help-23

[iron tulip]

that is 'greater than or equal to'

[languid crystal]

No i really mean the inplies arrow

I know it looks confusing like the way i did

[iron tulip]

2+2 is not a statement

[languid crystal]

But 2 + 2 implies 4

[iron tulip]

'implies' is just used for statements

2+2 is an expression, not a statement

[languid crystal]

Can you give a simple statement?

[iron tulip]

you could have [ x = 2 + 2 \quad \implies \quad x = 4 ] though

[flat frigate]

Lance

[languid crystal]

Ahhhh ty very much!!

[iron tulip]

[languid crystal]

.close

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[plucky island]

do u guys has the way to find x for x^2+5x+4 = 0 by not using the [x+4][x+1] = 0 but using the formular to find x ?

[bold ferry]

quadratic formula

have you learned about it?

[lean otter]

ya

quadratic formula

[flat frigate]

kihei

[plucky island]

nooooooooooooooo

[lean otter]

u can also formulate completing the square

[plucky island]

but imagine if there more number

[lean otter]

like?

cubics?

[plucky island]

x^2+2049x+1049600

b^2 imagine

[lean otter]

[bold ferry]

well you should still use it.. just a calculator is recommended

[plucky island]

like i just wanna know there another way or no

[lean otter]

yes

there is

[plucky island]

but if u in test

and it say just

the answer or x different is 1

can u do?

[lean otter]

u can max to max do

$x^2+bx+c=0$

$\implies (x+\frac{b}{2})^2+c-\frac{b^2}{4}=0$

[flat frigate]

Normal Zeta

[bold ferry]

what

[lean otter]

$\implies x+\frac{b}{2}=\pm \sqrt{\frac{b^2}{4}-c}$

[plucky island]

like answer of x1-x2 = 1

x1>x2

[flat frigate]

Normal Zeta

[lean otter]

$\implies x= \pm \sqrt{\frac{b^2}{4}-c} -\frac{b}{2}$

[flat frigate]

Normal Zeta

[lean otter]

thats a simplified version

for a is equal 1

[bold ferry]

I think you’d still need to find both values of x

unless I’m forgetting something

[lean otter]

[plucky island]

i has the equation 😄

i just think it last week

and its work

[lean otter]

well the difference of the roots is

[plucky island]

for almost all equation [except x2 > x1]

[lean otter]

$\sqrt{b^2-4c}$

[flat frigate]

Normal Zeta

[plucky island]

if u want just dm me and ill give

[bold ferry]

give what

[plucky island]

here

[plucky island]

for find this 10 second

[bold ferry]

what are you trying to say

[safe radish]

@plucky island Has your question been resolved?

[safe radish]

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[soft fulcrum]

Help please

[naive pivot]

hoooooooooooooolllllllyyyyy

[safe radish]

@soft fulcrum Has your question been resolved?

[soft fulcrum]

.close

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[lean otter]

How do i integrate this

[plucky elk]

why does that dx look like in the exponent

[lean otter]

It is

[plucky elk]

just seems like nonsense

[peak estuary]

first define what this is supposed to mean

then we can worry about actually calculating it

[lean otter]

The question simply states that you have to integrate underoot 2 times x raised to the power dx

[plucky elk]

can you take a picture

just seems like missing information after the word "power"

[lean otter]

Its not lol

[plucky elk]

alright then the question is just nonsense. you can ignore it

[final halo]

If you google it there's some quora answer where someone interprets it as an abuse of notation and assigns some kind of value to it

But it is nonsense

[peak estuary]

I have seen product integrals which also look somewhat like this

but still

pretty stupid notation

and needs to be defined

[lean otter]

💀

[plucky elk]

ah ya maybe something in General Relativity

[lean otter]

Saw that

Saw few yt vids too

They do something like this

[peak estuary]

jeez what kind of bullshit is that

[lean otter]

wtf lol

[peak estuary]

replace one instance of dx with h and not the other one?

