help-13

[glacial warren]

these have been the ones that had me concerned

i hope thats not a bad surprised face

[past canopy]

you can just use LH rule

pretty simple

[glacial warren]

which is just taking the deriv

[past canopy]

not really

you dont use u/v rule

you can write x^2 lnx as lnx/(1/x^2)

this is an infinite/infinite form

hence u can use lh rule

[glacial warren]

so why does it become lnx all over 1/x^2

[past canopy]

1/(1/t) = t

[glacial warren]

i just have to ask because normally when i see it go to the bottom its because it has a negative exponent

[native heath]

....what

[past canopy]

?

[glacial warren]

L'Hopitals rule

we were going over it in class but i understood it a little

[native heath]

at least tell them to rewrite it first

[past canopy]

i did tell it above

[glacial warren]

i had to google it but it makes sense now

(your formula)

[past canopy]

they teach fractions in like

3rd grade dawg

u doing limits without knowing a 3rd grade thing is

nice

[glacial warren]

i was misunderstanding what he was doing

i thought he was taking the derivative and moving it to the bottom

formal notation is hard to understand across texts rather than a worksheet

also take in mind i just started back in university after not going for 4 years to school period

sorry for wasting your time.

.close

[cedar kiln]

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[north osprey]

Prove that, if the non-singular matrices A and B commute, then so do A^-1 and B^-1

[last apex]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[north osprey]

1

[last apex]

what does it mean for A,B to commute?

[north osprey]

AB = BA

[last apex]

what does it mean for A^-1,B^-1 to commute?

[north osprey]

A^-1 * B^-1 = B^-1 * A^-1

[last apex]

yes

can you see how the second follows from the first?

[north osprey]

yes

[last apex]

so thats it

gj

[north osprey]

but how do i prove it

[last apex]

you just proved it

[north osprey]

did I?

i didnt even do anything

[last apex]

you said you understand how the second line follows from the first

[north osprey]

how do i write it out

[last apex]

explain to me what did you understand

[north osprey]

i dont really know how to say it

[last apex]

try

what is the justification for AB=BA --> A^-1 * B^-1 = B^-1 * A^-1

[north osprey]

oh thats the thing im stuck on

[last apex]

you said you understand why this is true

what is (AB)^-1?

[north osprey]

B^-1 * A^-1

[last apex]

do you see it now

[north osprey]

nope

sry

[last apex]

what is (BA)^-1

[north osprey]

same as this but switched around

[last apex]

notice anything?

[north osprey]

matrices in the brackets if the whole () is inverse, switches around when the bracket is removed

[last apex]

yes, but you know something about AB,BA

[north osprey]

they are commutative

[last apex]

they are the same!

[north osprey]

OHHHH

ong how did i not see it

thank you

how do i close this

[last apex]

.close

[north osprey]

.close

[cedar kiln]

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[pallid vector]

studying for the sat right now im pretty sure its possible to this question on desmos using regression but i cant quite remember how

[pallid vector]

i know its not 37 and its only showing that cause its making x1 be 1

how can i specify that x1 is any positive integer greater than 1

[last apex]

regression is unrelated

you should convert roots to powers by fractions

i.e $\sqrt[k]{x} = x^{\frac{1}{k}}$

[clear ember]

\sqrt[k]{...}

[wraith dagger]

ExpertEsquieESQUIE

[pallid vector]

so this isnt solvable the way im doing it?

[last apex]

its just not the topic, I don't know how desmos even got to it

its a question about power rules

[clear ember]

it is

[pallid vector]

oh

[pallid vector]

do u mind showing

[clear ember]

just add x1 = [2...5]

in the subsequent line

[pallid vector]

yessss thanks so much

thats what i meantt

i was doing x1 ~ 2

[last apex]

but its better to understand how desmos arrived to this

becuase you won't have access to it during the SAT

[clear ember]

you do have access to it

[last apex]

what??

[clear ember]

yeah it's weird

[pallid vector]

desmos is like the only tool u need in sat

[clear ember]

lol

[pallid vector]

thanks a lot mmmm7 ❤️

.close

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[boreal acorn]

I have this triangle, I gotta find HM through Pythagorean theorem, which is a^2+b^2=c^2 but in my notebook it's written as 13^2 - 12^3=HM. So I was wondering why it's with a minus

[ancient lodge]

If $\angle CHM$ is a right angle, then $13$ is the hypotenuse.

[wraith dagger]

Civil Service Pigeon

[boreal acorn]

Oh that's like a rule?

Oh righttt

It's something I never made a mental note of

Thank you Pigeon

Bye take care

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[hexed vortex]

Does this seem ok?

