help-13

[bold hazel]

oic ty

[cedar kiln]

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Use .reopen if this was a mistake.

[cedar kiln]

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[mild oyster]

logic

[mild oyster]

$$(\neg P \lor Q) \lor (\neg P \land Q)$$

[wraith dagger]

studying lang calculus

[mild oyster]

How can I simplify this?

There's some law I'm misunderstanding

Law of absorbation?

[obsidian coral]

Distribution

Use $$A = (\neg P \lor Q)$$ Then you can do $$A \lor (\neg P \land Q)$$

[wraith dagger]

dldh06

[mild oyster]

$$((\neg P \lor Q) \lor \neg P) \land ((\neg P \lor Q) \lor Q)$$

[obsidian coral]

Distribute, plug back in A and use other identities

[wraith dagger]

studying lang calculus

[mild oyster]

$$(A \lor \neg P) \land (A \lor Q)$$

[wraith dagger]

studying lang calculus

[mild oyster]

This is distributied

What else can I do?

[obsidian coral]

Associative law

[mild oyster]

??

That's for

[mild oyster]

A and (B or C) = (A and B) or C

[obsidian coral]

Associative law

[mild oyster]

We don't have 3 here

[obsidian coral]

$$((\neg P \lor Q) \lor Q)$$
Is three

[wraith dagger]

dldh06

[mild oyster]

Then there was no need to distribute A or (notP and Q)

[obsidian coral]

You needed distribution first

To get this $$((\neg P \lor Q) \lor \neg P) \land ((\neg P \lor Q) \lor Q)$$

[wraith dagger]

dldh06

[sacred grail]

you never wouldve gotten that expression had you not distributed first

[obsidian coral]

But then you can apply associative law

[mild oyster]

How? $$A \land (\neg P \land Q) = (A \land \neg P) \land Q$$

[wraith dagger]

studying lang calculus

[craggy oriole]

Notice that $\neg P\land Q\implies\neg P\lor Q$

[wraith dagger]

ralvrz

[mild oyster]

[obsidian coral]

Use $$A = (\neg P \lor Q)$$ Then you can do $$A \lor (\neg P \land Q)$$
Distribute, plug back in A and use other identities

[wraith dagger]

dldh06

[sacred grail]

oh god that green

[mild oyster]

my eyes

[obsidian coral]

$$A = (\neg P \lor Q)$$
Is suppose to help make distribution easier

[wraith dagger]

dldh06

[mild oyster]

But we're doing distributing we have $$A \lor (\neg P \land Q) = (A \lor \neg P) \land Q$$

[obsidian coral]

plug back in A

[sacred grail]

this is an or not an and

[wraith dagger]

studying lang calculus

[obsidian coral]

That is not how distribution works

[craggy oriole]

I don't see how distributing simplifies the formula 🤔

[mild oyster]

$$(\neg P \lor Q) \lor \neg P \land Q$$

[obsidian coral]

You did it right here

[wraith dagger]

studying lang calculus

[obsidian coral]

Then plug back in A

Use associative law

[mild oyster]

This is

A plugged back in

A = neg P or Q

[sacred grail]

the RHS of this is A plugged back in

[mild oyster]

$$(\neg P \lor Q) \lor (\neg P \land Q) = A \lor (\neg P \land Q)$$

[sacred grail]

[obsidian coral]

This is what you get when you distribute

[wraith dagger]

studying lang calculus

[obsidian coral]

When you do $$A = (\neg P \lor Q)$$ and distribute $$A \lor (\neg P \land Q)$$
You get $$(A \lor \neg P) \land (A \lor Q)$$

[wraith dagger]

dldh06

[obsidian coral]

Plug back in A

[mild oyster]

But we did that we've gone full circle @obsidian coral

[obsidian coral]

If you did properly, you have $$((\neg P \lor Q) \lor \neg P) \land ((\neg P \lor Q) \lor Q)$$

[sacred grail]

we havent gone full circle we just havent made any progress

[wraith dagger]

dldh06

[obsidian coral]

Use associative law

[sacred grail]

• commutative

[mild oyster]

But we already did

Why are we going back to zero

[sacred grail]

we have not yet used associative law

[obsidian coral]

Then your work is making no sense because you are jumping around

[mild oyster]

You

are

[obsidian coral]

When was associative law used?

[mild oyster]

the

one

jumping around

[craggy oriole]

As I said, idk how distributing simplifies it

[mild oyster]

why exchange A

[obsidian coral]

Is suppose to help make distribution easier

[mild oyster]

this is literally step 2

[obsidian coral]

Use associative law and commutative

[mild oyster]

but you're saying it's step 8

[sacred grail]

[mild oyster]

by turning neg P and Q into A

then A or (negP and Q)

[obsidian coral]

This was the original question $$(\neg P \lor Q) \lor (\neg P \land Q)$$

[wraith dagger]

dldh06

[mild oyster]

then plug back in and go back to the same place

Yes

[obsidian coral]

I was saying to do $$A = (\neg P \lor Q)$$ and distribute $$A \lor (\neg P \land Q)$$
You get $$(A \lor \neg P) \land (A \lor Q)$$

[wraith dagger]

dldh06

[mild oyster]

So after you distribute you get

[obsidian coral]