[lean otter]

Lol

I mean

U can replace the other one too

Then they say

U just solve the limit

,w lim h to 0 of (sqrt(2)x^h)/h

[flat frigate]

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[lean otter]

Their question was different

But

[dusty raven]

this is a cursed question

[lean otter]

The principle remains same ig

[final halo]

Swap limit and integral

[lean otter]

Integrate X to the power dx -1

[lean otter]

Agreed

I mean it kind of makes sense

As dx is something verrryyyyyyyyyyyyy small

That is approaching to 0

,w lim h to 0 of (x^h-1)/h

[flat frigate]

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[lean otter]

Hmm

So sus

,w lim h to 0 of (sqrt(2)x^h)/h

[idle parrot]

What am I looking at here 👀

[flat frigate]

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[dusty raven]

$x^{h}=e^{log(x)h}\approx{}(1+log(x)h)$

[flat frigate]

LayneTheAndroid

[lean otter]

A cursed question

[dusty raven]

It's clear that unless you cancel whatever terms don't have an h, then this limit will diverge

(this isn't how to actually find these kinds of limits, it's just a quick way to check)

[lean otter]

Oh

I just had a thought

[lean otter]

Can we assume dx to be 0?

[dusty raven]

In the context of this notation, it wouldn't make much sense

You'd just be integrating without a differential

[lean otter]

And then x^dx becomes x^0

[idle parrot]

I have a bad feeling about that dx/dx multiplication

[final halo]

You have to ignore every bad feeling you have if you want to do anything with this question lmao

[idle parrot]

Ig it’s like multiplying 0/0 but not multiplying it? Idk

[lean otter]

Then i am just left with root 2 * 1

[dusty raven]

since you're taking a limit anyway, it should be fine as long as the limit is well defined

[lean otter]

[dusty raven]

this doesn't make much sense ... why should we treat the dx in one place differently to the dx in other places?

where did this question even come from

[idle parrot]

It looks sus, you substitute dx=0 only for exponent and not other dx

[lean otter]

So sus

[idle parrot]

Yeah really

[lean otter]

I mean then we get 0/0

[dusty raven]

well situations like this is where limits are applied

But doing that makes it diverge so there really doesn't seem to be a sensible way to answer this

[idle parrot]

That’s why it is invalid statement

[lean otter]

Oh then limit route it is

[dusty raven]

If your question was $\int_{}\sqrt{2}(x^{dx}-1)$ then you could argue that $x^{dx}-1\approx{}log(x)dx$ and integrate the result

[flat frigate]

LayneTheAndroid

[lean otter]

[idle parrot]

Well we can add and subtract sqrt 2 right?

[lean otter]

Where tho

And why

[idle parrot]

Well because as layne mentioned above

[dusty raven]

I don't think this would be valid though

because then you'd be left with

$\int{}\sqrt{2}$

which I'd argue is even less well defined than what we had before because there's not even a dx

[flat frigate]

LayneTheAndroid

[idle parrot]

Right lmao

Where did you get this question anyways

[lean otter]

Tution mate gave me this question

[idle parrot]

Hmm, interesting

[lean otter]

But so sus it is hmm

Anyways ima head over to bed now

Already 3:20 in the morning

[idle parrot]

India?

Lmao

[lean otter]

Ji bahisahab

[idle parrot]

So jaao

[lean otter]

Haan bas jaa raha

स्कूल 7:30

Thanks for your time guys

Yall take care

.close

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[trim condor]

is this correct since its 12C3?

[trim condor]

a is just 10! i think

and then c is probability so no need for a combination

d is permutation i think

[white umbra]

yep you are correct

(it's 21C3 though)

[trim condor]

oh oops yea

read it wrong

im going to send any more hmwk questions i need confirmation on

just a sec

i have no idea if a is correct, i just know the other options seem wrong

like its definitely not b or d

[white umbra]

yep that's right

you can think about it like this

[trim condor]

i should stop doubting myself! hehe

[white umbra]

to choose the first book, you have 25 options; then to choose the second book, you have 24 options; then etc etc so you get 25 * 24 * 23 * 22 * 21 = 25! / 20!

but then, that overcounts by a factor of 5! because there are 5! ways to arrange the same 5 books in a different order

so the final answer is 25! / (5! 20!)

another way to write it is 25 C 5

[trim condor]

ohh

[trim condor]

i was thinking this

that is why i was confused looking at the options loll

but yea i understand

[white umbra]

yup yup they're the same thing

in general, n C r = n! / (r! (n-r)!)

by the same logic I just said

[trim condor]

exactly what i learned

[white umbra]

👍

[trim condor]

yea

i also dont know if this is correct

option d

it cant be a that is too high

30C5 doesnt consider the three students that must be chosen

[white umbra]

yup

do you know why your answer is correct though?

