help-42

[acoustic wraith]

what is your event A? what did you define it as?

if you defined it differently from me, then you'll need to check your definition

[weak vapor]

oh yeah, i didnt define it

but i also didn't for my other question and i got perfect on that

[acoustic wraith]

probably because there's only one event in consideration in those questions

[weak vapor]

yeah

[acoustic wraith]

but there isn't just one event here

you have an event and its complement - two events

so you need to define A first

[weak vapor]

okay, so a = hitting red and b = blue?

would that work

a' = hits red adleast once

[acoustic wraith]

if you want to do that, then you will need two letters per event, so I won't recommend this

[weak vapor]

would i just leave blue out/

[acoustic wraith]

if A is hitting red, A' is not hitting red at least once

honestly, let A be the event you want. that'd be simpler

[weak vapor]

so swap it?

a = not hitting red, a' = hitting red

[acoustic wraith]

since you want the probability of the caterpillar not being red on both attempts, why not define A as that?

[weak vapor]

okay, so a = not hitting red both attemps, a' = hitting red adleast once

[acoustic wraith]

yes

[weak vapor]

if i just define those variables, does it make my answer right

or do i still need to change

[acoustic wraith]

you'll still need to change your P(A')

P(A') itself is good, but the value is not

[weak vapor]

so swap p(A) = p(A')

and P(A') to p(A)

[acoustic wraith]

1/4 is not the value for P(A')

[weak vapor]

3/4

?

[acoustic wraith]

yes

oops

[weak vapor]

P(A)= 1 - P(A')
P(A) = 1 - 3/4
P(A) = 1/4

cuz we swapped a' and a

wouldn't this be right

[acoustic wraith]

first line has an extra '

[weak vapor]

i meant tor emove that

remove

[acoustic wraith]

P(A) = 1 - P(A')

[weak vapor]

A = not hitting red on both attemps
A' = hitting red on adleastt one attempt
P(A)= 1 - P(A')
P(A) = 1 - 3/4
P(A) = 1/4

[acoustic wraith]

there we go

it always helps to define your events

[weak vapor]

alright, thats all the questions I have

tysm for the help

[acoustic wraith]

nps

please remember to close the channel if you are done with it

[weak vapor]

alright, just confirming my therefore statement stays the same

[acoustic wraith]

change 3/4 to your new answer

[weak vapor]

so the caterpillar has a 3/4 chance of hitting red

?

[acoustic wraith]

3/4 chance of hitting red at least once in 2 attempts

but

that is not the event you're looking for, so

[weak vapor]

but my question is it never hitting red

[acoustic wraith]

^

that's why I said change 3/4 to your new answer, not change the event for which you are concluding

[weak vapor]

oh

A = not hitting red on both attempts 
A' = hitting red on adleast one attempt
P(A)= 1 - P(A')
P(A) = 1 - 3/4
P(A) = 1/4

Therefore the caterpillar has a 3/4 chance of hitting red adleast once in the two attempts

this all seems good?

[acoustic wraith]

Therefore the caterpillar has a 3/4 chance of hitting red adleast once in the two attempts

❌

[acoustic wraith]

^

[weak vapor]

3/4 chance of not hitting red?

[acoustic wraith]

A = not hitting red on both attempts
P(A) = 1/4

[weak vapor]

Therefore the caterpillar has a 1/4 chance of hitting red adleast once in the two attempts?

Therefore the caterpillar has a 3/4 chance of hitting red adleast once in the two attempts?

im a lil confused

[acoustic wraith]

you are interested in the event where the caterpillar never hits red on both attempts

your conclusion should be based on this

you don't really care about the probability of the caterpillar hitting red at least once beyond the fact that you can use it as a stepping stone to find the answer probability

[weak vapor]

ok, the catterpillars probability of never hitting red on both attempts is 1/4

therefore the catterpillars probability of never hitting red on both attempts is 1/4

last confirmation,

A = not hitting red on both attempts 
A' = hitting red on adleast one attempt
P(A)= 1 - P(A')
P(A) = 1 - 3/4
P(A) = 1/4

Therefore the caterpillar has a probability of 1/4 of never hitting red on both attempts.

[acoustic wraith]

at least*, not adleast, but yes

[weak vapor]

thanks for the help

sorry for being kinda slow lol

[acoustic wraith]

alternatively

you could have just realized that if the caterpillar isn't going red, then it has to go blue both times, so (1/2)^2 = 1/4

you could have checked your answer against this

but nps either way

[weak vapor]

!close

how do i close

lol

[acoustic wraith]

.close

[weak vapor]

.close

[calm coral]

Channel closed

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Use .reopen if this was a mistake.

[weak vapor]

.close

[acoustic wraith]

already

[calm coral]

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[winter elbow]

. @restive yacht post here

[restive yacht]

Can someone help me with this? Im not sure where to start.

[velvet osprey]

!noclopen

[calm coral]

Please don't repeatedly close and claim a new channel with the exact same question. This erases all previous progress made towards your problem and is confusing for helpers, making it more difficult to help you. Please be patient, even if your channel has not received much attention.

[velvet osprey]

<@&268886789983436800> spam

@winter elbow oh wait did you move channels bc of a spammer

[tardy plume]

no other guys trolling he actually need help

[winter elbow]

no phillip was at other ppl's channel, therefore I redirected them here.

