help-38

[vast geyser]

I need help filling in the boxes does anyone know how to do algebra polynomial division box method

Looks like this

[zenith nexus]

Fill in what you know first

Where does x-1 go

[vast geyser]

I think it goes there

[zenith nexus]

no

Think of it as reverse multiplication

the top row and left column when multiplied, obtains the whole rectangle when added

[vast geyser]

Like I’m pretty sure this right

[zenith nexus]

no

Here's an example of multiplication

you have the rectangle

You need to find the top row

[vast geyser]

Like this ?

[zenith nexus]

bro😭

Fill in x-1 first

[vast geyser]

im sorry i’m like really dumb

And I’m actually trying to try

[vast geyser]

Where 😭

[zenith nexus]

The blue are the two polynomials being multiplied, and the red is the answer when added

you have the red and 1 of the blue

P(x)(x-1)=4x³+6x²-11x+1

P(x) will be the top blue row

x-1 the left column

each square with the red rectangle is the top blue column times the left blue row

[zenith nexus]

That's not the answer that was just an example

[vast geyser]

😭

I figured

[zenith nexus]

For example here the first red square (top left) is X×X², (left×top of the square) and the next square is X×3X(left×top of the square)

[vast geyser]

Like this

,close

Close

<@&286206848099549185>

.close

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[vast geyser]

Question

.reopen

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✅

[vast geyser]

Question

Anyone

I think my helper fell asleep

<@&286206848099549185>

Can somebody help me please? I’ve been stuck in this chair for like a good two hours. No one has helped me.

I have tried

Many times

,help

[solid kiln]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[tawny sundial]

Did you understand the solution? @vast geyser

[vast geyser]

Can you do it, the box method

Like this

Help me fill in like the blanks

[trim jolt]

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.

[vast geyser]

.close

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[past heath]

do you know how to arrange n objects into n boxes but each box can also be empty?

[orchid wagon]

can each box contain multiple objects?

[past heath]

just one

and once it’s used it can’t be used again

[past widget]

Huh? So then suppose I put 5 objects into first 5 objects and leave the n - 5 empty, it also counts as a reasonable arrangement?

[past heath]

i’m not sure what you mean

like say you have 5 boxes, and each box can contain one of 5 unique objects or none at all

[past widget]

You said one box can contain only one object and cannot be used again. And each box can also be empty, so I were to arrange say 10 objects in 10 boxes, such that 7 are filled, 3 empty, I'll have 3 objects leftover

I'm asking if that's allowed

[past heath]

yes

also, can i ask a follow up question afterward if that’s ok?

[exotic pine]

are the boxes identical or no

[past heath]

yes

[exotic pine]

wait object and boxes are identical?

[past heath]

objects are unique, boxes are identical

[noble escarp]

wait but

if each box can only fit one object

does every object need to be put in a box?

[past heath]

no

each box can also have nothing

[noble escarp]

oh

[past heath]

and u can have leftover objects

[noble escarp]

then why is it not 2^n because each box either has a object or it doesnt

or does order not matter here

[past heath]

the objects are unique and order matters

wait let me be more specific with the actual problem

i have 22 spaces and 22 different letters to put in these spaces, but each space can also be empty

[past widget]

!original

[trim jolt]

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.

[past widget]

Send a pic of the question

[past heath]

i don’t really have an original sorry…

i can draw it if you’d like?

[past widget]

Your words lack reliability Try to recall the question, writing the question down on a piece of paper would help jog memory.

[safe loom]

consider only 1 object, there are n ways to put (because there are n boxes)
consider n objects, how many ways to put?

[safe loom]

what

[past heath]

it’s just this

[safe loom]

well, this works ig

it's simply 22^2

[noble escarp]

oh thats a lot more clear lol

the danger of abstractions

[past heath]

sorry it’s like a theoretical thing.. btw i love the sound euphonium pfp

[past heath]

so it’s 22^2?

[safe loom]

yeah, you count how many ways to put 1 object in 1 out of n boxes then do the same thing for the rest

[past heath]

there’s more than the case of a box having a letter or not though

the letters are different

like a b c d

[safe loom]

yes, there are cases a box can have multiple objects, and a case where a box can have nothing

[past widget]

Just uniquely counting, this can be written as $\sum_{i = 1}^{22} \binom{22}{i}\binom{22}{i}$

[past heath]

but there aren’t? a space can only have one of 22 letters or none

[safe loom]

what

[noble escarp]

i was thinking to first choose k letters to use, and then the ways to permute those k letters into the spaces, summing over k

[safe loom]

what exactly do you mean

[past heath]

ok let me do a smaller example

i have three spaces

and the letters a b c

the first box can have no letters or a b c

[past widget]

Quick question. Suppose you're sending letters A, C to one of the first 5 spaces.. A - 3, C - 5 and A - 5, C - 3 are identical or no?

[past heath]

if i choose one of those a b c

the next box can’t have that letter

but also none

and so on

[past widget]

?

