help-42

[old falcon]

some oscillate between a collection of values

[wild marten]

Ok so how would i know which ones work or which ones will not

[old falcon]

sometimes they exhibit chaotic fluctuations

[wild marten]

is it just luck of finding a close enough interval

[old falcon]

i mean its not luck

you can find where |g'| < 1 pretty easily

then, apply something as brainless as bisection to get an interval

or something like banach to argue one exists or something insane

[wild marten]

hmm ok

I think that makes sence

[old falcon]

but no, iteration methods can be absolutely insane

[wild marten]

so just a level of trial and error

[old falcon]

its experimental, for sure

maybe youve seen plots like this

[wild marten]

yeah

[old falcon]

not quite fixed point iteration directly

but they can be very screwy

[wild marten]

Yeah i see that

numerical analysis ftw lmfao

[old falcon]

its fun

difference equations are fun lol

and they dont get enough love

for how wacky they are

anyways, hopefully that answers ur question

[wild marten]

Yeah it helps lol

I think im starting to get it

ill shoot you a dm if anything else comes up

if thats ok lol

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[rain sequoia]

Can anyone help me with sig fig? I know this isn’t chemistry but please help if you can

[rain sequoia]

Number 3 at the bottom

[calm coral]

@rain sequoia Has your question been resolved?

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[upper sparrow]

There are multiple ways you could proceed, one step you may want to do is to get all the e^{...} terms together, maybe...

Well, I meant more by dividing so that you have a ratio of the things with e, and then using rules of indices

(it probably should come as no surprise that you'll need logs at some point, so you could instead take natural logs straight away and instead use log rules...)

Well-

You wanna use those log rules, on the original expression you have, and noting that e and the natural log cancel each other out

I'm saying that you can take logs of this here

Taking logs of both sides, and use those log rules on both sides, simplifying both $\ln(3e^{5x + 9})$ and $\ln(e^{4x})$

[potent lotus]

@upper sparrow

[upper sparrow]

Would you be able to work with something like $e^{y} = b$, solving for $y$?

[potent lotus]

@upper sparrow

[upper sparrow]

Do you also know how to simplify an expression like $\frac{a^p}{a^q}$?

[potent lotus]

@upper sparrow

[upper sparrow]

Yes and yes

Let's restart and go back to $6e^{4x} = 18e^{5x + 9}$, can you use those to solve it instead?

[potent lotus]

@upper sparrow

[upper sparrow]

You could do that, if you wanted

Put maths stuff in dollar signs, $a^{bc}$

[potent lotus]

@upper sparrow

[upper sparrow]

curly brackes {} for powers

$a^{bc}$ against, for example, $a^bc$

[potent lotus]

@upper sparrow

ccgul

[upper sparrow]

Well, I mean, you could take logs, do you know what $\ln(e^{4x})$ becomes?

[potent lotus]

@upper sparrow

[upper sparrow]

(that wasn't what I was hinting at, but anyway)

Yep, that's good

Now, do you know how you could simplify that $\ln(3e^{5x + 9})$?

[potent lotus]

@upper sparrow

[upper sparrow]

Sure

There you go, yep can you simplify that a tiny bit more?

[potent lotus]

ccgul

[upper sparrow]

I do and don't calculate anything yet, but yep, ln(e^{4x}) becomes 4x

Well wait one second

[upper sparrow]

Remember that it was $e^{5x + 9}$, but you basically have the right idea, $\ln(3e^{5x + 9}) = \ln(3) + \ln(e^{5x + 9}) = \ln(3) + 5x + 9$

[potent lotus]

@upper sparrow

[upper sparrow]

Which now gets us to the equation that $\ln(3) + 5x + 9 = 4x$, which hopefully you can solve for $x$ now?

[potent lotus]

@upper sparrow

[upper sparrow]

(the reason not to, is so you don't lose accuracy until you get x = ..., rounding errors can be annoying sometimes so better to keep things exact until the very last moment!)

[potent lotus]

ccgul

[upper sparrow]

you're pretty much there

[calm coral]

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[kindred timber]

The graphic of which of the following functions is shown in there?
I completely forgot how to do such tasks please explain me if its possible.

[lost quest]

where is y=0?

[kindred timber]

at the O on the painting

[lost quest]

no like at what value of x is y=0?

[lost quest]

what value of x would be at O

[kindred timber]

i dont know, the question is
"The graphic of which of the following functions is shown in this painting?"

as i said i completely forgot how to do these

[inner zealot]

The options given to you are all 2nd degree equations. That means that there will be two linear factors, x - a and x - b. a and b will be the points at which it crosses the x-axis, if it does so.

[kindred timber]

that`s it ?

[inner zealot]

For this type of problem, yes.

