help-0

[haughty oracle]

.close

[lone heart]

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[alpine sable]

I need help

[glass lichen]

Answer to your well worded question is 7

Glad I could help

Ask your math question in a clear, concise manner.
Show your work, and if possible, explain where you are stuck.

[alpine sable]

So, My teacher, Ms. Parry, told me to try and learn what polynomials are right, and so I need help trying to solve for them

I just need help on how to solve for them

[glass lichen]

I don't need to know your teacher's name.....

[alpine sable]

I know

[glass lichen]

Cause that's personal information

But anyway, I assume you've googled for information.

[alpine sable]

Yea

It takes me to some pdf

[glass lichen]

So..... what's your question?

[alpine sable]

Hold on

[glass lichen]

Cause there are plenty of resources online which can actually teach you.

[alpine sable]

She sent me

No context

[glass lichen]

yeah, that's factoring

[alpine sable]

I need help on it

[glass lichen]

Ok, post your attempt at factoring it

[alpine sable]

Soo

The problem is

Idk how to...

[glass lichen]

You looked up resources online

And taught yourself it

"How to factor quadratics" will more likely than not be very fruitful.

[alpine sable]

Is this a good video?

https://www.youtube.com/watch?v=VG1luiL_INQ&t=1s

[glass lichen]

probably

it's OCT

[alpine sable]

Yep

Imma be right back

So

I'm still confused

[glass lichen]

Damn, that's unfortunate

[alpine sable]

Ik

[glass lichen]

Too bad I don't know what you're confused about

[alpine sable]

This is the last day until report cards come back, last semester :((

Well

Thanks for your patronage!

Byeeee

[glass lichen]

@alpine sable close the channel then.

[lone heart]

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[dense plover]

can anyone help me?

i know that PC is 1 but i dont know what to do after

anyone?

[alpine sable]

AS * AB = AP * AC?

[dense plover]

ik

[hollow inlet]

Use chord theorem

[dense plover]

i got pc is 1

[alpine sable]

what would be another way to write AS and AP

in shorter segments

[dense plover]

?

[hollow inlet]

Sum of 2 segments?

[dense plover]

like as is ab plus bs?

[alpine sable]

what about ap

[dense plover]

ac plus cp?

[alpine sable]

(AB + BS) * AB = (AC + CP) * AC

now plug in know values

you get PC=1, and now you can write AP=PC+AC correct?

[dense plover]

yes

[alpine sable]

now using pythag theorem

what would SP be

[dense plover]

is sp the diameter?

because im trying the radius of the circle

[alpine sable]

just trust me

[dense plover]

oh

i get it

you can make another right triangle

[alpine sable]

triangle SPC comes will help you solve

exactly

[dense plover]

thanks

[alpine sable]

np

[dense plover]

.close

[lone heart]

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[alpine sable]

can aynone help me with this? huge thanks

[silver adder]

$\table$

[ocean seal]

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

[silver adder]

do you know what a truth table is

[alpine sable]

p | -'p | p ^-'p

T F F
T F F
F T F
F T F

like this?

i am just not sure about my answers

[silver adder]

kinda

but thats wrong

[alpine sable]

yea

:C

[silver adder]

well

its correct

u used too many entries

u only need 2

[alpine sable]

i tried number 1

[silver adder]

p | -'p | p ^-'p

T F F
F T F

[alpine sable]

oooo

for 1 to 5 i only need 2?

[silver adder]

huh

u only need 2 because p only has 2 states it can be in

p^-p in english means, p and not p, which is impossible

p cant both happen and not happen

[alpine sable]

ohhh

[silver adder]

p | -p | p V-p

T F ?
F T ?

[alpine sable]

TT?

T
T

?

[silver adder]

yup

p or not p is always true

[alpine sable]

ohhhh

thank u so much

[silver adder]

for c and d youll need a truth table with 4 rows

[alpine sable]

yea lemme try

p q ((p ∨ ¬q) → q)
F F F
F T T
T F F
T T T

is it

this

or wrong

[silver adder]

lemme check

hm

nuhuh

youll have to check the truth table for →

[alpine sable]

ohhh

[silver adder]

maybe its right actually

@alpine sabledo u think its right?

[alpine sable]

kinda

hahahaa

im not sure

[silver adder]

lets see

p q (p ∨ ¬q) ((p ∨ ¬q) → q)
F F T F
F T F T
T F T F
T T F T

seems right

good job @alpine sable

[alpine sable]

thanks man

now last onme

wait

[silver adder]

p q (p V q) (p ^ q) (p V q) → (p ^ q)
F F F F T
F T T F F
T F T F F
T T T T T

[alpine sable]

oh kinda hard

i was struggling hahaha

thank u so much for ur help @silver adder

u're the best

[silver adder]

np

what were u struggling on?

[alpine sable]

the last one but i figured it out

i'm kinda new to this because we have no math in our last sem

[silver adder]

i think just remember that T → F is false, and the rest is true

[alpine sable]

ooo

copy

[silver adder]

oh i mean T → F

its honestly confusing

[alpine sable]

yeah ikr

hahahaah

[silver adder]

something true cant imply something is true if its actually false

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[lone heart]

@silver tangle Has your question been resolved?

[lone heart]

@silver tangle Has your question been resolved?

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[half stream]

Anybody know how I can solve a problem like this?

