help-26

[neon iron]

@half pike I'm guessing you just didn't see the minus sign

Notice it says y=**-**2, not y=2

Pro tip: Close the dollar signs before writing text

[safe burrow]

huh

wdym

[neon iron]

The ", so " part

That's just regular text

[safe burrow]

oh right

tbh i dont care all that much its clear whats meant

[half pike]

ohh

wait what’s wrong abt 8c

[neon iron]

x^2y=1(-2)=-2

You once again ate the minus sign

[half pike]

whats Sox

OH

hang on

[neon iron]

Guess it wasn't clear what was meant after all

[safe burrow]

https://tenor.com/view/emoji-disintergrating-meme-emoji-disintegrating-meme-gif-21239978

[half pike]

for 8a

would it be 1^2 - (-2)^2 or 1^2 - (-2^2)

[neon iron]

The former

[safe burrow]

the first one

[half pike]

okk

[neon iron]

In case of doubt, when replacing a variable with its value, put parentheses around the value

@half pike You can now .close this channel

[half pike]

hang on

real quick

8a is -3
8b is 9
8c is -2

[neon iron]

Correct

[half pike]

yay

culd u help with this one

idk how to do this

like is it x + 5 < 0

?

[neon iron]

Yes

[half pike]

how do u do the other ones

[neon iron]

First of all do question a, it should be easy

[half pike]

idk how to do 1b or 1c

it doens tmake sense

[neon iron]

Do you know what the "solution set" is?

[half pike]

not rlly

[neon iron]

It's the set of all numbers that satisfy the equation/inequality

@half pike Does that make sense?

[half pike]

can u give an example

[neon iron]

x^2=1, the set of all solutions for this equation are {1,-1}

Because x=1 and x=-1 are the only solutions to this equation

[half pike]

ohh ok

how do i graph the set

[half pike]

how does x = -1 if x = 1?

[neon iron]

I'm not saying x=-1 and x=1

I'm saying that either x=-1 or x=1

[half pike]

ohh

[neon iron]

are the only solutions to this equation

@half pike What do you think the solution set is for the inequality x > 0?

[half pike]

-1

[neon iron]

?

That's not a set

[half pike]

-1,1?

[neon iron]

...?

x > 0

Are you still thinking about x^2=1?

[half pike]

yea

[neon iron]

Because that was an earlier quesiton

[half pike]

oh

[neon iron]

Read this again

[half pike]

-infinity?

[neon iron]

Again, that's not a set

That's a number

Not only that, but it doesn't even satisfy the inequality

-infinity is not greater than 0

[half pike]

whats a set again

[neon iron]

Just a collection of things

Although because we're talking about a solution set it must be a collection of numbers specifically

GG to the guy who just got automodded

[half pike]

oh ok

[lucid bridge]

Wait

[neon iron]

So do you know?

[lucid bridge]

Do u know what the form of a solution set is

Just yes or no

[neon iron]

form?

[lucid bridge]

Id call it a form

That's how I've always called it

[neon iron]

Any set of numbers can be a solution set

So I don't see what you're talking about

[lucid bridge]

It can be represented in multiple ways

Thus it's a form

(it's not actual terminology, I just think stuff like this)

[neon iron]

Yeah sure but I think if they knew this they'd be able to answer

@half pike So should I take that to mean that you don't know the solution set for x > 0?

[half pike]

yes

[lucid bridge]

Then you gotta teach them how to make a solution set

[neon iron]

What do you think I was about to do

[lucid bridge]

Fair

[neon iron]

Just testing their background knowledge first

[neon iron]

What is it?

[half pike]

what si what

[lucid bridge]

I think they mean take it as they don't know

Anyway

[neon iron]

Oh

[lucid bridge]

Also sorry if u find my tone rude or anything, I'm bad at communicating

[neon iron]

@half pike The set is (0,infinity)

[neon iron]

Crying and shaking rn

[lucid bridge]

What

Bruh wtf

That's interval notation

My guy

[half pike]

oh

[lucid bridge]

Nooooo

[neon iron]

An interval is a set of numbers

[lucid bridge]

It's interval notation, I learnt it as solution set is a set

Like

[neon iron]

Does it make sense that we're using an interval rather than {}?

[neon iron]

Okay?

[lucid bridge]

Idk if they learnt it like me so I would do it in the set notation

Not interval

[neon iron]

Ok...?

[neon iron]

@half pike What does the silence mean?

[lucid bridge]

I have no idea where the set thing went

[neon iron]

0 > infinity 📝

[lucid bridge]

${ {x \in \mathbb{R}; 0 < x <\infty} }$

[neon iron]

@half pike Last attempt, are you there?

[half pike]

yes

[lucid bridge]

Oh shot

Shit

[thorny flame]

Kai The Doge

[lucid bridge]

Finally

Omfg

[neon iron]

@half pike So does it make sense to you that for x > 0 we used an interval instead of the regular set notation with curly brackets?

