help-17

[vast shale]

why did it change to k+4 all of a sudden

[prisma linden]

right

[vast shale]

oh wait

.reopen

[vocal sleet]

✅

[vast shale]

this si wrong

what do you get after simplifying this ?

[prisma linden]

wait can I even do that

hold on

[vast shale]

no thats wrong

[prisma linden]

ok im back to here

imma try again

[vast shale]

alr

[prisma linden]

yeah I need a hint

ohhhh

is this it

[vast shale]

yes

also

while proving LHS = RHS

you should not write RHS for every step

[prisma linden]

what if I do

is it entirely wrong or is it just unnecessary

[vast shale]

Quite confusing tbh

[vast shale]

Simplify thjs

||hint :- a* c+a* b = (b+c)*a||

quite rude,apologies

[prisma linden]

what

[vast shale]

not sure

[vast shale]

its 1

not 2

[vast shale]

you have a as 2^k+3

[vast shale]

yes

do you understand why though ?

[prisma linden]

factorization

is it

I understand the concept but I usually don't think to use it

unless hinted

is this the end

[vast shale]

" Do not write RHS every time"

until its fine with your teacher

[vast shale]

yes

[prisma linden]

alright I appreaciate the help!

[prisma linden]

I'll check with her

[vast shale]

Okay

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[civic oracle]

i mean delta

[vocal sleet]

@civic oracle Has your question been resolved?

[flat whale]

delta = 1 doesn't work when M < 0 probably

[civic oracle]

so in what cases should i apply this in the future

[flat whale]

whenever delta doesn't work for all values of M

do a lot of problems to find out

[civic oracle]

ok lol

thanks

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[sturdy jolt]

I need help on part b of this question can someone explain how they get this

[sturdy jolt]

Its proof by induction btw

[heavy yoke]

it's assumed that you know already that [ \sum_{r = 1}^n r^3 = \ab(\frac{n(n+1)}2)^2 ]

[twin meteor]

cloud

[sturdy jolt]

Doesnt it equal 1/4 n^2 (n+1)^2

Or is it the same thing

[silk osprey]

that’s the same thing

(1/2)^2 = 1/4

[sturdy jolt]

So in the mark scheme what have they done at the top

[heavy yoke]

they set the formulas for the two sums equal

[sturdy jolt]

Ok i get it now

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[hardy crater]

I might be stupid but would this be correct?

[silk osprey]

yea

[hardy crater]

Well obviously it’s not correct but

That question looks really tricky to read, sorry

It looks better when you zoom in

[short bone]

is it 1, -12, -1

[hardy crater]

It might be, I’ll lose marks if it’s wrong though so if you could possibly explain what you did that would be greatly appreciated

Wait I’m an in idiot

I could have divided the 4 by 2

[hardy crater]

I agree with this

[short bone]

i got this for small value approximation of sin and tan, not really an expert tho

[hardy crater]

I got that

[short bone]

kk

[hardy crater]

I stupidly timesd it all by 2…

Instead of just dividing by 2

[short bone]

just converted everytg into sin and tan

[short bone]

happens to the best of us

[hardy crater]

Yea it was correct

Thank you so much

[short bone]

no problem

[hardy crater]

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[unreal sparrow]

The upper bound is pi/2, but if you set r = 2 equal to r = 4/sin(theta) you get 2 = sin(theta) and theta isn’t pi/2 at that point, what am i missing here?

[vocal sleet]

@unreal sparrow Has your question been resolved?

[wraith python]

The radius at (0,4) would be 4. 🤔

[vocal sleet]

@unreal sparrow Has your question been resolved?

[silk rampart]

Still need?

[unreal sparrow]

Ye plz

[silk rampart]

yk the equations and all, I can see that

you need the area bound by the curve

we won't get very far by integrating along the x-axis

try integration along the y-axis @unreal sparrow

@unreal sparrow also we can avoid integration altogether if we notice certain shapes which give us the area

[vocal sleet]

@unreal sparrow Has your question been resolved?

[unreal sparrow]

Ok thanks i’ll check in a few minutes,

[unreal sparrow]

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[livid void]

$-168\pi = -527.7$?

[twin meteor]

Ousel

[jagged cargo]

they probably want you to not approximate

[livid void]

So I should just assume not to include pi, when doing the equation, unless it says approximate?

[frozen bobcat]

it says 'exact answer', so you have to use the pi symbol

-527.7 is approximately -168pi

[livid void]

gotcha, didnt even notice the blue text honestly 😅

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[ivory sand]

Can someone help me with question 5

Better photo

[vocal sleet]

@ivory sand Has your question been resolved?

[formal yew]

Did you get part a

[ivory sand]

No I didn't

[formal yew]

A reflection over x axis is what

In simple terms

[ivory sand]

a mirror

[formal yew]

Horizontal or vertical reflection

[ivory sand]

vertical reflection

[formal yew]

Ye

[ivory sand]

I know that already

[formal yew]

Then vertical scale factor

[ivory sand]

can we do that with the equation though

[formal yew]

Yes it’s the exact same principle

The transformation is

y= -2/5h(x)

[ivory sand]

ok I gort that

got

[formal yew]

So now you just plug in h(x)

[ivory sand]

wait can you explain

[formal yew]

a=-2/5

f in this case is

h(x)

-2/5 (3(x-3)(x+2)(x-5)

[ivory sand]

ok thank you, but what about b

[formal yew]

B is nothin here

[ivory sand]

no like the question

[formal yew]

Oh

So to move the units all left 4

You just have to add positive 4 to each x intercept

[ivory sand]

is that it?

