help-39

[knotty prism]

This is the answer I got

My brain almost forgot what algebra was😭

[warm current]

Your algebra looks good

I'll trust that the calculation was right

[knotty prism]

Oke thank u for help

Is it okay if I can ask u one more question on related rates?

If ur unable to I understand

It's question number 3

.close

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[shrewd mist]

write r in terms of h, replace so you get V = f(h), differentiate and solve

[shrewd mist]

whoops

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[knotty prism]

.reopen

[shrewd mist]

.reopem

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[pearl pond]

✅

[knotty prism]

So u would basically solve for the rate of the radius?

[shrewd mist]

yeah, my bad, thought you wanted the rate of the height

same gist, write h in terms of r, and then replace h by that in the volume equation

differentiate everything wrt time and solve for dr/dt

[knotty prism]

So the equation would be
dh/dt = 1/3(3.14)(2.5)^2 (dr/dt)?

[shrewd mist]

you made a mistake in your derivative

[knotty prism]

Huh?

[warm current]

Sorry for missing your question

[shrewd mist]

Firstly, you want to get volume in terms of r, not height. This is because you are given dV/dt, but not dh/dt

[knotty prism]

It's oke bro ur good lol

[shrewd mist]

V = 1/3pir^2h, and h = 4r

subbing in gives you V = pi*r^3

[knotty prism]

Oooh that makes sense

Wait why would h=3r?

How did u get that?

[shrewd mist]

consider the triangular projection

lemme draw it out

[knotty prism]

excuse me!?!? What's the triangle projection?😭

Ooh wait nvm it just registered to my brain

[shrewd mist]

Basically as you fill the conical tank, the height and radius increase in tandem, looking at this from a horizontal angle (the "triangular" projection), you'll see a 2d triangle slowly growing. The height and base (radius) of this triangle is always proportionate

[knotty prism]

Oke I was low-key freaking out bc u said h=3r and I'm like where was the 3 from😭

[shrewd mist]

yeah, miswrote that haha, its 4r

[knotty prism]

Oke so the equation would be like this then?

[shrewd mist]

close, but you shouldn't plug in for r before getting the derivative

cause that'll just give you the volume at a given value of r, which isn't useful for you

[knotty prism]

Would I take the derivate of 1/3pi?

[shrewd mist]

No, the derivative of any constant (including pi) is zero. You would first write volume in terms of radius, so that would be V = 4/3 pi r^3. Afterwards, you would take the derivative of both sides of the equation with respect to time.

[knotty prism]

Ooh wait would it be 2= 1/3(3.14)(2.5)^2(4r)?

[shrewd mist]

Always take the derivative before plugging in any numbers (apart from cosntants liek 1/3 and pi) when doing a rr problem

[knotty prism]

So something like this then?

[shrewd mist]

Did you take the derivative (of both sides of the equation) before plugging in r = 2.5?

[knotty prism]

I low-key don't know if I did my brain is so slow rn

[shrewd mist]

You first need to find an equation for the volume in terms of the radius r

which you did correctly, V = 4/3 pi r^3

[knotty prism]

So would the equation be V= 4/3(3.14)(3r^2) before plugging in 2.5?

[shrewd mist]

close, chain rule.

Since you want to find the rate of change of the readius with respect to time, you want to differentiate both sides of that equation with respect to time

since both r and V are functions of time (they change as time passes, duh), the left hand side would become dV/dt (rate of change of volume with respect to time)

the right hand side will become (4/3)(pi)(3r^2)*dr/dt

[knotty prism]

Ooh yeah that rights

[shrewd mist]

the dr/dt appears because of the chain rule, since you're differentiating with respect to t, not r

[knotty prism]

Now I would plug in 2.5 for r and 2 for dv/dt and solve algebraically?

[shrewd mist]

yes, that would give you dr/dt, which is the rate of change of the radius when the radius is 2.5 and rate of change of volume is 2

[knotty prism]

Oke this is my final answer

Thank u for ur help!

Bye 👋

[shrewd mist]

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[tired tapir]

Prove that every countably infinite set has a well-ordering.

[tired tapir]

I have trouble to prove this problem.

[tropic saddle]

what does countably infinite mean

[tired tapir]

a countable set with infinite elements

I have a Theorem that: Let $\preq '$ be a well-ordering on the set $B$. Suppose that $h: A \rightarrow B$ is one-to-one. Define the relation $\preq$ on $A$ on by $x \preq y$ if and only if $h(x) \preq ' h(y)$, for all $x$ and $y$ in $A$. Then, $\preq$ is a well-ordering on $A$.

[jolly parrot]

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

[tired tapir]

gimme a sec..

Let $\preceq '$ be a well-ordering on the set $B$. Suppose that $h: A \rightarrow B$ is one-to-one. Define the relation $\preceq$ on $A$ on by $x \preceq y$ if and only if $h(x) \preceq ' h(y)$, for all $x$ and $y$ in $A$. Then, $\preceq$ is a well-ordering on $A$.

[jolly parrot]

Silly Cyn

[tired tapir]

This is a hint to prove that problem.

[tropic saddle]

what does it mean to be a countable set

[tired tapir]

A set A is countable if and only if there exists a one-to-one function $f: A \rightarrow \omega$.

[jolly parrot]

Silly Cyn

[proper dove]

So {1} is countable?

[tired tapir]

Since every finite sets are countable, yes.

[proper dove]

Oh

I mean yes

But we are talking about infinite now

[tired tapir]

true.

