help-13

[sand kayak]

i got 111

[radiant topaz]

562.5

[sand kayak]

your answer is wrong too

[radiant topaz]

It can't be 111 cuz its increasing

[radiant topaz]

What's the answer

[sand kayak]

535

makes sense why i didnt understand your solution

[radiant topaz]

Ok let me see

[radiant topaz]

Wdym

Your understanding has nothing to do with the solution

[sand kayak]

because your solution doesnt make sense

[radiant topaz]

To you perhaps

My solution is close to the answer for a reason and the case could even be that given answer is wrong

,w 250(1.5)^(15/8)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by it's Sam#5028

[radiant topaz]

See this

Either you translated bad or the question doesn't make sense

[sand kayak]

i translated with google translate

[radiant topaz]

250 is initial amount at t=0

Then this will be the case

By writing just one year it doesn't help

[sand kayak]

this is the exact translation

it says 1 year and 8 years

[radiant topaz]

That needs a modification as to start of year end of year or what

Year 1 to year 8 is 7 year period so it needs to be
Start of year 1 and end of year 8 in the statement to get the right answer

Anyway just continue over this you get a⁸ = 1.5
And you want to find 250 a¹⁵ = 250 (a⁸)^(15/8)

You get 535

[cedar kiln]

@sand kayak Has your question been resolved?

Channel closed

Closed by @sand kayak

Use .reopen if this was a mistake.

[sand kayak]

how do i solve y=220-200e^-kx

which one of them is a fctor

i dont know how to get rid of the 220-200

[dense wing]

wdym solve

[radiant topaz]

Factor to what?

[dense wing]

just isolate for e^(-kx)...

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sand kayak]

how

[dense wing]

well one, what's the actual question?

[bold vine]

^^

[dense wing]

cause you can't solve anything in a function

[radiant topaz]

^^

[dense wing]

unless you have a 2nd function

[sand kayak]

after 15 mins the temp is 53c. decide k

[dense wing]

???

[radiant topaz]

This is something about exponential growth

[sand kayak]

y=220-200*e^(-kx)

[dense wing]

Post the question verbatum, we're not mind readers

[radiant topaz]

What's initial temp koki

[sand kayak]

i posted the question

[dense wing]

You did not

[sand kayak]

just look up

[radiant topaz]

You did not

[dense wing]

what's y and what's x?

[radiant topaz]

This is your question not the book question

[sand kayak]

this is the whole question.. to decide k

[dense wing]

it isnt

[radiant topaz]

Ofc not

[dense wing]

cause we dont know what y and x are

[bold vine]

e^(-kt)?

[sand kayak]

y= c degrees x= minutes in the own

[dense wing]

wow, finally

[bold vine]

There we go

[radiant topaz]

I get one thing that it's exponential/continuous growth but full question

[dense wing]

so x=15 and y=53

[bold vine]

Please send the ORIGINAL problem whenever you're asking a question

The server really needs to put this in #❓how-to-get-help

[small bramble]

good evening

is this room free?

[dense wing]

clearly not

[radiant topaz]

What's with 220 and 200?

[sand kayak]

no idea

[radiant topaz]

Ok so finding k

I see now

[dense wing]

anyway... just sub in the values then isolate for k.

[radiant topaz]

Yeah

[sand kayak]

220-200e^-k15 =53

how do i get rid of the 220 and 200

[bold vine]

Basic algebra

[crimson sedge]

hello guys

i need help

[sand kayak]

i dont know which to do first

[crimson sedge]

with math

[bold vine]

@crimson sedge Not know

[crimson sedge]

please guys

i need help

[bold vine]

WSG

#❓how-to-get-help

Please.

@sand kayak $220 - 200e^{-15k} = 53$

[wraith dagger]

Shen

[bold vine]

Start by subtracting both sides by 220, and then dividing by -200

A natural log and another division step will be involved

[sand kayak]

okay that was what i wanted to know

[bold vine]

It's algebra

[sand kayak]

i got k=0,0120

[radiant topaz]

Ok

[bold vine]

Alright, if you're done, close the channel

[sand kayak]

i am not

i need to derive y=220-200*e^(-0,0120x)

[radiant topaz]

Yes thats the next step and it's over

You just replace k with its value

[sand kayak]

y'=220+2.4e^(-0,0120x)

is this correctly derived

[radiant topaz]

220 won't be there

[sand kayak]

Is this correctly derived?

[bold vine]

Forgot the constant of 9 in front

[sand kayak]

How do i know whats a constant and whats a factor

[radiant topaz]

You can't write y' on 2nd line

Constant is number or a letter used to denote constant

[cedar kiln]

@sand kayak Has your question been resolved?

[sand kayak]

how do i derive b)

derive y=Ca^x in regards to

b) a and decide dy/dx

(C and x are konstants)

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[sand kayak]

.reopen

[cedar kiln]

✅

[sand kayak]

i got y'=C*ax^(a-1) but its wrong

<@&286206848099549185>

[cedar kiln]

@sand kayak Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[sand kayak]

.reopen

[cedar kiln]

✅

[sand kayak]

i need to decide f'(4)

but i get 1 as the answer

[rugged pond]

How did you get to that answer?

