help-33

[smoky nest]

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[marsh citrus]

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[cyan cove]

how to find minimal solution (minimizing |x|+|y|) for linear diophantine equation (e.g. 17x+19y=524)

[cyan cove]

i have general solution to eqn:
x = 542(2)+11n
y= 542(-3)-17n

i just don't know how to find the n at which |x|+|y| is the lowest

[lilac siren]

Well the function |x|+|y| can be separated into at max three intervals since your equations for x and y are linear

|1084+11n| + |-1626-17n| is what you seek to minimize

so let's start by looking at the first term

when is 1084+11n positive and when is it negative?

[cyan cove]

when n < 1084/11 its negative

i mena

n < -1084/11

[lilac siren]

ys

and now what about the other term

[cyan cove]

and positive when greater

and for the other when n < -1626/17 its positive

[lilac siren]

exactly let me briefly calc which one is lower

-1626/17 is lower k

[cyan cove]

yeah

[lilac siren]

so you have three cases:

-X-Y if n<-1626/17
-X+Y if n<-1084/11
X+Y else

if you find the minimum of these three you found the total minimum

shall we do the first one together

or do you prefer to attempt yourself

[cyan cove]

hmm

i feel like i've done smthn wrong

wait nvm i havent

im getting the total minimum = -1626/17

[lilac siren]

do you mean at n = -1626/17

or the value of the minimum

[cyan cove]

yeah

at n = -1626/17

[lilac siren]

yop

[cyan cove]

i just used calculus for this tho

like first derivative test

i probably didnt have to do that

[lilac siren]

you don't need to

Its linear

which means the minimum of each interval

is one of the two boundary values

the minimum of a linear function on the interval [a,b] is either at a or b

[cyan cove]

true

wait i feel like i fucked this up tho

if i plug in -96

to my general solution it doesnt work 💀

maybe my general solution is wrong 1 sec lol

[lilac siren]

ok what was the initial question

[cyan cove]

o no it works

nvm

[lilac siren]

kk

[cyan cove]

wait

nvm 💀

i forgot to specify the problem correctly

so y has to be <= 0

[lilac siren]

What was the initial question :D

[cyan cove]

nvm i got it

[lilac siren]

😄

[marsh citrus]

@cyan cove Has your question been resolved?

[marsh citrus]

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[minor rivet]

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[lucid trail]

Does this seem to be right?

[steel sluice]

#bots message

[lucid trail]

oh thank you

[marsh citrus]

@lucid trail Has your question been resolved?

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[quaint crescent]

Can somebody explain to me what the "0 rule" is? I just don't get it...

[quiet anvil]

@quaint crescent is the "0 rule" the rule they used to break this up into cases?

x = 0 or x+3 = 0

[quaint crescent]

Yeah

[quiet anvil]

Ok

[quaint crescent]

I translated it literally

[quiet anvil]

Let's say you have two numbers

[quaint crescent]

5 and 7

[quiet anvil]

Let's call them y and z

[quaint crescent]

Okay

[quiet anvil]

And you know that y times z = 0

[quaint crescent]

How

One must be zero then right

[quiet anvil]

Let's just say that you arrived at this expression from something unrelated.

[quaint crescent]

Either z or y

Okay

[quiet anvil]

Or both

[quaint crescent]

Right

[quiet anvil]

But what about neither?

[quaint crescent]

Then it wouldn't be possible

[quiet anvil]

Correct

So either y or z (or both) must be 0

Now let y = x and z = (x+3)

[quaint crescent]

But with what logic does one cut apart x(x+3) into 2 seperate terms

[quiet anvil]

Well yz = 0 implies y or z is zero. That is exactly y = 0 or z = 0.

[quaint crescent]

Doesn't it imply that y*z is 0

So why is the multiplication thrown out and forgotten

There's some background stuff left untold or otherwise im just stupid

[quiet anvil]

Well, I think you might have a rather rigid understanding of math manipulations, so it's understandable that this might make you uncomfortable. So let's approach it a different way.

You have x(x+3) = 0

Divide both sides by (x+3), we are asserting x+3 ≠ 0 by doing this.

x = 0/(x+3) = 0.

We can verify that 0+3 ≠ 0.

What about the case where x+3 = 0? Well in that case we have x(x+3) = x(0) = 0. Which from the original expression is a true statement. So x+3 = 0 is also valid.

[quaint crescent]

What....?

Im kind of an idiot so...

[quiet anvil]

You are not. Learning is hard

[quaint crescent]

0+3≠0 yes

But where does 0+3 even come from I'm kinda lost

[quiet anvil]

We asserted that x+3 ≠ 0. Using this assertion, we found x = 0. Now we must check that our assertion was not violated by substituting in the result into the assertion.

That's how we got x + 3 becomes 0 + 3.

[quaint crescent]

Okay...

So when I see an expression where i have factored something

I can just take both of the terms and separate them

And make them equal 0

[quiet anvil]

Yes.

[quaint crescent]

Does the x+3=0 continue by turning into x=-3?

[quiet anvil]

Yup!