[last apex]

its very important to mention 11/(9n-3) is montonically decreasing

[hexed vortex]

why is that

[last apex]

xD why is that

I also don't know

mb

sry

[hexed vortex]

your good

[floral arrow]

Looks good to me

[cedar kiln]

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[young flume]

If I am using the limit law on lim z->9 z-2 , would it be( lim z->9 z ) - ( lim z->9 2 ) or would the 2 still be negative

[young flume]

or does the negative account for it

f(x) - g(x)

z - 2

z - (-2)

so im guessing i dont have to put the negative?

i just wanna make sure

[ancient lodge]

$\lim_{z \to 9} (z-2)=\lim_{z \to 9} z-\lim_{z \to 9} 2$ is true

[wraith dagger]

Civil Service Pigeon

[young flume]

so the negative accounts for the -2

so just 2

alright thanks

[cedar kiln]

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[wicked mantle]

.solved

[cedar kiln]

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[long beacon]

Hi! Can anyone explain to me how the transformation ended up like the answer??
First pic is my work and the second is the answer key

[cedar kiln]

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[true mist]

Can anyone just help me figure this out?

[true mist]

Forget that one I’m on this one now

[sullen cipher]

<2+<3=90

And u hv < 3

<3*

[cedar kiln]

@true mist Has your question been resolved?

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[hexed vortex]

does this mean bounded above and below when they say bounded?

[hexed vortex]

like I heard that if a set is bounded then it has a least upper bound and a greatest lower bounded

but x > 1 is bounded but it doesnt have an upper bound?/

[dire geode]

yes bounded means bounded from below and from above

[hexed vortex]

ok thanks

[cedar kiln]

@hexed vortex Has your question been resolved?

[wicked mantle]

Available

[fathom pasture]

Help

"If A, B, C and D are any four distinct points in a directed line, show that, for all possible arrengements of these points on the line, the equality shows"

[fathom pasture]

I've been trying to do it without doing all the 4! cases. Can I say that WLOG A<B<C<D and then any arrengements will work because you can label them however you want if the arrengement changes?

[tranquil oracle]

not really since the equality changes too

how is a directed line defined btw

[fathom pasture]

What I've also thought of is combining AB+BC=AC and then AC+CD=AD Holds because all the cases were covered in the textbook, and the same for BC+CD=BD and then AB+BD=AD is true

[fathom pasture]

Like AB=-BA

you start from A and go to B

[tranquil oracle]

yeah that sort of approach should work out fine

[fathom pasture]

Still I can't wrap my mind around how that can cover all possible cases

[tranquil oracle]

just AB+BC=AC and AC+CD=AD and you are pretty much done

[tranquil oracle]

that's because the equalities you use hold in all cases

[fathom pasture]

but doing that wouldn't be like ignoring a point? for example, take BC+CD=BD then it's like ignoring C?

[tranquil oracle]

think of it this way

A, B, C, D has some arbitrary configuration

AB+BC=AC holds in this configuration

AC+CD=AD holds in this configuration

[fathom pasture]

okay

but how can you tell they cover the 24 cases? I see it this way:
AB+BC=AC covers 6 cases, AC+CD=AD covers another 6 cases, wouldn't that total 12 cases?

instead of the 24 required

or am I missing something

Sorry, I'm not getting it

[fathom pasture]

I think I get it, so basically AB+BC=AC counts the 3! permutations of A, B and C, right? and then when doing AC+CD=AD it introduces D so that D has 4 possible positions and the positions that were counted in AB+BC=AC, which makes it 4! permutations, which is all of them right? Is my reasoning correct?

so I only need to show 1 (AB+BC=AC in this case) to prove the whole problem?

[tranquil oracle]

Maybe instead of counting, think about the fact it holds regardless of configuration

[fathom pasture]

it clearly does!

[tranquil oracle]

AB+BC=AC always holds
AC+CD=AD always holds

Then just use these facts to prove the result

[fathom pasture]

Okay, since AB+BC=AC always holds for any position of A, B and C, then AC+CD=AD also holds for any position of A, C and D, then AB+BC+CD=AD always holds for any position of A, B, C and D

oh wait

lol, it makes a lot of sense written likes this

is this right?

[hexed vortex]

Do they mean cyclic groups?

this chapter is about properties of cyclic groups

or do they actually mean subgroups if so idk how I'm supposed to find every subgroup of Z_20

oh wait let me look at the end of this chapter

havent checked that yet

nvm I need to use this

[fathom pasture]

.closed

.close

[tranquil oracle]

Yeah it's a direct application

[latent bloom]

Available

[foggy furnace]

Ok so how should I find a if I don’t have an x-int? Also, is y supposed to be 1 or is it the part where it hits the y line?

Disregard the random equations above the thing

[foggy furnace]

if no one answers by 10 ill just close it and go watch a yt video or smth

[latent bloom]

a is the scaling factor for the parabola. You can find it by plugging in a point and solving for a. The point doesn't have to be an x-intercept.

Also the y-intercept is the point which the curve hits the y-axis

[foggy furnace]

huh i thought that you could find y-int by just looking at the vertex and selecting the y in it

[latent bloom]

The form you wrote is the vertex form

[foggy furnace]

yeah

[latent bloom]

It doesn't include the y-intercept

[foggy furnace]

oh

[latent bloom]

The y-intercept appears when the function is in standard form

[foggy furnace]

so is this basically just any ordered pair?

[foggy furnace]

OHHH ok yeah i keep getting them confused together

[latent bloom]

Any ordered pair aside from the vertex

[foggy furnace]

ohh ok ok

that makes sense

tysm

.close

dang bot is still broken just-

ok..