Now apply associative law and commutative

[mild oyster]

Step 1 $$(\neg P \lor Q) \lor (\neg P \land Q)$$
Step 2 $$((\neg P \lor Q) \lor \neg P) \land ((\neg P \lor Q) \lor Q)$$

[wraith dagger]

studying lang calculus

[obsidian coral]

Because distributing $$(\neg P \lor Q) \lor (\neg P \land Q)$$ is a bit more difficult than $$A \lor (\neg P \land Q)$$

[craggy oriole]

You're overcomplicating it, your original formula is logically equivalent to $\neg P\lor Q$

[wraith dagger]

dldh06

ralvrz

[obsidian coral]

Don't give out answers

[obsidian coral]

Step three is apply associative law and commutative

[sacred grail]

and the logical equivalence should be proven anyhow

by this exact calculation

[mild oyster]

But for some reason you decided: Step 1 $$(\neg P \lor Q) \lor (\neg P \land Q)$$
Step 2 $$((\neg P \lor Q) \lor \neg P) \land ((\neg P \lor Q) \lor Q)$$
Step 3 $$A \lor (\neg P \land Q)$$
Step 4 $$(A \lor \neg P) \land (A \lor Q)$$
Step 5 $$((\neg P \lor Q) \lor \neg P) \land ((\neg P \lor Q) \lor Q)$$

[wraith dagger]

studying lang calculus

[sacred grail]

[mild oyster]

That's @obsidian coral steps

going circle

[sacred grail]

no they are not

[craggy oriole]

Nope, the logical equivalence follow from that implication

[sacred grail]

which ... you need to prove

[mild oyster]

Yes they are? Why did he end up with step 2 a second time?

[obsidian coral]

If you reread you never provided step two, I was starting from $$(\neg P \lor Q) \lor (\neg P \land Q)$$ and then after I stated, use $$A = (\neg P \lor Q)$$ Then you can do $$A \lor (\neg P \land Q)$$, you then provided your step two

[wraith dagger]

dldh06

[mild oyster]

I didn't provide step 2? #help-13 message @obsidian coral

[obsidian coral]

You shoved in your step 2 after I stated what to do from the original problem

[craggy oriole]

Conjunction elimination and disjunction introduction, qed

[obsidian coral]

You shoved in your step 2 after I stated what to do from the original problem

[sacred grail]

thats propositional logc

this is boolean algebra

[mild oyster]

Yeah which was to replace one term with A

[obsidian coral]

I was starting from the original problem and working step by step, then you shoved in your step 2

[mild oyster]

It's useless if we're going to plug back it in anyway

[sacred grail]

they are equivalent in some sense but you need to prove that

[mild oyster]

Yeah

Step 1 to 2 is much better

why 2 extra steps to unplug/plug A

[obsidian coral]

Here steps 1 and 2 are redundant

[obsidian coral]

.

[mild oyster]

How is it redundant we're applying distribution

[obsidian coral]

If you can easily distribute $$(\neg P \lor Q) \lor (\neg P \land Q)$$ go for it

[mild oyster]

??????????

[wraith dagger]

dldh06

[obsidian coral]

But I like distributing $$A \lor (\neg P \land Q)$$

[wraith dagger]

dldh06

[mild oyster]

That doesn't make it less redundant

[sacred grail]

distributing a massive parenthetical vs distributing a single symbol

[obsidian coral]

It makes it easier to deal with

[mild oyster]

"massive"

3 letters vs 1

[sacred grail]

comparatively

[mild oyster]

If it was 5-6 terms sure

But a single form? Bruh

[obsidian coral]

Still makes it much simpler to deal with

[mild oyster]

2 extra steps or 2 extra letters and 2 less steps

[mild oyster]

we arrive to the same answer

[craggy oriole]

And elimination, and Or introduction then

[mild oyster]

I clearly didn't mind not exchanging the term for A so maybe just go with that instead of forcing your convention idk

[sacred grail]

propositional logic ....

[obsidian coral]

I'm just saying if I was given $$(\neg P \lor Q) \lor (\neg P \land Q)$$, I would then do $$A = (\neg P \lor Q)$$ , and then do distribution with $$A \lor (\neg P \land Q)$$ because it's easier to deal with A rather than $$(\neg P \lor Q)$$

[wraith dagger]

dldh06

[mild oyster]

Yeah for YOU

To me it doesn't matter

I'm sorry It hough this was #help-13 and not #help-13 | dldh06

[obsidian coral]

And I was helping you

[sacred grail]

[mild oyster]

💀

[craggy oriole]

https://en.wikipedia.org/wiki/Boolean_algebra

[sacred grail]

the correspondence between boolean algebra and propositional logic needs to be proven ...

[mild oyster]

I'll just skip this exercise no one can explain I guess

[obsidian coral]

Through a method that has the same exact process, so what I have extra steps, the end goal is still the same

[sacred grail]

you cant just go from one set of axioms to the other without proof

[craggy oriole]

@mild oyster I can give you an alternative

[obsidian coral]

My process just has two extra steps because it's sometimes easier to use substitution

[mild oyster]

for you sure, but it doesn't help me

ill ask another channel

[sacred grail]

just continue the calculation from the distributed from bruh

[mild oyster]

.close

[cedar kiln]

Channel closed

Closed by @mild oyster

Use .reopen if this was a mistake.