[proper dove]

So there is a function

Oh no

That isnt even nessesary with the form the hint is written in

[tired tapir]

Surprisingly, only a textbook provides that theorem as main hint to prove.

[proper dove]

So with this

[proper dove]

And this

Omega is a set of natural numbers?

[tired tapir]

yes

[proper dove]

Do you have a well ordering on that?

[tired tapir]

which do you mean?

omega one?

[proper dove]

Yes

[tired tapir]

it does.

[proper dove]

So can you see how can you use it here?

[tired tapir]

It is still vague..

[pearl pond]

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[tired tapir]

imma close this.

.close

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[burnt storm]

We learned statistics a while ago in maths, and I can never remember the stuff because I feel like I can’t logically understand it like I can with the other topics.

For example, why is it that the normal distribution should be e^-x^2 (ignoring the constants, I figured out where they come from)? I saw a cool 3Blue1Brown video about it, and while it was interesting, I don’t feel like it gave me a wholistic understanding of where it actually came from. There are also formulae to do with like the error of a proportion and mean that I don’t fully understand, but I’m guessing I’d be able to figure them out if I knew where the e^-x^2 came from. Any help is appreciated, or resources you’d recommend!

[pearl pond]

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@burnt storm Has your question been resolved?

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@burnt storm Has your question been resolved?

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@burnt storm Has your question been resolved?

[burnt storm]

<@&286206848099549185>

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@burnt storm Has your question been resolved?

[versed mica]

https://notarocketscientist.xyz/posts/2023-01-27-how-gauss-derived-the-normal-distribution/

[burnt storm]

Thank you I will have a look :D

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[pearl pond]

@agile ridge Has your question been resolved?

[midnight haven]

those who know

[main oxide]

rude

[warm current]

what is rude, exactly? I'm confused by what omega said, but not seeing how rude.

[main oxide]

we'll never know 😭

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[warm elbow]

ok literally

how do you not get an erf when u try to solve the LHS

okay

i give up

lol

someone help me tmr

❤️

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[coarse harbor]

Well you get erf but they're asking for an upper bound for that.
Lhs is 1 - erf(a), of course

[warm elbow]

ack

how are we supposed to know that without a calcualtor

or are we just supposed to recognize it

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[pearl pond]

@rustic laurel Has your question been resolved?

[rustic laurel]

<@&286206848099549185>

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[midnight haven]

I was having an argument saying there’s a pretty easy way to do this

Is this handwavey at all?

I’m legit getting brainrot because one party’s saying it doesn’t work and to say Ax=0 has a non-trivial solution (vector of all 1s)

For reference, I got asked this, I have the solution to the matter I’m just being told my way doesn’t work and it makes 0 sense to me at all how it doesn’t work

[fringe raft]

it does work absolutely, well I would say that the columns are linearly dependent since you added up the numbers in each row to be 0, so each entry of the column sum becomes 0

[midnight haven]

That’s exactly what I was thinking

[fringe raft]

if you would use the same permutation in every row, then summing the rows does not give 0, but summing the columns does

[midnight haven]

That works. Thank you for your help!

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[runic rampart]

can someone explain how they get x^5/2

or like how x^3/x^1/2 = x^5/2

[plush bramble]

,tex .exp rules

[jolly parrot]

riemann

[plush bramble]

3 = 6/2 and 6-1 = 5

[runic rampart]

this is using the quotient rule?

[plush bramble]

[runic rampart]

ah i see it now

thanks

.cllose

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[lean crane]

Hello I wanted to ask if this is correct

[lean crane]

Expand the function 𝑓(𝑥)=𝑒^(𝑥^2−1); 𝑥∈[0,1] in the neighborhood of 𝑥_0 = 0 into a fourth-order polynomial and estimate the maximum error between the function 𝑓(𝑥) and the fourth-order approximation polynomial using the Lagrangian remainder.

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<@&286206848099549185>

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[manic sparrow]

why is b negative

[brave sluice]

if you move up parallel to the y-axis from point b you go from level set 5 to level set 4, which is a decrease

[midnight haven]

Yeah I was gonna say look at reasoning you did in question below that

Or what is D_yf(b) equivalent to

[manic sparrow]

huh

gradient . (0,1)

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[shell harness]

how would i solve this?

when i find the divergence, and sum up all of the partial derrivatives, they equate to zero

<@&286206848099549185>

[pearl pond]

@shell harness Has your question been resolved?

[ionic kestrel]

,w divergence [x^2z,4-2xyz-3y+x^2y,3z-x^2z]

[jolly parrot]

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Requested by tymelord14

[ionic kestrel]

You appear to be correct on the divergence

The divergence theorem only applies to closed surfaces, the surface you are asked to integrate over appears to be open. If you can relate the surface in the question to an appropriate closed surface, then you may apply the div theorem

[shell harness]

how would I make the surface in question closed?

[ionic kestrel]

By attaching any surface which shares the same boundary as the original, there's an infinite # of ways to do this but for example you could consider the whole sphere instead of just the hemisphere with y>0

Most ways of choosing this surface will not make the computation easier, but there should be an 'obvious' surface to chose which does make the computation easier

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[split berry]

why did the restirction on the integral change?

[woven garnet]

Those are the bounds in terms of u

They can also be written as x=pi/4 and x=0

[split berry]

oh i forgot about that

that makes perfect sense thank you

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[rapid halo]

how do I solve this?