[quaint crescent]

Okay

Thanks :)

ありがとう！

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[safe sparrow]

how do I work through these?

[bold ice]

You have 2 points

of the tangent line

(0,0) and (b, f(b))

From that you can find the inclination

that is equal to f'(b)

[safe sparrow]

i see, then i solve for b from the resulting linear graph correct?

[bold ice]

yeah

[safe sparrow]

how might I go about doing question 4b?

[bold ice]

same process

the derivative wont have k

[marsh citrus]

@safe sparrow Has your question been resolved?

[safe sparrow]

got the first one, working on the second

[safe sparrow]

if the derivative doesnt have k, how can i put k in terms of b using the following info?

so far ive worked out that the tangent line is y=x*2e^(2b)
and thus b = 1/2

[bold ice]

The tangent line has k in it

m = (f(b) - 0)/(b-0) = f'(b)

So e^{2b+1}+k = b*f'(b)

[safe sparrow]

ive got it now

thanks for the help @bold ice

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[calm harbor]

Completely 0?

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

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[ebon depot]

X is unknown matrix.
Determine X if we know XA = XC+B

Hint: Transpose and Gauss Elimination are your friend

[marsh citrus]

@ebon depot Has your question been resolved?

[marsh citrus]

@ebon depot Has your question been resolved?

[frail orbit]

@ebon depot

[ebon depot]

XA - XC = B
X(A-C) = B

[frail orbit]

this tells you what the dimensions of X must be

[ebon depot]

So the dimensions of X multiplied with (A-C) which is a 2x3 should result in a 2x3 matrix?

[frail orbit]

yes

[ebon depot]

So X is a 3x3 matrix?

[frail orbit]

no..

[ebon depot]

sorry

meant 2x2 matrix

[frail orbit]

yes

now you can either right multiply both sides by B^T or right multiply by (A-C)^T

[ebon depot]

I'm not sure what you mean

[frail orbit]

B^T=B transpose

AC is right multiplying A by C

so basically either X(A-C)B^T=BB^T or X(A-C)(A-C)^T=B(A-C)^T

now note the dimension of (A-C)B^T and (A-C)(A-C)^T in either case

[ebon depot]

But i'm trying to solve for X?

thank

FUCKKKKKKK chart is here

i've been trying to get help since 12 pm

7 hours and 42 minutes later ;-;

[glass silo]

Awww

[glass silo]

Anyways, if we take the latter of these

[glass silo]

Do you know what the dimensions of these would be?

[ebon depot]

I'm not sure why we would want to transpose

X(A-C) = B

This is what i have?

[glass silo]

Sure, but if you multiply by one of those transposes, the answer to the previous question will give you a hint of what you can do

[novel juniper]

one way of doing this, though that would be rather ugly, is to assume a general matrix X and then compute (A-C) and solve, but I wouldn't recommend that

[glass silo]

@astral trout

[astral trout]

hiiii

[ebon depot]

i don't ge tit

[novel juniper]

calm down

[ebon depot]

no

[glass silo]

What type of matrices are your favourite?

[ebon depot]

I just want to understand

what i am doing

and how to solve it

[glass silo]

Alright, say I gave you instead something like
[
\pmqty{2 & 1 \ 0 & 5} X = \pmqty{1 & 2 \ 3 & 4}
]
Would you know how to find $X$ in that case, or have any ideas?

[elfin berry]

@glass silo

[ebon depot]

Yes

Multiply both sides with the inverse of 2105

[glass silo]

Why does it have an inverse?

[ebon depot]

But we multiply with the inverse on my question

because it's a 2x2

[glass silo]

Sure, or because it's square

[ebon depot]

yea

[glass silo]

What are the shapes of these?

[ebon depot]

(A-C) = 2x3
B^T = 3x2

[glass silo]

So then if you multiply them together, you get...?

[ebon depot]

2x3

2x3*3x2 = 2x2

mb

[glass silo]

Ooohhh, a square matrix

So it might be invertible

If it is, we can just multiply by its inverse to get X by itself

[ebon depot]

$X(A-C)B^T = BB^T$

[elfin berry]

Merineth

[ebon depot]

since (A-C)B^T = 2x2 matrix?

[glass silo]

Yea, that's a 2x2 matrix (as per above, it might be invertible, though of course it's worth making sure that it is - being square is necessary but not sufficient for being invertible!)

[ebon depot]

So now i calculate (A-C)B^T and BB^T

?

[glass silo]

Yea, pain

Not the worst in the world but still

[ebon depot]

How does it become a 2x2 ?

oh nvm

Just tired

37 47
19 29

(A-C)B^T

and

182 22
22 62
BB^T

[glass silo]

[glass silo]

Looking kinda invertible

[ebon depot]

for sure

the calculation will be insane?