[latent bloom]

Dw we'll handle it

[wicked mantle]

@ivory goblet

[ivory goblet]

60

[wicked mantle]

Hint: Function a = Function b

[ivory goblet]

how do you know that

[wicked mantle]

Look, you want the function to be continuous

[ivory goblet]

correct

[wicked mantle]

Meaning that at x = a, the function should have the same value

[ivory goblet]

im confused what the x=a part means

and the x cant be equal a

[wicked mantle]

If it’s not continuous, the function would spilt into two parts at that specific point

[ivory goblet]

is a constant

or a point

[wicked mantle]

a is a constant

[ivory goblet]

if you set both equations equal to eachother

you js get a equation with a and x in it

that doesnt do us much good

[wicked mantle]

Do it first

[ivory goblet]

i did

8x-8a=x^2-a^2

[wicked mantle]

Alright, now replace x = a, this maneuver is called "plotting a spot into the function"

[ivory goblet]

ok

8a-8a=a^2-a^2

if you replace x with a

[wicked mantle]

Mhm

Solve for a

[ivory goblet]

0=0

[wicked mantle]

Wait, lemme check

[ivory goblet]

kk

[ivory goblet]

rip bro never got back lol

[wicked mantle]

Class 😭

[ivory goblet]

lol u good

i got it

i used ai and it actually made sense 😭

[ivory goblet]

lmfaoo

[ivory goblet]

its 10:30pm for me rn have a test the next day

[unique copper]

lemme help, for 60, you have to use the fact that the function needs to be continuous, i.e., lim x→a f(x) = f(a)

[ivory goblet]

i got it

i factored than got a equation

x+a

then i know that a point a,8 exsist

so i plugged it in

then i got a

[unique copper]

so what value of a you got?

[ivory goblet]

2

no 4

i lied

its 4

[unique copper]

yes

so what's next, is your problem solved?

[ivory goblet]

no

65

[unique copper]

you know lim x→a sina/a?

[ivory goblet]

ye

its 1

that 4x and 5x tho

not x and x

[unique copper]

you can multiply with something to make 4x

[ivory goblet]

?

idk

how to

[flat mica]

factor out 1/5 out of the limit first of all

what do you get

[unique copper]

you want 'something' in denominator so you can multiply and divide by that 'something' and seperate sin something / something and see what you get

[ivory goblet]

sin4x/x

[flat mica]

no limit?

[ivory goblet]

wym

[flat mica]

whered the lim and 1/5 go?

[ivory goblet]

ok sorry i write in better notation

lim as x apparoaches 0 from the right 1/5 * sin4x/x

[flat mica]

alright cool

so

if we substituted , let's say,

4x=u

then x->0+ becomes what in terms of u?

can we say as u->0+, x->0+ ?

[ivory goblet]

GHIMME 3 min

let me try on my own

is it 4/5 * sinx /x

is that allowed

can you do that

[flat mica]

sin u/u but yes

because

u->0+ and x->0+ are interchangeable

do you see why?

[ivory goblet]

not really

well

yeah actually

i got it

u=4x

[flat mica]

because 4*0 =0

[ivory goblet]

yeah

[flat mica]

gj

very common trick btw:

[ivory goblet]

ok

[flat mica]

if x->0+

let x=1/n,

then x->0+ is interchangeable with n->+infinity

keep that in mind for the future

[ivory goblet]

ok

[flat mica]

gn

[ivory goblet]

gn

.close

[flat mica]

bot is broken

[ivory goblet]

dang

alr

[hexed vortex]

am I starting this off right?

[cedar kiln]

Channel closed

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[hexed vortex]

.reopen

[cedar kiln]

✅

[cedar kiln]

@hexed vortex Has your question been resolved?

[cedar kiln]

@hexed vortex Has your question been resolved?

[cedar kiln]

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[sand abyss]

I want to express the result as a single power, if possible.

[sacred iron]

Excuse my stupidity, what do the colons mean?

[indigo lagoon]

What is this notation about?

[indigo lagoon]

Same doubt ngl

[sacred iron]

@sand abyss

[crimson sedge]

A ratio I believe

[sand abyss]

no no

divided

÷

@sacred iron @indigo lagoon @crimson sedge

[sacred iron]

strange notation

but it works It’s still a bit ambigious though

[indigo lagoon]

So it's $\left(\frac{10^{15}}{\frac{25^5}{25^5}}\right)^3$

[wraith dagger]

This is sad 😢

[sand abyss]

yes

[sacred iron]

ithe denominator cancels to *1

[sand abyss]

so its 10 at the power of 30

[sacred iron]

Not quite

[sand abyss]

mh

[sacred iron]

remember

[sand abyss]

oh yeah 15 times 3 is 45

[sand abyss]

yes

[wraith dagger]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[sacred iron]

how can you apply that here?

[sand abyss]

so imma do

10 at the power of 15 times 3

[sacred iron]

correct

[sand abyss]

if the denominator cancels to 1

[sand abyss]

yeah i didnt put it as fraction so i didnt' noticied

thanks mate 🙏

.close

[cedar kiln]

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[hexed vortex]

if a sequences converges it must be bounded from below and above right, how is a^1/n bounded above?