[past heath]

they’re not

order matters

[past widget]

Alr

[safe loom]

bruh

[past widget]

Is it fine if I leave the spaces empty?

[past heath]

yeah

but also, b- 1 c 2 is identical to b 1 c 3 if 2 is empty

[past widget]

Yeah that's what I was gonna ask

[safe loom]

I still don't get it, like, if you choose a, b, c, of course the next box won't have that letter

[past heath]

but if i choose none it can have all 3

[noble escarp]

that makes it more weird

[safe loom]

But it's already lying in the box you choose to put that letter in

why does it matter

[noble escarp]

so you're just stripping all the whitespace

[past heath]

i can choose to not put it anywhere

[safe loom]

???

[past heath]

i can leave it out

[safe loom]

that means i can throw all the 3 papers away and decide to throw anything i can in 3 boxes

[past widget]

Isn't it just $\sum_{i = 0}^{22} \binom{22}{i}\cdot (i!)$

[noble escarp]

oh so youre just summing $$\sum^k \binom{22}{k}k!$$ if im not wrong?

[past heath]

well no that’s just one possibility of nothing nothing nothing

[noble escarp]

oh lol

[past widget]

Lmao

[past heath]

would u mind explaining this if that’s ok?

[noble escarp]

since you're stripping all the whitespace you can just ignore them when permuting i think

[past heath]

sorry for all the hassle

[safe loom]

i wish you can remember the original problem because i feel like we are breaking the boundary of physics

[past widget]

Isn't what you're looking for basically Number of words you can form using {A, B, ..., V}

[safe loom]

😭

[past heath]

yeah actually

that’s it

[past widget]

So that is what it gives

sum of 22Ci . i!

[past heath]

thanks so much

[noble escarp]

[past heath]

can i ask something much more complicated that’s related

[past widget]

Choose i letters, permute them in i! ways

it'll be a heck of a big number tho

,w sum_{n = 0}^{22} 22!/n!

[solid kiln]

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[past widget]

there

[past heath]

what if instead of letters a group of 7 of those letters each had a different amount of colors for example a has 5 colors b has 6, but that group has to precede a group of four letters who all have 2 colors each, which also has to precede a group of 10 letters which have different color amounts

this is really stupid i don’t expect an answer

[safe loom]

i can do this if its asking probability

[past heath]

i meann if you can try

[safe loom]

P(4 letters that have 2 colors each) * P(10 letters that have different color amounts)
P(10 letters that have different color amounts) = 1 - P(10 letters that have same color amounts)

hope it works

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[magic eagle]

i see a+b+c+d=11, all are >0

so AM,GM,HM concept can be applied

also, 3750 = 1x2x3x5^4

also 1+2+3+5=11

hmm

so can we say a=5

b=3
c=2

d=1

or we cud interchange them also?

[safe loom]

nope we can't

[magic eagle]

like we cud find the possible cases

[safe loom]

your value is the only possible case

[magic eagle]

and match with the options

ye

i got 90

[safe loom]

correct

i would say, pretty nice guessing

[magic eagle]

lmao

i just saw the solution

even more crazy

[safe loom]

it looks fine to me

[magic eagle]

yeye

[safe loom]

they also guessed the equality just like you to get what looks like that

[magic eagle]

lol yeah

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[thorn copper]

ABCDE is regular pentagon
F and H are midpoints
C’ is C reflected about BF
G is the intersection of diagonals
show that C’G || CH

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[trim goblet]

Do you know similarity

[vast viper]

Yes@trim goblet

[trim goblet]

And median divide triangle into triangles of equal areas

[round imp]

You have that type of board

Man of the culture

[vast viper]

Median divide two parts

[vast viper]

What do you mean brother?

[trim goblet]

Then in triangle ABD and PBD

[round imp]

I have as well

[trim goblet]

height is same then find area ratio

[vast viper]

Actually I use roughs

So area of ABP:BPD will be 3:4?

So i got 4:14
2:7 yo

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[wind saffron]

Hello, I'm struggling trying to construct the production rules of the grammar; the condition j <= 2i is my problem :c

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<@&286206848099549185>

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[heady mesa]

need help with this one please

[lone basin]

You need to observe what happens to the surrounding piece of at x = 2

Because all three pieces have a domain that gets really close to x = 2

[limpid dawn]

you can also simplify x²-4 = (x+2)(x-2)

[heady mesa]

the 5 in the middle?

[lone basin]

The 5 doesnt matter, becasue the limit takes into account what happens around x = 2, not at x = 2
(otherwise, we would just be using f(2))

[past widget]

You're tasked with finding lim_{x to 2} so focus on that

[heady mesa]

ok

is it correct?

[lone basin]

You must calculate the other piece as well

$$\qty[\frac x{x - 3}]_{x = 2}$$

[solid kiln]

King Leo

[lone basin]

Because you care what happens on both sides of x = 2

[heady mesa]

so plug in 2 in x?

=-2

[lone basin]

So you got 4 for one piece

[lone basin]

And -2 for the other piece

[heady mesa]

thank you

[lone basin]

So do you know what the limit is?