[restive yacht]

Can I please get help

[stark sparrow]

I'm guessing no one else has dealt w this so uh
@thick ravine don't spam in help channels, #chill exists if you want to shit post

[thick ravine]

Okay so there are 6 turning points right as we counted

The left side goes down and the right side goes up

That makes your degree odd

N-1

[restive yacht]

Just @ me

I need help on this problem

https://cdn.discordapp.com/attachments/948770026914213909/1422088767220486184/IMG_0553.png?ex=68db66df&is=68da155f&hm=5a70f9fd156a2a844f317645ba2a3de647210d58ff928a22e583bab822dd8140&

[calm coral]

@restive yacht Has your question been resolved?

[restive yacht]

[restive yacht]

.close

[calm coral]

Channel closed

Closed by @restive yacht

Use .reopen if this was a mistake.

[potent smelt]

@restive yacht if you're here, I can help with this

.reopen

[calm coral]

✅

[potent smelt]

I'll start off, so that we don't waste a round trip.

For a) write a possible function for the graph. It will be easiest to write this in terms of the zeroes of the polynomial.

Any degree n polynomial can be written as f(x) = a(x-r1)(x-r2)...(x-rn), where the ri are the roots of the polynomial

(where the ri are counted with repetition, for instance the 3rd root from the left looks like a repeated root. Are there others?)

Ping me if you swing on back

And sorry about the trolling earlier on in your help channel

[calm coral]

@restive yacht Has your question been resolved?

[restive yacht]

Its okay I think I found the solution to f(x)= 1/4(x+1)x^3(x-2)^2(x-3)

[potent smelt]

@restive yacht you can omit the 1/4

(the question specified this)

But otherwise this looks good to me

Did you need further help?

[restive yacht]

No thank you very much though i genuinely thought no one was gonna help

[potent smelt]

Sometimes it can take a second.

[restive yacht]

ty very much though

.close

[calm coral]

Channel closed

Closed by @restive yacht

Use .reopen if this was a mistake.

[potent smelt]

Helpers are volunteers, and sometimes there just isn't anyone available to help.

But you're very welcome ❤️

[calm coral]

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[sullen loom]

i cant seem to figure out how to evaluate this

[sullen loom]

i get indeterminant forms when i follow along x=0 and y=0

and i dont think polar coordinates would make this better

i did make:
$$
(y^2)(1-cos^2(x))/(x^4+y^4)
$$

[potent lotus]

saini

[sullen loom]

which gets me no where anyways

[sour monolith]

What happens when you go along x=y?

[sullen loom]

wait i can do that

oh ym god yeah you can

[dire gulch]

Are you even sure that an answer exists?

[sullen loom]

i mean i checked along x=0 and y=0 and it seems to exist

[sour monolith]

spoiler

[sullen loom]

unless maybe i did ti wrong

[sour monolith]

It needs to exists on every path to (0,0)

including x=y

[sullen loom]

it doesnt exist

fawk

[safe lynx]

It doest

Doesnt

Limit depends on the slope of line

[sullen loom]

.close

[calm coral]

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[calm coral]

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[coral mulch]

Hey. I am playing Pokemon TCG. In this game, you sometimes have to flip a coin, heads or tails are the outcomes. How do I calculate the chance you flip 1 or 2 heads in 3 coin flips, as an example?

[ivory saddle]

,w binomial theorem

[potent lotus]

Results provided by WolframAlpha

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Requested by nyxzore

[marble pendant]

You can use probability here

Let's say we flip one coin

We can either get heads or tails

So total outcomes are 2

Now we want heads

So "favourable" outcomes is 1

P(event) = favourable outcomes / total outcomes

So it is 1/2 for one heads

@coral mulch do you follow?

[coral mulch]

I am sorry. I am very bad at maths. I am trying to follow. :).

So 50 percent chance for the correct outcome

[marble pendant]

Yup

Also one way to do such questions is to write all the outcomes

As it's coin tossing the number of outcomes will be less

So you have H or T

Now if we flip two coins we will have HH, HT, TH, TT

total four outcomes

Can you tell me the total outcomes for 3 flips?

[coral mulch]

3

4 outcomes

What do you mean with todal outcomes for 3 flips?

Total*

[marble pendant]

Do you get this?

[coral mulch]

Yes, because you flip 2 coins those are the possible outcomes. But in Pokemon you flip 1 coin each time. So it would be H, T per flip.

[marble pendant]

Yes

But flipping coins one after the other and flipping them simultaneously is the same thing

[coral mulch]

True

[marble pendant]

Hence

[coral mulch]

So there is actually only 1 outcome that is not good, if you want atleast 1 heads.

The TT

[marble pendant]

Yup

[coral mulch]

So does that mean it is 75 percent chance ?

[marble pendant]

Yes, 75% chance for atleast one head out of 2 flips

[coral mulch]

Ahh. Makes sense!

[marble pendant]

Nicee

[coral mulch]

So with 3 coins. It would be; HHH. HHT. HTT. TTT?

[marble pendant]

You have missed a few cases

[coral mulch]

Hmm

Like HTH?

[marble pendant]

We can see here that HT and TH are technically the same

[coral mulch]

Yes

[marble pendant]

But they are counted seperately

Because in this case order matters

[coral mulch]

Ahh. Yes. Sorry.

[marble pendant]

Npp

[coral mulch]

Is there a way to easily count how many probabilities there are?

Because with 3 coins and order, there are a lot.

[marble pendant]

Yes there is a way

2^n where n is number of flips

[coral mulch]

HHH
HHT
HTT
HTH
THH
TTH
THT

These are the ones I found. Is this correct?

Oh and TTT

[marble pendant]

You missed one

Yes absolutely

Now count the ones with one or two H

And you have your answer

[coral mulch]

Ahhh. I get it now.

[marble pendant]

Nicee

Good luck with your future battles 😂

[coral mulch]

7 for 1 heads atleast. Out of 8.

