help-23

[sharp owl]

so i thought this is the only releveant part with an amplitude

[desert juniper]

okay, i'm not explaining myself clearly enough.
A bound would be an approximate value, which is somewhat close to the actual value

for example, you know sinx takes values in the -1 to 1 range, and so does cosx

so the parenthesis must take values between -51 and 51

[raven heart]

they're using the good ole trick to write a sum of two sinusoids as one shifted sinusoid

[sharp owl]

honestly im not sure if the question wanted an approximation or exact answer

but if it ispossible to get an exact answer without a computer i think that would be best

[desert juniper]

the values seem kinda weird to be obtaining them exactly without a calculator tbh

[raven heart]

$\sin(x) - 50 \cos(x) = \sqrt{1^2 + (-50)^2} \left(\frac{1}{\sqrt{1^2 + (-50)^2}}\sin(x) - \frac{50}{\sqrt{1^2 + (-50)^2}} \cos(x)\right)$

[sharp owl]

i agree lol

[flat frigate]

aPlatypus

[desert juniper]

which as you can see doesnt stop people from doing them

[sharp owl]

nonetheless, i was curious if there is some simple way

[desert juniper]

that would depend on your definition of simple i guess

[sharp owl]

true

[raven heart]

now the two coeffs in front of sin and cos have (call them a and b) are such that a^2 + b^2 = 1

so you can interpret them as the sin and cos of some angle, call that angle d, sin(d) = a, cos(d) = b

[sharp owl]

alright

im more of a fart smella than a smart fella

where do we go from that expression

[sharp owl]

.

[raven heart]

so you get $\sqrt{2501} (\sin(x)\sin(d) - \cos(x)\cos(d))$

[flat frigate]

aPlatypus

[raven heart]

then pop in the addition formulas

you get $\sqrt{2501}\sin(x-d)$

[flat frigate]

aPlatypus

[sharp owl]

oh interesting

[raven heart]

so amplitude of sin(x) - 50cos(x) is sqrt(2501)

[sharp owl]

hmm ok

lemme think abt this a bit

[raven heart]

sure

[sharp owl]

ok i see what you did

very cool

thanks

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[pseudo verge]

my classmate's answer is a letter a, but i got

[vagrant ice]

ah it's a first-order linear differential equation

[pseudo verge]

yes, it is

[vagrant ice]

yeah okay so what was your integrating factor?

[pseudo verge]

e^x

[vagrant ice]

give me time while you keep waiting

[pseudo verge]

Why?

[vagrant ice]

I'm helping others

[pseudo verge]

Okay

Take your time

hi 🥲

[lean otter]

hes very busy rn

[vagrant ice]

,w integrate e^x cos(x)

[flat frigate]

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[vagrant ice]

yep this is correct

,w y' + y - cos x = 0

[flat frigate]

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[pseudo verge]

But it's not in the option

[vagrant ice]

yeah try rearranging option a then

[lean otter]

bro u indian ?

@vagrant ice

[vagrant ice]

$y = \frac{e^x \sin x + e^x \cos x - c}{2e^x}$

[flat frigate]

south

[vagrant ice]

nah

[lean otter]

how you know about JEE then

[vagrant ice]

not only Indians know about JEE....

[lean otter]

i mean mostly na

sorry if that was rude

[vagrant ice]

I'm the exception bhai

[lean otter]

damn

bhai

[pseudo verge]

Wait, let me rearrange it

is this correct 😊😊🙂‍↔️

🙂‍↔️🙂‍↔️🙂‍↔️

Hi

@vagrant ice

[vagrant ice]

you should have -c/(2e^x) but it doesn't matter

you can replace $\frac{-c/2}{e^x}$ with another constant $c_1$, so $\frac{c_1}{e^x}$

[pseudo verge]

it's failed to render

[flat frigate]

south

[vagrant ice]

arbitrary constant of integration

[pseudo verge]

Thank you so much South

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[marble thicket]

Hello I need help

[marble thicket]

Sorry for the blur

What is x2 and what does it mean and do

[cloud maple]

x² = x * x

[marble thicket]

So x the variable times something

[cloud maple]

x times x = x²

[marble thicket]

If I used x² + something would that number that I’m multiplying with x² be multiplied double the amount

Like x² + 5 = 10?

[cloud maple]

x² + 5 = (x * x) + 5

x times x, then plus 5

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[lofty wasp]

what is wrong with my solution?

oh I see it

fuck

nevermind haha

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[violet rune]

is there a 1/3 in front cause the question asked for average value over an interval that has 3 steps and wewant to find the average of function over that 3 step...? so multiply the function with 1/3?

[final loom]

Average = [sum]/(number)

in this case however, since you have an interval: Average = (integral)/(2 - (-1))

@violet rune

[hard kestrel]

Hey there

[violet rune]

I get it

I wanted to clarify if my understanding was right

thanks Arya 😄

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[arctic gate]

Let’s say my grandmother comes once in a week. And she came for past month at these times:

21:27
21:10
20:57
21:42

how to calculate on average what time she comes?