[half pike]

i guess?

[lucid bridge]

Both work*

[neon iron]

Like for x^2=1 there are finite solutions

[lucid bridge]

But if there are infinite, you cannot just give a few numbers or list them all, thus u have to use set notation

[neon iron]

Hence we can write them all down like {1,-1}

But for x > 0 there's an infinite amount of solutions

So instead of spending an infinite amount of time writing them all down we just use an interval

@half pike Does that make sense?

[half pike]

yea

[neon iron]

Great, do you think you can find the solution set for x+5<0?

[half pike]

-5, - infinity?

or -6?

[lucid bridge]

It's -5

Also is -infinity smaller or bigger than -5

[half pike]

uhhh

[neon iron]

Great, now can you put it in interval notation?

[half pike]

[-5, -infinity)

?

[neon iron]

Is -5 a solution to x+5<0?

[half pike]

it would be =

[neon iron]

Yes

So is it a solution or not?

[lucid bridge]

Is 0<0

[half pike]

no?

[lucid bridge]

Right

[neon iron]

Therefore -5 is not part of the solution set

So it should be (-5,-infinity), not [-5,-infinity)

[lucid bridge]

Also (-infinity,-5)

[half pike]

Ohhh

[lucid bridge]

Since smaller number...

[neon iron]

I don't think it's a rule to do that but it's the convention

[lucid bridge]

Follow convention, teachers may mark wrong

Better to be safe imo

[neon iron]

Yes, teachers hate students

So you gotta be careful

[lucid bridge]

Also is both or one correct? @half pike (-6,infinity] and (-6,infinity)

[half pike]

one

it’s the second one

[lucid bridge]

Why

[half pike]

cuz infinity is suppose to end with ) always

[lucid bridge]

Yep

Since infinity is a concept and not a number

It cannot be =

[neon iron]

Riemann spheres:

[lucid bridge]

Not relevant until like prolly a lot more years

[neon iron]

[lucid bridge]

[half pike]

wait so for my question

[neon iron]

POV: You're a student and spent infinite time writing all the solutions until Cantor comes along and shows you you missed one

[half pike]

1b would be that set right

how do i graph hit

[neon iron]

Mhm

Are you familiar with graphing intervals?

[half pike]

wait its (-infinity, -5)

[half pike]

nope

and idk how to

thr space is kinda small

[neon iron]

Yeah, it is

[neon iron]

Yes

[lucid bridge]

It's a line

[half pike]

okk

[neon iron]

I found this on Google

Does that ring a bell?

[lucid bridge]

Like a bad number line basically

[half pike]

ohhh

so how do i do that

[neon iron]

Draw a number line, then draw it just like they did

[half pike]

and do i jsut write it on the line?

[neon iron]

By the way because your interval goes all the way to -infinity you will only draw one parenthesis

[neon iron]

Yes

Well do it on another piece of paper or in paint (the program not actual paint) first

So that I can check what you did

[half pike]

ohh

idkk

i dotn get it

doesn tlook right

[lucid bridge]

I thought you were supposed to do it with dots like this

[half pike]

idk probs

do i use the third one?

[neon iron]

You really don't like different notation, do you

[neon iron]

Yes

[half pike]

okk

[lucid bridge]

Nope, I really don't Xd

[neon iron]

X*nophobia but math

@half pike Any progress?

[half pike]

yes

[neon iron]

Yep, that's correct

[half pike]

yay

[neon iron]

To remember this notation, you can think of a circle as an "open dot" and a point as a "closed dot"

[half pike]

yes

but it doesnt include -5 so

[neon iron]

Yes

Hence it must be open

[half pike]

open dot is excluded right?

[neon iron]

Hence it's correct

Yep

[half pike]

oh yay

[neon iron]

You may now .close this channel

Unless you have more questions

[half pike]

can we do the other one

[neon iron]

Oh I forgot about the other one

Right

[half pike]

x + 1 <= -2

is that right

?

[neon iron]

Yes

[half pike]

how do i write the soultion set

@neon iron help

[neon iron]

How's it any different than the previous question?

[half pike]

idk

uhhhh

[neon iron]

[half pike]

-3, infinity

?

[neon iron]

Yes

[half pike]

wait rlly

[neon iron]

Wait no

Not infinity

Think about it for a moment

[half pike]

oh what would it be

[lucid bridge]

Is infinity less than -3

[half pike]

no

[neon iron]

I don't really know how to make you realise this yourself

All I can say is wrong direction ⬅️ other way

[half pike]

oH

(-infinity, -3]

[neon iron]

Yes

Why'd you choose ] this time?

[half pike]

did i use brackets right

cuz it includes -2

because its <=

[neon iron]

-2?

[half pike]

not <

[neon iron]

Why are we talking about -2?