5 for 2 heads out of 8.

4*

[marble pendant]

How 5?

Yes

[coral mulch]

Typo lol

So 50 percent chance to hit 2 heads?

[marble pendant]

When you flip 3 coins

Yes

[coral mulch]

Amazing. Thanks! 🙂

[marble pendant]

Nice

!close

[calm coral]

@coral mulch Has your question been resolved?

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[sonic mason]

when do we use this

[gray python]

As in when that is < 0 or

[sonic mason]

yeaa actually the rule

and theorem

the name of it

[gray python]

It's just that 2 ln 2 > 0 and -1 < 0

So a positive number and a negative number multiplied becomes < 0 ?

[sonic mason]

yea i uderstand multiplying them will give me something negative

but like

hows it heloing my question

[gray python]

ohh

sorry give me a sec

[sonic mason]

yea

this the ques

[gray python]

I think they're trying to show that the function in [1,2] intersects the x-axis

[teal drift]

That's one of the hypothesis of Bolzano's theorem

• continuity
• opposite signs at a and b

[gray python]

Or sth

[sonic mason]

OH intermentent value theorem

[gray python]

I think the IVT they put means intermediate value theorem

[sonic mason]

olayyy aka there will be aroot

[gray python]

That's why they did the stuff above

[sonic mason]

and the root will be btw 1 and 2

because?

[rugged pine]

Bolzano’s theorem is like intermediate value theorem but only for f(x)=0

[sonic mason]

ahhh

[teal drift]

Exactly, if the function is continuous on [a, b] and f(a)f(b) < 0, then it's guaranteed f(x) has a root in (a, b)

[sonic mason]

okay so what will the f(1) f(2) being negative give me here

[sonic mason]

understood okay

[gray python]

maybe they're trying to prove they have opposite signs

As in F(1) and F(2)

[sonic mason]

ahhhh

okayy

makes sense

thank youall

[gray python]

No problem

[sonic mason]

.close

[calm coral]

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[calm coral]

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[wise moss]

hey i dont really understand a part of this solution, when they say the fundamental solns are the real & complex part of one of the solutions. how do we know this?

[sour monolith]

The space of solutions is a 2 dimensional vector space over R

And if you have a solution both the real and imaginary parts of it will be solutions (since the matrix has real entries)

[wise moss]

oh is that a known theorem i could look at

[sour monolith]

This should have been a theorem if you are in an ODE class

[wise moss]

yeah i get that, but why does that imply the real and imaginary parts are solutions themselves

[sour monolith]

Plug the solution in the form
Re(x) + i * Im(x) instead of x into the equation x'=Ax. And if you will equate the real and imaginary parts of your equation you will get that both real and imaginary parts are solutions.

[wise moss]

ohhh got you okay. so as long as theyre independant we're sorted

[sour monolith]

Yep

[wise moss]

great thanks

.close

[calm coral]

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[calm coral]

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[chilly galleon]

hi
i need a help i have a question wrote : 13,14,24,8,12,? find ?

[calm coral]

Please don't occupy multiple help channels.

[chilly galleon]

ok

[mellow harbor]

Since this is a duplicate, I'll close this for now

.close

[calm coral]

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[chilly galleon]

hi
i need a help i have a question wrote : 13,14,24,8,12,? find ?

[upper sparrow]

.(as a note, in case you're having trouble seeing the one you have open right now, it's #help-11, and make sure you have the Math Help (Occupied) section expanded to see all channels, including those with all messages read )

[calm coral]

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[desert matrix]

Maisy needs to be added to the number of girls first

since it is 15 OTHER girls

and the question says each child can only have 1 sibling

[calm coral]

@visual sigil Has your question been resolved?

[alpine cobalt]

yea you here?

its just 26-4-4-4=14

[visual sigil]

.close

[calm coral]

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[calm coral]

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[unborn garden]

Hello, I got an easy question; Is s*t considered to be a linear combination of s and t? or is it only addition that counts?

[sour monolith]

usually when we talk about linear combinations multiplication isn't even defined

and when it is, linear combination are sums with coefficients in your base field/ring

[unborn garden]

oh I see. So I can't use that to prove that if GCD(st, s+t, s-t) = 1 then GCD(s,t) = 1 right?

[unborn garden]

oke oke

[sour monolith]

wait

[sour monolith]

this was in the context of a vector space/module

[unborn garden]

I was only speaking abt naturals lol

[sour monolith]

yeah

you can use linear combinations here

just linear combinations of what

[unborn garden]

Of s and t

ive tried using contradiction but couldnt do it 100%

[sour monolith]

contradiction works great here

because if d divides both s and t then it divides st,s+t,s-t

[unborn garden]

I mean I suppose that (s,t)= 1 and i get that we have the negation, but why would it necessarily mean that that is the only case where st, s-t, s+t share a divisor?

[sour monolith]

contradiction would be to assume s,t share a positive divisor

i.e. that gcd(s,t) is not 1

[unborn garden]

yes thats what I started with

and i ended up that (st,s-t,s+t)=/ 1

[sour monolith]

yep

and thats a contradiction

[unborn garden]

but why would it mean that if (s,t)=1 then (st,s-t,s+t) also =1 always?

[marsh summit]

it doesn't

[unborn garden]

Im not good with logic

[marsh summit]

in fact that's not true

... actually maybe it is

[unborn garden]

lol dont stick your nose where you arent sure

[sour monolith]

but it doesn't follow from what we did before

[marsh summit]

but it's not what the problem was asking you to prove

[sour monolith]

he is right

[marsh summit]

so it's fine

[unborn garden]

ohhhh

I knew something was wrong with it

so what is the way to prove such a thing?