[rugged pond]

(Btw if it's easier you could take it in Swedish with me)

[sand kayak]

this makes no sense

because the top becomes 0

no matter what

it happens all the time i do the central differenskvot the top becomes always 0 to me

[rugged pond]

Well your function $f$ is given by
%
$$f(x)=\frac{200x}{x^2+1}$$
%
and you wanted to approximate the derivative $f'(4)$ using a central difference quotient. What that means is that your approximation will be given by
%
$$f'(4)\approx\frac{f(4+h)-f(4-h)}{2h}$$
%
correct?

[wraith dagger]

Lorago

[sand kayak]

what

its f(a+h)-f(a-h)/2h

[rugged pond]

Yes

[sand kayak]

f(x) to me is f(a)

[rugged pond]

And you're doing it at $4$

[wraith dagger]

Lorago

[sand kayak]

so i need to put in 4 first?

[rugged pond]

Well the general formula would be
%
$$f'(x)\approx\frac{f(x+h)-f(x-h)}{2h}$$

[wraith dagger]

Lorago

[rugged pond]

But you can plug in x=4 at any step here

[sand kayak]

what did i do wrong in my original equation

i dont get it

[sand kayak]

i did exactly just that

put in 200x/x^2+1 as f(x)

f(a) for that matter

and then with the h

[rugged pond]

Also the reason you're getting $0$ is because what you're writing is
%
$$\frac{f(x)+h-f(x)-h}{2h}$$
%
and if we just look at this we can simplify it as
%
$$\frac{f(x)+h-f(x)-h}{2h}=\frac{(f(x)-f(x))+(h-h)}{2h}=\frac{0+0}{2h}=0$$

[wraith dagger]

Lorago

[rugged pond]

So this will always give you 0 in the numerator regardless of your function

The important thing is that

$$f(x+h)\neq f(x)+h$$

[wraith dagger]

Lorago

[sand kayak]

what

so what am i supposed to do

i dont get it

[rugged pond]

Notice the big difference here. In your case you have:
%
$$f(x+h)=\frac{200(x+h)}{(x+h)^2+1}$$
%
but
%
$$f(x)+h=\frac{200x}{x^2+1}+h.$$
%
Notice that these are very different. The formula for the central difference quotient tells you to use the first one but you use the second one

[wraith dagger]

Lorago

[rugged pond]

$f(x+h)$ means you should plug in $(x+h)$ everywhere you have an $x$ in $f(x)$

[wraith dagger]

Lorago

[rugged pond]

Does that make sense?

[sand kayak]

Yea now it does

[rugged pond]

So do you see why it went wrong?

[sand kayak]

Like so?

[rugged pond]

Exactly!

[sand kayak]

Okay

[rugged pond]

Just keep in mind that it should be $f'(4)\approx$ and not $f'(x)=$. Remember that you're \textbf{approximating} it \textbf{at the point $x=4$}

[wraith dagger]

Lorago

[rugged pond]

Oh wait actually

I didn't read carefully enough

It's not entirely correct

What you wrote is what $f(4+0.001)$ should be, but remember that you want to compute

$$\frac{f(4+0.001)-f(4-0.001)}{2\cdot 0.001}$$

[wraith dagger]

Lorago

[rugged pond]

Does that make sense?

[sand kayak]

Okay now i am confused

Im lost

I need to start thinking over

We need to derive f(x)

Does making a central differenskvot the same as deriving?

[rugged pond]

Yes, you want to differentiate f, but you only want to approximate it using a central difference quotient

So the key thing is that the derivative is what you get as h goes to 0

But since your h is bigger than that, you're only getting close to the true value

I.e. you're approximating it

So the derivative of $f$ could be given by
%
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x-h)}{2h}$$
%
and so the central difference quotient is what you get when you remove the limit

[sand kayak]

What should i be doing nowp

[wraith dagger]

Lorago

[sand kayak]

I dont get it

[rugged pond]

You should plug everything into

$$\frac{f(x+h)-f(x-h)}{2h}$$

[wraith dagger]

Lorago

[sand kayak]

Put in 4+h into all the x’s?

[rugged pond]

Exactly!

[rugged pond]

No, that only gives you what $f(x+h)$ is

[wraith dagger]

Lorago

[sand kayak]

Then i dont understand what it wants from me

[rugged pond]

So what you should get is

$$f'(x) \approx \frac{f(x+h)-f(x-h)}{2h}$$
$$=\frac{\frac{200(x+h)}{(x+h)^2+1}-\frac{200(x-h)}{(x-h)^2+1}}{2h}$$
and then you can plug in your value for $x$ and $h$ and calculate it

uhhhh one moment

[sand kayak]

can we start over because i am lost

i can understand even less now

[wraith dagger]

Lorago

[rugged pond]

Take a look at this

And see if it makes any sense

[sand kayak]

but the we will get a 0

[floral thistle]

what the heck am i looking at

[sand kayak]

Im doing it like you said

But this is making even less sense

[floral thistle]

what they sent is not the derivative

[sand kayak]

Great

[floral thistle]

$f’(x) = lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$

[wraith dagger]

quantum

[rugged pond]

The question is not asking for the derivative

[sand kayak]

What kind of bs was that

[rugged pond]

It asks to approximate the derivative with a central difference quotient

[sand kayak]

I need to start over

I dont think this question is supposed to be this difficult

So why are we here for so long

With a bunch of numbers

[rugged pond]

It isn't that difficult either.