[heady mesa]

4?

[lone basin]

No

4 is the limit for only the left/negative side of x = 2

[heady mesa]

2?

[lone basin]

How did you get 2

(if you dont know the limit, its probably a concept that you havent been taught yet)

[heady mesa]

actually dont know

[lone basin]

Ok

Do you understand how the one-sided limits differ from each other?

[lone basin]

(if not, dont hesitate to tell me)

[heady mesa]

no

approach input 2

?

[lone basin]

So lets look at the limit on the left side of x = 2

That means we care about the domain x < 2

And the piece that matches with x < 2 approaches 4 as x approaches 2

Thus,
$$\lim_{x \to 2^-} f(x) = 4$$

[solid kiln]

King Leo

[heady mesa]

oh ok

i get it

[lone basin]

Now lets look at the positive/right side of x = 2

$$\lim_{x \to 2^-} \qty[x + 2]{x = 2} = 4$$
$$\lim{x \to 2^+} f(x) = \qty[\frac x{x - 3}]_{x = 2} = -2$$

[solid kiln]

King Leo

[heady mesa]

now i get it

appreciate it

[lone basin]

So the one-sided limits differ

So you cannot generalize a single limit

[heady mesa]

ok

[lone basin]

The question asks about $\lim_{x \to 2} f(x)$ which means it does not specify a side

[solid kiln]

King Leo

[lone basin]

So this limit only exists if both sides match each other

[lone basin]

Do both sides of x = 2 match each other?

[heady mesa]

no

DNE

[lone basin]

$$\lim_{x \to 2} f(x) = \mathrm{DNE}$$

[solid kiln]

King Leo

[lone basin]

!done

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[lone basin]

Np

[past widget]

The red curve hits +4 from left, blue hits -2 from right

[lone basin]

Its just a constant 5 at x = 2

[past widget]

But lim dne

Should be easy to see why

With the graph

[heady mesa]

.close

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[versed vortex]

Any idea on 5?

[lone basin]

$$\lim_{x \to \frac \pi 3} \frac{\sqrt 3 \tan(x) - 3}{3x - \pi}$$

[solid kiln]

King Leo

[ripe valley]

Lopitals?

[versed vortex]

I haven t learnt it

[lone basin]

:(

[lone basin]

Have you ever even heard the word "L'hopital"

If you have, its possible your teacher taught it

[versed vortex]

At class we use this formulas

[versed vortex]

We will study it later

I ve asked him about it and that is what he said

[lone basin]

,rotate

[lone basin]

Idk this

[versed vortex]

Oh,ok

[zinc ginkgo]

translate the instructions

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[past cargo]

given y = arcsin(x) then in expressing tan(y) we're technically not allowed to use a triangle, right?

[past cargo]

but why does constructing a triangle work?

first of all arcsin(x) requires -pi/2 <= x <= pi/2

[zinc ginkgo]

why do you think you can't construct a triangle

[past cargo]

cuz this i guess

for arccos(x) we have 0 <= x <= pi

this makes sense to have in a triangle

[zinc ginkgo]

triangles in the xy plane can have negative angles

see t'

[past cargo]

yes i get that but i don't see how it relates specifically to this

[zinc ginkgo]

Answer to this

[past cargo]

okay how to construct a triangle for sin(y) = x

given that y = arcsin(x)

so -pi/2 <= y <= pi/2

[past widget]

You don't get a unique triangle but a family of right triangles

[past cargo]

yeah that's what i meant by like

[past widget]

Draw a line segment, draw y angle on one end, draw perp on another end, intersection of both lines makes you a right triangle

[past cargo]

it's technically weird to use triangles then right

cuz let's say y = arccos(x) then it's triangle territory

there is a unique triangle?

[past widget]

No

[past cargo]

i think we have different definitions of unique then

😭

[past cargo]

this yeah

i don't see how that relates to sin(y) = x tbh given y = arcsin(x)

[past cargo]

hmmm

for cos(y) = x given y = arccos(x), we can draw above right triangle

and then yeah it's casework for x > 0 and x < 0

but how exactly are we emulating this for sin(y) = x given y = arcsin(x)

[past widget]

Point is, you take signed side lengths

You're not realising that arcsin has negative domain, so you'll require negative side lengths

[past cargo]

hmm i guess

[past cargo]

so i'd just draw something like this for arcsin(x) as well?

this is weird

because like that diagram itself

doesn't make sense to represent y in [-pi/2, pi/2]

idk something just feels weird 😭

[0,pi] is natural because y can be anything as long as it isn't pi in euclidean geometry

the negative angles just isn't represented well in that image

welp i have to go sleep

thanks for now

maybe i'll come back later to sort this out

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[past widget]

@past cargo it'd help to note arcsin and arctan functions are odd

So just figuring their values in one part does most of the job

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[glad patio]

ive tried to work it out but i just cant factorise the final bit

so im not sure if ive done something wrong

[robust wigeon]

I would just use the short quadratic formula

[glad patio]

oh put it into the quadratic formula?

