help-10

[timid silo]

then uv - integral(vdu) =
(t - tau)(-cos(tau)) - integral[-cos(tau) * -d(tau)]

[robust raven]

but u is inside of sin

[timid silo]

oh

oops

alright

thanks

.close

[obtuse pebble]

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[robust raven]

🙂

.close

[obtuse pebble]

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[full oar]

where do i start

[latent walrus]

say they meet xkm from toronto
then one train has travelled x km
the other has travelled (500-x)km

[full oar]

huh

so how do i write tu r formula

the

[latent walrus]

you need to find their speeds first really ig
mont to tor is v+50
tor to mont is v

express the above distances in terms of these speeds

you can then create 2 equations of v and x

recall distance = vt

[full oar]

what

what’s t

[latent walrus]

time

here thats just 2, since they meet after 2 hours

[full oar]

oh

so v is 50

ok h no it’s not

v is t+50

right?

@latent walrus

[latent walrus]

no

[latent walrus]

you have two distances

[latent walrus]

you have two speeds

for each train, match the distances and speeds
then distance = speed * time = 2* speed

for each train

then 2(v)=x

thats one equation

2v-x=0

[full oar]

oh

so is 2v=x the question

[full oar]

or is it this one

[latent walrus]

one of them

there is another one for the other train

[full oar]

yea ok@but which one are u talking abt

which is it

2v=x or this one

[full oar]

this

[latent walrus]

those are the same equation...

[full oar]

how

oh

ok

hang on is x toronto

or the distance

[latent walrus]

i created x as the distance from toronto

[full oar]

oh ok

and v is the distance to montreal??

[latent walrus]

v is a speed

I made it so that x is the distance from toronto

so the distance the train from toronto has travelled is x km
the distance the other train has travellled is 500-x km

its stated that the train from montreal is travelling 50km/h faster than the one from toronto

so if i say the train from toronto is travelling at v km/h
then the train from montreal is travelling at v+50 km/h

from distance = speed * time
i got the first equation for the toronto train as
2v=x

[full oar]

so isn’t the other formula just 500-x

which is montreal

[latent walrus]

thats not a formula

thats just an expression

but the distance the montreal train has travelled is 500-x yeah

you just need to make an equation using distance=speed * time

[full oar]

ohh

isn’t 500 v tho

no

ok wait

500 is distance

so 500=v*2

?

[latent walrus]

no

500 is the entire distance between toronto and montreal

and v is the speed of the toronto train, not the montreal one, we already did that

[full oar]

idk how to make the formula

[latent walrus]

for the montreal train, as i have defined stuff

distance travelled = 500-x
speed of travel = v+50
time = 2

[full oar]

yes

500-x=v+50*2

??

[latent walrus]

you have it in spirit

but brackets matter

v+50*2=v+100, which isnt what you meant

(v+50)*2

is what you want

[full oar]

Okkk

So is that the other fournuks

Formula

[latent walrus]

yeah
2v=x and 2(v+50)=x-500

time for some simultaneous solving

hol up

i may have done something dodgy

[full oar]

okk

[latent walrus]

oh right i just wrote this wrong here

i was writing it fine earlier though

it should be 2v=x and 2(v+50)=500-x

[full oar]

cna i write it as 500-x=(v+50)2

nvm

[latent walrus]

if you want

[full oar]

hang on

I got -2v+2v = 100-500

That’s like 0=-400

I think I did it wrong

[latent walrus]

howd you do that

[full oar]

I just plugged in x

since X = 2v right

so then it was 500-2v=2(v+50)

[latent walrus]

2(v+50)=500-2v
2v+100=500-2v
4v=400

but even then

x is the number you really want

this works too, but eliminating v would have been faster

[full oar]

Ohh ok

so the speed is 100

so how do we find the distance away from tontnto

100*2?

[latent walrus]

sure

[full oar]

so distance is 200

[latent walrus]

it seems to be so

[full oar]

Oh ok

200 km/hr

[latent walrus]

no

[full oar]

Oh shoot

Tahts speed

It’s just km right

[latent walrus]

yeah

[full oar]

Okk tysmmmm

[latent walrus]

🫡

[full oar]

.close

[obtuse pebble]

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[tranquil dirge]

How would I put this into polynomial form with the zeros?

[high lily]

identify the zeroes

then apply factor theorem

(just realised that's already mentioned in the hint)

[tranquil dirge]

Do I have to use synthetic division?

Cause we skipped over that section cause my professor said we wouldnt use it

[high lily]

no

synth division not needed

identify the zeroes
(at what values of x does the curve intersect the x-axis)

[tranquil dirge]

yes I have that part

but

where do I plug those zeros in

[high lily]

do you know what factor theorem states?

[tranquil dirge]

In my notes I have "if (x-c) is a factor of the polynomial f(x) then f(x)/(x-c) the remainder is zero"

[high lily]

that isn't what factor theorem states

that's the definition of what a factor is applied to polynomials

[tranquil dirge]

oh fk

[high lily]

factor theorem relates roots of a polynomial and its factors

if c is a zero of a polynomial, then (x-c) is a factor

[tranquil dirge]

So how would I apply this to the problem I showed? The main thing that im struggling to see is where to put the zeros to get it into a polynomial

<@&286206848099549185>

[iron juniper]

Can you identify the zeroes?

[tranquil dirge]

-1,0,1,2,3

[iron juniper]

I will give you a simpler, base case

Can you factorize the quadratic x²-5x+6=0?

From the factorization, tell me its zeroes

[tranquil dirge]

(x-3)(x-2) zeros are 3 and 2

[iron juniper]

So you notice that if a1, a2, a3 etc are the zeroes of a polynomial, then the polynomial is given by (x-a1)(x-a2)(x-a3) etc?