[safe radish]

Please don't occupy multiple help channels.

[hardy monolith]

!1c

[safe radish]

Please stick to your channel.

[hardy monolith]

ah, youve closed the other one

nevermind

[cold ridge]

21h 27min
21h 10min
20h 57min
21h 42min

convert all into min -> do average formula -> convert into hours

probably

[arctic gate]

How to do that in excel?

[cold ridge]

hours * 60 + minutes

https://www.myexcelonline.com/blog/1-5-hours-to-minutes/#:~:text=FAQ%3A Excelling at Excel Time Conversions,-What are the&text=To achieve this%2C you simply,the total time in minutes.

this should help

[arctic gate]

But from real time to hours and minutes

[median vigil]

if you want the average time in excel you can just use the average function directly

[arctic gate]

Oh really bby

Fine

Can you tell me what variance is?

[median vigil]

variance is a measure of spread

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[fiery portal]

Not again lol

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[foggy cloud]

need help with input please

[quiet fjord]

you can rewrite x^2 - 1 as (x-1)(x+1)

then you can split the modulus in |x-1||x+1| < 1/2

oh wait you need to use the graph

[foggy cloud]

yes

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[foggy cloud]

i got it

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[velvet mirage]

I recently had an elimination/substitution test. I got marked off for not stating my method of solving, and it was said I had to do it on a review day (which I wasn’t there for) who’s in the wrong?

[thin bridge]

,rotate

[thin bridge]

depends if they mentioned anything about stating it when you were initially taught

the method being applied is implied
no real reason to explicitly state it

unless they want to know if you know their definitions

how much is each question out of?

also in your work,
for Q9, eq3
ideally you'd have a multiplication symbol between the ) and -3

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[hoary knot]

Help

[hoary knot]

That’s my question..

I swear you use either lcm or hcf but I don’t know what you do after

[hard crest]

,rccw

[hard crest]

yes lcm

what you were doing would also work

just have to keep going, after about 11 of them they'll line up again

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[blazing palm]

?

Is this a troll

@hidden compass are you here to advertise or ask about math

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[outer river]

A poset is well funded (assuming the axiom of dependent choice) if there is no infinite descending sequence, I cant seem to understand why in lexicographic order and a finite set constructed on the alphabet say {a,b}* we have a<aa<aaa<... which is an infinite descending sequence. I feel like some element wont be in this set, because it is finite and the sequence is infinite. I think im misunderstanding the concept of infinity here, pls help

[outer river]

in short why can we construct an infinite descending chain if the set is finite?

[fiery merlin]

I'd guess that "a finite set constructed on the alphabet" means that it doesn't include every possible string produced by the alphabet.

[outer river]

oh

im so dumb

omg

ok

yeah probably

thank you

[fiery merlin]

No problem.

[outer river]

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[clear lantern]

So i have this RREF matrix

[clear lantern]

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[fathom adder]

!da2a

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

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[slow lily]

Did i do something wrong or is the angle rlly -0.35

[slow lily]

0.35 feels wrong

[tacit sparrow]

degrees or radians?

[hidden gyro]

0.35 in radians i think

[slow lily]

oh

ok that explains it

[echo bronze]

Yeah I think desmos is rad, or else it'd have the °

[slow lily]

Thx

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[steep magnet]

I kind of know this one but not really

[arctic vessel]

whats up gang

[steep magnet]

So first we look at the square root cause we want an x on the outside

[arctic vessel]

ok

[steep magnet]

So we can do that by turning the term 9 into (x^2)(9/x^2)

[fathom jewel]

7+9/x^2

[steep magnet]

Wym sry?

Ah ye

I get you

[arctic vessel]

pause

nvm

[steep magnet]

So sqrt(7x^2 + (x^2)(9/x^2)) becomes (x)(sqrt(7+9/x^2)

And we have our x outside the square root

[fathom jewel]

oh i see yes

well let me add on something

|x|

[steep magnet]

like (IxI)(sqrt(7+9/x^2)?

[fathom jewel]

yes

because it will depend whether it's x or -x on the limit you take

for +infinity you take x and -infinity you take -x

[steep magnet]

Why?

Wait

Yeah I get you

[fathom jewel]

nice

[steep magnet]

So for negative infinity

Does it tend to 0?

Or infinity?

[fathom jewel]

well right now you factored -x

[steep magnet]

Like in the denominator you have (-infinity) + (infinty)(sqrt(7))

[fathom jewel]

,, \lim_{x \to -\infty} \frac{x \left (-4+\frac{1}{x}\right )}{-x \left (-1+\sqrt{7+\frac{9}{x^2}} \right )}

[steep magnet]

But how is it that?

Cause of the IxI you cant factorize it out, no?

Wait nvm

Im still confused tho 1 sec

Figuring it out

[fathom jewel]

|| since we approach -infinity that means x < 0 which means |x| = -x ||

[steep magnet]

I thought that was the whole point of IxI? Was that it couldnt be negative?