[obsidian coral]

.reopen

[cedar kiln]

✅

[obsidian coral]

If you currently have $$((\neg P \lor Q) \lor \neg P) \land ((\neg P \lor Q) \lor Q)$$

[wraith dagger]

dldh06

[obsidian coral]

Use associative law and commutative

@mild oyster

[sacred grail]

use assoc + comm + idempotence

[mild oyster]

man why is this reopen I can't open new channel

[mild oyster]

.close

[cedar kiln]

Channel closed

Closed by @mild oyster

Use .reopen if this was a mistake.

[obsidian coral]

.reopen

[cedar kiln]

✅

[mild oyster]

please stop

[obsidian coral]

Because people are helping you

[mild oyster]

<@&268886789983436800>

[mild oyster]

he's not letting me make a new channel

[craggy oriole]

Are you?

[obsidian coral]

Let me reiterate, the original problem is $(\neg P \lor Q) \lor (\neg P \land Q)$, I stated use $A = (\neg P \lor Q)$ to assist with distribution, because in my opinion it is easier to distribute $A \lor (\neg P \land Q)$ than $(\neg P \lor Q) \lor (\neg P \land Q)$. If distribute, you get $(A \lor \neg P) \land (A \lor Q)$, plugging back in A, you end up with $((\neg P \lor Q) \lor \neg P) \land ((\neg P \lor Q) \lor Q)$, the exact same place you are in. I took two extra steps because it helps with the distribution, if you are not good at it

[wraith dagger]

dldh06

[obsidian coral]

The end goal is still the same

If you are at the step with $((\neg P \lor Q) \lor \neg P) \land ((\neg P \lor Q) \lor Q)$
Use associative law and commutative and idempotence

[mild oyster]

Bro is still trying to push his convention and opinion on me that I said is not relevant to me, at the same time holding me hostage

[wraith dagger]

dldh06

[mild oyster]

Man close this stop typing you are wasting time

I am not reading shit from you, hold someone else hostage

.close

[cedar kiln]

Channel closed

Closed by @mild oyster

Use .reopen if this was a mistake.

[obsidian coral]

.reopen

[cedar kiln]

✅

[craggy oriole]

Bruh

[mild oyster]

XDDDDDDDDD

<@&268886789983436800> nice )))))

[obsidian coral]

You want help, you got the help from me and snow, where we both stated use associative law and commutative and idempotence

That's the next step

[craggy oriole]

🤣

[mild oyster]

Maybe I'll develop stockholm syndrome to this channel

[slim hearth]

[mild oyster]

Let me close this channel please im held hostage

.close

[cedar kiln]

Channel closed

Closed by @mild oyster

Use .reopen if this was a mistake.

[slim hearth]

you are in the wrong here lol

but w/e if you don't want help you that's not my problem

but stop opening new channels then

[mild oyster]

I want help, just not from him and in this cluttered abomination @slim hearth

[slim hearth]

multiple people have now tried to help you and you have acted antagonistically to them

[celest seal]

genuinely dont see why you care this much about whether they use a substitution in their explanation or not

but youre being an asshole

[mild oyster]

couldn't read what they said cause of all his text

as I said too much clutter

better enter new channel imo

[celest seal]

how does it make a difference

[mild oyster]

3 people latexing and writing isn't easy to follow

[celest seal]

just post your "current" question, i.e. simplifying the form you currently have (whether you did it through the suggested subsittution or not is irrelevant)

opening a new channel removes context while contributing nothing

(and clutters the channel list)

(and makes it harder for helpers to keep track of what theyre working on)

[mild oyster]

At this point I don't need to write context I've gotten so confused from his jabbering I am better of reiterating the question

[celest seal]

you have the form $((\neg P \lor Q) \lor \neg P) \land ((\neg P \lor Q) \lor Q)$, right?

[wraith dagger]

Namington

[celest seal]

so your question should be "how do i simplify this"

[mild oyster]

Yeah

[obsidian coral]

Also, isn't this channel closed so it might disappear in a few minutes, right?

[sacred grail]

ping mods to get help

[mild oyster]

XD I was trying to close this, not complaining tho

[celest seal]

.reopen

[cedar kiln]

✅

[celest seal]

im not trying to offer help, just saying that you dont need to open a new channel

(though before that, maybe take a hard look at the form you have and see if you notice anything)

[mild oyster]

we'd just be arguing if that happened but ok

channel dead now I'll close it

.close

[cedar kiln]

Channel closed

Closed by @mild oyster

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[fathom condor]

Why doesn’t this have a solution

Can’t you plug the first solution into the second one

And do 1 = 3x - (3x +6)

[hollow minnow]

So 1=-6

[haughty wraith]

y=3x+6, y=3x-1

[fathom condor]

How

Ah I see

Nvm

I don’t distribute the 3x only the negative

[fathom condor]

?

[haughty wraith]

the 2nd equation can be written as y=3x-1

so the two equations are

y=3x+6 and y=3x-1

[fathom condor]

Oh same slope so it’s parallel

Meaning no point of intersection

[haughty wraith]

yes!

