]

try to derive it ,to find f, then use the expression of f in the integral for a

[left knoll]

use leizbnitz rule try to derivative it ,find f(t),Then integrate it

[loud igloo]

can anyone show me the steps to find the value of this

[peak gale]

find the value of sin(nPi) for any n; observe what happens when you sum up 1+(-1) 100 times; what is the value of cos(mpi) for any m? you will find

[fossil wedge]

thanks you guys

[stable ridge]

I Need help

[left knoll]

Can y’all help me with my pre cal exam review?

[frail nova]

I knew thr equation from veritasium
But i still have a hope that it is solvable

[weary arrow]

Good luck lol

[mental snow]

a = 9 and f(x) = x^3/2

[fossil wedge]

how did you find a?

[mental snow]

when we get f(x) = x^3/2 we put it back into the given expression

6 + integral a to x (x^3/2 / x^2 dx ) = 2 root x

[mental snow]

when we solve this integral with limits we can get a

[mental snow]

the square is over pi m or the whole cos pi m?

[fossil wedge]

Perfect, thank you

[mental snow]

😄

[cyan night]

what test

ah ok

which question do u need help with

8?

graph it out

then identify

@left knoll u graphed it out yet?

is the calc broken

[left knoll]

Can anyone solve it

[mental snow]

is it this?

-6 1 -6
4 -1 2
4 -1 4

[modern obsidian]

Eish matriciels is out of my ibises syllabus

[left knoll]

Thanks buddy for your help👍

[cyan night]

bezout's identity confuses me: my teacher said the identity states that if there exist x = a in the polynomial P (x) with a rational co-efficient then there co-exists a polynomial Q(x) such that P(x) = (x-a)*Q(x) @left knoll

[left knoll]

Yes that's correct

[cyan night]

i dont even understand

do u get it

[left knoll]

It's called fractional rational function. P(x)/Q(x)

Therefore the (x-a) which is one of the roots of the equation is multiplied with Q(x)

[cyan night]

🤔

😕

[left knoll]

The picture that I sent you had a question based on that

[cyan night]

how coincidental

[left knoll]

Like p(3)= 3* p(4)= 4
From here say p(x) =x

[cyan night]

wait i forgot

it said that

if there exist x = a

and a is the solution of the polynomial (the polynomial is = to 0

[left knoll]

P(x)= x becomes P(x)-x= 0 and is now a equation that has roots 3 and 4

[cyan night]

ok screw it i dont get it

[left knoll]

I will send you a picture with example

[cloud plaza]

Can someone explain me how it became 2/3 pie

[left knoll]

hi

anyone here knows about limits

[lofty crown]

yeah no problem, what's causing your headache?

[left knoll]

The problem is the way it is written is difficult, I will photograph it and send it

[lofty crown]

yeah, what grade you in? I know they can be a mindful but i think I can probably helkp

[lofty crown]

well it's pprobably known that YXZ is pi/3 so if you put it in the right hand of the equation then you aomplify pi buy 3 you get 3pi/3 wich you subtract from pi/3 so you get 2pi/3

[left knoll]

@lofty crown

[lofty crown]

aight so

[left knoll]

limit x-> 0

A,B >0

[left knoll]

The answer is in symbols A, B .

[lofty crown]

do you see the thread i created for this?

[left knoll]

YEAH

[cloud plaza]

Thank You , after cross checking it I found out ; didn’t do math for months

[steel plover]

that's a sheet of ratio to earth values. if I wanted the values of the ratio to the sun(or just a planet besides Earth), how should I go about calculating that?

[polar heron]

you are given ratio of a quantity, say X, of planet p, to earth, i.e. X_p / X_e. to calculate the ratio to any other planet t, divide the ratio to earth of the p by ratio to earth of t. i.e.
( X_p/X_e)/ (X_t/X_e) = ( X_p/ X_t).

[steel plover]

got it! thank you

[cyan night]

did i do this correctly

problem : Find the real numbers (a,b) that satisfy:
4x^5 + ax^3 + 13x^2 + bx + 8 is divisible by x^2 + x - 2
so
Let Q (x) = 4x^5 + ax^3 + 13x^2 + bx + 8
=> Q(x) is divisible by P(x)*x^2+x-2
=> Q (1) and Q (-2) = 0
then i substitute x = 1 and x = -2 in Q(x)
=> a+b = -25 and -8a-2b = 68
=> a = -3 and b = -22

[tough geyser]

nope

[cyan night]

that was expected

😦

[tough geyser]

yeah you did it correctly

[cyan night]

yes thank you

i meant the whole method not just the last calculation part

i used bezout's identity

[tough geyser]

yeah its right

[cyan night]

thank you

its actually wrong lol

a = -3

b = -22

xd

[celest geyser]

Do you use anki for math ?