[kindred timber]

thanks

[inner zealot]

You could also use the Quadratic Formula.

[kindred timber]

thanks for the help

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[rough arrow]

Factorize $$z^6-4z^3+4$$

[potent lotus]

Good

[rough arrow]

It could be factorized to $(z^3-2)^2$, but do I find the complex solutions to $z^3-2$?

[potent lotus]

Good

[tribal jetty]

Depends if you’re factoring it or finding the solutions to that equal to something (maybe 0)

[rough arrow]

The question asks "Find the complex and real factorizations of the polynomial"

[tribal jetty]

Then I would recommend finding the complex solutions and then the factorisation would be (z-a)^2(z-b)^2(z-c)^2

[rough arrow]

The answer is following, I got $\sqrt[3]{2}$ in the expressions of $z^3-2$ roots.

[potent lotus]

Good

[tribal jetty]

At least this would be the complex factorisation

[tribal jetty]

Yeah this would be the real factorisation

[rough arrow]

But I can't get the roots

$w_0=\frac{\sqrt[3]{2}\sqrt{3}}{2}+i\frac{\sqrt[3]{2}}{2}$

[potent lotus]

Good

[rough arrow]

Nevermind, I saw my mistake.

I got w1 and w2, but the imaginary parts have opposite signs.

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[drifting seal]

do you know the formula for elasticity of demand

[hallow rock]

yes

I got up to -6p/200-3p^2

wondering where I went wrong

$ -6p/(200-3p^2) $

[drifting seal]

$-6p/(200-3p^2)$

[potent lotus]

knief

[hallow rock]

yeah

[drifting seal]

so what did you get for p/q

[hallow rock]

p/200(-3p^2)

[drifting seal]

you know p=4

so what is that

quotient

[drifting seal]

^

[hallow rock]

plug it in?

[drifting seal]

yea

what did you get

[hallow rock]

24/25

[drifting seal]

$\frac{4}{200-48} = \frac{1}{38}$

[potent lotus]

knief

[drifting seal]

,w calc 4/152

[potent lotus]

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Requested by c.knief

[hallow rock]

nope

says its wrong

[drifting seal]

..

because that isn’t the answer

it’s $\frac{dq}{dp} \cdot \frac{p}{q}$

[potent lotus]

knief

[drifting seal]

we didn’t even find dq/dp

[hallow rock]

-6

[drifting seal]

so -6p = -24

not -6

$\frac{dq}{dp} = -6p$

[hallow rock]

can we start over

[potent lotus]

knief

[hallow rock]

1

[drifting seal]

$\varepsilon = \frac{dq}{dp} \cdot \frac{p}{q}$

[potent lotus]

knief

[drifting seal]

yes?

[hallow rock]

yes

[drifting seal]

and p=4

q=200-3p^2

so what is q

if p=4

[hallow rock]

you just said q

[drifting seal]

yes

if p=4

and q = 200-3p^2

[hallow rock]

I have

[drifting seal]

what does q=

[hallow rock]

$(-6)/(dp)*4/(200-3p^{2})$

[potent lotus]

LunaLoves❤

[drifting seal]

no

[hallow rock]

why not

[drifting seal]

dq≠-6

and just no

dq is not a value

it’s dq/dp

q’

and that isn’t -6 either

[hallow rock]

?

[drifting seal]

dq/dp is a derivative

dq is not a number

[hallow rock]

of

[drifting seal]

neither is dp

[hallow rock]

can you write out how to do this

[drifting seal]

i was

you weren’t answering me

[drifting seal]

i started here

let’s evaluate p/q first

[hallow rock]

Sorry man i've been awake since 3

[drifting seal]

no worries

[hallow rock]

its 7 now

[drifting seal]

we already know p = 4 because it’s given

so to find p/q all that’s left is to find q

which is given by 200-3p^2

since p=4 we know q is 200-3(4)^2

[hallow rock]

so its 4/200-3p^2

[drifting seal]

but the denominator is 200-3(4)^2

because p is 4

[hallow rock]

ok put in p

[drifting seal]

yup

[drifting seal]

this is 200-48 which is 152

then p/q is 4/152

[hallow rock]

ok

[drifting seal]

4/152 reduces to 1/38

now we need to find dq/dp

[hallow rock]

🙂

[drifting seal]

so we differentiate

differentiating 200-3p^2 gives -6p

because the derivative of 200 is 0

and we use the power rules

[hallow rock]

wait

so p/q is 1/38

so we're doing dq/dp

[drifting seal]