[sour monolith]

the other direction?

[unborn garden]

starting from (st, s-t,s+t)=/1?

[sour monolith]

this means there is some shared divisor

[unborn garden]

i think it would make sense

yes

but if we prove that that happens only if (s,t)=/1

[sour monolith]

not only if

we prove if it happens then gcd(s,t) is not 1

[unborn garden]

yes

so the negation must mean that (s,t)=1? or still no?

were tryna figure it out broski chill

[unborn garden]

that weird symbol is the negation right?

I think I get it

[sour monolith]

yeah

[unborn garden]

it makes sense to start from (st, s+t, s-t)=/ 1

[sour monolith]

yeah

and that means there is a divisor 1<d shared between st,s+t,s-t

[remote mural]

yall wrong

answer ts

[unborn garden]

thanks for the input

[unborn garden]

thank you for helping i was stuck at ts for three days

.solved

[calm coral]

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[remote mural]

.reopen

.reopen

[calm coral]

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[calm coral]

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[spare iron]

Could someone help me figure out what I am supposed to do here? Cause I somehow was able to solve a from ex.6 (resulting in D=arcsin (3/7.26) *(365/2pi) + 80) but when I do the same for question b and c it all goes south. I also have no idea how to solve ex.7, I hate trig with all my heart

[jovial granite]

I hate electromagnetic

[spare iron]

bro you how good are you in trig

[serene talon]

rambutan

[spare iron]

someone help

[jovial granite]

good

[serene talon]

wait, are there 2 people here trying to use the same chat

[spare iron]

yep lol

[spare iron]

gooood

[rustic osprey]

discord.gg/physics

[jovial granite]

why?

[spare iron]

could you help me a bit here

[jovial granite]

i have electromagnetic exam after hours

i cant think anymore

LOL

[spare iron]

uhhh good luck then, I've been stuck on this excericse for a week now

[jovial granite]

easy bro

[spare iron]

you are my only other hope

are you good at trig?

https://tenor.com/view/stan-twt-reaction-meme-funny-begging-crying-gif-11151243012681431164

[pure kayak]

how did you get 3/7.26 may i ask

[jovial granite]

i think ex7 a is 2π/3, 4π/3

cosx = -1/2

[pure kayak]

unless that was for b

[spare iron]

you are right it's wrong

but wtf I got the right result

[spare iron]

ok thank you so much I'm gonna check now

[jovial granite]

ok

best luck

[pure kayak]

,w solve 2pi/365 (D-80) =kpi for D

[potent lotus]

Results provided by WolframAlpha

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

Requested by aldrnari_

[pure kayak]

you got that?

[spare iron]

how do i use it?

[jovial granite]

d1= 80

d2=262.5

[spare iron]

btw I did a mistake that was the operation i tried for the b answer but I got wrong results

[jovial granite]

H = 15 so, sin θ = (15 − 12)/7.26 = 3/7.26 = 0.413

[spare iron]

for a I did D-80=kpi*(365/2pi)

[spare iron]

ohhh okok

[jovial granite]

Let alpha = arcsin(0.413) ≈ 0.425 rad.

[spare iron]

why the 15-12 in the brackets tho?

[spare iron]

here i mean

[jovial granite]

each year (two per cycle)

theta = α ⇒ D = 80 + (365/2π)·α = 80 + 58.091·0.425 = 104.7

theta = π − α ⇒ D = 80 + 58.091·(π − 0.425) = 237.9

[jovial granite]

easy

algebra

[spare iron]

and that's for b right?

[jovial granite]

That 12 is the baseline number of daylight hours in Edinburgh (roughly the equinox)

tthe 7.26 is the amplitude shows that maximum daylight swings about 7.26 hours above or below the baselinee

[jovial granite]

yaa

[spare iron]

oh right

[jovial granite]

easy example bro

[spare iron]

the last thing im missing is how you found 58.091

is it 0.425 rad?

[jovial granite]

number comes straight from the scaling between radians and days

theta=2pi/365 * ​(D−80)

365 / 2pi

[spare iron]

ohhh ok

[jovial granite]

= 58.091

what is that course?

[spare iron]

Could you please help with excericse seven pleasee

[jovial granite]

hhhhhhhhhhhhhhhhhhhhhh

am board

[spare iron]

I'm doing Math AA SL oat the IBDP

[jovial granite]

????

what is that

[spare iron]

it's alright, you already helped a lot thanks

[jovial granite]

no worry man

[spare iron]

the international baccalaureate

[jovial granite]

we are here helping each other

ohhhhhhhhhh

nice

do ur best

you can do more than you think

just focus

[spare iron]

thanks man, I hope one day I'll get trig

[spare iron]

I'll try

[jovial granite]

have some coffee

you will win the battle

hhhhhhhh

[spare iron]

Good luck for the exams

[jovial granite]

am done with EE

last semester

[spare iron]

you did the extended essay??

[jovial granite]

??????

[spare iron]

lol nvm we got different acronyms

what is EE for you

[jovial granite]

hhhhhhhhhhhhhhhhhhh

Electrical Eng

[spare iron]

oh damn

yeah I am running at light speed from everything with engineering in it

[jovial granite]

hhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

LOL

[spare iron]

mad respect and fear for anyone who does it

[jovial granite]

thx

see you later

brooooooo

[spare iron]

cya man

[calm coral]

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[scarlet saddle]

help?

[rugged pine]

!status

[calm coral]

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

[scarlet saddle]

dont know how to solve for a

like at all

[rugged pine]

Can you work out the number of minutes between each time that’s written on the question

[scarlet saddle]

0 mins, 90 mins, and 300 mins?