[half pike]

oh sorry

-3

[neon iron]

Yes

That's the reason

Now I think you won't have any problem graphing it

@half pike Did you graph it?

[half pike]

not yet

[neon iron]

That's correct

If that's it, you can now .close this channel

[half pike]

can i ask one more quesiton

[neon iron]

Phew, sure!

That's a lot of work you have there

[neon iron]

Or maybe I'm just not used to it because I'm lazy

[half pike]

the question is

[neon iron]

What in the-

[half pike]

Solve the linear inequality.

[neon iron]

I can't believe neither of us is paid to do this

Anyways

Tell me the first one where you get stuck at

[half pike]

i dont get what its asking for

[neon iron]

I assume it means isolate x

For example, something such as x <= 3 is considered solved

Because x is alone

@half pike Does that make sense?

[half pike]

idkk

[half pike]

hekp

[neon iron]

You can just say "no"

[half pike]

okk

[neon iron]

Basically in 2x <= 7, x is not alone

There's a 2 next to it

So you need to find a way to make it alone

Do you know how to do that?

[half pike]

yes

so is x = 14

?

wiat

sorry

my bad

[neon iron]

=

[half pike]

= 3.5

[neon iron]

14

=

How the frick did <= become =?

[half pike]

oh yea

so is x <= 3.5

[neon iron]

Yes

I think my teachers would accept this as an answer, not sure about yours

In theory both this and solution set form should work fine

[half pike]

oh ok

what abt the other ones

[neon iron]

Well same logic

Just do them all the same way and tell me if you get stuck at any of them

[half pike]

would it be 2x > 8

so x > 4

?

[neon iron]

Yes

[half pike]

oo yay

[half pike]

hwo do i do 7 - x >= 5

[neon iron]

If you rewrite it as -x+7 it becomes obvious

[half pike]

-x >= 5 -7

[neon iron]

Oh there you go

[half pike]

-x>= -2

but u cant have a negative varible like that right?

so is it

x >= 2

?

[neon iron]

No

If you're going to negate both sides the symbol flips

Like picture the number line in your head

Or even better, take a piece of string or a pen or pencil and think of it as the number line

Then flip it, which represents negating a number

Notice how everything that was on the left is now on the right and everything that was on the right is now on the left

That's why when you negative both signs of an inequality the symbol flips

[half pike]

ohh so its <=

[neon iron]

Yes

[half pike]

ohhhh

can u help me with the hrad ones pls

like the fraction ones

[neon iron]

Which one specifically?

[half pike]

21, 23, 33, 34, 32, 24, 22

i'd say those ones

[neon iron]

@half pike The fractions don't make it hard at all

Typical numerophobia or whatever the frick its called honestly

Students do be like that

It's just the same as what you already did but you have to write more stuff down

[half pike]

oh

okk ill try

[topaz sinew]

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[topaz sinew]

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[desert pulsar]

https://media.discordapp.net/attachments/903430334681608232/1109554715861196940/image0.jpg

[desert pulsar]

Is this correct?

@vernal matrix thanks

.close

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[rich rampart]

Does this series have a name?

[rich rampart]

I’m having a hard time finding the sum of this series but I can’t match it to any on Wikipedia

[neon iron]

could you show the original question

[ionic oar]

Seems to be something special

There is a pattern though

$T_n = \frac{n}{n(n + 1)}$

Bot broken

The nth term is n/[n(n+1)]

Which is just 1/(n + 1)

Which is a harmonic progression

[rich rampart]

On shoot that’s right

I forgot about the term out side the series

If it’s harmonic doesn’t that mean there’s no sum.

[ionic oar]

Well

There's no neat and tidy formula for it

[rich rampart]

So this series just goes to infinity

[ionic oar]

Yes it diverges

[rich rampart]

I probably did something wrong

this was the original question

for a I found it to be (1/k)(1/k+1)

Thanks for the help!

Really appreciate it

[molten raven]

I don’t think you did something wrong, I got the same

[rich rampart]

so this is basically saying my expected value of getting a white is an infinite number of tries?

[molten raven]

Yea this is weird let me see if I made a mistake

[rich rampart]

I guess it kind of makes sense for it to be infinite

cause the more black I choose the white basically just disappears

[molten raven]

I guess

[rich rampart]

to the point where i keep drawing which just hurts my chances of getting the white

[molten raven]

Yea that makes sense

[rich rampart]

Thanks for the help!!

.close

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[lyric mountain]

what would be a good hint for the anti deriv of tan x?
I'm trying to rewrite it but I got a product still in the end. Assume calc 1 knowledge

[pastel oracle]

try substitution

[lyric mountain]

ohhhhh

good idea

would you rewrite tan first as sin x / cos x

[pastel oracle]

yes

[lyric mountain]

then rewrite sin x / cos x as sin x * sec x

let's see

[pastel oracle]

that's not how I'd do it

[lyric mountain]

oh

you'd stick with sin x / cos x?

