help-49

[last slate]

like

[polar crystal]

use sohcahtoa

[last slate]

find your angle

[polar crystal]

helpful

[last slate]

and then the side the angle is looking at is the opposite

[polar crystal]

adjacent is 48

[polar crystal]

opposite is 14

and 50 is the hypotenus

using sohcahtoa

[last slate]

so just find where the angle is facing pretty much

and the adjacent is the remaining side

[little dagger]

so from 0 that points = opposite?

[polar crystal]

and if the angle has a line that coressponds with the 90 degree one, thats adjancent

[last slate]

0 that points?

what does that mean

[polar crystal]

she thinks thetha is 0

14/50=sinthetha

[last slate]

stop

[polar crystal]

48/14=costhetha

[last slate]

!nosols

[midnight plank]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[polar crystal]

oh

[slender walrus]

don't give wrong solutions either

[last slate]

can u explain what u meant here

[midnight plank]

@little dagger Has your question been resolved?

[little dagger]

sorry I'm back

[little dagger]

i was talking abt your drawing

so where the angle is facing its opposite?

[fallow scarab]

Opposite of theta is a side. Which one are you talking about

[midnight plank]

@little dagger Has your question been resolved?

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[nova yoke]

i don't see why the cs inequality is mentioned there; obviously if y=0 then <x - Px, y> = 0
(well maybe they invoke the cs inequality to prove that, but that's not really necessary)

[fallow scarab]

Show the entire statement and proof

[nova yoke]

|<x - Px, y>| <= ||x - Px|| ||y||, so if y=0 then the RHS is zero, hence so is the LHS, and from that it follows that Re<x - Px, y> = 0

you could show the same thing more simply by noting that if y=0 then <x - Px, y> = <x - Px, 0y> = 0<x - Px, y> = 0 i.e. just factor out the scalar zero

[inland patio]

yeah, somehow one must exclude the possibility for x-Px to be zero

[nova yoke]

do you need to exclude that here?

[inland patio]

no, true, it's not needed 😄

thanks for helping

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[twilit field]

how do I find alpha here using l'hopital

[twilit field]

I found beta to be -2

using l'hopital

I also would prefer not to use series expansions

so what do i do?

[fallow scarab]

How did you do this

[twilit field]

L'hopital's rule

[fallow scarab]

Show your work

[twilit field]

$\frac{\alpha cos(x) -\beta sin(x) + \frac{1}{1+x}}{3sec^2(x) tan(x)}$

[grand pond]

Why am. I here

[twilit field]

now on finding the derivative of the numerator and denominator again

$\frac{-\alpha sin(x)- \beta cos(x) -1/(1+x)^2}{3sec^4(x) + 6sec^2(x)tan(x)}$

[grand pond]

Why am. I here

[twilit field]

subbing x=0

we get -beta/3=1/3

oh

so beta is -1

[twilit field]

now to find alpha do I derive the numerator and denominator again?

<@&286206848099549185>

[fallow scarab]

You forgot a factor of 2 in the denominator

[twilit field]

oh, thanks

so beta is -2?

[twilit field]

wait, where?

[fallow scarab]

,w diff 3tan^2(x)

[grand pond]

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[fallow scarab]

Power rule

[twilit field]

the derivative of 1/(1+x) is just -1/(1+x)^2 though, right?

,w derive ln(1+x) twice

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[twilit field]

oo , nvm, understood

but how do I find alpha

using l'hopital's

<@&286206848099549185>

nvm, have to leave for school now. Thanks Dr matlab plot

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[strong lava]

Rip riemann

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[last slate]

I'm getting 8 but I have no idea if that is right

my steps are as follows

$\frac{56}{x-2}-\frac{45}{x}=1$

[grand pond]

mj

[last slate]

expanding, i get

$56x-45x+90=1$

[grand pond]

mj

[last slate]

simplifying, i get

$11x=-89$

[grand pond]

mj

[last slate]

my solution is

$x=-8.09$

[grand pond]

mj

[last slate]

this is a negative value

and thus is not valid

which is why im confused, I did the steps right as far as I'm aware.. is my formula wrong?

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[cedar vortex]

kinda annoyed I didn't get this q

[cedar vortex]

Given nonnegative integers $n$ and $k$, find the value of the sum:

[ \sum_{j=0}^{k} (-1)^j \binom{n+j-1}{j} \binom{n-1}{k-j} ]

where $\binom{a}{b}$ represents the binomial coefficient "a choose b."

[grand pond]

nosqldb

[waxen musk]

what did you do so far?

[cedar vortex]

well I'm a little confused on where to start

I think like

[waxen musk]

i didnt work it out, but it looks like the factorial definition of choose might help

[cedar vortex]

ah my professor wouldn't do that to me

I attempted to do this question on a quiz later today

and I was quite dissapointed that I didn't get this question

but I would like to learn how to do it

<@&286206848099549185>

[lethal holly]

use vandermonde

[cedar vortex]

what do I do about the (-1)^j

[lethal holly]

well, that would mean that the thing alternates signs

[cedar vortex]

mhm

[lethal holly]

did you know that if you do nCr, the sum of nCevens is equal to nCodds?

