help-17

[iron parrot]

good!

now try for h,and i

[shy walrus]

i will

i got

h= 0

i= -3

[iron parrot]

correct!

good job!

[shy walrus]

thank you

im doing this one now

can you help me through this one?

-18 divided by + 3 is -6

b-24 divide by -12 = 2

c+ 34 divide by -2 = -17

what next do i have to do?

would it be b+ 2 = -6?

[iron parrot]

exactly

[shy walrus]

i got b= -8 and c= 11

i got e= 24 and f= -9

i got h= 10 and i = 5

for question 3, i got b= 13 and c=-11

then e= 104 and f= 88

then h = -15 and i = -30

[vocal sleet]

@shy walrus Has your question been resolved?

[shy walrus]

for question 4 i got b= -1 and c=20

then e = 3 and f= 40

this was the full question btw

[vocal sleet]

@shy walrus Has your question been resolved?

[vocal sleet]

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[steep crescent]

given 2 positive sequences, a_n and b_n, lim (a_n/b_n) = L, prove that:

1. if L = 0 then sum(b_n) < ∞ imples sum(a_n) < ∞
2. if L = ∞ then sum(a_n) < ∞ inplies sum(b_n) < ∞

[tawny hollow]

What have you tried so far?

[steep crescent]

i dont know how to start

[tawny hollow]

Ok so for 1), then lim a_n/b_n = 0

Right?

[steep crescent]

yeah

[tawny hollow]

And we’re supposing here that sum of b_n is convergent right?

[steep crescent]

yes

[tawny hollow]

Alright as a start notice that for big enough n, then we have that a_n/b_n < 1

Do you see why?

[steep crescent]

thats from the definition of convergency isnt it?

[tawny hollow]

Yup

[steep crescent]

for any epsilon > 0

[tawny hollow]

I have chosen eps = 1 here

[steep crescent]

yeah

[tawny hollow]

Cool

Now try and work with that inequality

And see if you can draw the conclusion you want

Remember that sum of b_n is convergent

[steep crescent]

oh

a_n < b_n < alpha

oh wait

b_n has to converge to 0

[tawny hollow]

Remember also that these sequences are positive

[steep crescent]

and a_n < b_n

[tawny hollow]

Sure

[steep crescent]

so it also converges to 0

is that enough?

[tawny hollow]

No

You want to conclude the convergence of sum a_n,

[steep crescent]

wait isnt there a theorem

[tawny hollow]

Sure the condition that a_n converges to 0 is necessary, but does not at all imply convergence of its sum

[tawny hollow]

Yea, which?

[steep crescent]

if lim a_n/b_n > 0 and finite then sum(b_n) converges iff sum(a_n) converges

[tawny hollow]

Uh you can’t use that here

[steep crescent]

why

for a big enough then

the limit of that is 1

[tawny hollow]

Look at L

Nope

The limit is 0

[steep crescent]

oh right

[tawny hollow]

Alright, let me just nudge you in the right direction.

[steep crescent]

yeah then the implication is one way

[tawny hollow]

You want to use the comparison test here or whatever it’s called

[steep crescent]

if lim a_n/b_n = 0 and finite then sum(b_n) converges => sum(a_n) converges

[steep crescent]

whats that

[tawny hollow]

That’s what you want to show

• some other conditions

[steep crescent]

thats the convergence equivalence test

or whatever its called in english

oh wiat

[tawny hollow]

If 0 < a_n < b_n, and sum b_n is convergent, then sum a_n is convergent

[steep crescent]

thats what im supposed to prove

lmao

turns out thats a theorem

anyway

[tawny hollow]

Have u seen this?

[steep crescent]

i dont remember no

[tawny hollow]

This is pretty crucial for pretty much any stuff u do with series

[steep crescent]

good to know

[tawny hollow]

It’s basically the backbone of most of it, if not all results

Uh and it’s easy to show as well iirc

Not sure how you haven’t seen this

[steep crescent]

well this is the first problem about series' i do

lol

well, second

but it was supposed to be the first

anyway

the sum is convergent iff the sequence is convergent

so

if 0 < a_n < b_n then a_n is convergent

then sum(a_n) is convergent

thats enough no?

[tawny hollow]

I guess you can just prove this aswell within

[tawny hollow]

Huh?

[steep crescent]

theres a theorem

oh wait

[tawny hollow]

That’s not all right

[steep crescent]

i misread it

forget that my bad lol

[steep crescent]

so if sum b_n is convergent

[tawny hollow]

The crucial part here is by assumption sum b_n is convergent

[steep crescent]

then b_n converges to 0

[tawny hollow]

You have to use that

Sure, but that’s gonna mislead you

[steep crescent]

what do i do then

[tawny hollow]

So notice that you can take the partial sum of a_n and b_n and still maintain an inequality

[steep crescent]

the partial sum starting from a specific N right?