[next mantle]

bro

[formal yew]

Ye

[next mantle]

we've been doing this for 2 hours

[formal yew]

Makes sense why?

[ivory sand]

ya

thank you

[formal yew]

And the down part

You know right?

[ivory sand]

ya

[formal yew]

Final equation

[ivory sand]

THANK YOU

can you help us with some other quistions

[next mantle]

questions*

[formal yew]

The last one?

[next mantle]

6

[ivory sand]

ya

[formal yew]

Ok write down what you know

[ivory sand]

y intercept at -200

4 x intercepts

and x is negative when x<-4 and -1 <x <2

oh so x intercepts would be -4, -1, and 2

[formal yew]

ye

oops

what do u think u should do next

[ivory sand]

start writing equation with those x intercepts

[formal yew]

ye

something like this correct

we dont know A

[ivory sand]

ya

[formal yew]

its just a placeholder

now you got the next step

find A

[ivory sand]

so than you just set the all the Xs to 0 and make it so it equals -200

right?

[formal yew]

ye

[ivory sand]

I got 5 as A

Is that right?

[formal yew]

negative 5

A is negative or positive 5

when solving

then you have to take a step back

and then about

whether it is going to be (X-5) or (X+5)

[ivory sand]

ohh

wait how do I know which ones it going to be

[formal yew]

think about

if

A was 5

then we have an x int at x=-5

therefore the function is negative (-infinity, -5)

aswell

which isnt true

[ivory sand]

OH

that makes sense

wait so final equation would be (x+4)(x+1)(x-2)(x-5)^2

[formal yew]

ye

[ivory sand]

wait can you help me with another question or is that too much

Question 4

[ivory sand]

bro I'm not done yet

I'm still waiting for help

[ivory sand]

<@&286206848099549185> can I get help with question 4

[scenic holly]

Is q4 the whole photo?

[ivory sand]

No it’s just 4

[scenic holly]

Ok so the first thing I'd say is: since it's divisible by that quadratic, each of the factors of the quadratic is also a factor if the main polynomial

Based off of this , can you find for me two roots of the 4th degree polynomial?

@ivory sand Ping me when you text me back over here

[ivory sand]

Ok

[flat osprey]

!occupied

[vocal sleet]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[ivory sand]

It would be x=-5 and -3 right?

@scenic holly

[scenic holly]

Yes

Now, because 3x⁴+mx³+4x²+nx+75=0 has these two roots, the next thing you'd want to do is make a system of two equations in m and n, based off of the two roots you found, to be able to find the values of m and n

Do you know how you would do so?

[ivory sand]

wait can you explain more

[scenic holly]

3(-5)⁴+m(-5)³+4(-5)²+n(-5)+75=0
And
3(-3)⁴+m(-3)³+4(-3)²+n(-3)+75=0

Simplify these two equations for me

[scenic holly]

Also do you understand how we are able to write this?

[ivory sand]

yes you are inputting x with the roots

[scenic holly]

Yes

So @ivory sand What do the two equations become when simplified?

[ivory sand]

I'm doing that right now

wait can you help me simplify it

I want to make sure I'm doing it right

[scenic holly]

Ping me when you text back

[ivory sand]

@scenic holly can you help me simplify

[scenic holly]

Yes ofc

Here's the first one @ivory sand

You do the second one

[ivory sand]

Ok thank you

I got 354-27m-3n=0

is that correct?

[scenic holly]

Yo let me just double check if ite correct

@ivory sand Yes that's correct

Now with these two equations do you know how to solve for m and n?

[ivory sand]

No I don't

[scenic holly]

You didn't do systems of equations in grade 10?

Like substitution, elimination and all that stuff

[ivory sand]

I probably did but I forget

[scenic holly]

Ah ok

So

[ivory sand]

what exactly did you do here?

[scenic holly]

I multiplied the first equation by -3 , like I multiplied everything on both sides by -3. And then the second one I multiplied it by 5. The goal was to have 15n in one equation and -15n in the second, so that by adding them, the n's cancel and there's just an e2uation with only the m variable

[ivory sand]

wait so m is 18.25

[scenic holly]

Yes

But let's remember the entire question

We'd need to use this to find n, and now that we have the whole equation itself you should know how to find the roots

[ivory sand]

wait how would I find the n

[scenic holly]

354-27(18.25)-3n=0

[ivory sand]

oh ok

[scenic holly]

And then with the values of m and n will you be able to / know how to find the two other roots?