$-3(2)p^{2-1}$

[potent lotus]

knief

[drifting seal]

yes

[hallow rock]

so we do dq

[drifting seal]

dq/dp

it’s not a fraction

with a dq and dp being individual values

if it’s helpful we can write q’

pronounced q prime

to denote the derivative of q with respect to p

[hallow rock]

so we find the derivative of q

[drifting seal]

yes

[drifting seal]

^

[hallow rock]

its -6

[drifting seal]

how

it’s p^2

not p

[hallow rock]

oh its -6p

[drifting seal]

mhm

so p is 4 though

which means -6p is -24

thus our final result is (-24)(1/38)

[hallow rock]

wait is the bottom 0 because we do p prime

[drifting seal]

or -24/38

[drifting seal]

no

[hallow rock]

why

[drifting seal]

it’s not a fraction

[hallow rock]

then why is it there

[drifting seal]

dq and dp are not individual values

dq/dp is notation

used for the derivative of q with respect to p

it’s one singular value

dq and dp are not individual values

[hallow rock]

so its saying find derivative of q and put in p

[drifting seal]

dq/dp means the derivative of q with respect to p

so yes we evaluate dq/dp at p=4

[hallow rock]

idk what "respect" means in this situation

ok so now what we have -24/38

[drifting seal]

in other words dq/dp is the rate at which q changes as p changes

[hallow rock]

ok i have another problem

[drifting seal]

think of slope

[drifting seal]

i have to go

[hallow rock]

bye

oh yeah i have another one

.close

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[calm coral]

@errant knot Has your question been resolved?

[calm coral]

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[edgy elk]

I have a question about arithmetic sequences

[edgy elk]

Does the common difference stay constant or can it change after some terms? Let's say d is 2 and a_1 is 1

The sequence goes like 1, 3, 5, 7. After 7 the common difference changes to -2 so it goes 7, 5, 3, 1..

Is this possible?

[amber bolt]

it has to stay constant

[drifting seal]

wouldn’t be an arithmetic sequence

[edgy elk]

Ahh, alrighty thanks for the clarification

[quasi seal]

i forgot what kind of sequence that was called

[drifting seal]

melons are tasty

[quasi seal]

but it is technically a type of sequence

[amber bolt]

it's like triangles can;t have 4 sides, but shapes can

[edgy elk]

I'm not 😂

[drifting seal]

but melons are

[quasi seal]

i think it was like "tickster" sequence or something, idk 1s

[edgy elk]

Yep, indeed they are

Welp gotta close this now

Thanks again

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[halcyon ore]

hi

just to get a bit of understanding

VPW is the person

VWS AND VPS

VWS would be what and VPS?

both represent the shore?

oh

he explained it later

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[proven torrent]

It asks me to sketch the derivative

Doesn’t give me a function tho

This is what I put is it right???

[quasi seal]

you should be able to figure out the circle equation from that lol

[proven torrent]

Yeah but there’s other examples too

Where it’s harder to figure

[quasi seal]

like?

[proven torrent]

I don’t feel like wasting my thumbs rn can you just verify😭

[upper sparrow]

Probably should account a little bit for the rate of increase, your f’ should be “more curved”

[proven torrent]

Ohhh what about it being between interval -2,2

Does that like matter

[upper sparrow]

(And there are general examples where you aren’t given a function, and you can’t figure what it is, where you’re meant to sketch the derivative based on the behaviour of the graphed function)

[quasi seal]

like peicewises?

i guess so

then what a squiggly line? xd

[upper sparrow]

Well, you want your plot to be within that interval, but for example-

[quasi seal]

dang alright

[upper sparrow]

At -2, you’re like “infinitely increasing”, like you want the derivative to be undefined there, almost vertical

Then as you get closer and closer to 0, you increase at a smaller rate

[upper sparrow]

This almost implies that you have a quadratic like -x^2 or something as your original

[proven torrent]

Wait like this then?

[upper sparrow]

Yep

[calm coral]

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[icy mulch]

Anyone know why its K/M?

[leaden thunder]

If the balls are replaced, how many black and how many total balls are there

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@icy mulch Has your question been resolved?

[icy mulch]

K^n balls and M^n balls

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[icy mulch]

.reopen

[calm coral]

✅

[icy mulch]

so from that its (K/M)^n, how do u get rid of the n exponent?

[leaden thunder]

No

K black balls and M total balls

[icy mulch]

But why is it the ratio of black balls to total balls

can you explain using counting techniques?

because i was trying to find it using a priori definition of probability

[leaden thunder]

Probability of a black ball among all balls is just the number of black balls divided by number of balls

[icy mulch]

is that fr just it

[leaden thunder]

Yes

[icy mulch]

no need for counting techniques?

jesus

[icy mulch]

I guess its starting to make sense

Thank you!

