help-10

[ivory flare]

Ohmygod my pic isnt sending

But here they didnt write that it was continuous (same graph)

It was limit approaching 2 f(x+1)

Except they didnt write that it was continous

[obtuse pebble]

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[neat sparrow]

Is this a general law of working with Bayesian credences? I couldn't find it online, probably because I am using negation signs.
P(A) = P(A | B) * P(B) + P(A | ¬B) * P(¬B)

[crisp forum]

Law of total probability is what you're looking for

[neat sparrow]

can P(A) = P(A | B) * P(B) + P(A | ¬B) * P(¬B)
be derived from [picture] ?

[neat sparrow]

if so, how? (I appreciate the help already btw, thanks :) <3)

[neat sparrow]

thanks, does it use that 'P(A) = P(A∩B)+P(A∩¬B)' is true (always) ?

[crisp forum]

If they form a complete partition of the sample space yes

[neat sparrow]

shouldn't that be the case by definition? As they are eachother's negation?

[crisp forum]

I suppose

[obtuse pebble]

@neat sparrow Has your question been resolved?

[obtuse pebble]

@neat sparrow Has your question been resolved?

[neat sparrow]

in kane B's dogmatic skepticism video[1], the author defines the Pyrrhonian skeptic as someone who assigns a credence of 0.5 to every proposition, but is that a consistent thing to do? Like do you even get a probability distribution or are basic laws of probability violated then? Such as P(A)*P(B) = P(A&B) if A and B are independent.

[1] https://www.youtube.com/watch?v=DvcGW40risE

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[timid silo]

Hello whats a forumlua I can use to find the volume of this sphere

[worn yoke]

,w volume of a sphere

[warm shale]

Results provided by WolframAlpha

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

Requested by cloudedhorizons

[timid silo]

V=34​π(10)3
V=34​π(1000)
V=34​×3.14159×1000
V≈4188.79 cubic feet

Is that right

[worn yoke]

yes, although the notation is hard to follow

[timid silo]

4189 rounded right?

[worn yoke]

yes

[timid silo]

.CLOSe

.close

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[slow prawn]

if n---inf

appraoach 0

is that dereasing?

[warm shale]

Sadie Carnot (η > 1)

[old lily]

not necessarily

example an = 1/n , n odd and 0 if n even

in this case maybe ur right u'll have to check

[candid storm]

i cant math

yes

[ruby path]

wait i messed up the latex

other way round

😭

[unreal musk]

Always when there are two choices, they get mixed up

[obtuse pebble]

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[plucky beacon]

i need help understanding how to do this

[lone oyster]

If y is directly proportional to the cube root of x+3, can you think of an equation that links them?

$y \propto \sqrt[3]{x+3}$

[warm shale]

DJJ

[lone oyster]

Can you turn this into an equation you can use?

[obtuse pebble]

@plucky beacon Has your question been resolved?

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[outer spindle]

I’m trying to get this answer it’s due soon

[outer spindle]

If you zoom in, I don’t know what’s wrong with my equation

[lofty field]

You're missing a minus in the parenthesis

[outer spindle]

Where?

[lofty field]

-(-(-4)^3/3 + 16(-4))

[outer spindle]

@lofty field can I dm you what my professor said? Or I can say it here I prefer to keep it private

[lofty field]

Sure, you can dm me

[outer spindle]

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[dapper sandal]

yo

[dapper sandal]

how do I solve this

im confused if I used riemanns sum or not

wouldnt I need to be given a number of subintervals?

[timid silo]

D

[dapper sandal]

how

[lofty field]

Find the area manually using the graph

[dapper sandal]

got it

thanks

[obtuse pebble]

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[dapper sandal]

yo

[dapper sandal]

how would I solve this

wouldnt I just integrate f'(x)

[rich summit]

yes

[dapper sandal]

what is the point of f(2) = 6

[rich summit]

to find a particular solution

as in what your +C equals

[dapper sandal]

so after I integrate f'(x) add C to it

then set it equal to 2

to find C

[rich summit]

plug in x=2 and set it equal to 6

6 = x^2/x + x + C

where x = 2

[marsh geyser]

!nosols

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[dapper sandal]

ah ok got it

[obtuse pebble]

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[boreal palm]

how do u express this sequence as an explicit formula

i wasnt there in class

[heavy shore]

alr, loan = $150000.
Rate of interest = 6% yearly interest applied monthly at (6/12)% = 0.5%

do you know the simple interest of a particular amount after a given amount of time?

[obtuse pebble]

@boreal palm Has your question been resolved?

[boreal palm]

what does that mean

😭

no

idt so

[heavy shore]

basically the interest for a given time is given as Amount × Rate × Time / 100

[boreal palm]

oh ok

isnt that what i did in the 1st part

1500000.0055yr/100?

[heavy shore]

Yeah

But you have to multiply it with 2 in the second one, then by 3 in the third one and then by 4 in the fourth one

[boreal palm]

huh?

[heavy shore]

Because the number of months are changing in the next parts

[boreal palm]

oh

[heavy shore]

And the interest is directly proportional to the number of months/time

[boreal palm]

150000(0.005)*1/100 +

150000(0.005)*2/100+

until 5?