[cedar vortex]

yes

[lethal holly]

#❓how-to-get-help

[lethal holly]

well then you can change the summation

[cedar vortex]

so this would simplify to 0?

[lethal holly]

yeah

im pretty sure

wait

[cedar vortex]

it was a multiple choice q

and it was like (-1)^k

and stuff like that

[lethal holly]

mb vandermondes doesnt work here because of that -1 thing

[cedar vortex]

I'm not going to lie man

it really got me fucked up

[lethal holly]

what did

[cedar vortex]

it was the second quiz

and we had like 30 minutes for 5 questions and I wasn't able to get this q

for some reason

not smart enough

[lethal holly]

i got a 65 on my second most recent quiz 💀

because i forgot how to math

[cedar vortex]

I feel like this is my first brutal math course

I'm a freshman

[lethal holly]

what's the course called

[cedar vortex]

it's a umich course

umich 465

[lethal holly]

oh i see

ngl idt i can really help, but you can try to consider factorial definition as Cycadellic said

[cedar vortex]

<@&286206848099549185>

[last slate]

yes?

[last slate]

what seems to be the problem?

[cedar vortex]

^^

@last slate

[last slate]

oh crap i helo with a different type of math sorry 😭

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[mint flare]

what is the f beside x

[sinful trout]

u mean like f(x)??

that means taking a function of x

it's like a little machine that takes an input called "x" and gives an output in relation to whatever "x" is

[mint flare]

it look lik e that

[sinful trout]

so if f(x) = 3x, that means it will output whatever 3 times "x" is for whatever "x" you put in there

[mint flare]

oh

[sinful trout]

yea

[mint flare]

thanks

[sinful trout]

if f(x) = x^2, it will give the square of "x"

[mint flare]

k

[sinful trout]

for example, f(x) = x^2; if you plug -2 in there, so f(-2), that would be (-2)^2 = 4

[mint flare]

thanks i undestand now

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[final wedge]

1. (8-1)! = 7! = 5,040

2: 4!4! = 576

3. 4! = 24

Correct me if I'm wrong

[waxen musk]

1 is wrong

Think about it

The first person has 8 choices

The next has 7

…

[final wedge]

so the correct answer is B?

[waxen musk]

,w 8!

[grand pond]

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[final wedge]

circular permutation is (n-1)!

(8-1)!

7!

5,040

[pine thicket]

i agree with 7

the table can be rotated such that an arrangment is considered the same

[final wedge]

my number 2 and 3 is correct right

[waxen musk]

This wasnt assumed

[pine thicket]

i see why you think so, but based on the answer choices, it has to be 7!

[little sundial]

yeah

the first person has one choice only

because every chair is identical initially

but the next person will have 7 choices

[final wedge]

I don't really get number 2

but I'm correct right

[little sundial]

the second person?

[pine thicket]

yes, number 2 is 4!^2

[final wedge]

"if a certain 4 ppl seated side by side"

[final wedge]

Okay thanks

[little sundial]

we can assume the group of 4 to be one unit

[final wedge]

yeah

[little sundial]

isnt?

so (5-1)! * 4! for the internal arrangement

[final wedge]

Oh so D?

[little sundial]

wait which d?

[final wedge]

There are choices

[little sundial]

wouldn't it be a then?
4!*4! =24x24

[final wedge]

Oh okay thanks

[final wedge]

I was confused with the question because it said seated side by side so I assumed they would be counted as 1, 4!1!

But thanks for clearing it up

[little sundial]

yeah it happens

[little sundial]

my pleasure !!

[final wedge]

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[last slate]

slightly confused on how to get c, because O isn't at the point of intersection. I've managed to get the other 3 answers already, just struggling with part c.

[slender walrus]

draw the line connecting O and C

then apply inscribed angle theorems

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[last slate]

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[last slate]

What is the equation

[midnight plank]

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[grand silo]

You might want to explain the question

[midnight plank]

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[surreal moon]

@last slate you have to actually ask a question to get help.

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[unkempt rune]

what is a critical point

[unkempt rune]

could it be a local aswell as absolute min/max?

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[midnight plank]

Please don't occupy multiple help channels.

[agile snow]

i understood part 1

i dont understand part 2

because the answer i get is 0.5

but the textbook answer is 0.56

[midnight plank]

@agile snow Has your question been resolved?

[midnight plank]

@agile snow Has your question been resolved?

[midnight plank]

@agile snow Has your question been resolved?

[midnight plank]

@agile snow Has your question been resolved?

[fervent ember]

why are you occupiying more then one channel with the same question?

[midnight plank]

@agile snow Has your question been resolved?

[gloomy scaffold]

1. this is physics, 2) why are u using 2 channels lol

Draw the free-body diagram for both masses using f=ma

They have the same acceleration

if u need help I can give a short working

the lines aren't exactly proper but who cares

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[obsidian niche]

Topic: Particle motion, Calc
The only part I need help on is D. Looking at the key, I don't understand how they got from the marked step to the next. 2t^2-t^4 isn't the integral of 4t^3-4t, right? Can anyone help? Thanks in advance.