[naive otter]

If it is a tangent, then discriminant should equal to 0

[glad patio]

so i got x1= 3+root3

and x2=3-root3

[robust wigeon]

If $ y^2 + py + q = 0 $ then $ y = -\frac{p}{2} \pm \sqrt{ {p/2}^2 - q } $

welp

[glad patio]

ah

tyty

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[subtle canopy]

i have question here about step induction here on k which will be for k +1 how can i do that for the transtition states also what is the transition state of e if applied it only once ?

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[fair bison]

I don't know what you're asking. What part of the question are you working on? What are you trying to prove by induction? What have you done so far?

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[glacial crescent]

May someone help me with this

[hidden dew]

what specifically you need help with ?

[glacial crescent]

idk what angle I’m looking for

I have 90°

and cosx = 0

but at cosx=0

It’s not positive or negative

[hidden dew]

doesn't matter

[glacial crescent]

so what do I do

[hidden dew]

imagine the unit circle

and check when is the x coordinant is 0

[glacial crescent]

idk the unit circle…

[hidden dew]

cuz a point on the unit circle has coords
(cos(t),sin(t)) where t is the angle

[hidden dew]

you need to know it for trig equations

[glacial crescent]

isn’t that the unit triangle?

[hidden dew]

technically but the unit circle is more general for quadrants

[glacial crescent]

so how do I continue

what 2 answers am I looking for

[hidden dew]

90 is correct yea

[glacial crescent]

and 270?

[hidden dew]

yep

[glacial crescent]

but why 270

[glacial crescent]

ahhh

[hidden dew]

yep (0,-1) at 270

[glacial crescent]

is there an explanation to that or is it just like that

cos I honestly guessed 270

360-90

0+90

made sense to me

[hidden dew]

it is by defenition of the unit circle

[glacial crescent]

meh I’ll just go with it

but thank you for helping me 🙏🏻🙏🏻🙏🏻

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[serene crane]

How might I go about this? In my notes were just finding for example g(-10), but in this one we're supposed to be getting the piecewise funciton

[wanton rune]

remember the fundamental theorem of calculus

split your integral into parts as well, where the function changes

[serene crane]

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[verbal kettle]

may someone help me with this?

[wanton rune]

!status

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2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
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7. None of the above

[verbal kettle]

i know I have to divide but im unsure how that works

[lone basin]

First, its important to know that, when a polynomial divides another polynomial, only the term with the highest degree of each polynomial matters (if x approaches __+__infinity)

[verbal kettle]

So i only plug in infinity to the numerator?

[lone basin]

You cant directly plug in infinity

Instead, think about it like this

As x becomes a very large negative number, does the 4 have a meaningful effect on 3x^2 + 4

[verbal kettle]

No it does not

it just stays infinity pretty much?

[pure fiber]

group the x to understand this better

[lone basin]

So you can just evaluate $\lim_{x \to -\infty} -\frac{3x^4}{3x^2}$

And with some simplifications, this should be easy

[solid kiln]

King Leo

[verbal kettle]

Okay, so it becomes -x^2?

And then I plug infinity in?

[lone basin]

So what is $\lim_{x \to -\infty} -x^2$

[solid kiln]

King Leo

[verbal kettle]

• inf

[lone basin]

"plug infinity in" is the wrong terminology, but yes

[verbal kettle]

negative infinity

[lone basin]

Yes

[verbal kettle]

ah I see, my teacher explains it like that lol

[lone basin]

$$\lim_{x \to -\infty} -\frac{3x^4}{3x^2 + 4} = -\infty$$

[solid kiln]

King Leo

[verbal kettle]

Thank you very much, I believe I understand it now.

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[orchid wagon]

to approach this problem, do i first assume such map T exists, then prove the equivalence?

[left oriole]

for the ==> direction you would assume such a T exists and then prove the inequality

for the <== direction you would assume the inequality and show that such a T exists

[orchid wagon]

yes, that is my plan

but what i'm worrying is that my proof then would rest on the fact that such T exists in the first place

[left oriole]

why

(why is that a problem?)

[orchid wagon]

well, what if T doesn't exist?

[left oriole]

well in general, A <==> B holds even if A can never happen (then B can never happen either)

but that shouldn't be the case here

(provided the inequality holds)

[orchid wagon]

ah wait i see now, i was misinterpreting the question

it should be
(exists T such that kerT = U) <=> (inequality)

i was reading it as
exists T such that (ker T = U <=> inequality)

[left oriole]

oh i see

yea the first interpretation is what you want

[orchid wagon]

yes, thank you

[left oriole]

the second one would be crazy haha

because then T could have more than one kernel

[orchid wagon]

lol

the => direction should be trivial with rank-nullity

not sure about the <=, but i'll figure it out

thanks

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[merry crown]

I was wondering for the unit vector from A D how is it that the j direction is -cos30sin60 and k direction is cos30cos60

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[plain tendon]

hello

[plain tendon]

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[glad pumice]

Can someone please help me on this question

[glad pumice]

I'm pretty sure there is a theorem/principle with this question but i forgot what it was called

please help

<@&286206848099549185>

ple jelp

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[solid spoke]

hey guys how do we do part b and c of this practice frq? I know how to do part a but im confused by the explanation the answers give

[solid spoke]

wait guys i got it

i grasped the solution thanks guys

[midnight vessel]

👍

[solid spoke]

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[sullen swift]

I'm trying to test out this formula for finding the LCM of a string of numbers but I'm having some trouble, am I doing something wrong here?