[jade fractal]

at 0 mins the height of the candle was $x$, at 90 mins it was $x-3$ and at 300 minutes it was $0$

[potent lotus]

Médicis

[scarlet saddle]

yep

[rugged pine]

(So the question is asking if over 90 minutes the candle melts 3cm, how much does it melt after 300)

Use velocity=distance/time

Weird to use speed here but it is “moving”

You know the velocity is constant

[scarlet saddle]

so in 90 its 3

in 300 its 0

im confused

[rugged pine]

Sorry I meant over the 300 minutes how much it melts not after

[scarlet saddle]

i dont know because we dont know what the original height is

[rugged pine]

It melts 3cm in 90 minutes, how long does it take to melt 1cm?

[scarlet saddle]

30

[rugged pine]

Okay, so it melts at 1cm/30 minutes

[scarlet saddle]

,calc 300/30

[potent lotus]

Result:

10

[scarlet saddle]

,calc 10*1

[potent lotus]

Result:

10

[rugged pine]

Yeah that’s the answer

[scarlet saddle]

for original?

[rugged pine]

Yes

[scarlet saddle]

im a bit

let me see

wait im confised

confused*

in 30 min 1 cm melts

so in 300 min 10 cm melts

but in 300 min everything is gone

okay i got it

thanks

.close

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[calm coral]

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[steep yew]

can someone help out with these

[steep yew]

I know I got most if not all these wrong

and idk how to finish 15

I’ve got 14 a b and

C

I just don’t understand 15

[crisp sphinx]

do you know what the normal line looks like?

[steep yew]

perpendicular of the tangent line

I think I got this

[steep yew]

since this is just perpendicular of the tangent

we get -2

and that’s just the answer I’m pretty sure?

[crisp sphinx]

that is correct, but how do you know it's -2?

[steep yew]

just take the opposite reciprocal of 1:2

1/2

-2

and then -2 would just produce a straight line I’m 99% sure

or wait actually I’m not sure about that

since it’s linear wouldn’t it just have the same slope at each point

yesss because the slope remains constant at each point

[crisp sphinx]

sorry i disappeared

yes, thats good reasoning

you should also mention that, since the line is parallel to the normal, the normal also has a slope of 1/2

then you can do your calculation

[calm coral]

@steep yew Has your question been resolved?

[calm coral]

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[wintry shore]

what does Si represent?

[leaden thunder]

area under the curve between x_{i-1} and x_i

[wintry shore]

what does that mean?

[calm coral]

@wintry shore Has your question been resolved?

[fervent radish]

Si is just the naming of the strips

[calm coral]

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[crystal breach]

can someone explain this vector question to me

[crystal breach]

is this not the diagram

how do we find out the angles thats the part

and i dont get the questions answer why is it so different

for my part 1

[acoustic wraith]

your target vector is 20m/s south

[crystal breach]

i drew that similar diagram to what they gave but instead my current arrow was facing the other way

[crystal breach]

yes

[acoustic wraith]

this already looks wrong. this looks like your target vector is 20m/s at an angle

it's not. it's 20m/s due south

[crystal breach]

isnt my target the pink line

and the blue is the angle i need to find

[acoustic wraith]

if so, then the number 20 should be on the pink line

why is it on the sloped line

[crystal breach]

because like doesnt tehq uestion ask what would they need to compensate

to get to the pink line

is it not so that the 20 goes on the slope?

[acoustic wraith]

yes, but the target is 20km/h south

so the 20km/h belongs to the target

not to whatever speed you need to compensate with

[crystal breach]

hm so i switch that up

[acoustic wraith]

also

[crystal breach]

and then how do i find that angle

yes

[acoustic wraith]

the current is flowing from the southwest and not to the southwest

in your diagram your current vector seems to be pointing southwest

it should point northeast, as seen in the solution

[crystal breach]

wait u mena northwest right

isnt east like to the right

[acoustic wraith]

oh from the southeast and to the northwest

my bad

[crystal breach]

okay let me redraw it and do it

wait so basically the diagram is correct

the answers

[acoustic wraith]

the solution diagram you mean?

[crystal breach]

yes

[acoustic wraith]

yeah the solution diagram is perfect

[crystal breach]

my final one should look like that

ohhh okay then

ty ima go now

[acoustic wraith]

aye

[crystal breach]

ty

.close

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[clear sapphire]

I have trouble matching my answers from inertial and non-intertial frame

[sleek vessel]

Show your work?

[clear sapphire]

When I observe from the ground, displacement = 1/2 * (10) * (0.2* 0.2) = 0.2m
(because it would be in free fall)

When I observe from inside the elevator, I have displacement = 1/2 * (2) * (0.2 * 0.2) = 0.04m

[spice venture]

both answers are correct..its just that one is relative to ground and other is relative to elevator

[clear sapphire]

I have tried making sense of it by including the displacement of the elevator in that time

[spice venture]

ok what did u get the displacement of the elevator as

[clear sapphire]

by reasoning that everypoint displaces by -0.24m

because the whole elevator does

[spice venture]

yes so 0.24m-0.2m = 0.04..which matches what u got from inside elevator

[clear sapphire]

so what would have been the 0.04m point from the inside the elevator actually displaces by -0.24m as well

so net displacement would be -0.20m

is that line of reasoning correct?

[spice venture]

yes i think its correct but avoid the negative sign its confusing..we use the negative sign when the displacement happens in opposite direction to the acceleration

[clear sapphire]

Okay, i have another confusion

[spice venture]

ye?