[heavy shore]

Yeah as and what the question asks you to find out

Interest + Principle amount (loan) = total repayment so that's how you do b? part

[boreal palm]

o i already did b

i was looking at c

my friend gave me her formula but idk if its right or now to use it

[heavy shore]

Oh yeah she generalized it

[boreal palm]

hwo do i make it for 5 yrs?

wait would 150000 be p

[heavy shore]

Use R = 6%
T= 5
Same PRT = interest in 5 years and then proceed, now you can find out the money owed

[boreal palm]

where is t

in the formula?

oh wait

can i just do 0.005

[heavy shore]

but then you have to multiply by 12×5 = 60

[boreal palm]

150000(1+0.005) + (1-(1+0.005)/1+0.005)-5*1000

[heavy shore]

Monthly payouts of $1000 no? So it should be 5×12×1000

[boreal palm]

ohhh

150000(1+0.005) + (1-(1+0.005)/1+0.005)-5(1000)12

im still a little confused

[heavy shore]

Wait, also multiply the rate with time of 5 years in each bracket

[boreal palm]

how do i set it for 5yrs

[heavy shore]

150000(1+RT)

[boreal palm]

150000(1+0.005(5)) + (1-(1+0.005(5))/1+0.005*5)-5(1000)12

[heavy shore]

0.005 is the monthly rate, change it into yearly rate!

nothing but 12×monthly rate = yearly rate

[boreal palm]

ohhh ok

150000(1+0.005(12)) + (1-(1+0.005(12))/1+0.005*5)-5(1000)12

[heavy shore]

12×5 (5 years)

[boreal palm]

o

[heavy shore]

:)

[boreal palm]

am i supposed to use ^5 anywhere?

[heavy shore]

×5*, yes

[boreal palm]

150000(1+0.005(60)) + (1-(1+0.005(60))/1+0.005(60))-5(1000)12

[heavy shore]

YES, GOOD

[boreal palm]

THANK U

[heavy shore]

anytime, happy to help!

[boreal palm]

FOR BEING SO PATIENT I WAS RLLY CONFUSED 😭😭

[heavy shore]

yeahh no worries :)

[boreal palm]

the world needs more people like u

🩵🩵

ty!

.close

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[glossy warren]

help me idk how to continue 😭

[unreal musk]

What's the length OA?

[glossy warren]

7 😭 ?

[unreal musk]

Nope, it's not

[glossy warren]

its perpendicular BROO 💔

[unreal musk]

Well, it might be perpendicular, but OA is not 7

What is OAB again?

[glossy warren]

o i havent found the arch yet

is it just

WAIT

radius is 8

[glossy warren]

SO 8 - 6.9

[unreal musk]

Yep, there you go the length PA is 8 - 6.9

[glossy warren]

@unreal musk mind if i dm u i need help with math-

😭

MY FINAL BIG ASS EXAM IS TODAY

[unreal musk]

Awwww sure, I'm up for a while

[obtuse pebble]

@glossy warren Has your question been resolved?

[obtuse pebble]

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[twin bough]

can someone help with this question im not sure how to go about it

[teal turret]

Do u know PV of an annuity immediate formula

Well I assume it’s immediate

[twin bough]

I haven't done financial maths for a couple years, what is the formula?

[fallow vale]

Google exists

[teal turret]

PV = A(1 - v^t)/i

It might be a bit tough if u haven’t done this for a long time

Might wanna watch a video or two

[twin bough]

ok thanks

[teal turret]

Man I really gotta brush up on my stuff

Taking FM in August

Yikes

[obtuse pebble]

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[pale tusk]

Guys IQR is our new lesson I don't really know about it can someone help pls

[teal turret]

.close

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[ashen perch]

1. R is the midpoint of AU, <A and <U are right angles
2. AR cong. to RU

Just need help with #3 until #6

[obtuse pebble]

@ashen perch Has your question been resolved?

[ashen perch]

<@&286206848099549185>

[daring glacier]

which part?

!status

[obtuse pebble]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[ashen perch]

answer for 3-6

[daring glacier]

!status

[obtuse pebble]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[ashen perch]

the second one

[daring glacier]

see 3 is GIVEN

4. opposite angles

[ashen perch]

Oh alright

[daring glacier]

i am not quite sure what to write in 5, but in 6 i think we have to write hence proved

[ashen perch]

Oh thanks

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[magic escarp]

how to do it?

[obtuse pebble]

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[shell oracle]

.

Can someone help me solve a Pre Calc trig word problem, dont know where to start to solve it and need some guidance

[heavy summit]

send it

[shell oracle]

you think you can help?

@heavy summit

[heavy summit]

what do you have so far?

[shell oracle]

I dont know where to start to be quite honest with you

the 126ft part is sort of holding me back cause I dont know what I am suppose to do with that variable

[heavy summit]

that's for the last yes or no. leave it to the end

[shell oracle]

ok gotcha

I thought maybe if I did 126 - 78 I would get the left over fencing

and then I have to find the angle for it

[heavy summit]

so does the fence need to go all the way around?

[shell oracle]

that is what I assumed but I am afraid that isnt the case for whatever reason

wait I might i have figured it out

lemme see

sqrt 48^2 + 78^2 - 2 * 48 * 78 * cos(73) = 78.73195

Imma just chance it

If I get it wrong then oh well

[sharp moss]

do we know if it is isoceles

question doesn't seem solveable

[obtuse pebble]

@shell oracle Has your question been resolved?

[obtuse pebble]

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[rose ferry]

hey how do I graph this semi circle?

[slim lake]

Square both sides

$y^2=9-x^2\Rightarrow y^2+x^2=9$

[warm shale]

Max

[slim lake]

So we have a circle, but we have some restrictions from the original equation $y=\sqrt{9-x^2}$, namely that $y>0$

[warm shale]

Max

[bleak flare]

this is how you can plot the semicircle because y>0 just get 3 points using values of x.

