help-23

[round pawn]

√x+2 - √3-x = 1

[round pawn]

I don't get it, I've tried every method I know

Like

This question is genuinely just hard

[hollow pagoda]

whats the question?

to solve for x?

[quasi bison]

$\sqrt{x} + 2 - \sqrt{3} - x = 1$?

[flat frigate]

Ann

[quasi bison]

is this your equation?

[lean otter]

I think smth wrong with this equation

$\sqrt{x} - 2 - $\sqrt{3} + x = 1$?

[flat frigate]

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

[lean otter]

Gives x = 3

[quasi bison]

@round pawn please confirm or deny that i've written your equation correctly

if i have not, then correct me

[hardy sky]

just abandoned you lmao

[lean otter]

True

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[hollow elm]

sorry just mind blank what is the working out for 10i

.rotate

/rotate

:rotate

[potent seal]

,rotate

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[nocturne pond]

I was making a sign table of a function to find its inflexion points and concavity intervals, apparently I was missing 2 critical values

On an interval of [0, pi]
The first derivitave gave me -2sin(4x) so I found what gave sin(x) = 0 and divided by 4 which gave me pi/4

The second derivitave, -8cos(4x), I did the same thing but for cos(x) = 0 and got pi/2 and 3pi/2, divided them both by 4.

But I was missing 7pi/8 and 5pi/8, where did these come from?

[peak estuary]

sin(x)=0 doesnt only have the solution x=pi

[nocturne pond]

0, but I didn't put that in it since it's a border

[peak estuary]

when instead of looking at sin(4x)=0 you look at sin(x)=0, then you cant restrict yourself to the interval [0,pi] anymore

if x is in [0,pi], then 4x is in [0,4pi]

[nocturne pond]

Oh, so like that would add 3pi and 4pi as critical values then?

[peak estuary]

well, divided by 4

[nocturne pond]

Ok, I understand now, thank you

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[light ruin]

so i got this question i got -5/3w the answer was w^2-5/w(w+3), why couldnt you just expand the bottom and cancel out the w squared?

[dapper venture]

there's nothing to cancel

you can't cancel w^2 with air

[light ruin]

if you expand the bottom you get w^2 , its on the top and bottom

[dapper venture]

where is w^2 on top?

there only a 5

[light ruin]

so i timed the left one by 2 to get the same denominator

then its w^2-5 on the top

[dapper venture]

??

can you write that down

I've no idea what you are doing

[light ruin]

ok

[dapper venture]

ah I see

you can't cancel w^2 here

[light ruin]

how do i know when i can't

[dapper venture]

when you can isolate them

for example$\frac{w(w+2)}{w(w+3)}$

[flat frigate]

WhereWolf(ping if needed)

[light ruin]

if the question didnt have brackets in then could you?

like if the question was w^2+2w/w^2+3w

[dapper venture]

you need to factor them to check

to smiplify fractions we usually factor everything first

[light ruin]

ah ok

and we can only cancel out anything in the brackets if its the same?

[dapper venture]

yes

when you have plus or minus sign you can't cancel

everything must be in brackets

[light ruin]

ok thanks

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[north raven]

can't figure out what to do from here

[mortal sandal]

The gradient f should be <y,x,2z>

nvm

y,x,2

for some reason I thought you had z

Idk where you're getting y^2=2 but it doesn't look right

λ can be -1/2 here as well

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[lean otter]

Is this in echelon form?

no

So

it can't be brought to echlon form

I see

it doesn't have any solution

[peak estuary]

well thats bullshit

[lean otter]

So in matrices where u cannotr bring them to echolon form u just use whatever u get to see if linearly independent?

[peak estuary]

every matrix can be brought into echelon form

[lean otter]

Hm

[peak estuary]

and every system of equations where the right side is just zeros always has at least one solution

[lean otter]

actually that's what I thought after I sent it

because every matrix can be brought

Let me send the entire sum

[wild copper]

In this case, you can use R3 -> R3-4R2 to bring it into echelon form

[lean otter]

let me see

In all the other sums we reduced to echolon form

But for thiz we didnt

[peak estuary]

well we dont need to

we could, but we dont have to

[potent seal]

Wait is R3-R2 also multiplied?

[lean otter]

Yes but why didnt they do that in the problem?

[peak estuary]

we can just start with the last row

[wild copper]

I think in this case, it's just not necessary.

[peak estuary]

write out what the last row means

[wild copper]

You can already determine the determinant easily

[crimson ice]

linearly dependent so k1 and all should be 0

i wish i could send my notes

but its all dirty

[wild copper]

And if the determinant is nonzero, the vectors are linearly independent

[peak estuary]

you really dont need the determinant for this

dont just rely on the det

[lean otter]

does this equation's solution are 0, 0, 0?

ahh, okay

nvm

[lean otter]

Is there a wAy to tell

If its necessary or not

[lean otter]

But dont u determine that after finishing ur matrix

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[rotund turtle]

[\int_{0}^{\infty} {x^{-{1 \over 3}} \over x + 2} dx]

[rotund turtle]

the context is a complex variables class

[flat frigate]

epiphonically

[rotund turtle]

I have no clue what contour to pick

if I can even use residues

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@rotund turtle Has your question been resolved?