[glass silo]

annoying a bit but doable

[ebon depot]

37 47 | 1 0
19 29 | 0 1

Don't i just divide by 37 on the top and 29 on the bottom to get 1:s instead of 37 and 29

and then i add with a mutliple of something to get 0 on 48 and 19

[glass silo]

You could do it like that, but for 2x2 matrices, their inverses are quite easy to find

The inverse of $\pmqty{a & b \ c & d}$ when it is invertible is $\frac1{ad - bc}\pmqty{d & -b \ -c & a}$

[elfin berry]

@glass silo

[ebon depot]

ah right forgot

Okay there we go

that seems to be right?

inverted it to get it to the RHS
then did matrix multiplication

[novel juniper]

this is another question ?

[ebon depot]

Yes

[novel juniper]

so you have to find X is AX=3X?

[ebon depot]

"Determine all the vectors $X_{3x1} for which AX=3x if A = ... "$

[elfin berry]

Merineth

[ebon depot]

This is what i did on it

I can't figure out any other way it could've been solved

[novel juniper]

just a suggestion use \text{ the text }to enclose text when you're typing latex in the future

[glass silo]

More that put the dollars only around the math

[ebon depot]

Yeah i know :( I've been at it for 8 hours and 40 minutes now and i'm on question 3 so i'm kinda lazy about latex atm

[novel juniper]

looks like a good method to me

[ebon depot]

thank GOD

3 more

Hang in there mee!

do a inverse of an inverse cancel each other?

For example

I want to solve for A

Do i just take the inverse on both sides to remove the inverse on the LHS?

or sorry

I just remembered

If i multiply both sides with (A-6I) then LHS cancels out

[glass silo]

You can though

[ebon depot]

Help me :c

Can i do this

$(A-6i)^{-1} = M \
M(A-6i) = I \
MA - M6i = I \
MA = M6i + I \
A = 6i+I$

[glass silo]

Is M supposed to be the matrix on the left, this one?

[ebon depot]

yes

[glass silo]

And if so, no

[ebon depot]

why

[glass silo]

The right would be the identity matrix if you're doing it like that instead though, if you wanted

[glass silo]

Multiplying an invertible matrix with its inverse gets you the (multiplicative) identity, not zero

[ebon depot]

i just get the identity matrix not zero?

But why can't i do it like i did?

and instead of zero

i make it I (identity matrix)

[elfin berry]

Merineth

[astral trout]

@glass silo
what about this? expand the brackets and move over -6I or is this illegal too

[glass silo]

Yea you gotta be careful, you don't necessarily have (A - 6I)^{-1} = A^{-1} - (6I)^{-1}

[glass silo]

And apart from the last line, you could do it like that, the last line isn't quite like that @ebon depot

[ebon depot]

A = 7i?

[glass silo]

Not quite no (also better to have all the small i's as capital I's, to represent the identity matrix)

What you would have, if M was invertible (it better be!) is A = 6I + M^{-1}

(multiply from the left by M^{-1})

[ebon depot]

what about it is wrong thjen :c

[glass silo]

Explain what your thoughts are to get from the second last line to the last one?

[astral trout]

how did you get to that equation?

i think i did it wrong??

no latex sorry i got beat up by c++ today

[glass silo]

Remember you can't do $(A - 6I)^{-1} = A^{-1} - 6I^{-1}$, that's illegal

[elfin berry]

@glass silo

[glass silo]

Oh wait also with that said...

Actually anyways

[glass silo]

From $MA = M(6I) + I$, multiply on the left by $M^{-1}$ (as before it better be invertible!) to get $A = M^{-1} ( M(6I) + I )$

[elfin berry]

@glass silo

[glass silo]

Expanding that out gets you $A = 6I + M^{-1}$

[elfin berry]

@glass silo

[glass silo]

You can also notice that by inverting both sides of $(A - 6I)^{-1} = M$ to get $A - 6I = M^{-1}$ and $A = 6I + M^{-1}$

[elfin berry]

@glass silo

[astral trout]

ah alright, so you're just inverting the expression on lhs and as such you invert the rhs as well?

and we can solve for A if and only if M has an inverse basically

[astral trout]

i don't get how M appeared in this equation but maybe merineth will see it when she comes back

so if ur allowed to just invert the expression on lhs and do the same on rhs i feel like that would be the easiest?? if thats legal

[glass silo]

Yep you can invert both sides (it is legal to do that!)

[glass silo]

From the MA = M(6I) + I part?

[ebon depot]

I take the inverse of M on both sides

$M^{-1}MA = M^{-1}M6I$

[elfin berry]

Merineth

[glass silo]

you're missing stuff

[ebon depot]

I am?

$MA = M6I$

[elfin berry]

Merineth

[ebon depot]

this is right?

huh

[glass silo]

See your last line here, everything you've done up until the last line is fine

[ebon depot]

$(A-6I)^{-1} = M \
M(A-6I) = I \
MA - M6I = I \
MA = M6I + I$

My initial thought was that i want to "move" M to the RHS

[glass silo]

Capital I, not small i

[elfin berry]

Merineth

[ebon depot]

So now finally i want to "move" M to the RHS so i solve for A

and to move a matrix

i need to multiply it by it's inverse, right?