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[exotic cosmos]

cardinality of set containing 2 elements {2w, 3w-1} where w in Z

[limpid pilot]

what is cardinality

[exotic cosmos]

size

[limpid pilot]

and how do you find that

[calm coral]

@exotic cosmos Has your question been resolved?

[calm coral]

@exotic cosmos Has your question been resolved?

[exotic cosmos]

is it 2 and 1 or infinity?

[exotic cosmos]

<@&286206848099549185>

[limpid pilot]

its just 2

[exotic cosmos]

or 1

or infinity??

[limpid pilot]

2

"set containing 2 elements", cardinality is the number of elements in a set

so 2

[exotic cosmos]

it can contain only 1 element though

[limpid pilot]

how

[exotic cosmos]

x = 1/2

[limpid pilot]

still 2 elements

[exotic cosmos]

sorry wait

w = 1

then you have {2,2}

that’s only one element

[limpid pilot]

yeah in that case 1

not infinity tho

if its a set

it cant even have repeated elements

[exotic cosmos]

that’s what i said

{2,2} = {2}

[limpid pilot]

yeah

so 1 in that case

maybe write a case for that

[exotic cosmos]

someone said answer is infinite

[limpid pilot]

how can it be infinite

the set has max 2 elements

[exotic cosmos]

because {x | x in R} means all x right?

[limpid pilot]

the form u have here is set roster form

[limpid pilot]

this is set builder form

if u said {2w | w in Z} then its cardinality would be infinite yes

but if u say {w} w in Z then its not infinite

[exotic cosmos]

yes but do they not mean the same thing?

[limpid pilot]

no

A = {2w | w in Z} means A is a set of all integers

A = {2w} is just A = {2w} (u can w=1,2,3,4.... but A still has only 1 element depending on the input)

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[rich horizon]

?

[modern peak]

this is optimization

so you differentiate

and set equal to 0

then solve for x and input that into y

[rich horizon]

okay one second

[modern peak]

alr

uh

i can't really

understand

[rich horizon]

y become 0

[modern peak]

no

differentiate first

then set y' = 0

[rich horizon]

so -b/2a

which is 3

[modern peak]

what

no

show me ur working man i can't understand like this

ur answer is correct tho

x = 3

now input that back into y

[rich horizon]

I put 8/3 over -4/9(2) to get the rate of change which is 3

[rich horizon]

Okay Ill try'

you mean x right?

plug into x

[modern peak]

no

plug x = 3 into y

[rich horizon]

13

[modern peak]

yeah

that's your answer

[upper sparrow]

(I think they were using that it's a quadratic and e.g. completing the square?)

[rich horizon]

y=-4/9(3)^2+8/3(3)+9
y=-4+8+9
y=13

[modern peak]

yeah i got that later

[rich horizon]

sorry I was using that one formula

[modern peak]

but at first -b/2a was a little confusing

[rich horizon]

How do you remember to use this formula for this question, I just cant wrap my head around it

[modern peak]

its not a formula

[rich horizon]

am I just finding the y in the vertex or smt

[modern peak]

to find maxima/minima simply differentiate, set equal to 0, and solve for x.

you'll get the x-values for the extrema

[rich horizon]

How would I find minimum?

[modern peak]

same technique

[rich horizon]

what changes in the technique?

oh wait

is the minimum

the 3 I said earlier

[modern peak]

no

that's the maximum

because this is a downward-facing parabola

in R it only has a maximum

try it for some sort of quartic, you'll get 3-or-so values

[rich horizon]

wait so whats 13

[modern peak]

the maximum...

[rich horizon]

I am lost

you just said 3 was the max too

So are they both maxes?

[modern peak]

what???

no?

(3, 13) is the max...

3 is the x-value

13 is the y-value

that's it

[rich horizon]

ah okay

I think i get it

Thank you

have a nice night

.close

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[modern peak]

np

[modern peak]

you too

[calm coral]

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[green cosmos]

How do I start 39? I’m not sure how to begin except to find the derivative

[cobalt mason]

I have a question about related to the typical epsilon delta definition of a limit

[naive escarp]

What would be slope of the horizontal tangent

[fickle musk]

Horizontal tangent is parallel to the x axis (y=0)

[green cosmos]

y is 0

So

0 = 1 + 2cosx

Because that’s the derivative

[naive escarp]

Looks good

[fickle musk]

y=ax+b
y=0x+0

[green cosmos]

And x would be (?)

Not sure how to do that

[naive escarp]

2cosx = - 1
cosx = -1/2

[fickle musk]

Just explaining why the gradient is 0 😅

[green cosmos]

So just find where cos is at its max which is 1?

Actually I don’t know what I’m talking about

[naive escarp]

Just find x values, try known angles

[green cosmos]

To see when it equals 0?