[hexed vortex]

oh

[lavish vine]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[hexed vortex]

its bounded above by a?

im just wondering how a^1/n is bounded

I posted the question for extra info

[buoyant latch]

Bruh

Look at part a

[buoyant latch]

I think so

[chrome elk]

Oh above I'm dumb

Sorry

[hexed vortex]

ok

[chrome elk]

For a > 1 yes

[hexed vortex]

im trying to prove c with the squeeze theorem

oh wait sorry

I tried* to prove c with squeeze theorem

[pastel vault]

if n is positive, a^(1/n) < a given a > 1, yes

[hexed vortex]

didnt work

for me so now im using boundness

maybe theres some contradiction

[hexed vortex]

ah I see

ok thanks ill try reworking c

[pastel vault]

no worries

also note that parts a and b are the conditions of the monotone convergence theorem, also adding in that a^(1/n) is bounded above

well you don't need bounded-aboveness actually but it's not bad to include it "just in case"

[buoyant latch]

I would just use something like

[pastel vault]

with a lower and upper bound, you can relate it back to the general definition of boundness, which is there exists $M$ such that $|a_n| < M$ for all $n$

[buoyant latch]

For any 𝛆 > 0, suppose a^(1/n) = 1 + 𝛆

[wraith dagger]

south

[buoyant latch]

Then e^((1/n)lna) = 1 + 𝛆

[pastel vault]

oh well, <= M but anyways

[hexed vortex]

we can't use log

[buoyant latch]

e^(1/(n+1)ln(a)) < 1 + 𝛆

[hexed vortex]

someone mentioned using ln in office hours and he said we cant use it

[buoyant latch]

Wait that’s not right

[flat mica]

what exactly are we trying to do now

which part of the problem

[hexed vortex]

part c

[flat mica]

you showed it's bounded below in part a and that it's monotonic decreasing in part b, so it has a limit

you just need to prove that limit is 1

[hexed vortex]

ah

wait I think I see?

ok let me try

I didnt take in to the fact that it must have a limit

[flat mica]

decreasing and bounded below, or increasing and bounded above, means it has a limit

that's a theorem

[pastel vault]

.

[hexed vortex]

oo I see lol

[pastel vault]

yes you just needed bounded below and decreasing for the conditions

equivalently, bounded above and increasing

[flat mica]

if you'e not allowed to use log

then I would say

replace 1/n with x, then n->+infty as x->0+

do you see why that's true?

[hexed vortex]

yeah

[flat mica]

so whats the new version of the limit

[buoyant latch]

I don’t really see why ln isn’t allowed

My idea was take say 1.001 and raise it to the power of n

It’s either bigger or smaller than a

If it’s bigger, start with sqrt(1.001)

Try again with raising it to the power of n

If it’s still too big you can keep making it closer to 1

At some point, you’d have (1 + δ)ⁿ < a

Which means 1 + δ < a^(1/n)

[flat mica]

theres more than one way to it for sure

I have a different argument but im waiting for Branshi to answer my question

[hexed vortex]

I think

I mightve gotten it

I think lol it might be completely wrong

ok so if we have a_n > 1

then lets assume the limit is 1 for some epsilon > 0

we have |a^(1/n) - 1 |

we know a^(1/n) > 1

so |a^(1/n) - 1 | < |1 - 1| = 0 < epsilon

[flat mica]

no not quite

a^(1/n) <1

[hexed vortex]

oh does mine require a^(1/n) < 1 but we have 1 < a^(1/n)

[flat mica]

ah I see

then for a similar reason

it's not =0

try my method please,you'll see something nice happens

no epsilond deltas needed

[hexed vortex]

ok let me try

[hexed vortex]

so we have a^x

where 1/n goes to infinity

[flat mica]

so x->0+

but this function is continuous at 0

so no limits needed, just evaluate

[hexed vortex]

hm ok I see

[flat mica]

but of course, other ways to do this with epsilon delta

I just preferred a method that didn't

[hexed vortex]

ok I see thank you for your help

Ill move on to the last problem now and hopefully I dont need help again

[cedar kiln]

@hexed vortex Has your question been resolved?

[cedar kiln]

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[burnt fiber]

trying to solve 7th grade geometry problem
find X, a and b are parallel
have tried to connect them into all sorts of different triangles/shapes/whatnot and i cant solve it. can show another similar problem i solved thats on a similar basis

[cedar kiln]

@burnt fiber Has your question been resolved?

[late acorn]

sure, show the other problem you are talking about

[regal knot]

angle sum property ig

[burnt fiber]

was trying that aswell but i feel like ill get totally incorrect calculations lol

[sacred anchor]

70 / 2 = 35, careless mistake

[burnt fiber]

like i cant even say that 6x + 7x + 87 = 180 because i cant connect a goddamn triangle

whoops ur right

will correct that one

[sacred anchor]

does the dot in the angle mean 90 degrees?

[burnt fiber]

mhmm

[sacred anchor]

okay so you're on the right track with this idea

instead of trying to form a triangle

[burnt fiber]

feel like this also prob has 90 degrees somewhere but i just have nothing lol

[sacred anchor]

what if you tried to form a pentagon?