[tranquil dirge]

so do I just take the zeros and multiply them together?

[iron juniper]

Yes

(X+1)(x-0)(x-1)(x-2)(x-3)=0

Simplify, and you'll get the polynomial you need

[tranquil dirge]

ohhhh got it got it

thank you so muhc

.close

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[timid silo]

Prove that cosec2A + cosec4A = cotA - cot4A

[iron juniper]

What grade level?

Which curriculum?

[timid silo]

10th

[iron juniper]

Hmm

Ok

[timid silo]

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[dull void]

chain rule i dont know how from

$(2uv^{-3})(-5e^x) + (-3u^2 v^{-4})(15e^{3x})$
you get
$-5uv^{-4} [2ve^x + 9ue^{3x}]$

[warm shale]

odokawa

[obtuse pebble]

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[timid silo]

if you bind this as a definite integral

from 0 to t, with respect to x,

[timber island]

wdym?

[timid silo]

how on earth do they get this

[timid silo]

instead of treating it as indefinite

[timber island]

well thats not an integral

[timid silo]

treat it as definite with bounds of 0 to t

[timber island]

its what the integral spits out

[timid silo]

yeah thats the solution

ik

[timber island]

oh

i got what u mean

evaluate the solution from 0 to t

[timid silo]

ye just like plugging in 0 & t

idk how they got that

ye

wouldnt it be like, sin(4t - 2t) / 8

minus sin(0 - 2t) / 8

[timber island]

1 sec

[timber island]

well if u put in 0 in the solution u get 0, so 0 doesnt matter

combining the fractions

gives you

$\frac{2 \sin(2t) + 4t \cos(2t)}{8}$

[warm shale]

ItzKraken

[timid silo]

yeah im wondering how they get to that point

from this

by plugging in t & 0 to x

i just dont get it

[timber island]

$\frac{2 (\sin(2t) + 2t \cos(2t))}{8}$

[unborn vigil]

no

[warm shale]

ItzKraken

[timber island]

pls open a new channel. u can copy ur progress from the previous messages, if there was some progress

[timber island]

@timid silo cancel out the 2 and 8

[timid silo]

lemme see

[timber island]

if u still dont get it u need to go back to elementary school and learn how fractions work

jk

[timid silo]

bruh my problem is the plugging in part

like defining the integral solution from 0 to t

[timber island]

lemme plug it in for u

[timid silo]

sin(4x - 2t)/8

from 0 to t

the other part, the cosine, is already done

[timber island]

dont do it separately

[timid silo]

because its not with respect to x obvs

[timber island]

$$\frac{\sin(4x-2t)}{8} + \frac{x \cos(2t)}{8} = \frac{\sin(4x-2t) +x \cos(2t)}{8} $$

[timber island]

u have to sub for x no?

[timber island]

.

[timid silo]

yeah but i meant that cos(2t) part, no place to sub

thats all

[warm shale]

ItzKraken

[timber island]

now

putting 0

u have

$\frac{\sin(4(0)-2t) + 0 \cos(2t)}{8} = \frac{- \sin(2t)}{8}$

[warm shale]

ItzKraken

[timber island]

correct?

putting in t

$\frac{\sin(4(t)-2t) + t \cos(2t)}{8} = \frac{ \sin(2t) + t\cos(2t)}{8}$

[warm shale]

ItzKraken

[timid silo]

ye but i mean its from 0 to t so t should be first

no problemo tho

but yes that is right

also do you type that texit out or do you have some converter

ok so i see where the 2sin(2t) comes from

but what about the 2tcos(2t)

oh im silly

common denominator lol

im good now, thanks!

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[unborn vigil]

how can I do this?

[obtuse pebble]

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[unborn vigil]

Good. Is it possible to predict the result of the shift register with linear feedback using mathematical operations?

[timber island]

i type it

[timber island]

linear feedback, dont think i remember

mind enlightening me?

[unborn vigil]

I'm sorry that I'm giving you a link instead of an explanation, I think this way you'll understand what's going on better. https://en.wikipedia.org/wiki/Linear-feedback_shift_register

[timber island]

thats fine

ah so something like this:
let the original state of the shift register be a_0
Then a_1 (the next state) = a*(a_0) + b, correct?

a linear operation, right?

oh

a bunch of XORs is what u mean i suppose

[timber island]

i dont know if theres a way in standard operations like addition and multiplication, https://en.wikipedia.org/wiki/Linearity#Boolean_functions but this and https://en.wikipedia.org/wiki/Boolean_algebra should be helpful if a bunch of XORs is what u mean

[unborn vigil]

is xor responsible for ensuring that the previous number is not repeated further ?

[obtuse pebble]

@unborn vigil Has your question been resolved?

[unborn vigil]

yes

[timber island]

press the tick mark in that case

[timber island]

sorta

[obtuse pebble]

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[silver cove]

can someone help me with this? first is you're gonna write it in standard form right? so it's gonna be 2x^4-5x^3-10x+3. after that do I need to add "0x^2" ?

[silver cove]

<@&286206848099549185>

[tame dock]

split it

like factorise

[timid silo]

add "0x^2" ?

[high lily]

are you trying to do poly or synthetic division?