[clear sapphire]

If i were to use relative velocities and accelerations and set the elevator at rest and then find displacement, how would that work with these numbers?

This is what has me confused, it doesnt seem to work out

[spice venture]

huh so u want to work with relative velocity in a non inertial frame?

[clear sapphire]

yes

[spice venture]

okay lets try that

no wait

velocity with respect to what?

[clear sapphire]

wdym?

[spice venture]

in relative velocity..what will u take the velocity relative to

cause in acceleration our reference is gravity

[clear sapphire]

im not sure i understand

[spice venture]

ok see in the elevator frame here

u took acceleration as 2 right?

[clear sapphire]

Yes

[spice venture]

why did u do so

[clear sapphire]

net acceleration after considering pseudoforce

[spice venture]

yes so the pseudoforce occured because we have to consider gravitys acceleration right

[clear sapphire]

i dont understand. wouldnt that be included in the elevator's net acceleration (12m/s^2)

[slim horizon]

WAIT WRONG CHAT

[spice venture]

no

that is the elevators acceleration given to us in the question

but our object here is not the elevator it is the block

ok bro i have to go now sorry..

[clear sapphire]

its okay

thank you for your help

[spice venture]

i hope i helped

[clear sapphire]

ye it just struck me

.close

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[calm coral]

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[acoustic wraith]

[safe lynx]

Do you have any idea about yang mills?

[acoustic wraith]

what makes you think I have?

[safe lynx]

You help many people on different topics so i thought you might have some idea

[acoustic wraith]

so, so far I have done stats, combi, trig, algebra, calc. what makes you think I do QFT?

[winter elbow]

[safe lynx]

I just thought, nvm

[safe lynx]

Do you have any idea about yang mills?

[winter elbow]

nah

[safe lynx]

Alright

Makes me feel like closing the ticket

[acoustic wraith]

it's closed alrd

[calm coral]

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[somber swift]

My integration is very rusty and im having issues integrating this cdf to get the pdf

[patent phoenix]

hi can someone explain how to do this question

im very lost

[somber swift]

@patent phoenix we occupied the channel at the same time

[jade fractal]

try derivative $e^{-x^2}$

[potent lotus]

Médicis

[somber swift]

.close

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[obsidian cedar]

$\lim_{T\to0} \int_{-T/2}^{T/2} \frac{1}{T} \cdot x(t) dt$

[obsidian cedar]

How do you use the integral average theorem here?

[potent lotus]

Reginald Puddingface

[obsidian cedar]

x(t) Is continuos

[cedar oyster]

[=\lim_{T\to 0}\frac{\int_{-T/2}^{T/2}x(t)\mathrm{d}t}T]

[potent lotus]

Flip

[cedar oyster]

I think that helps a bit

[obsidian cedar]

I think I have solved

Thanks

.close

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[blazing coyote]

so $\sum_{i=1}^{n} i \frac{1}{2}^i$?

[potent lotus]

wai

[smoky spear]

yupp

lim n-> inf

[blazing coyote]

ofcourse

uh, lemme comput the sum and then the limit rq

so 2

[smoky spear]

yupp

[sleek vessel]

In general, 1/p I believe

Which makes intuitive sense too

[blazing coyote]

Yea

[blazing coyote]

you mean p?

[sleek vessel]

No, 1/p

Where p is the probability of what you want to happen

[blazing coyote]

ah, yes

then okay

[blazing coyote]

Here I'll just compute E(X_n) for now

[smoky spear]

um.. ig it'd be easier if you compute expectation of one step first and then multiply it by n

[blazing coyote]

I don't think it scales like that , does it

[sleek vessel]

It in fact does

[smoky spear]

if you dont think so, then try using summation of each step

[sleek vessel]

By linearity of expectation

[smoky spear]

you'll see it scales up

[sleek vessel]

The final position is the sum of what you do in n steps

[smoky spear]

yea

[sleek vessel]

So it's the sum of n identical random variables

[smoky spear]

um.. no

[sleek vessel]

You can calculate the expectation of one of them and then multiply by n

[sleek vessel]

iid random variables sorry

[smoky spear]

there are only two values your random variable can get

either left or right

[blazing coyote]

$\sum_{i=1}^{n} (\sum_{j=0}^{n} p^{n-j} (-1)^{j}(1-p)^{j})$

[sleek vessel]

I was doing 1 and -1

And then summing them up

[smoky spear]

same

[blazing coyote]

wait, this is wrong

I think part of it is right though

[potent lotus]

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

wai

[blazing coyote]

there we go, I think this works?

[sleek vessel]

Uh what does that calculate to

[blazing coyote]

I think the outer sum is unncessary on further thought, I'll compute the rest

yea, this is going to be messy

[sleek vessel]

Do what we suggested instead lol

Every step is -1 with probability p and +1 with probability 1-p

[blazing coyote]

so n(1-2p)

[sleek vessel]

Find the expectation of that

And then multiply by n

[blazing coyote]

this

[sleek vessel]

I've forgotten how to calculate bernoulli expectation

[blazing coyote]

I'm using this

[sleek vessel]

Yeah that should be it

[blazing coyote]

Cool

Thanks both of y'all!

That's it for now !( I'll solve my problem sheets now that I've read the theory and then go through this book, as it's not the easiest)

Thanks!