[fathom jewel]

yes

when x < 0 it means x is negative so you add a factor (-1) to make it positive

thats what |x| = -x iff x < 0 means

[steep magnet]

Ok I get it

So when its towards infinity its -4/(-1-sqrt(7)) and towards negative infinity its -4/(-1+sqrt(7))

Yessss I get it now

Thank you!!

❤️

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[fathom jewel]

i think you might wanna check your signs again

[steep magnet]

Ahhh yeye I mixed them up 😅

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[foggy cloud]

how to solve a

[blazing palm]

Do you have a graphing calculator?

[slim plinth]

Convert it to e^x form

Since it’s a 1^inf form

[foggy cloud]

yes but when taking exam professor does let use using one

[blazing palm]

Does or doesnt?

[foggy cloud]

doesnt

[foggy cloud]

there no graph in homework

[blazing palm]

Well thats pretty stupid, isnt it

[lime stone]

do you know how to find the lim x -> infinity to this?

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[blazing palm]

Bro

That was related lmao

[lime stone]

I was answering lol

[slim plinth]

oops

[lime stone]

no worries

[foggy cloud]

not this kind of example

that why i dont know

[lime stone]

it's that but instead of n it's x/(-3)

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[tawdry raven]

**I would like some help on this question: **A poker hand consists of 5 cards. If the cards have distinct consecutive values and are not all of the same suit, we say that the hand is a straight. For instance, a hand consisting of the five of spades, six of spades, seven of spades, eight of spades, and nine of hearts is a straight. Assume a standard deck of 52 cards. What is the probability that one is dealt a straight?

what i have currently is that there are 10 possible combinations of straights but unsure where to go from there

[safe radish]

@tawdry raven Has your question been resolved?

[tawdry raven]

@meager igloo

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@tawdry raven Has your question been resolved?

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@tawdry raven Has your question been resolved?

[safe radish]

@tawdry raven Has your question been resolved?

[quasi bison]

@tawdry raven still here?

there are indeed 10 different straights in terms of rank (A2345, 23456, ..., 10JQKA), so now think about the suits in each one. your only constraint is that you can't have all 5 cards being the same suit, bc that's called a straight flush and is the strongest hand in poker

so think about how many ways there are to select the suits of your cards once your ranks are locked in & subtract the 4 selections that result in straight flushes

[tawdry raven]

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[spark zinc]

Need help

[safe radish]

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

[spark zinc]

that question d)

some transformation to the original function: 6x^3-x^2-5x+2, where the exponents are reversed?

I did a, b and c all well but d is stubborn.

[fathom jewel]

Have you tried the rational root theorem?

[spark zinc]

idk what that is

[fathom jewel]

https://youtu.be/Iaq7z7reznM?si=4H-XfDV0zVNLNL1b

[spark zinc]

do you see any other methods here?

This is outside of my course, i'll use it if I have to but I want to try different methods if any exist

[fathom jewel]

other than guessing a root and doing a long division, idts

[spark zinc]

did you read the whole question

the basic concept of the first 3 parts was that you take the original function and each function of parts a, b and c had the x variable altered

For a, x=2x from the original function which I made a, solved for a, which was the same roots for x of the original function and divided them all by 2 to find roots for new function

for b, it was square rooting everything, for c it was multiplying by -1.

[fathom jewel]

oh i see

[spark zinc]

No doubt I could easily factor this out by inspection and then factor the remaining quadratic... BUT my teacher wants me to do it as the textbook intends

as it says, hence, solve... meaning I have to involve the original function of the question somehow

[fathom jewel]

ye

[spark zinc]

idk how tf to do it

[fathom jewel]

one question did you solve a,b and c by inspection or was there some way of doing it

[spark zinc]

literally ZERO polynomial expansion/facotrisation techniques used

I described it before

I'll tell you again

I'll use a. for example

[fathom jewel]

I think I got it

try

1/x

and multiply then by x^3

[spark zinc]

a. Original function: 6x^3-x^2-5x+2=0...
function for part 'a': 48x^3-4x^2-10x+2=0... = 6(2x)^3-(2x)^2-5(2x)+2=0

[spark zinc]

I was thinking this

about to write it down

not sure of bimdas will allow or not but we'll see

[fathom jewel]

if you look at the coefficients it's reversed

[spark zinc]

thats what I said lol

[spark zinc]

^

[fathom jewel]

ok i can't read lol i will just leave

[spark zinc]

doesnt this make it just x^2?

[fathom jewel]

if you plug in 1/x then the lowest you get is 1/x^3 so multiplying by x^3 makes it a polynomial again

[spark zinc]

plug in 1/x as x?

[fathom jewel]

yes

then
1/x^3 * x^3 -> 1
1/x^2 * x^3 -> x
1/x * x^3 -> x^2

[spark zinc]

hang on

nah doesn't work lol

when you multiply everything by x^3 you also do that to the constant, then it's wrong

this is the gay thing, it can only involve replacing the x, nothing to do with f(x)

[fathom jewel]

2(1/x)³-5(1/x)²-(1/x)+6=0
2-5x²-x²+6x³=0

[fathom jewel]

huh why?