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@rotund turtle Has your question been resolved?

[rotund turtle]

<@&286206848099549185>

[rotund turtle]

how

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[steep lily]

it's been years since I did complex analysis so I have no idea if this is even going to be helpful

but try a keyhole contour cut along the positive reals??

I seem to recall that after some limits they (sometimes???) end up giving those 0 to infinity real integrals

also try #real-complex-analysis because it does seem unlikely you'll get more help here by just waiting (given the evidence of the past 7 hours)

[rotund turtle]

yea i tried this but the problem is

the residues are on the real axis

[rotund turtle]

also yea ill try that

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[tight marsh]

5 and 6

i dont understand what they mean

do i plug them in??

[hallow night]

yes, these are of the form (x,y) so plug in the first value for x and the second for y and see if the inequality holds

[lean otter]

just plug

[tight marsh]

so see if the inequality is true

like 4 > 3

[lean otter]

yeah

so what would 5 be

[tight marsh]

false

because 1 > 1 cannot be possible

unless if its equal or more to

[lean otter]

yeah, so you can say it is not a solution of the inequality

[tight marsh]

k

and 6

it is a solution to the inequality

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[stuck dust]

How do we do this?

Find a, b, c so the solution for the equation showed is x=1, x=2 and x=3

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[stuck dust]

<@&286206848099549185>

[crimson ice]

sum of roots and product of roots

A=-5 or 5 idk

and all

u get the idea ?

[stuck dust]

Got lost there

sum of roots and product of roots

Oh i see

1, 2 and 3 are the roots

[crimson ice]

yep

[stuck dust]

Is that with the vietta theorem?

Mmm I'm still lost sorry, can you elaborate?

[crimson ice]

sure

1 min

@stuck dust

and in the question they gave roots

[stuck dust]

Right they did

[crimson ice]

should i send pic

or did u understand

[stuck dust]

A pic would be nice, thanks!

[crimson ice]

np

give me 2 mins

[stuck dust]

Thanks, appreciate it

[stuck dust]

The sum of roots isnt 6?

1+2+3

Product is 6 aswell

[crimson ice]

fuck

sorryyy

yes yes

[stuck dust]

No worries bro hahaha

[crimson ice]

did u get it ?

[stuck dust]

I think I do, i'm going to do it rn

Ill send you a pic once finished

[crimson ice]

okk

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[tender echo]

Can someone help me with a financial math problem

[tender echo]

I need the new balance and the total available credit

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<@&286206848099549185>

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<@&286206848099549185>

[devout fog]

what?

[tender echo]

I need the the answer to C and D for financial math

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[urban dock]

Can anybody explain me where do we get the $F_{2(n+1)-1}-1$?

[flat frigate]

Chrisfv97

[quasi bison]

For the n+1 case...

[urban dock]

Under inductive hypothesis

Wouldn’t it be $F_{2(n+1)+1}-1?$

[flat frigate]

Chrisfv97

[urban dock]

Because of our inductive hypothesis?

I’m not sure where that -1 comes from

[urban dock]

Still that does not make sense to me tho

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[quasi bison]

ah the -1

my bad i overlooked it

it's just a typo

[urban dock]

Thanks for the confirmation.

Have a good night.

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[stuck wren]

could someone help me with discrete math problem

[quiet plume]

Sort the edges by weight and add them in ascending order if it doesn't make a cycle

[stuck wren]

so to approach it does it have to be minimum weight?

[quiet plume]

The algorithm is as I explained

You sort the edges in ascending order and then add them to the graph one by one if they don't make a cycle appear

The result will be the minimum spanning tree

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@stuck wren Has your question been resolved?

[stuck wren]

so GF (5) + FE (6) HF(7) GH(8) all added is 26 but that isnt correct

im still lost

[quiet plume]

You can't add GH in your sequence

It creates a cycle

[stuck wren]

oh

i understand now

xd

can u help me double check why im getting a 50/100 on this

I feel like im not seeing the error on which box im getting wrong

[quiet plume]

Tree means that there is no cycle and it is connnected

Forrest just means that there is no cyclles

Is there one of the choices that you can see has a cycle?

[stuck wren]

corner right has a cyucle

[quiet plume]

It looks like it because it's tangled but there is no cycle

[stuck wren]

when i check both tree and forest

i get incorrect on all

[stuck wren]

are you still there by any chance?

[quiet plume]

Bottom left is wrong.