[obtuse pebble]

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[woeful talon]

how to simplify sqrt of x^2 all over absolute value of x

[woeful talon]

i don't know how to approach

√x^2 / |x|

[safe haven]

isnt sqrt(x^2) simply the absolute value of x

[versed pier]

Now you can sleep

Oops

That was the whole q

[woeful talon]

how about 1-x/|1-x|

how do i appraoch this

[high lily]

$1 - \frac{x}{|1-x|} ?$

[warm shale]

ℝαμΩℕωⅤ

[woeful talon]

no

1-x is in bracket

[high lily]

$\frac{1-x}{|1-x|}$

[warm shale]

ℝαμΩℕωⅤ

[woeful talon]

correct

[high lily]

that?

the value of that depends on the sign of 1-x

consider the two cases

[woeful talon]

if its negative and postive

if its negative x is larger than 1

if its positive x is smaller than 1

[woeful talon]

does 0 count as a case

[high lily]

it does, but its just undefined there

you can explicitly mention that if you want

[woeful talon]

im still cofnused though

i still haven't simplified the expression

[high lily]

you can do simplification for each case

how would you express |1-x| (without absolute value bars) if x>1?

[woeful talon]

1-x

if its when x < 1

[high lily]

incorrect

to both

[woeful talon]

the second one was a misclick

wait

if x < 1

its 1 - x

if x > 1

its x - 1

[high lily]

yes

for the purposes of simplification here, -(1-x) would be preferable

anyway

for $x<1$:
$$\frac{1-x}{|1-x|} = \frac{1-x}{1-x}$$
simplify that\
same idea for the $x>1$ case

[warm shale]

ℝαμΩℕωⅤ

[woeful talon]

alr

thanks

[short ore]

if you still find it tough its easier to do setting 1 - x = a

and that just becomes

$$\frac{a}{|a|}$$

[warm shale]

Spenny

[short ore]

for that its much easier to notice

when $a>0$
$$\frac{a}{|a|} = 1$$
and when $a<0$
$$\frac{a}{|a|} = -1$$

[warm shale]

Spenny

[woeful talon]

alr thanks

[short ore]

and plug in for x

[woeful talon]

just x and -x right

[short ore]

i dont understand?

[woeful talon]

when x < 1, its simplified to 1

when x > 1 its - 1

[short ore]

since a = 1 - x

a > 0 = (1 - x) > 0

minus 1 from both sides and multiply by -1

x > 1

and just flip the symbol for the other way

x < 1

[obtuse pebble]

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[latent knoll]

How do you find the 7th roots of unity for -i/2

[rocky goblet]

...i'm not sure what you mean by "for -i/2" there

if you're looking for a number x such that x^7 = -i/2, then that's not a 7th root of unity, it's a 7th root of -i/2

"of unity" in "root of unity" really just means "1", the 7th roots of unity are the numbers that when you raise them to the power of 7 you get 1

[kind hawk]

and with that being said, look at the polar form

then you can just compare length and angle quite easily

[obtuse pebble]

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[timid silo]

[
\f{x(x^2+1)}{(x^2 -x+1)^2}=2, \q x\in\R
]

[timid silo]

how do you solve this? Keep in mind no calc is needed

[ruby path]

Hmm

[timid silo]

expanding everything gets you a quartic, which we dont want

[ruby path]

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

[warm shale]

jan Nejon

[ruby path]

Maybe maybe maybe?

[timid silo]

explain?

[ruby path]

There's probably something doable such that it becomes a simpler polynomial in (x^3 + 1)

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

[warm shale]

jan Nejon

[ruby path]

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

[warm shale]

jan Nejon

[ruby path]

I don't know what I'm doing either lex

[timid silo]

[ruby path]

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

[warm shale]

jan Nejon

[ruby path]

This is futile

Was there anything interesting about the quartic

,w expand (x^2 - x + 1)^2

[warm shale]

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[ruby path]

$2x^4 - 5x^3 + 6x^2 - 5x + 2 = 0$

[warm shale]

jan Nejon

[ruby path]

lol

@timid silo you here mate?

Ah

[timid silo]

I have been trying to eyeball it for a while

[ruby path]

Wanna see something cool

[timid silo]

there is some cool trick deffo

yes

[bitter crater]

can't you divide by x^2 and group?

[ruby path]

$2\left (x^2 + \frac{1}{x^2} \right ) - 5\left( x + \frac{1}{x} \right ) + 6 = 0$

[warm shale]

jan Nejon

[ruby path]

Man you ruined the big reveal

$2\left (x + \frac{1}{x} \right )^2 - 5\left( x + \frac{1}{x} \right ) + 2 = 0$

[warm shale]

jan Nejon

[ruby path]

In general, when you have a quartic with symmetric coefficients like this one

You should divide by x^2

[bitter crater]

sry I got excied its rare that I am able to spot cool stuff like this

[ruby path]

Its alright

[timid silo]

oh I see so it be a quad

[ruby path]

Indeed

[timid silo]

,w 2x^2 -5x +2 =0

[warm shale]

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Requested by lexqa

[timid silo]

wait

so it's either 4 or 2.5?

[ruby path]

huh?

[timid silo]

,w x+1/x = 2

[warm shale]

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Requested by lexqa

[timid silo]

yeah right

so 1/2 is imaginary

leaves us with 2

thanks

.clsoe

.close

[obtuse pebble]

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[kindred basalt]

what did i get worng
i dont understand what the circle is fro

[twin sapphire]

3+2?

