help-0

[floral ermine]

is it

literally just the exact same

[violet urchin]

https://www.khanacademy.org/math/geometry/hs-geo-analytic-geometry/hs-geo-distance-and-midpoints/a/distance-formula

teaching new concepts is a little hard without a whiteboard to point at

[floral ermine]

true

[violet urchin]

so i'm going to have to let this teach you about distance formulas

sorry

[floral ermine]

no problemo

thats all for that section tho

well

iahve 1 section left

are you good with quadratic functions

[violet urchin]

im good with high school math yeah

why?

[violet urchin]

alright so do you sort of understand how distance works in the 2d plane

[floral ermine]

yeah'

[floral ermine]

see beacuse

here

i understand everything

except

the first one

i dont quite know how strecthing and shrinking works

[violet urchin]

oh ok

so basically

if we have a function f(x)

and that can be any function

[floral ermine]

mhm

[violet urchin]

then 2f(x) stretches it vertically

and f(2x) stretches it horizontally

[floral ermine]

like

i know how that works but i dont know

how to stretch or shrink

idk how to explain

[violet urchin]

uh

if you stretch it vertically

then it gets closer or farther from the x axis

[daring rover]

It becomes steeper i guess thats an okay explanation

[violet urchin]

if you stretch horizontally it gets closer or farther from the y axis

[floral ermine]

ohh

easy peasy

by the way if you have no time let me know

i dontwanna take up all your time

[violet urchin]

no its fine

i can study for midterms and do this 👍

[floral ermine]

awesome

[violet urchin]

well ok

is the coefficient of x^2 going to be positive or negative

[daring rover]

Literally me procrastinating my midterms by doing lower level math

[floral ermine]

💀

[floral ermine]

positive

[violet urchin]

right

so is the parabola facing up or down

[daring rover]

It scratches the productivity itch while still being useless to what i gotta do

[floral ermine]

up

[violet urchin]

bro thats so real i should be focusing on quotient groups not alg 2

but its whatever

[violet urchin]

exactly

so will the vertex be the max or min

[floral ermine]

alg 4*

[floral ermine]

min

[violet urchin]

right

and what is the y coordinate of the vertex?

[floral ermine]

I

what

[violet urchin]

what is the vertex

[floral ermine]

I don't know

[violet urchin]

look closely, this is still vertex form

its just that they made the coefficient look a little different

[floral ermine]

is it

-20

[violet urchin]

its just $\frac{1}{810}(x-20)^2+32$

[ocean seal]

大野雄大 👻

[floral ermine]

yeah

so -20

[violet urchin]

the vertex should be a point

[floral ermine]

OH

[violet urchin]

with an x and y coordinate

[floral ermine]

1, -20

holy shit my brain

[violet urchin]

?

a(x-h)^2+k

the vertex is h,k right

[floral ermine]

yeah

[violet urchin]

so what is h here

[violet urchin]

in this

[floral ermine]

oh im so fucking stupid

20, 32

is that the vertex

[violet urchin]

you arent

[floral ermine]

its just the fraction that confused me 😭

[violet urchin]

youre just learning

[violet urchin]

yeah

[floral ermine]

thanks ❣️

[violet urchin]

yes

so if the parabola is facing up

what is the minimum

[floral ermine]

my question is

its 32 but

why cant it be 20

[violet urchin]

well

that's the x value right

[floral ermine]

yeah

[violet urchin]

not the f(x) value

[floral ermine]

oh

f(x) = y ?

[violet urchin]

and we want the minimum f(x) valuye

[violet urchin]

yeah basically

[floral ermine]

oh so in these questions

i just

always take the y

awesome

i get it

oh wait no

hollon

nah dont answer this

shit is easy

[violet urchin]

wow

speedrunning fr

[floral ermine]

fr

[daring rover]

Since the left part is always non negative because it's squared. We want it to be as small as possible - What is the smallest non negative value?

Oh messages didnt load

[violet urchin]

those are right

[daring rover]

Yall are done lol

[violet urchin]

he did that

[daring rover]

Nice

[floral ermine]

😭

[violet urchin]

lol

[floral ermine]

32

[daring rover]

Yea yea i didnt load

[floral ermine]

but yeah

[daring rover]

Mb

[floral ermine]

macdonalds wifi

its ok thanks

[floral ermine]

so for this

i dont get why its

i dont quite understand

[violet urchin]

well

ok

[floral ermine]

is ti because x is squred

squared

so its srhunk

shrunk

[violet urchin]

when x=0, y=0^2

right

[floral ermine]

yeah

wait what

[violet urchin]

so then one of the points is (0,0)