[burnt fiber]

i thought about it so ill try

[sacred anchor]

lemme know if you need help finding which line to make

[burnt fiber]

there r 2 options i can do, the first one isnt a pentagon for certain but i thought id try lol

[sacred anchor]

a pentagon is just any polygon that has 5 sides, not necessarily equal

the first one is a pentagon

[burnt fiber]

i see, i figured itd make 90 at the bottom?

[sacred anchor]

that's one way to do it, yes

[burnt fiber]

looking over the pentagon on wikipedia i wasnt sure i could do that

[sacred anchor]

what's the sum of interior angles of a pentagon?

[sacred anchor]

you can definitely do that

[burnt fiber]

says each pentagon's angle is 108

sorry for being slow i have to translate some terms into my native language lol

[sacred anchor]

that's for a regular pentagon (pentagon with equal sides)

[sacred anchor]

it's okay

what's your native language? maybe someone else here might be able to help you in it

[burnt fiber]

bulgarian, certainly not anybody whos bg here LOL

its close to russian i guess

[sacred anchor]

you never know lol

[burnt fiber]

think its 540 for unequal side pentagons

[sacred anchor]

yup

[burnt fiber]

hell yeah i can do something now

[sacred anchor]

i hope you know why that is

[burnt fiber]

mhm i recall, theres a formula

[sacred anchor]

yup

[burnt fiber]

thank u twin

i figure this is the right way to do it so ill solve B) in a similar way

[sacred anchor]

✅

[burnt fiber]

god bless 🙏

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[sacred anchor]

there is one small thing i would like to point out though @burnt fiber

you dont necessarily have to choose 90 degrees for the two angles formed by the line joining a and b

their sum will always be 180 degrees (due to interior angle between parallel lines property)

[cedar kiln]

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[burnt fiber]

i see i see, will keep in mind

[burnt fiber]

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[edgy sun]

i’m asked to find the point of inflection of x^-2-x^-3

[native heath]

do it

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[indigo lagoon]

What have you tried in advance?

[edgy sun]

hold on i’m re doing my work to send a ss

[indigo lagoon]

Alright, take your time

[native heath]

looks fire

[edgy sun]

But i was told the answer is not x= 0 and x=2

[livid hound]

consider the domain of your function

[native heath]

oh hm i may be dumb

[edgy sun]

So you can’t have a point of inflection at an undefined point?

[livid hound]

it's undefined there,
there is no point

also what did you use for that sign graph

[edgy sun]

y’’=6x-12/x^5

[waxen quiver]

z’’=6x-12/x^2

help

[native heath]

now thats just rude

[edgy sun]

so uhm

did i plug it into the wrong equation?

or is it fine after i consider domain restrictions

[indigo lagoon]

yeah...not funny fam, consider going #discussion

[edgy sun]

i don’t understand the joke he said

[livid hound]

people usually use the first derivative to determine if it's an inflection point

[edgy sun]

ohhh because that’s what tells if the slope is positive or negative

[livid hound]

yeh

[edgy sun]

wait but how

inflection has to do with where concavity changes which is the second derivative

[livid hound]

I suppose both routes work

where is the slope remains the same sign in combination with f"(x) = 0, there'll be an inflection point

[edgy sun]

so for a point of inflection do i find where f”(x)=0 and undefined?

[livid hound]

that'll give you candidates

[edgy sun]

mmm but if one of them isn’t even defined in the original function then it cannot be a point of inflection

[livid hound]

and you'll have to investigate further to see if there are actually inflection points there

[edgy sun]

right?

[livid hound]

yrh

[edgy sun]

holy shit math is so cool

ty

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[umbral loom]

hello helps me pls

[umbral loom]

@upbeat musk here i am here

[wicked mantle]

Hello, any question today?

[umbral loom]

yeps

this is a cordinatesytem

[wicked mantle]

Can you post it so I can pin it for you?

[upbeat musk]

Hii

[umbral loom]

this are many boints (100 to be precise)

they are in the somewhere ig

b1 b2 and b3 is important

b1 distance to b2 is 1 Unit

b2 to b3 distance is also 1 Unit or smth

they are 90° also or smth ig

the boints 4-100 is somewhere in relation the b1,b2,b3

i want to move b1,b2,b3 to this

they must keep their relative position to each other

so they must be translated + rotated

and then after i done it

the b4-b100 must follow them

while keep the same relative position to each other and to b1,b2,b3

does this make sense

my brain hurt already

[upbeat musk]

Yes I think I understand but DAMN

[umbral loom]

you can bin it now

[upbeat musk]

What is this for??

[umbral loom]

i want to build a airplanes

[upbeat musk]

WOW

Are you in uni?

[umbral loom]

no i not allowed i have bad brain for it

[upbeat musk]

I’m gonna reread all that 1 sec

[upbeat musk]

Don’t let anyone stop you

You can do a foundation year if grades are the problem

[umbral loom]

no it ok i dont want to university

i just want solve this problem

[upbeat musk]

Okay honestly this is very hard i will try and understand tho

[umbral loom]

the real problem is even more problematic

one might say its a problem²

or problem³

[upbeat musk]

WHAT that’s not the whole thing??