[sharp coral]

they took the absolute value, so the signs were flipped when it's negative

[obsidian niche]

cloud I love you

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[summer tiger]

ive been constantly getting the wrong answer

so i first found u which was <850 sqrt(3),850>

and then v which was <500 sqrt(2), -500 sqrt(2)>

and then i used the formula u= w_1 +w_2

and so w_1 =

[sturdy spoke]

right

[summer tiger]

((u*v)/||v||^2)v

and then w_2 = u-w_1

lemme do the math again rq

[sturdy spoke]

i would just split the vectors into their respective x and y compents and combine them then find their magnitude from there

[summer tiger]

what

[summer tiger]

did i split it here

[sturdy spoke]

what is 850?

the previous numbers?

[summer tiger]

1700sin(30)

[sturdy spoke]

right yes

got it

[summer tiger]

wdym

[sturdy spoke]

no i was being dumb

[summer tiger]

oh

[sturdy spoke]

did you account for negative y in the 1000 force

ah i see you did

[summer tiger]

yeah

lemme calculate w1 rq

[sturdy spoke]

why do you have sqrt(3) and sqrt(2) next to 800 and 500

once you split it that should be fine

[summer tiger]

what

isnt it u*v

and u= <850 sqrt(3),850>

and v = <500 sqrt(2), -500 sqrt(2)>

so i multiply u_1 and v_1

and u_2 and v_2

and then add them?

[sturdy spoke]

what method are you using?

i only know the physics method

[summer tiger]

oh

[summer tiger]

this thing

[sturdy spoke]

i see

yea im unfamiliar with that method sorry man

but i got the answer using that ^ if it helps

[summer tiger]

oh yours worked?

[sturdy spoke]

yea try it

[summer tiger]

seems alot easier

than how i learned it

💀

maybe im looking at the wrong fomrula ort something

[sturdy spoke]

i have like 6 years of practice lol

and to find the angle

just use arctan

arctan ( sum of y / sum of x)

[summer tiger]

wait orthogonal is perpendicular right

[sturdy spoke]

yes?

[summer tiger]

i was using the wrong formula 💀

[sturdy spoke]

lemme check

[summer tiger]

oops

[sturdy spoke]

yes

orthogonal is indeed right angles

[summer tiger]

i was using one for when the 2 given vectors is orthogonal

[sturdy spoke]

ahhhhh

at least you have a back up method now good luck! 👍

[summer tiger]

yeah... thanks

maybe the teacher will let it slide if i use that

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[summer tiger]

i have no idea where to start with this one

[midnight plank]

@summer tiger Has your question been resolved?

[simple lynx]

do you know how to resolve forces

[summer tiger]

um

sort of

but not really

if the T_L and t_R was here i could solve it

[simple lynx]

ah

remember that at this level tension is constant in a rope

so the tension in the yellow part is the same in the black part

since its all one connected rope

think about pulling a bucket up from a building

it'll have the same tension throughout

[summer tiger]

by black part do you mean the part labled T_L

and T_R

[simple lynx]

ye

[summer tiger]

not done yet

um

i did something wrong

i switched T_R and T_L somewhere

[royal flame]

ik this aint much help but i can tell ur like almost there

i had to do this shit on a test like 1 week ago

and it looks right to me

so idk what u did wrong

[summer tiger]

nvm

i think i put it in my calculator wrong

and it was a coincidence my answer looked like the answer to T_L

[royal flame]

oh ok

[summer tiger]

finally i did T_1 and T-2

thanks both of you

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[cosmic dragon]

Hey guys, just wondering what the best approach to a problem like this is, whether theres a mathematical formula that i can derive to solve this easier

essentially, i am creating a tournament with 8 teams
each team has 4 team members (32 players altogether)
each game will have 4 teams (1 player from a team) competing per game

in one round, all 4 team members will be playing a game at the same time, therefore 8 games will run simultaneously as depicted in the image below
(the numbers are to represent each player in a team, and can be assigned however)

i want to work out the best way to randomise this type of bracket ensuring that:

• each team plays each other team as evenly as possible
• there will never be more than 1 player per team in the same game

if anyone would be able to give suggestions as to how to best work this out, that would be appreciated

[midnight plank]

@cosmic dragon Has your question been resolved?

[cosmic dragon]

<@&286206848099549185>

[cosmic dragon]

xD

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[chrome hill]

Domain of 2/3-{x}

[chrome hill]

What property should I use here?

[radiant beacon]

use the equal fraction theorem thingy

I forgot the name

[wanton shore]

AM >= HM

[radiant beacon]

like the one where 1/2 = 4/8 = 5/10

oh yeah theorem on equal rations

idk if it wojld work tjo

[radiant beacon]

this could also work..

[midnight plank]

@chrome hill Has your question been resolved?

[chrome hill]

How to use it?