[violet urchin]

y=x^2

so if x=0

y=0^2

=0

[floral ermine]

holy shit this is so easy

why is my brain wonk

so to solve this

i just pick values for x

and solve for y

[violet urchin]

yeah

but in this case we have a formula for y in terms of x

so we just plug in those values

[floral ermine]

yup

easy cheeks

last exam i had

basically

it was that

but there was a +2 in the end

so they tricked us

its basically that exact same thing

but shifted upwards

[violet urchin]

yup

[floral ermine]

oh

HERE

example

[violet urchin]

oh yeah

thats a little trickier since its shifted

but its roughly the same

[floral ermine]

its

pretty much teh same

yeah

okay

easy

shifted to the right one

and then the x^2

wait hold on

i know why its shifted to the right

but why is it x^2

[violet urchin]

?

[floral ermine]

how did we get that graph

[violet urchin]

oh

[floral ermine]

other than the fact its shifted right horizontally

[violet urchin]

ok so f(x)=x^2

and then f(x-1)=(x-1)^2

and f(x-1) is f(x) shifted to the right one

[floral ermine]

yeah

i get the shifted to the right

but why are the points plotted as if x is squared

[violet urchin]

because it's (x-1)^2?

its still the ^2 function

but with a different input

[floral ermine]

oh

right

okay

[violet urchin]

if that makes any sense

[floral ermine]

yeah it does

the rest of the questios

are basically teh same

now

i have 2 questions left

theyre tricky

[violet urchin]

well

we want to find the vertex right

[floral ermine]

yes

[violet urchin]

and the vertex is where (1/3 x -5)=0

[floral ermine]

first can i put 1 over 3 outside

[violet urchin]

that works yes

and then we get what

[floral ermine]

we get

1 over 3 |x+5 over 1 over 3|^2 + 14

if that makes sense

[violet urchin]

are you sure?

are we dividing by 3 or are we dividing by 1/3

[floral ermine]

1/3

i said 5 over 1 over 3 😭

[violet urchin]

oh sorry

[floral ermine]

no worries

[violet urchin]

its kinda hard to read if you type it like that

but yeah

youre right

so where is the vertex

if it's (x-5/(1/3))

[floral ermine]

it's

5/1/3

5 over 1 over 3 if that makes sense

[violet urchin]

right

what is that equal to though?

[floral ermine]

i just did it

on the calculator

15

oh

its

that easy?

[violet urchin]

yup

pretty much

[floral ermine]

last question

atleast

admit this one is hard

please

[violet urchin]

show it

[violet urchin]

well

its certainly a little trickier than the others

[floral ermine]

yes

[violet urchin]

lets start by using all of the information we have

we know that this is in the form y=a(x-h)^2+k right

what are h and k

(hint: we know the vertex)

[floral ermine]

1

1 and 9

[violet urchin]

right

so we have a(x-1)^2+9

what do you think we should do now?

[floral ermine]

hmmm

plot the points

-2, and 0 and 4, 0

[violet urchin]

maybe thats next

what do you think our first step right now should be

[floral ermine]

expand? idk

find a?

[violet urchin]

exactly!

[floral ermine]

find a

[violet urchin]

expand

[floral ermine]

oh expand

yup

[violet urchin]

so what do we get when we expand it

[floral ermine]

we get

a(x^2-2x+1)+9

right?

[violet urchin]

sorry for strandin ya

internet went out for a bit

[floral ermine]

no problem

oh lol

[violet urchin]

anyways now we should put in the points we were given right?

[floral ermine]

yeah

first -2 and 0

[violet urchin]

what does that get us

[floral ermine]

first one is

0 = a(4+4+1)+9

[violet urchin]

yup

[floral ermine]

second is

[violet urchin]

so now we can solve for a

[floral ermine]

oh o0kat

let me do that r

rq

first one is

-1 = a

a is -1 in the first one

[violet urchin]

well

a should be the same in both right?

since its the same quadratic

[floral ermine]

i dont know really

it should be

[violet urchin]

well

[floral ermine]

oh oaky

[violet urchin]

they should match the same quadratic

[floral ermine]

can i do

the 2nd one

for practice

[violet urchin]

so if they dont both work

[violet urchin]

sure

[floral ermine]

alr

one sec

actualy no

nvm

so -1

now i just

solve

it

without extended or anything

[violet urchin]

well

[floral ermine]

just y = -1|x-1| +9

right

[violet urchin]

we want the x intercept

right

or sorry

y intercept

[floral ermine]

y = -1|x-1|^2 +9

[violet urchin]

and the y intercept is at what x value

[floral ermine]

wdym by tjhat

[violet urchin]