[tawny hollow]

Yup! Now notice that you have an upper bound for the sequence of partial sums of a_n

[steep crescent]

right

[tawny hollow]

And since a_n is positive and goes to 0, we know that the sequence above is also positive and increases

[steep crescent]

why does a_n go to 0

[tawny hollow]

Now you can make use of the monotone convergence theorem and conclude that the sequence indeed converges, and the limit is just the sum of a_n

[tawny hollow]

You explained that earlier?

[steep crescent]

using this

[tawny hollow]

Ye

[steep crescent]

but i cant use it

or i can prove it first

how long is the proof

[tawny hollow]

That’s the proof

[steep crescent]

what?

[tawny hollow]

That’s the proof

[steep crescent]

what part

we proved the original question

[tawny hollow]

For 1)

Yeah

[steep crescent]

yeah but i need to prove this

oh

brain fart

[tawny hollow]

That’s what I proved within the proof of 1)

[steep crescent]

yeah yeah

sorry let me re read that

[tawny hollow]

Since you apparently haven’t encountered it before

[steep crescent]

so to write it formally

i can say

for any epsidon > 0, there exists N such that for each n > N, a_n/b_n < epsilon + 0

we choose epsilon = 1

so for every n > N, a_n/b_n < 1 => 0 < a_n < b_n

is that right?

[tawny hollow]

Well almost, the part where a_n and b_n are positive does not follow from this, you make it look that while it was just an assumption from the start

But maybe that’s clear

[steep crescent]

which part

[tawny hollow]

Typo

[steep crescent]

oh

yeah i can write that seperately

[tawny hollow]

But yeah that’s all correct

Good

[steep crescent]

thanks!

now for 2

[tawny hollow]

Uh but that’s just the start of 1)

Or was the rest clear?

[steep crescent]

yeah the rest is good

[tawny hollow]

Alright in such case, then 2) should be fairly obvious

[steep crescent]

why

[tawny hollow]

Well they’re almost the same

Just flipped in some sense

[steep crescent]

i dont see it

[tawny hollow]

Try harder, if you understood 1) then it’s worth trying something similar for 2)

[steep crescent]

do i start from inverting the definition of convergence?

[tawny hollow]

Uh not sure what that means

[steep crescent]

a_n/b_n doesnt converge

so

[tawny hollow]

Yup

Or not here

[steep crescent]

oh

[tawny hollow]

We’ve allowed infinity here

It’s a special type of convergence

[steep crescent]

thats considered convergence?

[tawny hollow]

Not the same as oscillating

[steep crescent]

so for any n > m > N, a_n/b_n > a_m/b_m

[tawny hollow]

Depends, I agree the way the statement is setup is vague

[steep crescent]

what confused me is that they explicitly say that for a series convergence is defined as a limit = finite number

anyway

[tawny hollow]

Well so let’s distill this more, take a sequence x_n which “converges to infinity”, then we can just say that for any epsilon > 0, there exists some N such that n>N implies x_n > epsilon

[steep crescent]

right

[tawny hollow]

So use this

[steep crescent]

so we choose epsilon to be 1

[tawny hollow]

Sure

[steep crescent]

there exists N such that a_n/b_n > 1

a_n > b_n

[tawny hollow]

Yea

And so you might see where this is going

It’s the exact same steps as before

[steep crescent]

oh then from here its the same as above

yeah

[steep crescent]

so this is called the definition of what

[tawny hollow]

Don’t recall if there’s a name but say we didn’t have this definition.

Then you could argue like this. So (by the vague statement of the question,) the sequence a_n/b_n should eventually blow up to infinity, so we know that for some n>=N we have atleast a_n/b_n > 1
…

And that’s sort of enough

[steep crescent]

but that doesnt prove that it doesnt go back down

[tawny hollow]

Wym prove? It’s given that it eventually blows up, so we can find some N where it doesn’t go down below a certain threshold and where it keeps getting larger. This N is surely finite

[steep crescent]

i mean that

dont we have to say that for evey n > N?

oh wait

oh yeah thats what that says

yeah makes sense i think thats more than enough

thanks a lot

[tawny hollow]

[steep crescent]

.close

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[waxen hawk]

How can I

[midnight panther]

.close

.reopen

[waxen hawk]

I

Have pictures

Of equation

I want my artificial slave help me solve it

Cause I’m too lazy to have my palm on the calculator

How

I want ChatGPT

Or any other artificial slave

Assistant

[sly sierra]

wtf

have you been taken over by a bot

[waxen hawk]

To calculate equation on a picture

[waxen hawk]

Wdym

I never trust a bot

[sly sierra]

why are you ranting

[waxen hawk]

I do not

[waxen hawk]

Someone did that before

[thin vale]

close this or ask a real question

[waxen hawk]

I wonder if it is possible for AI to solve equations on a picture

Like I send it a picture

And it returns with the result

[thin vale]

Does this sound like a math question to you?

[waxen hawk]

Yes

[drifting jackal]

That's not a mathematical question

[waxen hawk]

But it is math relevant

[thin vale]

not the point of the help channels

[drifting jackal]

No it's not, how is AI solving equations math relevant?