[fathom condor]

Technically

Ah I see

Thank you

.close

[cedar kiln]

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[crimson sedge]

how would I do b)

for #11

would I find how many revolutions they are in 5 mins and multiply it by the circumference

anyone

[zenith sail]

Yeah @crimson sedge that's right

[crimson sedge]

how would I find the number of revolutions in 5 mins?

[zenith sail]

Well you're told the number of revolutions in one minute

[cedar kiln]

@crimson sedge Has your question been resolved?

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[upbeat pond]

How did the teacher get 1/2 as k

If theres a negative sign here

Why would u get rid of the negative sign I dont get it

Idk what that means but yes

[lethal bison]

my guess would be that 2^(-1/2)=1/(2^(1/2))

[upbeat pond]

Im starting to think maybe it was an error the teacher made

.close

[cedar kiln]

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[upbeat pond]

ty

[cedar kiln]

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[limpid bane]

yeah so most likely no solutions unless theres a mistake

I reccomend graphing it

ill do that for you

[gusty forum]

no, it means something else

you might want to check a few x values and see what happens

[limpid bane]

ah yes sorry

i didnt mean no solutions

since the result is true

0x = 0

or 0 = 0

that means the equations will always be true reguardless of x and y values

or in other words

they are the same equation just rearanged

solution is that it is true for all values of x

??

IV what is wrong with it

[gusty forum]

try x=0, y=0

[limpid bane]

they are the same line though, I graphed it

[gusty forum]

do the equations hold for (x,y)=(0,0)?

yes, and is (0,0) on that line?

[limpid bane]

I see what you mean, how do you phrase it

[late tinsel]

https://www.desmos.com/calculator

[limpid bane]

A better way to put it is, there are an infinite number of solutions.

The answer you would write is, the two equations are functions of the same line, therefore there are infinitely many solutions.

Yeah that should be good enough

on the working out sheet you would have to show

yep thats right 🙂

you only drew one line though

yeah perfect

thats correct

[late tinsel]

https://tenor.com/view/theboys-perfect-gif-23821319

[limpid bane]

yep, if your plan is to then add both equations.

that would be a great example of the elimination method

mhm

then yes

[cedar kiln]

@crimson sedge Has your question been resolved?

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[tepid elm]

Hello, I am very confused.

[rain drift]

hi what are you confused about?

[tepid elm]

I don’t even know what I’m doing wrong but basically, I can do the basic parts of setting up the equation

like plugging in & mulitpling B by 5

[rain drift]

okay good

[tepid elm]

^ my work

Is that correct so far ?

[rain drift]

yep so far so good!

[tepid elm]

ok.

Then the inverses and the multiplication gets mez

I get so very confused.

[rain drift]

okay so the good news is that we know that A has an inverse

since det(A) is not equal to 0

[tepid elm]

yes

The determinate is 1/2 yes

[rain drift]

well be careful the det(A) = 2

[tepid elm]

Oh

[rain drift]

but to find the inverse we'll need the recipricol of that, 1/2 so you were almost there 🙂

okay so then for a 2x2 matrix, this is how you find the inverse of A

so you swap a and d and you negate c and b

[crimson sedge]

Is this gaussian

Oh nvm

Have fun guys

[rain drift]

nah

❤️

[rain drift]

so we know that $det(A) = 2$, so then $\frac{1}{det(A)} = \frac{1}{2}$, so then $A^{-1} = \frac{1}{2} \begin{bmatrix}
2 & 4 \
0 & 1
\end{bmatrix}$

[wraith dagger]

MellowDramaLlama

[rain drift]

yep perfect!

nice work

[tepid elm]

This scared me cause I didn’t understand lol

Ok cool perfect

[rain drift]

so now we need to multiply both sides by A^-1 on the left side

[tepid elm]

I am very bad at multiplying

[rain drift]

no worries

[tepid elm]

ok so,

how would I structure it which goes first ?

[rain drift]

so we get $AX + 5B = C \implies A^{-1}AX = A^{-1}(C - 5B) \implies X = A^{-1}(C - 5B)$

[wraith dagger]

MellowDramaLlama

[rain drift]

So we already know C - 5B

[tepid elm]

Yes

[tepid elm]

Yes

[rain drift]

so then we just do $X = \begin{bmatrix}
1 & 2 \
0 & \frac{1}{2}
\end{bmatrix}\times \begin{bmatrix}
-1 & 30 \
-2 & -21
\end{bmatrix}$

[wraith dagger]

MellowDramaLlama

[tepid elm]

Oh do we divide first ?

[rain drift]

either or

[tepid elm]

Or distribute ?

[rain drift]

it doesn't matter 🙂

I would say which ever is easiest

so I'll base it off of yours here

[tepid elm]

oh ok

thank you.

[rain drift]

then we get $ \frac{1}{2} \cdot \begin{bmatrix}
2(-1) + 4(-2) & 2(30) + 4(-21) \
0(-1) + 1(-2) & 0(30) + (1)(-21)
\end{bmatrix}$

[wraith dagger]

MellowDramaLlama

[rain drift]

then just simplify and distribute 🙂

[tepid elm]

woah Ok cute

[rain drift]

yep!

[rain drift]

let me double check 🙂

the math checks out!

nicely done

[tepid elm]

WOO

I’ve been.

doing.