[rose ferry]

ah okay

so when we find the y intercepts it never goes below 0 and thus a semi circle

i love this server

thankssss

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[bleak flare]

why cannot we solve this by plotting x and y as functions of t in a cartesian plane?

this is my method

im getting 8/5

[grizzled shore]

,rotate

[grizzled shore]

You’ve plotted y = x - x³ and y = 1 - x⁴

Isn’t this where you would use greens theorem or something

[bleak flare]

no can't we assume the axes to be t-f(t) instead of xy

if we have two functions f(x) g(x) we plot them in the same plane and find the areas

so we can do it that way too

[grizzled shore]

Yo

Move to your channel

I accidentally closed it but we should go there instead

[bleak flare]

yes

[obtuse pebble]

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[hallow atlas]

I'm struggling to understand what they've done on the dN side between lines 2 and 3

[fallow plank]

Partial fractions

[hallow atlas]

ah, that's what I had forgotten

thank you so much!

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[edgy needle]

Ezucse me how come question c is wrong

[cyan void]

It seems you forgor what b hat means

[edgy needle]

wait but dats not in the formula i used

[frail dome]

yeah your b, i just noticed

[edgy needle]

my teacher put the hat but in my formula theres no hat

[cyan void]

Then you forgor square

[edgy needle]

oh i squared it i just didnt write ti

wait

did i.

oh i did

its sqrt(100)

[frail dome]

you did?

[edgy needle]

then square

= 100

oops

tats why its wrong

[cyan void]

Yeah

[edgy needle]

thnx guys :D

[cyan void]

Np!

[frail dome]

yeah of course

[edgy needle]

wait can u help me with one more question?

i dont get question 3

[obtuse pebble]

@edgy needle Has your question been resolved?

[edgy needle]

<@&286206848099549185>

[edgy needle]

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[lone root]

i have a relation of a function: f(a+x)+f(a-x)=k.f(x) where a,k are constants. how do i find the periodicity of the function?

[deep skiff]

First try to analyze what that expression actually says

[lone root]

also prove is k≥2.. the function will never be periodic

[lone root]

f(a+x)+f(a-x)=k.f(x). this is everything i have

[deep skiff]

f(x) is that value of the function at some point x right?

[lone root]

thats it

yeah

f is a continuous function on R→R

[deep skiff]

so f(x+a) is the value of the function at a point that's a greater than x

[lone root]

not necessarily

a and k both belong to R

[deep skiff]

Well either way f(a+x) and f(a-x) are gonna give you 2 point which are one to the left and one to the right of f(x) equal distance away right?

[lone root]

okay to make it easier
try this f(x-4)+f(x+4)=sqrt(2) . f(x)

f is a function so when we say f(x-4) it means the whole graph/curve shifts to the rught by 4 units

my sir has given this question, ik the answer too

but i wanna know how to get to the answer

what he gave me was just a simple formula where i put in values of a and k and i get the fundamental period of the function

pls ping me

.close

[obtuse pebble]

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[lone root]

i have a relation of a function: f(a+x)+f(a-x)=k.f(x) where a,k are constants. PROVE that the fundamental period of the function f(x) is 2πa/cos⁻¹(k/2)

[lone root]

feel free to ping me

[timid silo]

what did you try

@lone root

[lone root]

idek how to approach this problem

my sir has given this as a trick to solve such questions quickly

[timid silo]

what trick

[lone root]

he says he's made it but wouldnt tell us how

[timid silo]

do you find the function

first

[lone root]

you dont need to

see if i have f(a+x)+f(a-x)=f(x) (note: the k is missing)
then i can do some substitutions and find the fundamental period

[deep skiff]

Could you try rewriting it as f(x) = f(a+x)/k + f(a-x)/k

Maybe that'll lead you somewhere

[timid silo]

yes i thought of using the first principle but that would be tedious here

i think there should be a shorter path

[lone root]

i dont this so cuz the k is outside the brackets so its gonna be difficult

for a situation like this substitutions work

here i got the ans as 18

[deep skiff]

Maybe define g(x) = f(x)/k

[timid silo]

no

[deep skiff]

they have the same period...

should have at least

[timid silo]

that's not a general way to find period

[lone root]

which?

[timid silo]

in this

[deep skiff]

I might be mistaken

[timid silo]

this

which you sent

[timid silo]

not you

[lone root]

see the first line was all i had been given in the question

how else could i do it

last weekend i had given a test in which too a question like this was solved the same way

[timid silo]

we may try defining g(x) = f(a+x)/f(x) lol

[lone root]

what about f(x-a)/f(x)

[timid silo]

then fundamental period of h(x) + h(x+a)= constant is just 2a

ive proved this before

[timid silo]

hmm

[lone root]

but do you understand what ive done here

[timid silo]

they do

[timid silo]

the question is different

[lone root]

once i replace x with x+3 and then the other time i replaced x with x-9

yeah ik

but i dont get any idea on how to go about it

[timid silo]

put x = a -x

you get another equation which is true

then eliminate f(a-x) from both equation

you may get something

[lone root]

2πa/cos⁻¹(k/2)
the most confusing thing here is how to get terms of cos and π

[timid silo]

this would be true for k belongs to [-2,2)

[lone root]

no i was told if k>2 then its impossible for the function to be periodic

[timid silo]

well

its not greater than 2

see what i wrote properly

also i got f(x) =f(a+x)/k-1

maybe we can use this

[lone root]

not even 2 cuz the arccos term becomes 0

[timid silo]

hmm

yes

[lone root]

i did what you said and got k.f(x+a)+f(x-2a)=(k^2-1)f(x)

doesnt seem like i made any progress

[timid silo]

fuck

apply

first principle

you get something better

[lone root]

doesnt seem to work

forget it, ive spent way too much time on this

.close

[obtuse pebble]

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[pliant canyon]

i’m so confused

[celest wind]

hi so i hve a question when adding and subtracting vectors.