I'm using 3, 4, 5 and 6 as the numbers im plugging in, and getting ((3)(4)(5)(6))/((1)(1)(1)) which just gives me 360, not the LCM, which is 60

[past widget]

if it's not working, then formula is wrong :o

[marble wharf]

thus concluding that the formula is wrong

[sullen swift]

ah okay, well thanks both of you for the answer :]

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[smoky marsh]

How do I solve this

[gilded pulsar]

!status

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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.
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6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[smoky marsh]

I kinda know where to begin

I just don't know how to do that

So basically I want to sketch the graph

But I don't know how to

[gilded pulsar]

You don’t really need to

[smoky marsh]

Ok how do I go on then

[gilded pulsar]

You might want to start by sketching 2 random graphs that form an area between them

To get an idea

[smoky marsh]

Because I put the curves in desmos and it was pretty well shown

But I have to do it without the help of anything

[smoky marsh]

How do I do that

[gilded pulsar]

Just draw two random graphs that intersect twice

[smoky marsh]

Just random?

That nothing to do with our curves?

[gilded pulsar]

Just to get the idea

[smoky marsh]

Ok I did that

[gilded pulsar]

Can you send it

[smoky marsh]

Ok

Sorry my drawing is kinda awful, like this?

Oh you meant intersect twice

[gilded pulsar]

Yeah also draw them in the same coordinate system

[smoky marsh]

I did that because you said 2 graphs

@gilded pulsar

[gilded pulsar]

Oh umm

And for visualisation purposes maybe draw them above the x axis

[smoky marsh]

Ok

Like this?

[gilded pulsar]

Well ones not a function anymore xd

[smoky marsh]

Well you said above x axis

[gilded pulsar]

Yeah but they should be graphs of functions

[smoky marsh]

I'm not sure how to do it then

[gilded pulsar]

No double y values

[smoky marsh]

Ok so no double y values above x axis

[smoky marsh]

So you want me to either make the curves on the negative side or positive side?

[gilded pulsar]

Just like that

[smoky marsh]

Oh ok

[smoky marsh]

And after that?

[gilded pulsar]

Now you see those Graphs have an area between the intersections

[smoky marsh]

Yeah

[gilded pulsar]

Well you dont know How to calculate the area of that yet

But you do know how to take the area between a Graph and its x axis

[gilded pulsar]

right?

[smoky marsh]

Well yeah but we don't know the coordinates

[gilded pulsar]

Its just to get the idea for your problem you know the coordinates

[smoky marsh]

Oh ok

Yeah

[gilded pulsar]

So because you know how to calculate the area between the x axis, can you find a way to express the area between the Graphs with the Areas formed with the x axis?

[smoky marsh]

Not really

[gilded pulsar]

You know the green and Blue areas

And you want the red area

[smoky marsh]

Yeah

[gilded pulsar]

Notice that green=blue+red

[smoky marsh]

Yeah

[gilded pulsar]

So how can you express red with green and blue

[smoky marsh]

Green-blue

[gilded pulsar]

Exactly

So now you know how to calculate the area

Just use this on your problem now

[subtle canopy]

first part (a) also i tried tp prove that nummeral experssion e transforemd in k steps to a e prime and i have done the base case for induction on k where k = 0 where no transformation will be done at this case and e1 will equal to e1 prime , now my question for inductive case after assume the IH that for k holds now i wanna prove it for k +1 so how can i prove this ? shall i make it seperatly one time for k and then from IH it holds and one time for 1 but in this case how can i prove it holds ? can i combine them after that ?

[smoky marsh]

Can you please open another channel

I'm currently using this

[subtle canopy]

sorry i did not mean to interrupt

[smoky marsh]

Well I'm still lost

How can I do this on the problem I have

[gilded pulsar]

So you calculate the area between two functions by subtracting the areas between the functions and the x axis

[smoky marsh]

Can't we also use integrals to find the area

[gilded pulsar]

Yeah

Do you know which bounds you should use?

[smoky marsh]

The coordinates

No?