Help

Sorry I cannot help u😭

[umbral loom]

how to summon more helpers

i dont even need a solution

i just need to know the name of thing i must to know

e.g it is vector algebra surely

but what of it

[wicked mantle]

Roger that

[umbral loom]

is this a advanced queston

[wicked mantle]

Not that I think of

[left juniper]

What do you mean by this?

They are 90 degrees?

Do you mean the three points make a 90 degree angle?

[left juniper]

I dont know if perhaps im looking at this the wrong way but isnt it just a simple translation?

[umbral loom]

see this blue

[left juniper]

Assuming you have a coordinate plane to work with

Ah

[umbral loom]

no it must involve two rotation

the rotation for the b1,b2,b3 should be different from the other rotation i think

the rotational point would differ

[left juniper]

This looks like it could be achieved with just one rotation.

[umbral loom]

hm yes you are right

[left juniper]

Why not just rotate it so that b3 is on the bery left, b1 on the very right, and b2 at the top?

[umbral loom]

idk what u mean

[left juniper]

Literally just this

Rotate it 90 degrees to the left

[umbral loom]

around which axis

[left juniper]

X axis, if im reading correctly

[umbral loom]

what is left then

math. negative here?

[left juniper]

Hold on. How are we doing rotations here exactly?

Like this is all just on a sheet of paper right?

[umbral loom]

idk i think not

[left juniper]

Hold on let me get some scratch paper

[umbral loom]

the coordinatesystem is (1,0,0),(0,1,0),(0,0,1)

[left juniper]

Ok 👍

[umbral loom]

the b1,b2,b3 are (xi,yi,zi) each

and are arbitrary values

only restriction i impose is they cant share a value

[left juniper]

Are these definitions agreeable?

Sorry for the bad handwriting lol

[umbral loom]

not sure

the arrow direction of each axis is the positive

the origin is (0,0,0) ofc

[left juniper]

Yeah shouldnt violate any rules I have

[umbral loom]

or after transformation the b2 sit on (0,0,0) too ofc

[left juniper]

But also

No yeah

Should be good

Ok

So where is b2 currently?

(Just pick a point at random)

I really just need one to see how to make the transformsiton, as we know the angle and distance

[umbral loom]

(20,10,5)

[left juniper]

Ok, that will be the location of b2

Ok so whatbthat means is if we move b2 to (0,0,0), then b3 will be at (0, 1, 0) and b1 will be at (1, 0, 0)

Assuming only one unit is only 1

Does that make sense? @umbral loom

[umbral loom]

what mean whatbthat

[left juniper]

If b2 is at (0, 0, 0), and the distance between b3 and b2 is 1, snd the distance between b1 and b2 is 1, then if we want to keep the same distance but have them be at these points, all we need to do is increase the y axis of b3 by 1 and increase the x axis of b1 by 1.

[left juniper]

Resulting in this

[umbral loom]

why

[left juniper]

B2 is at (0, 0, 0) right?

[umbral loom]

b1 and b3 can be in any (x,y,z) around the b2

in the original position i mean

[left juniper]

Ah!

Ok

So all you have to do

Is take the absolute difference between the y value of b2 and b3, and use it as the y value with the rest of the values being 0. Then take the absolute difference between the x value of b2 and b1, and use it as the x value with the rest of the values being 0.

[umbral loom]

[left juniper]

In reality though, youre just adding the absolute difference to the coordinates of b2

[left juniper]

Let me give an example

Im gonna write it by hand because its a lot of typing lol

And I think it would be best to write it so you can have a visual aid

[left juniper]

Does this make sense at all?

Sorry for the scribbles, made some errors and a pencil isnt readily available

[umbral loom]

no

[left juniper]

Really?

Is there any part youre particularly confused about?

[left juniper]

Look at how we got the values that we use for the coordinates

We get the original distance

[umbral loom]

yes but what about the rotation

[left juniper]

This is essentially the rotation

[umbral loom]

how

[left juniper]

If we were to plot these on a coordinate plane(well not these specifically but if you took the correct coords), you would be able to see it

It may be hard to conceptualize, but because we are using the absolute values it will use only a positive value, that being the distance between the points on the respevtive axis

[umbral loom]

i dont see how this invoke any rotation still

i think it too late now to understand too

thank yoru for trying tho

[left juniper]

I did plot them out in a coordinate plane for you to see btw

First 3 are correspondent to b1, b2, and b3 before the transformstion

I picked them out at random

As you can see, they end up being exactly where we want

[left juniper]

A rotation is really just a translation

[umbral loom]

they have different distances in relation to each other tho

[left juniper]

So you want all of the distances to stay the same?

I mean, im not really sure thats entirely possible with every configuration

[umbral loom]

ofc

the relative position of every boint must be exactly the same to all other boints after transformation

[left juniper]

This would not be possible woth every single configuration of points unless they were all at the same angle

[umbral loom]

why not it must be

[left juniper]

The points I just showed are proof

[umbral loom]

if the transformation ends up with points having different distances to each other its a incorrect one tho

[left juniper]

By your standards, yes. But also that would not be possible for every single configuration of points

[cedar kiln]

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[bronze linden]

Is this correct?