.close

[midnight plank]

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[obsidian flume]

I know this is likely a sort of odd question, but I'm just beginning to learn derivatives. f'(x) is the derivative of f(x), so what notation would denote the inverse of f'(x)? (i.e. f'^{-1}\left(x\right))

On a side note, would the inverse of f'(x) be equal to the derivative of the inverse of f(x)?

[obsidian flume]

$f'^{-1}\left(x\right)$

[grand pond]

./.../…

[vague remnant]

well

i think its (f^-1(x))'

[obsidian flume]

Integrals are the inverse of derivatives, right? Would that be applicable?

[vague remnant]

inverse in this thing and inverse in others are different

x and 1/x

but integral of 1/x isnt x

its lnx ig or whatever

they arent similiar

[obsidian flume]

Gross. If I had a function like $f'\left(x\right)=1$ how would I solve for x?

one money

moment

God my keyboard hates me

[vague remnant]

try this nootation

[grand pond]

./.../…

[obsidian flume]

there we go

[vague remnant]

ok

so you wanna integrate

[obsidian flume]

that would be where the tangent line has a slope of 1

[vague remnant]

$1=1x^0$

[grand pond]

Zemitrix

[vague remnant]

theres an inverse power rule

do you know

[obsidian flume]

I think I do

3x^2 -> x^3 right?

[obsidian flume]

yeah that's what I got

minus the constant at the end

[vague remnant]

so do it with 1

[obsidian flume]

it would be x+c

and now I would want to solve for c given f(x)

(I'm glad I finally know where the c comes from, a year ago I tried learning this and failed)

basically derivatives are a "lossy" function

[vague remnant]

in integrals you can get several solutions to one thing so theres an constant

ig

[vague remnant]

idk how to solve for this like

you can solve for c in terms of x ig

[obsidian flume]

yeah that's where I'm stuck then

shoving f'(x)=1 in Desmos does return a value, so how do we get that value?

[vague remnant]

x+c=0 x=-c am i wrong or wth

idk

[obsidian flume]

idk

what is c anymore

[vague remnant]

constant

some number

[obsidian flume]

cool let's just agree as a global math community that c=7 for every function no matter what

makes things easier

[vague remnant]

why 7

0 looks easier

[obsidian flume]

0 is weird

[vague remnant]

but actually its always different

[obsidian flume]

7

[vague remnant]

they say that you have to find c?

than it must be possiblr

[obsidian flume]

this is my own problem, so I need to find c

[vague remnant]

lol

[obsidian flume]

I've been stuck on this for years

[vague remnant]

f(x)=x?

[obsidian flume]

doesn't always work

[vague remnant]

ok lets say that its just c

[obsidian flume]

e.g. f(x)=1/x

[vague remnant]

we can do anything with c

idk

we need helpers

<@&286206848099549185>

[obsidian flume]

while we wait, who wants popcorn?

[bitter tide]

F(x) for anti derivative, (f'(x))^-1 is simply 1/(f'(x))

[vague remnant]

damn

i thoutght if we need to find the derivative of the inverse we have to inverse it firstly than derivative

[bitter tide]

depends what the question asks, i think

[obsidian flume]

that's ultimately my problem @bitter tide

[bitter tide]

integrate the function

u simply get the x=x

[vague remnant]

wheres +c

[bitter tide]

its 0

[vague remnant]

why 😦

[bitter tide]

because integrating left side gives u x, right side is x+c, but since x=x+c, it implies c=0

[vague remnant]

what is the left and right side damn

when integrating we always get +c

idk

[bitter tide]

f'(x)=1 is the question

[vague remnant]

ok

[bitter tide]

integrate both sides

[bitter tide]

yes, when indefinite

[obsidian flume]

Try that with f(x)=2^x and when f'(x)=1 then x=1 is the solution

[vague remnant]

so show how to integrate f'(x)

why its x

[bitter tide]

what?

[bitter tide]

FTC

[vague remnant]

FTC

[obsidian flume]

let me clarify one moment

[bitter tide]

taking integral on derivative cancels it

[vague remnant]

ok

then everything is clear

[bitter tide]

oh ok i think i understand what ur tryna say, the problem is its not the same function

the derivative of 2^x isnt 1

[bitter tide]

meaning its not the same

[obsidian flume]

If you have f(x)=2^x, then f'(x)=1 is where a tangent line of f(x) has a slope of 1. On the graph f(x)=2^x, that line is tangent at x=1. How would I solve for that last value?

wait

[vague remnant]

what is ' of 2^x

[bitter tide]

f(x)=2^x doesnt give f'(x)=1

[obsidian flume]

it's not tangent at x=1 it's tangent at like x=0.53

[obsidian flume]

No, I'm trying to solve for when f'(x) is equal to 1

[bitter tide]

well thats not possible, u need to plug a value for x

[obsidian flume]

Let me clarify with Problem v3™️

Let f(x)=2^x. The goal is to find the x value of that graph where the tangent line has a slope of 1. The problem boils down to finding which value of x results in f'(x) returning a value of 1. In the graph f(x)=2^x, the x value at which the tangent line has a slope of 1 is x=~0.53. How do you solve for that number?

(I think we're getting somewhere)

[obsidian flume]

I hope that answers that

f' of what value is equal to 1 given f(x)=2^x?