[gilded pulsar]

Which ones

[gilded pulsar]

You might want to look at the picture

[smoky marsh]

The ones from the x axis

[gilded pulsar]

Yeah the x coordinates of the intersection points

[smoky marsh]

Yeah

But I don't know how the graph will look

Because I'm not sure how to draw the curves of the given problem

[gilded pulsar]

You dont need to

[smoky marsh]

Ok

[smoky marsh]

What do I need to do then

[gilded pulsar]

You need to

1. calculate the intersection coordinates
2. calculate the areas under each function between them with an integral
3. subtract 1 area from another
4. take absolute value to make it positive (might be negative)

[smoky marsh]

Sorry but how do I calculate the intersection coordinates if I don't have a graph

[gilded pulsar]

Set the functions equal

[smoky marsh]

Ok

[gilded pulsar]

If the functions intersect, their y values are the same for an x value, so for that x value is f(x) = g(x)

So to find that x value you have to solve the last equation

Also if you got 3 or more intersection points, you have to do step 2-4 for each intervall between them and add all your solutions

[robust wigeon]

Sorry to interfere here, but I do think you should draw the graphs if you’re not sure about the whole procedure

[smoky marsh]

Ok so x=0 x=-1 and x=1

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@smoky marsh Has your question been resolved?

[gilded pulsar]

Have you defined the 5th Root for negative Numbers?

[gilded pulsar]

Omg Rony im so sorry…

I misread the question 😢

[smoky marsh]

Wdym

[gilded pulsar]

They are two questions

Wait

I still dont get the wording of the question

[gilded pulsar]

Do you agree that the question is to find the area between the two curves?

[smoky marsh]

That's the idea

[gilded pulsar]

But the wording of the question is so weird

[gilded pulsar]

Okay have you done steps 2-4 for the two Intervalls between them?

[smoky marsh]

I only did x^3 = 5√x

That's how I got 3 xs

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[gilded pulsar]

Then do step 2-4 for the intervalls-1,0 and 0,1

[smoky marsh]

Wait do I put the curves in the integral?

@gilded pulsar

.reopen

[trim jolt]

✅

[gilded pulsar]

You have to calculate the area under each curve

So yes you have to put each curve into an integral

[smoky marsh]

So i have to make 2 integrals?

[gilded pulsar]

Yeah

And then subtract one of them from the other

[smoky marsh]

But which interval goes to each integral

[gilded pulsar]

These

[smoky marsh]

I know the intervals

[gilded pulsar]

You have to do everything twice

[smoky marsh]

Ohhhh

[gilded pulsar]

The same

[smoky marsh]

So 4 integrals in total?

[gilded pulsar]

Yeah

[smoky marsh]

Alright

Since i gtg soon let's make this quick, so after i found the integrals

What do i do

@gilded pulsar

[gilded pulsar]

You need to

1. calculate the intersection coordinates
2. calculate the areas under each function between them with an integral
3. subtract 1 area from another
4. take absolute value to make it positive (might be negative)

[smoky marsh]

Yeah so we're going to have 4 results

From the 4 integrale

[gilded pulsar]

Repeat step 2-4 for each interval

So subtract integrals of the same interval

[smoky marsh]

Oh ok

In any order for the subtraction doesn't matter?

[gilded pulsar]

No if you get a negative answer just make it positive

[smoky marsh]

Alright

So I'm going to have 2 results then

Do I subtract those too?

[gilded pulsar]

No you add them

[smoky marsh]

Ok

Then that's it I'm guessing

[gilded pulsar]

Because you calculated two different areas

[gilded pulsar]

Thats why when you got 3 intersections you have to do everything twice and add

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[fiery frost]

Reading a paper about Geometric Algebra, and they give a quotient space definition for it. I am a bit confused about the ideal used here. Why is it in this form? How do you go about subtracting a quadratic form from a tensor product? What do elements of this ideal look like?

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[fair bison]

I'm confused, doesn't it explicitly tell you what elements of the ideal look like?

The end result is that v (x) v = q(v) in the quotient

Are you sure q is a quadratic form...? It says q(v) is a scalar. Unless that's shorthand for q(v,v) which would make sense.

[fair bison]

How do you go about subtracting a quadratic form from a tensor product?
You can subtract any two elements of T(V). It forms an algebra.

[fiery frost]

Yeah, Im moreso asking about the notation.

Like v x v is a tensor product so I imagine it looks like (v,v). q is a quadratic formm and the notation is shorthand for q(v,v). This is mentioned in the definition at the bottom of the image.

so if v x v is (v,v) what does it mean to subtract a scalar? feels like its saying something like (1,2) - 3

[fair bison]

Yes that's exactly what it's saying

If you quotient by the ideal generated by (1,2) - 3, then (1,2) = 3 in the resultant ideal

Scalars are in the 0th grade of T(V) and tensor products of two vectors are in the 2nd grade.

Think of it as similar to a polynomial. You can subtract x^2 - 5 without issue, can't you?

It just doesn't simplify further than that.

(Pinging @frank folio for GA)

[fiery frost]

hmm ok.

So then its just saying something like (3,q(v),1) is the same as (3,v,v,1)

[fair bison]

Just be careful not to confuse tensor product and direct product

Something like v3 (x) v (x) v (x) v1 is the same thing as q(v) (v3 (x) v1) in the quotient space, yes

(Using the fact here that the tensor product is associative so I don't need to add more parentheses)

[fiery frost]

Gotcha. The followup question I suppose, and perhaps the other person can chime in on this, but why is a GA defined using this specific ideal?