*(x-1)^2

[violet urchin]

well

[floral ermine]

yeah

[violet urchin]

it should look like (0,b)

right

[floral ermine]

yeah

so

[violet urchin]

so what is the x value

[floral ermine]

b = -1|0-1|+9

[violet urchin]

*^2

[floral ermine]

^2

yes

im so stupid

so id have

[violet urchin]

no you arent

this is all just a matter of practice

[floral ermine]

makin me blush fr

im not gay dw

bromance

[violet urchin]

💀

anyways

what do we get for b

if its -1(0-1)^2+9

[floral ermine]

9

i got

-1+1+9

= b

[violet urchin]

the -1 should be multiplied by the 1

right

[floral ermine]

OH YEAH

OH

now i get it

so expand, find a then just plot the points to find b

EASY

i'm done with mathematics now

tysm omg i love you

can i add you

[violet urchin]

mhm

[violet urchin]

sure ig

[floral ermine]

thankss

[violet urchin]

i dont really respond to dms but if you need math help again just @ me

[floral ermine]

awesome

tysm <3

.close

[lone heart]

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[lunar mica]

why is this plus

if

its - in the middle

for quotient rule

[violet urchin]

i actually can't provide an explanation for that, that seems like a simple error

not a typo though, since it stays the same on the next line...

[golden canyon]

Doesn't it change on the next line?

[tawny condor]

Yeah it does change

so final answer is still correct

[violet urchin]

oh it does

im stupid

yeah i guess its a typo

[lunar mica]

yeah ok teacher made a mistake

ty

.close

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[prime seal]

find all X for which

[prime seal]

holds true

is my solution correct?

and yea i should have wrote the solution as $\begin{pmatrix} a&b\c&d\e&f \end{pmatirx} $

[ocean seal]

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

[prime seal]

this is solution from textbook

[lone heart]

@prime seal Has your question been resolved?

[lone heart]

@prime seal Has your question been resolved?

[fickle heath]

Your last augmented matrix shows d=1

[lone heart]

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[vernal abyss]

Is it possible to find the decimal values of 7 using this limit, $\lim_{x\rightarrow 7....}x^2=51$? Or is there better way to approach this problem?

[ocean seal]

Bennxy

[exotic canopy]

what?

what are you trying to do

sqrt(51) is the solution

[vernal abyss]

O my days

I think I should rest a bit

[exotic canopy]

lol

[vernal abyss]

Thanks

Mate

[exotic canopy]

no problem

[vernal abyss]

.close

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[rose night]

Can i get help on a questipon here?

[shrewd carbon]

uh yes

[rose night]

i have this question here

[shrewd carbon]

Ok, type .close and retype the question

[rose night]

.close

[lone heart]

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[plain carbon]

Can someone help me with these questions

[shrewd carbon]

Oh god what's going on

[slow haven]

Ah make sure you read #❓how-to-get-help, somebody else is using this channel for help!

It says #help-0 so this is for the user mxp 🙂

[shrewd carbon]

@rose night

Create two cases.

First is the case where 9-2x is positive

the other case is where it's negative

You will end up with two linear equations you have to solve

The first is $(9-2x)<1$

[ocean seal]

diaas_(yt)

diaas_(yt)

[shrewd carbon]

Let me solve them for you

In the first equation you get x>4

In the second, you get x<5

So the answer will be the (intersection) of these

Answer is $4<x<5$

[ocean seal]

diaas_(yt)

[shrewd carbon]

broifimwrongimgonnabesomad

[rose night]

The answere is correct

but i don't udnerstand the working itself

[shrewd carbon]

See bro

Consider |f(x)|<3

[lone heart]

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[rose night]

okay thank you

[rose night]

.close

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[split magnet]

Not sure if this is how it works but can anyone explain how to find the domain and range of a piecewise function?

[mortal trellis]

do it for each piece separately

[split magnet]

.close

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[lone dagger]

why is the solution in this system of equations written with alpha and beta signs?

[vale crag]

that's because there's more than one solution to the system

[exotic canopy]

because there is no 1 solution, alpha and beta are parameters

[vale crag]

you put in any alpha beta you want, you get a solution

[exotic canopy]

if you choose any value for alpha and beta, you get a solution

[exotic canopy]

lol

[thick lynx]

Generally, there will be only one solution for n equations with n variables.
Here, you have 4 variables and only 2 equations

[lone dagger]

oh

aight

thank you

.close

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[grim compass]

if 3^(n+1) +3^(n+2) = a, then 3^(n+3)=? n>0

[exotic canopy]

hint: factor out 3^(n+1)

[grim compass]

cant get a right answer still 😦

[exotic canopy]

you got $3^{n+1}(1+3)=a$, right?