So

$M^{-1}MA=M^{-1}M6I+I$

[elfin berry]

Merineth

[ebon depot]

$IA = I6I+I$

[elfin berry]

Merineth

[ebon depot]

WHAT

how is that wrong

i'm multiplying both sides with

M^{-1}

oh..

HAHAH

okay hold on

$M^{-1}MA = M^{-1}(M6I+I)$

[elfin berry]

Merineth

[ebon depot]

$IA = M^{-1}M6I + M^{-1}I$

[elfin berry]

Merineth

[ebon depot]

Can LHS be written as A?

$A = I6I+ M^{-1}I$

[elfin berry]

Merineth

[ebon depot]

(caps lock wasn't on :c)

Isn't I^2 = I?

$A = 6I + M^{-1}I$

[elfin berry]

Merineth

[ebon depot]

Don't you dare leave chartbit <3

got 2 more ;-;

just calculating the matrix for A

[frail orbit]

lmao

[ebon depot]

me holding chartbit hostage for math questions

What does it mean when a matrix is perpendicular and symmetrical?

[frail orbit]

orthogonal?

[frail orbit]

orthogonal matrix means its columns dot products are all 0

the columns form an orthogonal basis

[ebon depot]

I'm supposed to determine B^{-1} to

$B= 21A^{105}+79A$

[elfin berry]

Merineth

[frail orbit]

orthogonal matrices have a nice property

symmetric matrices also have a nice property

[frail orbit]

so this is saying A is an orthogonal and symmetric matrix?

[ebon depot]

Yes

And that it is a 19x19 matrix

[frail orbit]

A is symmetric implies that A=A^T

A is orthogonal implies that A^T=A^-1

[ebon depot]

Doesn't that just mean that

$A = A^T = A ^{-1}$

[elfin berry]

Merineth

[frail orbit]

so A is its own inverse...interesting...

what does that tell you about $A^{2}$

[elfin berry]

MSC2020 55N31 (Moosey)

[frail orbit]

@ebon depot

[ebon depot]

I'm not sure

[frail orbit]

$A^{2}=AA$

[elfin berry]

MSC2020 55N31 (Moosey)

[ebon depot]

Yea

[frail orbit]

and if A is its own inverse...

what is AA

[ebon depot]

[frail orbit]

A=A^-1

so you can sub in A^-1 for one of the As, yes?

[ebon depot]

OOOh

so

$A^2 = AA = AA^{-1}$

[frail orbit]

no...

[elfin berry]

Merineth

[frail orbit]

yes

and AA^-1 is...

[ebon depot]

I

[frail orbit]

mhm!

so note you have a power between even power and odd power of A

[ebon depot]

The odd power makes it 21A ?

[frail orbit]

A, A^2=I, A^3=(A^2)A...

[frail orbit]

yes!

[ebon depot]

$21AI^{104}$ ?

[elfin berry]

Merineth

[frail orbit]

well, technically I^(52)

[ebon depot]

ah right

[frail orbit]

but still right

so now you can just apply inverse properties to find the inverse of B in terms of A

[ebon depot]

so essentially

$B = 21A+79A$

[elfin berry]

Merineth

[ebon depot]

$B = 100A$

[elfin berry]

Merineth

[ebon depot]

So now we just want the inverse of 100A?

[frail orbit]

ye

[ebon depot]

$B^{-1} = 100A^{-1}$

[elfin berry]

Merineth

[ebon depot]

Is this allowed?

[frail orbit]

note property 2

[ebon depot]

so we take the inverse of both 100 and A?

$B^{-1} = 100^{-1}A^{-1}$

[elfin berry]

Merineth

[ebon depot]

Based on property 2

[frail orbit]

yes

100^(-1)=1/100

[ebon depot]

Nice

And since we aren't given the A matrix

i believe we are done?

Last one for today

and it's diiiiiiiiiiiiiiiisgusting

(The old man in the drawer) A cubic drawer has a corner at the origin and edges OA, OB and OC with other points in the points A = (2,−2, 1), B = (1, 2, 2) and C = (−2,−1, 2). A dot-shaped old man is at the point with coordinates (0.9,−0.9, 4.5). Is the old man in the drawer? Tip: Transfer to a suitable new base. Rephrase the question 'lie in the drawer' i algebraic terms. What finesse does the change to a new ON base offer?

[still temple]

if the point isnt as far away as B in the B direction, not as far away as C in the C direction and not as far away as A in the A direction then its in the cube

[ebon depot]

@glass silo ;-;

comeback

.close

[marsh citrus]

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[marsh citrus]

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[azure trench]

Question is: find all 2x2 matrices A, with real entries such that A^2 = -I where I is identity matrix

[azure trench]

Is this remotely correct

Oh the circled things are my answers

Im just wondering how you’d know how many different combinations there are as well

Ohhh I should have assumed bc = 0

[frosty linden]

So what are the form of the matrices ? Sry I didn't see the answer

[azure trench]

What do u mean by form

[frosty linden]

If AD=0 we can't have bc=0

[azure trench]

There is no AD, is there?