[waxen hawk]

I should have asked it in discussion

Im sorry

.close

[vocal sleet]

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[sly sierra]

something like this? https://mathpix.com/handwriting-recognition

[vocal sleet]

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[waxen hawk]

What is the p

[simple carbon]

The percentile

[sly sierra]

probability of success on any given trial

[waxen hawk]

Is there a term for it

[simple carbon]

Looks like 3/4

[waxen hawk]

That’s a number

[simple carbon]

Sorry I must be confused

[simple carbon]

So what do you mean

[vocal sleet]

@waxen hawk Has your question been resolved?

[vocal sleet]

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[bleak whale]

the solution is correct but i cannot comprehend why x in the x(x+2 ) came from

[drifting jackal]

The common denominator

You multiplied by x(x + 2) on both sides

To get rid of the denominators on the left hand side

[bleak whale]

oh, I see

thank you

.close

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[reef isle]

Does the laplace transform of $e^{(t-1)^2}$ exist? No because it is not exponentionally bounded right?

[twin meteor]

Sonite

[vocal sleet]

Please don't occupy multiple help channels.

@reef isle Has your question been resolved?

[vocal sleet]

@reef isle Has your question been resolved?

[warm path]

I think it exists. but not in a nice form

[reef isle]

Whats a good way to prove if it does or does not? without some ugly evaluation

[warm path]

I think from what you have said earlier is one of the sufficient condition

but it is not necessary

you kinda need to show that the integral (to find laplace transform) converge

[reef isle]

the integral of e^(t-1)^2 dt converges? sorry which integral

[warm path]

$$\int_0^\infty e^{(t-1)^2-\sigma t - i\omega t} ; dt$$ for $s = \sigma + i\omega$

[twin meteor]

nonstationary nickname

[reef isle]

so provided that converges, the lapalce should exist? just seems like an ugly evaluation for the actual lapalce right? This was like a one point homework question lol

[warm path]

yes

I think this one exist in complex plane

[vocal sleet]

@reef isle Has your question been resolved?

[vocal sleet]

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[fossil jewel]

question: how are x and y-intercepts related to the slope of a line

[steel marsh]

wha?

wdym

like x and y intercept determines the slope of that line ..

[fossil jewel]

like how

like how are they related

is there a way to get slope using x and y intercepts

idk

[steel marsh]

there is $m = \frac{x1-x2}{y1-y2}$

[twin meteor]

≅ Σiλωυετ

[fossil jewel]

the formula?

[steel marsh]

since u have aleady have the 2 intercepts

[steel marsh]

ye

m is the gradient

y= mx + c

[fossil jewel]

yeah i see what you mean

cause two points

ok thank you

[steel marsh]

np ^^

<3

[fossil jewel]

.close

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[acoustic spindle]

our teacher gave us a challenge im so confused on how to answer it tho anyone help?

[acoustic spindle]

<@&286206848099549185>

[sly sierra]

!status

[vocal sleet]

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

[sly sierra]

also:

!15min

[vocal sleet]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[acoustic spindle]

oh crap sorry

[sly sierra]

nw, just making you aware

[acoustic spindle]

I don't know where to begin with

[sly sierra]

do you know in general how to prove that a subset of a vector space is a subspace?

[acoustic spindle]

i do not

[sly sierra]

you have to show three things:

let's call the subset S

• S is nonempty

[acoustic spindle]

yea

[sly sierra]

• S is closed under addition

• S is closed under scalar multiplication

[acoustic spindle]

okay

[sly sierra]

so the nonempty part first, that's easy

can you name an element of S?

[acoustic spindle]

wdym

?

[sly sierra]

give an example of an element of S

to show that one exists

[cobalt crypt]

V

[sly sierra]

V

[acoustic spindle]

we have two elements right

V and and w

[sly sierra]

what is w?

[acoustic spindle]

accident

V

right

[sly sierra]

what two elements are you referring to?

[acoustic spindle]

sorry i mean just V

im tweakng lol

[sly sierra]

yea so you have a set V

you want to show that it's a subspace of F^infty

you have to first show that V is nonempty

so can you find an element of F^infty that is in V?

[vocal sleet]

@acoustic spindle Has your question been resolved?

[acoustic spindle]

uhh

[cedar rune]

In practice the easiest thing is to prove (0) is in the subset

Since, if it's a vector space, it has to be in it

[vocal sleet]

@acoustic spindle Has your question been resolved?

[acoustic spindle]

right ok

[vocal sleet]

@acoustic spindle Has your question been resolved?

[acoustic spindle]

so v is an element

and an element of f^infty is uhh

[cedar rune]

?

Let $(u_n)$ be the constant sequence equal to $0$. Can you show that $(u_n) \in V$ ?

[twin meteor]

louise

[vocal sleet]

@acoustic spindle Has your question been resolved?

[acoustic spindle]

uhh

okay

[cedar rune]

Just a matter of understanding how V is defined

[vocal sleet]

@acoustic spindle Has your question been resolved?