[pearl pond]

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<@&286206848099549185>

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[copper stratus]

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[thorny ridge]

hi

[pearl pond]

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[unkempt yacht]

do you have any question?

[prime bramble]

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[knotty geyser]

how do you do this question? (correct answer circled) how do you find function f

[buoyant sandal]

do you know the average value of an interval formula

[knotty geyser]

is it mean value theorem

f(a)-f(b)/a-b = f'(c) theres this formula

[buoyant sandal]

it's basically the same as the mean value theorem for integrals

not for derivatives

[knotty geyser]

is it this thing

i dont see how that can help me solve the problem though 😅

[buoyant sandal]

yes

so what's the avg value of the function over [2,4]?

(using the formula ofc, we already know its actually 3)

[knotty geyser]

do you just do this

[buoyant sandal]

great

now what's the area of the region they describe?

[knotty geyser]

ohh wait the integral is just the area

[buoyant sandal]

yep

[knotty geyser]

why did they say f(x)≥0 for all x in [2,4] in the question though (is that relevant to solving the problem)

[buoyant sandal]

it's probably just to remove any unnecessary complications with sign of area

if f(x)≥0 for all x in [2,4] then we know the area is greater than or equal to 0

and we don't have to worry about signed area of anything ig

[knotty geyser]

ohh ok thanks :)

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[vast crystal]

Hey I’m fish. Im 21 and trying to study up for my Picat verification test. I need help with some factoring problems, specifically
5x^2+20x+30

Im literally on the verge of tears please help

[marsh moss]

Heyo, first to remember is deep breaths, you got this!

are you familiar with what a discriminant is in math?

[spare lark]

You can factor out a 5 to start well

[vast crystal]

No I am not

[marsh moss]

That's alright! Here's the formulas
D is Discriminant, using it we can solve most problems where there is x^2

[vast crystal]

Very cool so leave brother

[vast crystal]

Alright

[marsh moss]

Sorry it's so tiny I'm not sure what happened with the formatting)

[vast crystal]

No its all good

Here lemme show you where im at first

[marsh moss]

Sure

[marsh moss]

Alright gotchu

[vast crystal]

Will this method not work for this equation?

[marsh moss]

The first thing we gotta do with 5x^2+20x+30 is equal it to 0

[vast crystal]

Alright, whats the most efficient way to do this? Imma start on a new piece of paper

[spare lark]

||this has no roots||

[marsh moss]

No, since we have x and x^2
Basically an equation that has two unknowns, X and x^2
This means we have to use either Discriminant or another method which doesn't always work, so I'll show you how to use D

[marsh moss]

The most efficient way doesn't always work, so it's best to practice with D a bit to know how to do it quickly

[vast crystal]

Alright, I’m listening

[marsh moss]

So we've got this now
5x^2+20x+30=0

Another thing worth mentioning, we always put a x^2 first, then x, then the number (when dealing with such problems, where there is X and x^2)
This is called the commutative property

We've already done that so we're good to go

If you already know which value is a b and c, then please write them out here so I know we're on the same page
If not, lmk and I'll explain how to find them

[vast crystal]

So the B value would be 20x and the C would be 30 yea?

[marsh moss]

Yep!

[vast crystal]

Awesome

[marsh moss]

And value a iss

[vast crystal]

5

[marsh moss]

Yep!

[vast crystal]

Or would it be 5x^2?

[marsh moss]

Getting to that part, the important thing is you already know where to look to find out what is a b and c

Going back to this

We see that D=b^2+4ac

[vast crystal]

Alright, and the D value is unknown for now?

Wait no

Its 0

[marsh moss]

When dealing with a b and c, we only take the actual number, not x
So for example we have b=20

(If this was a case where nothing was next to x, not to worry, we just use/know that one is next to x; but this does not apply to our problem as each value has a number next to it)

[marsh moss]

Yes, we're looking for D

[vast crystal]

So the end bit of that equation would come out like
0=20x^2+4•5•30

Fucking asterisks bleh

[marsh moss]

No need to rush!

This is what we've got for now

Right now we're at putting in values (numbers) instead of a b and c

[vast crystal]

Oh! Okay so the equation we have no is literally just like that

[marsh moss]

Now we know that when taking a b and c, we don't take x, just the actual number

So what would we take for b again?

[vast crystal]

20

[marsh moss]

Yes!

One minute, gonna send a picture)

[vast crystal]

Okk

[marsh moss]

-4 is part of the formula, so we just rewrite it

We've got to find out what value a and c are

What does a=
And
c=

[vast crystal]

A is 5 and C would be 30

[marsh moss]

Yep!

[vast crystal]

And that would be multiplying the terms?

So

[vast crystal]

Yesyesyes

Okay so im not clueless, at least theres that

[marsh moss]

Yess, so we have to calculate 20^2 on its own, and then 4•5•30

What is 20^2?

[deep bison]

Or devide everything by five first

[vast crystal]

400

[marsh moss]

Yes, there are many ways of solving equations, this is the one I decided to show!)

[deep bison]

Ah sorry

[marsh moss]

Yep, and 4•5•30?

[marsh moss]

No worries, thanks for looking out)

[vast crystal]

600

[marsh moss]

Yep

[vast crystal]

So we wouldd shorten that to just 600?

D=20^2-4•600?

[vast crystal]

Ohh okay

[marsh moss]

Hopefully this makes sense

We've just calculated 20^2, and 4•5•30

The minus stayed from the formula of D

So what does D equal from here?

[vast crystal]

Yes yes, I understand where we’re getting that from

200

So our D value is 200

Sorry

-200?