[limber dawn]

⭐

[high coyote]

It is not

[fast osprey]

no
i think the answer is (c) cuz :
at x=4 the limit from right [3] does not equal the limit from left [2] so it DNE
at x=0 the limit is clearly exist and equal 2
at x=-3 the limit from right and left are equal so it exist and equal 2

[bronze linden]

Ohh ty

[fast osprey]

you'r welcom

[bronze linden]

.close

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[strong siren]

hey, im looking over a math test and i need help cuz idk what my teacher means by simplify

[ancient lodge]

They wanted you to combine like terms to get $3v-9v=-6v$

[wraith dagger]

Civil Service Pigeon

[strong siren]

can u elaborate if its ok im still a little confused cuz i got rid of 2x and -12 by doing -2x and +12

i also noticed i messed up by forgetting to set 2 as a - instead of + but i still got the wrong answer

[whole nexus]

Which one?

[strong siren]

the top onr

[whole nexus]

I guess 10 - 5x = 3(x - 4) - 2(x + 7)

So it's 3x - 12 - 2x - 14

So shouldn't you be substracting 2x from 3x and 14 from -12?

[strong siren]

but isnt 2x a negative so i should add it?

so it cancels out and goes to 0

and 14 is a negative too

or am i confused

[whole nexus]

Yes but you can write it as 3x - 2x - 12 - 14 right?

[whole nexus]

You add it when the number you're subtracting it from is negative too (like in -12 - 14)

Okay, imagine you had 3 apples, and I took 2 from you, how many would you have?

[strong siren]

1

so

[whole nexus]

So you had 3x, and 2x is being reduced from it, what should you be left with?

[strong siren]

1x

[flat mica]

1 apple

[strong siren]

or just x

[whole nexus]

Bingo

[flat mica]

but iu think you got it

[whole nexus]

Now imagine you had a stash of bananas, I'm an evil guy, I've taken 12 from the stash, and now I decide to take 14 more, how much bananas would you be down?

[strong siren]

i'd have -2

[whole nexus]

No, I mean I've already taken 12, and now I'm back to take 14 more, how many bananas would you be down

[strong siren]

-26

[whole nexus]

Exactly

[whole nexus]

So if it's -12, and you reduce even more -14 from it, you'd be left with -26, right?

[cedar kiln]

@strong siren Has your question been resolved?

[cedar kiln]

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[fading token]

What do I do now? I'm assuming I either have to cancel out the -7 or the 1+x from something in the numerator, bit I don't know how to get it from these factors

[sonic fossil]

,rotate

[fading token]

Thanks

[cedar kiln]

@fading token Has your question been resolved?

[fading token]

<@&286206848099549185>

[cedar kiln]

@fading token Has your question been resolved?

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[halcyon minnow]

Uhm guys

[chrome elk]

Hi do you have a question

[wicked mantle]

.solved

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[distant garden]

How do I solve this?

[fossil dawn]

what are you asked to do

[distant garden]

To find the cube of 2x-3y

[fossil dawn]

expand out normally

if it was a square, could you do it?

[distant garden]

(2x-3y)(2x-3y(2x-3y)

[distant garden]

Yes

[fossil dawn]

then do the square case

[distant garden]

Alright

[fossil dawn]

then, multiply that result by another copy of 2x - 3y

[distant garden]

What do you do if it’s cubed?

[fossil dawn]

here

it's the same as if it's being squared

just one extra copy to deal with

of course, if the power is too big, then make use of binomial expansion, but a cube should be decently okay to deal with by hand

[distant garden]

(A-B)^2

[distant garden]

What’s binomial expansion?

[fossil dawn]

if you need to ask, then your syllabus has not taught you that yet, and you probably cannot use it

but for knowledge's sake, binomial expansion allows you expand a binomial faster than you could by hand

you don't need it here though

[distant garden]

Alright

So this’ll be a^2-2ab+b^2

[fossil dawn]

this would be the general form of (a - b)^2, yes

[distant garden]

I’m getting 4x^2-12xy+3y^2

[fossil dawn]

wrong

last term

[distant garden]

9y^2

[fossil dawn]

better

[distant garden]

Do I multiply all of it now with (2x-3y)

[fossil dawn]

correct

[distant garden]

Is there a quick way of doing this?

Or is multiplying the whole thing the only way

[fossil dawn]

at this stage multiplying is the most straightforward and quick way afaik

[distant garden]

Alright

This is what I’m getting so far

[fossil dawn]

combine all like terms

[fossil dawn]

your last two terms are wrong

[distant garden]

I’m not sure what I’ve done wrong

[fossil dawn]

show your full working.

[fossil dawn]

i had a hunch you did not fix this

[distant garden]

Why’s that wrong?

[fossil dawn]

.

you identified the mistake yourself

[distant garden]

Oh yeah

I’ll just redo it again

[fossil dawn]

just redo from that step

[distant garden]

Alright

It’ll be 8x^3-36x^2y-18xy^2-27y^3

[fossil dawn]

i recommend putting some spaces

your last term is now correct but your second last term is still wrong

[distant garden]

Spaces in between the terms in the notebook?

[fossil dawn]

in here

it's confusing to read sometimes. i get what you meant here, but in more complicated expressions, consider adding spacing to make variables and exponents clearer

[distant garden]

Alright, will do

Should I redo it again, if the second last term is wrong?