[ocean seal]

artemetra

[grim compass]

yep

[exotic canopy]

this simplifies to $3^{n+1} = \frac{a}{4}$

[ocean seal]

artemetra

[grim compass]

yep

[exotic canopy]

here's the key idea

3^(n+3) = 3*3* 3^(n+1)

do you see it?

[grim compass]

yep

2,25a

thanks so much

[exotic canopy]

np

[grim compass]

c.lose

.close

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[alpine sable]

if (5x^4-1) * x^3- (x^5-x) * 3x^2
then the - in the middle belongs to the 3x^2 also and must be used to multiply with the parenthesis right
so not - (3x^2(x^5-x)) but rather (-3x^2(x^5-x))

[vale wigeon]

• (3x^2(x^5-x)) but rather (-3x^2(x^5-x))
  it doesn't make a difference.

$-(ab) = (-a)b = (-ab) = (-1) \cdot a \cdot b$ these are all the same

[ocean seal]

Ann

[hidden pine]

I don't know how to get i and -i to get 0

[alpine sable]

it did when I tried it because one made it into ... - 3x^7 -3x^4 whereas the other did -3x^7+3x^4

one was -3x^4 the other was +3x^4

[vale wigeon]

!occupied

[lone heart]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[vale wigeon]

-(3x^2(x^5 - x)) = -(3x^7 - 3x^3) = -3x^7 + 3x^3 though.

you would not get -3x^7 - 3x^3.

[alpine sable]

yeah I think I did some mixture of multiplying by 3x^2 but then not doing so with -1

parenthesis work can get confusing

do you have any tips on when to place parenthesis to avoid stupid mistakes like these?

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[smoky needle]

hello could anyone explain why the reflected vector about a normal can be calculated with this formula

v - 2.0 * (v DOT n) * n

[smoky needle]

I see lots of proofs but I do not quite understand because of the dot product involved

I do not quite see how the dot product works in helping this formula

here is a diagram I found just to show what the variables mean

v - 2.0 * (v DOT n) * n

[vale crag]

before talking about reflections

you know how projections work with the dot product ?

[smoky needle]

no, thats the part I am struggling with

[vale crag]

right ok

so you understand that the projection on u is the vector in the direction of v, closest to u right

[smoky needle]

not really, What do you mean by closest to u, I understand that it is parallel to v and is it the projection of u Onto V we are referring to here?

[vale crag]

no, u isn't parallel to v here

[smoky needle]

nono but the projection of it is right?

[vale crag]

yea

[vale crag]

closest in the usual (euclidean) distance sense

[smoky needle]

yes, as in the length of it right?

[vale crag]

as in the length of their difference

length (u - u_proj) is minimal

[smoky needle]

ok yes

[vale crag]

so yeah we can restate that problem, as minimizing <u-u_proj, u-u_proj>

<u-u_proj, u-u_proj> is just the square of their length

it doesn't matter if we minimize the length itself or its square

[smoky needle]

ok

[vale crag]

and we want u_proj to be in the direction of v

so u_proj = kv for some k

we wanna find that k

[smoky needle]

yes where k is a scalar right?

[vale crag]

yes

so <u-u_proj, u-u_proj> = <u-kv, u-kv>

you can expand that product and differentiate wrt to k to find the k that minimizes the square length

[smoky needle]

yes, but what do we need the k value for once we find it

[vale crag]

and you find that k = <u, v>/<v, v>

well if we know k, we know what the projection is

u_proj = <u, v>/<v, v> * v

[smoky needle]

yes

[vale crag]

now you see how projections relate to the dot product don't you

[smoky needle]

the length of the projection multiplied by the length of the other vector is equal to the dot product right?

[vale crag]

unclear what you're saying here

[smoky needle]

sorry, length of u_proj * length of v = u DOT v

[vale crag]

u_proj is a vector here

it's a multiple of v

[smoky needle]

oh

[vale crag]

<u,v>/<v,v> is a scalar

well I've literally shown you how to compute the projection itself (the vector) using dot products

if that isn't a relation, idk what is

[vale crag]

but hey that's also true I think

[smoky needle]

yes I understand the relation you showed, but I meant the implication of the relation

[vale crag]

yea

[smoky needle]

What do we now use the projection for?

[vale crag]

you mean in general or for reflections ?