[half stream]

I see

So I have 71p=(x-1)(x+1), what next

[rigid vault]

71p+1= 4a

[half stream]

Ahh that makes sense

wait this is 71 * p + 1 or 71(x)(p) + 1

72

no worries

73

[upbeat wing]

is 73 a prime

[half stream]

yes

ohhhh

interesting

and that fulfills the rule of x^2 = (x-1)(x+1) + 1

(72+1)(72-1)+1

that is sneaky

will the same process work for something like 71p+2?

assuming that there is a number that fulfills that condition

More general - something that is not +1

which is the constant?

I do don't worry

I'm a bit tired haha

But anyways, what would a problem in the same format of the one I sent look like for this problem

[upbeat wing]

errr

wait

thats a better example

[half stream]

would you work that out here so I can see another example? Just so I know what to plug in

[upbeat wing]

so we have 53 * p + 9

we square root the 9

we get 3

then x = 56

so we plug x = 56 into (x+3)

we get 59

if u calculate 59 * 53 + 9

you get 3136

which is 56^2

and also

59 is a prime

[half stream]

So 59 is p

Nice, you think you could give me an example to work out?

hmm im having trouble with the primes part, because 89-5=84 and 89+5=94

wait no

79=p

Thx

You think you could give me one more?

Sorry haha

Thank you so much for the help!

Hmm would 43 also work for that one? I calculated p for 23 and 43 and they both come out as a perfect square and they are both prime. Once again, thank you for the help

[lone heart]

@half stream Has your question been resolved?

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[tawny fable]

hey there

[tawny fable]

Ok sooo

I know its a pretty basic question

but please, if anyone could help me with that I would appreciate it

I guess I need to use the dimensions property that dim(U+W) = dim(U) + dim(W) - dim(UcapW)

[lone heart]

@tawny fable Has your question been resolved?

[tawny fable]

<@&286206848099549185>

<@&286206848099549185>

anyone? 🙂

[lone heart]

@tawny fable Has your question been resolved?

[mossy lion]

@tawny fable i think you can show that dim(W)=dim(U) by showing v1-w,...,vk-w are linearly independent. but I'm not 100% on that

if you do figure it out please ping or dm me, I'm curious to see if my solution is correct.

[lone heart]

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[opaque inlet]

How do I do question 5?

[vale wigeon]

have you made any progress so far?

[opaque inlet]

.close

[lone heart]

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[vale wigeon]

.reopen

[lone heart]

✅

[vale wigeon]

@opaque inlet you would do well not to just close the channel immediately on me. it's kind of rude.

if you do not need any more help then at least have the decency to say "sorry, i'm already done"

[lone heart]

@opaque inlet Has your question been resolved?

[opaque inlet]

Can anyone help with question 5?

[high badger]

try letting $x as the full price and build an equation base on that

[opaque inlet]

I’m not sure how to tho

Just with the variable x?

Wouldn’t we need y as well for the remainder

Wait

[high badger]

so the 50 wine are divided into 2 group
25 of them is full price ($x)
25 of them is 40% off
can you use a way to represent 40% off with x?

[opaque inlet]

25x + 25 * 0.6x=2260?

[high badger]

almost there

yes

40% off is (1-40%)=60%

[opaque inlet]

Sorry about the 0.4, I wrote 0.6 in workbook

[high badger]

thats correct now

[opaque inlet]

Yeah

Ahh ok

[high badger]

good job

[opaque inlet]

I got x=56.5

$56.50

Thank u

.close

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[rough salmon]

154x-135y=0 how should i find p and q if px+qy=0 assuming p is a prime number

[pale kestrel]

prime factorize

[rough salmon]

oh gg

hmm, still stuck tho... @pale kestrel

<@&286206848099549185>

<@&286206848099549185>

[lone heart]

@rough salmon Has your question been resolved?

[rough salmon]

<@&286206848099549185> someone plzz help

[twin haven]

prime factorize them

[pale kestrel]

maybe, just maybe post your work

we dont read minds

[twin haven]

do you know how to prime factorize

[rough salmon]

what work should i send?

[rough salmon]

after i prime factorize, what should i do?

sry for the late reply btw

[twin haven]

do you get prime no.s

[rough salmon]

heh nvm, the question i asked is derived from another question, n i got the nums wrong, soooo sry 😅

[lone heart]

@rough salmon Has your question been resolved?

[rough salmon]

.close

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[indigo crown]

How do I solve this system in terms of a and b?

[indigo crown]

I tried putting it into row echelon form and got y = 1/4 b, but this confused me more

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[grave matrix]

Vector calculus notation question: given F,G vector fields, does G(div F) mean G . (div F) [where . is the dot product]?

[grave matrix]

I've been trying to understand the expansion of curl (F x G)

[lone heart]

@grave matrix Has your question been resolved?

[vale sapphire]

div F is a scalar, so there's no need for a dot product

[grave matrix]

Ahhh

That explains my confusion

Cheers

.close

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[indigo crown]

I don't know what to do to the matrix to bring it to echelon form (if that's the right way to go), pls help

[zenith compass]

Do you know the determinant

[indigo crown]

no, what is that?

[zenith compass]

Then did you learn about matrix

[indigo crown]

Yes

[zenith compass]

Good

[indigo crown]

but i don't know what to do to the matrix to find k

[zenith compass]

Consider the square system of linear equations $Ax=b$ with $A=\begin{pmatrix}9&k\k&1\end{pmatrix}, b=\begin{pmatrix}9\-3\end{pmatrix}, x=\begin{pmatrix}x\y\end{pmatrix}$ where $k$ is an unspecified real parameter.

