help-23

[thin bridge]

you should definitely have done substitution before

[sand robin]

You’re an undergraduate student? You must have done something like this before

[thin bridge]

e.g.
given the equation p = q + 3
find the value of p when q is 5
doing that, you've done substitution whether you realise it or not

[marsh cloud]

p=8

[thin bridge]

doing that, you've done substitution whether you realise it or not

[sand robin]

Answering your question to why x^2 = y

[thin bridge]

its just replacing something with equivalent value and/or something you've defined to be equivalent to make things nicer/simpler

[sand robin]

Is because when you substitute x^2=y into the ugly equation

It becomes a factorisable quadratic

[thin bridge]

$y + \sqrt{y} - 20 =\underbrace{ \underbrace{\sqrt{y}}{x}\underbrace{\sqrt{y}}{x}}{x^2} - \underbrace{\sqrt{y}}{x} - 20$ \
using that substitution (replacing $\sqrt{y}$ with $x$)
$$x^2 - x - 20$$

[flat frigate]

ℝamonov

[thin bridge]

gets you something that looks much nicer. note that if you're comfortable working with $\sqrt{y}$ and have no issue with viewing
$$y + \sqrt{y} - 20 = (\sqrt{y})^2 + \sqrt{y} - 20$$
as a quadratic in $\sqrt{y}$ then doing a substitution like that is unnecessary

[flat frigate]

ℝamonov

[safe radish]

@marsh cloud Has your question been resolved?

[safe radish]

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[blazing jolt]

can someone help

[peak smelt]

ok so

you have

y = ___ x

the underlined is supposed to be a number

numbers connected to variables are called co-efficients

now, for lines

[blazing jolt]

yea ofc

[peak smelt]

we have this neat little formula

y = mx + c

the c number, is the y intercept

[blazing jolt]

that formula works for graphs?

[peak smelt]

as u can see, the line for steve goes through the y point 0

so, c = 0, and it disappears

yes graphs

straight lines

[shadow glade]

for graphs of LINES, yes

[peak smelt]

y = mx + c

[shadow glade]

so what is the slope and what is the y-intercept of Steve's line graph?

[peak smelt]

now, the letter m

is the slope of the line

how steep the line is

m will replace the underline (the part you are looking for)

m is the slope

and another way to write slope is

y/x

[blazing jolt]

Alright

[peak smelt]

rise over run

how much a line goes up for how much it goes across

so u have to measure the distance it goes up for how much it goes across

i would suggest u start at 0, and find where it crosses a gridline

and let me know what the x and y coordinates are on that point

[blazing jolt]

which point

[peak smelt]

so

measure how far the line goes up for however far across

you can pick 1 across, 2 across, 3 across, however much you want

[blazing jolt]

5, 200

[peak smelt]

but you gotta measure the height at the position across

[blazing jolt]

the graph says

[peak smelt]

great

[blazing jolt]

man this is hard

[peak smelt]

you couldve picked any point, but (5,200) works

i wouldve maybe done (1,40)

since its easier

can you see (1,40)

[blazing jolt]

yes

[peak smelt]

do you see how you couldve measured the height at 1 instead

5 works

but 1 is just easier (smaller number)

ok anyways

now

co-ordinates are given in this form

(x, y)

x coordinate, then y coordinate

right?

[blazing jolt]

yes

[peak smelt]

now

if u scroll up ull see i said

m = y/x

y over x

y divided by x

[blazing jolt]

so 40,1

40 divided by 1

[peak smelt]

yrp

which is?

m = 40/1

[blazing jolt]

40

ofc

[peak smelt]

great

so m = 40

so steve has

y = 40x

since normally

we have

y = mx

[blazing jolt]

wow

this is great

thanks man

[peak smelt]

so we can replace the m with the slope (y/x) that we found

no worries

hopefully the steps made sense

[blazing jolt]

yea

[peak smelt]

you should be able to follow in the future

and remember

that c = y intercept

so whereever the line crosses the y axis

y = mx + c

you will use that alot in the near future

good luck

[blazing jolt]

thanks you

.close

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[delicate glacier]

I need to demonstrate this (a, b, c are positive real numbers)

[delicate glacier]

It's clearly greater or equal to 64 but I don't know how to demonstrate it for 1/sqrt(abc)*64

[safe radish]

@delicate glacier Has your question been resolved?

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@delicate glacier Has your question been resolved?

[lapis shadow]

u can use the AM-GM inequality

when $a,b$ are two non-negative numbers, $\frac{a+b}{2}\geq\sqrt{ab}$

[flat frigate]

SilverSoldier

[delicate glacier]

Thanks, I'll try that

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[eternal sparrow]

Hey guys, I’m stuck on sketching this type of question

Can someone please guide me

[graceful tusk]

if you consider composing a function from simpler ones it can help

do you know what the graph of e^x looks like?

and its y-intercept?

because that will help you find the intercept and sketch of 5e^x

then if you negate the whole thing, you simply flip it vertically around the x-axis

[eternal sparrow]

Ok thanks

[graceful tusk]

alg, and once youre done try graphing it on desmos to see if it looks right

but try not to look there first otherwise you wont get better at it

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[stable tartan]