[bitter tide]

find the derivative of 2^x then solve for x

set the derivative u found=1

and simply solve

[obsidian flume]

Ultimately I want to have it solvable for an arbitrary function

[bitter tide]

yes its doable for every function

[obsidian flume]

For anyone wondering, the exact value is -ln(ln(2))/ln(2)

[obsidian flume]

Without knowing what the function is, just declaring f(x) is a function

Basically f' inverse

of 1

[bitter tide]

well i dont think thats doable, maybe it is but i have no idea, its too vague and some functions might never have a tangent with a slope of 1

for example 1/x

it has an imaginary slope i guess but

[obsidian flume]

oo fun

[bitter tide]

i think ur talking in R

[bitter tide]

imaginary solution mb

[obsidian flume]

imaginary slope sounded way cooler

[bitter tide]

¯_(ツ)_/¯

hope that helps, i gtg

[obsidian flume]

thanks, bye

I just don't like that I would have to manually solve the derivative for a function just to get the answer

there should be an f prime inverse function

g(x)=f'(x)
g^-1(1)=some number

[midnight plank]

@obsidian flume Has your question been resolved?

[midnight plank]

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[ivory plume]

Confused on this q?

[midnight plank]

@ivory plume Has your question been resolved?

[midnight plank]

@ivory plume Has your question been resolved?

[midnight plank]

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[zealous prairie]

is this d

[winged gust]

Ok

take it as D

It ain't possible for higher values

so yeah

it is D

[zealous prairie]

ok

[winged gust]

.close

[zealous prairie]

.close

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[raven frost]

I need an explanation of this notation

[jagged saffron]

https://www.cuemath.com/algebra/cartesian-product/

[raven frost]

Im not sure what it means in this context

[jagged saffron]

it is a set of pairs (a,b) where a is in R and b is in [1,inf[

[raven frost]

because which pair is it referring to

uhh i mean

which piece of the pair

[jagged saffron]

its all possible pairs

[jagged saffron]

of this form

[raven frost]

So its saying that all pairs must be 1 or higher ?

up to but excluding infinity

[jagged saffron]

the second element in each pair must come from [1,inf[

[raven frost]

why the second element

how is that seen

[jagged saffron]

because [1,inf[ is on the right

[raven frost]

oh

@jagged saffron its the image for this function

What I dont know is why its bound to the second one and not the first.

why does x2 have to be 1 to infinity ?

why cant x2 be -1000

[jagged saffron]

it can, its talking about the image, not the domain

[raven frost]

yeah

[jagged saffron]

you're looking at what possible points you can obtain with that function

[raven frost]

mhm

[jagged saffron]

those points are R x [1,inf[

[raven frost]

so like

(x, y > 1)

[jagged saffron]

= 1 but yeah

[raven frost]

right

but why y > 1

[jagged saffron]

well how would you make something less than 1?

[raven frost]

if we input: f(1, -100)

[jagged saffron]

thats going to be bigger than 1

[raven frost]

oh i didnt even see the ^2

because the font is all wonky

[jagged saffron]

yeah font is unfortunate

[raven frost]

x >= 1 as a result of ln(x)

[jagged saffron]

the output x can be anything

[raven frost]

yeah

sorry i meant x1

[jagged saffron]

x1 > 0 is what you mean

[raven frost]

right

x1 > 0

x2 >= 1

[jagged saffron]

no x2 can be anything

i think youre confusing inputs and outputs

[raven frost]

yeah sorry

my bad

x1 > 0

x2 can be any value in R

as for the points

(x,y)

y >= 1

x1 * 0 + 1 = 1

since thats the lowest for y

x can take on any value

[jagged saffron]

yeah bc x1*x2^2 + 1 is adding something nonnegative to 1, so you're always going to be 1 or higher

[raven frost]

right

and so because of the placement of the cartesian 'x' it means that it's y that must be [1, inf[

makes sense

[jagged saffron]

yeah its just notation for "pairs of points from each set"

and in this case is describing the image of the function

[raven frost]

Nice, thanks a lot. I just have a proof left and them im done with the coursework for the day

Do you have time? 😄

I have a lot of questions

[jagged saffron]

sure

[raven frost]

I get the overall picture. Basically for it to be continuous it must hold that for any point, it should be the case that when x goes towards x0, f(x) must also go towards f(x0)

[jagged saffron]

mhmm

[raven frost]

my question is how to define epsilon

[jagged saffron]

you dont define epsilon

you fix an arbitrary epsilon

and look for a delta

so you say "fix eps > 0. I am now looking for a delta such that if |x - x_0| < delta, then |f(x) - f(x_0)| < eps"

[wary thorn]

Really you’re defining delta , in terms of an arbitrary epsilon

[raven frost]

and that arbitrary epsilon

its just made up?

as in you just give it a go and then take it from there?

[jagged saffron]

we're playing a game, i give you an epsilon, as small as i like, and you can in return give me a delta that makes the above true

f is cont if you can always give me a delta no matter what eps i give you

[raven frost]

right, okay

and eps is a number or?