[naive escarp]

To see where given equation is true

[fickle musk]

You should be given a range smh

[green cosmos]

He said between 0, and 2pi is fine since it didn’t specify

Sorry

Should’ve said that

[naive escarp]

They can do
$\alpha + 2k\pi$

[fickle musk]

Ok

[potent lotus]

MetuMortis

[fickle musk]

Yeah

That's the general form

[green cosmos]

The first one is pi

Idk about the second one

Oh

Nvm

[naive escarp]

It is not pi

[naive escarp]

Is everything good till here?

[green cosmos]

Its 4pi/3

[naive escarp]

Thats one answer, there is one another also

[green cosmos]

2pi

Over 3

[fickle musk]

Second quadrant too

,calc 2/3

[green cosmos]

Can’t I just do
$\alpha + 2k\pi$

[potent lotus]

Light

[fickle musk]

Yes

[green cosmos]

Then I wouldn’t need to find the other

[fickle musk]

You will have to find both

One will be alpha the other beta

[potent lotus]

Result:

0.66666666666667

[green cosmos]

1pi/6

And 11pi6

Over

[fickle musk]

Wha-

[naive escarp]

Are you just guessing

[green cosmos]

Yea

[fickle musk]

2π/3 was correct I think

[naive escarp]

,w solve cos(x degrees) = - 0.5

[potent lotus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by metumortis

[green cosmos]

I mean pi over 3

And 5 pi over 3

Those are 1

[fickle musk]

Try it again.

[green cosmos]

How are they not one

2cos (pi/3)

Is one

[fickle musk]

Cosx=-1/2

[green cosmos]

I already covered both then

Because cos is negative at 4pi/3 and 2pi/3

[fickle musk]

Yes

Then just do + 2kπ

For both

Aaaaaaaa

[green cosmos]

Can I ask why did I need to find those two points

I don’t understand how they correlated with me finding the horizontal tangents

[fickle musk]

They both satisfy the solution

[fickle musk]

When the gradient is zero

[green cosmos]

Oh ok

[fickle musk]

[green cosmos]

Very nice thanks

.close

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[west geode]

why is the answer 104??????

im putting it into desmos and im getting -76. since it says determine the positive angle theta against the positive x axis, surely the answer would be 360 MINUS 76 (which is 284)??

can someone draw a diagram of this or something this is baffling me to a ridiculous extent

[pastel edge]

If you add 180 to -76, you get 104

[west geode]

why is it 180?

i have vague notions of "tan has a period of pi" but im not sure what that has to do with this problem at all

[pastel edge]

Pretty much that notion

Adding multiples of 180 or pi to the argument of tan preserves the value of tan

so tan(-76) = tan(104)

[west geode]

?!

[pastel edge]

What's confusing about that lol

You literally just said that

"tan has a period of pi"

[west geode]

why does that change the basis against you weigh an angle?

[upper sparrow]

(also be very careful, that you should be using radians in problems like these)

[west geode]

like regardless of whether or not its tan the x positive axis stays the same

[pastel edge]

See the red/blue line? Against the x-axis, it has angle -76 or 104 depending on how you look at it

[west geode]

if you draw a line its gonna have a particular degree against that axis no matter what

[west geode]

What?

[west geode]

i thought it was being measured against the line y=0

[pastel edge]

It's more a matter of preserving the clockwise orientation I suppose

[pastel edge]

You're going against the positive x-axis with the positive angle theta

[west geode]

why?

what the scallop

but it just says x axis

[pastel edge]

How the question is worded

[west geode]

?!

[pastel edge]

Relax lol

I'm trying

[west geode]

I understand

i just despise trig

[pastel edge]

Those are the angles against the positive x-axis

Those are the angles 104 and -76

[west geode]

Okay

[pastel edge]

But then it specifies that it wants the positive angle

So then it's 104

[west geode]

man this is NOT something we tackled in class tf

[pastel edge]

Basic idea is just that tan(-76) has a negative argument and you want to make it positive by adding 180 to -76

[west geode]

okay...

how would this change if it was sin instead

idk im not good at trig

Im having a really hard time understanding this concept

[pastel edge]

Uh sin is kind of different

I'd like to see a problem with that before I make any claims

But I'd probably just add 360 to any negative angle?

[west geode]

why does tan theta equal the angle of the tangent line against the positive x axis in the first place?

[pastel edge]

tan theta is the slope of the line

The slope here:

and the slope here:

are the same

It's just a simple way to visualize that tan(x) = tan(x + 180)

because flipping around half-way preserves the slope

https://www.desmos.com/calculator/tro9sjnlyu

[west geode]

ohhh

[west geode]

but why is it the slope?