First do you understand such notation

[indigo crown]

Yes I understand

[zenith compass]

Good

[indigo crown]

But what do you have Ax = b?

[zenith compass]

Oh wait

[ocean seal]

Trenton

[zenith compass]

It looks better now

The focus is

The matrix A

[indigo crown]

Yes I understand that

[zenith compass]

The determinant of $A=\begin{pmatrix}a&b\c&d\end{pmatrix}$ is denoted by $det(A)$ and is defined by $det(A)=\begin{vmatrix}a&b\c&d\end{vmatrix}=ad-bc$

[ocean seal]

Trenton

[indigo crown]

so is that 9y-k^2?

[zenith compass]

When there is no solutions or infinitely many solutions, you have $det(A)=0$

[ocean seal]

Trenton

[zenith compass]

Not really, y is not included

[indigo crown]

oh ok

9-k^2? for a unique solution?

[zenith compass]

Well if you want unique solution for the system, you should set $det(A)\not = 0$

[ocean seal]

Trenton

[zenith compass]

Otherwise, $det(A)=0$, which implies there are no solutions or infinitely many solutions.

[ocean seal]

Trenton

[indigo crown]

oh i get it now

[zenith compass]

Good

Btw after you find k

[indigo crown]

but what does a determinant mean?

[zenith compass]

You should try to compute such k into the system

Cuz u dun know whether it is the case of infinitely many solutions or no solutions, unless you put back into the system

[indigo crown]

I just did that and I got -9 = 9

[zenith compass]

You can think it as a characteristic of the square matrices

[zenith compass]

?????you mean k=9 or -9?

[indigo crown]

I got k = 3 and computed it into the system which gave me -9 = 9

[zenith compass]

Ic

[indigo crown]

Is that proving that there's no solution? Or did i make a mistake?

[zenith compass]

Alright you should get k=3 or -3 if you are correct

K=3 for no solutions

K=-3 for infinitely many solutions

[indigo crown]

why are you also saying -3

oh ok

[indigo crown]

What do you equate the determinant to for finding k = -3

[zenith compass]

Well

What is the determinant?

9-k^2 right?

Set it equals to 0

$9-k^2=0 \iff 9=k^2 \iff k=3$ or $k=-3$

[ocean seal]

Trenton

[indigo crown]

Ok

how do you know that it's just 3 for no solutions and -3 for infinitely many solutions?

[pale kestrel]

When there is no solutions or infinitely many solutions, you have det(A)=0

Did you understand this?

If not then it will less easy to explain

[indigo crown]

I just accepted it. I don't really get why that's the case though.

[pale kestrel]

Have you done linear transformations

[indigo crown]

Yes, but i forgot most of it

😅

[indigo crown]

would you please explain it anyway?

[pale kestrel]

So you have seen different types of matrices

and what geometrical transformations

they are meant to represent?

[indigo crown]

Yes

[pale kestrel]

Ok, so take your 2x2 matrix.

Oh wait --- are you familiar with the geometrical interpretation of determinant

If I take a 1x1 grid square in my original plane

The area of this square after a linear transformation on the plane is the determinant of the 2x2 matrix

This generalises to higher dimensions (1x1x1 cube in 3d, etc).

[indigo crown]

ok

[indigo crown]

and then what's next?

[pale kestrel]

Are you familiar with the word 'bijection'

'kernel'

either of these wrods

[indigo crown]

no

[pale kestrel]

Did you know a matrix has an inverse if and only if the determinant is non-zero

[indigo crown]

i knew this:

[pale kestrel]

If a matrix is invertible, this tells you the transformation is one-to-one

Right and if the determinant is 0, this is undefined

If a matrix is invertible it tells you that you can 'undo' the transformation

This is because no information is lost. Each point maps to exactly one point in R^2 and no two points map to the same

[indigo crown]

What R^2? Row 2?

[sterile idol]

This sounds very familiar to the conditions required for there to exist a proper inverse of a function

Maybe it's the same thing

[sterile idol]

Two R

(R, R) I believe, might be wrong lol

[pale kestrel]

$\bR^2$

[ocean seal]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[indigo crown]

What's 2R

[pale kestrel]

This is the plane

[sterile idol]

R^n means nth dimension

[pale kestrel]

Have you seen this notation?

Its all the (x, y) coordinates

in the plane

[indigo crown]

nope

[pale kestrel]

Sorry I haven't got a good grasp on what you have or haven't seen

😅

$$\bR^2 := {(x, y) : x,y\in\bR}$$

[ocean seal]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[pale kestrel]

I don't know if you have seen set notation either

if not, then this won't help

It is the set of all points written as (x, y)

Or alternatively all 2-dimensional vectors

[indigo crown]

I've seen the notation but not the R^2 (I understand what you are trying to say though, I just haven't come across the concept of R^2 as the set of...)