[marsh moss]

No need to apologise

[marsh moss]

Yep

[vast crystal]

Okay, awsome

[marsh moss]

Now going back to those formulas

x1 and x2

Can we find them knowing what D is?

[vast crystal]

Lemme jot it down and find it, one moment please

[marsh moss]

Well, just a moment^^

We see that the formulas for x both require the square root of D

And our D is..

[vast crystal]

14?

14.14

[marsh moss]

Let's go back to here

D=400-600

So D=

[vast crystal]

-200

[marsh moss]

Yep

Is that less or more than 0?

[vast crystal]

Less than

So would we have to make it a positive?

200 flat?

[marsh moss]

Mhm! So we know that our D is less than 0
When D turns out to be less than 0, the problem has no solutions, since we cannot find a normal root for a negative number

And we need the square root of D to find x1 and X2, which we cannot do (get the square root)

So in the answer, we'd write like so:

This symbol means that there are no solutions

Sometimes yes, there are problems that don't have solutions!

If this was a bit confusing, we could try using the other method, but we'll end up with the same outcome (no solutions)

[vast crystal]

Well

There wont be an option for no solution on my test, I took the practice and the answer option that it shown as correct came out as

5(x^2+4x+6)

[marsh moss]

Ah I see
Our job was to simplify?

What is it that we have to do with x^2+20x+30

[vast crystal]

I’m guessing it was to simplify 😦 I’ve been told by like 20 people I have to factor the equation to get the answer and ive been so lost

No wonder I couldnt figure it out

[marsh moss]

Alright, so you don't get how we got this, right?

[vast crystal]

Not at all

Clueless on that part

[marsh moss]

Let's look at our equation
5x^2+20x+30

This part is a bit tricky
What do 5 20 and 30 have in common
(Clue: the smallest number apart from 1 they're divisible by is..)

[vast crystal]

2

The GCF

[marsh moss]

we've got to make sure it's a whole number
So if we divide 5:2=2,5 (not a whole number)

[vast crystal]

Oh shit yea youre right

Well then it would be 5 correct? Since 5 isnt a prime number

[marsh moss]

Yep!

From here, we can put 5 in front the brackets
Doing so means everything in the brackets has to be multiplied by what's outside of the brackets

[vast crystal]

Alright so

Its the smallest GCF and then we start the brackets

Gotcha

[marsh moss]

So everything inside the bracket has to be divided by 5)

5x^2:5

What do we get?

[vast crystal]

25x:5=5

Because we still have to multiply 5x by itself right?

[marsh moss]

Remember the ^2 only applied to x

[vast crystal]

Oh alright so

Its 1?

[marsh moss]

Yep

[vast crystal]

Okay cool

Lemme make sure im on the same page rnn

[marsh moss]

So 5x^2:5=
Basically if we take away 5 from both sides what are we left with

[marsh moss]

Everything is correct apart from 5(5x^2)

[vast crystal]

Hm okay so can we rewind a second. Im a little lost

[marsh moss]

How would we go about solving this?

[vast crystal]

We’d just divide 5 by itself

[marsh moss]

Yep
So we realise that all the numbers are divisible by 5, meaning that's why we put 5 in front of the bracket

[vast crystal]

And thats part of the answer and NOT what we’re currently calculating

5(

[marsh moss]

Now that we did that we have to divide everything by 5
So we have 5x^2 which we need to divide by 5
(See picture I attached)

[vast crystal]

Ohhhhh

[marsh moss]

Yes! That leaves us with...
(If we completely remove 5 from what I wrote we are left with...)

[vast crystal]

Okay I see what you mean

It leaves us with 1

I’m going in the complete wrong direction arent I?

[vast crystal]

X^2?

Okay cool

[marsh moss]

It leaves us with 1x^2
We usually don't write the one, since we know that multiplying anything by one just gives us the same value

[vast crystal]

So itd look like this now

5x^2:5=1x^2

[vast crystal]

Which would mean

[marsh moss]

Yep
And by putting it into our bracket it looks like so (since we just take away the one):

[vast crystal]

Oh alright yea

[marsh moss]

Onto our next value

20x

What do we have to do with it?

[vast crystal]

Divide it by 5

[marsh moss]

Yep!

[vast crystal]

4x

[marsh moss]

Yess

[vast crystal]

Lemme take some notes, I’ll read the next but momentarily

[marsh moss]

Alright no worries!

And our last number in the bracket would be...

[vast crystal]

6!!

[marsh moss]

Yep^^

Hopefully that explains how we simplified to this answer

[vast crystal]

It does! A lot actually

[marsh moss]

Glad I could help then

[vast crystal]

And will this method work with every equation I have to simplify?

[marsh moss]

Yep! You just have to figure out what all the numbers/values have in common

[vast crystal]

And if they dont all have something in common then it is unsolvable! Awesome thank you so much

[marsh moss]

Well you've got to be careful with that one
I'll set an example one sec

[vast crystal]

Okay

[marsh moss]

Let me know if there's a part that confuses you here

What I mean to show by this, is that it doesn't always have to be a number
It could be a value, such as x
In this case, x^2

[vast crystal]

Hm

Lemme look it over

Okay so

X would be set to 1x^2

Since you dont show the number when you have a lone variable

So it would be 1x^2•3=3x2

My A value would be x^2

[marsh moss]

=3x^2 yes

[vast crystal]

B would be 3

But c wouldnt be 3x^2

[marsh moss]

What are you trying to solve it by?