[spark zinc]

wait nvm i'm blurring out, it's meant to do that

nvm nvm'

So.... what is the "a" value?

I get it now but how do I get the roots

[fathom jewel]

also you said for c) multiplying by (-1) but i think it's actually plugging in -x for x

[spark zinc]

ref to this

[fathom jewel]

well you know the transformation is 1/x so x = 1/root

[spark zinc]

yeah, since -x=x, the solutions for x must be divided (same as multiplied) in this case by -1 to get the actual solutions

[spark zinc]

1/root what?

oh the roots

[fathom jewel]

yea of the very first polynomial

that you had to solve anyway

[spark zinc]

So all the solutions for original function to the power of -1?

I swear

[fathom jewel]

yes

[spark zinc]

how does that connect to this?

seems random af

its wrong...

[fathom jewel]

in c you said it was multiplying by (-1) because x -> -x
so in d since x -> 1/x = x^(-1) you do root^-1

[spark zinc]

original solutions are x=-1, 2/3, 1/2... so....

[fathom jewel]

yea so for d) you'd expect -1, 3/2 and 2

[spark zinc]

it's wrong lol

d= -1/2, 1/3, 2 in the book. I trust book

[fathom jewel]

,w 2x^3-5x^2-x+6=0

[flat frigate]

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[spark zinc]

the book has to be wrong

[fathom jewel]

looking at it, it seems they have a typo

[spark zinc]

nvm i read solutions for question 10 like a dumbass

can you run me through it again, sorry

[fathom jewel]

what we did for d?

[spark zinc]

yes

[fathom jewel]

so basically you already noticed the coefficients, which was the key

[spark zinc]

yes

[fathom jewel]

then you'd replace x with 1/x basically

[spark zinc]

sure

[fathom jewel]

and multply by x^3 to get the original equation 6x^3-x^2-5x+2=0.

[spark zinc]

nope

you deleted it

nvm

[fathom jewel]

yes

lol

was wrong

but the rest still stands

[spark zinc]

it just keeps RHS=0 so hte trasnformation of x doesnt change right"?

something like that

[fathom jewel]

yes

[spark zinc]

like if you multply f(x) by 2 or some shit and f(x)=0, 2f(x)=2(0)=0

[fathom jewel]

yes

[spark zinc]

how am i gonna go up and explain this in front of the class

[fathom jewel]

equivalent steps

[spark zinc]

gonna waffle

[fathom jewel]

you say that you noticed the coefficents were reversed

so you want to recreated 6x^3-x^2-5x+2=0 by plugging in 1/x

[spark zinc]

the teacher will check who has the answer with working, when she notices i'm the only one I'll have to go up

I don't mind it just want to be able to explain

[spark zinc]

yes

I did trasnformations last year, forgot everything

[fathom jewel]

and that new equation 2(1/x)³-5(1/x)²-(1/x)+6=0 is equivalent to 2-5x²-x²+6x³=0 if you multiply by x³ basically

[spark zinc]

exactly

[fathom jewel]

that's it

[spark zinc]

WAIT... then 2(1/x)³-5(1/x)²-(1/x)+6=2-5x²-x²+6x³

so... all we've done is prove equivalence! NOW we can just back track and now that we've proven they're equivalemt, we know the solutions to the original equation, and we know that the only thing we need to change to impose equilvalence is applying -1 exponent to all roots

I cracked the code

or fucked up

[fathom jewel]

yea basically they have the same solutions

[spark zinc]

so or dont

do*

[fathom jewel]

because 2(1/x)³-5(1/x)²-(1/x)+6 = 0 is equivalent to finding 2-5x²-x²+6x³ = 0

[fathom jewel]

yes

also say suppose x is not 0 to really ensure that you are allowed to multiply/divide by x^3

because x = 0 is not a solution anyway

[spark zinc]

why would that interfere?

[fathom jewel]

your teacher might ask you why you are allowed to simply multiply by x^3

and what happens if x^3 = 0

so that you know

because multiplying by 0 is not allowed

[spark zinc]

sure

still dont really get that x^3=0 bit sorry

[fathom jewel]

i am just saying that the possibility of x being 0 is out of question

so multiplying by x^3 is harmless

and leads to solution you want

[spark zinc]

how did you arrive at this

[fathom jewel]

you know that x = 0 is not a solution of the 6x^3-x^2-5x+2=0 and a not a solution of 2x^3-5x^2-x+6 = 0 as well

if you plug in x = 0

then both equations give a false statement

so you are free to divide and multiply by x

[spark zinc]

d starts on 3rd line

[fathom jewel]

because for any other value of x they are not

they are not for all x equivalent

[spark zinc]

Yeah you're right

does everything else look good?

[fathom jewel]

yep

[spark zinc]

I feel like that bit to let 1/x=a is a bit lacking

idk

do you feel like it's enough proof

[fathom jewel]

what is it lacking

it's just a try based on the observation that the coefficients are reversed

and that try worked out

[spark zinc]

fair enough

almost done hang on

Alright man

Thank you so much, I'm grateful you came here and spent your time helping me!