AB + BX -> ?

what would I name this resultant vector as?

similarly,

AB - BX -> ?

[fluid snow]

This channel is occupied, use e.g. #help-6

[celest wind]

Oh, im sorry! Thanks

[warm shale]

faiyrose

[pliant canyon]

4x+12 sqrt x +9?

4x+9

[warm shale]

faiyrose

[pliant canyon]

i’m doing it correct so far right

[pliant canyon]

now I check if it’s extraneous or not right

alright thank you it does

[obtuse pebble]

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

[pliant canyon]

.close

[obtuse pebble]

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[whole dirge]

Can someone confirm to me whether this is right or miss smth

<@&286206848099549185>

[gray sun]

pls domt ping me good sir

[whole dirge]

Mb but can u quickly tell me if i made a mistake here

I completely forgot how to do these

[obtuse pebble]

@whole dirge Has your question been resolved?

[drowsy coral]

Did u find the area of these 2 triangles?

[drowsy coral]

@whole dirge Has your questions been resolved?

[whole dirge]

No i just automatically

Times them together

Which is why i thought i stumbled on an error

[drowsy coral]

Did u use a height of 6?

[whole dirge]

6?

[drowsy coral]

The height of the triangle

For base times height

[whole dirge]

Nope

[drowsy coral]

10^2 - 8^2 = 36

Square root of 36 is 6

[whole dirge]

Oh Insee

I see

[drowsy coral]

W

[whole dirge]

Wait where’d u get 8?

[drowsy coral]

16/2

For the base

Of each triangle

[whole dirge]

Oh

[drowsy coral]

I split em in 2

[whole dirge]

Okay so 16/2 = 8 which is the base

[drowsy coral]

Yes

Then Pythagorean theorem for the height since u have the hypotenuse

[whole dirge]

And then u use the hypotenuse ^2 and minus it by the base ^2

[drowsy coral]

Yes

[whole dirge]

Sorry i havent done this topic in a second and i just tend to forget the work i do after i pass it

[drowsy coral]

Yeah I get it. The same thing happens to me

Hope this helped tho 👌

[whole dirge]

Any tips on how i can keep these kind of things even after finishing the topic?

Im looking to really start putting work in my studies since i have been slacking off significantly

[drowsy coral]

What I do is help on this server lol. Since there are so many people at different levels of math here, I can refresh my memory by helping out someone with something I learned a while ago.

[whole dirge]

Oh, thats smart, i feel like i just forget about it because i didnt think id have to do something i did from the last grade to the next grade again

Even tho ive had to do the same thing since the 6th grade and still do it now in the 10th grade

[obtuse pebble]

@whole dirge Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

@atomic sequoia Has your question been resolved?

[obtuse pebble]

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[meager palm]

how do you compute the coefficienct of x^3y^2

[meager palm]

with binomial thrm

[bleak flare]

hint : multinomial theorem

[meager palm]

no clue what that is

[bleak flare]

or rather binomial will work here

see

Tr+1 = nCr x raised to r multiplied by (2y) raised to (5 - r)

[meager palm]

latex work

[bleak flare]

set r = 3

[meager palm]

does it matter if you do r = 3 or r = 2

[bleak flare]

yes here x raised to r so r should be 3

if you write it as 2y raised to r then you can use r = 2

and x will be raised to 5 - r = 3 in that case

5c3 = 5c2 so remains unchanged.

[meager palm]

how do we know that x is raised to r and not y

[bleak flare]

it is a way of writing, in general use this as a working rule you'll run into fewer errors
(a + b) raised to n
the r + 1 th term will be ncr a raised to r and b raised to n - r
a and b in the order as specified

this helps in avoiding errors

[meager palm]

oh shoot i did not know thus

this

[bleak flare]

yeah, it's a good method.

.close

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[obtuse pebble]

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[neat sparrow]

**Bayesian epistemology / statistics **
so we have a prior credence of 0 in theory 1 & theory 2, but theory 1 and 2 do not contradict each other, they actually cohere very well. Does this say anything about what our prior in theory 1 and theory 2 should be?

[glass dagger]

If theory 1 and theory 2 are the same theory then they will "cohere" very well despite us not genuinely gaining any new insight into the theory

Maybe you should post the full assignment or give more context

I would imagine that the main challenge here is about posing the problem formally, which probably gets pretty philosophical

[obtuse pebble]

@neat sparrow Has your question been resolved?

[neat sparrow]

what if they are not the same theory?

[neat sparrow]

the context is that theory 1 is that the universe expands indefinitely (i.e. that we get a heat death, which is predicted by our best cosmological theories (LCDM-model) and data and theory 2 is that large scale quantum fluctuation are possible such that a brain could quantum fluctuate into existence (which follows from quantum field theory). Curious to hear your thoughts! Appreciate the interest :)

[neat sparrow]

does that help? The problem is that if you take both together you get 'the Boltzmann brain' paradox and Sean Carroll proposes to assign a credence of 0 to any theory that entails Boltzmann Brains (which is the conjuction theory of theory 1 and 2) and I'm wondering what the consequences would be for our credence in theory 1 and our credence in theory 2 independently

nvm, I got it

[glass dagger]

yeah that's getting pretty philosophical

Personally, I think people put way too much faith into the leading scientific theory, like for example I would think of the assumptions of statistical mechanics similarly to how you might think about giving every belief a prior uniform distribution

like, clearly the actual probability distribution of any clearly articulated belief is that it's either true or false with probability 1 but giving stuff uniform prior distributions etc. is a useful concept in the presence of uncertainty

and in a similar vein I think assuming that a system spends equal time in each of its microstates is a good assumption when we don't perfectly understand how the system behaves but that doesn't mean that it would actually necessarily be possible for a Boltzmann brain to form

but unfortunately I'm not a physicist so I'm not really qualified to enter that discussion and take that with a hefty amount of salt

[obtuse pebble]

@neat sparrow Has your question been resolved?