[smoky needle]

for reflections, but if you know what it is for in general that would help too

[vale crag]

well with that formula we get that u_perp = u - u_proj = u - <u, v>/<v, v> v

and in that context u_refl = u - 2u_perp

you get everything for free from the projection

@smoky needle

[smoky needle]

I see it now

[vale crag]

and in general, well the one big thing you can relate to projections and orthogonality is fourier series

but that's maybe a bit too advanced for you

that's the one application of that that I find somewhat interesting

[smoky needle]

ok

[vale crag]

other questions? @smoky needle

[smoky needle]

not at the moment, thank you have a great day!

.close

[lone heart]

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[sage herald]

would 0.4140 be correct? its my final check & i rlly dont wanna get this wrong

[ivory pivot]

try to get it wrong and see what happens

[grizzled gulch]

Bruh

[lone heart]

@sage herald Has your question been resolved?

[sage herald]

nvm i figured it out

so helpful….

.close

[lone heart]

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[grizzled gulch]

Sry i dont know how to solve that

[lone heart]

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[frail grove]

Why can't we do this xd

[vale crag]

why tf is everybody bringing this today

did someone post a video or something?

[frail grove]

Saw it on yt shorts

[echo socket]

Because it's wrong Pretty sure you can't just let
[ \dv{}{x}{\sum_{i=1}^xf(x)} = \sum_{i=1}^x\dv{f}{x} ]

[ocean seal]

A Lonely Bean

[worn fox]

Try it with x = 1.5

[echo socket]

Because there's x as the upper bound for i

[frail grove]

Fair enough

Yeah it's a dumb idea anyway

.close

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[dawn crest]

hello! Im a bit rusty and struggling with this series. I was trying to apply the absolute convergence theorem but didnt get anywhere

[dawn crest]

its probably dumb but im struggling with this limit

[ivory pivot]

hai il visetto angelico ?

[dawn crest]

i got jumpscared because i forgot what my bio was

[ivory pivot]

ahah however maybe you can use that th sum is alternating ?

since $\sin((2n+1)\pi/2)=\pm1$

[ocean seal]

everg

[ivory pivot]

so you only need to prove that $\frac{\ln^4 n}{e^n(n-1)}\to 0$ monotonically.

[ocean seal]

everg

[lone heart]

@dawn crest Has your question been resolved?

[lone heart]

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[gritty pond]

Can someone explain this solution?

[ivory pivot]

.close

[lone heart]

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[ivory pivot]

sorry

[gritty pond]

The heck

[ivory pivot]

by bad..

[gritty pond]

Hey, do u know this problem?

[ivory pivot]

yes

[gritty pond]

My only confusion is how we get to area(S1) formula

Idk why we get that starting formula

from the graphs

.reopen

[lone heart]

✅

[ivory pivot]

mh...

interesting

[ivory pivot]

if you compute this one without 1/1-x2-y2 then you get the area of D ...the rectangle

[gritty pond]

How do u know that?

[ivory pivot]

who do i know that $\int \int_D 1 dA=area (D)$?

[ocean seal]

everg

[gritty pond]

Oh, it's just an integral of 1? Which is

Yeah

Sure

[ivory pivot]

ok ..now your surface is a graph ..

its like (x,y)->(x,y,z(x,y))

[gritty pond]

Sure

[ivory pivot]

in your case f=1 because you want area(S)

[gritty pond]

Yes that looks like the right formula

[ivory pivot]

r(x,y)=(x,y,z(x,y))

[gritty pond]

But how do I do thatr when there are 3 variables?

don't i need to change it to 2 variable then?

[ivory pivot]

so that one

could be this one

[gritty pond]

how

how to get to that step?

[ivory pivot]

because $d r(x,y)/dx=(1,0, \frac{x}{\sqrt{1-x^2-y^2}})$

[ocean seal]

everg

[ivory pivot]

and $d r(x,y)/dy=(0,1, \frac{y}{\sqrt{1-x^2-y^2}})$

[ocean seal]

everg

[gritty pond]

Whaa

What are the original graphs

without deriving first

[gritty pond]

I meant what is the original vector before taking derivative*

to get the 2 components?

[ivory pivot]

$r(x,y)=(x,y,z(x,y))=(x,y,\sqrt{1-x^2-y^2})$

[ocean seal]

everg

[gritty pond]

How did we know to do that from the original z equation?

Is it just a formula?

[ivory pivot]

because you can change x and y as you want then z is fixed

so this is the graph of $\sqrt{1-x^2-y^2})$

[ocean seal]

everg

[gritty pond]

how to visualize?

i typed the same thing in to desmos 3d and it didnt work

[ivory pivot]

its just the upper sphere ..naturally you have to set constraint to x and y

[gritty pond]

why is urs different

[ivory pivot]

its pretty the same

[gritty pond]

mine have negative

has*

yours dont

[ivory pivot]

since d -x^2=-2x

by bad

[gritty pond]

Whisch one is correct?