Take care.

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[tawny beacon]

is this properly the way u derive this?

[thin bridge]

differentiate, but yeh
also don't forget the =

[tawny beacon]

yes lol

but bro, what im confused with is

why in this second case

the constant goes before eulers number, while in the first one, nothing goes infront of it

[thin bridge]

looks nicer written that way

since 2 is a constant

considered the simplest way to express that

e.g. you'd ultimately simplify $p \cdot 2$ to $2p$

[flat frigate]

ℝαμOmeganato5

[tawny beacon]

so ur saying
2e^2x

is the final product after finding the derivative?

[thin bridge]

yeh

[tawny beacon]

ohh so these is a process behind that

[thin bridge]

same process,

the rest is simplification

[tawny beacon]

damn

thanks alot bro

ur saving me on this calc exam

[thin bridge]

$e^{2x} \cdot (2x)' = e^{2x} \cdot 2 = 2e^{2x}$

[flat frigate]

ℝαμOmeganato5

[tawny beacon]

yeahhh ur right lmao

thanks bro

i understand it now

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[gleaming cove]

r(x) = 7 / (x2– 1) Find functions f(x) and g(x) so the given function can be expressed as f(g(x)).

[gleaming cove]

bro im sooo confused

i asked gauth

it gave me an answer

but im confused on how to get it

uh

[lean otter]

what is r(x)

[gleaming cove]

wdym

7 / (x2– 1)

?

[lean otter]

ok then what is g(x)

[gleaming cove]

thats what im trying to figure out?

lol

hello?

[tidal imp]

There are multiple possible answers to this

Do you know what f(g(x)) means?

[gleaming cove]

yes

@tidal imp

[tidal imp]

So you know that g(x) is “inputted” into f to give r, yeah?

[gleaming cove]

mhm

[tidal imp]

So in your function r, can you spot another function that looks to be “plugged in?”

[gleaming cove]

hold on can you make the image like big

using the like

$r(x) or whatever

[tidal imp]

ah you want $r(x)=\frac{7}{x^2-1}$

[flat frigate]

CST (please ping when replying)

[gleaming cove]

yes ty

x^2 -1

probaby?

@tidal imp

[tidal imp]

That works

So okay, g(x)=x^2-1

[gleaming cove]

oh wait

u can do 7/x

riught?

[tidal imp]

For f, yeah

[gleaming cove]

as f

ohhh

i see

wow

that was very easy

thank you

soi much

[tidal imp]

There are multiple answers actually, that one was the most obvious

[gleaming cove]

oh kk

imma keep this open as i have like 7 of thesed so in case i have trouble with another imma keep it open is that fine

[tidal imp]

I would close it until you find another problem where you have trouble

[gleaming cove]

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[snow stag]

PLEASE help me understand

how do we find the displacement

how do we draw picture of this

[blazing palm]

Heres one part for you

[snow stag]

i just dont know how to solve these problems

how do I go about

it

[blazing palm]

What is the bottom leg (or x-displacement) of the triangle i drew

Hint: ||sin(34 deg) = x/117||

[snow stag]

i genuinely dont know where your getting this info

[desert juniper]

from here

[snow stag]

what is the bottom leg

how do I solve that is that something with sohcahtoa

or no

i see the hint though why would it not be cos theta if its has hypotuse in the bottom too

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[lean otter]

hello, can someone explain why its -3 for the amplitude?

[hardy monolith]

the amplitude is 3, but its -cos(ax)

so it becomes -3cos(ax)

[lean otter]

oh what makes it cos?

[hardy monolith]

you can look at x=0

but you can make it sin aswell

but then you need a shift

[lean otter]

yea idrk

i thought sine was positive like this and cosine would be this just flipped across the x and shifted so im confused

[hardy monolith]

,w graph cos(x)

[flat frigate]

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[hardy monolith]

,w graph sin(x)

[flat frigate]

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[lean otter]

oh i see... ok thanks

i think i got it now

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[grave ridge]

I do not really understand how this is reflexive (discrete math)

[neat patrol]

a=a satisfies less than or equal to

[grave ridge]

thank you

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[languid cobalt]

can someone please help me on this problem

[languid cobalt]

is there any faster way to doing this

[plucky elk]

No

[languid cobalt]

im not really sure what to do though

oh wait

hold on

im a genius

synthetic division

ok bye thanks

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[silent sandal]

yes horner

or just use calculator kekw

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[languid cobalt]

could someone please help me on this problem

[languid cobalt]

i forgot how to find the range of the functions

[tidal imp]

note that the given function is quadratic in form

[languid cobalt]

sorry, im not really sure what to do next

do i like find the derivatives and stuff

[tidal imp]

I mean, you could, but it’s not necessary

Do you know how parabolas work?

[languid cobalt]

yeah

[tidal imp]

Does this parabola open upward or downward?

[languid cobalt]

this isnt a parabola tho right?