[sand kayak]

I dont know what to do

And this doesnt make sense

Nothing ever makes sense

[rugged pond]

I think you should work a bit more on understanding the concepts here instead of just plugging in numbers into formulas of that's the case

[sand kayak]

I understand the concept

Its just not workingo

And i dont know why

And no one can help

[long badge]

@sand kayak has a boyfriend

[rugged pond]

So do I, what's your point? lol

[surreal hemlock]

@sand kayak are you familiar with what a derivative is?

Sorry if it sounds like a stupid question, I want to try a different angle and just checking your understanding

[sand kayak]

Yes

[surreal hemlock]

So let's pretend you had no idea how to find the derivative of a function

How would you approximate the derivative?

[sand kayak]

Using one of those 3 rules with quotes

[surreal hemlock]

Well that's really vague

Instead of relying on rules, can you explain the concept to me as if you were teaching me?

[sand kayak]

F(a+h)-f(a-h)/2h

[surreal hemlock]

Use what you understand about what a derivative is

Ok that's a formula

[sand kayak]

Okay so a derivate decribes the speed of something changing

[surreal hemlock]

Do you understand what this formula is doing?

Where it comes from?

Right

[sand kayak]

Okay so

[surreal hemlock]

So there's another word for it

[sand kayak]

There is a thing in calculators where they put in bad values when you try to read points on a line

Thats where this quote comes from

[quiet flint]

close this one

[surreal hemlock]

Forget the quote

[sand kayak]

We use it to be able to more accurately find a point between 2 other points

[quiet flint]

he already has two channels

[surreal hemlock]

Should close the other one

The other one poorly defined the problem and was open later

So not quite

Another word for rate of change

Is slope

Are you familiar with this?

[sand kayak]

Yes

[surreal hemlock]

Ok so imagine you drew the function

Graphed it out

How would you use the slope definition to find the derivative at x=4

[sand kayak]

whats the slope definition

[surreal hemlock]

You don't know how to calculate slope?

[sand kayak]

oh so k= y2-y1/x2-x1

[surreal hemlock]

Right

Think of it as rise divided by run

Rise being the difference in y values, the vertical change

[sand kayak]

yea

[surreal hemlock]

Ok so

Now look at your graph

And see if you can visualize

How you would pick two points

To calculate the slope

[sand kayak]

i dont have my graph pictured

[surreal hemlock]

It doesnt have to be this exact graph

Imagine any graph

It's the concept that matters

[sand kayak]

i would pick any 2 points on the line and put in the coordinates to get k

[surreal hemlock]

Ok and if we are trying to calculate f'(4)

What 2 points would you pick?

What would be pretty close to f'(4)?

[sand kayak]

i would pick x=4 and then idk i would have to put in x=4 into the y=kx+m thing

i cant pick y when i dont know where it is

[surreal hemlock]

Ok let's take this one step at a time

X=4 is good

[sand kayak]

i dont understand how this is helping

[surreal hemlock]

If you know x =4

Where is the graph at x=4?

How would you find it

[sand kayak]

put in x=4 into the equation

[surreal hemlock]

Yep!

In other words

The coordinate would be at f(4)

Er

(4, f(4))

That is your first point (x,y)

So you have one point right?

[sand kayak]

yes but how is this helping me

[surreal hemlock]

https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcR8JW0AXRviubZpm0ypHscT3wnJZlEFXd6SmA&usqp=CAU

Look at this diagram

[sand kayak]

i need to derive my original question

[surreal hemlock]

I'm getting there

Please be patient

[sand kayak]

okay

[surreal hemlock]

Notice how if I want to find the slope at x

I can make this secant line

[sand kayak]

yea

[surreal hemlock]

And it's not perfect but

It's pretty close, right?

[sand kayak]

yea

[surreal hemlock]

And if it's not close enough

Just pick points that are closer together

So what would your second point be then?

[sand kayak]

i dont know.. they have to be in between the point you want if you cant derive it

[surreal hemlock]

Why not just pick something really close to 4?

Like 4.1?

x=4.1

[sand kayak]

yea

thats what i said

2 points and the point you want inbetween

[surreal hemlock]

Right

But notice that the gap between 4 and 4.1

Was arbitrary

I could've picked anything right?

[sand kayak]

yea

the smaller you go the more accurate it would be

[surreal hemlock]

Exactly

And so they want that gap to be a particular width

That's the h

Which was what, 0.001?