[ivory pivot]

yours..but its not important because you have to do compute the norm of the vectorrial product

so minus signs will vanish

[gritty pond]

i did the computation but i didnt get the same as the problem

here is my input for wolfram alpha cross product

my result:

Then I did magnitude calculator

my result:

but, expected value is

[lone heart]

@gritty pond Has your question been resolved?

[lone heart]

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[dreamy swallow]

Hi, this is not something I got for homework. I just want to know this purely out of my own curiosity. Our teacher said we haven't learned the stuff necessary to solve question 38.2.1, but I am curious as to how you would you solve this. Is there anyone that can maybe just guide me with this question?

[granite pilot]

You can realize that the point P for any x looks like (x, f(x)) and the point Q looks like (x, g(x)), so PQ is just the distance between those points (for the same x!!!) so is just f(x) - g(x)

Note this only works because they're saying PQ || y - axis i.e. you're using the same x for both

[dreamy swallow]

Ohh, I see! Thanks for the help

[granite pilot]

no worries!

[dreamy swallow]

.close

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[distant citrus]

How do i get Wn = … in my ti 84

Cause i cant generate table, how i did it

[lone heart]

@distant citrus Has your question been resolved?

[distant citrus]

Hello??

.close

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[teal magnet]

i need help with the last one im not sure how to lay it out

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@teal magnet Has your question been resolved?

[teal magnet]

.close

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[real gazelle]

.close

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[celest briar]

Can anyone help me with these please 🙂

[pseudo ice]

Do you have any ideas of how to start?

[celest briar]

I know something about setting (1+x)^n = (The Summaation) then I believe it ends up going from 2^r --> x^r ? But not sure why for any of it

[pseudo ice]

Well do you know how to work with (1 + x)^n in general?

[celest briar]

Honestly, not really, I remember doing it in previous years for normal binomial expansion but honestly mostly just remember seeing its form not much else

[pseudo ice]

Fair, well if you've seen normal binomial expansion you could make use of that, notably you should have seen that $(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k$

[ocean seal]

@pseudo ice

[pseudo ice]

Basically making use of that, it should hopefully be easy to compare the sums you have and all

[celest briar]

Okay give me a second to digest and remember latex so I can write my first line

[celest briar]

$(1+x)^n=\sum_{r=0}^n\binom{n}{r}2^r$

[ocean seal]

CoshamGames

[celest briar]

Would this be my first line ?

[pseudo ice]

You could do it like that if you'd like

[celest briar]

Okay so from here where do I go ?

I assume i turn 2^r --> x^r?

$(1+x)^n=\sum_{r=0}^n\binom{n}{r}x^r$

[ocean seal]

CoshamGames

[pseudo ice]

That's a fair idea

[celest briar]

Now my next intuition says differentiate but idk why that's what im telling myself

[pseudo ice]

The idea is that you should compare this sum here to this, and think about what choice of x would turn that general formula into this

[pseudo ice]

Not for this one you don't need to

[celest briar]

Honestly, I'm baffled

[mild quail]

Yeah I’m kinda baffled as well

[celest briar]

Following along? XD

[mild quail]

Yeah 😂

I’m kidding I have absolutely no clue what I’m looking at I’m just a kid in high school 😭😭

[pseudo ice]

tl;dr replace x with 2 here, that gets you the sum they're asking you for, right?

[celest briar]

I think so

[pseudo ice]

Cool, but then you have this here, so if you replace all the x's with 2's, what does that become?

[celest briar]

$(1+2)^n=\sum_{r=0}^n\binom{n}{r}2^r$

[ocean seal]

CoshamGames

[celest briar]

$(3)^n=\sum_{r=0}^n\binom{n}{r}2^r$

[ocean seal]

CoshamGames

[pseudo ice]

There you go

[celest briar]

Okay cool, so then where does the calculus come in because the question says "By using calculus on (1+x)^n and setting x equal to an appropriate number"?

[pseudo ice]

The calculus comes in for the next one

For the (b) you don't need any calculus really, but for the other one you would

[celest briar]

so $(3)^n$ is the final answer for b?

[ocean seal]

CoshamGames

[pseudo ice]

Yep, 3^n is the answer for it

[celest briar]

Surely it can't be that easy

[pseudo ice]

For this one it is

The suffering is with the next one

[celest briar]

oh god

xD

$(1+x)^n=\sum_{r=0}^n\binom{n}{r}r^2$

[ocean seal]

CoshamGames

[celest briar]

First step again ?