[tidal imp]

$f(x)=-2x^4+x^2+6$

[flat frigate]

CST (please ping when replying)

[cloud maple]

It is a quadratic if u=x²

[languid cobalt]

ok well i know it should open downwards

[languid cobalt]

ok thanks

[languid cobalt]

it opens downward

sorry i forgot to repy

ok so i should find the vertex right

[tidal imp]

Yeah

Try that

[languid cobalt]

sorry someone called me to do something

so it should be like around 6

i got some like ugly numbers

ok thanks i got D

cya

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[chrome heath]

is the slope -1/2 or -2/1

[ripe lintel]

-2

-2x

[chrome heath]

i know thats what the slope actually is but do you graph it down 2 or to the left 2

im guessing down 2

like this

[cedar owl]

rise/run

[ripe lintel]

down 2 right 1

[cedar owl]

and it seems right

[chrome heath]

alr

thanks

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[balmy cloak]

Hi I don't understand the fundamental aspect of permutations, for example if I had 4 items and 4 places to put them, it would be 4!

What I don't understand is why multiplication is used as 4x3x2x1

Why is multiplication used?

[cloud maple]

4 ways to place item 1
3 ways to place item 2
2 ways to place item 3
1 way to place item 4

The process is done in one event, so its multiplication

[fathom jewel]

you can visualize this too drawing a tree diagram might help

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[balmy cloak]

I guess that makes sense thank you

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[safe radish]

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[hollow vale]

are my equations for x and y correct?

also, does tension B equal to 2mg?

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[cosmic kayak]

help with question 28

im going to show you my work can you help me through how to do it correctly

[novel pasture]

What did u do?

[cosmic kayak]

idk where to go from here

but this is where ive gotten

[lean otter]

Since f(x) il already written as Taylor series you should check for the (x-2)^2 coefficient

You already have it

[cosmic kayak]

-4/3/2!??

[lean otter]

Do you know how to express factorials?

[cosmic kayak]

3!=321

3!=3x2x1

[lean otter]

Yes

So 2! Is equal to 2 right?

[cosmic kayak]

yeah

[lean otter]

Then f"(2)/2 is equal to what the problem already give to you

[cosmic kayak]

ohhh

thanks

[lean otter]

🫡

[cosmic kayak]

wait

but how could i get -4/3(x-2)^2 by plugging in the nth term into the general term?

wouldnt the nth term be equal to 3?

how do you know that -4/3(x-2)^2 is the (f''(0))

ohhh

[rigid hazel]

taylor series yea

$\frac{f'}{n!} (x-c)^n$

[flat frigate]

Ryse

[rigid hazel]

make sure to get rid of the factorial thing

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[placid dawn]

Would this be right? The way i was taught in class was to rewrite and express something into terms of h if its vertical slices but it wouldnt make sense here

[plucky elk]

no

find a, b, x_i, and f(x)

looks like you can pick either left or right since they don't specific

[placid dawn]

I don’t think thats what they’re asking for. What we did in class was like just using sigma with the limit and then transforming it into an integral

Like that

[plucky elk]

yea you did the riemann sum part wrong

[plucky elk]

this is how you do it

[placid dawn]

I’ll try. Thanks

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[steep magnet]

I dont understand, the period of x(t) is pi, isnt that smaller than 2pi

So we take the smaller period, no?

[loud wren]

Is the period of y necessarily pi?

[loud wren]

No. You take the period that is synonymous with both functions.

It works similar to finding the period of 4sin(x)+3cos(2x)

[steep magnet]

Thats what I meant, like pi is smaller so its synonymous with both no?

[loud wren]

No. That's not how period works. If you want to test it out for yourself, plug in (t+pi) into y(t) and see what happens.

[steep magnet]

So we need like a 'lowest common denominator' kinda?

[loud wren]

More like a least common multiple.

Wait, no, wrong analogy.

Nevermind, correct analogy.

:headache:I'm doing a project right now and the deadline is coming up soon, sorry if I seem a little out of it.

[steep magnet]

Haha its past midnight for me my brain is also fried

[loud wren]

But you know what a least common multiple is, right?

[steep magnet]

Yeye I think I get it now

I was thinking the other way around

But ya I tried this method and it works everytime now, no matter what the numbers are

Thanks!!

❤️

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[native kernel]

i got help on this earlier and i finished solving it an my answer is wrong

can someone see if i did something worng

[fathom jewel]

can u send a sharper pic

[native kernel]

ya let me try again

r these better?

[fathom jewel]

yessss

how

[tidal imp]

Yeah slight issue with the integration of sec there

[native kernel]

shit it should be + not * right?

[fathom jewel]

yes

[tidal imp]

Yes

[fathom jewel]

otherwise i dont see an error

the -7ln(4) can be absorbed by the constant

[native kernel]

nope all it needed was the + thanks!

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[fathom jewel]

yea it was optional!