[vocal sleet]

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[hidden kelp]

I geniuenly have no clue how to get started. I tried a couple things but nothing‘s working. $\int_{0}^{\frac\pi 2} \ln\left(x^2 + \ln^2(\cos x)\right)\dd x$

[twin meteor]

𝔧𝔪𝓣𝛾𝜑𝜽

[gaunt sparrow]

It's giving contour?

[hidden kelp]

Oh bruh I haven‘t tried that :despair:

[hidden kelp]

Wait how would I make a contour on (0, pi/2)? Usually I work with integrals over the real line or (0, inf)

[vocal sleet]

@hidden kelp Has your question been resolved?

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@hidden kelp Has your question been resolved?

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[sharp comet]

so I tried to use secants theorem

and got 4

[half imp]

!show

[vocal sleet]

Show your work, and if possible, explain where you are stuck.

[sharp comet]

ok

[sharp comet]

bruh

why do I keep using blue

for a black board

[half imp]

Looks right

[sharp comet]

ok

idk why calculator said 7.5

[half imp]

What calculator?

[kind light]

huh

[sharp comet]

https://www.omnicalculator.com/math/circle-theorems

[kind light]

doesent the formula states A×B=C×D

[half imp]

7.5 is what you'd get if you misremembered secant theorem as 6*10 = 8*x

[sharp comet]

I checked my answer on this calculator

maybe I iddn't put numbers in right

[steel snow]

Isn’t this for intersecting chords

[half imp]

Which is what we have here

[sharp comet]

I selected intersecting secants theoerem

[half imp]

The chords just intersect outside of the circle

[half imp]

Probably you interpreted the lines the wrong way

[half imp]

Use intersecting secants theorem

[steel snow]

What is this formula then, cuz I swear it exists

[half imp]

Yeah, PC isn't 10

[sharp comet]

a(a+b)=c(c+d)

[kind light]

wtf i swear i was teached A×B=C×D

[sharp comet]

?

[steel snow]

But what is it for if it’s not for this

[sharp comet]

what I thought it said BC

[half imp]

Look at the diagram and compare it to your example

[sharp comet]

that is secants theorem

[sharp comet]

I trie

d

[half imp]

Well seems like you've figured it out

[sharp comet]

this calculator

[half imp]

^

[sharp comet]

idk if it isn't right

or

[half imp]

PC is 16, not 10

[sharp comet]

ye

lemme try that

[sharp comet]

like this

?

[half imp]

Yes

[sharp comet]

I got PD length 12

so

PD - PA

[half imp]

right

[sharp comet]

ok

thanks

[half imp]

np

[sharp comet]

kinda got confused when checking my answer with calculator

.close

[vocal sleet]

Channel closed

Closed by @sharp comet

Use .reopen if this was a mistake.

[sharp comet]

.reopen

[vocal sleet]

✅

[sharp comet]

lemme get my work

nvm

I lost it

and I don't wanna do the work rn

so I'ma just see if I get it wrong or not

.close

[vocal sleet]

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[vocal sleet]

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[last moss]

Hey guys I have this fourth-degree polynomial which is x^4-x^3-x^2-x-2=0. I know the roots are x=-1,2,i,-i but I don't know how you factor this polynomial into (x+1)(x-2)(x^2+1)=0. Any help would be appreciated :)

[manic atlas]

It is a theorem

fundamental theorem of algebra

[oak magnet]

If a polynomial P(x) has a root named R then it can be factor like (x-R)Q(x) where deg(Q(x)) < deg(P(x))

[manic atlas]

a n-degree polynomial has n roots

And it can be expressed as:

a(x-x1)(x-x2)…(x-xn)

So in this question

(x-(-1))(x-2)(x-i)(x+i)

[last moss]

Yes but how do you go from the original polynomial of x^4-x^3-x^2-x-2=0 to the factored form of (x+1)(x-2)(x^2+1)=0.

[manic atlas]

You dont need to from original

you dont even need to know the original polynomial

you just

know it is 4 degree

so it has 4 roots

and you know the coe of the highest x is 1

if you have a poly:6x^4+……

[last moss]

Oh ok thank you for your help

[manic atlas]

If you already have 4 roots,You dont need to know what is in ……

You can write in: 6(x-x1)…(x-x4)

[last moss]

Ok thank you

.close

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[steep berry]

What would my initial process be? I took the gradient of F to get <1,1> and then I took the dot product with <-y,x>. I took it with <-y,x> as a general solution that I saw in the "help me with this" part of the last problem, but this was not helpful.

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[steep berry]

this one

[steep berry]

<@&286206848099549185>

[gaunt sparrow]

(-y,x) is perpendicular to the vector field everywhere.
It's useful because you can use it to see where the actual vector field is tangent to C (since we can only test for orthogonality using the dot product).

So you can do it like that. Where would (-y, x) be orthogonal to C?

Or if we go a step before, what would the dot product between (-y,x) and some point on C be?