[trim condor]

30 - mandatory 3 students

27, and then the 2 extra random students

[white umbra]

yup!

[trim condor]

perfect

i thought it was a but my friend said b

since its all the options divided by what the people want

[white umbra]

lemme ask you first: is this question a permutation or combination question

[trim condor]

uhm combination? theres no order

[white umbra]

yup

[trim condor]

oh

so its not a

its b

[white umbra]

yup exactly, do you see how they got b

[trim condor]

I thought 12C3 plus 12C2 but i just did that and got 286

oh wait

no

5c3 plus 7c2

[white umbra]

close

if you have two independent choices, what do you do

do you add?

[trim condor]

multiplyyyy

[white umbra]

yup!

[trim condor]

i have like 2 more

my friend said d

3 black, face cards

and divisible by 5 is the 2 black 10s

so thats 5 cards total

[white umbra]

I think there are 6 black face cards

[trim condor]

oh right yea

k q j x clover and spade

3x2

6

[white umbra]

and the cards that are divisible by 5 don't have to be black I think

also, 5 is also divisible by 5

[trim condor]

so 6 +4

10

[white umbra]

you're forgetting the 5s I think

[trim condor]

oh yeaa

14

[white umbra]

this is a weird question haha

[trim condor]

true true

[white umbra]

yeah

[trim condor]

well 6C4 plus 8C4 is 14C4

so i guess d is correct

[white umbra]

wait I don't think that's right

[trim condor]

oh

[white umbra]

because 6C4 + 8C4 = 85 while 14C4 = 1001

oh wait

[trim condor]

ohhh

[white umbra]

oh wait I get it nvm

[trim condor]

i just calculated that

[white umbra]

I interpreted the question wrong

[trim condor]

collect like terms

i think

[white umbra]

it's either they are all black (6C4) or they are all divisible by 5 (8C4)

so yeah the answer is 6C4 + 8C4

that is not the same as 14C4 though

[trim condor]

i see that now yea

gotcha

okay last one i didnt know for sure

i dont even remember if i sent this earlier or not

but im gonna double check

[white umbra]

hmmmm

well this is equivalent to picking 5 marbles right

if there are 2 blues, 6 greens, and 7 reds

[trim condor]

yea you pick 5

you must pick 3 blue

[white umbra]

I don't see a super clean way to do this, but you could split it up into cases, so either you pick 0 blues or 1 blue or 2 blues on top of the 3 you already have

if you pick 0 blues, then how many combinations for the remaining 5 marbles are possible?

[trim condor]

look at this answer my friend sent me

[white umbra]

alright

[trim condor]

5C36C57C0

5C36C47C1

5C36C37C2

5C36C27C3

5C36C17C4

5C36C07C5

Number of ways 8 marbles can get picked from the bag

= 60+ 1050 + 4200+ 5250 + 2100+ 210

= 12870 ways

seems correct? like his work makes sense

oh there are multiplication symbols between some parts

[white umbra]

wait are you counting each different marble as distinct

like for example if I pick green marble 1 and green marble 2, is that different from picking green marble 2 and green marble 3

[trim condor]

no

all the same

he treated them as distinct it looks like

[white umbra]

yeah

all the answer choices here seem way too high

so maybe the person who wrote the question made the same mistake

haha

[trim condor]

right

if its not distinct then it has to be the lowest option

[white umbra]

there are three color options, and five marbles to choose, so the max is 3^5 = 243 different options

so all these answer choices seem very wrong to me

unless I did something wrong

[trim condor]

yea honestly my teacher said for the homework if it says 3 yellow tiles or something, that means they are identical tiles

so that means the marbles are identical

not distinct

[white umbra]

yeah this question might just be wrong then

[solar hazel]

I haven't worked it out but just from the answer choices it seems like the marbles should be distinct

[white umbra]

yeah the answer choices don't make sense if all the marbles of the same color are interchangeable

[trim condor]

if they are distinct then my friend is correct i think.. but if not (which is what my teacher was telling us) then not only is my friend wrong, but all of the answer options are wrong aswell

lol

[white umbra]

but if all the marbles of the same color are distinct, then the colors.. literally don't matter? because all the marbles would be distinct

[trim condor]

exactly

[white umbra]

and then the answer would just be 15 choose 3 = 455

[trim condor]

confusing! ill talk to my teacher about it

[white umbra]

so yeah, this question makes no sense lol

hope it goes well

[trim condor]

thankyou guys for helping me!