[vast crystal]

The same method you showed me, so that way I can understand why it wont work

[marsh moss]

We can't use Discriminant here, since we don't have a 'c' value

This is just for simplification

Not solving/finding solutions/roots

[vast crystal]

Okay, so how would you solve the third equation? That seems the easiest to solve

Simplify *

[marsh moss]

Oh those aren't the equations they're just steps to show how I solved what's next to 1)
So sorry if that caused confusion!

[vast crystal]

Ohhh

Okay

[marsh moss]

1. is that base equation (first step, rewriting the equation)
2. second step (putting what's in common in front of the brackets)
3. third step (calculating)

[vast crystal]

Would you be opposed to a voice chat? I feel like it might be easier for me to grasp some of this if I heard it

[marsh moss]

I'm not opposed to it however Im not sure for how long I can chat

[vast crystal]

Thats okay! Id just need a little to understand

[marsh moss]

Alright one second

[vast crystal]

Take your time I’m around

[marsh moss]

You can use .close to close this channel for now)

[vast crystal]

.close

[pearl pond]

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[vast crystal]

.reopen

[pearl pond]

✅

[vast crystal]

Okay so we back, were struggling we absolutely braindead

Gotta solve linear equations, and the one im working on now is
5n-3=7n-5

If anyone can help me please ping me. Voice calling preferably, but text will work too

I dont know where to even start, and I know if I look it up on google it will just tell me how and not explain how and why to do it that way

[marsh moss]

could also help here but i only got abt 7 mins

[vast crystal]

Nono, youre all good!! You did plenty for me and I appreciate your time so much

You get some rest or get ready to go to work or whatever you gotta do

[pearl pond]

@vast crystal Has your question been resolved?

[vast crystal]

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[rapid halo]

can someone help me how did they get that last statement ?

[pearl pond]

@rapid halo Has your question been resolved?

[rapid halo]

<@&286206848099549185>

[fallen forge]

man, x^6 gives you 1. revise complex numbers

[rapid halo]

no ik that but like where are the rest of the terms and how did they take the 6 common

[compact ridge]

from this 3B1B video btw, https://www.youtube.com/watch?v=bOXCLR3Wric

[rapid halo]

thank you so much

r.close

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[torn sky]

So I have trouble understanding this inequality

(6x-x^2) >= 0

I tried to factor it as

x(6-x) >= 0

and then proceeded to draw this to see how I can control x, but the book says that the left side is positive when x is an element in [0, 6], and not the other way like I did

[torn sky]

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[agile ridge]

Why AB and BA are similar if one of them is invertible? Is there any intuitively explanation? Why is it only "one of them" instead of " Both of them"?

[fringe raft]

recall the definition of similarity

[agile ridge]

I know the definition🥲

[fringe raft]

then which one is it?

[agile ridge]

Hmm wdym which one?

[fringe raft]

S^-1 P S = Q?

[agile ridge]

Yes

[fringe raft]

ok now looking at AB and BA

$S^{-1}ABS = BA$

[jolly parrot]

rbit

[fringe raft]

suppose B is invertible, what would you choose for S?

[agile ridge]

$B^{-1}S^{-1}ABS= A$

[fringe raft]

is that if A is invertible?

[jolly parrot]

Cryolite

[fringe raft]

but no I mean, choose some value for S so that the equation holds

[agile ridge]

Ooh uhm S might be B^-1

[fringe raft]

right

then it cancels the B on the right and introduces a B on the left

and if A is invertible instead?

[agile ridge]

S will be A^-1

[fringe raft]

almost

[agile ridge]

$S^{-1}ABSA^{-1} = B$

[jolly parrot]

Cryolite

[agile ridge]

Ok S will be A

[fringe raft]

right

so in either case, be it A invertible or B invertible, you can satisfy the equation

if both of them are invertible, it doesn't really matter which one you choose

[agile ridge]

Btw is there any geometric interpretation for this maybe?

@fringe raft where did you go😭

[fringe raft]

not that I can think of rn

[agile ridge]

Ok I see, thanks!!❤️❤️

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[jolly parrot]

prograce

[inland ivy]

!xy

[pearl pond]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

@rapid wagon Has your question been resolved?

[rapid wagon]

It's not in English

"Prove using the definition of boundedness and limit only that the sequence $a_{n}=n^{((-1)^n)}$ is not bounded and yet does not approach an infinite limit"
Is the exact translation

[tropic saddle]

but what even is a_n

give an image of the original problem

[rapid wagon]

Oh omg

[jolly parrot]

prograce

[rapid wagon]

Mb

I edited it

[pearl pond]

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[rapid wagon]

Omg

Anyeays I proved it doesn't converge to infinity I'm struggling wjth boundness please

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[magic knot]

I'm not 100% clear on what they're asking me to do, could somebody possibly walk me through the first row in the table? :o

[tulip marsh]

in the text, the 2 solutions of f'(x)=0 are calculated.

Those points are where the function f changes monotony.

[magic knot]

oo right, I remember the professor talking abt that

[tulip marsh]

The points are 5 and 9; that means between -inf and 5, for example, function f has same monotony.

either increasing or decreasing.

f is increasing if and only if f' is positive on that interval.

decreasing <=> f' < 0 on interval.

you take an x within interval; compute f' (x); see sign (+ or -); say the behaviour.

since there's 2 roots, there's 3 intervals. if there were n roots, there would be ... intervals

[magic knot]

aaaa ok, I can work with that. Thank you :)

[tulip marsh]

You're welcome.