[ivory sand]

No I don't

[scenic holly]

Now the equation becomes like this. So, to find the roots of this equation, first divide it by (x+5) and then divide it by (x+3) knowing that both of these are factors of the polynomial. Do these two divisions and then ping me

Or if you're confused about how polynomial division works , then ping me

@ivory sand

[ivory sand]

wait isn't n -46.25

[scenic holly]

Yes you're actually correct

I had put 18.5 instead of 18.25 into the calculation by accident when I was finding n

[ivory sand]

wait can you long divide those for me

I think I made a mistake

@scenic holly

[scenic holly]

You do the long division

You'll be the one doing it in the test

[scenic holly]

Send a photo of your work

[ivory sand]

Oh I used the wrong n value

@scenic holly I understand this now thanks for the help

I'm going to call it a night

[scenic holly]

No problem @ivory sand

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[vocal tusk]

Lemma 8.2.7 and Theorem 8.2.8 in the attached images come from Analysis I, by Terence Tao (I attached the lemma since it's referenced in the proof of the theorem). I need help proving the second bullet point in the proof of Theorem 8.2.8: "Case 1 occurs an infinite number of times,...".

Here's my attempt for showing it for Case 1 (Case 2 should follow a similar argument):

---

Suppose, for sake of contradiction, Case 1 occurs a finite amount of times, say $k$ times. This means we set $n_j := \min {n \in A_{+} : n \neq n_i \ \forall i < j} \ k$ times. Hence, the set

$$
B = {m \in \mathbf{N} : m = \min {n \in A_{+} : n \neq n_i \forall i < j} }
$$

is finite with cardinality equal to $k$. The set $A_{+} \setminus B$ is a countably infinite subset of $\mathbf{N}$ -- thus, by the well-ordering principle, it has a minimum element, say $m'$.

---

So this is where I got stuck. First of all, I'm not sure if I defined set $B$ correctly. Second of all, I want to somehow show that $m'$ is not equal to any element in $B$. but also that it must be in $B$, contradicting the cardinality of $B$. Can anyone help me fix this?

[twin meteor]

pash

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[vocal sleet]

@vocal tusk Has your question been resolved?

[vocal tusk]

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[still mica]

How to integrate
dx/(6sinx + 10)

[tribal cedar]

u=tan( x/2)

sinx =2u/1+u^2

[still mica]

I see

Thanks

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[unborn beacon]

Why is the area of a rectangle = length x width?

[unborn beacon]

And How do you prove it? (Ping me)

[potent beacon]

I think it depends how you define "area" and "rectangle"

[unborn beacon]

Area is generally defined as "How much space a 2D shape has Occupied"

[potent beacon]

This is some intuition

Not definition to prove anything

[unborn beacon]

Then, what would be the needed definition to prove this statement?

[potent beacon]

I guess you could define area in terms of Lebesgue measure and then take integral from constant function

[brittle cipher]

I suppose the distributive law might be useful to know if you chop a rectangle into two rectangles, then it's consistent

[unborn beacon]

My first idea was to define the Rectangle as Rectangles with width so short that it is almost consider a line and follow up with some stuff.

[potent beacon]

But then you need area of rectangle anyways

To define Riemann integral logically

[unborn beacon]

Ok.

[potent beacon]

What you could do

[hard atlas]

well the lebesgue measure is defined using that the area of rectangles is length times width

[potent beacon]

Is prove that Lebesgue integral is equal to Riemman if it exists

[unborn beacon]

But my second Idea is to make points Nth demonsional Hyper cubes.

[potent beacon]

But its universal

And you can define it

In search of a property

That satisfies things we want to be obeyed

And it works in n dimensions

[unborn beacon]

So @potent beacon What are your plans?

[potent beacon]

And than you can take integral of constant function

[unborn beacon]

But don't integrals require the definition of rectangles?

[hard atlas]

you could also show that the determinant is the only multilinear map

[potent beacon]

You can just use Lebesgue measure

[unborn beacon]

I know nothing about that topic.

[potent beacon]

That doesnt suprise me

[unborn beacon]

I did think of a simplier model

I let a point of a cube shaped brick

[potent beacon]

What?

[unborn beacon]

and the *magnitude of the length is how many bricks there are

and the magnitude of the Area of any shape you make with the brick will also be the amount of bricks you placed

[potent beacon]

Umm

[unborn beacon]

and you can keep going

But it sound a little bit weird

[potent beacon]

But what if ratio of sides is irrational?

[unborn beacon]

That's a great question

I haven't think of it before.

Ok, I guess it's Lebesgue time. Can you explain how you would proof it with the Lebesgue measure?

@potent beacon

[potent beacon]

If you dont know what is it

There isnt really a way

Youd just take integral

Of a constant function

[unborn beacon]

So, this is an axiom?

[unborn beacon]

Yes, but isn't integrals derive from adding the areas of infinite rectangle? is it?

[potent beacon]

Riemman integrals

Lebesgue integral is generally a Lebesgue measure

[unborn beacon]

Ok, thanks anyways.

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[lone igloo]

Right, to find acceleration, Force / Mass? So 25 + 50 = 75N then 75N/5 kg is 15?

15m/s^2 going right is the acceleration right.

?

[heavy yoke]

yes

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[cunning olive]

Hey, just recently came across logarithms. I understand them at the basic level but need help understanding something please

[cunning olive]

If a log doesn’t have a base, what do we assume the base to be?

[hard atlas]

e

[cunning olive]

Eulers number?

[warm vine]

in high school they teach 10

and ln for natural log

[cunning olive]

Ok thanks!

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[quiet gazelle]

Considering a function f: R -[−1, 1[→ R given by f(x)= x² −1, the image is given by the interval:
a) [1,+∞[

b) [0,+∞[

c) [−1,0]

d) [−1,0[

e) [−∞,−1[

[quiet gazelle]

I didn't understand how it works

Posting it again btw

[oak magnet]

When does f(x) is negative ?