[pseudo ice]

Well this one is a bit of thinking

[celest briar]

uh oh

[pseudo ice]

Also this isn't the first step at all

From the binomial $(1 + x)^n = \sum_{r = 0}^n \binom{n}{r} x^r$, you want to somehow make something like $\sum_{r=0}^{n} \binom{n}{r} r^2$ show up

[ocean seal]

@pseudo ice

[pseudo ice]

This is where calculus on $(1 + x)^n = \sum_{r = 0}^n \binom{n}{r} x^r$ shows up \catthink

[ocean seal]

@pseudo ice

[celest briar]

so why do we go from r^2 --> x^r ?

[pseudo ice]

We don't here...

[pseudo ice]

try differentiating this a few times maybe

[celest briar]

this is really making no sense

[pallid scarab]

Maybe twice ;)

[celest briar]

im very lost

I'm stil baffled on how the hell the first one is so simple

[pseudo ice]

that one is really just taking note of the binomial expansion and comparing it to what you have, it's really not too bad at all(!)

[celest briar]

maybe I don't understand the point in the question at all

what are we finding

how is $(3)^n$ useful at all

[ocean seal]

CoshamGames

[pseudo ice]

You want to find what the value of the sum is, what we've found is that the sum here evaluates to 3^n

You want to find out what each of the sums end up turning into

[lone heart]

@celest briar Has your question been resolved?

[lone heart]

@celest briar Has your question been resolved?

[lone heart]

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[final parrot]

How to do 53?

Pre calc 12 conics

[final parrot]

can someone help me

@mellow saffron

<@&286206848099549185>

[lone heart]

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.

[vapid shuttle]

don't troll

[final parrot]

<@&286206848099549185>

[near chasm]

Basicly this

[vapid shuttle]

Did you not read the previous message? Only ping the helpers once

[final parrot]

you told me to do it after 15 minutes

so i did lmao

@vapid shuttle

[vapid shuttle]

Don't ping me for no reason. You pinged them too early, you're supposed to wait 15 minutes. Additionally, it is supposed to only be once, so pinging them again is just being unnecessarily difficult.

[final parrot]

What u gonna do

@near chasm thanks

[near chasm]

Np but #rules pls

[final parrot]

.close

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[final parrot]

how to do 61

pre calc 12- conics

[tacit arch]

!noclopen

[lone heart]

Please don't repeatedly close and claim a new channel with the exact same question. This erases all previous progress made towards your problem and is confusing for helpers, making it more difficult to help you. Please be patient, even if your channel has not received much attention.

[final parrot]

bro its not the same quesiton

i meant it is but i was using 2 channels at the same time so i closed this one

now im reopening cuz the other channel got answered

[tacit arch]

!1c

[lone heart]

Please stick to your channel.

[final parrot]

i finished my channel

do i have to use a diff channel or something?

.close

[lone heart]

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[lone heart]

Please don't occupy multiple help channels.

[alpine sable]

ok so first i made 4 into one

4^3/2 ( 1 + 1/4 * x ) ^ 1/3

but idk how to find set of values of x

plz help

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

WHY IS NO ONE HELPING ME

<@&268886789983436800> hi can u ping some helpers?

[tacit arch]

!patience

[lone heart]

Please wait patiently, and do not interrupt other channels with your question. Helpers in this server are volunteers, and the server cannot guarantee that someone will be able to help you. By being impatient or begging, you will only turn potential helpers away.

In the meantime, please make sure your channel contains the original question, clearly describes what you have already tried, and states exactly what you are having trouble with. This increases your chances of getting a good response.

[sly mantle]

you can ping helpers yourself

[alpine sable]

oh last time i did... i got timed out

[rapid thistle]

Pinging helpers once after 15 minutes will not get you timed out

[sly mantle]

no you were timed out for being rude to a helper

[rapid thistle]

^

[alpine sable]

oh ok

[alpine sable]

ive been very patient for last 20min...

[sly mantle]

go ahead and ping helpers. just know it does NOT guarantee someone will help. the server is run by volunteers

[tacit arch]

Doesn't matter

!volunteers

[lone heart]

Helpers are just people volunteering their time to help you. Be polite and patient.

[alpine sable]

ok thanks

[alpine sable]

could u give me helping hands?

🫶

[tacit arch]

No, stop pinging me

[alpine sable]

<@&286206848099549185>

can someone help me with the question?

[alpine sable]

can i make another channel if they dont reply?

i think people dont care about me cuz im at the bottom

[sly mantle]

no

[fringe inlet]

Sure!

[sly mantle]

helpers can see every channel whenever theyre available to help. plus youre near the top

[fringe inlet]

Do you know binomial theorem?