[fossil dawn]

show full working again

unless you know where your mistake is

[fossil dawn]

the sign of the term in red is wrong because of the two terms in blue

[distant garden]

It’ll be +36xy^2

My bad

Is 8x^3 - 36x^2y + 48xy^2 -27y^3 the correct answer?

[fossil dawn]

no

your expansion is now entirely correct, however

assuming you fixed the sign error

[distant garden]

I did fix that

[fossil dawn]

then check your xy^2 term again

[fossil dawn]

so how did you get 48

[distant garden]

Oh yeah, it’ll be 54

[fossil dawn]

there we go

that would be the correct coefficient for xy^2

and after that all four terms are correct

[distant garden]

Nice

I had another question as well

Could I send it here again?

[fossil dawn]

sure

[fossil dawn]

same as earlier

like, exactly the same concept and method would get you the answer

[distant garden]

Alright, I’ll try it then

I’m getting x^3 - x^2 -xy^2 - y^2

Is this accurate?

[fossil dawn]

let me check

correct

[distant garden]

Alright

How do I solve this one?

[fossil dawn]

can you solve the example on the right?

[distant garden]

Sure

[fossil dawn]

how would you do it?

[distant garden]

You’d find the area of the square and then the area of the rectangle

[fossil dawn]

and then?

[distant garden]

And then add them together

[fossil dawn]

add?

[distant garden]

I think so

[fossil dawn]

well

the area of the rectangle already includes the square, and i think you would agree with me?

[distant garden]

Yes

I guess you’d have to subtract then to find the value of shaded area

[fossil dawn]

mhm

same strat with the variables

[distant garden]

It’d be 60-9=51cm^2

[fossil dawn]

right

would there be anything else?

[distant garden]

Is this accurate so far?

[distant garden]

Should I redo it again then?

[fossil dawn]

well this is on lines 2 and 3, so redo from there i suppose

[distant garden]

Alright

I’ll redo it again and send it

[fossil dawn]

the other terms are correct (those that aren't the x term and the constant term)

(correct for the expansion of the area of the rectangle)

[distant garden]

Alright

So everything except the 28x is accurate?

[fossil dawn]

and the lack of a constant term

but technically yes

that 28x should really just be a 28

[distant garden]

What’s that?

[fossil dawn]

every other term in the initial expansion is correct

[fossil dawn]

the term without any variables attached to it

[distant garden]

Oh ok

That’d be 28 here, if I’m getting it right

[fossil dawn]

yes

[distant garden]

Is this accurate?

[fossil dawn]

where did the blue term go?

[distant garden]

I didn’t write it properly

I’ll redo it again

Is this accurate now?

[fossil dawn]

did you mean to share what looks like a sheet of mostly blank paper

[distant garden]

No, my bad

[fossil dawn]

exact same mistake

[distant garden]

I’ve added it directly in the end

I didn’t edit the middle part

[fossil dawn]

try not to do that (or at least add it in by putting a ^ and writing a -7y where it's missing)

but anyway

why did you multiply the area of the rectangle with the area of the square

[distant garden]

Oh yeah, I should’ve subtracted it

If it were to be multiplied though, would it be accurate?

[distant garden]

Will remember

[fossil dawn]

no

[distant garden]

Alright

I’m going to stop here for now

Thanks for helping me out, it helped a lot

Bye

[fossil dawn]

nps

!done

[cedar kiln]

If you are done with this channel, please mark your problem as solved by typing .close

[distant garden]

.close

[cedar kiln]

Channel closed

Closed by @distant garden

Use .reopen if this was a mistake.

[cedar kiln]

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[daring hill]

Can someone help me with what I'm doing wrong? I can't seem to get the same answer from the textbook.

[normal cipher]

!original

[cedar kiln]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[livid hound]

you have an extra ^2 on the 4

[daring hill]

omg thank you, I didn't even see that

[sacred iron]

the little dash through the circle lmfao

My sense of humor is beyond repair i dunno even why I found that funny

[daring hill]

.close

[cedar kiln]

Channel closed

Closed by @daring hill

Use .reopen if this was a mistake.

[cedar kiln]

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.

[vapid grotto]

Could anybody help me with the circled problems we never went over this in class

[sacred iron]

I’m assuming (𝑓±𝑔)(𝑥) = 𝑓(𝑥) ± 𝑔(𝑥)

?

[vapid grotto]

I have no idea I’m sorry

My teacher gave this to us after a quiz I’ve never done this stuff before

[sacred iron]

I think that’s what it means

[vapid grotto]

For which number?

[sacred iron]

Oh wait nvm

sorry

okay

so (𝑓 +-×÷ 𝑔)(𝑥) = 𝑓(𝑥) +-×÷ 𝑔(𝑥) i think

for any of the four functions

[cedar kiln]

@vapid grotto Has your question been resolved?

[sacred iron]

@vapid grotto sorry if i wasn’t much of a help. You can ping helpers now

[weak niche]

f(x) + some function = g(x)

[cedar kiln]

Channel closed

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[cedar kiln]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[paper karma]

Help me with 49,50 n 51

[paper karma]

It tried well and got solutions but was told wrong