[neat sparrow]

thansk!

[neat sparrow]

thansk for your input anyway

[obtuse pebble]

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[acoustic rover]

not sure where to go from here :(

<@&286206848099549185>

maybe there's something wrong with the problem?

omg wait

that looks much more simpler ill try to do it

is this product rule or chain rule

or both

[obtuse pebble]

@acoustic rover Has your question been resolved?

[obtuse pebble]

@acoustic rover Has your question been resolved?

[acoustic rover]

Hm

[obtuse pebble]

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[indigo ridge]

hi i need help w logarithms

[indigo ridge]

i have no idea what the question is asking me

nor how to solve it

the above one

and if im not helped i will explode 😤🤯

[obtuse pebble]

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[obtuse pebble]

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[halcyon gyro]

I am trying to solve a geocaching mystery. The title of this mystery is "Decimal primes" so I believe it has something to do with prime numbers although I do not understand the decimal part as demical numbers are not primes. I have tried to factorize the numbers but so far I haven't gotten anything out of it. The solution should be N 60° xx.yyy E 024° aa.bbb where I believe xx is 10 or 11 and aa is between 51 and 53.
The given sequence from which I should find the coordinates is 1 497 608 686 475 023
link to the task: https://www.geocaching.com/geocache/GCV5R0
I don't need the full solution but merely a bumb to right direction.

[pseudo swift]

have you tried just finding the prime factors of that huge number?

[brazen gorge]

it's 24884603×60182141

two primes

[pseudo swift]

@halcyon gyro

[halcyon gyro]

my goodness

that was fast

I didnt even think about that

[brazen gorge]

wolframalpha 🤙

[brazen gorge]

oh and they match the coordinate requirements

awesome

[halcyon gyro]

oh stupid me, should have done that. thank you so much

that is really neet solution

[brazen gorge]

no problem

[halcyon gyro]

yeah it is almost there. I can work out the rest of it with this

need to just do the conversion

[brazen gorge]

https://www.google.com/maps/place/60°10'55.7"N+24°53'04.6"E/@60.1823857,24.8841512,17z/data=!4m4!3m3!8m2!3d60.182141!4d24.884603?entry=ttu

[halcyon gyro]

yup 🙂

got them already, place looks promising

[brazen gorge]

have fun!

[halcyon gyro]

it is a fun hobby. I love solving the mysteries and it has outdoors/health benefit as well. I recommend 🙂

[obtuse pebble]

@halcyon gyro Has your question been resolved?

[obtuse pebble]

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[static relic]

my first thought is (x-5)(x+5/3)(x-2/3)

[pastel wren]

yeah i think its correct

[static relic]

if submitted similar answers and it said it was wrong

[old lily]

is this a test 🤨

[static relic]

no

[pastel wren]

maybe try in expanded form?

[static relic]

its a extra credit thing

[old lily]

still kinda cheating

whatever

i'll let it slide

[static relic]

ive asked my teacher if using discord for homework is allowed and she said it was

[old lily]

she knows discord ?

[static relic]

it says factored form

[static relic]

she knows that i am getting help from strangers on the internet if i cant get help from a tutor or from her directly as fast

[old lily]

right.

[pastel wren]

oh right, it should be correct then

[static relic]

the only problems i have been asking on here for help i dont get any points on i have already gotten the max extra credit

[static relic]

it was wrong

[high lily]

what comes up when you press preview

[static relic]

nothing i think the button is broken

i think i need to have a number infront of th equation

[high lily]

they want the leading coefficient to be 1

[high lily]

so you shouldn't

[static relic]

so what else do i need to do after i get (x-4)(x-4/3)(x-3/2)

[high lily]

nothing

[static relic]

but when i submit that it says i am wronf

[high lily]

bring that up with your teacher

[static relic]

ok

[high lily]

the only other thing i could think of is that they want integer coefficients/constant for each linear factor

factor out 1/3 from (x-4/3)
and factor out 1/2 from (x-3/2)

[static relic]

ok i will just wait to see what my teacher says

.close

[obtuse pebble]

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[sweet spear]

I need help with part c)

[sweet spear]

I know I can create a bijection between P(N) and infinite strings of 0 and 1

I have thought about constructing a set of antichains where each string as a unique 1 in the nth position

This is the extent of my thinking and I'm stuck at this point

[obtuse pebble]

@sweet spear Has your question been resolved?

[mossy fjord]

So is this trying to do $f:\mathbb{R}\Rightarrow\mathbb{N}$ for the first one?

[warm shale]

Narutoes

[mossy fjord]

So an uncountably infinite set to a countably infinite set?

Or is it $f:\mathbb{Q}\Rightarrow\mathbb{N}$, which would be two countably infinite sets

[warm shale]

Narutoes

[obtuse pebble]

@sweet spear Has your question been resolved?

[obtuse pebble]

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[trim pelican]

What does this mean?

[trim pelican]

here's the proof

[dark stirrup]

It's breaking a number down into its digits

$a_k$ is the $k$th digit of $n$.

[warm shale]

SWR

[obtuse pebble]

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[fallen inlet]

Hello

[fallen inlet]

Can someone help with this?

Ik that 3 is the root and 2 is the exponent

But idk where to put them

And where the 2 goes

Inside or outside of square root

[surreal forge]

both are correct... but normally you write exponents inside the root

[hidden compass]

The 2 is multiplied with $(x^{\frac{2}{3}})$

[warm shale]

Alberto Z.