.close

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[finite mortar]

for 8c

[finite mortar]

i dont get why u need the line after this one --> i thought u can find probability that B scores first from there

[pastel oracle]

what do you mean?

[finite mortar]

i dont get why theyve done the last bit

[pastel oracle]

to find that probability

[finite mortar]

so theyve done the first part to get enough information for the last part

this question always confuses me

[pastel oracle]

yes

[finite mortar]

oh mb you can't say that the probability of A winning = 0.3 because that has the condition of B scoring first?

ohhh so u consider A winning if they do score first and if they dont

ohhhhhhhhhhhhhhhhh

thanks

[pastel oracle]

they might be using the law of total probability in the denominator

[finite mortar]

whats dat lol

[pastel oracle]

google it, there should be plenty of resources on it

[vernal matrix]

Tl;dr this, those make up all the possibilities of A winning

[finite mortar]

h

*oh

yh

oh i see

ill keep that in mind

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[gusty bane]

!status

[topaz sinew]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[lucid oak]

so look

2^x=5

how do i know what is x?

[gusty bane]

do you know what a log is

[lucid oak]

no with out it

can we?

[gusty bane]

we know x is between 2 and 3

[lucid oak]

okay

[gusty bane]

2^2 = 4, 2^3 = 8, and 5 is between those

[lucid oak]

so how do we know exacly?

[lucid oak]

do we have to know what is it exactly or close?

[gusty bane]

know what

know what exactly or close

[lucid oak]

the value of x

[gusty bane]

you just need to be able to estimate it is between 2 and 3

[gusty bane]

based on this

[lucid oak]

what about the rest of the q

how do we write x?

[gusty bane]

they gave you that x + 2y = 6

the smallest value to consider is x = 2

and the biggest is x = 3

plug them in and see the largest and smallest values for y

[karmic monolith]

huh

[lucid oak]

huh

[karmic monolith]

You have to learn what logarithms are

It only takes like an hour

[lucid oak]

i will

[gusty bane]

@karmic monolith this is an estimation question

they want you to use the fact that x is between 2 and 3

[karmic monolith]

I don care

[gusty bane]

?

[karmic monolith]

Logarithms make it easier

[gusty bane]

what is log_2(5)?

without using a calculator

[karmic monolith]

okay

let me open libre math

[gusty bane]

i don't understand why you are being combative when the intended solution is clear

[gusty bane]

this is again more complicated than it needs to be

[topaz sinew]

@lucid oak Has your question been resolved?

[topaz sinew]

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[wispy mulch]

Can someone give me the steps to solve this one please, my teacher only wants us manipulating one side.

[wispy mulch]

It’s the one question I haven’t been able to figure out

[radiant tapir]

if you expand out tan(X) as sin(x)/cos(x), then simplify, you'll see that becomes a trig identity

you were on the right track

here's a hint

$\frac{tan(x)}{sin(x)cos(x)} \\= \frac{\frac{sin(x)}{cos(x)}}{sin(x)cos(x)} \\ = \frac{sin(x)}{sin(x)cos^2(x)}$.

[thorny flame]

MellowDramaLlama

[radiant tapir]

now simplify and bob's your uncle 🙂

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@wispy mulch Has your question been resolved?

[wispy mulch]

Tysm man I love you

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[clever cloud]

Are vector fields just the gradient of some level curves?

[knotty ledge]

not all

ones that are are "conservative"

or are you not on about vector fields that are gradients of just some function?

[clever cloud]

By conservative, do you mean that whose partial derivatives of g and f with respect to x and y respectively?

Damn, I guess i'd have to go way back and read the textbook.

[knotty ledge]

conservative vector fields are those whose line integrals over them only depend on the end points

[clever cloud]

Yeah, I've learned the same concept in Calc-based Physics as well.

[knotty ledge]

yeah those are precisely the vector fields that are gradients of some function

[clever cloud]

Okay, so if those "some functions" are non conservative, then there can't be a gradient of any sort?

[knotty ledge]

if the vector field is not conservative, it is not a gradient of a function

[clever cloud]

Those are "vector fields," am I correct?

I know these are the gradients but

[clever cloud]

Got you

[knotty ledge]

those arrows are some of the vectors in the vector field yes

and because this vector field comes from the gradient of phi, all the vectors are normal to some level curve

[clever cloud]

Got ya...

Thanks.

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[static jacinth]

I want to prove that the space of convergent sequences in $X$ is closed in $(X^\mathbb{N}, d_{\infty})$. $X$ is a metric space.

[thorny flame]

Casiel368

[topaz sinew]

@static jacinth Has your question been resolved?