[pale kestrel]

Ok, (am thinking a bit how to explain sry)

So if the determinant is 0

The image of the transformation is either a point (the origin) or a straight line

Is this familiar or no

[indigo crown]

yes vaguely familiar

[pale kestrel]

Right and this is where the justification for infinite or no solutions is

We have

Ax = b

b is a point on the plane after we applied transformation A

Let's say the determinant of A is not 0

Then we know exactly one point maps to b

and we can invert the transformation by multiplying by the inverse of A

$$A\bs x = \bs b$$
$$A^{-1}\left(A\bs x\right) = A^{-1}\bs b$$
$$(A^{-1}A)\bs x = A^{-1}\bs b$$
$$I\bs x = A^{-1}\bs b$$
$$\bs x = A^{-1}\bs b$$

[ocean seal]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[pale kestrel]

The last line gives you an exact 'formula' to get the solution x

it is A^-1 b

This happens whenever A is invertible. This means there is exactly 1 solution

When A isn't invertible, then det A = 0. This tells you the image of the transformation is either a line or a point

[indigo crown]

What about when there are infinite solutions?

[pale kestrel]

Right that's what I'm trying to say

$$R^2=\left{\begin{pmatrix}x\y\end{pmatrix} : x,y\in\bR\right}$$
$$\tn{im }A = \left{A\begin{pmatrix}x\y\end{pmatrix}: x,y\in\bR\right}$$

im A means 'the image of A'

[ocean seal]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[pale kestrel]

That's everywhere A can map to

[indigo crown]

Thank you, but I don't understand how to use that to solve this:

[pale kestrel]

So imagine this picture

[pale kestrel]

If the image of A is that line, then how many solutions there are depends on where b is

If b isn't on that line, then there are no solutions

If b is on that line, then there are ??? solutions

[pale kestrel]

This is the infinite case.

(I wish I explained this better )

[indigo crown]

Yeah i understand the graphic explanation. But how do I show this in my calculations?

[pale kestrel]

So firstly you need to calculate the determinant of A

that is.........

(9 x 1) - k^2

[indigo crown]

9-k^2

[pale kestrel]

9 - k^2

So if this thing isn't 0

[indigo crown]

thank you anyway 😄

[pale kestrel]

There is a unique solution

Is that good?

So that solves part c)

If it is 0, then you either have no solutions, or infinite solutions

[indigo crown]

Does that mean I equate it to not zero (for a unique solution)?

[pale kestrel]

yes

$$9 - k^2 = 0$$

[ocean seal]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[pale kestrel]

Or you can equate it to 0

then unique solutions is whenever this is false

So k = +- 3 gives 0

everything else gives not 0

So k isn't 3 or -3, then we know there is a unique solution

[indigo crown]

OHH okay

[pale kestrel]

Next, if k is 3 or -3, you have to do some thinking

It relates to this picture in a way

The image of A is either a line or the origin

If you have a 0 determinant

It is the origin only when A is the zero matrix (matrix with all entries 0)

Otherwise you get a line

Does that make sense?
So in this case, you don't have the 0 matrix for sure, so im A is a line

You then need to figure out if the image of A includes or doesn't include b. Which is (9, -3) here

You do this by solving the equation system

[indigo crown]

I substituted k = 3 into the system and ended up with -9=9

[pale kestrel]

Right

So there must be no solutions

$$\begin{pmatrix}9&k\k&1\end{pmatrix}\begin{pmatrix}x\y\end{pmatrix} = \begin{pmatrix}9\-3\end{pmatrix}$$

[ocean seal]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[pale kestrel]

Well you don't need matrix you can work with linear equations

(probably easier)

then you sub in k = -3

[indigo crown]

how do you say that? Is that based on the answer that -9 cannot equal 9?

[pale kestrel]

You can say the system is inconsistent

if k = 3

because you end up with 9 = -9

This is impossible, therefore there are no solutions

[indigo crown]

Right okay 😄

[pale kestrel]

9x + 3y = 9
3x + y = -3
=>
9x + 3y = 9
9x + 3y = -9
=>
9 = -9

Therefore there are no solutions for x, y

Next, you solve the system for -3

9x - 3y = 9
-3x + y = -3

[indigo crown]

yup and x=1

[pale kestrel]

no no

Can you see the 2 equations are the same?

If you multiply the bottom one by -3

you get exactly the top one

(or divide the top by -3)

9x - 3y = 9
-3x + y = -3
<=>
-3x + y = -3
-3x + y = -3
<=>
-3x + y = -3

So the system of equations is equivalent to just one equation

So now you know (x, y) is a solution to the original equation if and only if -3x + y = -3

[indigo crown]

oh okay

[pale kestrel]

There is 1 equation, but 2 variables

[zenith compass]

Once you find out that the equations are the same, geometrically there would be two overlapping straight lines

[indigo crown]

and that means there are infinite solutions!

[pale kestrel]

!

👌

[zenith compass]

Ya

[indigo crown]

That means for a unique solution, k cannot be +3 or -3

Thank you so much @zenith compass @pale kestrel !!

[zenith compass]

Note that geometrically, no solutions is just two parallel lines

[pale kestrel]

And also you know what happens when k = 3. and k = -3

[zenith compass]

You are welcome

[indigo crown]

when k = 3, there are no solutions, when k = -3 there are infinite solutions (based on substituting each k value into the system)

[pale kestrel]

yes

There are 2 geometrical pictures you can consider. The linear transformation of the matrix itself. And the individual rows as linear equations y = mx + c
(I think they coincide in some way but uhh don't confuse them)

[indigo crown]

so next time, I have to equate the determinant to 0 and see how it goes

[pale kestrel]

Basically yes.

[indigo crown]

what happens when I don't have a square matrix?

[pale kestrel]

This is a bit more tricky

we cannot talk about determinant

But you do the same thing

You attempt row reduction

[indigo crown]

into row echelon form?