[rain drift]

https://tenor.com/view/celebration-celebrate-yay-gif-24947639

[tepid elm]

20 problems.

In this genre of question

For 5 HOURS

thank you so much

[rain drift]

kind of makes you cross eyed doesn't it? lol

yeah np!

[tepid elm]

I started crying

anyways thank you

[rain drift]

linear algebra is your transition into adulthood

yeah of course! Best of luck to you 🙂

[tepid elm]

:)

.close

[cedar kiln]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• 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.

[polar quail]

Can someone help me with this please?

[storm glacier]

for the vertical asymptote imagine a vertical line that the graph doesnt touch

that vertical like would be at x=0

same thing for horizontal asymptot

asymptote

the graph approaches y=1 on either side

[polar quail]

I got that one.

But what about this?

[storm glacier]

where do you think the vertical symptotes are?

@polar quail

[polar quail]

0,-1?

[storm glacier]

no

@polar quail

[storm glacier]

if this was coordinates then yeah thats one of them

[polar quail]

The only answer I can think of is 0

[storm glacier]

i sent a picture with dotted lines

the two vertical asymptots at at x= -1 and x=3

and the horizontal asymtote is at y=0

[polar quail]

I put them as answers. One of them is not correct

[storm glacier]

which one>

[polar quail]

It doesn't tell me which one.

It just tells me one of them is incorrect

[storm glacier]

do you have a clue as to what the function is

[polar quail]

Wait, I see it now. The 0 is incorrect.

I got it. Turns out the horizontal asymptote was -1

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[clever lark]

Create a finite state machine with binary input and output that outputs 1s when the input symbol it reads is equal to the previous input symbol, and 0s otherwise.
For example, the input string 0001100 should output the following: 0110101
How am i supposed to do this ?

[tropic oxide]

the machine only needs to keep track of what its last-read symbol is

construct states to reflect this

[clever lark]

so the last last-read symbol here would be the inpout string ?

[tropic oxide]

no, you're overthinking it.

[clever lark]

uhmm, okey

could you elaborate ? ^^,,

are you still here .. ? :_)

[cedar kiln]

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[zenith sail]

You're taking one digit of the input string, and reading it, then another, then another

If the current digit you're reading matches the one from the previous step, add "1" to the output string, otherwise add "0" to the output string

[clever lark]

okey so i read the first letter from the input one ?

i start there ?

[zenith sail]

yeah

The example you gave, do you understand it?

[clever lark]

so 0

which is still 0

then the next is 0 as well, so i get a 1 ? because it matches the previous

[zenith sail]

yes

[clever lark]

okey i think i got it now.

!

thank you 🙂

[zenith sail]

👍 No problem!

[clever lark]

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[velvet pond]

Hello, I'm lost. How do I calculate a determinant of this Matrix?

[foggy merlin]

consider small cases and try to see a pattern

[velvet pond]

wdym

[foggy merlin]

calculate the determinant of this matrix for dimension n=2, then 3, etc. and see a pattern

[velvet pond]

oh i see

Thanks

[foggy merlin]

spoiler :
||the determinant is (n-1)^{k-1} (n+k-1)||

||with k the dimension of the matrix||

when you find the result, you prove it by induction

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[clever lark]

Draw the state diagram for a finite automaton (finite state machine with no output)
with input alphabet I = {0,1} which recognizes all binary strings containing the string 010. its asking me to draw the finite state, i know how to convert the input string to what its supposed to be but, not how i should draw it.

[cedar kiln]

@clever lark Has your question been resolved?

[clever lark]

s0 (0) --> s1 (1) --> s2 (0) --> s3 so i draw 3 point with each number on the lines

but i do not understand what makes it acceptance and where am i supppoosed to put the other lines it supposed to have...

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[amber epoch]

i need help with getting the exact value of this

i'm stuck at this

[zenith sail]

Instead of doing sin(tan^-1(2)), draw a right-angle triangle with an angle theta such that tan(theta) = 2. Then find sin(theta) from the triangle

same for cos(tan^-1(2))

[amber epoch]

oh thanks

not sure why i didn't think of that before

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[crimson sedge]

How did u change from -1+1to+1-1

[vagrant elbow]

Uh

[bitter reef]

a + b = b + a so
(-1) + (1) = (1) + (-1)
^^ a = -1, b = 1

[vagrant elbow]

Real numbers are commutative in addition

Yeah that

[bitter reef]

1 + (-1) is just 1 - 1

[crimson sedge]

No, I mean how to change-to+

[bitter reef]

you had -1 + 1, addition is commutative so you change their order, you get (1) + (-1) = 1 - 1

[crimson sedge]

Thanks

I just got it

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[twin ivy]

Hello

[twin ivy]

Do you have any idea about geometric sequence?😞

<@&286206848099549185>

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[vagrant elbow]

substitute (2, 2) in place of x and in place of the function

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[floral cedar]

How can I prove question 4b. 2q³ = 2α³

[south tundra]

Any further information?

[floral cedar]

The stuff in the part above

4a

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[dire geode]

Write each as a geometric series

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[nova snow]

Hi, I need help with part B. Do I just do the same thing I did for part A?