[vocal sleet]

@steep berry Has your question been resolved?

[bitter pilot]

Also I think gradient of a vector field f : R² -> R² is not vector but a 2 by 2 matrix

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[vast shale]

!help

[vocal sleet]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[tawny nacelle]

do you have a question?

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@vast shale Has your question been resolved?

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[desert edge]

Did my prof make. A mistake here or am I lost??

[desert edge]

I’m confused why C11 wouldn’t be a11

C is cofactor here and M is minor pretty sure

[gaunt sparrow]

Does it not say a22?

[desert edge]

Yeah

I’m totally lost on why that is

The a are positions in the matrix okay, I get that, I know how to find minor and then cofactor too

[gaunt sparrow]

Well the minor is the determinant of the matrix you're left with once you delete the rows and columns that contain a11.

[desert edge]

Ohhhh wait

[gaunt sparrow]

So you're just left with a22

[desert edge]

Okay since it’s 2x2

There’s nothing but one value left

In the first case a22

I see

[gaunt sparrow]

Yeah, the determinant of a 1x1 matrix is just the value itself

[desert edge]

Yeah okay that’s what got me, I just didn’t get how it was just one value

[gaunt sparrow]

Although with this definition I feel like c22 would be a11. Just checking my definitions a minute

[desert edge]

Yeah that’s the thing, I was just confused even more that two of them could be the same

Unless prof made mistake here

[gaunt sparrow]

I think C22 would be a mistake then

[desert edge]

So just to be clear since we’re using equals signs here etc

When we say C

It’s just awkward to me that we’re saying C11 but then it’s equal to a22

But I get that

Basically first we have to find the minor

Being the determinant of what’s left

So what were really saying is that for. a11

The matrix left is only a22

So it’s determinant is itself a22

And then C is -1 to the power of ij times M

and ij here is 1, 1

So -1^2

Is positive 1

So 1 times M

Which is just a22 itself

[gaunt sparrow]

Yes

[desert edge]

Right.

OK OK

Just awkward to me how it begins with C11 cause my eyes naturally go to a11

I am confused by this though if you can help

Multiplying the a by the c??

Ahh nvm okay

Thank you

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[stuck girder]

Quick question can someone help me solve this

9x - 7i > 3 (3x - 7u)

i is not an imaginary number btw

[stuck girder]

I’m solving for i

[white tartan]

sure, so do you have any idea how to start?

[stuck girder]

uhh

Distribute

And I’m kinda stuck

[white tartan]

yeah you could start by distributing

try that first, and then we can work from there

[stuck girder]

OK

9x - 7i > 9x - 21u

[crimson inlet]

You use sin, cos, and tan

[white tartan]

check your negative/positive signs

[crimson inlet]

thats easy. You distribute 33x and 3-7u

[stuck girder]

I fixed it

[stuck girder]

What’s the answer

[white tartan]

!noans

[vocal sleet]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[stuck girder]

Sorry

[white tartan]

so, now you have 9x - 7i = 9x - 21u

[crimson inlet]

@stuck girder thats the anwser

[white tartan]

!noans

[vocal sleet]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[stuck girder]

No homo

[crimson inlet]

oh sorry

[crimson inlet]

alr srry dude u asked me

[stuck girder]

no worries

[crimson inlet]

for the anwser

[crimson inlet]

I will teach u

[stuck girder]

no need

[crimson inlet]

I can teach just ask me

[stuck girder]

ok

[crimson inlet]

ok so..?

[stuck girder]

Uhh i need help with inequalities

[crimson inlet]

yes ofc

Inequalities are the mathematical expressions in which both sides are not equal. In inequality, unlike in equations, we compare two values. The equal sign in between is replaced by less than (or less than or equal to), greater than (or greater than or equal to), or not equal to sign.

Rules. Inequalities follow many of the same rules as normal equations: Adding or subtracting the same quantity from both sides leaves the inequality symbol unchanged. Multiplying or dividing by a positive number on both sides leaves the inequality symbol unchanged.

[stuck girder]

yes

I don’t get it

[crimson inlet]

give a question

and we solve

[stuck girder]

Ok

[crimson inlet]

First distribute -5 to u and -19

[stuck girder]

Ok

-5u - 95 < -6 + 2u

[crimson inlet]

remember the line under the inequalty means its less then or equal to

[stuck girder]

Ok

[crimson inlet]

but now

[stuck girder]

I don’t know how to make the line

[crimson inlet]

Oh I see

When theres a line under the inequality u shade in the dot

since it can be equal to that number

[stuck girder]

What dot?

[crimson inlet]

you have to make the line right

want me to send picture

[eager nexus]

it's supposed to be +95

[crimson inlet]

yes

@stuck girder

[stuck girder]

Ok

[stuck girder]

Oh

-5u + 95 < -6 + 2u

I’m sorry I have to go now

Thank you for teaching me

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@stuck girder Has your question been resolved?