24 Minuten zum Befüllen von 6000 Flaschen Mineralwasser.
Die Funktion f fordert einer Anzahl n solcher gleichzeitig arbeitender Abfüllmaschinen die Dauer f(n)
zu, die für die Befüllung der 6000 Flaschen benötigt wird n (n Element aus Natürlichen Zahlen\{0} und f(n) in Minuten).
Stellen Sie eine Gleichung der Funktion f auf.```
Translation:
```If four filling machines working at the same speed are used at the same time, they require
24 minutes to fill 6000 bottles of mineral water.
The function f demands the duration f(n) from a number n of such simultaneously working filling machines
that is required for filling the 6000 bottles n (n element from natural numbers\{0} and f(n) in minutes).
Write an equation of the function f.```
Please explain it to me, I know the answer but I dont know how to solve this in another case

[stable tartan]

Anyone? :/

[safe radish]

@stable tartan Has your question been resolved?

[wooden garnet]

How long does 1 filling machine take?

@stable tartan

[stable tartan]

6 mins?

[wooden garnet]

Think logically

I have 2 people doing a task, it'll take them 30 mins

If I say that I take one person away, you are now stuck trying to do it alone

How long does it take you?

[stable tartan]

wait, gimme time

omg I

am so stupid

ofc it takes longer

indirectly proportional

[stable tartan]

96 min

[wooden garnet]

Inversely proportional but yeh

Ok

Good

F(n) takes the number of filling machines

If I have 2 filling machines how long will it take?

[stable tartan]

double time?

[wooden garnet]

Double?

[stable tartan]

4 => 24
2=> 48?
but wait it is inversely

[wooden garnet]

Other way right

If I have two times the amount of water flowing it'll take half as long to fill

[stable tartan]

thats right

*2 =^ /2

[wooden garnet]

Ok so if I have 2 filler machines how long ?

[stable tartan]

is 96 f(x)?

[wooden garnet]

?

[stable tartan]

if 1 takes 96 min (!!and not 6!!)
96(=^f(x))

[wooden garnet]

I've never seen that notation

[stable tartan]

I know, i see you making a big facepalm lmao :,)

[wooden garnet]

Normally functions are written f(n)=n^2

[stable tartan]

=^ should be "stands for"

it was just a note sorry I did a mistake

[wooden garnet]

96 = f(1), when you have one machine, it takes 96 minutes

What is f(2) equal to?

Ie how long does it take for 2 machines

[stable tartan]

96/2

omg? is that correct?

[wooden garnet]

Good job. F(3)?

[stable tartan]

yes omg

ok

96/3

so like 3 is the parameter

[wooden garnet]

F(n)?

[stable tartan]

and now its 96/x omg

tysm!!!!!!!

[wooden garnet]

No problem

[stable tartan]

:DDDDDD

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[untold canopy]

Determine the cartesian equation of the plane containing the point (-1, 1, 0) and perpendicular to the line joining the points (1, 2, 1) and (3, -2, 0)

[untold canopy]

can someone help me with this?

please ping me when you have a response

[glass carbon]

find normal vector firstly (use the points on the line)

then just use formula for the plane which is perpendicular to the vector and passes through the point

[untold canopy]

how can i find the normal vector of the line

[glass carbon]

not of the line, of the plane

you know line is perpendicular to the plane, so direction vector of the line have to be perpendicular to the plane as well, this is why it will be normal vector

[untold canopy]

so how do i find the normal vector

of the plane

@glass carbon

[glass carbon]

don't you know how to find vector when two points (ends) are given?

[untold canopy]

usually we have the cartesian equation

Ax + By + Cz + D = 0

and the normal vector would just be n = (A, B, C)

maybe i'm having trouble visualizing the problem

[glass carbon]

just subtract appropiate coordinates

here it's better to use A(x-x1) + B(y-y1) + C(z-z1) = 0, but your formula will work also (you have to find D by plugging coordinates of the point in)

[untold canopy]

alright thank you

i figured it out

.close

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[warped hawk]

.open

.ask

[pallid hearth]

Does anyone know how I can show that $\binom{n}{x}$ is an integer for any $0 \le d \le n$ given that $\binom{n}{x} = \binom{n-1}{x} + \binom{n-1}{x-1}$ for all $ 1 \le d < n$?

[stray socket]

Does anyone know how I can show that $\binom{n}{x}$ is an integer for any $0 \le d \le n$?

[flat frigate]

Umbraleviathan

[stray socket]

There

[pallid hearth]

Thank you!

[flat frigate]

Modus

[glass carbon]

other way

[flat frigate]

lynnie

[pallid hearth]

Thanks!

[flat frigate]

lynnie

[pallid hearth]

For context, I did part (a) of a problem which successfully showed the condition that $\binom{n}{x} = \binom{n-1}{x} + \binom{n-1}{x-1}$ for all $ 1 \le d < n$, but I am stuck on even how to proceed the second part which is problem above. I tried doing induction which I feel like might be the right path, but im unsure how to induce on something with two variables.

[flat frigate]

lynnie

[pallid hearth]

<@&286206848099549185>

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[vapid maple]

Fu

[tall bough]

you missed 2 letters

[worthy hemlock]

.close

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[pallid hearth]

Reasking this question here again: Does anyone know how I can show that $\binom{n}{d}$ is an integer for any $0 \le d \le n$ given that $\binom{n}{d} = \binom{n-1}{d} + \binom{n-1}{d-1}$ for all $ 1 \le d < n$?