[stuck wren]

dang that took forever

xd

thanks

i've been struggling on this problem as well its my final problem

68 is incorrect but i feel like its right using kruskal's alg

would you guys know how to solve

[quiet plume]

Draw it out as you go

Make sure you don't create a cycle

[stuck wren]

sheeesh

37 was the answer

gg

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[deep dagger]

I tried to factor the numerator using the quadratic formula to get this answer, but I can't find any result other than -3 and 1, which would ruin what my teacher said

[deep dagger]

omg typing that out made me realize the issue, nvm 😭

ty anyways <3

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[grizzled topaz]

proof using t formulae

ngl im not really sure how to even incorporate t formulae into these. ik the formulae tho don't really understand the application

[vale oriole]

barely used it, i guess just express cos2x and sin2x in their double angle identity form, then replace every cos and sin with its t formulae counterpart

[grizzled topaz]

wait nvm i got it

[grizzled topaz]

yeah ig the only reason it exists is so that if you can't make conceptual leaps and see identities you have something to use

thanks though

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[frozen bloom]

Hey so I have a problem that I'm not quite sure about how to approach

It's a true/false question

Demand is q(p)=-6.8p+78.2, if p=6, is price elasticity around -1.1?

[frozen bloom]

Frankly I think the thing I struggle to understand is how am I supposed to calculate price elasticity without any given target price? Isn't price elasticity the change in demand divided by change in price?

<@&286206848099549185>

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[lean otter]

Q) Area of the shaded region as kπ cm^2

[lean otter]

do you know about the area of a sector

yeah

state it

theta/360 * pi r^2

i tried and got 24

right so first let's try and find the area of the outer donut entirely

what would you say that is

the ring?

yeah

all of it

27pi

84.8

yeah right

you want to exclude the portion that makes 60 degrees in the outer ring but want to include that same region in the inner circle

yeah we can use major sector with theta=300

treat both the inner circle and outer circle as one circle. can you find the area of the sector of the combined circle

6pi ?

nope

the sector

the shaded one would be 3/2 pi

then the none shaded would b 6-3/2

9/2 pi?

the combined circle would have a raidus of 9 cm

why

the combined circle has radius 6

6 + 3 = 9

oh wait I misinterpreted

right okay my bad

yes

i use major sector

[lean otter]

so yeah that's right

that major sector theta is 300

so area is 300/260 * pi r^2

360*

which is 30 pi

than we see the small major sector

which has radius 6

3*

we get 15/2 pi

then we subtract

45/2 pi as shaded donught

this is what i did

and added the small sector as 60/360* pi(3)^2

and got 3/2pi+45/2 pi

and got 24pi

but i think this is wrong

[willow gust]

1. figure out the area of the big circle
2. Figure out the same for the smaller one
3. Minus those two
4. Get the area of the entire circular sector i.e the 60 degree one
5. Minus the sector's area from the previous
6. Then add the smaller circle's sector

[lean otter]

did i do it right?

[vale oriole]

xd

[lean otter]

my teacher does not allow radians

[willow gust]

Not a big deal use this
Theta/360 * pie * r^2

[lean otter]

did i do it right?

[willow gust]

Theta is in degree

[lean otter]

is this okay?

[willow gust]

Yea u r good to go

[lean otter]

i got 24 my majority friends got 23 or 22.5

does not that mean i am wrong

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[dreamy nacelle]

can someone help me prove this

[dreamy nacelle]

this is what i have done myself

is this a complete proof

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[summer bay]

Hello, i need some help with a transfer function

[summer bay]

I know that the answer should be in the form Ti*K < some value

And this is my try so far:

The issue here is i dont know how to solve this for Y(S)/R(S) without having R(S) in the right side of the equation

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[shrewd timber]

if infinity doesn't belong to real number system then to which one it belongs?

[lean otter]

∞ or −∞ are not elements of R

[west haven]

well as far as I know I'd read this once upon a time on math stack exchange and someone convincingly said that infinity (+ and -) are not a member of R but there's an extended system of R to which they belong

[lean otter]

R∪{−∞,∞} which contains ∞ and −∞as its elements.

[west haven]

yes, that

[lean otter]

it belongs to the extended' real number system

[shrewd timber]

but you have to extend the number system to add them. By default inf doesn't belong

[lean otter]

no

[shrewd timber]

yes

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

[lean otter]

they dont belong to the standard real number system

[shrewd timber]

yup

[lean otter]

we extend it by union to add the infinitity element'

hope this helped

[shrewd timber]

but we had to extend this number system so it means that they didn't belong to it earlier

[lean otter]

yes

[shrewd timber]

and my question is: where did they belong earlier

[lean otter]

they bargain to be added

does not belong anywhere

[shrewd timber]

oh

[lean otter]

it did not had a home

[shrewd timber]

thx

[lean otter]

it gave rent to real number system

then got in

[shrewd timber]

thx

[lean otter]

np

[shrewd timber]

.close

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[safe radish]

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[wise pebble]

Help

[wise pebble]

How do I go about this

Do I just use the decay rate formula?

Is there another way of doing this? Because my teacher never went over that formula

[quasi bison]

what is "the decay rate formula"?

[wise pebble]

@quasi bison

[quasi bison]

this smells like chatgpt

but anyway no you do not need this

you want the hourly decay rate

i.e. by what percentage does erbium-165 decay in 1 hour?