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[wheat harbor]

My current solution process is wrong, and the right one is
476 = (15 - 2W)(H)(25 - 2L) and I don't understand why there's a multiplication of 2

I understand that my current solution provides the area of two of the squares for one of the equations (15 -2W) or (25 - 2L) but I don't understand how multiplying L and W by 2 give me the solution

[wary granite]

assume the side of the square you're cutting is x, using that you can make an equation that only has a variable of x instead of whl

[wheat harbor]

so 476 = (15 - x)(25 - x)(H)?

[wary granite]

pretty close, but you're cutting 2 squares from each side, so for example k in the picture will be 15-2x and so on

[wheat harbor]

i'm still having a little trouble understanding

[wary granite]

ok let's break it down a bit more

you have a 25 x 15 sheet, and you're cutting 4 squares with the same side length from each corner to make a box

[wheat harbor]

yes

[wary granite]

you don't know the side length of the square yet

so let's call that x

[wheat harbor]

yes

if i take away k from 15 I get 2 times the square

so 2 of x/the side length

[wary granite]

that is correct

let's write what you just said in math

15 - k = 2x

[wheat harbor]

oh i see what i've done wrong

[wary granite]

now change that a bit so k is on one side of the equation

[wheat harbor]

i was looking for the area of the square instead of the area of k

15 - k - 2x = 0

15 - 2x = k

[wary granite]

yep

now use the same logic to calculate x and j

[wheat harbor]

yeah i was assuming too quickly ahaha im going to get to solving x now and ill let you know what i come to

thanks for helping me solve this

[wary granite]

cheers, good luck

[wheat harbor]

i spent a little more time thinking

the height could just be x right

since its being folded up I think that one side length would be the height and 2x is in the equation

[wary granite]

that's correct

[wary granite]

the height in the equation can just be x, 2x would calculate two times the volume haha

[wheat harbor]

yeah

[vocal sleet]

@wheat harbor Has your question been resolved?

[wheat harbor]

I got 4cm as the area of the square

the reason it took me so long is because I was looking for ways other than factor theorem to solve it

but the assignment's following the theme of long polynomial division

[wary granite]

you got it

[wary granite]

yeah usually a pain to solve cubic ones without guessing numbers lmao

[wheat harbor]

splitting the variables has never been fun

thank you for helping, have a good night

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[finite hatch]

what points do I need to graph this

i have dis rn

[halcyon ice]

wdym?

[finite hatch]

likeee

when i draw it

is it just asymptotes and one or 2 points?

[halcyon ice]

yeah

[river kettle]

Bum chicken

[finite hatch]

.close

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[hybrid gull]

Need help

[white whale]

!status

[vocal sleet]

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

[hybrid gull]

4

[gentle thicket]

What did you get?

[hybrid gull]

Number 3

[gentle thicket]

Yeah, it is indeed correct

she missed the multiplicity for that 0 root

[hybrid gull]

Thank you

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@hybrid gull Has your question been resolved?

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[hybrid gull]

.reopen

[vocal sleet]

✅

[hybrid gull]

Need help with this

I got 2x

[vocal sleet]

@hybrid gull Has your question been resolved?

[vocal sleet]

@hybrid gull Has your question been resolved?

[vocal sleet]

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[young falcon]

ohh thats how the channel works

[young falcon]

video instructor didn't explain how to find the ml of the desired solution

.close

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[finite hatch]

how to find common ratio

for part a

um..

hi…

[sharp walrus]

double

that means 2

[finite hatch]

its not

it doubles in 8 days

[sharp walrus]

oh then root_8(2)

2^(1/8) (positive real)

[finite hatch]

why?

[sharp walrus]

because it should satisfy that r^8 = 2

[finite hatch]

oh

can u pls explain that a lil bit more

i get it a bit

[gentle thicket]

why common ratio?

It does not grow in the form of a Geometric progression

It is exponential growth

[finite hatch]

it worked

[finite hatch]

how would i do exponential growth tho

[gentle thicket]

what you did is exponential growth formulation

[finite hatch]

what

i used geometric sequence!

[sharp walrus]

isn't exponential function sampled at every integer just a geometric sequence?

What's the difference?

[finite hatch]

If it says in the 10th year

It means n-1

And if it says in 10 years

It means just n?

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[lost trail]

the problem being that the least value of the quadratic is negative

i expected it to give me one single value which has to be positive

either am making a calculation mistake or i chose the vectors wrong?

kindly ping me while answering, thanks!

[loud walrus]

The closest distance occurs when the angle is 45 degrees, and because both have the same speed that happens when they travelled 5 kilometers

There you can draw a triangle with both sides 5, and find the other side

@lost trail

[lost trail]

i mean yeah that's true

but i want to know the flaw in this method

A(t) and B(t) here represent the distance travelled by them in t seconds

<@&286206848099549185>

[loud walrus]

Where did you get 320t^2 from?

[lost trail]

oh shit

im a dumbass

nvm it's just a calculation mistake

[loud walrus]

Happens to everyone

[lost trail]

yeaaaa, thanks anyways

.close

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[random ice]

How to ans 1.)?