[magic knot]

I shall close the channel now :D

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[harsh nymph]

any idea on what method to do this?

[strong relic]

whem in doubt use l hopital

[harsh nymph]

my school doesnt allow lhopital 😭

[strong relic]

um

try a substitution to get rid of all the roots

and see if you can factor amything

[harsh nymph]

alright

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[stiff mauve]

im struggling with normal distribution

[stiff mauve]

question 52b

[plush bramble]

3 + 1/3 is not the same as 3.3

,calc 3+1/3

[jolly parrot]

Result:

3.3333333333333

[stiff mauve]

omg 🤦‍♂️

thanks😭

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[river lodge]

Solve the trigonometric equation sin(3x) - sin(6x) = 0 within the domain of [0, 2pi]
can someone help me w this
i got to

3sinxcosx - 6sinxcosx = 0
i can factor 3sinxcosx

3sinxcosx(1-2)
so u left with -3sinxcosx
what exactly do u do next

[tulip marsh]

well, if that's the result you got, then that happens when either sinx or cosx is 0.

since it's a product

-3sinxcosx=0 if and only if sinx=0 or cosx=0 (since -3 is definitely not 0)

although I am not 100% sure the result should look like that...

did you write sin(3x) as sin(2x+x), used the formula, and switched every sin(2x) and cos(2x) with the correct formulas?

[river lodge]

oh

isnt sin(3x) = 3sincos

💀

[tulip marsh]

afraid not. sin(2x)=2sinxcosx, but that's simply because of the formula sin(x+y)

[river lodge]

oh right

[tulip marsh]

it doesn't happen to look exactly same for sin(3x)

but it can be deduced.

eventually you should be able to express sin(3x) and sin(6x) using only sinx and cosx.

[river lodge]

so for sin(6x) i need to do sin (3x+3x)

[tulip marsh]

yea, that's the quickest approach, since you'll already find formula for sin(3x)

will also need for cos(3x) before sin(3x+3x)

sin(a+b)=sin(a)cos(b)+sin(b)cos(a)

[river lodge]

yea

alr

i see

[tulip marsh]

cos(a+b)=cos(a)cos(b)-sin(a)sin(b)

[river lodge]

thanks for ur help

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[spring crystal]

So for i1 I make y and z all the opposite signs?

And i2 all y and z the opposite sign?

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[split crag]

Is this the correct notation?

[pine jay]

Yes

[split crag]

thank you

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[olive arch]

guys how is 7pi/5 - pi = to 2pi/5

[sharp vigil]

pi = 5pi/5

[nimble osprey]

pi=pie

[regal herald]

3=6

[olive arch]

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[spring crystal]

Why mine wrong

[plush bramble]

did you substitute back in u = e^x

[strong relic]

and its 1/2 u not 1/2 x

[brave sluice]

how did you get rid of cos^2?

[strong relic]

its correct, its cosine double angle

[spring crystal]

Ok so e^x + sin(e^x) all divided by 2

?

[strong relic]

yup

[spring crystal]

Thanks

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[gusty prism]

heres my work:

thnxs ):

like even if i converted that answer to kj its still wrong 😦

the answer issupposed to be 79.8 kj/mol

[mossy halo]

can you rotate and send?

[gusty prism]

Ok sry about that

[pearl pond]

@gusty prism Has your question been resolved?

[gusty prism]

😭

[lucid anvil]

you calculated the energy of a single photon

to get the energy of a mole of photons, you have to multiply by avogadro's number

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[stoic imp]

If a number ends in zeros, the zeros are called terminal zeros. For example, 520,000 has four terminal zeros, but 502,000 has just three terminal zeros. Let $N$ equal the product of all counting numbers from 1 through 20:$$
N=1\times2\times3\times4\times\dots\times20.
$$How many terminal zeros will $N$ have when it is written in standard form?

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[summer imp]

So you’re given lots of factors of N.
What kind of factors do you expect might give a 0 at the end?

[stoic imp]

the ones that are divisible by 10?

I might be tripping

[versed stirrup]

hi guys, Could someone help me with this exercise?

[pearl pond]

@stoic imp Has your question been resolved?

[brave sluice]

yeah maybe find the prime factorization of 20!

[stoic imp]

how?

[brave sluice]

okay... that's one way to do it

lol

[stoic imp]

help me raphaelo

[brave sluice]

i mean sure, why not?

oh

[stoic imp]

donatello

[brave sluice]

well let's see, we can write
1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*17*18*19*20

[stoic imp]

ye

Agreed

[brave sluice]

actually it will be pretty annoying to find the whole prime factorization

but we only need to collect factors of 2 and 5

[stoic imp]

how to do it

[brave sluice]

how many factors of 2 in all these numbers?

how many in 1?

how many in 2? in 3?

[stoic imp]

in 1, there are zero factors of 2

[brave sluice]

yes

[stoic imp]

in 2, only one

[brave sluice]

yes

[stoic imp]

in 3, zero factors of 2

in 4, two factors of 2

in 5, zero factors of 2

in 6 one factor of 2

in 7 zero factors of 2

in 8 three factors of 2

[brave sluice]

yes

how many in total?

[stoic imp]

do I need to count up to 20?