[safe radish]

@wise pebble Has your question been resolved?

[wise pebble]

it is

[wise pebble]

i forgot that i can plug any value in for a

as the starting value

that's what i forgot

[quasi bison]

!nogpt

[safe radish]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[wise pebble]

are you saying i'm breaking the rules

i'm not answering people's questions with ai bruh

[quasi bison]

i am not saying you are breaking the rules

if you were breaking the rules, i would be modmailing this, and a mod would show up.

i am simply doing this

[wild cape]

not talking about that. its this

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors.

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[ornate pagoda]

How do I prove a linear transformation is injective if and only if its kernel=0?

[vast obsidian]

Did you already try something? Or do you have an idea?

[odd zenith]

oops sorry

[ornate pagoda]

I think it has something to do with there being all linearly independent rows.

[vast obsidian]

Yeah, I think you can do it like that, but I wouldn't go this path

Let's say the kernel isn't just {0}. Can it still be injective?

[granite idol]

you could try using the definition of injectivity to show the kernel is 0 (I assume that means it only contains the 0 vector). then go in the other direction

[wild cape]

theres no matrix so no rows to talk about. all we have is a linear map

[ornate pagoda]

I don't know.

[vast obsidian]

Ah, yes. I think this equivalence still works in infinity dimensional vector spaces?

[wild cape]

ya

[granite idol]

ooooh, hadn't considered that

[wild cape]

the proof has 0 dependence on dimension

[vast obsidian]

We know that 0 is definitely in the kernel, because our function is linear. So for the kernel not to be {0} there needs to be another non-zero element inside it. So is our function still injective?

[ornate pagoda]

I guess not, because 0 would map to more than one element?

[vast obsidian]

Nah, then the function wouldn't be well defined

The kernel is the set of elements being mapped to 0

Therefore, we would have f(a) = 0 = f(0), but a isn't 0 => not injective

[ornate pagoda]

I was right then?

[vast obsidian]

Well, what you basically said is f(0) has two values. I think you meant the right thing, but it was expressed a bit misunderstandable

[ornate pagoda]

Oh I meant 0 in the codomain 😳

[wild cape]

to move the convo along ill label stuff

let $V$ and $W$ be vector spaces and let $f:V\to W$ be linear. prove $f$ is injective iff $\ker(f)=\brc{0_V}$

[flat frigate]

Roketsune Janiku

[vast obsidian]

Do you want to take over? Obviously you are far more qualified...

[wild cape]

if you know the proof then feel free to continue

[vast obsidian]

Yeah, I took Lin-A 1 before

[wild cape]

its good to ask guiding questions before trying the proof, but judge how long is appropriate

[vast obsidian]

So we have one direction done. So only one more left 🙂 For the other one suppose f wasn't injective.

What would that mean using the definition of injectivity?

[ornate pagoda]

That each element mapped in the codomain didn't map to exactly one element in the domain?

[vast obsidian]

No! A function is already not injective if two different elements of the domain are mapped to the same value

Can you express that with formulas?

[ornate pagoda]

But I thought you were asking about if it wasn't injective.

I don't think so.

[vast obsidian]

Not (not every x in the domain is mapped to a different value) <=> There exists a ≠ b so that f(a)=f(b)

[vast obsidian]

Ok?

[ornate pagoda]

Oh, I didn't know that...

[wild cape]

you should have at least one rigorous definition of injective memorized

[ornate pagoda]

I thought the domain could map to the same value in the codomain.

Or an element in the domain.

The definition I had was that every y value mapped to exactly one x value.

[wild cape]

wdym? can you state that formally?

[ornate pagoda]

Not really. That's just the definition I have. I guess it means f(a)->a and only a.

[wild cape]

heres the definition most students learn

f is injective given f(a)=f(b) implies a=b

[ornate pagoda]

Okay.

[wild cape]

now that i put a formal phrasing heres the intuition behind it

do you know contrapositives?

[ornate pagoda]

No.

[wild cape]

an implication is a statement of the form "P implies Q"

the statement is defined as true when, in each case P is true, Q is also true

[ornate pagoda]

Oh wait.

Yes I do.

[wild cape]

i was building up to contrapositives

do you know this along with the above stuff?

[ornate pagoda]

Yes.

I know most formal logic rules.

[wild cape]

ok go back here

take the contrapositive of this

f(a)=f(b) implies a=b

a!=b implies f(a)!=f(b)

so this gives the intuition for injective functions: different inputs have different outputs

[ornate pagoda]

Okay. Makes sense.

[wild cape]

f(a)=f(b) implies a=b
tho this phrasing is used more often when proving a function is injective

[ornate pagoda]

0 can only map to 0 in the kernel then?

[wild cape]

now return here

i believe Error tried to walk you through proving the reverse implication, assume ker(f)={0_V} then show f is injective

so do that using the definition of injective i gave you

[ornate pagoda]

If ker(f)=0v, it means f(a)=...f(n)=0 implies a...n=0?