[alpine sable]

oh i see

[alpine sable]

yh

[alpine sable]

FINALLY ❤️

[alpine sable]

ok thanks

[alpine sable]

4^3/2 ( 1 + 1/4 * x ) ^ 1/3

I can pull out a factor like this to make it 4 into 1

and then use 1 + nx + n(n-1)x^2 / 2

to solve it

[fringe inlet]

Ye basically you can do that

That'll work I think

[alpine sable]

but how do i find the set of x?

[fringe inlet]

?

Wdym the set of x

[alpine sable]

set of values of x

its on the question

[fringe inlet]

I'm assuming that's just asking for domain?

[alpine sable]

i think so

like

< x <

[fringe inlet]

So greater than or equal to -4?

[alpine sable]

yh thats the answer

how did u get that

[fringe inlet]

There's a 3/2 power

So there is a cube then square root

So 4+x is nonnegative

[alpine sable]

wait u can't do negative power?

-2 ^ 3 = - 8

[fringe inlet]

You can square root negative

-2 to the 3/2 is -8 square root

Which isn't defined

[alpine sable]

oh ok

can u explain why it can't be higher than 4?

[fringe inlet]

It can?

[alpine sable]

this was the answer

idk

[fringe inlet]

Ok ngl idk where the less than 4 came frome

What expression is it talking about lol

[alpine sable]

lol

this was the full mark scheme

i dont get it lol

[alpine sable]

do u get this mark scheme?

hi?

[fringe inlet]

it seems like that expression is defined over all x

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

What does that mean

[fringe inlet]

it's asking which values give a valid expression

which is basically the domain

but it seems the domain is all reals

bc it is quadratic

[alpine sable]

so how does this make x lower than 4?

[fringe inlet]

ye that's what im wondering too

[alpine sable]

hmm alr i will ask my teacher

and tell u

idk

.close

[lone heart]

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[vapid steppe]

how to do

[lone heart]

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[molten wagon]

Can eigenvalues and singular values coincide? Is there a group of matrices that have singular and eigenvalues equal besides the identity nxn?

[neat arrow]

help

[molten wagon]

I had a test yesterday about svd and more stuff. One question (true or false) was

Consider a square matrix with eigenvalues 4 5 and 6. Then, its singular values are 2, sqrt(5) and sqrt(6). Thats obviously false but it would be fun to find a matrix that eigenvalues and singular values coincide

i know the test question is not the point of this, but i was thinking about this cause that

[lone heart]

@molten wagon Has your question been resolved?

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[spare fern]

How do I integrate this?

[spare fern]

$$\int _{ }^{ }\frac{2x-1}{x^3}$$

[ocean seal]

Lex1729

[spare fern]

with respect to x

[jagged cobalt]

split the fraction

simplify

[spare fern]

$$\int _{ }^{ }\frac{2x}{x^3}-\frac{1}{x^3}$$

[ocean seal]

Lex1729

[jagged cobalt]

yeah

[spare fern]

$$\int _{ }^{ }2x^{-2}-\frac{1}{x^3}$$

[ocean seal]

Lex1729

[jagged cobalt]

sure, why not do the same to the other one

[spare fern]

$$\int _{ }^{ }2x^{-2}-x^{-3}\ dx$$

[ocean seal]

Lex1729

[jagged cobalt]

you alright from there?

[spare fern]

Yeah, I can just use power rule now

Thanks for explaining 👍

[jagged cobalt]

np

[spare fern]

Also one more question

[jagged cobalt]

sure

[spare fern]

How would I do this question?

[jagged cobalt]

0 to 1 is the first second, 1 to 2 is the second... 3 to 4 is the fourth second

youd integrate from 3 to 4

[spare fern]

But why can't I just integrate and substitute t = 4?

[jagged cobalt]

that would just give the displacement at t=4 rather than how far it travels during the 4th second

also you dont know what C would be since you dont know any conditions

[spare fern]

I'm still a bit confused though, like for this question

Would I apply the same logic, as in do I need to find the definite integral?

[jagged cobalt]

you would use an indefinite integral to get v(t), use the fact v(4)=7 to find C,
then do a definite integral on v from 0 to 9

[spare fern]

Because when I do the steps you listed, integrate, find c, etc except I don't find the definite integral towards the end - instead I just substitute t = 9 I still get the correct answer of s(9) = 83.16m

[jagged cobalt]

i believe that works solely because s and v are strictly increasing, so the displacement is the distance travelled

wouldnt always work necessarily

if v was negative at some point, then s(t) would be lower than the actual distance travelled

[spare fern]

Oh, I see, so you are saying for these types of questions I should always find the definite integral from 0 to a value of t, rather than just substituting the t value in the function.

[jagged cobalt]

yeah