[twilit thicket]

looks like you multiplied instead of adding

[kindred basalt]

so the 6 that is in the circles is supposed to be a 5

[twilit thicket]

yes

[kindred basalt]

ok thx

.close

[obtuse pebble]

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[gilded drift]

help why is it A and D here

[sage geode]

For 3 you can reduce 2023 modulo 15 and also apply Euler's theorem

[gilded drift]

for 3.
=> summation of n=1 to 2023 (-2)^n
which would then be
-2+4-8+16-32+64-...+2^2022
=2+2^3+2^5+...+2^2021
mod 15
wouldn't it be
2+8+2+8+2+... +8(406 times)
4060
10

oh wait

-2^2023

it would be -2 though

[sage geode]

Remainder is defined to be nonnegative, add 15 to -2

[gilded drift]

23*88=2024
so 2(87+22)

[gilded drift]

10-2 '<'

[sage geode]

Hmm wait

[gilded drift]

It went through mod and there's also the 10 taken from the 4060

but apparantly the answer is 4

[gilded drift]

which is 218, but apparantly it's 2 less at 216

[sage geode]

,w remainder of the sum of 2023^n from n = 1 to 2023 when divided by 15

[warm shale]

Results provided by WolframAlpha

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Requested by alonelybean

[gilded drift]

PMO 26th is making no sense 😭

[sage geode]

Alright, wolfram doesn't help, let me see

,w phi(15)

[warm shale]

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Requested by alonelybean

[sage geode]

So from n >= 8 upto 2023 it's just 1's

$\sum\limits_{n=1}^7(-2)^n + \sum\limits_{n=8}^{2023}1$

[warm shale]

A Lonely Bean

[sage geode]

,w -2 + 4 - 8 + 16 - 32 + 64 - 128 + 256 + 2015

[warm shale]

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[sage geode]

,w reduce 2185 modulo 15

[warm shale]

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Requested by alonelybean

[sage geode]

[gilded drift]

this random guy just hit us with the "Part II No.3
Reduce the expression by applying mod to each of the terms in the expression.
2023 mod 15 = 13
Thus the expression can be reduced to 13^1 + 13^2 + 13^3... + 13^2023
Note that
13^1 ≡ 13 mod 15
13^2 ≡ 169 ≡ 4 mod 15
13^3 ≡ 2197 ≡ 7 mod 15
13^4 ≡ 28561 ≡ 1 mod 15
13^5 ≡ 371293 ≡ 13 mod 15
13^6 ≡ 4826809 ≡ 4 mod 15
Notice that for every four terms, the remainder repeats itself(13,4,7,1), thus we can add all of them giving us a "remainder" of (13+4+7+1) = (25)
Then we can find all the pairs of four in the expression by
2023/4 = 505.75 or just equal to 505
Then we can easily just do the expression as follows
(25)(505) + (13+4+7)(this came from the modulo of 13^2021+13^2022+13^2023)
We can then find it equal to 12649, then we can divide by 15 showing a remainder of 4"

[sage geode]

Ah, should be 2016, still doesn't make sense though

Ah wait I see why I'm wrong

[gilded drift]

Wait, oh
i did
-2+4
when it's 15-2+4+15-8+...

[gilded drift]

ohhhhhh

OOH MY GOD

I hate PMO

[gilded drift]

this one

WAIT THERE COULD BE AN OCCURANCE WHERE A MULTIPLE OF 23 AND 88 ARE CLOSE TOGETHER BUT NOT THE SAME-

Ohhhhhhhhhhhhhhhhhhhhhhh

Oh

And I thought I could get a double digit score

.close

[obtuse pebble]

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[vernal pagoda]

Can any amount of 300 and 500 produce any number in hundreds?

[vernal pagoda]

Like above 800

[brazen gorge]

wdym by "any amount"?

300 + 300 + 500 = 1100

[spice citrus]

https://www.cantorsparadise.com/chicken-mcnugget-theorem-1df48d273a57

*note, this is serious

[obtuse pebble]

@vernal pagoda Has your question been resolved?

[vernal pagoda]

Like 1000, 1.1k,1.2k,1.3k,1.4k and etc

[vernal pagoda]

Like 1000, 1.1k,1.2k,1.3k,1.4k and etc

[brazen gorge]

look at what jelle posted

[obtuse pebble]

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[vast willow]

q/q=1

(x+1)/(x+1)=1

2(x-1) was originally on the bottom right ?

and nothing changed to it so it should stay there

$\frac{x+1}{2(x-1)(x+1)}=\frac{x+1}{x+1}\frac{1}{2(x-1)}$=1\cdot\frac{1}{2(x-1)}=\frac{1}{2(x-1)}$

[warm shale]

calculus is fun
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[final thunder]

,, \df{x+1}{2(x-1)(x+1)} = \df{1}{2(x-1)} \cd \cancelto{1}{\mr{\df{x+1}{x+1}}} = \df{1}{2(x-1)} \cd 1

lol

[warm shale]

Pure

[obtuse pebble]

@timid silo Has your question been resolved?

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[night prism]

part b)

[night prism]

no idea where to start

[obtuse pebble]

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[dapper venture]

Can someone explain to me what happens on line 6,7 and 8 please

I don't get how sin2xsin2x = (1-cos4)/2 and I don't get how cosxcos4x goes in to -1/2cos5x -1/2 cos3x

[hollow solstice]

do you know this thing?

[hollow solstice]

Well you have cosx * cos4x you multiply and divide by 2 you get 1/2 *(2cosx cos4x) now Apply
2cosAcosB formula

[obtuse pebble]

@dapper venture Has your question been resolved?

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[verbal niche]

is $\leq$ reflexive ?

[warm shale]

Joshii

[tardy epoch]

Try a few numbers

[verbal niche]

i assume its not but yeah

[tardy epoch]

You don't need to assume

You just try a few numbers

[verbal niche]

oh sorry

i meant symmetric

[tardy epoch]

Same thing

[verbal niche]

3 and 1

not 1 and 3

[obtuse pebble]

@verbal niche Has your question been resolved?

[obtuse pebble]

@verbal niche Has your question been resolved?

[merry sable]

1+6(-2)+1(6-8)

[obtuse pebble]

@verbal niche Has your question been resolved?