[fair bison]

So that the vector squared with itself gives you the length squared of the vector

You get a nice formula for the geometric product (the tensor product in the ideal) this way

[fiery frost]

I think I'm starting to see how these things connect. Thanks for the help!

[fair bison]

Namely a (x) b = a · b + a wedge b

in the quotient

I think it's something like that

But the idea is that the tensor product should capture both the correlation between the two vectors and the area the two vectors form, simultaneously

If a and b are parallel, then a (x) b is just equal to q(a,b)

If a and b are perpendicular (so q(a,b) = 0), then you would just have a (x) b = a wedge b

I might be getting some of the details wrong, but that's the basic motivation

[frank folio]

Yes

[frank folio]

The difference v \otimes v - q(v) can be treated as a formal difference, same thing you do when you look at complex numbers

Like wtf is 1+i?

It just is

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[barren tinsel]

Why is my solution wrong?

[barren tinsel]

this is the problem

someone plz help

[zenith nexus]

what is the full question

[barren tinsel]

calculate area

[zenith nexus]

Because I see integrals when what you posted only has 2 eqn

[barren tinsel]

find area

[barren tinsel]

2 eqn?

[zenith nexus]

what bounds

[zenith nexus]

equations

[barren tinsel]

Yes only 2 equations

two functions restrict area

I have to find that area

using integrals

I wrote down this integra

l

and timed it by 2

and the solution should be 4

but I got 8 :(( @zenith nexus

[zenith nexus]

at the extremes?

[barren tinsel]

I mean this should be the area that I have on the photo in my notebook

the same thing is in answers

@zenith nexus

[zenith nexus]

What is the original question

Because what you posted isn't clear whether you need to calculate both areas or just 1

[barren tinsel]

both areas

@zenith nexus

it means draw the area restricted by these functions and then find the area

THIS

[zenith nexus]

Seems like the answer is wrong

[barren tinsel]

thank you

I was so confused

🙂

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[pure epoch]

i dont know where to begin.

the gradient

[limpid barn]

Ok, so we have the information that a = pi +pj has a magnitude of 6.5 and the tangent of the angle (theta) that the vector makes with the positive x-axis is tan (theta) = -5/12

[pure epoch]

yes

sqrt(x^2+y^2)

ty

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[grave dove]

what did i do weong here?

[grave dove]

,rotate

[grave dove]

,rotate

[grave dove]

so the function is 4cos(1/2x-pi)-2

and i did substitiution

but i think i did something wrong

[mossy epoch]

@grave dove Still there ?

[grave dove]

yes

[mossy epoch]

Cos(u)=1/2 => u=60°

Can you tell me how you proceeded on obtaining this ?

[grave dove]

by doing the arccos

like

the opposite of cos

[mossy epoch]

So if you are using an inverse function, do you know its domain and codomain ?

[grave dove]

uhhh

my main language isnt english but

ill look it up rq

[mossy epoch]

I'll give you an example

[grave dove]

okay

[mossy epoch]

The domain of cos is

R

Because you can input any real number inside the function and it will make sense, it will give you a result

[grave dove]

yeah

uhh

[mossy epoch]

The codomain of cos is [-1;1], because there is no other values cos can reach

[grave dove]

oh okay

i didnt know

[mossy epoch]

Right

So maybe the use of the arccos function isn't to your best advantage since you don't seem to know it so well

Maybe what would help you more is to see in a trigonometric circle, what are the possible angles u for which cos(u)=1/2

Pretty sure you have done that in the past already

If not, then I'll need to know what you've been taught

[grave dove]

wait

[mossy epoch]

Does a trigonometric circle ring a bell already ?

[grave dove]

u mean the

sinus to cosinus to -sinus

[mossy epoch]

Yeah that thing is probably the same thing I am talking about

[grave dove]

wait so

what did i do wrong

like did i forget to turn 60 degrees into radians

[mossy epoch]

Yeah pretty much

[grave dove]

okay

wait i need u dont go

the last part is rly rly confusing

[mossy epoch]

Mostly I wanted to make sure you knew about the unit circle basically

Because using an inverse function involves certain things you might not know

Like what I told you about

So about that last confusing part, please tell me

[grave dove]

okaa

wait im nearlyt here

[mossy epoch]

Welp ping me when you're here

I'll do something else while I have some time

Law is the one who needs help

And I'm waiting him to ping me when he's done

Np np

[grave dove]

okay so

,rotate

[solid kiln]

Couldn't find an attached image in the last 10 messages.

Jan 28, 2025, 8:34 PM

[grave dove]

,rotate

[grave dove]

so

i did the u = 1/2x-pi to u+pi/1/2 = x

and put the stuff inside

and got the zeropoints yeah

[mossy epoch]

You've expressed x with respect to u

[grave dove]

uhhh

[mossy epoch]

Just a second

Did you have any sort of interval in which you needed to study f ?

[grave dove]

wdym

like a range?

from where the zeropoints go

[mossy epoch]

Yeah

?

When you're mentioning the zeropoints, you are mentioning the x values you want to get, right ?