[pale kestrel]

yes

And split into cases whenever you want to divide by something that might be 0

[indigo crown]

how though?

I tried doing that for the same question and I didn't know how to go about

[pale kestrel]

Let me write an example

 3x + ky = 3
 kx +  y = 4
-kx + 3y = -2```

might be a stupid example 😂 let me change some numbers

Ok, so row reducing what might your first step be

[indigo crown]

one moment

[pale kestrel]

https://i.imgur.com/tO1vlCb.png

I would reference a list like this that tells you the legal operations

[indigo crown]

row 2 + row 1

to make -k = 0

[pale kestrel]

are you sure?

(3 + k)x + (k + 1)y = 7

[indigo crown]

i mean row 2 + row 3

[pale kestrel]

ah sure

[indigo crown]

sorry 😅

[pale kestrel]

So you are applying rule 3

[indigo crown]

ok

[pale kestrel]

 3x + ky = 3
 kx +  y = 4
      4y = 2```

[indigo crown]

yeah i understand

[pale kestrel]

Ok, then next is a tricky part

You want to eliminate either of the x

If you divide by k, you need to split into 2 cases, k = 0, or k not 0

[indigo crown]

is it row 2 x 4 subtracted from row 3?

nvm we already have enough zeroes right?

[pale kestrel]

Uhhhhh I usually would be systematic about row reduction

although you could

I would usually first eliminate all the x's possible

then all the y's

[indigo crown]

ok

[pale kestrel]

and so on

(note that sometimes is not a good idea when there is some shortcut)

So you want to eliminate either x

[indigo crown]

ok

[pale kestrel]

I would advise using rule 3 again

 3x + ky = 3
 kx +  y = 4
      4y = 2```

what if you first multiply row 2 by 3 (rule 2)

 3x + ky = 3
3kx + 3y = 12
      4y = 2```

And then can you apply rule 3?

(You could have divided row 2 by k, but you want to avoid that basically - splitting cases early on means more math)

[indigo crown]

but row 3 has a 0

so how are you getting 3kx

[pale kestrel]

Row 3 has no x

[pale kestrel]

I just multiplied row 2 by the scalar 3

R2 -> 3R2

[indigo crown]

ohh

okay

[pale kestrel]

Applying rule 2

(i didnt have to, but to make things easier to see)

Then you want to kill the x's of one row

Can you add a scalar multiple of row 1 to make this happen?

[indigo crown]

k?

[pale kestrel]

-k

[indigo crown]

Ah okay

[pale kestrel]

 3x + ky = 3
3kx + 3y = 12
      4y = 2```
R2 -> R2 - kR1

[indigo crown]

Yup i understand that

but doesn't that make 0?

[pale kestrel]

   3x +         ky = 3
        (3 - k^2)y = 12 - 3k
                4y = 2```

I may have written it incorrectly

just checking first

ok

[pale kestrel]

yes k could be 0

but rule 3 is very useful

because you are allowed to the scalar multiple to be 0

$R_n \to R_n + 0R_m$

[indigo crown]

i just multiplied R1 by -k

[ocean seal]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[pale kestrel]

This is allowed

[indigo crown]

so i have -3k k and 3k

[pale kestrel]

No, you cannot do that

You cannot do R1 -> 0R1

(so if you multiply by k, you could need split cases)

[pale kestrel]

But this operation is allowed without needing to split cases

$R_2 \to R_2 - kR_1$

[ocean seal]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[pale kestrel]

So you are allowed to do this

[indigo crown]

are you saying that if i multiply by k and then subtract, it is allowed? But if i just multiply by k, it is not allowed?

[pale kestrel]

There is no restriction on the 'scalar multiple' for rule 3

I am saying you cannot do
$$R_1 \to -kR_1$$
and then
$$R_2 \to R_2 + R_1$$

[ocean seal]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[pale kestrel]

Instead, you should do

ie. leave R1 unchanged

Eliminate the x's in R2 utilising rule 3

Basically, I am following these rules exactly

And rule 2 has a condition that the scalar cannot be 0

[pale kestrel]

So the first step requires me to split cases if I do R1 -> -kR1 first

[zenith compass]

The reason why you cannot multiply both sides by zero is that, if u do so, you always get 0=0 which is always true (even originally it is sth nonsense like 0=1)

[indigo crown]

ok

[pale kestrel]

2x + 2y = 1
1x + 1y = 2

~

2kx + 2ky = 2k
 1x +  1y =  2

~ (if k = 0)

 0x +  0y =  0
 1x +  1y =  2

Its something like this

if k is 0

you kindof kill an equation

[zenith compass]

And it will be ridiculous if u get 0=0, totally useless to your calculation, so we define that we can only multiply a row by non zero scalar

[indigo crown]

ok

[pale kestrel]

But rule 3 is different.

[indigo crown]

now i have 0 (3-k^2) and (12-3k)

[pale kestrel]

   3x +         ky = 3
        (3 - k^2)y = 12 - 3k
                4y = 2```

You should have this

yes?

So all the x's are eliminated

Next you need to try eliminating the y's

Again, rule 3 is recommended

Try
$$R_1\to R_1 + \lambda_1R_3$$
$$R_2\to R_2 + \lambda_2R_3$$

[ocean seal]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[pale kestrel]

I write lambda1 and 2 for the scalars you need to figure out

It is possible to eliminate the y's in Rows 1 and 2 this way.