[jagged saffron]

its a real number bigger than 0

lets try an example: if i give you $\varepsilon = 1$ can you find a $\delta > 0$ such that if $| x - x_0| < \delta$ then $|f(x)-f(x_0)| < 1$?

[grand pond]

ΣAC

[jagged saffron]

i encourage you to sit with this for a while

[raven frost]

Ill write something down and then post it

one moment

I dont remember the command to flip it

xd

[jagged saffron]

,rccw

[jagged saffron]

oh brilliant you skipped the example and went straight to general epsilon

what you've written is perfect

[raven frost]

srs or joke?

[jagged saffron]

srs

[raven frost]

okay, nice

[jagged saffron]

so now if i give you an epsilon you just give me back epsilon/3 for delta and it makes everything work

[raven frost]

The thing is though that this is how a teacher processed a similar question before

i assume this would still make sense if we said that delta = epsilon * 3

[jagged saffron]

no that would not work

because then at the end you would get thats its all less than 9*epsilon

which is not less than epsilon

[raven frost]

right, okay

[jagged saffron]

any delta less than eps/3 would also work

[raven frost]

But how did he from the beginning find that delta = eps / 3

[raven frost]

so delta = eps / 4

[jagged saffron]

because your goal is to get 3 * |x - x_0| less than epsilon

you know that 3 * |x - x_0| < 3*delta

so now you just have to solve 3*delta = eps

[raven frost]

right, okay

I forgot the B here

The domain for the function listed earlier with (ln(x1), x1*x2+1) would be a subset of R ^2 where x1 > 0 right ?

(question B)

[jagged saffron]

yea

[raven frost]

ty!

.close

[midnight plank]

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[glossy sleet]

what is the probability of getting a hand in poker, from a standard deck of cards, of a king queen jack of hearts, and two other cards that aren't hearts?

[glossy sleet]

the number of ways to get that hand is 52C3 x 39C2

the total 5 card hands is 52C5

[next rover]

it's just 39C2

there's one way to get those 3 cards

[glossy sleet]

bruh

you're right

.close

[midnight plank]

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[ocean sable]

9^x + 3^(X+1) = 27^ (3^x + 3)

[ocean sable]

can someone help me find X

[red crow]

yes

just multiply the coefficient behind the brackets by the the numbers inside the brackets

[ocean sable]

wdym

what i did was i factored 3 out of the first part

so it was 3(3^x + 1^(X+1)

then divided 3 from 27

to get 9

then i was confused what to do

@red crow

[red crow]

AAAAA

I READ

WRONG

[ocean sable]

lol

[red crow]

sorry man its 11pm

HAHAH

[ocean sable]

all g

[red crow]

write 9 as 3^2

[ocean sable]

wouldnt i say 3^2(x)

[red crow]

yeah, that's the next step

[ocean sable]

i did the earlier and my dad said it was wrong

and i was so confused

[red crow]

don't let your dad cook

xd

btw

yeah, can i have the solution? i might try to solve it and then explain.

[ocean sable]

i dont have it 😭

[red crow]

damn

[ocean sable]

ive looked everywhere for it

[red crow]

Ok, nvm

[ocean sable]

ill send u what i did earlier

[red crow]

sure

[ocean sable]

but my dad said it was wrong

one sec

dont mind the handwriting

[ocean sable]

did it load for u

[red crow]

yes

[ocean sable]

bro ilooked on google used chat gpt bing ai google bard everythibng had a diff answer

i was so confused

[red crow]

no

try to replace 3^x = t

[ocean sable]

?

wdym

[red crow]

lemme writeit for you

[ocean sable]

Wait

NVM what i sent was wrong

i wrote it wrong

[red crow]

oh

yea

can i see the correct?

[ocean sable]

this

[red crow]

bro

your solutions are imaginary

[ocean sable]

💀

HOW DO I EXPLAIN THAT TO MY DAD BROO

[red crow]

i think you wrote was wrong

[ocean sable]

thats what we wrote for me

[red crow]

yeah he probably invented it

[ocean sable]

9^x + 3^x+1 = 27(3^x+3)

[ocean sable]

sum fckin random guy from facebook did

[red crow]

nah bro, this aint possible in R

if you didnt study complex number you cant do it

[ocean sable]

GNg im in 8th grade 😭

I dont understand anything the image u sent even means

[ocean sable]

can u help with another 1 my dad gave

(3x-17)^2 = 28

[red crow]

Sure

Just calculate the brackets

[ocean sable]

i got to 9x^2 - 102x + 261 = 0

idk how to factor this tho

[red crow]

I dont think it is factorable lmaooo

Yeah

You use the quadratic formula

[ocean sable]

yeah

so i got (102 +- sqrt 10404 + 9396) / 18

[ocean sable]

so how would i do that?

we never rlly went over quadractic formula in school

[red crow]

Nah you cant do it then without quadratic

[ocean sable]

but how would i do it then

wit quadratic

@red crow

[red crow]

Justv solve it

[ocean sable]

How 😭

we never did quadratic formula in school 😭

` im confused on the sqrt 10404 + 9396 part

wait

i got it to

x= (51 +- sqrt 252) / 9

[midnight plank]

@ocean sable Has your question been resolved?