Or something

[sand kayak]

yes

[surreal hemlock]

So one of your points is

x= 4-0.001

And the other one is

x= 4+0.001

You find the slope between

[sand kayak]

yea

[surreal hemlock]

(3.999, f(3.999))

And

(4.001, f(4.001))

And that's it!

That's your answer

[sand kayak]

hmm not really

i am supposed to use the formula

and not whatever you wrote

and my problem is that its not working

[surreal hemlock]

Well, you are using the formula

This is literally where the formula comes from

That's the part I'm trying to explain

If you're getting lost doing pages and pages of arithmetic and work

Try taking it piece by piece

[sand kayak]

and yea i understand it, my problem isnt understand the formula it is the fact that its wrong

[surreal hemlock]

By understanding where the formula comes from

[sand kayak]

i used the formula and no matter what i do i get 1 as the answer

i sent pictures of my solutions before

and i been talking to people for one hour and no one solved it

[surreal hemlock]

Please compute f(4.001)

I doubt you did this correctly

,w 200x/(x^2+1) where x=4.001

[wraith dagger]

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

Requested by CosmoVibe#7735

[sand kayak]

Here are my calculations on it

[surreal hemlock]

I saw all of them

You're not listening

Please calculate f(4.001)

You are calculating this value incorrectly

[surreal hemlock]

Why are you writing 4001

x=4

h=0.001

[sand kayak]

X+h is 4.001

I am doing it accorsing to the formula

[leaden otter]

People also write 4.002 as 4,002

[surreal hemlock]

Oh wait is that just how you write decimals

Ah I see ok

My bad

But why did you round

You should be keeping quite a lot of those decimal places

[surreal hemlock]

Good

[sand kayak]

So you wanted decimals?

[surreal hemlock]

Keep those decimals

Yes

Because those decimals make a HUGE difference in the final answer

You now have one point

[sand kayak]

Isnt this the final answer already?

[surreal hemlock]

(4.001, 47.04845)

No

You need the other point

Calculate f(3.999)

Keep decimals

[sand kayak]

Why?

[surreal hemlock]

.

[sand kayak]

I am supposed to calculate the value on f’(4).

And we already did it with those quoatas

[surreal hemlock]

By approximating using a secant line slope with h=0.001

No you havent

If you got 1 then your answer is wrong

Please calculate f(3.999)

[sand kayak]

i dont understand why

[surreal hemlock]

This is how you solve the problem

We are essentially doing the formula without doing the formula

You've made mistakes in between that people are trying to highligjt

[sand kayak]

i dont understand why we do this central quote thing for then

[surreal hemlock]

Because that's literally what the question asks for

It is asking you to do it

We are doing it

[sand kayak]

so we already did decide f'(4)

[surreal hemlock]

No we havent

You don't have the second point yet

How are you going to find the slope without a second point

[sand kayak]

so central quote doesnt = derivative?

[surreal hemlock]

It is a method to approximate the derivative value

Approximate

It's not exact

[sand kayak]

so why isnt it enough?

[surreal hemlock]

Why isn't what enough?

[sand kayak]

we do the central thing when we cant derive, so i did the central thing now so it should be derived

[surreal hemlock]

But you did it wrong

So I'm showing you how to better understand it

So we can break it down into more manageable parts

So it is easier to see where your mistake is

[sand kayak]

okay

[surreal hemlock]

I'm literally walking you through this step by step

We are like 70% there

And you seem intent on going back to the formula you dont understand well enough

If you have questions like that, we can go over that afterwards

But I would like to walk through to the solution first

[sand kayak]

okay

[surreal hemlock]

Do you understand that we are looking for two points

And we have to find the slope between them?

[sand kayak]

sure

[surreal hemlock]

What are those two points

[sand kayak]

4,001 ; 47,04844502

[surreal hemlock]

That's one point

That's the x,y of one of them

What's the other one?

[sand kayak]

4;47.1

[surreal hemlock]

I believe you wanted to use f(x-h) in the formula?

If so, you should use x=3.999 instead of x=4

[sand kayak]

i dont even know anymore

[surreal hemlock]

Do x=3.999

Just in case we will need it

In case that's what the problem wants

[sand kayak]

wait im lost

so we got 2 points

[surreal hemlock]

Yeah

[sand kayak]

what do we do now

what do we do with those points

[surreal hemlock]

Find the slope k between them

Derivative is just slope

[sand kayak]

0,01/0,001

that is the k

which is 10

wait i thought derivative =k

[surreal hemlock]

When h is infinitely small

If h is finitely small, then it is only an approximation

You got rid of a lot of the decimals again

Don't do that, use a couple of decimal places

Also I think you computed the slope wrong

Make sure if you do y1-y2 you're not doing x2-x1

You're like so close

All you have to do is punch the correct numbers into the calculator without rounding your decimals

And youve basically got it

[sand kayak]

okay wait

So this is wrong?