[cyan night]

are u asking me

[glad sierra]

I’m assuming you don’t need me to check anymore 😂

[cyan night]

it was correct xd

[cyan night]

@near umbra what ya need

hihihihihi

hihi

hihi anyone

uh

😕

anyone

[near umbra]

Why me lmao

[cyan night]

because u told me to go here

[left knoll]

Hello guys

Can any1 help me

?

[cyan night]

hihihiihi hi

me

[left knoll]

With this

Hi

[cyan night]

im math helper

[left knoll]

Do u know how to do this?

[cyan night]

i believe

wait does it not give me a variable for the width

[left knoll]

No

[cyan night]

oh length

ok

uh

[left knoll]

R u solving it?

[cyan night]

yes

[left knoll]

Ok

[solid anvil]

which grade question is this?

[cyan night]

its 6 grade

[left knoll]

9

Quickly answer me

Pls

Are u guys solving it.

?

[cyan night]

p-r-q-r

lol

[left knoll]

Lol

[weary arrow]

r-(p-q)

[cyan night]

i meant that

[left knoll]

How?

[cyan night]

i messed up my keyboard

[cyan night]

p-q is that white part

and r is white part + green

just subtract

[left knoll]

I just did r/2=p/4

[cyan night]

noone where trig

yes it is

noone where trig

yess it is 6 grade why 9

[vestal spoke]

@cyan night if,7sin²A+3cos²A=4 prove that tanA=1/(root 3)

9th grade

[weary arrow]

Correct

[cyan night]

what

tan a = 1/root 3 when a is 30 degree

[left knoll]

Thx @weary arrow @left knoll

[cyan night]

that isnt 9th grade!!!

omg im gonna rage!!

[vestal spoke]

We never learned trigo in grade 6 sooryyy...

[cyan night]

uh

[mental snow]

4sin^2x + 3(sin^x + cos^x) = 4
4sin^2x = 1
sinx = 1/2 that is x = 30 degrees
so tan x = 1/root3

[left knoll]

Was a trik question????

[vestal spoke]

It is!!!!!

[cyan night]

give different question

i want question relate to geometry

i cant do formulae trig 😦

!!!

[mental snow]

😮

[vestal spoke]

OK...sorry faygo... I will select some ques including geometry later...I have exams ahead...

[cyan night]

okay no problem but that wasnt 9 grade!!

[left knoll]

I thought did I miss something

[cyan night]

wait until i learn all the formula and then i go big brain

[vestal spoke]

Yeah sure...I will tell that to our education minister👍 👍 😤 😤

[cyan night]

its 10th grade for vnmese too

after this class i spend about 1 hour to study econ then ima go to sleep

[vestal spoke]

Sleep well ques will be ready tomorrow promise

[cyan night]

hmm noone is typing

ah ok

good luck on exam prep noone

[vestal spoke]

Best of luck...I am sure u will

[cyan night]

yes i go big brain

anyone need my trig notes dm me they are basic trig notes

[left knoll]

What?

[cyan night]

from basic in right triangle -> law of sines and cosines goes thru unit circle and introduce to identities with examples and elucidations

dm me for my trig notes for free

just pls dont sell them

[stable ridge]

i need help

[vestal spoke]

@cyan night ur ques ...they are mostly geometry based i think u can solve them👍 👍

[cyan night]

uh ok

but idk formulae

[vestal spoke]

They are geometry based

[cyan night]

for 2 i cant i just input

[vestal spoke]

Correct

[cyan night]

ok i do

[left knoll]

ye to prove u gotta do that but later when u study more trigo u will have to learn these lol cause these are identities

[cyan night]

but given a and b then can input

wait im in taekwondo class ill do later

[vestal spoke]

Yeah,they are formulas when u study more advanced trigonometry

[cyan night]

what

[left knoll]

<@&717391911132069929>

[cyan night]

is that abot

@signal sable

@signal sable ugly

[vestal spoke]

Ok....i m gonna go back to studies

[cyan night]

ok

[cyan night]

maybe ill do them later i cant do 1 yet\

wait let me try

[cyan night]

@vestal spoke

i got 1

easy

i think im done

all

pog

let me send u

pog

can u read it clear

[vestal spoke]

Oh...good good ...congrats👍 👍 👍

[cyan night]

can u read the paper

im writing down 2

[vestal spoke]

U just did 9th grade maths

[cyan night]

2 is just pure calculation do i have to do it

[vestal spoke]

Ok

[cyan night]

wait how do i handle tan^2 theta

noone

[vestal spoke]

Where is tan²theta ??