[pastel edge]

sin / cos

rise / run

[west geode]

ahhh

[pastel edge]

(sin is a rise, cos is a run)

[west geode]

That makes sense

thanks!

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[tribal marsh]

how do i graph lim x approach plusminus ifinity. sorry am lost

[late torrent]

its + and -inf

so make x approach inf

and -inf

[tribal marsh]

Like this?

Hmm I guess like this?

[calm coral]

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[queen citrus]

solve each equation:

[queen citrus]

tbh! im new to this stuff so im gonna need a thorough explanation of at least one and i should be able to pick things up

ive figured u should log both sides tho correct me if im wrong

[graceful dust]

seems reasonable

[small shuttle]

yea

[queen citrus]

just stuck on what to do after

[small shuttle]

important rule is $\ln(x^a) = a \ln(x)$

[potent lotus]

b

[calm coral]

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[queen citrus]

.close

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[north copper]

Solve the right triangle ABC, where C =90. Give angles in degrees and minutes. a=17.00 c=21.00

[north copper]

b rounds to 12.33

A is 35 degrees 57 minutes

B is 54 degrees 3 minutes

For b it says round to nearest hundredth

And for both angles A n B it says to round to nearest minute

I don’t understand how I’m getting it wrong?

What am I missing

[calm coral]

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

[kindred estuary]

,rotate

[kindred estuary]

You swapped the two results

You drew that AC is Opp and AB is Hyp and to find angle A, it's using sine

[north copper]

Ohhhhhhh

🤦

Thank you so much

Sorry for taking up your time bro

[kindred estuary]

Did that fix it all? Is it correct now?

[north copper]

Yeah I’m just going to do it again rn and enter it in rn

Wait I’m confused

I did the equations again and I got the same answer

Here’s my re workout

[kindred estuary]

Yeah the results should be the same, you just swapped the results of A and B

A is 54 degrees

[kindred estuary]

Here you said B was 54

[north copper]

Oh

Ok

I see

Alright!

Thank you!

👋🏽

I appreciate you helping me!

[kindred estuary]

Good now?

[north copper]

Yup!!!

Finally!!!

.close

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[crystal vigil]

Is there any dimension at which it becomes impossible to take the Riemann Integral? Meaning, we can handle R^2 very easily, but R^3 already becomes very hard. At what point is it not longer just "hard" but more-or-less impossible (or actually impossible)? Is there an R^n that makes Riemann integration not possible?

Yes, we can then use Lebesgue Integrals, but I'm curious about the limitation with Riemann

[crystal vigil]

For example, I'm not sure what it even means to take a Riemann Integral in R^4 space. Sure, with R^3 we have cuboids, but what do we have in 4d?

[leaden thunder]

Just use Cartesian product of 4 intervals

[calm coral]

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[calm coral]

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[tulip kindle]

i need help asap

[tulip kindle]

please help i dont understand any of it

please someone help

i have a quiz tommrow

and i dont get any of this

[fossil orchid]

Where are u struggling?

[tulip kindle]

all of it

i pointed the cordinates

but i dont know how to find cordinates of p

or what ever partition

for example 1

[sullen dock]

You can use the formula.

a = (mx2+nx1)/(m+n)
b = (my2+ny1)/(m+n)```

[tulip kindle]

can you show me

this not homework

its review for quiz

[fossil orchid]

Use section formula

[sullen dock]

Just put the points from your first question, i have written in that way , you need to find (a,b) . Points A(x1,y1) and B(x2,y2) are given and ration m:n as well

[tulip kindle]

so i pointed (3,4) and (6,10)

now whay

what

[fossil orchid]

P = (m₁x₁ + m₂x₂) / (m₁ + m₂), (m₁y₁ + m₂y₂) / (m₁ + m₂)

[tulip kindle]

.

.

your confusing me

[sullen dock]

Is it m1x1 or m1x2?

I might be wrong

[tulip kindle]

just tell me how to do first question bro and ill do rest

[fossil orchid]

No, u are right. It's just the point of view...which are you assuming

[tulip kindle]

tell me how to do first question and ill do the rest

[fossil orchid]

P(x) = (1(3) + 2(6)) / (1 + 2) = (3 + 12) / 3 = 15 / 3 = 5
P(y) = (1(4) + 2(10)) / (1 + 2) = (4 + 20) / 3 = 24 / 3 = 8. Tell me is it correct. Or is it same with your formula?

@sullen dock

[sullen dock]

Yeah you took 1:2 , I will take 2:1

It's same

[tulip kindle]

:/

[sullen dock]

You can use the formula.

a = (mx2+nx1)/(m+n)
b = (my2+ny1)/(m+n)```
Do you understand this @tulip kindle

[tulip kindle]

is y2 t

is y2 the cordinate

[sullen dock]

(a,b) are the cordinate you need to find

[tulip kindle]

?

im finding p tho

[sullen dock]

A-----------P-----------B
A(x1,y1)
P(a,b)
B(x2,y2)

Do you understand now

[tulip kindle]

idk

[sullen dock]

I have drawn the line as well

[tulip kindle]

A(3,4)?