For context, I did part (a) of a problem which successfully showed the condition that $\binom{n}{d} = \binom{n-1}{d} + \binom{n-1}{d-1}$ for all $ 1 \le d < n$, but I am stuck on even how to proceed the second part which is problem above. I tried doing induction which I feel like might be the right path, but im unsure how to induce on something with two variables.

[drowsy moss]

what's d?

and inducting on n is probably the correct path.

[pallid hearth]

d and n are both integers

but it's not fixed. We want to show that for all cases s.t d and n are integers and 0 < d < n, the above is true

[drowsy moss]

ok, but d isn't in any of your formulas? is d supposed to be x?

[pallid hearth]

Oh sorry hahah yes d is supposed to be x im an idiot

[drowsy moss]

no worries, just wanted to make sure

[pallid hearth]

I fixed it ... mostly

[flat frigate]

lynnie

[pallid hearth]

Do you have any idea whether i should induct on n or d?

[drowsy moss]

induct on n. Your statement for the hypothesis should be "assume (n-1 choose d) is an integer for all 0 <= d <= n-1"

[pallid hearth]

Thank you so much!

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[dreamy meadow]

I need help with a geo question

[dreamy meadow]

Usually this type of problem is easy but I think its just been a while I need help om where to start

[sharp tiger]

It says the triangle is isosceles, so the third angle will be equal to 3x+20

Now u just need to use the sum of angles

[dreamy meadow]

got it. I was just blanking

thanks

.close

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[dreamy meadow]

.reopen

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✅

[dreamy meadow]

where do i start here? why are the equations outside?

[tall bough]

No space inside?

[junior smelt]

Angles in a triangle add up to...?

[dreamy meadow]

180

oh well if they are supposed to be inside then i know. I've just never seen em outside thought it might mean something else ive never seen

[tall bough]

And u can see an angle is given

Nice

[dreamy meadow]

can I have some help with this

I dont know where to start

.close

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[dreamy meadow]

<@&286206848099549185>

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[dull sequoia]

what have you tried?

[safe radish]

@lean otter Has your question been resolved?

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@lean otter Has your question been resolved?

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[warm tulip]

i don't know what to do after evaluating f(-2), f(-1), f(0)

[warm tulip]

@final bay sory i need help again

<@&286206848099549185>

[safe radish]

@warm tulip Has your question been resolved?

[safe radish]

@warm tulip Has your question been resolved?

[safe radish]

@warm tulip Has your question been resolved?

[junior smelt]

Oh hey @warm tulip, guess who’s waking up

Check which one is the largest and the smallest

In your case 13 is the max, -4 the min if I’ve read your work right

I come bearing gifts
[note the “in the interval” for (a), that’s what your teach said earlier]

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[warm tulip]

oh my good morning

[junior smelt]

,time

[flat frigate]

The current time for chartbit is 12:01 PM (GMT) on Wed, 11/01/2023.

[warm tulip]

so do i just put like max = x=13

and min = x=-4

[junior smelt]

More that the maximum is 13 [happens when x=-1], and the minimum is -4 [happens when x=-2]

[safe radish]

@warm tulip Has your question been resolved?

[warm tulip]

o

[warm tulip]

ohhhhhhh rite

and then when i justify do i have to say the points or can i just say the x

[junior smelt]

I think you just have to say what the maximum and minimum are tbh

But I accept no liability if your course is different and all

[warm tulip]

🤔

idk my teach always writes like a whole sentence for the justification

could u help me wit my related rates hw 😭

[junior smelt]

Sure send it through

We'll take a look

[junior smelt]

oh yeah did you like my gift btw

[warm tulip]

yas

[junior smelt]

Oh yeah @shut inlet what time did you get up?

[warm tulip]

i kinda don understand it doe 😭

leik wat is ck

[junior smelt]

That's basically the process of finding the maximum and minimum on an interval

[shut inlet]

12:00 PM

[junior smelt]

Find where the derivative is zero on the interval, then work out what what the value of the function is for those points

[note how on the interval is bolded, the ones outside you ignore them]

Then if it's a closed interval, evaluate the endpoints to see what they are

Because you could e.g. have something which has a turning point, but then increases, if you get me?

Basically they did it in symbols haha

[junior smelt]

Ah, very good, you actually rested! Look at the time I messaged egg

[warm tulip]

ooooohhhhh ok

🥚👍

oka oka

here is da hw

[shut inlet]

i can't woke up if i didn't sleep 7-8 hours

[junior smelt]

Right, I would quite like to start with 5

You know volume of a cone yea?

[warm tulip]

oke

naw

😳

no omg i have memorize that right

😭

oka

pir^3(h/3)

[junior smelt]

Well kinda, it's $\frac{\pi}{3} r^2 h$

[flat frigate]

chartbit

[junior smelt]

...if I remember correctly

[warm tulip]

wat tha google sayin somthing diff

[junior smelt]