[outer warren]

a graph by itself is not a question

[gentle thicket]

What's the problem?

[random ice]

find the domain and range

mb

[gentle thicket]

are you aware about the definition of domain and range?

[random ice]

yea

Im not really sure how it works if its two graphs

[outer warren]

the same definitions apply

[gentle thicket]

or you may apply separately on both curves and then take the union of the final answer

[random ice]

So it'll be

Domain and Range for the first graph :

Domain : {-3}
Range : [2, -inf)

Domain and Range for the second graph :

Domain : {2}
Range : [2. inf)

?

[gentle thicket]

yeah
so now take union of both cases for complete answer

[random ice]

Domain : {-3, 2}
Range : [2, +- inf)?

[gentle thicket]

yeah

but it is better to write negative infinity to positive infinity for range

[random ice]

I see

For graphs like this

How do you determine whether ones just going to go infinity ( I know these graphs are all going infinity )

But like for the Domains and Ranges, I'm very confused

Like for 14.)

Domain is probably (-inf , inf)
Range is (-inf , 4]

Just a guess

[civic otter]

Yep it's correct

At least for the part of the graph we can see

[random ice]

I see

For 15.)

Domain : [0, inf)
Range : (-inf, inf)

16.)
domain : （-inf, inf)
range : (-inf, inf)

[gentle thicket]

Mistake here

[random ice]

Where?

[gentle thicket]

in 15th

[random ice]

yea

Domain or Range?

[civic otter]

Both

[random ice]

ohj shit

[civic otter]

Can you give the definition of domain you've been taught?

[random ice]

Domain : Input, set of all x-values
Range : Output, set of all y-values

[civic otter]

Nice, so why did you exclude negative values of x in the domain (input)?

[random ice]

I have no idea, I saw the 0

And I just assumed

Though for correction

Its (-inf, inf) I think

[civic otter]

Yep

[random ice]

Though for the range

I have freaking no idea

[civic otter]

Yeah, would y = -5 be reached?

[random ice]

reached?

[civic otter]

Yes, reached

I mean, does your function give -5 as output for some values of x?

Or, in other words, does the horizontal line y = -5 intersect the graph?

[random ice]

yea

it does

[civic otter]

Perfect, that means -5 must be included in the range

Now, is y = -6 reached by the function?

[random ice]

Yes

[civic otter]

And y = -7?

[random ice]

No

[civic otter]

Therefore...

The range is

[random ice]

[-6, inf)

[civic otter]

You nailed it

[random ice]

So always lowest point right

I'm still confused a bit

[civic otter]

Wdym?

[random ice]

Range : lowest to highest point

Same for Domain

[civic otter]

Well, of course 😅

Numbers are in increasing order

[random ice]

ahh, yeah, that makes sense

[civic otter]

If you watch the directions of the x and y axes you'll see why

Unfortunately here they're not shown

[random ice]

I see

[random ice]

For 2.)

Domain : (-5, 5]
Range : (-2, 2]

I think?

[vocal sleet]

@random ice Has your question been resolved?

[unkempt owl]

Range is the lowest point to the highest and so is domain but range refers to the Y axis while domain refers to the X axis.

[unkempt owl]

That’s really close. In a range where there’s a closed and open point with the same y value but different x values, there would still be brackets because if the open point means every value on the graphed segment other than that point and the closed point just means every value on the graphed segment, then as long as there’s a closed point at the lowest y value then there will be brackets. Not parentheses

If that’s hard to understand, here’s a visual reference

[random ice]

because if the open point means every value on the graphed segment other than that point and the closed point just means every value on the graphed segment What does this line mean? I read this for three times already, I still didn't get it

[unkempt owl]

Wait wrong one

There

See, the red represents the values included in domain and range on the graph

Even though the red doesn’t overlap the open circle

It does overlap the closed one

Therefore the lowest value would be -5

Rather than being the point before it

[unkempt owl]

I’m not great at explaining without visuals. Sorry

[random ice]

This is domain right?

[unkempt owl]

No the range is the y axis.

[random ice]

Oh this graph

I thought you meant my graph I sent, I was confused

[unkempt owl]

Ohhhhhhh haha

Sorry ;-;

[random ice]

No thats mb

[unkempt owl]

Does this help tho?

[random ice]

oh then for the graph I sent it would be [-2, 2] then? wait is it?

I get what you mean

[unkempt owl]

Yes!!

That’s the idea

There would be 2 brackets

[unkempt owl]

You’re getting it!

I’m happy to help if there’s others too

[random ice]

What about for x's?

Does the same logic apply also?

[unkempt owl]

Yup

Hang on lemme show you

[random ice]

alrighty

[unkempt owl]

This one. What you think the domain would be?

[random ice]

(-1, 6]?

[unkempt owl]

Ok so if the same logic applies, it wouldn’t be the parentheses, remember? Think of it like this:

The closed and open circles are playing tug of war, where the rope is the line of the graph connecting the two. The closed circle will win because there is more space being filled by it. Just like how the heavier person will win in a game of tug of war where both people have equal strength.