[indigo crown]

One moment I have a question

[pale kestrel]

(Ah the pains of explaining through text linear algebra 😂 --- much easier in person)

[indigo crown]

See where we multiplied by just 3 for R2? How is that allowed?

[pale kestrel]

because 3 is not 0

[indigo crown]

But for multiplying a row by k, it has to split?

[pale kestrel]

because k might be 0

Also, instead of multiplying R2 by 3

and then R2 - kR1

I could have done something else

[indigo crown]

ohhh

ok ok

[pale kestrel]

$$R_2\to R_2 - \frac{k}{3}R_1$$

[ocean seal]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[pale kestrel]

I multiplied R2 by 3 just to avoid fractions

[pale kestrel]

This step alone would achieve the elimination of x

[pale kestrel]

So you could have done this first and then R2 -> 3R2 to get rid of the fractions

So technically, the 2 steps we did is this
$$R_2\to 3R_2 - kR_1$$

[ocean seal]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[pale kestrel]

This is allowed, because this applies rule 2 (multiplying by 3) and then rule 3 (adding -kR1 to R2)
OR
rule 3 (adding -(k/3)R1) and then rule 2 (multiplying by 3)

$$R_2\to 3R_2 - kR_1$$
$$R_2\to 3\left(R_2 - \frac{k}{3}R_1\right)$$

[ocean seal]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[pale kestrel]

Either works (which represents 2 steps)

[indigo crown]

ok ok

Now is it kR3-R1?

to eliminate y?

[pale kestrel]

So I suggested this

let me see

   3x +         ky = 3
        (3 - k^2)y = 12 - 3k
                4y = 2```

I don't think so

[pale kestrel]

You should apply row transformations in this form

[indigo crown]

oh yeah i remember 😄

[pale kestrel]

The reason I am doing it like this is to avoid multiplying by k...

I don't know if that is clear to you

Why I decide to leave R_3 unchanged

and eliminate y from R1, R2

If I tried the other way round, case splitting would be necessary.

[indigo crown]

o

ok

What is lambda

[pale kestrel]

some scalar constant

you need to find

So let's think about the first one

R1 -> R1 + ? R3

What multiple of R3 would eliminate ky from R1

   3x +         ky = 3
(R2)
                4y = 2```

[indigo crown]

4

[pale kestrel]

R1 -> R1 + ?R3

This is

ky -> ky + ?(4y)

We want this to be 0

? isn't 4.

[indigo crown]

-1

[pale kestrel]

ky - 4y makes (k-4)y which isn't 0

ky -> ky + a(4y) = 0
ky -> ky + 4ay = 0
4a = -k
a = -k/4

Would this thinking help?

The top is what we want. The bottom is what we need in order to make it happen

[indigo crown]

ah yeah i understand

Thank you

[pale kestrel]

So just one last thing

You might have examples where splitting cases is unavoidable

1x +  ky = 3
     2ky = 1

In this case, to eliminate the y's

You have to divide by k no matter what

[indigo crown]

ok

[pale kestrel]

So you split into cases.

[indigo crown]

how

[pale kestrel]

Suppose k = 0

1x +  0y = 3
      0y = 1

Ok, this was kinda a dumb example, so this automatically eliminates y

I will change the example to this

1x + (k+1)y = 3
        2ky = 1

Because both y's have a coefficient involving k in them

You will have to split cases in order to eliminate.

Suppose k = 0

1x +     1y = 3
         0y = 1

(Ok there was no point changing the example, nvm)

Then this gets you the answer

Suppose k isn't 0

1x + (k+1)y = 3
        2ky = 1

Now you can divide by k

1x + (k+1)y = 3
         2y = 1/k

And then I can subtract (k+1)/2 of R2 onto R1

R1 -> R1 - (k+1)/2 R2

1x +     0y = 3 + (k+1)/(2k)
         2y = 1/k

Then this gives you the solutions

Maybe there's a good video online/tutorial explaining this

[lone heart]

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[wispy urchin]

Solving steps, please?

[wispy urchin]

Here’s my attempt.

[lone heart]

@wispy urchin Has your question been resolved?

[wispy urchin]

<@&286206848099549185>

[harsh swallow]

apply the formula again to the cos^2(2x)

to get 4x

@wispy urchin

[lone heart]

@wispy urchin Has your question been resolved?

[wispy urchin]

Thank you!

.close

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[glad sluice]

hmm that's weird

[swift locust]

mhm

[wispy urchin]

?? I’m not the best at trig identities

[wispy urchin]

Yo mamas weird :/

[glad sluice]

yeh i agree

[wispy urchin]

@swift locust

Does this help?

[swift locust]

thats the answer i put

[wispy urchin]

Oh, I didn’t see the incorrect part!

[swift locust]

and it said that i was wrong >:C

[wispy urchin]

LMAO

[coral moss]

very weird indeed

[swift locust]

the machine is drunk or some

[coral moss]

so many different ways to show that the answer you have put in is right yet the "system" says it's wrong

[swift locust]

anyways

i also thought this answer was correct but system says otherwise

this problem takes more steps so i prob made a mistake but i solved it three times and got same answer

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@swift locust Has your question been resolved?