[oak magnet]

You want to know the least value possible for f(x) and the biggest possible

So if you have x² -1

The roots are -1 and 1

[quiet gazelle]

And then what?

[oak magnet]

As coefficient in front of x² > 0, then its positive on ]-inf, -1] U [1, +inf[

With the restriction, we have that f(x) is always positive, and so it is 0 in one value, 1

So 0 <= f(x) < +inf

(You could said that x² -1 is an even function too, and so only works with the positive interval)

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[icy stirrup]

hi can anyone explain what step (3) is doing when it puts the sigma sign in front of the integral? cuz it seems to suggest that its the summation from - to + infinite but i thought that only applies to normal distribution curves that tend to y=0 at both ends, the definite integral would be a slice of that so no asymptotes so im confused as to the notation here

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[sharp jetty]

Can u help me to understand this problem ?

[short bone]

this is considered one of the toughest problem to come in jee advanced, u can look it up in youtube, there are lots of vids explaining it if u dont find any help here

[sharp jetty]

Hmm i watched one but didnt got anything helpful

[short bone]

u can find more

havent really seen or tried this, out teacher advised us to do this after a few months n not even look at this question

[sharp jetty]

okh no problem
And thank you for your time 🙂

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[dry yacht]

Use binomial.

[indigo karma]

To find the 6th term?

[flat whale]

whatever term that gives you x^5

[indigo karma]

126.16.243x

[outer warren]

wrong power for the x, that term would have x^5

and then you only need the coefficient of that

[indigo karma]

9C5 * 2^4 * (3x)^5

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[left portal]

\textbf{Exercise 5.} Let ( f: \mathbb{R}^3 \rightarrow \mathbb{R}^2 ) be the linear transformation ( f(x_1, x_2, x_3) = (x_1 - x_2, x_2 + x_3) )

and let ( \mathbf{v} = (2,3) ); ( S = \langle (1,2,1) \rangle ); ( T = { \mathbf{x} \in \mathbb{R}^2 \mid 3x_1 - 2x_2 = 0 } ).

Find ( f(S) ), ( f^{-1}(\mathbf{v}) ), and ( f^{-1}(T) ).

[twin meteor]

938c2cc0dcc05f2b68c4287040cfcf71

[kindred epoch]

Matrices are the key for inverse linear transformations in R. Have you found the standard matrix yet/know how to?

[left portal]

i did

can u check my shit

shit

I did not find the matrix

let me try again I misread your shit

[kindred epoch]

U good I actually think this is also a viable way to find the inverse but I didnt check it to thoroughly

I just recommend the matrix bc its so methodical so it makes you much less prone to mistakes vs expanding out a whole system of equations

[left portal]

ye

f(1,0,0)=(1,0)

f(0,1,0)=(-1,1)

f(0,0,1)=(0,1)

[kindred epoch]

hold the phone

I might be tweaking

Is this even an invertable transformation?

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@left portal Has your question been resolved?

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[round bone]

.open

lol it is no command

[round bone]

can anyone explain 2x^3-18xy^2

factorise

and 1+x-y-xy

[twin meteor]

amitt

[raven zephyr]

2x(x+3y)(x-3y)

[round bone]

how to get it

[raven zephyr]

take out 2x, 2x(x^2 - 9y^2)

use a^2-b^2 = (a+b)(a-b)

[arctic mantle]

yeah you can just post a message

[round bone]

lol i forgot

hehe

[round bone]

a is ?

2x^2?

no

yes?

lol idk

[raven zephyr]

yeh look at signs....

[round bone]

huh

[raven zephyr]

yeah u r right

[round bone]

x(2x^2) right?

?

[raven zephyr]

whaatt

[round bone]

x(2x^2)=2x^3

?

[arctic mantle]

yeah

but he's suggesting you pull the 2 out as well

i.e.

[round bone]

2x(x^2)?

[arctic mantle]

\begin{align*}
2x^3 - 18xy^2 &= 2x(x^2) - 2x(9y^2)\
&= 2x(x^2 - 9y^2)
\end{align*}

[round bone]

the last one is the final answer right?

[arctic mantle]

no

[twin meteor]

ren

[arctic mantle]

use the identity a^2 - b^2 = (a+b)(a-b)

[round bone]

what is a here

oooooooooooooo

[arctic mantle]

my guy we're not going to hand you everything on a silver platter

what do you think a is

[round bone]

so we need to make it so it is ^2

both terms right

[arctic mantle]

yeah

[round bone]

oh

x^2 = a

9y^2 = b

?

[arctic mantle]

no-

think again

close though

[round bone]

bruh i cant

imma go take a break

and comeback

my brain ain't braining🤣

i am sorry to waste ur time

.close

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[sturdy jolt]

For question 2 a i dont get how they simplified it

[sturdy jolt]

.

I need help

[flat whale]

which part first

[sturdy jolt]

Wdym

2a

At the third = sign

[flat whale]

from this

to this?

[sturdy jolt]

Yh

[flat whale]

13 = 7 + 6

[sturdy jolt]

So they seperated the 13(13^k)

[flat whale]

yes

[sturdy jolt]

I get

Ty

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[vast shale]

so is this channel used for help?