.close

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[visual sigil]

why is it not

t = d/s

10/60 = 1/6 hrs
15/60 = 1/4 hrs

1/6 + 1/4 + 5 min = 6/12 = 1/2 hrs, so 12:30pm

probably something related to the post train

but it asks for the express arrival so i'm confused

correct answer is D

[rugged pine]

You have to wait for the post train to get to Layton station

[visual sigil]

but why

i thought this too at first

but then it says laydon has two seperate tracks

[rugged pine]

Just the station

Between stations there is only 1 track

[visual sigil]

ah so they can arrive at the same time

ok

thanks

.close

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[patent phoenix]

hi can simeoine heklo me with this

[patent phoenix]

idk how to prove

@simple musk sorry i went iut but tysm for your help i rlly appreciate it

[vagrant oak]

what part

[patent phoenix]

C

[vagrant oak]

general = homogenous + particular, have you ever heard this?

[simple musk]

say "y" is the solution to the homogeneous system (Ay = 0), and say "z" is a solution to the system Az = b, then A(z + λy) = b because
A(z + λy) = Az + A(λy) = Az + λ.Ay = Az + 0 = b

[simple musk]

oh, mb

[vagrant oak]

oh

[vagrant oak]

this is waht part a) says

think about how you could apply (a)

the set of solutions of Ax = 0 consists of infinitely many xh lying on that line

ill let u figure out the rest

[patent phoenix]

never

[patent phoenix]

ius this to c

[vagrant oak]

it's basically what part a) does, try applying it (perhaps to find at least one other solution of Ax = b)

[patent phoenix]

can u guys help w c first

i got a and b

im genuineluy confused w the wording for c

[simple musk]

the set of all solutions of the system Ax = b, is given by the line λ.y + x
where x is the particular solution to Ax = b and y is the homogenous solution , Ay = 0, so basically, we have a line that passes through x, but we move in the direction y, but for this you have to see the tip of the vector x1 as a point call it P, where x1 is a vector made out of P - (0,0,0)

[patent phoenix]

i really dont understand

sorry

im so confused

whyre we making it as point p

[simple musk]

a line is made out of a point and a direction vector, in this case we want that the whole set of solutions is λ.y + x right? where y is the homogenous solution and x is the point the line passes through

[patent phoenix]

i thouht A(x1) would look smth like this

because if A is a matrix isnt it just a transformation (scalar multiple) of x1

[patent phoenix]

idk what a homogenous solution is

😭

[simple musk]

just, the solution to a system Ax = 0

[patent phoenix]

is it the x vector?

so like xh?

[simple musk]

yeah xh is the solution to our homogeneous system

A. xh = 0

[patent phoenix]

okay

that makes sense

okay so that part i understood

so A(xh) = 0

just dont get what it means by A(x) = b

what is b and what is x omg

[simple musk]

x is the solution to the system Ax = b

[patent phoenix]

what is b exactly

😭

does b = 0??

[simple musk]

so if A is 2x2, then x is 2x1 and b is 2x1

[simple musk]

b is not zero

otherwise x is a solution to the homogeneous system

[patent phoenix]

oh

wait

can you come back to this in a sec

my friend needs help

on this

😭

b

lemme try to digest everything you said abt c first

[simple musk]

you want to find the matrix representation ?

[patent phoenix]

i dont get how to prove it

but yea

i thought i just

solve the matrix

solve for x and y

@simple musk

[simple musk]

both lines needs to pass through the origin

and be non parallel

what I am saying is that, if this two non parallel lines pass through the origin, such that their direction vectors are non parallel (aka not linearly dependent), then we can form a basis of R2 out of them

where this basis of R2 is made out of the direction vector of L1 and the direction vector of L2

[calm coral]

@patent phoenix Has your question been resolved?

[simple musk]

for b) if you see the line L1 as a parametrization we get
x = t
y = 3t + 5
then we get
L : (x,y) = (t, 3t + 5) = t(1,3) + (0,5)

[patent phoenix]

but that only considers one of the lines

tho

also like

is there no way of getting 2 lines into matrix form

like n x m matrix

@simple musk

[simple musk]

well a line that passes through the origin (0,0)=(x,y) can be represented as a subspace, say for example L : (x,y) = λ(1,2) + (0,0), is represented as the subspace <(1,2)>

also, we are not trying to get the lines into matrix form, but like, we are trying to find the matrix representation of the projection T(v) = v1

[calm coral]

@patent phoenix Has your question been resolved?

[simple musk]

ok, so for a) we have two lines, L1 and L2 that literally, have a direction vector and a point that they pass through, say for example L1 : (x,y) = t.(1,3) + (0,5), the point they passes through is (0,5), so if a vector v1 is in L1, then it must be a multiple of (1,3), now say L2 : y = 2x + 7 , such that x = t and y = 2t + 7, then (x,y) = (t, 2t + 7) = t.(1,2) + (0,7), so we found that L2 : t.(1,2) + (0,7), then if v2 is in L2, then v2 is a multiple of (1,2), having said that, since we know {(1,2), (1,3)} forms a basis because they are linearly independent (they are non parallel, both are not multiples of each other), we get a basis of R2, so we can represent every vector in R2 as a linear combination of this two, right?

[simple musk]

for b), we will reuse the parametrisation of both of this lines, so
L1 : (x,y) = t.(1,3) + (0,5)
L2 : t.(1,2) + (0,7)
Now, they are talking again about v1 and v2, where v1 is a vector in L1 and v2 is a vector in L2

basically we need to re use the approach from a) that we have a basis of R2

call this basis of R2, call it, B = {(1,2), (1,3)}

@patent phoenix

we are tasked to find [T]_{std->std}

so that matrix is findable by doing

[T]_{std -> std} = [id]_{std->std} * [T]_{B -> std} * [id]_{std -> B}

which is then simplified to

[T]_{std -> std} = [T]_{B -> std} * [id]_{std -> B}

which is then simplified to

[T]_{std -> std} = [T]_{B -> std} * ([id]_{B -> std})^(-1)

so we need to find the inverse of the matrix whose columns is the basis vectors of B placed as columns

and we need to find the output of T(b1),(Tb2) and place them as columns

@patent phoenix

you multiply those two together and you have your matrix rep of T wrt std -> std

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[calm coral]

@chilly palm Has your question been resolved?