I mean yeah there's not much other possibility now that I'm thinking about it

So yeah I'm talking about that

I can see a bit of an interval here at the top right for example, which makes me think that you didn't even really show me what you were supposed to do

I sort of deduced that you needed to solve the equation f(x)=0 for x

But that means I thought you were solving it for all real numbers

[grave dove]

yeah

[mossy epoch]

You don't seem to be doing that for all real numbers here

[grave dove]

yes

wait

okay

this is the interval

,rotate

[mossy epoch]

4pi included or excluded ?

[grave dove]

included

[mossy epoch]

Alright so

[grave dove]

but did i do the things correct

[mossy epoch]

You did correct things, in the way that all you did is correct

But did you miss on certain possibilities, you certainly did

[grave dove]

uhhh

wdym

[mossy epoch]

Once again, draw that unit circle you didn't draw

You'll realize there why

Okay, just imagine you have multiple solutions for the equation g(x)=k

[grave dove]

yes

[mossy epoch]

For k being whatever value

Then let's extract these solutions, let's say there are 3

x1, x2, x3 are then all the different solutions of the equation g(x)=k

Can you tell me what is the "arcg" of k ?

[grave dove]

arcg?

oh

[mossy epoch]

Just like you did with your arccos(1/2)

[grave dove]

90?

[mossy epoch]

You know cos is an even function, right ?

For any x, cos(x)=cos(-x)

[grave dove]

yeah

[mossy epoch]

Which means

Let's say our g takes the place of that cos function

So g is an even function

[grave dove]

yeah

[mossy epoch]

you know that g(x)=g(-x) for any x

[grave dove]

yeah

yes yep

i gotchu now

[mossy epoch]

So that means, you could say x1 is a solution

[grave dove]

bcs of the symetry

[mossy epoch]

But -x1 would be one as well

So in that case

What's the value of "arcg(k)" ?

x1 ? -x1 ?

What about x2 ? x3 ?

The moral of the story basically is :

When you have multiple solutions in such equations, there is a problem in using an inverse function

[grave dove]

yeah

[mossy epoch]

Because a function, by its nature, only gives one single value as a result

You are missing out on aaaaaaall of the other ones

[grave dove]

yeah

[mossy epoch]

May it be by missing out on the parity, as I explained here

Or by missing out on the periodicity as well !!

So yeah you goofed

Now try to work with a unit circle

I promise you that's gonna be a delightful tool when you want to look at all the possible solutions

@pastel drum

[grave dove]

yeah but

what did i do wrong

in short

[mossy epoch]

<@&268886789983436800> This @pastel drum is doing advertising in unappropriate channels

Sorry if you've been told multiple times

[dull temple]

banned like two minutes ago 😎

[mossy epoch]

Hell ye

Thanks

[mossy epoch]

In short, what you did wrong is to only consider one value of u to be solution of cos(u)=1/2

[grave dove]

no??

i said x2 is -x1

[mossy epoch]

That's for the parity, good

But what about the periodicity ??

A cosine function is 2pi-periodic may I remind you

So you might have missed on certain solutions out of just that

[grave dove]

oh

[mossy epoch]

And even though you didn't, you would NOT have proven that you just had all solutions all along

[grave dove]

okay man should i be honest

[mossy epoch]

Which wouldn't have completed the resolution of the equation

Yeah ?

[grave dove]

i still dont understand what i did wrong

[mossy epoch]

<@&268886789983436800> Yet an other one

[mossy epoch]

Maybe a vc with a stream would help better on explaining

<@&268886789983436800> Raid I guess ??

Good luck 💀

[grave dove]

i just checked on geogebra

i did it correct

for the first 2

but now

how do i go from the first x1 and x2 to the other x3 and x4

this was the confusing thing

i wanted to ask

[mossy epoch]

Maybe a vc with a stream would help better on explaining ?

Because I basically already explained it but it didn't work

[grave dove]

okay but uhh

[mossy epoch]

Images might make the trick

[grave dove]

i cant talk rn

[mossy epoch]

No problem I don't ask you to talk

[grave dove]

okay

[mossy epoch]

I only need to be heard so the help improves lmao

[trim jolt]

@grave dove Has your question been resolved?

[trim jolt]

@grave dove Has your question been resolved?

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[dusk summit]

am i right with D here?

[lone basin]

✅

!done ?

[trim jolt]

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

[dusk summit]

THANK YOU sm

[dusk summit]

yess

.done

.close

[trim jolt]

Channel closed

Closed by @dusk summit

Use .reopen if this was a mistake.

[trim jolt]

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.

[severe vigil]

I'm not sure how these answers are wrong. For (a) and (b) I initially put 1 for both, which was marked incorrect, then I changed it to the number that would be in the set. The question is not asking for the value of the number in the set, but rather the amount of numbers within the set. The questions I got wrong I was not able to do the work around as there were multiple numbers in the set and it is only accepting one number as the answer

[sharp heart]

With (A n B)^C, you first find which part is A n B.

Then, you only look at every other part.