[nova snow]

But what would the line y = -1/2x + 1 change anything formula wise

[nova snow]

btw ignore the = sqrt20 on part A ik it’s just 20

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[median bay]

$\lim _{x\to \infty :}\left(\sqrt[x]{x}\right)$

[wraith dagger]

『Marius』

[median bay]

is this x^(1/x)

[modern compass]

yes

[median bay]

and just x^0 ?

so 1?

as in the limit

or inifnity to the power of ^0

ok thx

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[dim terrace]

can someone explain how this works to me please

[dim terrace]

how do you get from the lhs to rhs

[gusty forum]

for the inner sum: observe that |i-j| = |j-i|, so it doesn't matter which one of i, j is larger

therefore your sum counts each distinct |i-j| twice, for example, when i=2, j=1 and i=1, j=2 the inner expression is identical

so the second expression fixes it so that j is always smaller than or equal to i (helps to notice that when i=j your inner sum is 0) and you can pull the 2 out the front

[dim terrace]

ohhh

but then how do you get to the n^2-1/3n

[gusty forum]

treating 1/n^2 as a constant (which it is) for the inner sum wrt j, can you rewrite the sum as an expression?

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[long arrow]

on a) just conjugate pair

[crimson sedge]

yeah i figured A

was just curious

for part B
would I use the cubic rule
ab + by + ay = c /a to find gamma ?

[gusty forum]

where's gamma in your question?

[crimson sedge]

wdym where is gamma ? I know alpha and beta are the two complex pairs and to find p & q i must find y

[long arrow]

rather use Vieta

[crimson sedge]

im referring to roots as alpha , beta and gamma if that clears up any confusion 🙂

here is what I did i'm just curious as to whether or not it's valid

will send a photo

[gusty forum]

so that should work, but in context of the question i guess the question maker intended for it to be solved with long division

[crimson sedge]

will look into that method , we have only been taught to solve based on the rules

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[crimson sedge]

ye

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[mystic reef]

Does anybody know the degree symbol shortcut for web assign

[pine gazelle]

should be on the tools tab or something on the right

under "symbols"

[long arrow]

°°°°°°°°°°°°°°°°

Alt + 0176 (Num Lock)

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[fleet kernel]

what am i supposed to get from the row reduction?

and why wouldn't i just follow the usual leontief model

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[cinder zenith]

ive watched the video provided and im still getting it wrong

[cinder zenith]

square based pyramids formula is 1/3 x base area x height

so it should be 1/3 x 36 x 7 ?

so its 84

cylinder is pie x radius squared x height

so pi x 9 x 6

which is 54 pie which needs to be halfed as its only half a cylinder

27 pie + 84 is 168.8230016 or whatever

and i round it to 1 dp and it says its wrong

💀

idk what happened but it changed question i did it again and it was correct so im just confused but i got it correct and i didnt really get help so f

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[grand raptor]

this is from a powerpoint about PORE (propagation of random errors). Can anyone help me understand how they got the variance from this table?

[grand raptor]

the arrows are kinda vague

<@&286206848099549185>

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[barren hinge]

Can someone explain how this is wrong? My thought process for the first one was 2 sqrt root 3(x-2) -4

then I got 2 sqrt root 3x-6 -4

And for the second one I did 2 (sqrt root -3x) -4

then 2 sqrt root (-3(x+2)) -4

and got 2 sqrt root (-3x-6)-4

[weak lake]

@barren hinge which one is wrong

i get the same answer for both

[barren hinge]

both

is there a way to simplify it further?

[weak lake]

i mean you can write 3(x-2)

sqrt(3)*sqrt(x-2)

but i dont know that program is weird

[barren hinge]

For the second one can I write -2 sqrt root (3x+6) -4?

Sometimes if you get one wrong it marks both wrong

[barren hinge]

And is this one correct also?

[weak lake]

no you cant take out the negative sign

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[dim coral]

f(10)-f(1)/10-1

[dim coral]

for me is 1/18

but i don't know if that's correct

because i got a value of c that i don't think is correct

[runic garnet]

well 1/18 isnt the answer, but its a step towards finding it

@dim coral did u find f'(x)

[dim coral]

yes

[runic garnet]

whatd u get

[dim coral]

it's

x*(-1(x+5)^-2)+(x+5)

[runic garnet]

u think u could write it out and send a pic? its kinda unclear from the text

or latex

[dim coral]

$x\cdot (-1(x+5)^{-2})+(x+5)$

[wraith dagger]

cynicalcola

[runic garnet]

so (x+5)-(x/((x+5)^2))

or are u trying to say the x+5 - x is all part of the numerator of the fraction

just simplify it down, it will make it a whole lot easier

[dim coral]

yeah i just gave you like the original derivative without simplifcation

to avoid mistakes

[runic garnet]

ah ok

[dim coral]

but i think this is basically same thing

[runic garnet]

yea i rewrote what u had

but just simplify now

[dim coral]

it's the same number i got before

5.053381905

it doesn't feel right?