[wild cape]

i need to head out now. maybe @vast obsidian is still on

what i can say is take a bit more care in notation, and recall the proof goal, in particular how to prove implications

[vast obsidian]

Sorry, got to solve a system of differential equations

[wild cape]

ODEs, always fun...

[ornate pagoda]

I actually have to go too, but I think I understand the idea.

Thank you!!

.close

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[wild cape]

ok if you have a proof attempt later and want it checked, try #linear-algebra

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[lean otter]

are there any cases where the geometric multiplicity of an eigenvalue is not 1?

[quiet plume]

Yes there is no reason why it has to be 1

[tall bough]

^

[quiet plume]

It's just the dimension of the eigenspace

It won't, however, exceed the algebraic multiplicity

[lean otter]

may you explain multiplicity to me

i dont really understand any of it 😭

geometric, algebraic or the other one idk

[tall bough]

U can google them

There are plenty videos

[lean otter]

i did i still dont rly get it aaa

[quiet plume]

Me explaining it here probably won't be any clearer though

The algebraic multiplicity is just the 'power' of the root in the characteristic polynomial

The geometric multiplicity captures how large the space of vectors with this eigenvalue is

Since it is the dimension of the associated eigenspace.

[vocal kiln]

Do you understand linear independence?

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

ye

[lean otter]

wouldnt algebraic multiplicity always just be the degree of the characteristic polynomial?

is there only one algebraic multiplicity for a matrix or is it for like each eigenvector

[quiet plume]

The matrix has a characteristic polynomial. The eigenvalues are its roots.

[quiet plume]

If the characteristic polynomial is x^2(x-1), then the eigenvalues are 0 with multiply 2 and 1 with multiplicity 1.

[lean otter]

oh i think i understand

[safe radish]

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[safe radish]

Please don't occupy multiple help channels.

[limber silo]

but your aren’t answering

@safe radish

[final halo]

the bot doesnt magically help with questions, people do, close one of your channels and wait for someone to come along and help

[limber silo]

alright sorry my friend

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[molten hawk]

How do you solve this?

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[molten hawk]

<@&286206848099549185>

[vagrant oracle]

@vagrant jolt

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

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@molten hawk Has your question been resolved?

[subtle oriole]

.reopen

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[tawny vessel]

So can someone help me on that

[tawny vessel]

i dont understnad that

where is the ln|x| from

the first term is the tan

isnt ln x only for 1/x

where is 1/x in that original equation

.close

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[twin prawn]

question, if the limit is undefined (1/0 for example), why do we indicate infinity or negative infinity when looking for the left hand or right hand side?

[sinful basalt]

it trends towards that but doesn't go to a specific point

It gives you an idea of the type of behavior you're looking at, which is occasionally relevant

[dapper venture]

Infinity here just means 1/x keeps growing without bounds as x approach 0+

[twin prawn]

so if the limit of a right hand or left hand is infinity, that means as x is apporaching a certain value, it will stop (y value) at infinity?

but

wait

[dapper venture]

Infinity is not a number you can't stop at infinity

[twin prawn]

ohhh

itll never stop

until it reaches it (which will not)

but in this case, why cant the right hand limit of that be the limit of the simplified form of the fraction

[dapper venture]

?

[twin prawn]

instead of the limit being +infinity, why cant the answer be like 2.01 / 0.0401

or a further simplified answer of the latter

[dapper venture]

2.01/0.0401 is the value you get when you plug in x=2.01

[twin prawn]

yeah because 2.01 is the

[dapper venture]

The limit means you keep getting closer to 2

[twin prawn]

mhm

but this one, is like what to do when the limit is undefined

ie 1/0, 2/0

[dapper venture]

Imagine a sequence of number that keep getting close to 2

2.1 2.01 2.001 2.0001

[twin prawn]

it reminds me of

that guy telling me to like

walk half a step

then half of that

then walk half of what you just halved

and so on

thats how it worked

[dapper venture]

That's a geometric series, also closely related to limits

[twin prawn]

oh, mb

[twin prawn]

but yeah, since this is limits where the initial value is undefined, why did we consider the positive fraction as positive infinity

[dapper venture]

So as we keep inputting these x values we get larger and larger result

[twin prawn]

yea

[dapper venture]

We say the limit approach infinity if the result keep getting larger without bound

Another way to say this is that no matter what number you give me, I can get a larger number by inputting a small x value

[twin prawn]

on both sides? (left and right)

[dapper venture]

Yes

[twin prawn]

cause i think the limit diverges or DNE if one of the sides do not match the other

however tho, if the limit is undefined

we have to look at the left hand and right hand

[twin prawn]

this is the right hand part of it,
the technical answer is 2.01 / 0.0401; but my question is, why did we consider this as positive infinity and not the former?