[white umbra]

no problem!

[trim condor]

.close

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[foggy tapir]

How would you set this up in a triangle?

[foggy tapir]

<@&286206848099549185>

[unkempt solstice]

Your first step should be reading the rules

!15min

[safe radish]

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[worthy hemlock]

!15m

[safe radish]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[foggy tapir]

Apologies, but this is just a quick question

[unkempt solstice]

Doesn’t matter

You need to wait just like anyone else

[foggy tapir]

Ok

[white umbra]

do you see where the relevant triangle is here

in this image

[foggy tapir]

Yea, I just don't know where the angle would be?

Since the angle decides if it's tan, sin, or cos

I'm assuming the angle is in the bottom right of the triangle, but I'm not sure?

[white umbra]

the bottom right of the triangle is 90°

[foggy tapir]

Left, I mean

[white umbra]

the angle of elevation would be the bottom left angle

yeah

that's right

[foggy tapir]

Alright, thanks

I'll let you know if I have any other questions later

but I'll close this for now

.close

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[orchid finch]

does sin(90-a) == cos(a)?

[feral linden]

Yeah

[orchid finch]

ok ty

.close

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[worthy hemlock]

Complete the square

[thin bridge]

who's saying it's wrong

[safe radish]

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[foggy tapir]

@feral linden Hey, if you're still here, how do I solve this?

[feral linden]

(16π)(5/12)

[foggy tapir]

The question asks me to use 3.14 for pi

[feral linden]

I know

[foggy tapir]

So switch?

[feral linden]

Yeah

[foggy tapir]

Alright, how about this?

[feral linden]

Compare π(6^2)/6 and π(5^2)/4

Little tired answering those, they are exactly same type of questions

You always ask questions of the same type

[foggy tapir]

My teacher always gives us the most arduous stuff

[feral linden]

So far you have only asked three types of questions

[foggy tapir]

Well I actually answer most of the stuff pertaining to distance formula and everything

[feral linden]

But you asked, maybe 10 for each one of them

[foggy tapir]

Pi is just very difficult

[safe radish]

@foggy tapir Has your question been resolved?

[worthy hemlock]

Pi is just a number

[foggy tapir]

It's a number I don't know how to apply in a lot of scenarios

[worthy hemlock]

It's a number that relates to circles

[foggy tapir]

Like this

[worthy hemlock]

Do you know what formula you need to apply?

[foggy tapir]

X/2 pi r = central angle/360

But I don’t have a central angle

[worthy hemlock]

But you have the arc angle

[foggy tapir]

Do I use that?

[worthy hemlock]

Central angles concept

[worthy hemlock]

Yes

[foggy tapir]

I got 1308

[worthy hemlock]

First off, it's arc length

Part of the circumference

[worthy hemlock]

This relates to area of a circle

Because that's pi * r^2

So you need to use the circumference formula instead

[foggy tapir]

Ok, sure

What is it?

[worthy hemlock]

What's the formula for circumference?

[foggy tapir]

@worthy hemlock Also, what's that image you sent earlier?

[worthy hemlock]

Secant tangent theorem

[foggy tapir]

2pi(r)

[worthy hemlock]

So use that instead

[foggy tapir]

For this?

Ok

@worthy hemlock PA is 24?

[worthy hemlock]

Show work

[safe radish]

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[foggy tapir]

.reopen

[safe radish]

✅

[foggy tapir]

Gives me 24

[worthy hemlock]

No

Not quite

It's PA * PB = PC^2

What's PA?

What's PB?

And PC?

[foggy tapir]

PA is unknown, PB is 6

and PC is 12

[worthy hemlock]

Recall that PA, is PB + BA

[foggy tapir]

Yea

[worthy hemlock]

What's PB + BA?

[foggy tapir]

24

[worthy hemlock]

?

[foggy tapir]

Since BA is 18, no?

Wait, let me run this back for you

So we know 6 is PB

We know 12 is PC

PA is x

So

[worthy hemlock]

No

[foggy tapir]

PA is 18?