[cyan night]

2c

oh and can u see the paper for 1

[vestal spoke]

Just input the value of B and find out value of tan B ...then squre tan B

[cyan night]

oh ok

[vestal spoke]

I saw it its correct

[cyan night]

ok

thank you for problems

[vestal spoke]

👍

[covert anchor]

hi

question 9

[cyan night]

zoom in and see lol

[worn mist]

I’m not sure because its been 2 years since I’ve done this maths

[left knoll]

Tree 1: I will be twice for Tree 2 if it will lend me 1 bird.
Tree 2: I will be equal to tree 1 if tree 1 lends me 1 bird.

Question: how many birds are there on tree 1 & 2 initially?

[worn mist]

But for the first part I think you have to make sure to get the x’s on one side and then you have something like x=… for the one equation and x=… for the other. You know that both x’s of both equations will be the same so you can say those are equal and find y

[left knoll]

Would the answer be:

Tree 1: 3
Tree 2: 2?

[worn mist]

Because y is also the same that will be your first coördinate. You can then do the same with y=.. but you dont have to do all the work to find the x coördinate, just fill in the y you found in one of your equations to get the x and you’re good since they will be the same

To find out if they are collinear, theres a formula I think, not sure its been a while

[worn mist]

Tip: write it our with constants

[left knoll]

Tried , can to confirm the answer?

You*

[worn mist]

Umm i’d have to write it down but im in bed🥲

Also does tree1 have 2*tree2 of the amount tree2 has after giving the 1 bird or before

Cuz that would make your equations a lil different

[left knoll]

If it is before then I believe the answer is 3 & 2. However, idk if it's after then what would be the answer

[ember kernel]

9.1 : solve the equation

[ember kernel]

There are two solutions for x, and these correspond to the values for x where y are the same for the curve and the linear function

9.2 : Find the linear function that goes through both (x,y) solutions. Obviously it'll be something along f(x) = y = a x + b
Plug in (0;0.5) and see if 0.5 = 0a + b uf verified. If so, that means that (0;0.5) belongs to the graph of f(x), and thus that the three points are colinear (aligned)

(Mistake in the tex formula : it's 8x in the denominator, not 8y)

@covert anchor there you go

[ivory pumice]

No , 3&2 is wrong, answer would be 7 (tree 1) , 5 (tree 2)

x+1 = 2(y-1)
x-1 = y+1

solve these , you will get x= 7 , y = 5

[frail nova]

Anyone help

1+x=3

[left knoll]

Square both sides and solve, x should come pi, pi/4

[ember kernel]

⬆️ There you go. From there, you can just use the quadratic formula to find the roots of the last polynomial

[frail nova]

Thanks

Why its that long

I wish i wanna die now

[ember kernel]

That's just a joke, don't worry

Here's the real answer :

Substract both sides by 1, and you get :

[ember kernel]

(Click for full answer)

@frail nova

[frail nova]

I have a question

Why you did add -1?

[ember kernel]

Yup

[frail nova]

Is that a rule we can use in these situations or what?

[ember kernel]

When trying to solve an equation (like this one 1+x = 3), your ultimate goal is to isolate x in one side, so that you get x = ..., which directly gives you the value(s) that x can have to verify the equation

That's why I chose to substract 1, so that x is isolated in the left side

[frail nova]

Ok ok

Thanks alot man

[ember kernel]

No problem ! (And sorry for the little joke :p)

[cyan night]

wait why 3-x=1 is hard

[ember kernel]

It doesn't have to be hard to be asked about. Some math concepts, however basic, may require some time and alternative ways of learning to be clearly understood

[slow cipher]

how would you derive f(x)=2/x^3 ?

[weary arrow]

@slow cipher lmk if you want me to explain further

[slow cipher]

yeah, like I understand that 1/x^3=x^-3, but I don't know how to apply this rule with 2/x^3

[weary arrow]

You just apply normal derivation rules after this

This is an explanation for 1/x^3, so your expression is just twice that

[blazing socket]

guys i if you have to give someone tips to start to study the highschool math entirely, what do you recommend?

where should i start?

[flat blaze]

Algebra 1 Algebra 2 Geometry Pre calc

Thats why i did

review all the past concepts to prepare for ap calc

@blazing socket

if that answers ur question

also thats the for American school standards

i frgt about British School

[blazing socket]

ok. thanks i will take this to my notebook

whats does mean ap calc?

is like rial calc?