[frosty linden]

What are the matrices A such that A² = -I

The product ad

Because you Say a=0 and d=0

So the product is 0

[azure trench]

Yeah but if the product isn’t in the matrix why is it relevant

I’m not doubting u im just lost

[frosty linden]

Yes sure but we know that A is invertible so his determinant is non zero so with ad=0 we already know that bc is not 0

[azure trench]

Ohhhh determinant didn’t cross my mind for this

Fair enough

Ohnshit yeah

[frosty linden]

I don't read all the calculation but if you did that correctly it seems to be the good reasonning

[azure trench]

Oh yeah I said bc = -1

To make it into identity matrix

Okay I don’t think this is wrong

But there’s so many more answers ffs

[frosty linden]

👍

[azure trench]

Okay thanks

.close

[marsh citrus]

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[marsh citrus]

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[hasty lotus]

quick SUVAT question

[hasty lotus]

I have confirmed question a to be 1.22s to its max height then that will also be the same for time taken to reach starting point

[brittle coral]

[frozen delta]

that is correct

[hasty lotus]

with part b would it be -1.5 x 1.22s or would it be for the full 2.44s

[brittle coral]

full 2.44

[hasty lotus]

okay so -3.66m to the left of original position

thanks all

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[vocal cloud]

can i please get help with this

[vocal cloud]

i dont exactly see how they are going from first line to second

[harsh falcon]

think about it visually

im gonna draw something

[vocal cloud]

i mean i do get it visually when i take an example of 2 venn diagrams

but like if i were to show it mathematically

[harsh falcon]

hmm ok

[twilit arrow]

if A and B are disjoint, then P(A) + P(B) = P(A U B)

A and A' are disjoint

and so A n B and A' n B are as well

from there it's just undoing distributivity

[vocal cloud]

we get something like this right

[twilit arrow]

yes

the inside is equivalent to the inside in the original pic

it is undoing distributivity

[vocal cloud]

how are we doing that i dont get that

[twilit arrow]

(P n Q) u R = (P n R) u (Q n R)

[vocal cloud]

i dont get using that 😭

[twilit arrow]

no

im saying undoit

[vocal cloud]

oh

[twilit arrow]

this

[twilit arrow]

is the right side of this

identify P, Q, and R

then plug them into the left hand side

[vocal cloud]

oh

got it

thankyouu so much

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[dawn elbow]

I neeed help again

[glad kernel]

isn’t this kinda like physics

[dawn elbow]

im confused here on what he did with the bounds to turn them into 2 and 0

from 20 and 16

[hazy lion]

looks like a usub

so, you either need to switch the bounds or keep track of the old variable

[left wadi]

when you do a sub the bounds change

[hazy lion]

since the sub is u=x-16 here

the 16 becomes 16-16=0

[dawn elbow]

but its root x-16

[hazy lion]

20 should be 20-16=4

not 2

but whos counting

[left wadi]

@hazy lion you forgot to root

sqrt(20-16)

is 2

[hazy lion]

oh, youre right

[dawn elbow]

so if there are bounds and i have to do u sub then i need to plug in the bounds to the u sub formulas to see what they changed to?

[left wadi]

yes

[hazy lion]

or, keep track of them

and undo the substitution on the back end

[dawn elbow]

where did he get u=0 from

[hazy lion]

if you convert the bounds, you dont have to undo the substitution

[hasty lotus]

Yeah

[hazy lion]

if you dont convert the bounds, you do

so you just have to decide which one is less work

[dawn elbow]

where did he get u=0 from?

[left wadi]

on the right side?

[dawn elbow]

oh oh nvm u^2=16-16

okay ty

[left wadi]

sqrt(16-16) = 0 yeah

[dawn elbow]

so the upper bound becomes ur new x and ur lower bound becomes ur new U?

like why did he chose to sub in x=20 at the start and then u=16

did he mean if x=16 u = 0?

[left wadi]

i think

[dawn elbow]

i have

another question

how come B is just B and not Bx+c

[marsh citrus]

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[young swan]

Can anyone check if I've calculated this correctly? I guess this is physics but it seems appropriate to ask here

[young swan]

please ping on reply, will be heading off to bed 🙏❤️

[graceful owl]

So equivalent resistance is (20R)/(20+R) + 11

[graceful owl]

Then V=IR so 13=0.35 times that mess

Then solve for R

@young swan

[young swan]

oh god wait is my shit wrong

So this ain't enough to just get the resistance for R 😭?

That's a shame cause that is one beast of an algebraic question idk if i got the patience for

[graceful owl]

R in ohms law isnt 11

It’s 11+the parallel eqn of 20 and R

[young swan]

well, I was gonna head off but I kinda wanna solve this first.

2am math time?

but btw will this work?

0.35A isn't like

[young swan]

the whole circuit, which i think was implied here

It's just through 20 Ohm resistor

[graceful owl]

Ohh wait im being dumb the amperage is out of the 20ohm

Not the whole circuit

Sorry im also really tired

[young swan]

It's alright this is one of three servers I've asked in, so even if one strikes out it's ok by me

You should head to bed if possible

I have 8 hours before i need an answer, bcs that's when my exam is. I'll be heading off to bed

AS YOU SHOULD !!!