[tardy epoch]

Does this answer your question

[verbal niche]

yeah it does i think

[tardy epoch]

Sooooooo

[obtuse pebble]

@verbal niche Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[glossy lintel]

Can $sin^2theta + cos^2 theta = 1$ be
$-1 = -sin^2theta -cos^2theta$

[warm shale]

stuffy

[glossy lintel]

also

2nd question, do you derive other identities ONLY from the original identity like

$sin^2theta + cos^2 theta = 1$

[warm shale]

stuffy

[glossy lintel]

or can you derive them from the other derived identities

[pulsar quarry]

sin^x+cos^x=1, it is an proven identity and yes you can also write in this negative form

and no not all identies are proved from sin^2x+cos^2x=1

ping when reply pls

[glossy lintel]

so...?

[glossy lintel]

fr, hate it when people dont ping

[pulsar quarry]

abey tum 10th me ho na, 3 toh identies hai 10th mei, wo sari sin^2x+cos^2x=1 se hi aati hai aur baki 11th me jo hoti hai unke new derivations hote hai

[glossy lintel]

are tu wahi hai na

nam kab change kar lia

[pulsar quarry]

yahi mera naam tha phle bhi😞

3:30 baje 10th wale trigno padh rhe hai, yahi dhekna baki tha

[glossy lintel]

ok but this #help-10 message was the question. when i say other dervied i mean the identities derived from the 3 main identities in 10th

[glossy lintel]

nahi bhai, pura saal procrastinate kia isilye

[pulsar quarry]

nahi 11th me aur bhi hoti hai jinka naya derivation hota hai

10th me jitni hai question wali wo sari inse hi karni hai

like jo question wgera hai wo inhi sse hi prove karne hai

its like

jo hazaro question rhte hai wo sari identity hi hoti hai bas koi unhe yaad nhi karta

aur wo sari sin^2x+cos^2=1 se prove ho jati hai

but 11th me yahi sin^2x+cos^2x=1 jaise naye standard formulaes aa jate hai

[glossy lintel]

for example

$cos^2 theta - 1 = -sin^2 theta$

[warm shale]

stuffy

[pulsar quarry]

maine toh 11th feb se padhna chalu kia tha 10th ke liye aur shyd se 21 feb se boards the

[glossy lintel]

ye nahi use hoga kya?

[pulsar quarry]

are ye sab toh rhte hi hai ye toh basics hai

[glossy lintel]

kud se padh rahi hu isilye

[pulsar quarry]

but like jaise $\sin(A+B)=\sin{A}\cos{B}+\cos{A}\sin{B}$

[warm shale]

ഴajat

[pulsar quarry]

aisi nayi identities aa jati hai

[glossy lintel]

so that identity is correct? if it is, is it required? since NCERT doesn't give this derivation of the identity

[pulsar quarry]

ye wali?

[glossy lintel]

yep

[pulsar quarry]

ha ye toh bas uss hi identity ko thoda idhar udhar kari dia hai isme kya naya hai

ye bhi use kar skti ho

[glossy lintel]

accha..

but why is it not mentioned in NCERT ?

will questions requiring this derivation come?

[pulsar quarry]

it is not mentioned in ncert because that's just like something about common sense

ki agar A+B=C hai

toh A=C-B hoga hi

[pulsar quarry]

ha honge hi, practice karogi toh dhikne lag jayega ki kab kya karna hai aur identity lagake simplify karna hai

[glossy lintel]

thanks 👍

[pulsar quarry]

np

ho gya ho toh band kardena channel

[glossy lintel]

.close

[obtuse pebble]

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[untold minnow]

I don't know where to begin

[glossy lintel]

can i just say i love ur name & aesthetic

[untold minnow]

tyty 🙂

[daring sorrel]

what does it mean for derivative to be negative/positive on an interval? How about the 2nd derivative

[untold minnow]

if the derivative is negative on an interval it is concave down and vice versa??

[daring sorrel]

that is 2nd derivative

[untold minnow]

ooh

i don't really understand what the second derivative provides me with beyond inflection points

[daring sorrel]

like u said it tells u the concavity, which is also the rate of change of the rate of change

so 1st derivative is just rate of change

[untold minnow]

rate of change at a given point correct??

[daring sorrel]

yeah'

[untold minnow]

ok

[daring sorrel]

if you have a positive rate of change at some value, what can you say about values slightly above it in the original function

[untold minnow]

the rate of change will decrease becauase if x >4 f'(x) < 0

[daring sorrel]

Lets think of an actual situation. Imagine you are running on a track. There is a function f that gives your distance from the start after some time. A function f' that gives the speed you are running after some time, and a final function f'' which tells you how fast you are accelerating after some time.

If after 10 seconds, you have ran 50 feet, and are running 5ft/s (f(10)=50, f'(10)=5>0), you would expect your position at 11 seconds to be greater than 50 feet right? You would NOT expect it to suddenly decrease to 45 feet. What this is showing is that when 1st derivative is positive, the original function is increasing. On the flip side when its negative, it would be decreasing.

To sum it up, first derivative sign tells whether your original function is increasing/decreasing

And as mentioned before, 2nd derivative sign tells you concavity.

[untold minnow]

ok so a slight increase in x of f'(x) would imply rising rate of change

which would also imply an increase in both f(x) and f''(x)

[daring sorrel]

if your f'(x) was already positive yeah

but if your f'(x) was negative (you were running backwards), and it increased a bit (you are running backwards slower, lets say walking backwards now), your position would still be decreasing, it would just be decreasing at a lower rate

[untold minnow]

i think i got iy

it

ill draw it out

[daring sorrel]

yeah looks right, could probably make the right side look more concave down though

[obtuse pebble]

@untold minnow Has your question been resolved?