[lone heart]

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[candid torrent]

hey so i'm just wondering how to factor polynomials of degree three. For example, x^3 + 2x^2 -x -2

[candid torrent]

i was told that all the possible factors of this polynomial is the factors of the final term

do i just use synthetic division with all of the possible combinations until i find one?

please ping me when you have an answer

[vale sapphire]

well, what people probably meant by this is that the rational roots of this polynomial must be divisors of the last term

[candid torrent]

yes, that's what I meant

[vale sapphire]

Since the last term is -2, the only possible integer roots of the polynomial are 2, 1, -1 and -2

[candid torrent]

wait so it's +-

?

[vale sapphire]

Yeah

2 divides -2, does it not?

[pale kestrel]

Assuming we have rational roots then sure

[candid torrent]

ok so i try synthetic division with 2, 1, -1, -2?

[pale kestrel]

indeed.

or maybe dont do that

just try factor theorem

much quicker

[candid torrent]

sure

so uhhhh

let f(x) = , x^3 + 2x^2 -x -2

f(2) = (2)^3 + 2(2)^2 -2 - 2

so i just check if this is zero? @pale kestrel

[pale kestrel]

yes

[candid torrent]

ok i can take it from here

[pale kestrel]

if this is 0, then x - 2 is a factor

[candid torrent]

thanks man

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[unkempt umbra]

how do I solve (ii)?
there are n! bijective functions between M and M so we have 6 functions in this case
But how should the operation table look like?

[oak perch]

A 6 times 6 multiplication table…

Just do it by definition…

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[alpine sable]

Is this correct?

[dense compass]

the 1/2 is multiplying (v_0+v)t so I don't think you can just subtract it to the other side

if that makes sense

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[slate kayak]

If a graph intersects the x-axis then the coordinates used to formulate the tangent will only be (x,0) right?

[slate kayak]

This is not a textbook question or anything i just want to know for sure before solving an exercise

[merry depot]

If the tangent is at that point, sure

[steady echo]

yes, I think so

[slate kayak]

Yes, thanks!

[slate kayak]

Thank you!!

Have a nice day you both

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[austere karma]

Why do we ignore the square root sign when integrating

[steady echo]

What do you mean?

[austere karma]

I mean why do you not take the absolute value when taking the square root when integrating

[tacit arch]

Give a concrete problem or example

[little drum]

?

"when taking the square root" ?

[austere karma]

When we integrate $\int \sqrt{x^2} dx$ we get $\int x dx $ not $\int |x| dx$

[ocean seal]

S̴̤̃e̶̪͗e̵͉̋

[austere karma]

why do we ignore the absolute value

we get told to ignore it for some reason

[steady echo]

Ah that, so afaik (other ppl may add more), that's just mathematical convention I think. Square root in general refers to only the positive roots

[austere karma]

I don't think mathematical convention explains that

You need the modulus sign when square rooting something that has been squared in general

[steady echo]

There's a deeper reason relating to functions too, I think. So, a function always has to have only one y value per x, if you took both possible roots, it'd mean there's more than one y value for a given x so you just take the positive ones I believe

[austere karma]

But to just take the positive root, you need the modulus sign

As x may be negative

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[stoic hound]

$\int_{-\infty}^\infty \frac{\sin^2x}{x^2},dx=??$
Apparently the answer is $\pi$. How?
The antiderivative is non-elementary, so there must be some sort of 'trick'. But the most common trick for integrals like these, i.e. the residue theorem, doesn't work since there aren't any simple poles. So I don't know how to tackle this now. Please help!

[ocean seal]

idioticbaka1824

[tacit arch]

Umm Fourier Transform?

[alpine sable]

#bots

[tacit arch]

Complex analysis works too

[stoic hound]

i totally forgot about the fourier transform. lemme look up how it works again real quick

wait a sec, isn't fourier transform mainly for solving DEs?

[tacit arch]

Fourier transform is more versatile than that

But you're probably thinking of Laplace transforms

They're nearly identical anyway

[stoic hound]

oh wait yeah there was like parseval's theorem or something

the FT preserves inner products, so let's see if we can rewrite this as an inner product

oh man, i spent a while puzzling over the FT of sinx/x and eventually found an answer on stackexchange - you need to IFT the sinx/x to get a little box-like function. who would've thought of that?

anyways thanks a lot!

.close

[lone heart]

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[dire epoch]

Hi, I'm currently doing a recap on sequences (I did this two years ago, but it's completely slipped out of my head and I feel idiotic)
I just want to get a better understanding of formulas
I'm supposed to 'Write down the formula for the nth term of the following sequence: 1, 3, 9, 27, 81

[ebon condor]

I mean like there are potentially infinite sequences

[dire epoch]

How so?

[ebon condor]

I mean like you can construct any n > 5 degree polynomial

[vagrant raptor]

It's obvious what they are looking for though.

[ebon condor]

but there is a simple solution

and it's geometric

[dire epoch]

Ah right, Arithmetic, Geometric, Polynomial sequences, I believe this is geometric (Multiplication)

[vagrant raptor]

Yes, what's happening between each element?

[dire epoch]

It's multiplying by 3 (1x3 = 3, 3x3 = 9, etc..)

I was going to put 3n as the formula, but since it starts at 1, it can't be that

[vagrant raptor]

you mean 3^n?

[dire epoch]

no I meant 3n because it goes 3 then 9, but like I said it can't be 3n since it starts at 1

[vagrant raptor]

3*n would mean the sequence would go 3, 6, 9, 12

3^n would mean element 1 is 3, but... it's close

[dire epoch]

oh. my mistake sorry

[vagrant raptor]

no worries at all!

when writing the formula for the nth element we completely ignore what the element before was etc.