[sturdy sage]

uelp please

[silk osprey]

yea

just post your question

[silk osprey]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

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[heady vine]

How can I graph equations in the from
|y|=|x|+c

[gentle thicket]

nvm didn't saw that y

[heady vine]

It’s ok

[gentle thicket]

it will be still a similar process though

[heady vine]

I understand the simple y=|x| graph and it’s transformations
But the problem with me is for
|y|=|x|+c cuz for some reason in desmos when I graph it
the positive part is moved in the y direction by c units
And the negative part by -c units

[gentle thicket]

if I were to do this problem, I might first plot y=|x|+c roughly
then just transform it into your one

[gentle thicket]

oh okok wait

[gentle thicket]

but what's your problem with it?

[heady vine]

Also the when y=|x|-c
when y is less than zero the values just disappear in |y|=|x|-c

gets revalued *

[heady vine]

My problem with it after experiencing with it in desmos
Is in some cases the graph would like |x| and -|x|
and other case where it would like |y| and |-y|
I want to know simply how can I assume what would the graph look like

[gentle thicket]

,w plot y=|x|+5

[twin meteor]

Results provided by WolframAlpha

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

Requested by seventysevensquare

[gentle thicket]

see the graph of y=|x|+c would be similar to this except the y intercept

[heady vine]

I think I found it out after experimenting a little
|y|=|x|+c
When c is positive the graph would look like |x|+c and -|x|-c
and when c is negative
it’ll look like x=|y|+c and x=-|y|-c

[gentle thicket]

okay that's fine too

[heady vine]

But I want to find mathematically tho instead of graphing it

[gentle thicket]

what else do you want to do with it?

[gentle thicket]

you just mentioned graph the equation thing
but what else do you want to do?

[heady vine]

I don’t think I need it for any other thing tho lol

Just want to find it that’s all

[spiral inlet]

|a|=b is equivalent to a=±b, b≥0

or in this case

|y|=|x|+c is equivalent to y=±(|x|+c) for |x|+c ≥ 0

So you can find the values of x such that |x|+c ≥ 0, and graph y=±(|x|+c) for those values of x

[spiral inlet]

If c is positive (or 0), that will be all real numbers, if c is negative, there will be a gap in the middle, like you noticed here

[heady vine]

I see

[heady vine]

ngl that makes it clearer in the mathematical sense

Thanks everyone

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[fallen sandal]

idk how do

[mild rapids]

dy/dx = 0

find the derivative, then just = 0

[fallen sandal]

y

[grim lotus]

what

[tired condor]

are you using limit definition

[vocal sleet]

@fallen sandal Has your question been resolved?

[dull bear]

Remember that you wanna work out $f'(x) = \lim_{h\to 0} \frac{f(x + h) - f(x)}h$ for the $f(x) = x^2 - 6x + 2$, and once you have that limit, as above set it to zero

[twin meteor]

@dull bear

[fallen sandal]

wat

like the h

[outer warren]

what about the h

[vocal sleet]

@fallen sandal Has your question been resolved?

[fallen sandal]

wat does liek where the tangent line is horizontal mean

[vocal sleet]

@fallen sandal Has your question been resolved?

[vocal sleet]

@fallen sandal Has your question been resolved?

[tidal salmon]

When the f'(x) = 0

So assuming ur talking about the definition of a limit

Which uses the f(x) as x->h

H is the difference between a different point

And x

[tidal salmon]

So x plus h as shown here

Minus f(x)

Which would be a secant

But as h approaches 0

The difference between the two values also approaches 0

Making it tan

And tan is the slope at any point

Which means for it to be horizontal the slope has to be 0

So do u get what that equation means now?

What h is, what horizontal tan is and why you set it to 0

[fallen sandal]

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[vast shale]

Dont mind the horrible english (blame the person who made this lol) but i assume a) is 0.9 correct?

[vast shale]

my idea is that you have $\6\P{H\given F} = 0.5$ and $\6\P{H\given NF} = 0.8$ so [
\6\P{\8{H\given F}\cup \8{H\given NF}} = \6\P{H\given F} + \6\P{H\given NF} - \6\P{H\given NF}\2\6\P{H\given F} = 0.9
]

is this crankery

[urban edge]

there is also the possibility of being told heads when the answer is tails

[vast shale]

oh and that shouldnt be accounted into what P(H) is i assume

ok maybe i should get my notation straight first

[urban edge]

this feels very wrong to me

[vast shale]

same

uhh

P(H): probability that B is told that coin landed on its head
P(aH): probability that the coin actually landed on its head
P(F): probabiity A forgor
P(nF): probability A didnt forgor

i think this is what i need to deal with here

ok

[urban edge]

personally i would go by cases

[vast shale]

oh thats smart actually

wait but there are like

8 cases right

[urban edge]

4

actually 6

[edgy gulch]

maybe try a tree diagram

does that work actually idk

[vast shale]