[worthy hemlock]

No

PA = PB + BA

What is PB?

[foggy tapir]

6

[worthy hemlock]

What is BA?

[foggy tapir]

6 as well??

[worthy hemlock]

No

Can you draw where BA is?

[worthy hemlock]

So what's BA?

[foggy tapir]

Is there something I'm missing?

BA is a variable

[worthy hemlock]

You understand that variables can exist

You can use variables to label an unknown value

[foggy tapir]

Yea

So BA is x

[worthy hemlock]

Yes

So what's PA?

[foggy tapir]

OHH

6x?

[worthy hemlock]

No

PA = PB + BA

[foggy tapir]

6+x?

[worthy hemlock]

Yes

[worthy hemlock]

Yes, solve for x

[foggy tapir]

18

[worthy hemlock]

Plug in that value and check

Does the left side equation the right?

[foggy tapir]

It gives me 144

So checks out

[worthy hemlock]

So there

[foggy tapir]

Thank you

Alright

I don't want to inconvenience you, but I have one final thing

Are you up for that?

[worthy hemlock]

Just ask, if I'm not available someone will come

[foggy tapir]

Alright

[safe radish]

@foggy tapir Has your question been resolved?

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[velvet peak]

Hi, if I have this transitive relation R1, if I have R^-1, is it transitive? How can I exactly prove that?
I did this but dunno if its enough

[velvet peak]

$aRb \wedge bRc \rightarrow aRc$\
in $R^{-1}$:\
$cRb \wedge bRa \rightarrow cRa$

[flat frigate]

Neo

[velvet peak]

I also had to prove that given R1 and R2, if both of them are reflexive, the union is reflexive

I split in cases, but I dont know how can I prove case 2 and 3

[safe radish]

@velvet peak Has your question been resolved?

[lean otter]

this is basically it, just written a little more formally

like suppose $(x,y)\in R^{-1}$ and $(y,z)\in R^{-1}$ then $(y,x)\in R$ and $(z, y)\in R$ so $(z,x)\in R$ thus $(x,z)\in R^{-1}$ and it is transitive

[flat frigate]

jswatj

[lean otter]

reflexive just means $\forall{x}(xRx)$, here it looks like symmetry

[flat frigate]

jswatj

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[drowsy heath]

how do I know if something is a linear function

[plucky elk]

What class is this for

[drowsy heath]

based on the given equation

algebra I

I’d say that is one

[plucky elk]

If you can put the equation into y=mx + b

[drowsy heath]

Ax + by = c?

[plucky elk]

For m and b real numbers

[drowsy heath]

is that one or no

[plucky elk]

Yes that works too

[drowsy heath]

is that example one?

[plucky elk]

But a function is typically written y=f(x)

Yes. m is 2 and b is 6

[drowsy heath]

I don’t think so

[plucky elk]

Find m and b

[drowsy heath]

nope

[plucky elk]

Or a, b, and c if you prefer this way

[plucky elk]

Try again

[plucky elk]

.

[drowsy heath]

really?

can’t I just plug them into a graph

[plucky elk]

What would that accomplish

[drowsy heath]

I guess you’re right

[final halo]

y = 100x + 0

Does that look linear now?

[drowsy heath]

yeah

some of these equations are confusing

[final halo]

How so

[drowsy heath]

I figured it out

just applying the basic forms

how would you find the slope of each line based on an equation?

[final halo]

What do the letters in y=mx+b mean

[drowsy heath]

uhm

that’s a good question

m is the slope

b is the intercept?

[final halo]

Indeed

So get your equations in that form

And you can find slope

[drowsy heath]

how’s this look

[final halo]

Looks good

[drowsy heath]

what’s the slope

just the 4/3?

[final halo]

.

[drowsy heath]

4/3

.close

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[safe radish]

Please don't occupy multiple help channels.

[lean otter]

Is it correct guys?

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

<@&286206848099549185>

The second one is a D ig

a.D b.D c.D d.D e. D f.I

?

.close

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[vagrant sand]

The exterior of a cube is painted and now cut into the smallest pieces (27 pieces for example 3x3x3). Randomly select a piece and toss it on the table. The 5 sides you can see are all colorless. Find the probability of the last side being colored

[vagrant sand]

if the exterior or in other words the outside of the cube is pained, than how come the problems states that the 5 sides (left, right, back, front) is colorless?