[fallen inlet]

Ok so 2 has a secret exponent of 1?

[hidden compass]

Yep

$a^1 = a \ \ \forall a$

[warm shale]

Alberto Z.

[fallen inlet]

What is that triangle thing

[hidden compass]

It means "for all" or if you prefer "for every"

[hidden compass]

Meaning that no matter what value of a you have, it will be always the case that a¹ = a

[fallen inlet]

.close

[obtuse pebble]

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[timid silo]

This is a quick one

[timid silo]

Ignore my workijg

Ii

ii

[obtuse pebble]

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[timid silo]

.solve

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[gray kindle]

Can someone explain how I can prove that a tangent line to any point(x1, x2) on the parabola x^2 = 4py is x1/2p?

[gray kindle]

Without derivatives

I don’t know where the p comes from because I don’t think I can use the focus or the directrix in any way

[obtuse pebble]

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[gray kindle]

.close

[obtuse pebble]

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[tight basalt]

i think what ive learnt is math should only be done by doing problems right?

[tight basalt]

you shouldnt watch youtube vids

or like read a textbook

you have to physically do as many problems as possible

idk if any1 agrees?

[left vault]

not necesarily no

[tight basalt]

why

[pseudo swift]

what do you do if you have no idea what the problem asks tho

[left vault]

how will you understand the concepts without learning them from somehere

[tight basalt]

look at the solutions

[normal cairn]

i think visualisations are the best way to truly understand how something works or why its true

but if you want more hands on knowledge then you have to get in there yourself

[tight basalt]

how do we visualise

[normal cairn]

have you ever seen a 3blue1brown video

[frosty river]

This is not a specific maths question, ask on a regular chat

[versed stratus]

use graphs

[tight basalt]

No

[versed stratus]

do see one

[tight basalt]

kk

[versed stratus]

https://www.youtube.com/c/3blue1brown

[tight basalt]

so basically combine visualising with a tonne of problems

?

thats how we get a good grade

[stuck grove]

My dude is just trolling

[pastel wren]

there is no one objective way

everyone is different and some things might help you more than others

[tight basalt]

yh i think for me just as many problems as possible

[pastel wren]

repetition is important tahts true

[tight basalt]

but then repetition is one layer, if the question is slightly different u need to apply a tactic to solve it

how do we get that tactic

[pastel wren]

i watch a lot of math stuff and read theory too

[pastel wren]

some tactics you just learn other you get intuitively and others come from understanding of theory

always take time and think about it too

if you dont get it instantly

[tight basalt]

yh like a 1 mark application question usually takes a couple mins

to solve

[pastel wren]

dont force yourself to be able to do everything and instantely

there will be things youll have problem with

and will take you a lot of time to solve

[tight basalt]

yh i remember not understanding a problem for hours

[pastel wren]

thats normal

trust me

be patient

[tight basalt]

i wanna do a 24 hour study session

because im a bit behind

[pastel wren]

i discourage this

[tight basalt]

alr

[pastel wren]

your health and sleep is most important

[tight basalt]

yh apparently doing like all nighters reduces life expectancy

[pastel wren]

it does

[tight basalt]

farkk

[pastel wren]

and you gonna feel awful

be consistent thats it

[tight basalt]

but im gonna die anyway right?

someday

[pastel wren]

id prefer dying later and happier tbh

[tight basalt]

alright i agree

[pastel wren]

if you gonna learn everyday for even an hour it will eventually benefit you much more than doing an all nighter once a week or even trying to force yourself to do like 5-10h everyday

it will benefit you short term thats true if you wanna get good grades here and now

[tight basalt]

but the way i think about these things is if i put myself through a tough 24 hour study session i will have done more than most ppl have done in months

[tight basalt]

this is the way my brain works

[versed stratus]

I also suggest you don't look at answers immediately , spends a lot of time on each problem

[tight basalt]

yh

i do that

[pastel wren]

if you gonna do 1h for 24 days you will do more than other ppl in a month

[tight basalt]

i have to push it up

survival of the fittest

[pastel wren]

its gonna ruin you in a long run

[tight basalt]

i love math

which is the main thing

[pastel wren]

thats great

and also the more reason to not push yourself over your limits

[tight basalt]

yh true

[teal turret]

unless youre superman, your returns will increase only for a few hours, then they will start to drop

[tight basalt]

thats where caffeine comes in

and adderall

[teal turret]

dont get on that stuff

[pastel wren]

yeah, no

dont do that

[tight basalt]

alr soz

nice talking to you guys

@pastel wren @teal turret thanks for the more health beneficial revision approach

at the end of the day its about being happy

[pastel wren]

yeah

take care

[tight basalt]

thanks!

you too

.close

[obtuse pebble]

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[pastel wren]

[obtuse pebble]

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[warm matrix]

(x,y) = (3,2)

[pulsar gate]

how

all im getting is xy=6

[pastel wren]

show your work

also from xy=6 you get x = 6/y

and you can sub this into one of the original equations

[zenith raft]

!nosols

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

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[pulsar gate]

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[ember yew]

hello. why is random variable X plus a constant "a" is a thing???? isnt X a function? how are we allowed to sum a function to a constant ?
plus it says E(X + a) = E(X) + a
how ???? since we know E(X) = Σ xi p(X = xi) ---> it should be that E(X+a) = Σ (xi + a) p(X = xi + a) how is p(xi+a) in that sum is not 0 ???????

[frosty river]

Real function plus a number is a function

[fossil crag]

(f+a)(x) = f(x) + a

[frosty river]

If X is a function X(x) then X+a is the function (X+a)(x)=X(x)+a

[ember yew]

oh so i thought it was f+a not f(x)+a.

thats clear. what about the other

[frosty river]

f+a is the name of the function whose values are f(x)+a

[ember yew]

the other would be true in my mind only if the p(xi + a) remains same as p(xi) somehow

[ember yew]

yep i see that. never knew we could do that.