❗

I'll be heading off legit rn, so once again, if anyone do reply, please ping ! ❤️

[marsh citrus]

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[spring stirrup]

how did they get -1539 ?

I plugged in 2 and i keep getting random numbers

[marsh citrus]

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[plucky blaze]

im understanding it cant be D and E because those would hacve the same magnitude

and i think its not A because the charges cancel from both things

and then my multiple choice brain gives me the answer B but it could be C and idk why it would be B

i think it would be B and not C cuz its closer to any charge

but if there was just an E and not a D i wouldnt know if it was B or E

and the second one im fully confused what its saying

[marsh citrus]

@plucky blaze Has your question been resolved?

[plucky blaze]

i think the 2nd one is A actually now bc there should be no force on that one itshoudl all cnacel out

[marsh citrus]

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[still temple]

find f^-1(2)

you dont need f^-1(x)

[lilac siren]

hm wouldn't that still let me solve 2² = 2 + x + x^4 for x?

[still temple]

why 2²

[hushed egret]

are you one of those people that write R^+ for [0, infinity)?

[lilac siren]

since f(x) = sqrt(2 + x + x^4)

[still temple]

oh yeah right

[lilac siren]

I do both

[still temple]

Well you can get x=1

pretty easily

[hushed egret]

wait nvm lol

[lilac siren]

yeah but how would I properly calculate it

[still temple]

can you get f'(x)

[lilac siren]

nup

[still temple]

what

[lilac siren]

idk how to get f'(x)

ah sry f'

ofc

[still temple]

so f^(-1)(2) = 1

can you find f'(1)

[lilac siren]

am aware, but how would I determine x=1 in the first place :D

[still temple]

this

[lilac siren]

without guessing

eg if the task were instead:

I'd have 5² = 2 + x + x^4

[still temple]

calculator i guess

[lilac siren]

oh so there's no direct approach?

[still temple]

theres no direct approach to solve this

but once you get x

[lilac siren]

aw that's boring

k thank thee then :)

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[lilac siren]

was the closure timer increased?

[marsh citrus]

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[rough trellis]

dibs

[rough trellis]

what is m<EDC

i have no idea what to do

[civic moss]

Do you see that angle 1 is equal to angle 3?

[marsh citrus]

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[marsh citrus]

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[rigid breach]

Does anyone know to formula for question 11 b

[dry prawn]

the maximum value of a parabola occurs at its vertex

do you know how to find that?

[marsh citrus]

@rigid breach Has your question been resolved?

[rigid breach]

No I don’t😭

I don’t even know what a vertex is haha

Oh ok highest point

But idk how to find that

I think there’s a formula but I completely forgot

[dry prawn]

you don't know what the vertext form is?

[rigid breach]

Nope

[dry prawn]

y = a(x-h)^2 + k

not ringing any bells?

[rigid breach]

That reminds me of completing the square

Of a quadratic

But idk how it’ll help in this case

[dry prawn]

it helps because the vertex is precisely (h, k)

that's why it's called vertex form

[rigid breach]

OHHHH

I didn’t know that this form was equal to that

And so

Is h =x and k=y

[dry prawn]

what?

[rigid breach]

U said the vertex is h and k right?

So like, what do I do with them to get my answer

Cuz the answer is 159m

[dry prawn]

(h, k) is an ordered pair like (x, y) if that's what you're asking

[rigid breach]

Yaaa

Ok thank you

[marsh citrus]

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[rigid breach]

.close

[marsh citrus]

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[ebon depot]

(The old man in the box) A cubic box has a corner at the origin and edges OA, OB
and OC with others in the points A = (2, −2, 1), B = (1, 2, 2) and C =
(−2, −1, 2). A dot-shaped old man is at the point with coordinates
(0.9, −0.9, 4.5). Is the old man in the drawer?
Tip: Transfer to a suitable new base. Rephrase the question 'lie in the drawer' in ¨
algebraic terms. What finesse does the change to a new ON base offer?

I assume i want to make a base with OA, OB and OC?

[sand fable]

(somewhat interesting how a different person posted this exact problem 2 days ago)

[ebon depot]

It's probably because we go to the same school and we aren't getting any help from the teachers on these questions

[sand fable]

makes sense

[sand fable]

OA/OB/OC define the shape so ye you want to transform them

[ebon depot]

transform?

@glass silo Sleeping? ;-;

<@&286206848099549185>

[marsh citrus]

@ebon depot Has your question been resolved?

[ebon depot]

Anyone?

I found a similar answer online but i dont' know how to interpret it

dead discord 😔

.close

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[steel sluice]

what is this function

[fathom hornet]

hii, im a grade 6 student and i need help with percentages convertion since its our test tomorrow, is tutoring allowed here? to be specific, I need help with decimal-fraction & vice versa, also fraction to percent. thanku!