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[gusty aurora]

Find all integer values of m for which this polynomial has 3 real roots with 1 of them being the average of the other 2

[gusty aurora]

I know that uhh for there to be 3 real roots the derivative has to go have 2 roots I think

With one extrema being above the x axis and the other below it

But the derivative is so complicated I can't do it

[fathom jewel]

why would you need the derivative

try to factor the left cubic

[gusty aurora]

Oh

I knew there must've been another way 😭

[gusty aurora]

I'll try to do it that way ty

[fathom jewel]

as a hint it has one integer root zhe cubic

[safe radish]

@gusty aurora Has your question been resolved?

[gusty aurora]

Yes -1

I have another problem now thoughh

That second cubic with the parameter

It has to have 2 roots right but neither of them can be equal to -1 how do I do that?

This is for yk

So that one root is the average

[safe radish]

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[fathom jewel]

.reopen

[safe radish]

✅

[fathom jewel]

have you tried to find the roots in terms of of the right quadratic

[fathom jewel]

so you have basically three options

this one didnt work out as you didnt get an integer

so investigate (x_1 + (-1))/2 = x_2 and (x_2 + (-1))/2 = x_1

||x_1 = m and x_2 = m+1 when u solved the right quadratic||

[safe radish]

@gusty aurora Has your question been resolved?

[cloud maple]

x² - (2m+1)x + m² + m = 0 seems like can be immediately solved using quadratic formula

[gusty aurora]

Well ya

I tried to calculate the delta and got 1 > 0

[cloud maple]

excellent, use quadratic formula and find the roots

[gusty aurora]

Okay I'll try to find them cuz maybe I messed up

[gusty aurora]

Yes I've got that

Soooo now we havee

x1 = m, x2 = -1, x3 = m + 1 yes?

[fathom jewel]

these are your roots yes

now there are three possiblities to contruct that the mean of two roots is equal to the third root

[gusty aurora]

What does contruct mean 😭 sorry

[gusty aurora]

Can we just do this except plug in the values this time?

[fathom jewel]

mean formula is (a+b)/2 = c

and what i mean by construct is you rearrange a,b and c

like you can also have

(b+c)/2 = a or (a+c)/2 = b

[gusty aurora]

So we just do

(m + m + 1)/2 = -1?

I've done that earlier though 😭

[fathom jewel]

for example that didnt work out

then try

(m+(-1))/2 = m+1

and

[gusty aurora]

Ohh

Okayy

[fathom jewel]

(m+1+(-1))/2= m

[gusty aurora]

I see

I've gotttt

m = -1/3 and m = 0

!

Soo we only want the m = 0 one

[fathom jewel]

hmm

how did you get -1/3?

[gusty aurora]

Wellll (m-1)/2 = m + 1

Soo m - 1 = 2m + 2

Soooo

[fathom jewel]

[gusty aurora]

Ah it might be -3 😭

Oopss

[fathom jewel]

might?

[gusty aurora]

WILL BE

IT'S -3

Butttt we have to account for the fact that neither x1 nor x2 can be equal to =-1 right?

I don't think we've done that

[fathom jewel]

hm?

x_1 = m and x_2 = m+1
and m is free to be anything until we consider the mean constraint

[gusty aurora]

Wah wait now that I think about it I don't think that's possible because our delta is always > 0

[gusty aurora]

Well I just thought they can't be equal to =-1 because yk

[fathom jewel]

well m != m+1 for all m

[gusty aurora]

If our root of the cubic was -1

[fathom jewel]

so you always have two distrinct roots of the quadratic

[gusty aurora]

Okayy

[fathom jewel]

it can

question is do they satisfy the mean thingy

[gusty aurora]

They're also asking for 3 roots

[fathom jewel]

but we just found out our solutions anyway so no

[fathom jewel]

yes but that doesnt mean necessarily distinct

or does it

[gusty aurora]

Well uh yeah they did say that 😭

[fathom jewel]

well maybe u got a point 😄

[gusty aurora]

Hahah

Soo how do we make them distinctt?

[fathom jewel]

well they are

you got two solutions

[gusty aurora]

That are distinct from -1?

[fathom jewel]

(m+(-1))/2 = m+1 -> got you m = -3 so {-1,-3, -2}
(m+1+(-1))/2= m -> got you m = 0 so {-1,0,1}

[gusty aurora]

Ohh okayy

I think that is all thenn

[fathom jewel]

yes these were all possibilities

[gusty aurora]

Okayyy tysmm byee!

[fathom jewel]

bye roblox girl

[gusty aurora]

.close

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[soft lava]

For the problem in the left bottom corner I don’t understand why the answer key says to subtract not add. We are going from s to d. If we take the x values we are going from 125 to 5 which is a left translation on the graph since we are subtracting. So why would it be addition

[soft lava]

This graph is not accurate but gets my point accrosss

[safe radish]

@soft lava Has your question been resolved?

[blazing palm]

This translation has been performed incorrectly

$f(x - h)$ translates $f(x)$ to the \underline{right} by $h$ units.

[flat frigate]

King Leo

[soft lava]

Which expression is f ?