[fossil crag]

btw

"E(X+a) = Σ (xi + a) p(X = xi + a)" is wrong

it's

[frosty river]

E(X+a) is not Σ (xi + a) p(X = xi + a)

[fossil crag]

E(X+a) = Σ (xi + a) p(X+a = xi + a)

[frosty river]

E(X+a) is Σ (xi + a) p(X + a= xi + a)

and you can substract a both sides

[fossil crag]

so it's Σ (xi + a) p(X = xi)

[ember yew]

ahhhhhhhhhhhhhhhhhhh

yea

now it makes all sense

thanks so much

[frosty river]

Prepare yourself for linear and even nonlinear combinations of random variables

[ember yew]

all g. notation kinda confusing though

thanks anyway.

how do i close this?

.close

[obtuse pebble]

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[ember yew]

again thanks guys

[obtuse pebble]

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@fluid snow Has your question been resolved?

[fluid snow]

<@&286206848099549185>

[obtuse pebble]

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[golden oriole]

Why does this not converge at 1 ?

[golden oriole]

The formula for the sum is given in the mark scheme right there aswell

Wait im trying to send another picture but its not loading

I tried looking in the fraction for any place where x = 1 could give a /0 but that also doesn't seem to be the case

[obtuse pebble]

@golden oriole Has your question been resolved?

[obtuse pebble]

@golden oriole Has your question been resolved?

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[pure nacelle]

This is probably super simple but I was watching this proof for real analysis on youtube

and I am really confused why he said ofcourse for all n n is greater then or equal to 1

why is this true?

I understand the rest of the proof just not how he got that

[brave bramble]

For all n in N, n ≥ 1

Where N is the "set of natural numbers", 1,2,3,4...

[pure nacelle]

Ok so what symbol means the set of natural numbers

[brave bramble]

N

[latent walrus]

1/n starts at 1/1=1
then gets smaller as n increases 1/2, 1/3, 1/4, ... tending to 0

1 is the highest it ever gets to be

[pure nacelle]

So the big N means the set of natural numbers which is non negatives 1 2 3 4 etc

[latent walrus]

yeah

[pure nacelle]

ok

ty ty thats all i needed

[latent walrus]

$\mathbb{N}$

[warm shale]

AℤØ

[obtuse pebble]

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[lusty badge]

Only after 15 minutes

[obtuse pebble]

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✅

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[flat vortex]

why does tan have more then 2 values, unlike sin and cosine

[teal turret]

wdym unlike sine or cosine

give some more context

[flat vortex]

so lets say cosine x= 60 degrees, then the values would be 60 and 300 degrees

but for tan it has 4 values

but why since it is negative in the other 2 quadrants,

and shouldnt this answer be 60 degrees and 240 degrees

[teal turret]

tan^2 (theta) = 3

tan(theta) = +-sqrt(3)

theta = arctan(sqrt(3)), arctan(-sqrt(3))

[flat vortex]

where you get the +- from

[teal turret]

when u sqrt both sides, you have to put the +-

for example

$x^2 = 4$

[teal turret]

what values of x satisfy this?

[flat vortex]

2

[teal turret]

what else

[flat vortex]

-2

[teal turret]

$x = \pm \sqrt 2$

[flat vortex]

ohhhh

[teal turret]

correct

[flat vortex]

ok now i get it, thx

[teal turret]

np

[flat vortex]

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[plain jungle]

Q. sum of 2 irrational numbers is always irrational?

[plain jungle]

Is this statement TRUE or FALSE

[dusk widget]

false

e.g take sqrt(2) and 1 - sqrt(2)

both are irrational, but their sum is clearly 1

@plain jungle

[obtuse pebble]

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[shut citrus]

i know this is U sub but i dont know how to get rid of that extra t in the numerator?

[brazen viper]

Well, if t+1 = u then it stands to reason that t = ...

[shut citrus]

u-1 got it, ive been brain farting recently... calc 2 final on monday 😭
thank youuuu

[brazen viper]

Yw

[obtuse pebble]

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[indigo folio]

how do i solve -8x^-1.5

[indigo folio]

this is my working out

[brazen viper]

You can't solve that. There's nothing to solve, it is just an expression involving x, not an equation.

[indigo folio]

uhh

not solve sorry

how do i simplify it

[brazen viper]

Well, I'm not sure what simplification can be done, actually

[indigo folio]

here is what it says

[brazen viper]

As written, that's more or less in lowest terms.

!original

[obtuse pebble]

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

[indigo folio]

“Write each of the following in simplest surd form with positive exponents”

[brazen viper]

Ah

So recognize that x^-1.5 = √(x) / x^2

And use that

[indigo folio]

wait

i dont get that

im getting this far

[brazen viper]

Ok, so: you remember that x^(a+b) = x^a * x^b right?

[brazen viper]

And that x^(-a) = 1/x^a

[indigo folio]

yes

[brazen viper]

Ok, now consider that -1.5 = 0.5 - 2

[indigo folio]

you mean -2?