[drowsy gust]

!occupied

[marsh citrus]

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

[drowsy gust]

mod is the modulo function

$\mod(x,y) = x \mod y$

[elfin berry]

AlphaNull

[steel sluice]

oh ok

thanks

.close

[marsh citrus]

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[fathom hornet]

hii, im a grade 6 student and i need help with percentages convertion since its our test tomorrow, is tutoring allowed here? to be specific, I need help with decimal-fraction & vice versa, also fraction to percent. thanku!

[steel sluice]

ask in a new available help channel

[fathom hornet]

how?

[steel sluice]

https://discord.com/channels/268882317391429632/486844829209198613 aassk herre

[fathom hornet]

oh, thanks

[marsh citrus]

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[dense lynx]

hello for this question, I always seem to get only on epossible answer for a and b (3 and 18)

[drowsy gust]

i assume you plugged in a + ai into x^2 - 6x + b

you should have gotten $b = 6 a - 2 i (-3 a + a^2)$

[elfin berry]

AlphaNull

[drowsy gust]

and using the fact that b must be real, its imaginary component must be 0

and the imaginary component on the RHS is a quadratic which has two solutions

[dense lynx]

i used the sum and product of roots rule

so we know the two roots are a-2i and a+2i so

a-2i+a+2i = 2a (sum)

(a-2i)(a+2i) = 2a^2 (product)

x^2-(sum of root)x + (product of root) = 0

x^2-2ax+2a^2

-2a=-6, a=3

2a^2=b

b=18

@drowsy gust thats whati did

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[mighty swift]

find g^-1(x)

[mighty swift]

i got ans f^-1(x)=x-5/8

am i correct

[glossy flint]

No

[hard gull]

f(f^-1(x)) = f^-1(f(x)) = x , u can verify urself using this

[glossy flint]

It's $\frac{x+5}{8}$

Ups

[mighty swift]

why +..

[elfin berry]

Alberto Z.

[glossy flint]

Because of this

[mighty swift]

i didnt get it.

can u write it like that pic thing

[hard gull]

show ur working

[mighty swift]

ok

one sec lol i was drinking water

Here

[hard gull]

when u send 5 to lhs. It becomes
y+5

[mighty swift]

we should replace y place with 8x , no?

[hard gull]

$$y = 8x - 5 $$
Adding 5 on both sides
$$y + 5 = 8x - 5 + 5$$
$$ y + 5 = 8x \cancel{-5} \cancel{+5}$$

[elfin berry]

JustToPro

[hard gull]

$$y + 5 = 8x$$

[elfin berry]

JustToPro

[mighty swift]

oh

ok thanks i get it noww

[marsh citrus]

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[mighty swift]

.reopen

[marsh citrus]

✅

[mighty swift]

i got ans 2/-5

[glossy flint]

No, pay attention to the sign

When you square something you can't get a negative thing

x² = x•x
If x < 0, you get neg•neg = pos
If x > 0, you get pos•pos = pos

In both cases the result is positive

[mighty swift]

..

shouldnt we replace x^2

by -5

??????

[glossy flint]

Why??

f(-5) means replacing x with -5

[mighty swift]

but x^2 is still x

tho

[glossy flint]

x² is x², x is x

[mighty swift]

exponent stays same?

[glossy flint]

In fact x² = x only for 0 or 1

[glossy flint]

Wdym?

[mighty swift]

like (-5)^2

[glossy flint]

Yes, that's x² when x=-5

[mighty swift]

oh

so final

ans

is

2/25

[glossy flint]

Yes exactly

[mighty swift]

ohh

okkk thanks

[marsh citrus]

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[marsh citrus]

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[rocky scaffold]

Stuck after this

[rocky scaffold]

Used kings rule

[crystal wraith]

whats that ive circled

[rocky scaffold]

Split 0 to pi/2 into 2 parts basically

[crystal wraith]

No.

[rocky scaffold]

Same thing written with different limit

[crystal wraith]

Is it sin(pi/4-x)

[rocky scaffold]

Yes

[crystal wraith]

It's incorrect.

It should be a+b-x

[rocky scaffold]

Sin (pi/4 -x) + (cos pi/4-x)

[crystal wraith]

pi/2+pi/4 - x

[rocky scaffold]

Oh right

Youre right

Thanks

That was a horrible mistkae

.close

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[ebon depot]

The matrix $A_{3x3}$ has the determinant, $det(A) = 2$ \
Calculate: $det(3A^2)$

[elfin berry]

Merineth

[ebon depot]

Do i form an equation or how do i go about solving this?

[spark crystal]

this 2 equation would help u i think

for n is matrix n x n

[ebon depot]

Hmm

[spark crystal]

wait a sec

lemme recheck

ye its ok

try solve it @ebon depot

[ebon depot]

I'm not sure what i'm trying to solve

$det(kA) = k^n det(A) \
det(A^m) = (det A)^m$

[elfin berry]

Merineth

[ebon depot]

this?