Also note here they've told you that $h = \frac{3}{8}(2r)$ yeah?

[flat frigate]

chartbit

[warm tulip]

umm i think

how did the r become 2r

[junior smelt]

"the height of the pile is always three-eights of the base of the diameter"

[warm tulip]

diamte ok

oh ok

[junior smelt]

Yep, so now we can express the volume purely in terms of the radius, yes?

[warm tulip]

u sayin we finding r?

[junior smelt]

I'm saying to write $V$ in terms of $r$ only

[flat frigate]

chartbit

[junior smelt]

As in use this, and...

[warm tulip]

🤔

[junior smelt]

...this together

[warm tulip]

i am so confus

[junior smelt]

Alright, I'm saying put $h = \frac{3}{8}(2r)$ (simplify that) into $V = \frac{\pi}{3}r^{2}h$

[flat frigate]

chartbit

[warm tulip]

uhhh

how to simplify

or do move h to other side

[junior smelt]

What's 2/8?

[warm tulip]

1/4

[junior smelt]

Yep, so does that help in simplifying $\frac{3}{8}(2r)$?

[flat frigate]

chartbit

[warm tulip]

i dun kno

[junior smelt]

$\frac{3}{8}(2r) = \frac{3}{4}r$

[flat frigate]

chartbit

[junior smelt]

Ya agree with that?

[warm tulip]

wot

o

yes i think

it become (6/8)r and den u simplify?

o oke

[junior smelt]

Yep, now can you do the next bit for me, put that $h = \frac{3}{4}r$ into $V = \frac{\pi}{3} r^{2} h$?

[flat frigate]

chartbit

[warm tulip]

oook

and then do i multiply all dat an den find derivative?

[junior smelt]

[note that with what we've done, we would be doing b first, but then I've been freestyling tbh

• careful of units, I can help you with that if you'd like me to]

[warm tulip]

😨

ya i tink my teach did d that

i đón understand when he does the dv/dt thingy

[junior smelt]

Basically, the question says this, this is rate information

[warm tulip]

yes yes

[pure agate]

They are actually solving a parametric equation with time as the dependent variable but the complete explanation is for Calc 2 or Calc 3. For now, you only need to concern yourself with solving with respect to time.

[junior smelt]

In equations, that's them saying $\dv{V}{t} = 10$ if you represent $V$ in $m^3$ and $t$ in mins

[flat frigate]

chartbit

[warm tulip]

o ok

[junior smelt]

Are you at least familiar with the chain rule though?

[warm tulip]

YAR

[junior smelt]

Basically we're using that here but it might not be as explained well I guess

[junior smelt]

Does this make any more sense, have I explained it well at all?

[warm tulip]

umm i tink

😭😭

u sayin he usin da speed thingy

[junior smelt]

Basically they're telling you the rate of change of the volume with respect to time

We chose to represent volume here, in m^3 with V, and we'll choose to represent time here, in minutes, with t

[warm tulip]

ok ok

[junior smelt]

Then in differential equation form, that's saying dV/dt = 10

[warm tulip]

otay

[junior smelt]

Yep, are you happy to accept that?

[warm tulip]

yar

[junior smelt]

Cool cool, now let's take that, what did you get as your V in terms of r?

[junior smelt]

This right here

Or if you want, you can have V in terms of h instead?

Whichever you like

[warm tulip]

wuh

i don kno

whichever

[junior smelt]

Alright, if I put $h = \frac{3}{4} r$ into $V = \frac{\pi}{3} r^{2} h$, I get $V = \frac{\pi}{3} r^{2} \left( \frac{3}{4} r \right) = \frac{\cancel{3} \pi}{\cancel{3} \cdot 4} r^{3} = \frac{\pi}{4} r^3$

[flat frigate]

chartbit

[junior smelt]

Do you agree with that?

[warm tulip]

oh yes

[junior smelt]

Could do the opposite, write $r = \frac{4}{3} h$, then you'd get $V = \frac{\pi}{3} \left( \frac{4}{3} h\right)^{2} h = \ldots$

[flat frigate]

chartbit

[junior smelt]

Those are what I wanted you to do, you can pick either one you wanted

[junior smelt]

This one better for part a

[warm tulip]

did u switch h and r for dat

OKE

[junior smelt]

Yep yep, so are you happy with what we have so far?

[warm tulip]

yes 🫡

[junior smelt]

Yaaaayyy

[warm tulip]

[junior smelt]

Alright, cool, so for part a, we want to find the rate of change of the height with respect to time, when the height is 4m

So what do you think we'll need to do?

[warm tulip]

ummmm

🤔

use rate of change formula

?

[junior smelt]

This is a differentiation question, yea? Maybe we should differentiate something?

[warm tulip]

ok tbh idk wot differentiate mean

☠️☠️☠️

[junior smelt]

Wait didn't you do differentiation in the question you sent us before?

[warm tulip]

oh is that when u do the calc

leik derivate

😭😭😭😭😭

o ok

ok les derivative v=pi/4(r^3)

[junior smelt]

And what do you get?

[junior smelt]

What are you deriving with respect to?

[warm tulip]

uhhh

time

[junior smelt]

Not yet no

We have a V and an r

[warm tulip]

uhhh

v respect to r

oke

is dat dv/dr den

ooooooooooooook

me got dv/dr=-pi/16(r^3) should i use chain rule for da r

[junior smelt]

Show your work please!

[junior smelt]

Remember that $\frac{\pi}{4}$ is a constant so you can leave it as is, you don't change the constants

[flat frigate]

chartbit

[junior smelt]

And also

,w diff (pi/4)*r^3 wrt r

[flat frigate]

Results provided by WolframAlpha

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

Requested by chartbit#3171

[junior smelt]

You should have this here?

Do you see why?

[warm tulip]

ummm i tink

why pi/4 constant

[junior smelt]

Because it doesn't change, pi will always remain that 3.14... whatever

Would be something if pi could change