[sullen dock]

Yes

B(6,10)

[tulip kindle]

b(6,19?

now what do i do with them

[sullen dock]

10 not 19

[tulip kindle]

ye

what do i do with it now

[sullen dock]

m:n = 2:1

[tulip kindle]

so?

[sullen dock]

So put the values of m,n,x2,y2,x1,y1

And calculate

You can use the formula.

a = (m * x2+n * x1)/(m+n)
b = (m * y2+n * y1)/(m+n)```

[tulip kindle]

first cordinate is x1 and then the second x2?

like 4 and 10

4 x1 and 10 x2?

hello?

[sullen dock]

a = ( 2 * 6 + 1 * 3)/(2+1)

[tulip kindle]

thatss it?

(5,8)

[sullen dock]

Then you need to find b as well

Oh you got b as 8?

[tulip kindle]

what>

?

[sullen dock]

How did you get 8?

[tulip kindle]

i answered the other one

[sullen dock]

Yeah

That's it then

[tulip kindle]

but

[sullen dock]

(5,8) is cordinate

[tulip kindle]

it would not make a stright line

it makes a angle

a----------- -------------------b

                    p

like the line is up

but the p cordinate

is below the line

not on the line

does it have to be on the line?

of line AB

[sullen dock]

Try to draw on graph and show me

I don't have a graph paper or pen paper unfortunately

[tulip kindle]

might have to get extra day to study

from teacher

i think i figured it out

[sullen dock]

Looks straight line to me

[tulip kindle]

yea

the teacher printed it ass

thats why

it looks bad for me

if point P(a,b) lies on A(x1,y1) and B(x2,y2) with a ratio of m:n , then
a = (m * x2+n * x1)/(m+n)
b = (m * y2+n * y1)/(m+n)

i use this for all?

[sullen dock]

Yep

[tulip kindle]

are you sure?

[sullen dock]

3,4

Are different

Be careful

Same formulas will be there but in 3 A and B will be swapped because ratio is given in BA , not AB

And in example 4 P is given , you need to find B

[tulip kindle]

i just swap things

?

for 3

a= (1 * 6+2*-3)/(1+2)?

[sullen dock]

If ratio is given m:n along AB then you use my formula with A(x1,y1) B(x2,y2)
But 3 says m:n ratio is along BA, so now B(x1,y1) and A(x2,y2)

[tulip kindle]

is a correct?

a= 8?

[sullen dock]

Nope

[sullen dock]

Read this

[tulip kindle]

dude im confused

[sullen dock]

Understand the concept rather than formula

[tulip kindle]

i would if you provided me the setup

[sullen dock]

You can use the formula.