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[untold minnow]

thank you I really appreciate the help

[obtuse pebble]

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[hard minnow]

Could someone help me with this problem?

[hard minnow]

Been struggling with it for some time.

[clear kraken]

Which part?

[hard minnow]

A...having trouble all around. I was told that I'd be implicitly working on the level curve z - sinx * cosy = 0, so I'd just compute the gradient for the function f(x,y,z) = z - sinx * cosy, and use that as my directional derivative...

[clear kraken]

no don't...

z is fully determined by (x,y)

so just treat it as a function of 2 variables

sinxcosy is clearly continuous everywhere

so you can use the directoinal derivative theorem

dot product between the gradient vector and the unit vector point in the direction (4,-3)

[hard minnow]

But my professor told me "...In this problem z=f(x,y), so the third coordinate is f(x,y).

This is just like in single variable calculus where we might refer to the function f(x)=sin(x) and refer to the point (pi/6, 1/2)."

[clear kraken]

yes

so g(π/3, 2π)=3

[hard minnow]

So basically what you're saying is that I disregard what I say, calculate the partial derivatives with respect to y and x, and when it comes to z, just calculate the partial derivative with respect to z and it'll be 0 because there's no z defined in the original function?

[clear kraken]

no

you need the gradient vector for g(x,y)

so find x partial, y partial

then evaluate with (x,y)

z is already determined, so no need to plug in z=3

[hard minnow]

right, but I was told to not interpret it that way, I was told to let z = g(x,y), and then do g(x,y)-z, which would be z-sinxcosy

[clear kraken]

well you could also do that

[hard minnow]

They are one in the same?

[clear kraken]

yes

but your way makes more sense here actually

becuase c) requires considering a function of 3 variables

[hard minnow]

Your way seems easier though, could I do either way or is the way I mentioned earlier more efficient and I should 100% do it for some reason?

[obtuse pebble]

@hard minnow Has your question been resolved?

[hard minnow]

Kalgar?

[hard minnow]

.close

[obtuse pebble]

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[hard minnow]

Could someone help me with this problem?

.close

[obtuse pebble]

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[sacred jetty]

<@&286206848099549185>

[sacred jetty]

Pls anyone help ne

me

[high lily]

!15m

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[pulsar quarry]

!15m

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[twin sapphire]

ask lol

[latent walrus]

why ask for help if theres nothing to help with

[high lily]

don't ping especially if you haven't even asked a question

[sacred jetty]

omg

[pulsar quarry]

pinging helpers and not having any question

[sacred jetty]

how many people

[pulsar quarry]

yea because that's a crime

[sacred jetty]

yep lemme get the question

[latent walrus]

you caused a ruckus

[pulsar quarry]

3 years of jail is there for just an act

gtg

[sacred jetty]

Need help with q4

[drifting wraith]

helpers should wait 15 seconds before pointing out that there was no question

[sacred jetty]

sorry guys

[twin sapphire]

lol

[sacred jetty]

im kinda new

[twin sapphire]

its ok

[sacred jetty]

anyways is anybody free to help me?

[twin sapphire]

so what did you try?

[sacred jetty]

well i know average speed is total distance/total time

[twin sapphire]

ok

so lets do this

calculate total distance first

[sacred jetty]

how tho?

[twin sapphire]

well by reading

[sacred jetty]

would the distance be 10+21?

so 31?

[twin sapphire]

yes

[sacred jetty]

ok

[twin sapphire]

that wasnt that hard cmon

now we want the time

how much time did the person spend on the first 10 km?

[sacred jetty]

do i do 10/4

?

[twin sapphire]

yes

[sacred jetty]

2.5

hrs

[twin sapphire]

yes ok

[sacred jetty]

and then 3.5 hrs

for 21/6

[twin sapphire]

ok

so all in all?

[sacred jetty]

so is it 31/6

the answer?

[twin sapphire]

,w 31/6

[warm shale]

Results provided by WolframAlpha

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Requested by b__n

[sacred jetty]

Ok thanks so much

got the answer

Do you mind helping me with 6?

@twin sapphire

.close

[obtuse pebble]

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[twin sapphire]

i could

[obtuse pebble]

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[steel goblet]

So i know how to do this question. Here is the instructor's solution as well. My only question is what does it mean by "state at which points the extreme values occur

does that just mean to list the points (0,0), (1,1), (1,-1), (-1,1), (-1,-1)?

[obtuse pebble]

@steel goblet Has your question been resolved?

[steel goblet]

<@&286206848099549185>

[obtuse pebble]

@steel goblet Has your question been resolved?

[steel goblet]

<@&286206848099549185>

[steel goblet]

.close

[obtuse pebble]

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[modern basin]

Hi!

[modern basin]

I'm trying to derive the formulas for propagation of uncertainties.

When I take the differentiate
y = a/b,
I get
dy/y = da/a - db/b
Even though the formula is
dy/y = da/a + db/b

I wonder why this is. Is it because "db" is negative when it's in the denominator and you have to take the absolute value?

[obtuse pebble]

@modern basin Has your question been resolved?

[obtuse pebble]

@modern basin Has your question been resolved?

[tardy epoch]

What do you mean "the formula"?

[obtuse pebble]

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[sonic glacier]

need help

not sure if I'm doing it correctly

I have a few more questions that I also don't really know

[obtuse pebble]

@sonic glacier Has your question been resolved?

[glossy basalt]

hello, I'd wish to help, but I'm not quite available right now 😦

here's some hint for you:
y=asin(b(x-c))+d
where
a = amplitude
2π/b = period
c is the phase shift
d is the vertical shift

[sonic glacier]

alr thanks

I'll try

[sonic glacier]

I got the first one

and the one after

but idk how to do the word problem

oh wait

read the rules

.close

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[dull void]

for

$f(x) = ln(\sqrt{x+1} + x)$
I have to use
$f'(x) = u'/u$
or
$f'(x) = 1/x$

[warm shale]

odokawa

[dull void]

hum

to solve a derivaite

.close

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[hard minnow]

Could we go over a specific problem. It's asking to find the critical points, saddle point, local max, and local min.