[static jacinth]

<@&286206848099549185>

[fringe kernel]

wth

i dont know about this

wait for another helper

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[prime meteor]

.reopen

[topaz sinew]

✅

[prime meteor]

I saw this wasn't solved yet so I reopened it

<@&286206848099549185>

[empty sail]

Please don't do that nor ping helpers. People are capable of doing both themselves

.close

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[neon iron]

How do I do part a?

[loud oasis]

use newton's third law

[half edge]

I’d consider it as a system

[neon iron]

What do you mean?

[neon iron]

The topic is vectors

How would I relate that to vectors

[loud oasis]

draw a free body diagram showing the forces acting on each block as a vector

[half edge]

The outstanding forces are the gravitational forces

The tension is equal and opposite as cloud posited

[thorny flame]

Stephen

[neon iron]

I’m still confused

[half edge]

https://youtu.be/N6IhkTjWrd4

[neon iron]

Does tension in the string equal downwards force+upwards force?

[brittle elk]

yes

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[thorny ice]

Don't know how to find the non-trivial solution for this to get Kn and thus K3

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@thorny ice Has your question been resolved?

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@thorny ice Has your question been resolved?

[halcyon slate]

this seems too advanced for help channels try #odes-and-pdes

[thorny ice]

Alright

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[neon iron]

Why didn't i need to flip the intergral to have the lowest value on the bottom.

[neon iron]

oh

nevermind

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[magic night]

Uh I attempted but don't know how to get it to it's antiderivative

[magic night]

My attempt led me to 1/2 x (x-5)^-1

Now I'm gonna take a guess and say that's not it

[neon iron]

Let u = 2x-5 and du = 2 dx

[magic night]

Oh

Wait what do I do with du

keep in mind I'm very new to antidifferentiation 😎

[neon iron]

If you let u = that, then differentiating u gives du = 2 dx

So your integral becomes

1/(2x-5) dx
= 1/u * 1/2 du

Integral of 1/u is ln|u|

[magic night]

so

I let u equal the denominator

and then there's 1/u (from the x-5) and 1/2 from taking out half, multiplied by du

Then by antidiffing it, I get ln|x-5| x 1/2 x 2?

[neon iron]

its 2x-5 not x-5

[magic night]

Whcih gives 1/2 x 2 ln|x-5|?

Oh ye

[neon iron]

And where are you getting the 2 from?

we get the 1/2 because

du = 2 dx
dx = du/2
But we are integrating 1/(2x-5) dx

dx = du/2

Or 1/2 du

[magic night]

Uh

I think I'm a bit confused sorry-

I get defining u and then getting du as 2

[neon iron]

du is not 2

du = 2 dx

[magic night]

wait what 💀

oh

[neon iron]

u = 2x-5

D(u) = D(2x-5)
D(u) = D(2x) - D(5)
du = 2 dx - 0

You can YouTube search "u substitution for integration"

To learn more about this technique

But essentially

We make a dummy variable by substituting the expression as something which we can easily integrate

Its all just algebraic manipulation of the expression

We know integral of 1/x is ln|x| but
1/(2x-5) is annoying to deal with

But if we could somehow mold this expression

Into something which does look like

1/x dx we would be able to easily integrate it

So we let u = 2x-5 and check if we are transforming from the x world to the u world

How will that affect dx and the integrand as a whole

Any integral has 2 parts

Integral of f(x) dx
The two parts are f(x) and dx

We know that 1/(2x-5) becomes 1/u

So replacing that we get

Integral of 1/u dx

Uh well we cannot really integrate this

variable in u but integrating in the x world

So we have to change dx into something du as well

u = 2x-5
means du = 2 dx
Or, dx = du/2

Now replacing that

We get

Integral of 1/u dx
Becomes
Integral of 1/u * 1/2 du
Or,
1/2 * Integral of 1/u du

Does that make sense?

If not, I highly encourage checking YouTube

U substitution

[magic night]

Uh just to check but this is just saying du is equal to whatever du is times dx right? which is why it's 2 times dx?

I follow everything before that though

and I get rearranging the equation to make dx the subject

[neon iron]

If I understand you correctly, right

Try to solve similar problems like integrate sin(3x-9) to see if you have really understood it

[magic night]

Oh that'd be -1/3 Cos (3x-9) right

[neon iron]

Right

[magic night]

I think I get the usual antidiff for ex, basic variables and circular, I just can't grasp the du changing thing

[neon iron]

,w integrate sin(3x-9)

[thorny flame]

Results provided by WolframAlpha

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

Requested by Masala Oats#7677

[magic night]

9-3x?