[glossy sleet]

@ocean sable

you have to do quadratic

9^x can be written to 3^2x

then let 3^x be a variable like a

[ocean sable]

what abt the sexcond half

with the 27(3^(x) + 3)

[glossy sleet]

27 = 3^3

base is common

so it's 3x

nah wait

forget that

[ocean sable]

k

[glossy sleet]

27(3^x + 3) remember we said that 3^x is "a" so just expand simplify, bring to the other side

it's just 27a + 81

[ocean sable]

im so confused im sry

ive never dun anything like this

[glossy sleet]

why are they giving you questions like this in school then lol

[ocean sable]

my dads giving them

[glossy sleet]

oh

[ocean sable]

😭

and he refused to help me

[glossy sleet]

ok well 9^x is 3^2x

and 3^(x+1) is 3^x times 3

[ocean sable]

wait what

i wrote 3^2(x) + 3^(X+1)

then wait

if i made 27 3^3

since there all 3s in the bottom

cant i do

2x+x+1=3(x+1)

or am i jsut stupid

@glossy sleet

[glossy sleet]

i don't think so

[midnight plank]

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[radiant helm]

i thought its b

[midnight plank]

Please don't occupy multiple help channels.

[radiant helm]

but the answer said its d

i thought x has to be less than 10??

[tacit rose]

When you multiply or divide by a negative, involving inequalities, you flip the sign

So -3x > -30 is the same as x < 10

[radiant helm]

OHHHH

whoops.

thans

[midnight plank]

@radiant helm Has your question been resolved?

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[midnight plank]

Please don't occupy multiple help channels.

[radiant helm]

for a), do they mean to find the average rate of change of 1 and 5?

cuz i just thought its instantaneous rate of change at 5 but thats b) lol

[hot canopy]

SOME HUMAN HELP ME

PLEASE

I REALLY NEED THIS

PLZ

[prime hornet]

the average velocity for the 1st 5 seonds means [h(5)-h(0)]/5-0

[hot canopy]

the fu

HELP ME

[radiant helm]

wait

[prime hornet]

please stay in your own channel

[radiant helm]

does it have to be at 0?

not 1?

[prime hornet]

first 5 seconds means from 0 to 5 lol

[radiant helm]

ohhhhhh

[prime hornet]

0s to 1s, 1s to 2s, 2s to 3s, 3s to 4s, 4s to 5s

[radiant helm]

is that like self implied, so next time i see first wtv i do 0- wtv?

[prime hornet]

thats 5 phrases seperated by commas

so thats the first 5s

[radiant helm]

okay thanks

[signal sigil]

make sure you do it with the derivative though cuz it's asking about average velocity not the average height

[radiant helm]

i dont know calc tho

[midnight plank]

@radiant helm Has your question been resolved?

[radiant helm]

hi, how do i do this

<@&286206848099549185>

do i just multiply by the sqrt orrr

[midnight plank]

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[soft burrow]

You can get rid of the square roots on numerators by using the idea that
(a-b)(a+b) = aa+ab-ba-bb = a^2 - b^2

[midnight plank]

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[soft burrow]

@radiant helm did you figure it out already or should you .reopen ?

[radiant helm]

i dont get it sorry 😭

[radiant helm]

yea hold on

.reopen

oof

[soft burrow]

Applying that to the first case, if you multiply numerator and denominator with "sqrt (16-h) +4" (here your a =sqrt(16-h) and b = 4), something great should happen.

You can try it out. Or if you are still not sure what's happening here, take a look at this:
https://youtu.be/OmaFy0NKTss?si=Uf0kd3AgNXZXMXMm and come back to me if you still need help.

Sorry, wrong link

https://youtu.be/OmaFy0NKTss?si=Uf0kd3AgNXZXMXMm

Now that's the not advertisement, but the actual video.

[radiant helm]

LOL

thanks

wait so

wat they want me to do is to get rid of the square root in numerator, so by applying what we know w difference os squares, we multiply by the conjugate term?

[soft burrow]

Yes, exactly.

[radiant helm]

and we also have to multiply it to the deno,mimator as well?

[soft burrow]

Correct, otherwise we would be changing the value of the expression.

[radiant helm]

okay

tysm

i think i got it

i have another question

do yk rate of change

[soft burrow]

Sure. Ask away and we'll see if I can help.

[radiant helm]

well its not rlly like a solving problem lol

im just confused what to do here

ive nevre seen it set yup this way

[soft burrow]

This looks like a prequel to derivatives.

So, is the a part clear, or that's exactly the confusing part?

That was my sketch drawing of the first row of P Q and PQ-slope.

So the values of P = (2,f(2)), Q = (3,f(3)) which are P=(2,8) and Q=(3,27).

[radiant helm]

mb i was afk

[radiant helm]

uhhh

im in precalc 😭

[radiant helm]

im confused what to fill in the brackets, do i plug the x balue into f(x)?? and im just overall confused on the instructructions for b-d

[soft burrow]

What you put in the first row is P=(2,8) and Q=(3,27).