How am i supposed to know how many decimals to use

[surreal hemlock]

Well you're dividing by 0.001

So very small changes in the numerator results in a VERY different final answer

[sand kayak]

4.001-4 is 0.001

[surreal hemlock]

You rounded the hell out of y2

The denominator is fine

That's perfectly accurate

The numerator you rounded y2 like crazy

Keep as many decimals as you can because the result is so sensitive

[sand kayak]

Would 47.01 work?

Okay

[surreal hemlock]

No it would not work

[sand kayak]

Since the first y was only 47.1 i thought i should be using only one decimal..

[surreal hemlock]

Keep as many decimals as you can

The final answer is VERY sensitive because you are dividing by 0.001

You want to use at least like 4 decimal places

[sand kayak]

Then i need to start over because i rouded up in the very beggining

[surreal hemlock]

Yeah start from beginning then

[sand kayak]

I am lost

[surreal hemlock]

This first point is good

,w 200x/(x^2+1) where x=4

[wraith dagger]

Results provided by WolframAlpha

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

Requested by CosmoVibe#7735

[surreal hemlock]

Your second point is not good enough

[sand kayak]

So i have the 1st point what do i do now?

[surreal hemlock]

Should be like

4, 47.058823529411

[sand kayak]

How do i get the 2nd point again

[surreal hemlock]

So let's say you have x=4

Plug that into your f(x)

f(x) is your y

[sand kayak]

We already have x and y

I dont know how to get the 2nd point

[surreal hemlock]

You need two points

The second point is a different x and y

So your first point here

Is like your x1 and y1

Your second point you would do the exact same thing

[sand kayak]

Yea and how do we get the other point?

[surreal hemlock]

With x2 = 4

Plug 4 into the function

[sand kayak]

Why 4?

[surreal hemlock]

Because that's your second point

[sand kayak]

F(4)=800/17

Okay

[surreal hemlock]

Or you could use x=3.999

Now that you have that

Find the slope between the two points

Just as you did before but

Round as little as possible

Do. Not. Round.

[sand kayak]

Okay

Like so

[surreal hemlock]

Good!

Now compute that and that should be your answer

And if it's not right

It sure as hell is close

[sand kayak]

i got k=103,7850941

[surreal hemlock]

That doesnt seem right

,w 47.05882353-47.04844507

[wraith dagger]

Results provided by WolframAlpha

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

Requested by CosmoVibe#7735

[surreal hemlock]

,w 0.01037846/-0.001

[wraith dagger]

Results provided by WolframAlpha

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

Requested by CosmoVibe#7735

[surreal hemlock]

How did you get 103

[surreal hemlock]

First of all

One too many 0's

You divide 0.001 not 0.0001

Secondly you missed the sign

4-4.001 is -0.001, not positive 0.001

You punched into the calculator incorrectly

[sand kayak]

4-4,0001 is 0,0001

[surreal hemlock]

It should be 4.001

I missed the mistake earlier

That's what we used before

In your problem it says h= 0.001

[sand kayak]

Ph

[surreal hemlock]

But also you missed the negative sign

[sand kayak]

I dont see where

Y2-y1/x2-x1

[surreal hemlock]

4-4.001

What is that

Isn't that a smaller number minus a bigger number?

How can that be positive

[sand kayak]

Oh

I got the right answer now

K=-10

[surreal hemlock]

There you go

[sand kayak]

Thank you

[surreal hemlock]

You have a lot of gaps that need filling

This problem was probably quite frustrating but

Remember to be patient

Make sure you try to understand the how and why more

Don't just think that math is all about plugging and chugging formulas

You won't get very far that way and this will only make the whole process harder for you in the long run

Be patient

[sand kayak]

Yes that is very hard to do on my own

I have been studying lots at home but i still dont understand anything

And its been 2-3 months by now

Im supposed to take my test soon

[surreal hemlock]

Well I would also personally recommend you go back to foundations a little bit when you get the chance

Typically students taking this level of coursework will be very comfortable with a lot of number sense concepts, significant figures, slope, etc

This might be a bit hard for you because it might just be too hard for you

Do what you need to get through the course but consider some remedial classes

There's no shame in it

[sand kayak]

I did well on my previous test on polynomials and 2nd degree functions and counting rules

I cant take anymore classes

There is only one 2 hour class available per week that i go to

Everything else im supposed to do on my own and finish the course by christmas the entire 4-5 chapters

Im still on chapter 2

And i will have to redo it again

[surreal hemlock]

I mean in the future

Do what you can to get through this one

[sand kayak]

I doubt i will pass tbh

[surreal hemlock]

But in the future consider maybe you're taking courses that are less difficult to brush up

[sand kayak]

After that i wont be able to retake it

[surreal hemlock]

Anyways that's my personal recommendation

[sand kayak]

I dont know

I passed the previous maths course

So i took the next one

There is nothing i can do tbh

The teaches only gives lecture for 2 hours and since im a whole chapter behind i am not getting lectures on chapter 2

Because my whole class is on chapter 3

i am stuck again

what is the function it asks

<@&286206848099549185>

[cedar kiln]

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<@&286206848099549185>

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[slim birch]

1. You are rolling 5 dice at once. You need to roll at least 2 fours to win. What is the number of expected trials to win?