[hard minnow]

I went over one recently, and there's another one I'm focused on and I'm not getting it right..

[obtuse pebble]

@hard minnow Has your question been resolved?

[runic void]

do you have the question with you ?

[obtuse pebble]

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[dull void]

why
-x^2 - 6x - 6 is equal to (x+3)(x-2)

3 * -2 = -6
3 + -2 = 1 //so this doesnt make sense

[robust raven]

and where did you get it?

[dull void]

can i multiply
$-x^2 - 6x - 6$ to -1

[warm shale]

odokawa

[dull void]

how?

[robust raven]

yes what is precise text of yoru task

is it an equation? or some equivalent form?

[dull void]

well now i have
$x^2 + 6x - 6 = 0$

[warm shale]

odokawa

[robust raven]

ok so it is an equation

[dull void]

yes it is

i have to find f'(x) for real values which values are f(x) = 0

f(x) = 6 - 6x - x^2 is the task

[robust raven]

you should rather write in a such way:

you have to find real values of x, for which f(x) = 0

better to say:

you have to find zeros of the given function f

then it is udnerstandable

now all is clear

[high lily]

is there a pic of the original problem? there are conflicting parts here.

[dull void]

upside part translated is determinate f'(x) and the real values of x for those which f(x) = 0

[high lily]

ahk

[robust raven]

right odokawa

[dull void]

right

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[ember frost]

$\left(\frac{x}{p}\right)^{2}+\left(\frac{y}{q}\right)^{2}=\sec\left(\arctan\left(\frac{x}{p},\frac{y}{q}\right)-\frac{\pi}{2}\operatorname{floor}\left(\frac{2}{\pi}\arctan\left(\frac{x}{p},\frac{y}{q}\right)+\frac{1}{2}\right)\right)^{2}$

[warm shale]

FungusDesu

[ember frost]

i was playing around in desmos and i found this, which is the equation for a rectangular graph with sides of 2p and 2q. for the most part, i dont understand anything, but what i dont understand most is why there is a second parameter in arctan?

most explanations i found online were complex, so it would be helpful if you explain to me as if im 5

[late totem]

Imagine you have a rectangular graph, like a rectangle on a piece of paper. The sides of this rectangle are not necessarily the same length. Let's call the length of one side "2p" and the length of the other side "2q". So, the width is 2p, and the height is 2q. Now, let's say you want to find an angle inside this rectangle. But here's the trick: the angle you find will depend on where you are in the rectangle. The arctangent (arctan) function helps you find an angle when you know the sides of a right-angled triangle. In this case, the sides are 2p and 2q. However, the arctan function has two parameters because it needs to know which side is which. It's like saying, "Hey, I want to find the angle, and I'm looking at this side (2p), and this side (2q)." So, the two parameters in arctan(2p, 2q) are like telling the function about the two sides of the triangle. It's saying, "Here are the lengths of the sides, now tell me the angle." In simpler terms, the second parameter in arctan helps to specify the sides of the triangle so that the function can give you the right angle based on those sides. It's like giving the arctan function the information it needs to do its job correctly.

[ember frost]

ah i got it now, so in a way it creates an implicit triangle, and the two parameters specify the measurements of the adjacent and opposite sides, right?

[late totem]

bingo

The arctan function helps you find an angle in a right-angled triangle, and the two parameters specify the lengths of the sides of that triangle. One side is like the "adjacent" side, and the other is like the "opposite" side, and by using those lengths, you can find the angle inside the triangle. So, in the context of your rectangular graph, you can think of the sides of the rectangle as the adjacent and opposite sides of the implicit right-angled triangle, and the arctan function helps you find the angle based on those sides.

[ember frost]

right, thank you

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[timid silo]

How can i prove that the order of the subgroup of simmetric group S(n) that contains only even permutations is n!/2?

[timid silo]

What grade is this ?

[safe haven]

this is some undergrad level stuff

[timid silo]

in my country 11th grade

[safe haven]

oh damn mb then

[timid silo]

Exactly what I thought

Proving

[timid silo]

What country is this?

[zenith raft]

one thing you try is proving there are as many odd permutations as even permutations

so that half of the n! permutations are even

[timid silo]

Is that question really from 11th grade

[zenith raft]

you can make a bijection between the even and odd permutations for this

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

<@&286206848099549185>

[timid silo]

can i say that sign of permutation A is different from sign of permutation A' (inversion of A) so that means bijection?

[timid silo]

Ukraine

Have to check the curriculum man or maybe we have it easy

[zenith raft]

no

going to bed soz

[timid silo]

i've solved already, thank you for help!

.close

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[timid silo]

what have you consider so far

[spring steeple]

Well the common difference of this AP must be positive

[timid silo]

okay

[spring steeple]

Otherwise it will not have infinite terms

[timid silo]

but it can tend to -infinite

so it can be infinite

either way

[spring steeple]

Perfect squares

[pulsar quarry]

roots of negatives arent defined

it should just have equal differences

start by writing some of them

and if a_2-a_1=a_3-a_2=a_4=a_3 then you can say that its an AP

and if not then it wont be an ap and therefore no infinite arithmetic progression

[spring steeple]

The terms are unknown here so you can't figure out that

[spice citrus]

Suppose the common difference is some constant c

then you can show that at some point, all the squares will have bigger differences than c

[spring steeple]

how do you show that if you're defining c as the common difference?