[neon iron]

cos(x) = cos(-x)

And wolframalpha is weird

[magic night]

Oh ye, I thought that'd only give the -1/3

oh wait

Nevermind that comment I just realised what that meant

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[magic night]

uh ye

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[magic night]

I figured it out

Sorry for the trouble 💀

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[craggy shoal]

we got value of PR from a previous question, which gives us 8

[dapper lantern]

Find PR first

And then use trigonometry

[craggy shoal]

we have s = 60

[shut vigil]

That's pythagorean theorem

For the 60 degrees

that is a special angle

[craggy shoal]

i have PR, i dont have ps or pr

[muted spire]

use sin 60 = PR/RS

[craggy shoal]

ok

sin 60 is root 3/2

[shut vigil]

If you do something like this, that is an equilateral triangle so all sides are the same

[shut vigil]

and you'll find this

[craggy shoal]

?

i didnt quite understand

just started trigonometry this year

sin s = pr/rs

sin s = 60

sin 60 = root 3/2

understandable

then what?

8/2?

4?

[muted spire]

ok so you have the equation
(root3)/2=8/RS

[craggy shoal]

yes

[shut vigil]

And try pythagorean theorem here

[craggy shoal]

that just makes it look more complicated bruh

tf is this

[muted spire]

cant you solve the equation

its just some multiplication

[shut vigil]

Lemme explain 1 by 1

[shut vigil]

This picture shows an equilateral triangle, a triangle with the same three sides

[shut vigil]

because angle SRP = 30

if we add two of these together we have

[shut vigil]

this

[craggy shoal]

so 16?

[shut vigil]

ah no

[shut vigil]

16 = b frome here

so sub it

x whatever I wrote = 16

find for x

[craggy shoal]

16/root 3

[craggy shoal]

still dont get why u did all this

but yeah

ty

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[shut vigil]

It's wrong

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[craggy shoal]

16/3 * root 3?

[shut vigil]

nah

its 16 = x sqrt(2)/2

so it's 32/sqrt(2) = x

or 32 sqrt(2)/2 = x

or 16 x sqrt(2)

18.48

[craggy shoal]

that aint even a option bruh

[shut vigil]

Ah bruh I made a mistake there

[craggy shoal]

16/root 3 which makes it 16 by root root 3 * root 3 / root 3 which is 16/3 * roott 3?

[shut vigil]

AHHH

I know where the mistake is

It's here

Maybe I'm too tired
It's x^2 / 4

so 3x^2/4

or xsqrt(3)/2 = 16

so 16/sqrt(3)

yeah it's D

cause that's just the same but rationalized

so 16sqrt(3)/ sqrt(3)^2 or 16 sqrt(3)/3

Yah

I'm just too tired maybe I should play rest first

[craggy shoal]

yep

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[severe bramble]

Integrate x^(-a/b)*sin(x) dx from 0 to 1. Do it with partial integration (Done) and then by variable substitution with x=y^p for a well choosen p. Now whenever I try to select, 1, -1, 2, -2, +-5, +-10, +-9, +-10/9 I keep getting a thing I do not want when I get the dx/dy part. I think I'm missing something obvious. It's ok if we need to solve the integral with numerical methods after the substituion.

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@severe bramble Has your question been resolved?

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@severe bramble Has your question been resolved?

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@severe bramble Has your question been resolved?

[brazen plover]

What if you chose p=-b/a so the first part becomes simple

[severe bramble]

Ok so I think I solved it but still get a wrong answer when I calculate it. So dx = py^(p-1) so $integral_0^1 p y^(2p-19/10)cos(y^p) dy, if I take p = 19/20 lots of the things cancel out and I can numericaly solve it in matlab with '''matlab integral(@(y) (19/20).*cos(y.^(19/20)),0,1)'''

so if I integrate, from 0 to 1 of (19/20) .* cos(y.^(19/20)) I get 0.7943

so if I integrate, from 0 to 1 of (x.^(1/10)) .* (sin(x)) I get 0.4369

But they should give me the same value

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[rugged idol]

I struggle with the following math exercise:
You can have friends from town A and Town B.
You can have at max 4 friends from A and unlimited friends (like N) from B
X is a random variable representing the number of friends you have from B
You can only have friends from B if you have 4 friends from A.
X is Zipf distributed.
Whats the probability that you have at least one friend from A

My approach was we introduce Y that represents the number of friends I have from A and calculate
P(Y=4 & X>0)
But now I struggle with calculating Y. I dont have any probability given for someone from A being my friend.
I just know that X is Zipf distributed with parameter 4.
Can someone maybe give me a hint or something like that ?

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[rugged idol]

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[lyric mountain]

Assume Calc 1 Knowledge.
For questions like these, aren't I just finding the first deriv and C, then I just plug in -1 into f'?