Case 1: coin lands on head, A forgets, B is told it is head
Case 2: coin lands on head, A forgets, B is told it is tails
Case 3: coin lands on tails, A forgets, B is told it is heads
Case 4: coin lands on tails, A forgets, B is told it is tails
Case 5: coin lands on tails, A remembers, B is told it is tails
Case 6: coin lands on heads, A remembers, B is told it is heads

ok yeah it is 6 cases

ok so the probability that the B gets told it is heads is 0.5 then

well no

P(H) = P(aH | F) + P(aH | nF) + P(T | aH)

i think this is it

,,
\P(H) = \4{\P(aH\cap F)}{\P(F)} + \4{\P(aH\cap nF)}{\P(nF)} + \4{\P(T\cap aH)}{\P(aH)}

[vast shale]

ok are all of those independent? thonk

ok i think so

[
\begin{aligned}[t]
\P(H) = \4{\P(aH\cap F)}{\P(F)} + \4{\P(aH\cap nF)}{\P(nF)} + \4{\P(T\cap aH)}{\P(aH)} = \4{\P(aH)\P(F)}{\P(F)} &+ \4{\P(aH)\P(nF)}{\P(nF)} \ &+ \4{\P(T)\P(aH)}{\P(aH)}
\end{aligned}
]

[vast shale]

Actually im not sure if P(T cap aH) is independent?

ok this is horribly wrong

can someone guide me?

[cobalt crypt]

holy

[cobalt crypt]

might i recommend this approach

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

,, 2\20.4 \2 0.5 + 0.6 \2 0.8

[vast shale]

did it in my head, but i think it is this then?

and b it would be

,, 0.4 \2 0.5 + 0.6 \2 0.8

[vast shale]

so far so good?

[cobalt crypt]

i didnt check the numbers

[vast shale]

:xd:

[vocal sleet]

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[dull bear]

,rccw

[hollow raft]

I’m not sure where to go from:

Y = mX + c

Where those are the logs of y and x

[hollow raft]

Plug what in exactly?

[edgy gulch]

whats your Y and whats your X

[dull bear]

Put in your Y = log(y) and X = log(x) and then manipulate for an equation y = …

[edgy gulch]

or take the log for each option and see which one results in the correct form

[edgy gulch]

ok good

so thats the form that you want

now take the log for each option

[hollow raft]

Thank you!!

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[vast shale]

im working on proof of lim x -> a x^3 = a^3

for all e > 0, there exists del > 0
Let x in R 
|x-a| < del => |f(x) - a^3| < e

|x^3 - a^3|
|x-a||a^2 + b^2 + ax|
|x-a||(x-a)^2 +3ax|```

[vast shale]

i can stipulate the del <= 1
get the value of |x-a| by triangle inequality
but im stuck with the 3ax lmao

how deal with it!

[vocal sleet]

@vast shale Has your question been resolved?

[vocal sleet]

@vast shale Has your question been resolved?

[bronze osprey]

go back a step, so $|x^3 - a^3| = |x - a| |x^2 + ax + a^2|$

so if we assume $|x - a| < 1$ or $a - 1 < x < a + 1$, we have $ |x^2 + ax + a^2| < x^2 + a|x| + a^2 < (\max(|a - 1|, |a + 1|)^2 + a \max(|a - 1|, |a + 1|) + a^2$

so we can let delta equal min{1, epsilon / (that nasty right-hand side)}

you could also simplify $\max(|a - 1|, |a + 1|) \le |a| + 1$ if you wanted to

[twin meteor]

south's secret twin brother

[vast shale]

oooooo

now i realize i could expand |x|

|x-a+a| <= |x-a| + |a|

[twin meteor]

south's secret twin brother

[bronze osprey]

yeah techniques like that

this bound isn't particularly neat but it works

[vast shale]

yeah

thank you!!!

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[vast shale]

gonna try similar approach for

higher powers

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[sand forge]

is b= 1/1*2 + 1/2*3 + 1/3*4 +... + 1/99*100 in the interval [51/100, 11/10]

[devout kindle]

can you tell me general term of it?

[sand forge]

I know it simplified somewhere and it ends up being something like 1- the last term or something

but I dont know exactly how to do it

[devout kindle]

well if you look closely its written as:

[raven zephyr]

i remember this one

[sand forge]

...Yeah...?

[devout kindle]

which can be further simplified as
1/n - 1/(n+1)

[sand forge]

wait what

[devout kindle]

uhm?

[sand forge]

1/ab == 1/a-1/b???

I didnt know this

[devout kindle]

no

what is n + 1 - n?

[sand forge]

1

oh

[devout kindle]

there you go

rewrite numerator

as that

and then seperate it?

[sand forge]

yey thx

I got it

[devout kindle]

alr

now write some terms

and maybe i think you will be left with first and last term

[sand forge]

(I remembered something from class but I wasn't quite sure)

[devout kindle]

everything gets cancelled out

[sand forge]

.done

[devout kindle]

!done

[vocal sleet]

If you are done with this channel, please mark your problem as solved by typing .close

[sand forge]

.close

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[devout kindle]

lmao

[sand forge]

lol

wait no

.reopen

[vocal sleet]

✅

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[sand forge]

wait no

[sand forge]

OMFG im dumb, sorry...