[spice citrus]

At some point you must encounter a square that's bigger than c^2 right?

[spring steeple]

How do you show that?

[spice citrus]

In your sequence there should be infinitely many squares that keep getting bigger, so at some point the squares will be bigger then c^2

[spring steeple]

That's not a valid argument

[spice citrus]

Let $a_n$ be your sequence, by the problem statement $$a_1 = x_1^2 < a_2 = x_2^2 < a3 = x_3^2 ....$$ because the sequence is increasing you can show with induction that $$x_n \geq n$$, so $$a_c = x_c^2 \geq c^2$$

[kind hawk]

worded differently, there are only finitely many squares below c^2

[warm shale]

Jelle

[spice citrus]

yeah, should have said that. But I thought he wanted a more formal proof

[spring steeple]

Alright got it

Thanks

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[earnest dagger]

how would i do b and c?

[earnest dagger]

,rotate

[civic zealot]

C is because the constant term of a polynomial is always the product of roots of the polynomial.

[earnest dagger]

oh but how would i do b?

[inland matrix]

Have you done a?

[earnest dagger]

yes

[inland matrix]

For b, you can substitute x^2 = t

[earnest dagger]

but how would i use a cos it says hence

[obtuse pebble]

@earnest dagger Has your question been resolved?

[obtuse pebble]

@earnest dagger Has your question been resolved?

[earnest dagger]

<@&286206848099549185>

[earnest dagger]

.close

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[long hedge]

How did they get to the last step ?

[rich plume]

property of logarithms

[long hedge]

holy fuck i forgot

[obtuse pebble]

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[plush egret]

Hi

[plush egret]

Omg why the b

In a. the radical is multiply both numerator and denominator

The sqrt of 2x is multiply both numerator and denominator

[rich plume]

what is your question?

you seem to be unclear

[plush egret]

But why in b is multiply with this

[rich plume]

that is called the rationalising term

[rich plume]

it would not be same for two different fractions

[plush egret]

Not this😢

[rich plume]

It is generally the term which in under the sqrt in the denominator

[plush egret]

Wait how do we do it

[inland matrix]

Hi beard

[plush egret]

Cuz after I remove the nth power of the radical

[rich plume]

sup

$\frac{1}{\sqrt{x}}\cdot\frac{\sqrt{x}}{\sqrt{x}}=\frac{\sqrt{x}}{x}$

[warm shale]

B-eard

[rich plume]

This is the main idea of rationalisation

i.e. removing the surd from denomninator

[plush egret]

It come up with this and I thought after you remove the nth power of the radical in denominator you multiply it by both denominator and numerator but the exponents is different

[rich plume]

this won't work

[plush egret]

Why

[rich plume]

because you need cube of each part

so that its cube root gets cancelled

[plush egret]

So like the power rule something like to add the exponents that is equal to cube?

[rich plume]

$\frac{1}{\sqrt[3]{x^{2}}\sqrt[3]{y}}\cdot\frac{\sqrt[3]{x}\sqrt[3]{y^{2}}}{\sqrt[3]{x}\sqrt[3]{y^{2}}}=\frac{\sqrt[3]{x}\sqrt[3]{y^{2}}}{\sqrt[3]{x^{3}}\sqrt[3]{y^{3}}}=\frac{\sqrt[3]{x}\sqrt[3]{y^{2}}}{xy}$

[warm shale]

B-eard

[plush egret]

Ohh omg it makes sense

Ill make the exponents 3..

[pulsar quarry]

hi B-eard

[rich plume]

you need to multiply it by such power so that it transform in the form $\left(x^{n}\right)^{\frac{1}{n}}$ which would be just $x$

[warm shale]

B-eard

[plush egret]

Yess

[rich plume]

Hii

tell me you opted for CS in 11th

[pulsar quarry]

no im painting

fine arts/ arts whatever

[rich plume]

then what are you doing on VSC

[plush egret]

Look so this is the full solution?

[pulsar quarry]

learning python out of curiosity, sorry for the interruption milonmeh

[rich plume]

yeah

[plush egret]

No problem

Omgg

Yeheyy

Ill try the c

[rich plume]

damn you are actually learning it out of curiosity

[rich plume]

sure

[pulsar quarry]

yea because ik im a waste i would not be able to qualify for jee so that's just a backup

that's a joke btw

[rich plume]

this can't get more real

[pulsar quarry]

ikr

[plush egret]

Is this correct?

@rich plume

[rich plume]

yep, it seems to be right

however, you can simplify it

[plush egret]

How!

*?

@rich plume

And what does this mean fractions in the radicand?

Where’s the fraction?

[plush egret]

Ohh yess the second step we already remove the nth power in both numerator and denominator right?

I forgot that the radicand is in the form of fraction

Right?

@rich plume

@waxen cove

[waxen cove]

yup....

[plush egret]

Yeheyy

[waxen cove]

i can see a lot of factor there so...

you just divide them so it will be soon solved

[plush egret]

Yes thank you so much

@waxen cove

.close

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[ember frost]

The set of all the real roots of the inequation $3^{x^2} - 9 \times 3^x < 0$ can be expressed as an interval (a; b). Calculate a+b

[warm shale]

FungusDesu

[ember frost]

i honestly have no idea of what to do, it would be way easier if its (3^x)^2 instead of 3^x^2

[brazen gorge]

well

$9\cdot 3^x = 3^{x+2}$

[warm shale]

artemetra

[brazen gorge]

so we get $3^{x^2}<3^{x+2}$

[warm shale]

artemetra

[inland matrix]

Are a and b whole numbers?

[ember frost]

they are real

[brazen gorge]

which is pretty easy to solve