It's an "unusual" constant sure, but it is constant nonetheless

Same with 4, that will never turn into like 5

[pure agate]

$\frac{d( c\cdot f(x))}{dx} = c \cdot \frac{d( f(x))}{dx} = c \cdot f'(x)$

[flat frigate]

Kookiemon

[pure agate]

If you have some function multiplied by a constant, c, you can factor out c, evaluate the derivative, multiply the derivative by c and get the same result.

[warm tulip]

oh ok

so only derive thingies without variable?

[pure agate]

$\frac{d \left(\frac{\pi r^{3}}{4}\right)}{dr} = \frac{\pi}{4} \cdot \frac{d(r^{3})}{dr} = \frac{\pi}{4} \cdot 3r^{2} = \frac{3\pi r^{2}}{4}$

[flat frigate]

Kookiemon

[pure agate]

And it's not just non-variables that are treated as constants, any variable that is not "changing" can be treated as a constant as well.

[warm tulip]

[pure agate]

If you had x^2 yz and differentiated with respect to x, its derivative would be 2xyz.

If you differentiatd x^2 yz with respect to y, its derivative would be x^2 z.

[warm tulip]

i see

🤔

[pure agate]

It makes sense after awhile. 😉

[warm tulip]

okaa

so since it respect to r then r get derive

[pure agate]

Yes.

[warm tulip]

oke

[pure agate]

What problem were you working on?

I got an hour.

[junior smelt]

5 from here

There's also the others but that's the one we started on

[warm tulip]

yes yes

[pure agate]

And did you figure it out?

[warm tulip]

naw

[junior smelt]

We're at the point where we have $\frac{dV}{dr}$ from the above

[flat frigate]

chartbit

[junior smelt]

Are you happy with why it's the way it is from the explanation above, @warm tulip?

[basically doing (b) first]

[warm tulip]

i hav no idea

yes

[junior smelt]

So we basically are at the point where we have $\dv{V}{r} = \frac{3\pi}{4} r^{2}$, and as we said before, $\dv{V}{t} = 10$

[flat frigate]

chartbit

[junior smelt]

Up until this, do you follow?

[warm tulip]

yes yes

wait a min

y we doing b first

[junior smelt]

Because we ended up rearranging for the radius, if we wanted to do a first we would have needed to do the height instead

Either way, both ways are relatively similar

In fact, let's do a first, that will test whether you understand

[warm tulip]

oh ya

[flat frigate]

chartbit

[warm tulip]

ummmm

oke

plug in h

uh oh

[junior smelt]

You want to eliminate r, yea?

[warm tulip]

wat yes

[junior smelt]

So we should really put in r = "something", yea?

[warm tulip]

waaaa

o k

h-3/4

[junior smelt]

Let's take it in steps together 🧑‍🤝‍🧑

[warm tulip]

wat i mean r=h-(3/4)

[junior smelt]

We have $h = \frac{3}{4}r$

[flat frigate]

chartbit

[warm tulip]

😭😭😭😭😭

yes yes

[junior smelt]

And we want r

What's the first thing we should do?

[warm tulip]

4h/3

multiply da 4

on both side

[junior smelt]

There you go, you got it 🥳

[warm tulip]

yayayaay

[junior smelt]

So $r = \frac{4h}{3}$

[flat frigate]

chartbit

[junior smelt]

Now put that into $V = \frac{\pi}{3} r^2 h$, and simplify as much as possible please?

[flat frigate]

chartbit

[warm tulip]

yes ok

🫡

[junior smelt]

Remember you are not alone 🧑‍🤝‍🧑

[warm tulip]

[junior smelt]

Well done!

Now derive that for me, with respect to h?

[warm tulip]

[junior smelt]

Remember you have this if you get stuck

[warm tulip]

okee

[junior smelt]

Mehdi back

[warm tulip]

o helo

[junior smelt]

,w diff (16pi/27)h^3

[flat frigate]

Results provided by WolframAlpha

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

Requested by chartbit#3171

[junior smelt]

,calc 1296/729

[flat frigate]

Result:

1.7777777777778

[junior smelt]

,calc 16/9

[flat frigate]

Result:

1.7777777777778

[junior smelt]

You are correct, though you can simplify that down a bit more

[warm tulip]

[shut inlet]

,w simplify 1296/729

[flat frigate]

Results provided by WolframAlpha

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

Requested by Mehdi_Moulati#1210

[warm tulip]

will i have to do dat on da ap exam

without a calc

[junior smelt]

Well let me give you a simpler way to have done it

$V = \frac{16\pi}{27} h^{3}$ right, and that $\frac{16\pi}{27}$ is a constant, right?

[flat frigate]

chartbit

[warm tulip]

yes yes

[flat frigate]

chartbit

chartbit

[warm tulip]

ok ok ok

[junior smelt]

[ping me if you need me]

[warm tulip]

sory i in class now so it might take a little

[warm tulip]

would that make it 3h^2

?

okeeee

[junior smelt]

Yep, so basically we take that, and multiply that by the constant we had, so we get the final answer to be $\dv{V}{h} = \frac{16\pi}{27} \times 3h^2 = \ldots = \frac{16 \pi h^2}{9}$

Which saves quite a bit of work

[flat frigate]

chartbit

[warm tulip]

ok ok

[junior smelt]

Yep, so does that help out?

[btw if you want, you can come back later if you're busy now?]

[warm tulip]

idk i kinda wanna finish my packet

actually i'll try to make time in a little

i think u simplified wrong

[junior smelt]

Hmmm did I?

[warm tulip]

idk man

maybe i did

😭😭😭😭

[junior smelt]

Might have tbh

[warm tulip]

i got 16pi/9

[junior smelt]

So what I have then

[warm tulip]

okaaaaa

o wait do u multiply

[junior smelt]

Not that it matters much tbh, but I did

[flat frigate]

chartbit

[warm tulip]

sory it kinda hard to focus on math in clas

[junior smelt]

Yep yep, ah you also said, so don't worry

Come back later when you can, and if you want, you can ping me