[floral bison]

Do you know the FTC

[lyric mountain]

idk what FTC is 🙂

[silver trench]

i’m assuming it stands for fundamental theorem of calculus

[lyric mountain]

OHHH

Yeah

I tots do

[winter egret]

The idea is that f(x)=F(x^2)-F(2) for some F that has the property that F'(t)=t/(t+1)

But if f(x)=F(x^2)-F(2), then certainly f'(x) = F'(x^2) - 0 = x^2 / (x^2 + 1)

[lyric mountain]

this solution doesn't make sense to me

it looks like he subbed the upper and lower limit

like

when I do my u sub, I know I have to change the upper and lower limit

but he used those to sub his integrand??

idk why my Professor would include parts of a solution where he hasn't taught us how to do

[floral bison]

Ok

[lyric mountain]

Like, I did u sub on t + 1

[floral bison]

$\frac{d}{dx}(\int_{a}^{f(x)} (g(x)) dx)$
=$g(f(x))(\frac{d}{dx} f(x))$

Oh lord

Hold up

[lyric mountain]

sure

[thorny flame]

% Openglobe %

[floral bison]

There it is

[lyric mountain]

do I have to do it that way

[floral bison]

That’s an alternate way

[lyric mountain]

that looks like chain rule

[floral bison]

But much easier

It’s the reverse chain rule

[lyric mountain]

LOL

yeah

mf never taught us about reverse chain rule

ty

at least now I know what he did

[floral bison]

This “u” stuff is confusing

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[lyric mountain]

Yeah

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[neon iron]

what are we talking here about ? it is permutation right?

[alpine mist]

no, the n things are all the same. So the number of values of r is n+1

so there are n+1 ways to choose r things from n identical things

[neon iron]

okie

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[neon iron]

p & q are alike objects and s is distinct object... in how many ways can we select object in this ?? i know when one group are alike the eq is (f+1)(h+1)-1 and when distinct it is (p+q)!/p! * q!

[neon iron]

.close

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[elfin dagger]

did i do this right?

[topaz sinew]

@elfin dagger Has your question been resolved?

[elfin dagger]

<@&286206848099549185>

[visual wedge]

0

It should be 0

[elfin dagger]

why 0?

[visual wedge]

Wait

Cross product gives you a vector that is perpendicular to the plane of the crossing vector i.e p and r here

[neon iron]

Shouldn't it be up the plane?

[visual wedge]

In my diagram, no

[neon iron]

Why not?

[visual wedge]

Taking dot of 2 perpendicular vectors give 0

[neon iron]

Oh nvm its the right hand rule

Im dum enough to repeat the same mistake over n over again

[elfin dagger]

im not just doing cross product the question asks me to do a dot product of r too

[visual wedge]

Happened with me couple of times.

Practice made it go away

[visual wedge]

Yes

[visual wedge]

Here

[elfin dagger]

oh so is there no calculations needed then?

[visual wedge]

No

Simple observation, will save you some labour

These are trick question

[elfin dagger]

so i did my cross product wrong then?

[visual wedge]

Let me check

[elfin dagger]

since if i do the dot product with r it has to equal 0

[visual wedge]

I got 2,2,1

🤔

Do it like this, it is cleaner

[elfin dagger]

is my method bad? it's what was taught in my lesson

[visual wedge]

Determinant

[visual wedge]

It looks good but you can space out to avoid confusion

[ivory swallow]

actually it would be 3i+2j+k

[visual wedge]

Can you kindly point out the mistake in my image?

[ivory swallow]

(4-1)i

[visual wedge]

Right

My silly mistakes are still there then

[ivory swallow]

you have done right . you just did minor mistake in j part

[elfin dagger]

ok what was it?

[ivory swallow]

1 X 0-(-2 X 1) that will be 0+2=2

[elfin dagger]

ohh i see it thanks

real dumb mistake my bad

thanks for the help guys have a good night

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[velvet phoenix]

I'm going through some solutions to a past paper (at uni) and I don't understand the solution given

[velvet phoenix]

Let $\chi$ be a non-trivial Dirichlet character modulo an integer $m > 1.$ Prove that the sum of the values of $\chi$ at the first m non-negative integers equals 0

[thorny flame]

LeftySam

[velvet phoenix]

Solution:

Since $\chi$ is non-trivial there exists an invertible $b$ such that $\chi(b) \neq 1$

[thorny flame]

LeftySam

[velvet phoenix]

(this is the first step, not the whole solution); I don't get this? Why is this true?

[odd pagoda]

well thats what non-trivial means

trivial sends all invertible elements to 1

[topaz sinew]

@velvet phoenix Has your question been resolved?

[velvet phoenix]

Why is n non-invertible if gcd(n,m) > 1?

[odd pagoda]

bezouts lemma

or well, thats the other direction

if gcd(n, m)>1 then it is clearly a zero divisor

[velvet phoenix]

Idk what that means

I'm confused what we're even talking about with regards to invertible here too

[odd pagoda]

has a multiplicative inverse

[velvet phoenix]

But what field are we talking about here?

$\mathbb{R}$?

[thorny flame]

LeftySam

[odd pagoda]

not a field

a ring

mod n

Z/nZ

[velvet phoenix]

It's mod m no?