[dapper venture]

I think you meant if the function is not defined

[dapper venture]

As I said the limit means you keep getting close to 2

2.01 is not "close enough"

[twin prawn]

ohhh

[twin prawn]

because there is a sequence like this thats why its a positive infinity

[dapper venture]

No the sequence is not important

[twin prawn]

oh

[dapper venture]

The important part is the result we get after plugging the sequence

[twin prawn]

okok

[dapper venture]

The result keeps getting larger without bound

[twin prawn]

plugging the sequence of 2.01 reaching 2

[dapper venture]

You keep going

After 2.01 you do 2.001

Then 2.0001

You get closer to the limit

[twin prawn]

but you dont reach the limiut

thats why its infinity

[dapper venture]

Each number down the sequence

[dapper venture]

Not quite

[twin prawn]

okay dumb question

[twin prawn]

oh

[dapper venture]

The limit could be like 1 for example

For another function

We still don't reach 1

[twin prawn]

asymptotic?

[dapper venture]

Infinity just means growing without bound

?

[twin prawn]

yeah but, the sequence that youve shown

[twin prawn]

wait

is infinity also decreasing without bound

[dapper venture]

We call it negative infinity

[twin prawn]

yea

but wouldnt your sequence be like

decreasing

without bound

2.01

2.001

2.0001

2.00001

etc etc

[dapper venture]

It's bounded below

It's never less than 2

[twin prawn]

Ohhhhhh

decreasing without bound would be

2

1

0

-1

-2

-3

-4

-5

[dapper venture]

Ya

[twin prawn]

-6

etcetc

[safe radish]

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[glossy elbow]

Im doing finals review, and this is the last problem I have and im completely stuck

[dapper venture]

fill in ¬p and ¬q first

[glossy elbow]

im not sure how Im looking at my notes and it just doesnt ring a bell

[granite idol]

just negate p and q

[dapper venture]

do you know what the symbols mean

[glossy elbow]

it means not

so Not P is true when p is false right?

[dapper venture]

yes

[glossy elbow]

okay

okay its starting to come back

[dapper venture]

not true is false and not false is true

[glossy elbow]

let me do the first two

the third means if not q then not p?

but what does that mean

so its true when both is not true?

wait

if not q, which is when q is false, then not p is true when p is false

[dapper venture]

A->B is the same as not A or B

[glossy elbow]

so its only true if either p or q is false?

[buoyant shadow]

(not A ) or B

[dapper venture]

sorry for the bad notation

[glossy elbow]

why isnt it not B as well?

since it has the symbol

[dapper venture]

what symbol

[glossy elbow]

wait

so for the 6th column, I would only need to look at the 4th and 5th column correct?

[granite idol]

you should review these things

[mossy bobcat]

if not q then not p or r

the arrow (->) is a conditional, and conditionals only apply if the initial condition is true

[mossy bobcat]

i'd recommend memorizing when it's false because in all other cases it will be true

[glossy elbow]

sorry it went up too high

[mossy bobcat]

it's pinned don't worry

[granite idol]

also note that ~q -> ~p is the contrapositive of p -> q

[glossy elbow]

im going to quickly watch a video because I keep looking at my notes and they just don't ring any bells for me

[mossy bobcat]

🫡 .

[glossy elbow]

okay

[glossy elbow]

-If I guessed RIGHT then answered RIGHT, it make sense(it is RIGHT)
-If I guessed RIGHT then answered WRONG, it doesn't make sense (it is WRONG)
-If I guessed WRONG then answered RIGHT, it still make sense (It is RIGHT)
-If I guessed WRONG then answered WRONG, it still make sense (It is RIGHT)

[mossy bobcat]

yep!

[glossy elbow]

okay I understand that part

I also understand the negation part as well

but Im still confused about what the -> arrow is trying to tell me I know it means if then but

[mossy bobcat]

just plug the negated values into this

you don't have to think too hard about it making sense in english, just take the truth table for the conditional and the values you got for ~q and ~p

[glossy elbow]

so for example, I guessed wrong, got it wrong, meaning it is wrong

for the first one

same with the second

for the third one, I guessed wrong, then answered right, it is right

same with the fourth?

oh wait

I need to be looking at q first

[mossy bobcat]

yeah

[glossy elbow]

F F is F

[mossy bobcat]

also this would still be true because the conditional is only false if the initial condition is true and you fail that initial condition

[mossy bobcat]

^ do you see in this truth table F F -> T

[glossy elbow]

ohhh

so thats what you meant about memorizing when its false

[mossy bobcat]

yep

[glossy elbow]

okay let me look at it again

But in my case it would just be the opposite

because of negation

[mossy bobcat]

yeah, but since you have the values for that negation already in the truth table, you don't need to try and apply that in your head and just look at the column values that you have

[glossy elbow]

oh im just overthinking it

[mossy bobcat]

F F = T
F F = T
T F = F, so on and so forth

yeah you are very much overthinking it 😅

[glossy elbow]

I see

[mossy bobcat]

truth tables aren't that bad

[glossy elbow]

😭

[mossy bobcat]

it's alright, you're learning

[glossy elbow]