And for PQ, you calculate how much y-cordinate changes compared to x. So in this case (27-8) / (3-2) = 19.

For other rows in a, you do the same, but you punch in different y-cordinate for Q. The P stays the same though the exercise.

[radiant helm]

OHHHHH

is this part a or

[radiant helm]

okay !

[soft burrow]

Yes, the "slope of PQ" section.

[radiant helm]

okay so if i do all that then what am i supposed to do for b-d?

[soft burrow]

The part b is where you compute/estimate the derivative of f(x) = x^3 at point (x=2) we just don't call it that, to avoid math anxiety. 🙂

[radiant helm]

um

does derivative mean instantaneous rate of change

like at 2??

[soft burrow]

On the columns of PQ slope, you should get left-hand-side some numbers approaching from 19 to something close to 16 (but a bit more). And on right hand side you should get numbers growing from 7 towards 16 (but staying below).

[soft burrow]

Yes, exactly!

Sorry, towards 12, not 16.

[radiant helm]

oh ok

wait so

if we're just finding iroc at 2.. why is there 3 steps............

[soft burrow]

Well basically to give you the sense that this number grows from 7 to 12 on one side and shrinks from 19 to 12 on the other side.

However, the scary thing: if you try to calculate this the same way at exactly 2, you compute zero divided by zero, which is undefined.

[radiant helm]

wdym by the second part?

[soft burrow]

So if you make P=(2,3) and Q=(2,3) and compute the slope PQ now, you get (3-3)/(2-2), which is an error/undefined.

[radiant helm]

hmm why is that

[soft burrow]

Why division by zero is undefined? Because the computation is absurd. Like a question of "if you have zero cookies and you split those with zero people, how many cookies each of them gets?" You can't ask that because there was no-one there.

[radiant helm]

lmao you're funny

okay im gonna do that question later than, hopefully i get the right answer lol

[soft burrow]

Even the slightly easier question of "having 5 cookies and splitting that with zero people" is impossible. You could argue that it's infinitely many cookies, but even then it depends if we have positive zero people or negative zero people.

What mathematicians do in this situation (this is not precalculus anymore), is compute the limit by making marking the distance between P and Q with something like h and then compute the slope when "sliding" h to zero. That's the derivative.

[radiant helm]

very interesting way to see this

[radiant helm]

woah

[soft burrow]

I.e P=2,3 and Q=(2+h, f(2+h)), so slope PQ = ((2+h)^3 - 2^3 ) / (2+h-2) =
(2^3 + 34h+32h^2 + h^3 - 2^3) / h
= 12 + 6h + h^2, which is 12 when h is close to zero.

[radiant helm]

ohhhhh

thanks for explaining =0

[soft burrow]

Actually your c&d question asks to compute what I just computed.

You could take it further though. The P doesn't have to be at 2. You can use the same math with keeping the x coordinate of P at x (and computing this with x and x+h). The result would give you the immediate slope at any point of that function.

I think I'll head to bed now.
Thank you for sharing these problems.

.close

[midnight plank]

Channel closed

Closed by @soft burrow

Use .reopen if this was a mistake.

[radiant helm]

oh okay

[radiant helm]

thank you so much for helping and explainingg

sleep well !

[midnight plank]

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[livid crystal]

Hello,
Where did I go wrong in factoring this? Im confused about the negative sign :( do I ignore it, do I apply it only to one number, does it make all of the numbers negative? Idk bro

[olive matrix]

18(2-x) is right

[livid crystal]

The answer sheet says -18(x-2)

[olive matrix]

oh yeah those are equal

[livid crystal]

Oh

[olive matrix]

sometimes it's helpful to have x with a positive sign

uh to see what you could do you could look at 36 as - (-36)

[livid crystal]

Also they used double negatives ?

In the answers

Oh ok

[olive matrix]

it's also just really useful to know that $\gsq - \psq = -(\psq - \gsq)$

[grand pond]

hayley!

[olive matrix]

like that shows up all the time

[livid crystal]

.close

[midnight plank]

Channel closed

Closed by @livid crystal

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[midnight plank]

Please don't occupy multiple help channels.

[radiant helm]

for identities, if i had to simplify this side, does the 3 belong to tan^2x only? like if i combined tan^2x and -1, does it get distributed to the cos^2x that was originally in the numerator as well?

[last slate]

No.

[radiant helm]

i thought you dont

so

i didnt distribute, but i didnt end up getting the correct answer

[last slate]

Let me check

Wait

It is distribution

[radiant helm]

yhuhhhhh

hold on

[last slate]

It's essentially a plug in

[radiant helm]

can you just check my work lol

[last slate]

Think of tan as x

[radiant helm]

yea?

[last slate]

And the identity you plug in is what x equals

[last slate]

What the heck

[radiant helm]

😭

sorry its a little messy

[last slate]

Where did you get the 4sin^2 theta from

[last slate]

yeah I'm lost sorry 😭

[radiant helm]

😭

is there