2. You are drawing 5 cards from a deck at once. You must have all 5 cards be the same suit to win. What is the number of expected trials to win?

3. You are spinning 9 color wheels at once. The probability of landing on red is 90%, and the probability of landing on blue is 10%. You must get red for the first 5 wheels, get blue the 6th wheel, and get red on the last 3 wheels to win. What is the number of expected trials to win?

[slim birch]

So, the part I'm having trouble with is I'm confused by what to use to calculate expected trials.

My answers (probability only)

1. 1 - (5/6)^5 - (5/6)^4 * (1/6) * (5C1)

2. (13C5) * (4C1)/(52C5)

3. (9/10)^5 * (1/10) * (9/10)^3

[slim birch]

I'm confused with the different kinds of distributions (binomial, geometric, and hypergeometric) and I'm not sure which to use for each of the 3 questions? How to calculate expected trials in each question?

[elfin hemlock]

Once you have probability of winning its a “lose/win” setup now with prob p of winning

So bernoulli trials

[slim birch]

Sorry, I'm confused. We didn't learn bernoulli trials. I searched it and I'm not sure how it helps me find number of eexpected trials...?

[elfin hemlock]

Like 99% sure you did, how else were you introduced to binomial or geometric distribution?

But anyway just means a trial has outcome A with prob p and outcome B with prob (1-p). For example yes-no questions (which is what we have here - did we win on this trial?)

You can derive expected number of trials from this not too hard or look up geometric distribution

Exactly what this models

[slim birch]

Well, what I found was that Expected no. of trials is 1/p, so that is what I used for all 3 questions

Just wondering. Are all 3 questions using geometric distribution?

[elfin hemlock]

Yes indeed its 1/p

[slim birch]

And how do we know that for sure?

[elfin hemlock]

Yes

Because bernoulli trials

[slim birch]

Hmm, and in a nutshell, you use binomial distribution if you have a fixed number of trials, and you want to know probability of an event happening in n trials?

[elfin hemlock]

Binomial is repeated bernoulli trials

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[crimson sedge]

I am not sure if this is multiplied correctly using natural logarithms- in addition, not sure what to do next

[timber onyx]

it is mtiplied correctly

solve for x then u can simplify the solution with log rules

[crimson sedge]

okay lemme see

how do you simplify from here 🤔

3-ln4?

OR

(ln4^3/12)/ln4

I think that makes more sense

[neat oak]

(3ln4)/ln4 -ln12/ln4

then simplify

as 3 - (2ln2+ln3)/2ln2

=2+ln3/2ln2

[timber onyx]

ln12=ln(4*3)=ln(2^2 * 3)= 2ln2+ln3

ln4=ln(2^2)=2ln2

[timber onyx]

you can expand right term again like he did in first

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[crimson sedge]

hello

[crimson sedge]

can you please help me

with differential equations

this is my problem

I have 2 actually

and

this ones a inverse laplace

but has to be done with convolutioon

<@&286206848099549185>

pleaseeeeeeeee help

[timber onyx]

read this

[crimson sedge]

my bad im sorry

um anyone can you help me? <@&286206848099549185>

its been 15 minutes so I am pinging again sorry

nvm I dont think anyones able to help

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[lament flint]

I am confused on how I can generate and graph the Gini index using the following data. Can anyone help me understand how to proceed?

[crimson sedge]

no

no

not at all

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[lament flint]

my channel now boi

lmao

[crimson sedge]

lol

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[rose sand]

I have no clue where to even begin with this. Everything I try seems to come up incorrect

wait

im a dummy

its just -1

.close

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[abstract perch]

In a lightbox experiment, why is a semicircular shape used?

[abstract perch]

ray box*

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[fickle trellis]

Hi

[abstract perch]

hi

<@&286206848099549185>

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[narrow glacier]

hey

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[long badge]

WHAT IS 2+1

It is 3

[cedar kiln]

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[atomic viper]

Help

[atomic viper]

.close

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[atomic viper]

.reopen

[cedar kiln]

✅

[atomic viper]

Help

.close

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[versed lava]

I’m getting A as the answer

How is it bijective

It’s for natural nos so it’s one one

And then if we look at onto then we know that some values in the range are not going to have a soln. Like 3 or 2

[pure quiver]

The function is surjective if its range is the same as its codomain. Here the codomain is defined as X. Now check the range of f and see if the values you get all belong to X

[versed lava]

Yea I checked

Idk how I’m wrong :(

[pure quiver]

so does the range match the codomain?

[versed lava]

[versed lava]

Co domain is x, and range is x^2

And

[pure quiver]

the codomain is X, defined as x^2 for all x natural

[versed lava]

We are looking at natural numbers

Yep

[pure quiver]

so in other words, 1, 4, 9....

[versed lava]

Mhm

[pure quiver]

now for f

when you plug in x = 1, 2, 3..., what are the corresponding f(x) values?