[if I'm still free then, I'll help you out ]

[warm tulip]

oka oka thank u :D

[warm tulip]

oke i dunno wat to do after finding the derivative

do i plug it back in

?

[junior smelt]

Right, back to where we were

[warm tulip]

into the other equation

which one

[junior smelt]

"it"?

[warm tulip]

da derivative

[flat frigate]

chartbit

[junior smelt]

Can you work that out for me please?

Put the h=4 into dV/dh

[warm tulip]

i got

[junior smelt]

🇾 tho

[warm tulip]

i đi derivative of h squared

[junior smelt]

You don't need to differentiate it again though

All you need to do is replace h with 4 here

[warm tulip]

oh okk

and then u multiply that with 16pi?

[junior smelt]

Yep, do $\frac{16\pi \times 4^{2}}{9}$ if you'd like

[flat frigate]

chartbit

[warm tulip]

okkk

[warm tulip]

and what do u do after

[junior smelt]

So you know that when $h = 4$, that $\dv{V}{h} = \frac{256\pi}{9}$, and also that at the same time, $\dv{V}{t} = 10$, and we really want $\dv{h}{t}$, yeah? You agree that's what the question asks for?

[flat frigate]

chartbit

[junior smelt]

Now it's time for chain rule ⛓️

[safe radish]

@warm tulip Has your question been resolved?

[warm tulip]

chain rule?

where do i do that

[junior smelt]

We want to somehow get $\dv{h}{t}$ out, but we should already know by the chain rule that $\dv{V}{h} \dv{h}{t} = \dv{V}{t}$, yes?

[flat frigate]

chartbit

[junior smelt]

We know the values of two of those, so rearrange to get $\frac{\dv{V}{t}}{\dv{V}{h}} = \dv{h}{t}$

[flat frigate]

chartbit

[warm tulip]

o oke

imma try to do dat

[safe radish]

@warm tulip Has your question been resolved?

[warm tulip]

this is so hard 😭😭

[junior smelt]

It takes a bit of experience, I’ll say

Have to work at it a bit

[junior smelt]

Are you happy with at least here though? @warm tulip

[warm tulip]

ummmm to be honest i only learn that chain rule is like derivative outside times derivative inside

so dis is sumting new

[junior smelt]

Ahhh fair fair, yep, they really haven’t helped if they hadn’t explained this

Basically these are like two equivalent ways of stating the chain rule (take the bottom one in the second pic, top is particular case)

Assuming you’ve seen the second one but not the first?

[flat frigate]

chartbit

chartbit

chartbit

[junior smelt]

Does all the above make sense?

[warm tulip]

hold on i am destroying my lunch

[warm tulip]

no 😳

is that like some formula or did u use ur noggin

[junior smelt]

Do you agree with this step here? @warm tulip

That if you have $y= f(g(x))$, that $\dv{y}{x} = f’(g(x))\cdot g’(x)$

[warm tulip]

iz dat product rule

[flat frigate]

chartbit

[warm tulip]

o

[junior smelt]

Literally chain rule ⛓️

But are you happy with that?

[warm tulip]

ohhh yea outside inside

yes yes

🫡

[junior smelt]

Cool cool, now, let’s write $y= f(u)$ and $u=g(x)$. Are you happy to see why? Basically we have that $f(g(x))=f(u)$

[flat frigate]

chartbit

[warm tulip]

ooo ok

yes i am happy

[junior smelt]

Perfect (btw please ping me each time so I see the messages )

So now, if we have $y= f(u)$, then we have that $\dv{y}{u} = f’(u)$, you agree?

[flat frigate]

chartbit

[warm tulip]

how come it become dy/du

[junior smelt]

Because we have a function in terms of u, so we decided to differentiate y with respect to u

[warm tulip]

oh ok

[junior smelt]

Yep, so are you good with me so far?

[warm tulip]

yaa

[junior smelt]

Cool, now we have that $u = g(x)$, so if we differentiate that, we get $\dv{u}{x} = g'(x)$, you agree?

[flat frigate]

chartbit

[warm tulip]

YAAAAAAA

sory i keep forgetting 2 reply

😑

[junior smelt]

Yep, now if we do $\dv{u}{x} \times \dv{y}{u}$, that's basically $f'(u)\cdot g'(x)$, yea?

[flat frigate]

chartbit

[junior smelt]

Ouch, you forgot me

[warm tulip]

no sori i am distractful person

[junior smelt]

[I've just finished something else so coincidentally came here to peek as you sent your reply ]

[warm tulip]

[warm tulip]

YES 🫡

[junior smelt]

Yep, now, remember we said that $u = g(x)$, right!

[flat frigate]

chartbit

[warm tulip]

YA

[junior smelt]

Put that back into we have, we get that $\dv{u}{x} \times \dv{y}{u} = f'(u)\cdot g'(x) = f'(g(x))\cdot g'(x)$

[warm tulip]

[flat frigate]

chartbit

[warm tulip]

oh my if all da same

it*

[flat frigate]

chartbit

[junior smelt]

Happy with all of that?

[warm tulip]

yes yes

happy

[junior smelt]

Perfect, now that is a nicer form of the chain rule we can use in questions like this

Basically replace variables as appropriate

[junior smelt]

Are you happy with this statement now?

[warm tulip]

OKAAAAA

yes yes

[junior smelt]

Cool cool, now...

[flat frigate]

chartbit

[warm tulip]

ummm

why is 10 on top

[junior smelt]

What do you mean? As in why $\dv{V}{t} = 10$, or why does the fraction $\dv{h}{t} = \frac{\dv{V}{t}}{\dv{V}{h}}$ look as it does?

[flat frigate]

chartbit

[warm tulip]

oh nvm

i was saying why dv/dt is on top of fraction but i realize da dv/dt equation when isolating height with respect to t makes it that way

[junior smelt]

Okay cool cool, so you happy with me?

[warm tulip]

yes yes

i am sleepy doe

[junior smelt]

Haha fair fair, wanna at least push through to the end of this question?

[warm tulip]

meh oke

it ok if my paper in my backpack

😭

i am in bed rn 🤣

[junior smelt]