[brave sluice]

or use wolfram

[stoic imp]

in 9 zero factors of 2

in 10 one factors of 2

in 11 zero factors of 2

in 12 two factors of 2

in 13 zero factors of 2

in 14 one factors of 2

in 15 zero factors of 2

in 16 four factors of 2

in 17 zero factors of 2

in 18 one factors of 2

in 19 zero factors of 2

in 20 two factors of 2

[brave sluice]

there are 2 in 20

[stoic imp]

my badd

I messed up the last one, sorry raffaelo

[brave sluice]

i'm raphael haha

anyway i counted 18 in total

[stoic imp]

,calc 0 + 2 + 0 + 1 + 0 + 3 + 0 + 1 + 0 + 2 + 0+ 1 + 0+ 4+0+ 1 + 0 + 2

[jolly parrot]

Result:

17

[stoic imp]

u sure?

how can we check?

[brave sluice]

i think you missed the first two

[stoic imp]

yes, my badd

so 18 factors of 2

[brave sluice]

yeah phew

and how many factors of 5?

this will be quicker

[stoic imp]

1 has 0 factors of 5

2 has 0 factors of 5

[brave sluice]

they will all have 0 except 5, 10, 15, and 20

[stoic imp]

okay, good

[brave sluice]

how many in 5, 10, 15, and 20?

[stoic imp]

5 has 1 factor of 5

10 has 1 factor of 5

15 has 1 factor of 5

20 has 1 factor of 5

[brave sluice]

four in total

so 20! = 2^18 * 5^4 * x
with some number x that isn't divisible by 2 or 5

[stoic imp]

,calc 2^18 * 5^4

[jolly parrot]

Result:

1.6384e+8

[brave sluice]

so 20! = 10^4 * y
with some number y that isn't divisible by 10

[stoic imp]

?

[brave sluice]

$20!=2^{18}5^{4}x=2^{4}5^{4}2^{14}x=(2\cdot5)^{4}y=10^{4}y$

[jolly parrot]

Axe

[stoic imp]

where did 2^14 go

[brave sluice]

y ate it

[stoic imp]

[brave sluice]

there are no factors of 5 in y

so y isn't divisible by 10

[stoic imp]

ok

,calc 10^4

[jolly parrot]

Result:

10000

[brave sluice]

yes so there are 4 zeros

[stoic imp]

[brave sluice]

👍

[stoic imp]

.solved

[pearl pond]

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[pearl pond]

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[spring crystal]

Am I doing this right so far

Separable differential equations

[plush bramble]

Yea just keep solving

[solemn marsh]

Yep, just make sure you add the +C *right after *the integration

[spring crystal]

? I have divided by 0

[plush bramble]

1/arctan does not equal arccot

Use arctan arcsin and arccos from now on so you didn't confuse yourself

[spring crystal]

I did the reciprocal

[plush bramble]

Yes I see. 1/thing is reciprocal of thing

[spring crystal]

So where I do that

[plush bramble]

Oh now that I zoomed in it's more legible

Looked like cot instead of tan

[spring crystal]

So mine is right?

Y= -1/arctanx +c

@plush bramble

[plush bramble]

You're missing the +C here

[plush bramble]

You didn't do this appropriately

[spring crystal]

i added it after

but when im doing y(0) im getting 1=undefined

[plush bramble]

.

Fix that before evaluating

[spring crystal]

i added the step after that

[plush bramble]

That's incorrect then

[spring crystal]

?? what

[plush bramble]

Read this more carefully

[spring crystal]

ok so its -c

[plush bramble]

What is "it"

[spring crystal]

instead of +c

y=-(1/tan^-1(x)) - c

?

[plush bramble]

You need parentheses

[spring crystal]

now what

im still getting -1/0

[plush bramble]

Then you're not doing this

[spring crystal]

bruh i dont get it

[plush bramble]

That's not correct

This was your last correct step

Just plug in the initial condition there

[spring crystal]

whats wrong with the recriprocal

also i thought i had to get y by itself so i did that

[plush bramble]

3 = 2+1. Does 1/3 = 1/2 +1?

,calc 1/2+1

[jolly parrot]

Result:

1.5

[plush bramble]

Use parentheses to fix this

[plush bramble]

And the identically placed parentheses to fix this

[plush bramble]

You can do that after finding C

[spring crystal]

im confused what u mean

[plush bramble]

Did you understand this

[spring crystal]

so i change c to 1/c?

[plush bramble]

No

[plush bramble]

Do this

[spring crystal]

ok its y=-1/(arctanx+c)

so its 1= -1/c?

c=-1

so the answer is y= -1/(arctanx -1)

??

[plush bramble]

,w solve dy/dx = y^2/(1+x^2), y(0)=1

[spring crystal]

Is that the same

[jolly parrot]

Results provided by WolframAlpha

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

Requested by rie.mann

[spring crystal]

Mines different answer?

@plush bramble

[compact ridge]

they're the same actually!

try multiplying top and bottom by -1

(-1)/(-1) = 1 so you are multiplying by 1

[plush bramble]

Just because functions look different, doesn't mean they are

Use plotting software to check your answers if you're not familiar with the algebra

[compact ridge]

Desmos and GeoGebra my beloved

[compact ridge]

but this appears really really often

[pearl pond]

@spring crystal Has your question been resolved?

[pearl pond]

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[lusty pier]

Hi do i have any errors in my calculations?

[plush bramble]

I don't see how anyone can answer your question without much more context

[lusty pier]

Oh like in my conversions - are my conversions good.

im getting an error in sig figs i think

i mean scientific notation

[fiery charm]

i think wolfram or something of that sort would do you better.