[left knoll]

Can anyone help me solve this without using calculator?

Log 18 base 3sqrt2

[flat blaze]

AP calc is like Pre Uni Standard

Something like IGs for u maybe

[ember kernel]

⬆️ There you go

[left knoll]

Ouuu just like that? My gad I'm so stupid😂😂 thanks

[ember kernel]

No worries ^^

[slow cipher]

so 2/x^3= 2*(x^-3) ?

[weary arrow]

Yes!

[cyan night]

pro

Given parallelogram ABCD. Let M,N be the midpoints of BC and AD, respectively. AM intersects BD at P. NC intersects BD at Q.
a) Prove that BP = PQ = QD
b) Let I be the intersection of AM and BN, K be the intersection of DM and NC. Prove that AC,BD,IK and MN all intersect at one point

ABCD parallelogram means AB // DC

go clockwise

[red spruce]

this is badass?

this is flatass

cmn

[cyan night]

try

b is hard

[red spruce]

later kid

[cyan night]

stop calling me that

its grade 8 math

why?

define kid

why are u saying this

[slow cipher]

ty

[prime imp]

Hi I'm a 10th class CBSE student
So I wanted to ask that is trigonometry of ncert would be enough for boards 2021-22or I have to do from more books

@prime imp Indian will understand this

[left knoll]

@prime imp I dunno

[timber scarab]

So I would recommend must do ncert (2 times atleast) and yes please go for rd or rs as it will give you wider range of questions

[cyan night]

@red spruce can u solve math

[timber scarab]

Mayb I can!

[cyan night]

can u try doing it

he said its flatass

[timber scarab]

Yss

[cyan night]

do u know how to do it yet

[timber scarab]

Questions?

[cyan night]

Given parallelogram ABCD. Let M,N be the midpoints of BC and AD, respectively. AM intersects BD at P. NC intersects BD at Q.
a) Prove that BP = PQ = QD
b) Let I be the intersection of AM and BN, K be the intersection of DM and NC. Prove that AC,BD,IK and MN all intersect at one point
ABCD parallelogram means AB // DC
go clockwise

its up there btw

[timber scarab]

Okay...gimme some time

[cyan night]

ok

[cyan night]

got it yet

[timber scarab]

A part done B part doin'(

[cyan night]

B is the hard one ://

[timber scarab]

INTERESTING ONE!

[cyan night]

u need me to give u hint

[timber scarab]

Yes if it has

[cyan night]

so first u call the intersection of ik and nm

u wanna prove that imkn is a parallelogram

thus that intersection is the midpoint of both diagonals

and then u consider the parallelogram amnc

[timber scarab]

Ypp...was goin' in right direction!😼

[cyan night]

in that parallelogram ac and mn are the diagonals

and then the big parallelogram ac and bd are diagonals

so they intersect

wait that wasnt a hint

but u have to show me how u can prove imkn is a parallelogram

[timber scarab]

Stopp!! Lemme complete

[cyan night]

okay

[timber scarab]

Listen! Suddenly my teacher has announced a meeting...so wait ..I'll do this question today but later

[cyan night]

okay

[gentle kraken]

Guys pls solve this one. I m not getting the right ans 😿

[cyan night]

what

what question

[zenith quarry]

Can someone explain me , y the period of cos theta is 2π ??

[cyan night]

Period of everything is two pi

Cuz two pi is going another roujd around the unit circle

360o

[zenith quarry]

Then what would be period for sin theta ? Will it be same

Is this concept from functions ???

[cyan night]

are u talking about on the unit circle or a graph @zenith quarry

[zenith quarry]

Graph

[cyan night]

same

the graphs are standard

[zenith quarry]

Okay thanks for the help

[cyan night]

np

[ember kernel]

What's the question ???

If it's about the number of cubes, there are :
9x8x3 + 7x6x1 + 5x4x1 + 3x2x3 = 216 + 42 + 20 + 18 = 296 cubes

[cyan night]

can u tell me how u calculated it

[flat blaze]

p sure he used the volume theorem Length X Width X Height for each section

[cyan night]

in a quadrilateral do the diagonals always intersect

[astral berry]

i desperate quick

[cyan night]

period is 180 degrees amplitude is -3

[twilit crane]

Rule means formula ?