[brazen viper]

Sorry yes

[indigo folio]

ah yes

i get that

how do i do it for -8

you cant square root a negative number

[brazen viper]

No need, -8 doesn't have the exponent applied to it

[indigo folio]

OHHHH

sorry

[brazen viper]

No worries

Also

[indigo folio]

i knew something was wrong

cause no brackets

[brazen viper]

√(x) / x^2 = 1/√(x^3)

And I'm not really sure which one they want

I personally consider the first to be more simplified than the second

[indigo folio]

the second one is written in the book

[brazen viper]

But there are probably some sort of rules they want you to specifically follow

[indigo folio]

how does that turn into the second part

[brazen viper]

Well, instead of following my method instead use yours where you take advantage of x^(ab) = (x^a)^b

You used a = 3 and b = 1/2

[indigo folio]

oh wait

[brazen viper]

With a negative somewhere in there

[indigo folio]

(x/x^4)^1/2

[brazen viper]

Yes exactly

[indigo folio]

makes sense

alright thank you

.close

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[neat sparrow]

LOGIC AND PROBABILITY PROOF (slightly advanced)
I want to prove ¬( P(B | ¬A & C) ≈ 1 )
from the premises

• P(A | B & C) ≈ 1
• P( C | A & B) ≈ 0
  we don't have precise numbers, just approximate one's.
  can this be done? If not, what other (approximate) premise would be required to prove this?

[neat sparrow]

the proof:

Essentially, the total Event Space is split up into 8 possible cases, because there essentially 3 coin flips.

A & ~A
B & ~B
C & ~C

We can pair these three objects into different permutations like A & B & C or A & (~B) & (~C), etc.

We know that the probability of this event space has to be equal to 1, but we don't necessarily know any of the individual probabilities of the 8 states.

---

Now, we are making the claim that P(A | B & C) ≈ 1. By definition, this means:

1.) P(A & B & C) / P(A) ≈ 1. Moving the P(A) to the other side, we get:

2.) P(A & B & C) ≈ P(A).

What this tells us is that P(A) is heavily conditional in B and C. A most likely isn't going to happen unless B and C are also happening.

---

Now, for the other given approximation, we get P(C | A & B) ≈ 0. By definition, this means

3.) P(A & B & C) / P(C) ≈ 0.

Since P(C) has a upper limit of 1 (Probabilities are always less then 1), this means that the only way for the above to be approximately zero, is if the numerator is approximately zero.

4.) P(A & B & C) ≈ 0, which means P(A) ≈ 0, due to the approximate equivalence in sentence 2 above.

What this means is that almost all the probability is in ~A, or, P(~A) ≈1.

---

This essentially limits the event space into two Subsets of the event space that hold to the following:

E1 = {All events containing ~A}: P(E1) ≈1
which includes:

~A & B & C
~A & ~B & C
~A & B & ~C
~A & ~B & ~C

and

E2 = {All events containing A}: P(E2) = 0

Which looks like the above, but replace all of the ~As with As.

A & B & C
A & ~B & C
A & B & ~C
A & ~B & ~C

Now, this has an interesting result. Although we don't know anything about the probabilities containing ~A, we do know that any probability that contains A ≈ 0. This is because the sum of all the part have to be approximately zero.

What we want to know is the approximate probability of the event P(B | ~A & C). By definition, this expression is equivalent to:

P(~A & B & C) / P(B)

The problem is, this question is about ~A, which we know nothing about it's probabilities. We were able to reduce the event space, but we weren't able to reduce it enough. In order to find the approximate probability, we need to get an approximate probability of the remaining 3 probabilities.

So, let's assume that what we want to prove, and see what needs to be the case.

So, the assumption given is the following.

5.) P(B | ~A & C) ≈ 1.

By definition, this is equivalent to the following:

6.) P(~A & B & C) / P(B) ≈ 1

Which moving the P(B) term to the other side gets us:

7.) P(~A & B & C) ≈ P(B)

So how does this get us closer to a solution?

---

Right now, P(B) is split up into two components (due to the truncated event space).

P(B) = P(~A & B & C) + P(~A & B & ~C)

But since we know Equation 7, we know that the latter term in this equation must be approximately zero, as that is the only way that the first two terms can be approximately equivalent.

What this means is that we can work backwards towards the solution, and make our solution what we desire to find.

If we know that P(~A & B & ~C) ≈ 0, then we can say for certain that P(B | ~A & C) ≈ 1.

However
we wanted to prove that ~(P(B | ~A & C) ≈ 1), would the only way that can be done be if we know that ~(P(~A & B & ~C) ≈ 0)? Or would it be doable from the following three additional premises that stem from the context of the problem (first I thought them to be unnecessary)

• P( ~A | ~B & C) ≈ 1
• P(C | ~A & B) ≈ 1
• P(C & A | B) ≈ 0
  would these allow me to prove that ~(P(B | ~A & C) ≈ 1) or do we have an inconsistent set of premises now?

curious to hear your thoughts :)

[obtuse pebble]

@neat sparrow Has your question been resolved?

[neat sparrow]

actually, the first step in this proof is confusing to me:
premise 2: P(A | B & C) ≈ 1
definition of conditional probability yields
P(A & B & C)/P(B & C) ≈ 1
not:

P(A & B & C) / P(A) ≈ 1.
or is a different definition than the definition of conditional probability being used in this first step?

[obtuse pebble]

@neat sparrow Has your question been resolved?

[neat sparrow]

@trim portal I think the proof that showed that it cannot be proven is actually wrong [reason below], but I found a very simple proof after all. See #help-35 message

Reason: The first step in this proof is confusing to me:
premise 2: P(A | B & C) ≈ 1
definition of conditional probability yields
P(A & B & C)/P(B & C) ≈ 1
not:
"P(A & B & C) / P(A) ≈ 1."
or is a different definition than the definition of conditional probability being used in this first step?

.close

[obtuse pebble]

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[rocky citrus]

Hi

[rocky citrus]

Hello, for my question. I may need a mass group amount of answers to certify it.

There is a task, that asks us to use a dataset from the answers from a google form that 28 students did.