[pearl pond]

@lusty pier Has your question been resolved?

[pearl pond]

@lusty pier Has your question been resolved?

[pearl pond]

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[livid spire]

Equilateral triangle ABC, AH is perpendicular to BC at H. M is a point in ABC. MP, MQ MR is perpendicular to BC, AC, AB at P, Q,R. Proves that MP + MQ + MR = AH.

[livid spire]

I don't have an idea on how to do it.

[strong relic]

this is vivianis theorem

[latent fjord]

i have an idea but im not sure if its the solution

[strong relic]

therss many proofs online

[latent fjord]

nevermind you can do vivianis theorem then

[strong relic]

one way is by segmenting the large trisngle into 3 smaller ones, with EQ ER EP as their heights

[livid spire]

I think I got it

.close

[pearl pond]

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[pearl pond]

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[fierce snow]

could someone explain how to do this question ?

[fierce snow]

i did part a and b im having trouble with the last two

[pearl pond]

@fierce snow Has your question been resolved?

[fierce snow]

.close

[pearl pond]

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[sharp smelt]

Let $a<b$ be real numbers, and consider $T= \Q \cap [a,b]$. Prove $sup(T)=b$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

One way would probably be to do cases

if b is rational and if b is not rational

[cunning comet]

why should you need cases?

[spare lark]

Q do not verify sup propriety

[cunning comet]

what is the sup propriety?

[sharp smelt]

I know, I mean if b is in Q than the largest element in the intersection is b

[sharp smelt]

That was just a thought

[spare lark]

Least upperbound proprety

[sharp smelt]

So Here I just attempt to use the defn of the supremum ?

[fringe raft]

yes

[cunning comet]

rationals dense in reals.

[sharp smelt]

noted

planning on using that here

thanks

$Sup(T) <b \implies x<sup(T) ,\forall x \in \Q \cap [a,b] < sup(T)$.However, as the rationals are dense in $\R$, , there exists a rational, between $sup(T) and b$ , which would lie in the intersection. Therefore $sup(T) \geq b$.

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

Not sure what to do from here

I mean I know I now have to show all upper bounds are more than 2

just not too sure of how to do that

<@&286206848099549185>

[fringe raft]

did you show is an upper bound yet?

[sharp smelt]

yeah, here

[fringe raft]

that doesn't show b is an upper bound

[sharp smelt]

hmm

[fringe raft]

don't think too complicated about it

[cunning comet]

b is by definition an upper bound.

[sharp smelt]

hmm, okay

fair

So I now show if if there's any other upper bound, it's bigger

[fringe raft]

yes

[mint wave]

using this idea, you can also do a contradiction. assume that b is not a supremum and that there is a different supremum, say c. then c<b. then use the density to show that there is a number in between c and b, contradicting that fact that c is a supremum

[sharp smelt]

oh yeah, forgot that

Let $c$ be the suprmeum, $c<b$, but by the density of rationals, a rational can be inserted between $c$ and $b$ , but as this rational is less than $b$, it's in the interesection, contradicting our assumption that $c$ is the supremum, thus there is no upper bound less than $b$, and b is teh supremum

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[cunning comet]

this.

[sharp smelt]

I used that

did I not

[pearl pond]

@sharp smelt Has your question been resolved?

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[stoic imp]

\textbf{3.-} Find all $z \in \mathbb{C}$ such that
$64z^3 = (1 + \sqrt{3}i)^7 \overline{z}^2$.

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[hallow remnant]

Do you know angle and magnitude?

[stoic imp]

angle is argument

[hallow remnant]

ok so suppose $z=(r,\theta)$

[jolly parrot]

Arnavutköy

[hallow remnant]

then can you tell me what $z^3$ is?

[jolly parrot]

Arnavutköy

[hallow remnant]

in terms of $r$ and $\theta$

[jolly parrot]

Arnavutköy

[stoic imp]

we prolly need to use this for that, no?

[hallow remnant]

yes

[stoic imp]

,, z^3 = |z|^3 \left(\cos(3\theta) + i\sin(3\theta)\right)

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

,, z = a + bi \ |z| = \sqrt{a^2 + b^2}

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

right? or am I missing something?

[hallow remnant]

yes

i will use coordinates to represent this to keep track

so if $z=(r,\theta)$, where $r$ represents magnitude

[jolly parrot]

Arnavutköy

[hallow remnant]

Then, $z^3=(r^3,3\theta)$, right?

[jolly parrot]

Arnavutköy

[stoic imp]

sure

[hallow remnant]

And $64z^3=(64r^3,3\theta)$

[jolly parrot]

Arnavutköy

[hallow remnant]

Btw using these parenthesis is called polar coordinates and is considered common notation

Okay, can you tell me $(1+\sqrt{3}i)$ in terms of angle and magnitude?

[jolly parrot]

Arnavutköy

[stoic imp]

magnitude of that is $\sqrt{1^2 + 3} = 2$

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

the argument, idk

[pure rapids]

Think of inverse tan

[stoic imp]

,, \theta = \arctan\left(\frac{\sqrt{3}}{1}\right)

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

ty

[pure rapids]

Np

[stoic imp]

Re(1 + sqrt(3)i) = 1
Im(1 + sqrt(3)i) = sqrt(3)
both Real and Imaginary parts are in first quadrant, so this is correct

[hallow remnant]

do you know what $\arctan(\sqrt{3})$ is?

[jolly parrot]

Arnavutköy

[stoic imp]

π/3 + 2kπ with k in Z