[spark crystal]

that 2 equation is Properties of Determinants

using that 2 equation to solve ur problem

det(3A^2) = k^n (det(A^2))

[ebon depot]

Determinants has so many properties i sadly don't know them by heart

[spark crystal]

these 2 is 2 of the properties

[ebon depot]

These are the properties we are given

I assume the two you gave are the first two

Or nvm

They don't seem to be the properties you gave

[spark crystal]

for the Rakneregler or what ever

it prove this one

[ebon depot]

Oh okay

So how do i go about using these two equations

what is m and n also

[spark crystal]

let me give u and example

it might easier

[ebon depot]

Ok!

[spark crystal]

for this is just say for the properties of power
Ex. det(A^2) = (detA)^2

[ebon depot]

Ah oki

Can't we say that

$det(3A^2) = (det(3A))^2$

[elfin berry]

Merineth

[spark crystal]

yes

[ebon depot]

any way we can apply a property to remove the 3 from inside the det?

[spark crystal]

for this

Its Matrix 3x3 so n =3

[ebon depot]

ohh

[spark crystal]

det(3A) = 3^3 det(A)

[ebon depot]

$(3^3det(A))^2$

[elfin berry]

Merineth

[spark crystal]

$3^3(det(A))^2$

[elfin berry]

Flwo : 004

[spark crystal]

this one

[ebon depot]

why does 3^3 move outside the ()?

[spark crystal]

wait me a sec lemme get my book

for me ill say idk

my master just me teach me like that

[ebon depot]

haha well it's ok i'll look it up in my own time

tysm for the help

🫶

.close

[marsh citrus]

Channel closed

Closed by @ebon depot

Use .reopen if this was a mistake.

[ebon depot]

.reopen

[marsh citrus]

✅

[frail orbit]

oh boy

[ebon depot]

@spark crystalSince you know determinants.
Do i need to find A^-1 here if i want to determine det(A^{-1})

[spark crystal]

if it help some way

[novel juniper]

what's the question?

[ebon depot]

Yeah i was just about to mention that

I rewrote it as 1 / detA

[novel juniper]

oh, find $A^{-1}$

?

[elfin berry]

Why am. I here

[frail orbit]

oh wait part of the question is cropped out

wait is it

[ebon depot]

yeah so sorry will fix it

There we go

It says "Calculate det(A^-1)"

[frail orbit]

AHHH

thank god

[spark crystal]

so find det(A) then do 1/det(A)

[ebon depot]

Funny enough he writes "Do you need to find A^-1?"

[spark crystal]

i hate to inverse matrix

its too long

[novel juniper]

IMO applying some row transformations to get as many 0s as possible may be a good idea

[ebon depot]

I've never done det on anything greater than 3x3 matrix.
Does the method for 3x3 apply for 4x4?

[novel juniper]

which, the det formula or transformations?

afaik, the formula applies for a nby n matrix

I could be wrong though

[ebon depot]

https://youtu.be/fWzUwrt1Z0s?t=574

I was thinking something like this

WIll take an eternity to calculate lmao

[ebon depot]

I think i have to do this first right?

Because trying to apply the method as he does in the video

will leave me prone for error

with so much calculations

[novel juniper]

yeah, but I don't see any obvious transformations

oh

nvm,

I do now

[ebon depot]

Am i allowed to manipulate it as much as i want?

[novel juniper]

I think so

there is one transformation that will give you 2 0s striaght away

[ebon depot]

Hmm let me see if i also see it

first row times -1 and adding that to row 4?

[novel juniper]

hint:- ||look at the top right ||

the 4 elements there to be precise

[ebon depot]

the 4th column are all multiples of each other?

I'm very bad at identifying stuff like that

[novel juniper]

the 4 elements on the top right are identical

I;ll show what I mean

just a mo

\begin{pmatrix}
1 & 2 \
1 & 2
\end{pmatrix}

[elfin berry]

Why am. I here
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[novel juniper]

these elements

[ebon depot]

so if i take the first row * -1 and add it to the second row

[novel juniper]

or you could subtract the second row from teh first

[ebon depot]

oh right

So it'd become something like this?

[novel juniper]

uh, you want 0s in the first Row IMO

so do it the other way round, atelast I always expand a det wrt the first row, so that's what I'm comfortable with

[ebon depot]

Hmm

Afaik we wont get any 4x4 matrixes on the exam

so i probably will just use a online det calculator

This is a lot of calculation which i don't know lol

[spark crystal]

u should try to know

[ebon depot]

I want to but i've been at it since 10 am

[spark crystal]

i mean

[ebon depot]

mentally exhausted so i'm not sure i'll be even able to

[spark crystal]

it in my test at high school

[ebon depot]

Ok :)

[spark crystal]

but it quardratic form

like x+y+z = 15
and 2 more equation

[ebon depot]

So now i apply the method?

[novel juniper]

yes

[ebon depot]

Now i only get 2 3x3 matrices ?

since two of the constants are zero

?

[novel juniper]

yeah

the sum of 2 3by3 dets

[ebon depot]

i got it to be 55

which is wrong

i knew i'd do it wrong

Any idea where?

[novel juniper]

what's the answer?

[ebon depot]

I got mine to be 55