[dire epoch]

ahh okay I wasn't aware of that

[vagrant raptor]

yeah, we want some expression that works on its own

if that makes sense?

[dire epoch]

yeah I think I understand

[vagrant raptor]

okay but let's say we try 3^n

what would some element 0 be?

[dire epoch]

since you ignore the element before, would that be 1?

[vagrant raptor]

yeah, it's a completely standalone expression

1 is correct

now, one could argue that the first element is no. 0

but lets assume they want us to solve it as if the first element is no. 1

[fluid eagle]

programmer moment

[vagrant raptor]

yes ^^

(in computer science, counting often starts at 0)

[fluid eagle]

also polynomials and many nice summations

[dire epoch]

hm okay, that's good to know thanks!

[vagrant raptor]

but you see how 3^n is correct, just shifted one step?

[dire epoch]

sort of, yes

[vagrant raptor]

so where we have element n right now, we want element (n-1)

i'd write out the sequence to visualize it better

[dire epoch]

wait sorry I'm kind of confused what do you mean by this?

[vagrant raptor]

well, if we write out the sequence for 3^n we get

[dire epoch]

element 0 is 1 (3^1 = 1)
element 1 is 8 (3^2 = 8)
element 2 is 27 (3^3 = 27)

[vagrant raptor]

yes, good

but if we want the sequence to start from element 1, and element 1 to be 1

any idea what we could put?

or wait wait wait a sec that's not correct

element 0 is 1 (3^0 = 1)
element 1 is 3 (3^1 = 3)
element 2 is 9 (3^2 = 9)

[dire epoch]

3^0 = 1?

0 x 0 x 0 = 0

[vagrant raptor]

3 times itself 0 times is 1

it's not 0 x 0 x 0

3^4 is 3x3x3x3

but 3^0 is just 1

[dire epoch]

oh

ohh

[vagrant raptor]

that's just how it is

[dire epoch]

I thought 3^1 is 1x1x1
3^2 is 2x2x2

sorry

[vagrant raptor]

no worries

3 is base, n is exponent

first element is 1

second element is 1x3

third is 1x3x3

fourth is 1x3x3x3

[dire epoch]

got you

[vagrant raptor]

perfect

first element n=1, but we want some exponent which is 0

i don't want to give you the answer directly but i think i will, just make sure you look at it properly anyway, ok?

[dire epoch]

that's fine

I just want to understand this better

[vagrant raptor]

okay so the general expression is 3^(n-1) if we want the first element to be 1

[dire epoch]

oh I get it

[vagrant raptor]

if the expression is 3^n you get
n | 0 1 2 3
3^n | 1 3 9 27

if the expression is 3^(n-1) all elements shift to the right
n | 1 2 3 4
3^(n-1) | 1 3 9 27

does that make sense?

[dire epoch]

actually yes I didn't understand something for a second but I got it

yeah

[vagrant raptor]

great!

just remember any number to the power of 0 is 1 :)

[dire epoch]

will do, thank you, if you don't mind, can you explain why that is?

[vagrant raptor]

a to any other power is simple:

a^3 = a * a * a
a^5 = a * a * a * a * a

right?

[dire epoch]

yeah

[vagrant raptor]

maybe you'd expect a^0 to equal 0, but 0 is special

1 is kind of like a factor of every number

[dire epoch]

yeah

[vagrant raptor]

so when you times by no a, it's still 1?

i think there'd be examples to show you why it is this way. but it is just a definition.

[dire epoch]

hmm ok

[vagrant raptor]

i found this which might help

http://scienceline.ucsb.edu/getkey.php?key=2626

[dire epoch]

well I searched up an analogy for it,
"How many ways are there to place a penny, a nickel, and a quarter on the table such that no coins are on the table? Again just 1."

If there are no coins on the table, there is only 1 way to arrange them

I think that's a good way to remember it

[vagrant raptor]

that's a good way to think of it!

[dire epoch]

well, thank you very much for all the help! I really appreciate it

[vagrant raptor]

i'm glad you found a satisfying answer

no worries at all :) good luck, have fun!

[dire epoch]

you too! :)

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[lone heart]

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[ancient sedge]

hello, help me please with this IQ question, i can't seem to find the pattern in any way.

[ancient sedge]

nevermind i found it, what about this one

[echo socket]

It's an IQ test

The answers should be purely yours

No?

[vagrant raptor]

they're testing OUR iq.. ?

[fluid eagle]

||(2+6)*(5+7) = 96, (5+8)*(3+4) = 91||

[lone heart]

@ancient sedge Has your question been resolved?

[ancient sedge]

Hey, not really universities in turkey require students to study for IQ and math exams for university entry, so we literally take ideas about IQ questions lol

thank you guys

[lone heart]

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[restive pilot]

Need a little help finding the area of this thing!

[bright gust]

You have a big rectangle and 4 triangles

[restive pilot]

oo just add triangles to make a rectangle?

[bright gust]

In this case yes

[restive pilot]

OHH adding 6cm to the base and height! or?

wait no

nvm

[vagrant raptor]

find the sides of the triangles

first of all

[restive pilot]

find the area?

[vagrant raptor]

no, the sides first

what are these

[restive pilot]

6cm

[restive pilot]

like the sides are 6cm

[bright gust]

How exactly are you finding 6

[restive pilot]

Because this side is 6

figured they're the same length