[versed lava]

1,4,9

[pure quiver]

yes

[versed lava]

So the domain and range is x squared

What the shit

[pure quiver]

so we can conclude that the range and codomain are exactly the same

hence f is surjective

[versed lava]

Oh okay rip

[pure quiver]

and now combined that we also know that f is injective

f is bijective

[versed lava]

But doesn’t N approaching X mean that

The set of natural numbers is approaching the Relation?

In this case the relation is x squared

:(

[pure quiver]

It says N -> X

not N -> R

[versed lava]

Yea sorry

[pure quiver]

it means the set of natural numbers mapped onto the set X

[versed lava]

Mhm

[pure quiver]

and X is defined as x^2, where x is natural

[versed lava]

Yes

What about the domain

[pure quiver]

the domain is just N

[versed lava]

It’s defined as the set of natural nos right

Exactly

So shouldn’t it be one one only

[pure quiver]

yes, we already concluded it is injective (one-one)

just that you had issues figuring out that f is also surjective

[versed lava]

Oh waittt

Oh no

I figured out where I was wrong

It’s not

Ugh

Okay sorry

My bad

Thanks

[pure quiver]

np

[versed lava]

.close

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[lilac bison]

can an exponential function have a negative common ratio?

[lilac bison]

i know the form that an exponential function takes is a(r^x) but can r be negative?

[crimson sedge]

You mean r^x?

[lilac bison]

yeah

can the r be negative?

[crimson sedge]

Yes

[livid hound]

it won't be well defined for negative r

[lilac bison]

i don't really get that

[crimson sedge]

You mean the oscillating between negatives and positives? @livid hound

[lilac bison]

that's what i thought would happen

[crimson sedge]

If x is an integer it would be easier

Otherwise you'll be taking the root of a negative number

[lilac bison]

i tried inputting it into desmos but it didn't graph it

[lilac bison]

so it won't be defined for real numbers if x isn't an integer then?

[livid hound]

there are some values where it'll still be defined

[crimson sedge]

@lilac bison because it would need two graphs

One of the imaginary part and one of the real part

[lilac bison]

i see

oh i see

[lilac bison]

like if x is an integer?

[crimson sedge]

When x is even

It will be purely real then

If it is odd it will also be real

[lilac bison]

so it will be real numbers if x is an even and odd integer?

[crimson sedge]

If it's pair it will be positive otherwise negative considering x is an integer

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[exotic junco]

please help

[exotic junco]

I've been trying to understand from the maths book for about half an hour and my test is tomorrow and it's getting late

I don't understand any of it

Could somebody please tell me how to use graphs to solve linear equations?

I need help ASAP

<@&286206848099549185>

[obsidian coral]

You have to wait at least 15 minutes before pinging helpers

[exotic junco]

ok

sorry

could somebody please help though?

[livid hound]

post a specific problem

[bold vine]

Man's didn't read #❓how-to-get-help and it couldn't be more obvious

[obsidian coral]

Also, what did you try/get stuck on?

[exotic junco]

pretty much the whole of the last chapter

I honestly can't understand any of it

[bold vine]

@exotic junco Read #❓how-to-get-help

[livid hound]

yes i know what book you are reading

[bold vine]

Lmao

[exotic junco]

idek how to ask

[livid hound]

without a specific question,
my response would be to graph the (two) lines and their point of intersection will be the solution

[bold vine]

Wait what

[exotic junco]

those are examples and i dont understand

[bold vine]

🤨

[exotic junco]

I don't understand the explanation

thanks for the help

[dense wing]

what dont you understand about the explanations?

[exotic junco]

Pretty much everything

How they got the answer

They don't explain it clearly

[dense wing]

Read the explanation for how they got the answer.

They do

[exotic junco]

bruh can you just reword it

[dense wing]

Nope, what's written is exactly how I'd explain it

[obsidian coral]

The explanation is clear

[exotic junco]

ok

wait nevermind

i think i just didnt read right/skipped over it

you can now close this chat

[dense wing]

It's your channel..

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[vague crescent]

Help

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[dusty oriole]

In Minecraft, you can barter with Piglins to receive different items. These items have different probabilities of occurring with the important item for speedrunners being Ender Pearls with approximately an 2.18% chance of being dropped per trade. You make 3 trades in a row.

⦁ What is the probability getting Ender Pearls all three times? (2 marks)

[dusty oriole]

<@&286206848099549185>

i need help with this

.close

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[pure brook]

hey guys, I'm having a hell of a time figuring out what divergence test to use here, could I please have some assistance?

[pure brook]

I checked it through a calculator and it technically equals to 0, but I'm not too sure how to manipulate the equation to get there

[elfin hemlock]

(-1)^n

So always alternating

[pure brook]

Even with the 1/n portion?

[elfin hemlock]

Oh wait ((-1)^n-1)/n=(-1)^n-1/n

Compute first few terms

Then rewrite the series when you notice a pattern

[pure brook]

kk let me take a look

my apologies friend I'm not seeing how I can algebraically manipulate it beyond

[elfin hemlock]

Well what are the first like 5 terms?