[devout kindle]

just close it, if its a mistake

[sand forge]

uhh no

help

n+1-n / n(n+1)

this is just 1/n - n/n+1

OHH

nvm thx

.close

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[sand forge]

(the problems be solving themselves)

[devout kindle]

fr

[sand forge]

(the memes be making themselves)

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[ruby aspen]

Let $ABC$ be a triangle, the angle bisector of $\angle BAC$ intersects $BC$ at $D$, and the midpoint of $BC$ is called $M$. If $\angle MAC = 40^\circ$ and $AM = MC$, find the value of $\angle DAM$."

[twin meteor]

ZiXO

[ruby aspen]

this is what i have done so far

[signal fox]

Is AD perpendicular to BC?

[ruby aspen]

that's my problem , the problem doesn't specifically state that it's perpendicular , but it is a bisector though

[wraith walrus]

Nope

Make it straight

Using scale

[ruby aspen]

sorry , don't quite understand what you mean by that

what should be straight

[wraith walrus]

So put a scale the scale edge should touch the point A and it should be perpendicular, by doing that you will get a 90 degree perpendicular line

[wraith walrus]

AD

[ruby aspen]

but does AD being a bisector typically means that its perpendicular ?

that's the issue

[wraith walrus]

Bisector means cutting a line equally perpendicular

I mean that if AD is perpendicular bisector of BC

[hexed kestrel]

not necessarily

[wraith walrus]

That means that line AD should cut BC in half

[hexed kestrel]

No

[ruby aspen]

from what i have read so far, an angle bisector doesnt necessarily bisect the opposite side

[signal fox]

Heres your answer

[ruby aspen]

excuse my ignorance here, i dont know how to continue with that

[signal fox]

Look at triangle BAM

What kind of triangle is it?

[ruby aspen]

well. i think you mean if MC = MB and we know that MC = MA , this means that MA = MB meaning an isosceles triangle

that tells me that angle MAB and MBA are equal , which means each one is (180-80)/2

50 degrees

MAB is 50 and MAC is 40 which means MAD is 10 ?

wait i think my last conclusion is wrong

does that mean MAD is 0 ? meaning point D is actually at the same spot as M ?

oh never mind i am dumb , its indeed 10 , correct ?

[signal fox]

Yes

[lyric hamlet]

sorry but shoudlnt mad be 5?

[ruby aspen]

well , thanks alot my friend

[lyric hamlet]

also ur diagram shoudnt angle bam be 50 not bad

[lyric hamlet]

my reasoning for this is that length BM = length AM, thus angle MBA and angle BAM should be equal

right....

[ruby aspen]

correct

[lyric hamlet]

that saying we know angle BAD is equal to angle DAC, from the qn

[lyric hamlet]

yes that saying the entire angle of BAC is 90 degrees

[lyric hamlet]

so BAD and DAC must be 45 degrres

[ruby aspen]

you're correct actually

[lyric hamlet]

which means DAM is 5 degrres

because angle BAM minus angle DAM must equal 45, hence DAM is 5 degrres

[ruby aspen]

lets wrap this up again i will draw the diagram again with the information we have

maybe i made a slight mistake there

you are correct

BAM is 50 and MAC = 40 , and we know that AD bisects the angle A, meaning DAB = DAC = 45 , and since MAC = 40 , that leaves DAM to be 5

that makes sense to me

[lyric hamlet]

yuhhhhh

[ruby aspen]

That's great @signal fox and @lyric hamlet , i love yall

Good luck and thank you

[lyric hamlet]

no worries goodluck to u too!!

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[mental night]

Am I correct A cylindrical tank has a radius of 4 meters and a height of 10 meters. Water is being pumped into the tank at a rate of 2 cubic meters per minute. However, the tank is leaking water at a rate of 0.5 cubic meters per minute.

How long will it take to fill the tank if it starts empty?
What will be the water level in the tank after 3 hours? that is the question this is my answer 270=π(4) ²h=π(16)h h= 270/16π​ ≈ 270/50.27​ ≈5.37 meters my process is The cylindrical tank holds 502.65 cubic meters of water in total. Water is being pumped in at a rate of 2 cubic meters per minute, but since it’s leaking at 0.5 cubic meters per minute, the net rate at which the tank fills is 1.5 cubic meters per minute.502.65/1.5 ​ =335.1minutes (about 5.6 hours).So, it will take around 5.6 hours to fill the tank completely if it starts empty. After 3 hours (or 180 minutes), the water level in the tank will be about 5.37 meters high.

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[vast shale]

yeah

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[drowsy cypress]

i dont get the last step why is it 61 and 67

[bronze osprey]

it's better to say add 32 to both sides

in algebraic equations, you always do the same thing to both sides

so you have $f = 32 - \frac{99}{5} = \frac{170}{5} - \frac{99}{5}$

[twin meteor]

south's secret twin brother

[bronze osprey]

make a common denominator

then simplify

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[drowsy cypress]

thanks

[vocal sleet]

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[fading cove]

can someone help me understand how to find the angle between two vectors ?

[fading cove]

it is a more theoritical question

[lusty parrot]

dot product

[twin horizon]

I got Sniped

[lusty parrot]

[fading cove]

no but like what i dont understand is if i put them in some configuration the angle will be different than in another configuration

[twin horizon]

Can you give an example of what you mean?

[fading cove]

yes

i'll draw something a sec

[brittle cipher]

dot product should give you the correct answer regardless

[lusty parrot]

Do you mean it gives complementary angles?