[blazing palm]

neither, i just gave you a generic rule

[soft lava]

I just don’t know what point I went wrong

Because I’m order to get 5 from 125 your decreasing

And I know if I am going left it’s addition and if I’m going right it is subtraction, opposite of what you think

Are we going to d to s instead jm so confused

Because that would contradict all my right answers

Did I get my x and y values mixed up. Is distance actuallt x and Time y? I got the x and y values from the info part

Like time would x because d(t) is inside the () so it’s actually d(x) and then the other value has to be y.

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@soft lava Has your question been resolved?

[safe radish]

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[gleaming pollen]

could someone pleaes summarize what this is saying. explain like im 5 please

[quasi bison]

it's telling you how the expectation and variance of random variables interact with addition and scaling

[gleaming pollen]

so its synonymous to stretching and translations in graphing?

[quasi bison]

ok they say mean instead of expectation but still

[quasi bison]

i would not say so

random variables are a different kind of beast than function graphs

[gleaming pollen]

addition and scaling is like basically what stretching and translations are in graphing terms tho?

[quasi bison]

hhh

i mean

i guess? but we aren't working with any graphs here

[gleaming pollen]

yeah i know that but

[quasi bison]

so you would just be confusing yourself this way

[gleaming pollen]

oh maybe im confusing u

[quasi bison]

you're trying to bring in something irrelevant

[quasi bison]

both of us

[gleaming pollen]

LOL

okay maybe

[gleaming pollen]

reword this a litlte

u explained like im 18

but im only 5 so

i dont quite understand

[quasi bison]

ok wait

[gleaming pollen]

okie

[quasi bison]

i hate that i have to ask this

[gleaming pollen]

ye

[quasi bison]

but you're not literally 5 years old are you

like that would violate discord terms of service if you were under 13

[gleaming pollen]

i said explain LIKE im 5

[quasi bison]

ok right no im just making sure

[gleaming pollen]

yeah ofc

[quasi bison]

we get the occasional standup comedian

who thinks it's funny to pretend to be underage

anyway

moving on

[gleaming pollen]

all g i understand

[quasi bison]

so you know how the mean of a random variable can be found by adding up all the values it takes, weighted by their probabilities

[gleaming pollen]

yeah

[quasi bison]

so

going from X to X+b means adding b to every single value

without changing any probabilities

the net effect is that you add b to the mean as well

that make sense?

[gleaming pollen]

YEs

i understnad

[quasi bison]

and going from X to aX means multiplying all of the values by a

which by the distributive law amounts to multiplying the entire weighted sum by a

so the expectation also does the same thing

[gleaming pollen]

i see isee

is that it?

[quasi bison]

well that covers the expectation bit

the other line tells you what happens to variance

it stays the same if you just add a constant to your variable

but if you multiply it by something, it scales by the square of that

[gleaming pollen]

i see

is that it now

the expectation part IS like graphs

or at least thats how i think of it

[quasi bison]

the other bit is about how expectation and variance interact when you add two random variables

[gleaming pollen]

alr i got it thanks

.close

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[desert niche]

can someone please explain to me why the intervals are not pi/2 and pi/6?

[desert niche]

don’t you input the intervals into u?

[quasi bison]

i think you mean "endpoints" not "intervals"

[quasi bison]

they don't format this as a u-substitution at all

[desert niche]

wait so you don’t use u substitution?

[quasi bison]

they're taking you through their worked solution and having you fill the steps in

and their solution doesn't involve one

doing this with a u-sub would look different

(even though the answer would be the same ultimately)

and then like, u := 2x would get you your recalculated endpoints as pi/6 and pi/2 just as you said

[desert niche]

so u substitution would still work for this problem?

ok thanks for clarifying

.close

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[gleaming moon]

how can i prove this ?

[quasi bison]

do you know how to expand (1+x)^p when p isn't necessarily an integer

[gleaming moon]

yeah

1+px + p(p-1)x^2/2! +...

[quasi bison]

yeah so do that here

with p=-1/2

and all x's replaced by (-4x)

[gleaming moon]

but how do i go from the lhs to rhs

cuz finding the value of summation 2nCn (1/6)^n was the actual question

[quasi bison]

hmm

well hm i guess there really isn't any trick to that unfortunately

[cloud maple]

$\left( 1-4x \right)^{-\frac{1}{2}}=\sum_{n=0}^{\infty }\binom{-\frac{1}{2}}{n}\left( -4x \right)^n$

[flat frigate]

TargetVN

[cloud maple]

... well

(-4)^n * (-1/2)Cn = ?

[gleaming moon]

(-1/2)Cn ?

how do i write that

[cloud maple]

isnt it just binomial coefficient

$\left( -4 \right)^{n}\binom{-\frac{1}{2}}{n}=\text{?}$

[flat frigate]

TargetVN

[cloud maple]

simplify it

[gleaming moon]

coefficient of x^n in (1+x)^(-1/2) ?

[cloud maple]

yes

[gleaming moon]

is there a direct way to write it?
i just open the expansion like 1 + nx + n(n-1)/2!x^2 and try to find it

[safe radish]

@gleaming moon Has your question been resolved?

[safe radish]

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