[cyan night]

equation

[twilit crane]

-3sin(2x*pi/180)

[cyan night]

Given trapezoid ABCD (AB//CD) with AB = 3cm, CD = 5cm. M is a point inside the trapezoid. Draw parallelograms AMDN and BMCK. Find the length of NK.

so my thinking was that i would draw a midline of the trapezoid

and the midpoints where i use to connect the midlines are also the midpoints of the paralellogram's diagonal

but i have no direction in finding nk

[glad sierra]

ye i mean you need to create those parallelograms first right?

[cyan night]

u draw a trapezoid, pick point M inside the trapezoid and draw parallelograms

but i apparently dont have direction in finding nk it seems irrelevant

[glad sierra]

can you share what you've drawn?

[cyan night]

ok

i added qf which is = 4 and the midline of the trapezoid

@left knoll

[left knoll]

This looks correct

[cyan night]

its a different problem

but nk doesnt look relevant

[glad sierra]

bruh
didn't you say trapezoid? ;_;

[cyan night]

abcd is trapezoid

[glad sierra]

im def sure that this nk isn't what they had in mind

[left knoll]

Yup there it is

[cyan night]

the drawing is correct

but that is the nk

what can i relate nk to

[glad sierra]

also,. if m is a point, what is n?

[cyan night]

N is a point of the staged parallelogram

amcn is parallelogram, i just have to draw that satisfy it

[glad sierra]

Draw parallelograms AMDN and BMCK --> ye but you haven't done this?

[cyan night]

wait sorry

[twilit crane]

He just switched C and D in the drawing

[cyan night]

the points are wrong order

yeah

i have to switch n and k

doesnt really make a diffrence

[twilit crane]

Just switch C and D

[cyan night]

no im not allowed to do that

[glad sierra]

more like this s

[cyan night]

oh wait

tf

i have to switch c and d

[glad sierra]

duh that's what we're saying :p

[cyan night]

guys im a little high sorry

[twilit crane]

[glad sierra]

too much math

[cyan night]

so where should i start

guys

😦

[left knoll]

@cyan night ??

[twilit crane]

So N, K and M are alligned?

[left knoll]

If it has to be inside then why not?

[cyan night]

i dont think so

[cyan night]

wdym

[left knoll]

Then if it's not alligned then you will have to find NM MK and then you get NK

[cyan night]

u mean if it is alligned

[left knoll]

If it's not

[cyan night]

then u cant do it

[left knoll]

Mhm

[cyan night]

can u prove that they are alligned

[left knoll]

If M is a midpoint?

[cyan night]

it isnt

random M

[left knoll]

Can't actually then

[cyan night]

then how can i do it

if n,m,k are alligned

then the intersection of ad and nm is the midpoint of nm

the distance from the intersection of ad and nm to the intersection of bc and mk is 4

so i pretty much have the answer which is 4x2 = 8

but with that sense m also has to be a point in the midline of trapezium abcd so its incorrrect

thus n,k,m are not alligned

[glad sierra]

it says a point inside
doesn't say where 🤷‍♂️🤷‍♂️

[cyan night]

What parallelogram could nk potentially be a diagonal of

[dim warren]

yo could anyone help me with a question

[left knoll]

So the coefficients are 7 0 6 -8
So Why did he added zero
And can we add this zero to the other sums in this condition

rip English skills

[ivory pumice]

Adding 0 does not effect anything , yes you can add 0 in any sum it won't affect , here the coefficient of x² is 0

[ember kernel]

Just to show all the polynomial coefficients

For clarity, but as @ivory pumice said it doesn't change anything

[left knoll]

OKAY

THANK U THANK U THANK U

[left knoll]

figure?

@cyan night

[cyan night]

Given quadrilateral ABCD. M,N are midpoints of AD and BC respectively. H and G are respective intersections between MN with AC and BD. Prove that AC = BD if and ONLY IF ∠AHM = ∠BGN

do u need me to draw for u

[left knoll]

naa

[cyan night]

ok

[left knoll]

ill try

one sec

[cyan night]

so i developed an idea

fron n i drew a NP that is collinear to NA such that NA = NP

so i got that ABPC is a parallelogram since their quadrilaterals intersect at the midpoint of each other

but i dont know how to use ∠AHM = ∠BGN

[left knoll]

send your figure

@cyan night

[cyan night]

ok

@left knoll

[left knoll]

WTH

[cyan night]

wuat

its my best attempt at it

[left knoll]

U solved this?

[cyan night]

not yet

i added NP

such that NP is collinear to NA and is equal to NA

so i get a parallelogram

u there

@ember kernel quentin are u there

help

[left knoll]

hey

i think

[cyan night]

hi

[left knoll]

h and g

shouldnt be at each others position?