I used to speed run through these but idk whats wrong with me rn

F F is T
F F is T
T F is F
T F is F
F T is T
F T is T
T T is T
T T is T

[mossy bobcat]

yep

[glossy elbow]

sweeet

let me quickly plug them in

[mossy bobcat]

also for the conditional it might also help to think of it this way. "if i have a banana, then i eat it." (p = having a banana, q = eating it) if you don't have a banana, you can't eat a banana, so it doesn't matter if q is true or false because either way you're still banana-less. however, if you do have a banana, and then you don't eat the banana, you've failed the conditional, therefore it's false.

that, along with memorizing the only possible time it's false (you have a banana and you choose not to eat it, disobeying the 'if i have a banana then i eat it') will probably help. also it's just funny i mean cmon

[glossy elbow]

Let me try to use that while I try to mentally answer the truth table

[mossy bobcat]

🫡 don't take me and my ramblings too seriously, if you think it's hurting your understanding of conditionals just stick to the truth tables because those won't lie to you

[glossy elbow]

I think it probably is a good way to think about it but I think trying to think about all these bananas is a little difficult for me at the moment

lol

[mossy bobcat]

HAHA

real and i had a feeling it was like that

i think you've figured out the worst of it though (overthinking) so

[glossy elbow]

Its probably because ive been studying all of decreit math for the last few hours im just drained atm

okay sweet

[mossy bobcat]

yep that looks good

[glossy elbow]

so i know v means or (disconjunction)

[mossy bobcat]

yes

[glossy elbow]

so same concept of -If I guessed RIGHT then answered RIGHT, it make sense(it is RIGHT)
-If I guessed RIGHT then answered WRONG, it doesn't make sense (it is WRONG)
-If I guessed WRONG then answered RIGHT, it still make sense (It is RIGHT)
-If I guessed WRONG then answered WRONG, it still make sense (It is RIGHT)

but I can include r if it is not true

[mossy bobcat]

ehh kinda

i don't think guessing really applies to ors unfortunately

there's not really a concept of 'guessing' that you can apply to or, it's just either one or the other is true, therefore it's true overall

[glossy elbow]

I have written down "because the logic symbols represents or, in order for the statement to be true, only one must be found true

[mossy bobcat]

ohhh

yes that's very correct

[glossy elbow]

okay now im going to try to understand it

[mossy bobcat]

or is very straightforward 👍

[glossy elbow]

okay since they have the first part that Ive already completed

All the T I have is T

[mossy bobcat]

yep

[glossy elbow]

so in total theres only 1 false

[mossy bobcat]

yep!

you got it

[glossy elbow]

jeez

i really

did overthink this

[mossy bobcat]

a bit 😅

[glossy elbow]

okay thank you so much 😭 thanks for dealing with all of it

[mossy bobcat]

it's all good mate

good luck with your work!

[glossy elbow]

thank you

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@glossy elbow Has your question been resolved?

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[lean otter]

Hii, so I have a final for stats tm but basically what is happening is I currently have an 81.64% (370.6456/454) in the class . However, my professor is dropping our lowest exam AND quiz score from the gradebook. My lowest exam score was a 76/100 and my lowest quiz score was a 0/40. The exams are weighted at 40% of our grade and quizzes at 15% of our grade. I'm trying to figure out a way that I can calculate my new grade after my lowest exam and quiz score are dropped from the gradebook

[lean otter]

btw if you need more information id be happy to provide

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@lean otter Has your question been resolved?

[pine bronze]

tbh i gotta go right now but it sounds like chatgpt might know his way around this, or have you tried that already?

[lean otter]

i tried it alr lol

[pine bronze]

ah dangit hahha, theres def someone in here w some spare time, goodluck

[hard crest]

pretty sure more information is needed, mainly how many quizzes and exams you've taken

but it might be easier for you to do this yourself

calculate your quiz score on its own (take the average of all of the quizzes except the 0/40), and multiply it by 15%
calculate your exam score on its own (without the 76/100, like the quiz score) and multiply it by 40%
calculate the rest by taking your 81.6% and multiplying by 45%
add all of these up and that will be your new score

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@lean otter Has your question been resolved?

[lean otter]

ok thanks!

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[dreamy nacelle]

can someone check my proof. I feel like I am missing some cases at step(rule 2). And I am not sure about my step(rule 3) part.

[safe radish]

@dreamy nacelle Has your question been resolved?

[fickle pendant]

I'm slightly confused by the headers

[dreamy nacelle]

What would you like clarification for?

[fickle pendant]

oh my bad - I see, each of the Step (rule x) is part of the inductive case

[dreamy nacelle]

yes exactly

[fickle pendant]

eesh, yeah I typed out a few paragraphs and realised how awful this is to try and write explicit cases for

I think your approach is good - the only place we could reach more than 2 Rs is the join of c and d. the applications of rules 1 and 2 give you at most 1 external R and any application of rule 3 will produce a string surrounded in B
so the most you can get is xxxRRxxx

[dreamy nacelle]

yes, this is what i wrote down right?