fair enough I guess, can't argue with that, tried working in stuff while in bed earlier haha

[warm tulip]

[junior smelt]

But basically the next step would be you put the numbers we have already into here

[warm tulip]

sleep is th best

ok ok

so plug n chug

[junior smelt]

So then we would get $\dv{h}{t} = \frac{\dv{V}{t}}{\dv{V}{h}} = \frac{10}{\left( \frac{256\pi}{9} \right)} = \ldots = \frac{45}{128\pi}$

[warm tulip]

yar

and then u multiply by reciprocal?

[flat frigate]

chartbit

[warm tulip]

okk

and is that the height then?

[junior smelt]

That's the rate of change of height with respect to time, but in m/min

So the rate of change of the height is $\frac{45}{128\pi} m/min$

[flat frigate]

chartbit

[junior smelt]

Butttttt....

So you need to convert the m/min into cm/min

When you do, you'll get what they have!

[warm tulip]

is that 1125 or 11.25

[junior smelt]

1125

How do you convert m to cm?

[also I thought you went to sleep ]

[warm tulip]

yea i fell asleep but my mom got so mad so i woke up

[junior smelt]

Ah, one of those ones damn sorry for that

[warm tulip]

yea......

she wants to kick me out of the house

[junior smelt]

Oh no, that's actually terrible :/

Not that great

Hope things get better for you, know how difficult family stuff can be at times honestly

Not even amusing at all

But anyways, do you want to continue with these? If not then that's perfectly fine, I understand

[warm tulip]

i mean i kinda have to but i don't think i want to

[junior smelt]

Hmmm, the question I think I have is do you think you'll be able to understand things if we do work on them? If not then it would be better to pass these until another time where you'll be in a better place

I mean, while it's good to work hard, I would want you to understand what we're doing, so either way, whatever you would like to do

[warm tulip]

ok let me just quickly eat dinner

[junior smelt]

Cool cool, ping me when you're back!

[warm tulip]

hi

umm

sory

umm

and is 5b the same but wit radius

[junior smelt]

Yep it kind of is, but we can use the 5a to get the answer straight away with not a lot of work

Remember we know that the height and radius are related, by $r = \frac{4h}{3}$

[flat frigate]

chartbit

[junior smelt]

So we can derive this with respect to t now, and we will be able to cheat

[warm tulip]

o so dr/dt=umm

uhh

[junior smelt]

I'll help you here, don't worry 🧑‍🤝‍🧑

It's a bit confusing if you aren't familiar

Differentiate the left hand side with respect to $t$, we get $\dv{r}{t}$, differentiate the right hand side with respect to $t$, we get $\frac{4}{3}\dv{h}{t}$, so then $\dv{r}{t} = \frac{4}{3}\dv{h}{t}$

[flat frigate]

chartbit

[junior smelt]

Does that make sense as to why?

[warm tulip]

oh yeaaaa

leik for these problem when u derive da letter it become the respect to thingy

?

[junior smelt]

Yep, so if it doesn't have the letter shown in it and it isn't a constant, you become like that

[junior smelt]

So like in here, the r didn't have a t in it but it depends on it, same with the h

[warm tulip]

mehdi stalk chat

[junior smelt]

Well known lurkers, me and Mehdi are

[warm tulip]

wat u mea n letter not shown

?

o leik if u derive and letter disappeare?

ohhhhh

wow

ish u math major ?

[shut inlet]

[junior smelt]

Hmm what I meant was that, for example

[warm tulip]

wut

?

[junior smelt]

Let's take $r = \frac{4}{3} h$

[flat frigate]

chartbit

[junior smelt]

Now $r$ and $h$ are not constants, they both depend on $t$

[flat frigate]

chartbit

[warm tulip]

hm i see

[junior smelt]

So to differentiate that with respect to $t$, we need to replace them with $\dv{ \text{[thing]} }{t}$

[shut inlet]

since the h and r change when t change

[flat frigate]

chartbit

[warm tulip]

okee

so watever variable u put in thing

cus u derivitave it

derivative

?

[junior smelt]

Yep that's basically the idea, but make sure you know whether it's a constant or variable first!

[warm tulip]

yes yes

🫡

[junior smelt]

Anyways, from here, we know $\dv{h}{t}$ from the first half, so put that in and work again [the one in cm/min, please!]

[flat frigate]

chartbit

[warm tulip]

okaaaaaa

me me gots ummmm 375/8 pi

cm/min

?

i wana work on me study guide for test :D

[junior smelt]

Let me check that for you, check the numbers for you

[junior smelt]

Actually give me a min (ha! ) but it doesn’t look right for cm/min

[warm tulip]

WA

uh oh

[junior smelt]

I think I know what you did

[warm tulip]

o no pls

[junior smelt]

You should have $\frac{1123}{24\pi}$ if I'm right?

[flat frigate]

chartbit

[warm tulip]

WAT

[junior smelt]

Hmm hang on how did you get that?

[warm tulip]

ummmmmm

lets just say i used sum resources

[junior smelt]

This is $\dv{h}{t}$

[flat frigate]

chartbit

[junior smelt]

Resources...

Please do not say something like "chatgpt", please

[warm tulip]

LMAOOOOO ISNT THAT FOR WRITING ESSAYS?

NO I USED THE ANSWER KEY 😭

CUS I GOT LAZY

U GOT ME

[junior smelt]

Hmmmmm then what am I doing wrong...

[warm tulip]

maybe teacher is wrong

[junior smelt]

Oh no it's me, never mind!

[warm tulip]

oopsie

[junior smelt]

tf did I do here then

Oh ye did the wrong way round, me dum

[junior smelt]

Yep you were right, but oi, no cheating

[warm tulip]

no no u smartie

soree

[junior smelt]

Basically you put these together and when you multiply them you should get $\frac{375}{8\pi}$