a = (m * x2+n * x1)/(m+n)
b = (m * y2+n * y1)/(m+n)```

```if point P(a,b) lies on A(x1,y1) and B(x2,y2) with a ratio of m:n along BA then 
a = (m * x1+n * x2)/(m+n)
b = (m * y1+n * y2)/(m+n)```

[sullen dock]

Find the difference of words in question 1 and 3 . Read it properly

[tulip kindle]

P(x,y) = (mx1+nx2/m+n, my1+ny2/m+n)

is this correct form i can do

@sullen dock

[sullen dock]

For question 3 yes

[tulip kindle]

whats different about these bro

how do i know which formula

to use

[sullen dock]

Visualise the points or atleast draw them

Then you will have a better idea

A----m----P----n----B
This was first question

A----n----P----m----B
This is 3rd question

Do you see difference @tulip kindle

[tulip kindle]

where is the n and m coming from

[sullen dock]

That's the ratio given

[tulip kindle]

oh ok

[sullen dock]

Now it's always m:n given but you need to read the question whether RATIO is given to AB OR BA

That's why I asked you to read question 1 and 3 again and see difference

[tulip kindle]

can i solve it without doing formulas vishually>

?

[sullen dock]

I doubt that 😅

[tulip kindle]

im not taking the quiz tmr i hope my teacher gives me time

whats the formula for 4?

[sullen dock]

Well same formula but you don't know what's B(x2,y2)

[tulip kindle]

so p instead

[sullen dock]

So you will get 2 equations in terms of x2 and y2 which you will have to solve

[tulip kindle]

same formula for 4

but its instead

using a and p?

[calm coral]

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[calm coral]

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[honest swallow]

i got the maximum , no clue how to get the minimum

[stoic oyster]

well how did you get that max then ?

@honest swallow

[honest swallow]

i found out the value of x^2+ y^2

and substituted it

i got r = 53-4(x+y)

so when x+y is zero we get the max as 53

[gentle stream]

but x+y could be negative

[gentle stream]

hint for this would be ||to think geometrically||

[honest swallow]

i have tried using circles but im not sure how to apply it here

[honest swallow]

i got 53 by taking x=y too ( taking both the terms to be equal usaully gives either the minimum or maximum)

[calm coral]

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[worn heath]

hiii have i done this correctly?

[limpid pilot]

yes

[worn heath]

thank you!

.close

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[mossy geode]

how to evaluate further

[mossy geode]

nvm got it

sinc - sin d

.close

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[lyric flame]

Hey can someone help me with setting up this problem, im really struggling

[pure kayak]

what was your process

[lyric flame]

I tried to set it up as dv/dt = -g - air resistance

tried to double with chat gpt,
I struggled integrating it into velocity and then position

[calm coral]

@lyric flame Has your question been resolved?

[strange tinsel]

from the 2nd red line to the 3rd, you divided by the mass

you forgot to divide the 3rd term as well, its v/(30m)

as a result, the numbers below it change

[strange tinsel]

also, double check your work, you cant have e^(t/30) in the expression of the velocity

otherwise the system is unstable and the velocity explodes to infinity as t increases

its e^(-t/30)

when you divided to separate the variables, you should have left the minus sign in the dx side!

[strange tinsel]

but overall its good, its the method

Newtons' 2nd law and signs are correct

[strange tinsel]

$\mathrm{\int e^{at} , dt = \frac{1}{a} e^{at} + c}$

[potent lotus]

Emily

[strange tinsel]

also, dont forget the initial conditions, theyre not there for decoration!

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[rain bay]

help

[rain bay]

my teacher never taught us how to read functions and whatever

[calm coral]

@rain bay Has your question been resolved?

[remote mural]

what are u asking

@rain bay

Are you needing to translate the function?

[rain bay]

how do i graph cuberoot x-5 - 5

yea basically graph it

[remote mural]

five down and five to the right

[rain bay]

how

[remote mural]

green circle down 5 units and right 5 units

[rain bay]

yea but howd u get that from theequation

does the square root not affect anything

[remote mural]

-5 inside the root means right 5

-5 outside the root means down 5

[rain bay]

what abt this one

if its multiplying the root

[remote mural]

Did you get the first one right?

[rain bay]

yea

[remote mural]

okay

so now u need this one?

[rain bay]

mhm

[remote mural]

pay me

jk

so basically you multiply root 3 by 2

Do you under

stand

[rain bay]

no 😭

[remote mural]

Okay so

what level grade is this

[rain bay]

alg2

is it over for me dawg

[remote mural]

Sophomore

[rain bay]

yea

[remote mural]

this is easy

okay so

you see the first blue dot to the right of the green one

raise it up 1

and the first blue dot to the left of the green one move it down 1

Do u understand

[rain bay]

how do u know what inputs to put

[remote mural]

coordinates?

[rain bay]

yea how do u know what 2 cube root x becomes

like i legit dont know what the numbers x and the sqrt does

[remote mural]

square root squares the number

so basically

[rain bay]

so what would that graph be

[remote mural]

it would compress the 3rd root by 2

so it multiplies whatever point you got on the regular root by 2

[rain bay]

so u dilate by 2

[remote mural]

x2

so it grows

not shrink

[rain bay]

o

[remote mural]

here

ignore the sloppy

[rain bay]

ohh i see it now

[remote mural]

it makes it grow by 2

cuz multiply by 2

lmk if it workesd

this is basic algebra

[rain bay]

oh ok

but is it over if i dont even know this

[remote mural]

do u have an exam approaching

@rain bay

[rain bay]

kinda

but my teacher doesnt teach anything she just assigns stuff

and yea i have no idea how to do anything

[remote mural]

is that homework

Do you know what base functions mean?

[rain bay]

no

[remote mural]

average highschool teacher lol

[rain bay]

being so fr i dont think imma pass this class

imma beg counselor to switchme out

[remote mural]

it cant be that bad

blud its algebra 2

[rain bay]

its been 2 weeks since school starte

she hasnt taught a single thing

no grades up at all

its that bad dawg

[remote mural]

ciooked

cooked

can i have an idea of whats she assigning?

[rain bay]

i can figure these out with common sense

except for reading functions

[remote mural]

what are you struggling with

[rain bay]

just reading functions and graphing

[remote mural]

i can help

[rain bay]

could i just dm u when i need help

[remote mural]

ye i added u

[calm coral]

@rain bay Has your question been resolved?