[runic garnet]

whatd u get after simplifying

the derivative

[dim coral]

i got it on the first try

[dim coral]

oh same thing you did

it could be many things

but the one you said is true

so i used it

[dim coral]

eitherway i get this as c

[runic garnet]

wat

[dim coral]

when i substitute x with 1/18

that's what i get

[runic garnet]

ok hold up, lets take this 1 step at a time

first completely simplify the derivative, so we dont have to worry abt mistakes

[dim coral]

$-x(x+5)^{-2}+(x+5)$

[wraith dagger]

cynicalcola

[dim coral]

$-\frac{x}{(x+5)^{2}}+(x+5)$

[wraith dagger]

cynicalcola

[runic garnet]

no thats not right

[dim coral]

$(x+5)-\frac{x}{(x+5)^{2}}$

[runic garnet]

thats why i was confused before abt what u wrote

[wraith dagger]

cynicalcola

[runic garnet]

how come the x+5 is outside the fraction

[dim coral]

from finding the derivative

$x\cdot (-1(x+5)^{-2})+(x+5)$

[wraith dagger]

cynicalcola

[runic garnet]

no, try finding it again

[dim coral]

product rule

x*derivative of x+5 + vice verca

but i flipped the original equation to apply product rule

[runic garnet]

so u did x * (x+5)^-1 ?

[dim coral]

$x\cdot (x+5)^{-1}$

yes

[wraith dagger]

cynicalcola

[runic garnet]

ur derivative is wrong

try it once more

product rule works, but id recommend quotient rule here

[dim coral]

i'll try quotient rule

but do uknow what is wrong?

doesn't seem like anything is wrong for me

did it multiple times

[runic garnet]

yes, ill show u how to do it with product rule then.

u = x, v = (x+5)^-1, u' = 1, v'= -(x+5)^-2

uv'+vu'

=

x(-(x+5)^-2) +((x+5)^-1) *1

[dim coral]

$\frac{x+5}{(x+5)^{2}}$

[wraith dagger]

cynicalcola

[dim coral]

is that the derivative with quotient rule?

oh wait

[runic garnet]

no, but close

[dim coral]

$\frac{(x+5)-x}{(x+5)^{2}}$

i forgot that

[wraith dagger]

cynicalcola

[runic garnet]

there you goooo

now simplify completely the numeraotr

[dim coral]

hmm

i don't know to do that

[runic garnet]

(2+2) - 4 = ?

[dim coral]

0?

[runic garnet]

correct

associative property

[dim coral]

i can just plug 1/18 in

[runic garnet]

(2+2) - 4 = (2-4) + 2 = 2 + (2-4)

[runic garnet]

well u could, but thats not how u get the answer

do u see what the numerator becomes

associative property of addition

[dim coral]

i don't really understand what we're doing here

[runic garnet]

at the moment, we're completely simplifying the derivative of f(x)

do u not know how to do (x+5) - x ?

the parentheses dont even matter

[dim coral]

so you mean it's just 5?

[runic garnet]

yes

[dim coral]

oh ok

[runic garnet]

so its 5/(x+5)^2

[dim coral]

sorry i have some gaps in algebra

[runic garnet]

no worries

now, remember our value from earlier, the 1/18?

set that equal to what we just got for f'(x)

[dim coral]

1620/8281

that's the value

is that c?

[runic garnet]

howd u calcualate it

[dim coral]

on the calculator

i plugged 1/18

in

5/(x+5)^2

[runic garnet]

no, thats not what i said to do

set the 1/18 = f'(x)

then solve for x

[dim coral]

oh ok

wait i'll try

$\pm 2\sqrt{10}-5$

[wraith dagger]

cynicalcola

[dim coral]

that's what i got

[runic garnet]

hmm

u sure abt that 2?

[dim coral]

idk why do you suspect that

but that step before that was

40=(x+5)^2

[runic garnet]

howd u get 40

[dim coral]

i moved (x+5)^2 to the other side

5/(x+5)^2 = 1/18

here

then 1/18 to the side of 5

[runic garnet]

whats 18 * 5

[dim coral]

oh it should be actually 90

[runic garnet]

yea

[dim coral]

i guess it+s

$\pm \sqrt{90}-5$

[wraith dagger]

cynicalcola

[runic garnet]

yes, now check which of those falls in our interval

[dim coral]

with original function?

[runic garnet]

wdym? check if these +- values we just got fall into the interval [1,10]

[dim coral]

the positive one does

negative doesn't

[runic garnet]

correct

so theres our c value

[dim coral]

ok got it

thank you so much

[runic garnet]

np

[dim coral]

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[crimson sedge]

Why is the general solution (pi)n and not 2(pi)n?

[wintry violet]

Not sure how my an and bn values are wrong

[crimson sedge]

.close

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[patent leaf]

Haiya! I'm being asked:

If 3x - y = 10, what is the value of 6x - 2y
I know that the answer is 20, because x should be 4 and y should be 2, but I feel that there would be a better way to answer the question with a better solution- Is there any?

[dense hornet]

x and y are not specifically 4 and 2 respectively. If not given any further constraint on what x and y are, they could be just anything. For example, x=1 then y=-7

the answer is 20 is because if 3x-y=10, then if we multiplied both sides by 2, we would get 6x-2y=20

which ideally is what the question asks

[patent leaf]

ohh, I see

[dense hornet]

or in another way, you can view it as because 3x-y is 10, then 6x-2y must be 20 in order to satisfy the first condition for any pairs of x and y

[patent leaf]

i did not see the question that way, thats cool

so any value for x and y would give a 20, as long as it satisfies the condition

[dense hornet]

yes, any (x, y) pair that satisfies 3x-y=10 would make 6x-2y=20

[patent leaf]

thats the answer i needed, thank you very much!