[flat frigate]

chartbit

[junior smelt]

Heheh next time we will do it the long way

But are you happy with all the work we've done?

[warm tulip]

D:

okaaaaaaaaaaaaaaaaaa

[junior smelt]

Sure, 100%, promise me you're happy with what we've done? 💯

[warm tulip]

👍

🫡 yes yes

[junior smelt]

Perfect, now, we have finished 5!

,calc 5!

[flat frigate]

Result:

120

[warm tulip]

yayayayaayyayaya

wat

[junior smelt]

Oh wait, no, question 5, my bad

[warm tulip]

🤣

[junior smelt]

Silly math joke

Do you wanna move on to another one?

[warm tulip]

]

oke hold on leme looks

i tinks i ok for now i going to try sum out on me own

🫡

tank u teacher

🫡

[junior smelt]

Hahah no worries, best of luck with them!

[warm tulip]

.close

[safe radish]

Channel closed

Closed by @warm tulip

Use .reopen if this was a mistake.

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.
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• Be polite and have a nice day!

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

[molten owl]

at one point in a cirle how many tangents are possible ? it's 1 right??? 👀

[patent vine]

in?

or on?

[gentle shale]

If on, yes one tangent (one tangent per point, but infinite amount of tangents on the circle in total)

[molten owl]

on

[molten owl]

aight thenks

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[stark tulip]

stuck on IRC part

this too

<@&286206848099549185>

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[stark tulip]

<@&286206848099549185>

does the role even do anything..

[frank glen]

Yes

You just wait till someone helps

[stark tulip]

but does it really tag people

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[cyan elbow]

Hi I had a problem at math and I solved the big part I only need to prove that my algorithm is true. Here is the problem:

On a board are written 3 distinct natural numbers 2 by 2, we choose 2 numbers and replace one of them with the arithmetic mean of the 2 numbers, the next step is the same and so on. Prove that whatever the numbers originally written, there is a sequence of steps by which we get 2 equal numbers
Here is my algorithm:

In the problem says that we need to prove that if the numbers that we put ANY NUMBERS ON THE BOARD but the numbers that we make the arithmetic mean we chose, so based on my algorithm we have only 4 forms of groups with 3 numbers : (e-even number, o-odd number) 1. e,e,e 2.o,o,o, 3o,e,e, 4e,o,o. It dosen t matter how we mix the numbers, we can only have 2 or 3 same numbers (odd or even). So the rules are like this, we don t make the arithmetic mean on 2 different types on numbers, only same type (odd+odd or even+even NOT odd+even) so that we will always have a even number when we add them so the arithmetic mean will work. The result of the aritmetic mean can be or even or odd, if it is odd, then we swicth any of the 2 numbers that we made the arithmetic mean with and we will continue this algorithm till we will get 2 equal numbers. If we respect this rule we can make it. The other rules dose not exist bc with only this rule we can make it work. So I can play with the arithmetic mean however I want, till I will get that 2 equal numbers, so it will always work
How to porve it works?

[cyan elbow]

Using math

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[hardy lion]

how many weeks are in a year

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[fiery elbow]

Can someone explain how to convert from radians to values

[fiery elbow]

degrees*

i get that, pi = 18

180*

but how do i effectively convert back and forth

[plucky elk]

Use 1 = pi rad / 180 deg as a conversion factor

Same as you would use any other conversion

[fiery elbow]

whats rad?

radical?

ohh radian

so do i write in the rad?

like

when im actually converting

what would it be

instead of rad

what goes there i mean

?

lets say i have 45 degrees

and i need to convert it

so i would first try pi/4 ?

Okay, but when its a strange degree which is more confusing, is there a way to quickly find it?

yea okay

makes more sense

but what if its a strange degree

[plucky elk]

Memorize all the divisors of 180

[fiery elbow]

Whats the difference between divisors and factors

Okay, ill try

Thanks guys

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[tall pumice]

Calculus
So I'm trying to find the following's limit when x -> 0+
I figured I could just say:
lim(sin(x)) = 0

if a = 0:
f(x) = sin(x) / sqrt(x)
(And figure it out from there, I know how its just a bit long to type)

if a != 0:
lim(sqrt(x + a^2)) = lim(sqrt(a^2)) = A
so
lim(0 / A) = 0

But I was told it's not so simple. Is one of my statements here wrong?

[tall pumice]

As in, did I make an illegal move?

[pearl rock]

Seems right to me

[tall pumice]

Yeah?

[tall pumice]

Like, I can just break down the problems into smaller functions as long as everything is defined, right?

I can't remember if this ties to consecutiveness or not

[pearl rock]

Yup

[tall pumice]

Does this happen regardless of it being consecutive? As long as the limit is defined?
Intuition tells me I don't need the function to be consecutive but I can't recall

[pearl rock]

What do you mean by consecutive

[tall pumice]

Oh did I mistranslate it?

[pearl rock]

Seems so

[tall pumice]

Continuous* my apologies

[pearl rock]

Yes