[fickle pendant]

yes

[dreamy nacelle]

The thing that i was doubting about is if I should account for a colorsequence made with rule 2 was put pack into rule 2 that it still satisfies the claim. So if for example some sequence created with rule 2 .....Br is c and used to create .....BRBR. If you understand what i mean

[fickle pendant]

for rule 2, you've proven a more general case

you've shown that no matter what c is, as long as it satisfies the no RRR constraint, then rule 2 will always produce a valid string

even if that original c isn't made from rules 1, 2, and 3 (e.g. RR)

[dreamy nacelle]

should i consider c = RR since this is still is a color sequence that satisfies the claim?

[fickle pendant]

you don't need to

in fact, it would break if you consider arbitrary but satisfactory sequences for rule 3

it just so happens that the general case is easier for rule 2

[dreamy nacelle]

I guess so

then I won't change anything about my proof.

thanks for your time 🙂

.close

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[eternal willow]

how do i approach this

[potent seal]

For continuity, left and right hand limits are equal

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[lean otter]

<@&268886789983436800>

[hard crest]

they left the server i think

[lean otter]

oh you did see it tho right?

the invite Link

[hard crest]

yea

[lean otter]

ok great

[hard crest]

i deleted it

[west haven]

.close

ok

sad

@hard crest could you help close this

[hard crest]

.it is closed already, it'll update any minute now

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[light dagger]

Help

[safe radish]

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

[light dagger]

1

[pulsar pecan]

for part a, is it asking for just the floor, or the surface area of the whole tent?

Part b is easier than part a

[light dagger]

we need the surface area of whole tent

how do u do it

[pulsar pecan]

Add up the areas of each of the faces

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@light dagger Has your question been resolved?

[light dagger]

.close

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[vale estuary]

use trigonometric substitution

assume a right triangle with hypotenuse = 1, and the other 2 sides to be x and y and one of the angles to be theta

now find x and y in terms of theta

and find dx/d(theta)

and find dy/d(theta)

and divide them

you will get dy/dx in terms of theta

yeah

one second

instead of doing it like this,
x = sin A

y = cos A

do this

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[mint raven]

help with reference to previous question

[mint raven]

with reference to a prior question

#help-34 message

would the answer 5y-2x<=20 also be correct

[granite idol]

looks like it, yes

[mint raven]

thats unusual

the website i got this from seems to reject that answer

and thats what got me stuck

does the requirement of it being in standard form the reason for the discrepancy

.close

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[loud cedar]

help

[flat frigate]

OccamsChainsaw

[lapis wren]

this is your equation correct?

[main mural]

i am pretty sure "4." is just number of the question lol

[glacial cairn]

Pretty sure 'z' is a 2

But first I'd ask:

[glacial cairn]

what is the question?

[loud cedar]

Oh its y + 2

[loud cedar]

Find the value of the unknown variable

[loud cedar]

Yes its just the number

[loud cedar]

Yes

@glacial cairn

[main mural]

condider crossmultiplication

[loud cedar]

am I gonna get the LCD?

Of denominators

[glacial cairn]

$\frac{y+2}{y^2-1} = \frac{2y}{2y^2-y-3}$

[flat frigate]

Nel

[loud cedar]

Yes this

Its actually 12 am here hrudushshsh

[glacial cairn]

First, determine the values that y cannot take

[lapis wren]

do you know about criss-cross multiplying with equations w/ fractions?

[loud cedar]

Nope

But Ik the getting of Lcd

[glacial cairn]

That's not very useful here

[lapis wren]

basically, in equations like this you can criss-cross multiply and get an equation without fractions

(this only works if equations on either side are in their "final" form, so not for 1/2 + 2/3 = 2/4 + 4/5)

[loud cedar]

final form?

[loud cedar]

Wait???

What does it means

[glacial cairn]

That's a pretty bad way to look at it

[lapis wren]

simplest i guess?

i dont know of a good translation

[glacial cairn]

Just multiply both sides by (y^2-1), and then by (2y^2-y-3)

[lapis wren]

but basically, if you dont have another fraction to add or subtract on either side, then you can do this

[lapis wren]

why..? its much faster to criss cross than LCD if youre dealing with bigger numbers

[loud cedar]

Why

[glacial cairn]

I'm not talking about LCD. I'm saying that your "criss cross" method is a shortcut to multiplying both sides by both denominators, except it's not really a shortcut, and you can lose sight of the fact that y cannot equal some values (that make the denominators zero)

[loud cedar]

But will I get the same answer?

[glacial cairn]

It's the same thing if you're careful about it

[loud cedar]

Also I need to do the checking

[loud cedar]

Afterwards

Hm which one should I do?

[loud cedar]

Cross multiply or this

[lapis wren]

whichever one is easier for you

[loud cedar]

how

Do i write it

@glacial cairn Nel

@lapis wren

Guyss

@lapis wren Ill just cross multiply

[snow haven]

hello

still need help?

[loud cedar]

yes pls

Its almost 1am ahahajauwha im so sleepy