[chilly palm]

.close

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[hexed grove]

.reopen

[calm coral]

✅ Original question: #help-42 message

[hexed grove]

ur back

remember my previous questions now

1. what is the difference between permutations and combinations
2. what is the difference between combinations with and without repetition
3. which formulas correspond to which way of counting

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[night lion]

.close

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[unique juniper]

can anyone please tell me what on earth this is 😭

[sly brook]

completing the square skills are necessary here

watch a few vids on it

[unique juniper]

I know how to do completing the square

[sly brook]

then?

[unique juniper]

wait I need help with part c

the rest is fine

[warm warren]

What is troubling you in part c

[sly brook]

i or ii?

[unique juniper]

both

[sly brook]

i ->can be done using time of flight
ii ->can be done by finding the time of flight for half of the trajectory, that corresponds to the total time of flight divided by 2 and the max height can also be dealt with using formulae

[unique juniper]

alright thanks

.close

[calm coral]

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[sour snow]

How can I expand (1+x+x^2)^4

[sour snow]

is there any beautiful formula?

[odd fulcrum]

Unless you feel like writing (1+x+x²) four times then probably not

[sour monolith]

the coefficient of x^n will be the number of ways to choose a,b,c,d that take values in 0,1,2 and sum to n

[glass heart]

there is the multinomial theorem which generalizes the binomial theorem

whether you wanna call it beautiful is up to you

[odd fulcrum]

@sour snow has your question been answered? Is there anything else we can help you with?

[sour snow]

So you are saying i will have to check for each powers

Multinomial theorem was in bucket

If I have terms like

(3x^2+3y^2+8xy)^4@glass heart @sour monolith

@odd fulcrum

Will the formula still works?

[odd fulcrum]

!noping

[calm coral]

Please do not ping individual helpers unprompted.

[odd fulcrum]

It’s not a big deal for me, but many other helpers take this seriously.

[sour monolith]

multinomial works for (a+b+c+...)^n

[sour snow]

so a,b,c can be any thing

I meant in any power

Not linear

[sour monolith]

yeah its like binomial is for (a+b)^n

[glass heart]

just like you can use the binomial theorem for situations like (2x^3+7x^4)^17 you can also use the multinomial theorem for any situation (sum)^power

[sour snow]

So i will have denominator factorial

(4,0,0,)(0,4,0)(0,04)
(1,1,2)(1,2,1)(2,1,1)(2,2,0)(0,2,2)(2,0,2)

So via multinomial i found there will be 9 terms

I am just checking roughly i don't need to find all the terms. I only need to count exactly

,w (3x^2+3y^2+8xy)^4 expansion

[potent lotus]

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[sly brook]

usage of multinomial theorem is needed

[calm coral]

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[atomic notch]

Hi is my proof correct, I think it's correct

[atomic notch]

cause each step is reversible equivalence

.close

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[opal sinew]

need help for geometry hw

[opal sinew]

exercice 12

[marble pendant]

Translation?

[fallow haven]

english?

[opal sinew]

yea gimme 1 sec

[velvet osprey]

c'est quoi la médiatrice ?

[opal sinew]

jcrois cest la ligne qui passe par le milieu des deux points

[mossy wigeon]

cest quoi ta demarche pour le moment

ou tas genre aucune idee

[opal sinew]

aucune idée

💔

[mossy wigeon]

pour le 1

att faut que je reflechis mdr pardon

[opal sinew]

prend ton temps c correct lol

[mossy wigeon]

genre je peux dire la reponse mais pour le prouver cest tuff lol

genre dans C1 et C2

[opal sinew]

ouais le cours au complet c juste des preuve😭

[mossy wigeon]

tse O1A + O1B cest = r

et puis O2A O2B = r

puisque A et B sont les deux point dintersection du cercles C1 C2 et centre O1 O2 avec le meme radius

[opal sinew]

ca derange tu que les cercle ne sont pas alligné

[mossy wigeon]

enft

sils sont alignee

je pense pas quils forment un triangle

genre ils faut quils soient pas aligne

[opal sinew]

okokok

[mossy wigeon]

mais estce que tu vois que par ca A B ont meme distance de O1 et O2

[opal sinew]

att je vais faire un petit brouillon

[mossy wigeon]

okay okay

[opal sinew]

Comme ca?

ah merde je les ai alligné

[mossy wigeon]

2 ca va

faut juste que

taligne pas tout les 3

[opal sinew]

mais comme la je fais quoi

[mossy wigeon]

si tajoute un troisieme cercle

estce que tu peux visualizer que

AB ce que tas dessine

est la mediatrice de O1O2

CD

sera mediatrice de O1O3

EF

sera O2O3

donc OA = Ob et O'A = O'B

tu vois quils sont equidistant de O et O'

[opal sinew]

oui

[mossy wigeon]

donc avec ca estceque tu peux determiner quils sont perpendiculaire tu penses

[opal sinew]

2 sec jvais le dessiner pour le voir

est ce que le troisieme cercle doint avoir un point dintersection avec les deux cercle

[mossy wigeon]

ouiii

tse ton C1 coupe C2 en A et B

C1 doit couper C2 en C et D

donc C2 coupe C3 en E F