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[solid mulch]

im rlly confused on this

[short blade]

draw a triangle

remember sohcahtoa

[solid mulch]

yes

[short blade]

then label one of the angles theta, and the opposite side should have length sqrt3, and it's adjacent should have length 7

[wraith dagger]

What the hell am I doing here?

[solid mulch]

what about the negative

is the negative adjacent?

it has to be i think

[dusty hazel]

It doesn't matter.

[solid mulch]

ok

[dusty hazel]

This works, without you having to do the negative and shit. Because there's tan^2

[solid mulch]

oh

thx

that makes sense

ur a good person brother

@dusty hazel

i got 46 as the answer

is that right?

[dusty hazel]

Holy shit

that by no means is even close.

[solid mulch]

damn

[sacred grail]

lmao

[short blade]

remember that cos(theta) is between -1 and 1

[sacred grail]

cosine should be between -1 and 1

[dusty hazel]

Exactly.

[sacred grail]

[dusty hazel]

Look Eddie, you brought so many people here.

[solid mulch]

i dont understand what i did wrong

[dusty hazel]

Show your working that lead to 46 as the answer?

[solid mulch]

k

there

[short blade]

how did you get 49 for cos^2

[solid mulch]

cos is -7

[short blade]

no

first label your theta

then recall sohcahtoa -> cos = adj/hyp

[dusty hazel]

yeah.

Not simply adjacent.

[solid mulch]

oh

wait so i wanna solve it with this equation

[short blade]

you can

whether you want to is up to you

[short blade]

you could also use this

[solid mulch]

i cant i wanna learn what my teacher taught

[dusty hazel]

You'll have to find the hypotenuse using Pythagorean theorem. To be able to use this one.
Which is something your teacher should have taught you.

[solid mulch]

wait so i find the hypotenuse for these sides?

[solid mulch]

can u pls explain it to me again using my teachers equation

i think i drew the triangle wrong

[dusty hazel]

Can you find the hypotenuse?

[solid mulch]

yes

its sqrt 51

52

[dusty hazel]

yeah.

then what's cos theta

(adj/hypotenuse)

[solid mulch]

cos theta?

-7/ sqrt 52

[dusty hazel]

yeah.

[solid mulch]

omg

[dusty hazel]

how about sintheta

[solid mulch]

holy shit

i got cos(2theta) = 46/52

is that right

pls tell me thats right

[dusty hazel]

yeah. It is.

[solid mulch]

YES

🎊 🙏 👏 😱 🥹

@dusty hazel i am proud to say that u are my mentor

ur the teacher of teachers

the master of masters

you created every art known to mankind

[dusty hazel]

...

[sacred grail]

[solid mulch]

when you speak, angels sing about you and praise your name

[dusty hazel]

Yes, but are you done?

With the problem.

[solid mulch]

yes

[dusty hazel]

Cool.

.close

[solid mulch]

🥹

its been nice knowing y-

.close

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[cedar kiln]

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[fervent vortex]

Find the value of x (correct to 3 dp) for which 3arctan(x) = pi/2 - arctan(3x), where 0 =< arctan(x), arctan(3x) < pi/2
So far I let a = arctan(x), giving 3a = pi/2 - arctan(3x)

[fervent vortex]

nvm

.close

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[blazing zephyr]

the mu symbol is a variable, can i just substitute it to get (say mu=x because i dont have mu on my keyboard) x^1/3=1/3

[drifting marlin]

If you're asking if you can set $x = \mu^{\mu^{\mu^{\dots}}}$ then yes

[wraith dagger]

Steakanator

[blazing zephyr]

it is asking what mu is and the mu isnjust a vqriable so imagine it is x because i dont have mu on my keyboard

[drifting marlin]

Oh ok

[blazing zephyr]

can i still subtitute it?

[brave aspen]

Mu is just a name

You can say let mu = x

Then rewrite it all as x

[blazing zephyr]

yeah fair

[drifting marlin]

In that case yes

[blazing zephyr]

ok

so x=27?

wiat no

1/27

[drifting marlin]

Better

[blazing zephyr]

ok

.close

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[wanton adder]

is |x|²= x²?

[wanton adder]

cuz like even if its negative the is ² makes it postitive

[vagrant elbow]

yes

[wanton adder]

okay thanks

how do i close it

[vagrant elbow]

.close

[wanton adder]

.close

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[rough lodge]

Can anyone help me with the (e) part

[crimson sedge]

Hard to see what you wrote in the second picture

[rough lodge]

uh

basically the aswers for the first page

ima try and take another one

here are the questions

in the second page ive done the working for (a) and (d)

and plotted the graph

need help with this ques

[cedar kiln]

@rough lodge Has your question been resolved?

[rough lodge]

.close

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[restive plover]

I do not understand why the point m is given here

is m just a name given to the point or is it (0,m)

[crimson delta]

is that really such a big difference? if the point is called m, then m=(0, n) for some number n

dont hang yourself up on that

[restive plover]

The next question is like this

and m is in this equation which i am supposed to prove, this is what is bugging me

in the first part of the sub question, I got $y= \frac{sin \theta}{1+cos \theta} \times x+1$