[odd pagoda]

same thing

[velvet phoenix]

I don't get what you mean here

Or why that implies the result

[odd pagoda]

if m=dk with d=gcd(m,n) and n=dt, then nk=dtk=mt=0 mod m

an element can only be a zero divisor or invertible

exclusive or

[velvet phoenix]

Why wouldn't this method work for gcd(n,m) = 1. I.e. if m = k and n = t then nk = tk = tm = 0 mod m

[odd pagoda]

then k=m=0

so you are multiplying by 0

which isn't interesting

but if d>1 then k is nonzero so you are multiplying by a nonzero element and get 0

which means zero divisor

[velvet phoenix]

Okay, the next is

As $a$ runs through all residue classes modulo $p,$ so does $c = ab$ as $c = ab$ has a unique solution in $a$ for every given $c.$ Hence $$\sum_{c=0}^{m-1} \chi(c) = \sum_{a=0}^{m-1} \chi(ab) = \chi(b) \cdot \sum_{a=0}^{m-1} \chi(a)$$ shows that $\sum_{a=0}^{m-1} \chi(a) = 0.$

[thorny flame]

LeftySam

[velvet phoenix]

So a i'm assuming is what? A generator?

[odd pagoda]

you have (thing)=(something not 1)*(thing) so (thing)=0

just the sum index

[velvet phoenix]

No I get that, but I don't get the wordy bit

What does it mean that "a runs through all residue classes modulo p" anyway? A generator? Contained in the set {1,2,...,p-1}?

[odd pagoda]

if you let a be every element mod m, then also c=ab will hit every element

not sure why you switched to mod p now?

[velvet phoenix]

I meant m mb

Bruh this is crazy I can't believe this was an exam question

[odd pagoda]

this is an easy question

[velvet phoenix]

Seems unnecessarily difficult

[odd pagoda]

no offense but you dont seem to know any basics

[velvet phoenix]

I'm in third year

It's just "a runs through all residue classes" is a very vague statement to me at least

[odd pagoda]

it's a very classic math statement

just look how its used in the sum

you also say that sums run over some numbers

every number from 0 to m-1 is used

for a

and then also for c because b is invertible

[velvet phoenix]

Does it mean a generates the group mod m {1,2,3,...,m-1}?

[odd pagoda]

a is not a fixed element

it is a running index

[velvet phoenix]

So what does it mean, you have a in {1,2,3,...,m-1} mod m and you sum over each possible a?

[odd pagoda]

do you know how sum notation works

[velvet phoenix]

Yes

[odd pagoda]

you keep excluding 0

a runs from 0 to m-1

just like in the sum

[velvet phoenix]

Sure, then include 0 in that, is that correct?

[odd pagoda]

yes

[velvet phoenix]

.close

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[hallow garden]

6. Water flows through a cylindrical pipe of internal diameter 7cm at 5m per second. Calculate:
   i) The volume in litres, of water discharged by the pipe in 1 minute.

Here's what I'm doing

π * 300 * 3.5
11545.35/1000
11.545

[hallow garden]

Am I doing correct?

I don't have any marking scheme which I can check against

[mellow arrow]

consider volume of water that flows in one second
It should be a cylinder of definite length

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[neon iron]

i need help with this

im kinda stumped

i did ratio thing

for first one

i got mod x <1

so r is 1

?

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@neon iron Has your question been resolved?

[neon iron]

help

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@neon iron Has your question been resolved?

[neon iron]

help

[sour bloom]

yeah looks like left one is r=1

[sour bloom]

what do you reckon right one is

[neon iron]

i did that one too

can u look at my solutions

ill send them

@sour bloom

[sour bloom]

main ideas are there, presentation can be more formalised

[neon iron]

of course

i will do

is it correct?

@sour bloom

[sour bloom]

yeah seems so

[neon iron]

ty

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[ivory cradle]

ok I have geometric equations, I'm at Tn=6600(0.825)^n-1 to model the value of a machine after n years of depreciation, how much was the machine originally worth

[ivory cradle]

after looking I am just stuck idk what I'm doing to start

<@&286206848099549185> ?

[fickle oracle]

It hasn't been 15 min yet

[ivory cradle]

yes it has ....

I'm not quite that bad at math

[fickle oracle]

Oh sorry, I didn't see the time

It's 11:28 already

OK let's see}

Is this your equation

[pastel oracle]

find the value when n=0?

[fickle oracle]

$T_n = 6600 \cdot (0.825)^{n-1}$

[thorny flame]

Max Hetfield

[ivory cradle]

yes

[fickle oracle]

What is the n?

[ivory cradle]

i mean for the initial value 1 ??

[fickle oracle]

n is the years of depreciation

[ivory cradle]

or is that for after 1 year of depreciation

okay

[fickle oracle]

If the machine is new, how many years of depreciation have passed?

[ivory cradle]

so then its just 0

yeah

[fickle oracle]

Yes

n = 0

Replace and solve