That’s how I picture it.

[unkempt owl]

Got it?

[random ice]

That makes a lot of sense

[unkempt owl]

That’s a relief

I have autism and I think of things differently than the average person lol

I was expecting you to have no idea what I meant

[random ice]

haha, I have really bad logic, but I can understand weird analogies

[unkempt owl]

That’s perfect hehe

A kindred spirit!

Feel free to dm me any time you need help!

[random ice]

Alrighty

.close

[vocal sleet]

Channel closed

Closed by @random ice

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[vocal sleet]

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[fallen sand]

pls help

why in the question they did < 1/6!(.1)^6

and not <=

[fallen sand]

this is the question

.close

[vocal sleet]

Channel closed

Closed by @fallen sand

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[vocal sleet]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[serene mesa]

I hate mafs

[serene talon]

nobody asked

[serene mesa]

Shut up Rudoph I'm gonna touch u

😭

[serene talon]

u dont even know where i live

[serene mesa]

I do

125 Timothy lane

How are we supposed to answer that geometry question without answering it

[tidal dock]

!status

[vocal sleet]

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

[tidal dock]

hey, no touching ❌❌❌❌❌

[serene mesa]

I'm sorry he or she was being a meanie

I got ahead of myself

[worldly mulch]

probably 1?

[tidal dock]

what do these symbols mean if anything

[worldly mulch]

just some chinese words

[serene mesa]

Fr

[worldly mulch]

if , and , find

but in chinese

[tidal dock]

ah

[serene mesa]

You dummy

Losa

That one really got me

[worldly mulch]

can I know what property is this?

[serene mesa]

Wait so I'm a friend now?

Wow

Add me lets be frens

I have no friends

Meanie

I'm 4 years of age

You mean unless you're Asian?

[worldly mulch]

why am I asian but dumb asf

[serene mesa]

I'm 40 percent Asian and 60 percent whitey

[worldly mulch]

I still can't get that ratio...

[serene mesa]

What grade y'all going too

This is geometry right??

[worldly mulch]

we call it secondary 6 here

[serene mesa]

I'm in no shape to answer these questions cuz I lowkey almost failed geo

Got a 5 on the eoc because of the question curve and It saved my grade

[worldly mulch]

my grades in calculus class is better than my grades in maths class when the math is geometry

[serene mesa]

Nerd

[warm path]

<@&268886789983436800> maybe this is out of hand

[serene mesa]

Hi nick

[worldly mulch]

what about 2:3?

[serene mesa]

4

😔

Who's that guy

Stop whipping out YouTube shorts

[worldly mulch]

which one is the base?

[serene mesa]

The base is the one that is the base

[urban edge]

<@&268886789983436800> can we get this channel in check please

what

im not a mod

[vocal sleet]

@worldly mulch Has your question been resolved?

[worldly mulch]

<@&286206848099549185> can someone help me.

I want the steps to find the ans

[vocal sleet]

@worldly mulch Has your question been resolved?

[vocal sleet]

Channel closed

Closed due to timeout

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[vast shale]

help

[vast shale]

I just did a real analysis midterm and there was a question that was like f is cts over [a,b] except at some point c in (a,b) show that f is integrable over [a,b]

we're allowed to use "if f is cts over a closed interval f is integrable"

i wrote something like let epsilon >0 consider the bounds [a, c-epsilon], [c+epsilon, b]

and showed it was piecewise integrable and thus integrable over the whole thing using partitions

is this right?

[sly sierra]

well it's integrable on [a, c-epsilon] and [c+epsilon, b] for any eps>0

so choose an epsilon such that any rectangle of width < epsilon that contains c, has only a small contribution

(less than some arbitrary fixed delta)

[vast shale]

i used riemann integrablility condition and for [a, c-epsilon], [c+epsilon, b] made the upper and lower less than epsilon/2 for some partition 1 and partition 2 respectively

then took p = p1 union p2 as a refinement

[sly sierra]

yea, and that introduces one new rectangle, namely the one with base [c-epsilon, c+epsilon]

[vast shale]

then since one partition exists for all epsilon as epsilon goes to 0 the integral approaches the "true integral"

okay thank you

.close

[vocal sleet]

Channel closed

Closed by @golden delta

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[sly sierra]

btw, did they also have the hypothesis that f is bounded

[vast shale]

LOL the prof forgot

[sly sierra]

because if not then the claim is false

oh haha ok

[vast shale]

he said that like assume it's bounded when someone asked if he can just write can't riemann integrate cuz not bounded

[sly sierra]

good good

[shy wagon]

Did the question ask if it's integrable or riemann integrable?

[sly sierra]

so yea just use the boundedness of f to bound the height of your new rectangle

[vast shale]

riemann we haven't introduced other integrals like lebesgue yet

.close

[vocal sleet]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[bleak prawn]

The dot product is cos(theta) = a • b/IIaII • IIbII