[fading cove]

just a sec

if you look at this

the angle is different

and i dont understand the rule to be sure that i'm using the right angle

[twin horizon]

Those vectors are different

[fading cove]

when i use the scalar product

[lusty parrot]

By definition a line extend to infinity. so in reality the rightmost pic would give the same angle as the left one

[twin horizon]

The orientation matters

[fading cove]

like if i want to do !!u!! x !!v!! x (u;v)

[twin horizon]

Wait

[fading cove]

IIuII x IIvII x cos(u;v)

[twin horizon]

Okay I misread 😞

[fading cove]

when i want to use this formula then angle can have multiple configurations

[brittle cipher]

that's not true for the sake of using the dot product formula with vectors

[heavy yoke]

should be the one where both vectors share the same base/starting point

[fading cove]

so i dont understand how to be sure that i use the right angle

[fading cove]

are you sure ?

because i did an exercise where it was not the case

[heavy yoke]

the scalar product-based formula will return that angle

[fading cove]

wait i'll do another scheme

[fading cove]

so they must share the same base ?

now that i drew it it seems to make sense

[heavy yoke]

you can calculate other angles from that where they don't share the same base, but the formula giving "angle between vectors" is always the angle where they share the same base

[fading cove]

ok thank you very much

[pallid jewel]

wsg

whats the Q

[fading cove]

got answered

[pallid jewel]

demn nooce

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[wraith walrus]

Can anyone check my answer please

[civic oracle]

,rotate

[vast shale]

,ROTATE\

FIRSTER

bruh

your answer is correct

[vast shale]

i believe they don't need to tell it's the center

yes obviously

[prisma sage]

How can I reduce to the simplest form?

[vocal sleet]

@wraith walrus Has your question been resolved?

[wraith walrus]

Thanks,

[vocal sleet]

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[rancid yacht]

how can I find the amount of numbers in this sum? (1-4)(2-5)(3-6)...(70-73)

[hybrid flicker]

first of all this looks like a product and not a sum

and second of all what does "the amount of numbers" mean? the value of the product?

[rancid yacht]

what i meant with the amount of numbers is that how many -3's are in this product

nevermind i found the answer.

it was -210

we have -3*n so n will be 1,2,3,70 which that will be n=70

and 70* -3 = -210

.close

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[hybrid flicker]

wait what

-3 * -3 * -3 * ... is not -210

that would be -3 + (-3) + (-3) + ...

[rancid yacht]

so?

we have 70 -3's

which is -210

[rancid yacht]

.

[vocal sleet]

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[rocky wyvern]

Am I looking for an integral? If so what do I do with it?

[silk osprey]

do you know the formula for average value of a function

[rocky wyvern]

No I don't think so

[tall lagoon]

if your function is like water level in a tub, the average is when the water levels become the same

[silk osprey]

$\frac{1}{b-a} \int_a^b f(x) dx$

[twin meteor]

knief

[silk osprey]

not sure that’s helpful sir

[rocky wyvern]

Oh that's not too bad ok

[tall lagoon]

yeah sure give the unmotivated formula helpful person 👍

[silk osprey]

brother what you stated had no relevance whatsoever to the question

[short bone]

unrelated to the question, how do yall type like this

[silk osprey]

latex

go to overleaf.com

[short bone]

any guide?

[silk osprey]

free

yea hold on i’ll send a link

[silk osprey]

https://www.overleaf.com/learn/latex/Learn_LaTeX_in_30_minutes

[short bone]

thanks

[silk osprey]

read through that article

and some more

then watch some youtube

and just practice

[rocky wyvern]

So (5^3-1^3)/5-1

[silk osprey]

there is a latex testing channel in this server

[silk osprey]

yup

[rocky wyvern]

Got it, thanks!

[silk osprey]

youre welcome

question

doggimal

[rocky wyvern]

Yes?

[silk osprey]

did the bathtub analogy help you

😹😹😹

did you find it relevant

or just pure unrelated common sense

[tall lagoon]

no idea why you like arguing

[rocky wyvern]

Er... Not in really but maybe if had time to I think about it it could make sense

[silk osprey]

dope pfp

[short bone]

sorry to ask, but where is it

[rocky wyvern]

Sometimes I'm kinda blind when it comes to analagies

[silk osprey]

https://discord.com/channels/268882317391429632/844681108473118750

[rocky wyvern]

Thanks

[silk osprey]

you’re welcome

[tall lagoon]

I hope you get it sometime. you're a kinder person

[short bone]

ohh didnt have it before, thanks

[silk osprey]

no worries

there also a latex help channel

if you see that too

[short bone]

cant see it

[silk osprey]

https://discord.com/channels/268882317391429632/840667252793802752

[short bone]

can u send a link , will add it to channel list

[silk osprey]

learning latex shouldn’t take more than a few hours

[short bone]

thanks once again

[rocky wyvern]

Ok bye guys, thanks for the help!

.close

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[vernal radish]

if anyone uses matlab, can someone explain to me why this script gives this output ? I really don't understand it it's completely different from what I need, gpt keeps telling me that the output should be different...

[vernal radish]

expected output :

[twin gorge]

yo another person with a genhin pfp

[vernal radish]

😭

yea

[twin gorge]

XD

[vernal radish]

i just want help with matlab man

[mild flower]

diff looks like it's going by column