Its more series manipulation

[pure brook]

oh ok I see, let me crunch those out

oh I see the pattern, it's every even number is 0, every odd -2/n

oh is this a telescoping series?

[elfin hemlock]

Yes indeed

[elfin hemlock]

No

So if you define a new sequence which for n=1 takes n=1 of old sequence, n=2 takes n=3 of old sequence…

And so on (so a sequence of the odd terms in our old sequence)

Then sum from n=1 to 2k of old sequence is equal to n=1 to k of our new sequence, agree?

[pure brook]

pretty much

?

[elfin hemlock]

Ye but kinda annoying you only sum over odd integers now right?

[pure brook]

yes

[elfin hemlock]

Hence I said redefine so it sums over all

Find a formula that works

[pure brook]

hmm

[elfin hemlock]

$\sum_{n=1}^{2k} \frac{(-1)^n-1}{n}=\sum_{n=1}^{k}-\frac{2}{2n-1}$

[wraith dagger]

ScapeProf

[pure brook]

oh

[elfin hemlock]

For the inf series take lim k->inf on both sides

And now you have a series you can deal with right?

[pure brook]

interesting way of dealing with the series, never seen that in class before

well the right equation diverges, which means that since it's equal to the left it must also diverge, no?

[elfin hemlock]

Si

[pure brook]

ah, the limit comparison test

alrighty, thanks

[elfin hemlock]

I mean they are straight up equal

No test

I just rewrote the series

[pure brook]

oh, well that also works

I appreciate the help, this one was weird for me

.close

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[harsh pagoda]

lets say you have a 1m path. if you go half of it every single time, would you reach the end? u go 1/2 of that path then another half which means 1/4 then another path thats is 1/8 and goes on. would you reach end? im curious of this question for a while and like to see u think

[normal wraith]

you would never reach the end in a finite ammount of time, but you are getting arbitrarily close to it

https://en.wikipedia.org/wiki/Zeno's_paradoxes

[harsh pagoda]

hm

wait

zeno think was that i was thinking

i didnt know that lol

anyway thx

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[crimson sedge]

no clue how to do this

[thorny pebble]

Remainder = -16/(x-2)

so -16

[thorny pebble]

Do u want explanation?

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[crimson sedge]

Thank you

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[crimson sedge]

Sus

[crimson sedge]

Are 1 and 2 true?

I don't think 2 is true

nvm

.close

[cedar kiln]

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[cedar kiln]

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[sand kayak]

f(x)=4x-3x^2 i tried deriving 4 times and failed

[sand kayak]

i need help

to derive this

f(x)=4x-3x^2

[cedar kiln]

@sand kayak Has your question been resolved?

[crimson sedge]

by deriving, do you mean finding the derivatives?

if so, look up "sum rule for derivatives" and "power rule"

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[cedar kiln]

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[dense wing]

For a, I got $y(t)=C_1e^{3t}+C_2\cos(t)+C_3\sin(t)$, however b we were never really taught (This is from an old exam) and just want to check if I'm correct.

$y_(t):=y(t)\implies \abs{y_(0)}=\abs{C_1+C_2}$

$\forall\delta >0, \exists C_1,C_2\in\mathbb{R}$ st $\abs{C_1+C_2}<\delta$

$\lim_{t\to\infty}y_*(t)=\infty$

Therefore $y(t)=0$ is not asym. stable

[wraith dagger]

Mosh

[dense wing]

is that a valid proof for b?

and existence of C_1 and C_2 is given since C_2:=delta-C_1-epsilon for small eps

[cedar kiln]

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[olive sun]

I'm supossed to find the scalar equations of 3 planes in 3D space given that they intersect at point (6, 5, -6) and that plane 1 and plane 2 are perpindicular and plane 1 and plane 3 form an angle of 30 degree

[olive sun]

I have no idea how to solve this question as there are 3 variables I'm supossed to solve for when I use dot product

I have two equations that intersect at (6, 5, -6)
x + y + z - 5 = 0 2x + 3y - 5z - 57 = 0

But I can't find a third due to what I mentioned above

[cedar kiln]

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[olive sun]

<@&286206848099549185>

[cedar kiln]

@olive sun Has your question been resolved?

[nova geyser]

make x+y+z=5

2x+3y-5z=57

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[cedar kiln]

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[sonic moth]

Hello, is ASA postutate a valid answer here?not just AAS

[bright surge]

well
you can point out that ABC = DEC and then use ASA

but that would just require extra steps

[livid hound]

AAS would be better

[sonic moth]

Sighed... the answer key only accepts AAS,but not ASA...I just find it irritating

They are both the same answers anyway

But they only accepts aas

[livid hound]

you'd need an extra step to justify it with ASA

[sonic moth]

We'll see how it goes ,I appealed to my teacher

Thanks...

[bright surge]

normally, the answer would be the postulate that you can use to prove that the triangles are congruent in the fewest amount of steps
in this case, it's AAS

(also, here in my country, AAS isn't even a postulate)