[sharp crane]

with the 3x3x3 example: if you cut out a smaller cube from the middle of a face of the cube (i.e. it doesnt touch any edges of the big cube), only 1 of the 6 sides will be painted, as you didnt paint the inside of the cube

[vagrant sand]

why would 1 of the 6 sides be pained if the question states "The exterior of a cube is painted"?

[sharp crane]

there are only a few possible subcubes where only 1 side/face is painted

why not?

[vagrant sand]

since we only painted the exterior of the cube

[sharp crane]

you only paint the big cube, not the smaller ones

[vagrant sand]

yeah

[sharp crane]

so whats the problem?

[vagrant sand]

sorry i meant to say the interior of the cube is going to be colorless

[sharp crane]

right

[vagrant sand]

oh this is easy

what does the question mean about the last side?

[sharp crane]

only the "middle" pieces have 5 colorless sides, right

[vagrant sand]

yeah

[sharp crane]

if you divide into 3x3x3, there are 6 out of 27 such subcubes

now you select randomly from those 6, and its now basically 1/6 that the colored face is on the bottom/invisible

[vagrant sand]

please dont help me i wanna do this by myself just answer my questions

wait wait wait

i said 1/6 before

the question asking whats the probability of selecting the middle cube from the cube since the middle cube is the only cube that has its last sided being colored

[sharp crane]

its a conditional probability problem i think

the question asking whats the probability of selecting the middle cube from the cube since the middle cube is the only cube that has its last sided being colored
@vagrant sand but yeah i think thats what its basically asking

[vagrant sand]

yeah so 1/6, my answer before was correct but reasoning was off

[sharp crane]

wouldnt it be 1/7 since theres also the center cube?

which has 6 colorless sides

[vagrant sand]

wdym by center cube

[sharp crane]

if you divide by 3x3x3 theres also a subcube in the middle of the big cube, completely interior

[vagrant sand]

wouldnt all those sides be colorless though?

[sharp crane]

yes

[vagrant sand]

so how would that have the last side being colored than

[sharp crane]

but given that you see 5 colorless faces, there are 7 possibilities, either its one of the 6 cubes with 1 side colored or the one with 0 sides colored

[vagrant sand]

wouldnt ALL sides of the center cube be colorless

since we pained the exterior of the cube

[sharp crane]

yep

[vagrant sand]

so how would that center piece have a side that is colored

[sharp crane]

it doesnt

[vagrant sand]

so why did you say 1/7?

[sharp crane]

because you are given that you see 5 colorless faces, 1 of the faces is invisible as its on the bottom. given that fact, we know that i can only be one of these 7 subcubes. but only 6 out of 7 have 1 face colored

you have to consider that possibility to get the right probability

[vagrant sand]

can you draw the center piece of me please?

[sharp crane]

imagine you remove the front 3x3 layer of cubes, then its the middle one of the next layer

[vagrant sand]

so if the exterior is painted, how would any side of that center piece be pained?

<@&286206848099549185>

[sharp crane]

it isnt painted at all...

you might want to read up on conditional probability

[vagrant sand]

yeah so how would that be part of the 6 sides that have their LAST side be painted than?

you added it along with the other 6

i dont get that

i know what conditional probability is

im just confused how its not 1/7 and not 1/6

[sharp crane]

its neither 1/7 nor 1/6. without the condition that you see 5 colorless sides the probability would be 6/27. but they say you already know that 5 sides are colorless. that reduces the possibilities from 27 down to 7 because for all other subcubes you would always see less than 5 colorless faces.

so, given that you know that already, the probability will be something out of 7

[vagrant sand]

because we know the five sides of the center cube is colorless that fits the constraint of "The 5 sides you can see are all colorless" the problem states for us right?

[sharp crane]

conditional probability always has do something with shrinking the amount of possibilities given that you know something else.

because we know the five sides of the center cube is colorless that fits the constraint of "The 5 sides you can see are all colorless" the problem states for us right?
@vagrant sand right! it has 6 colorless sides, but that fits the "we can see 5 colorless sides" too

[vagrant sand]

oh so as long as i have 5 sides that are colorless which the center cube does it should be added among the probabilities i see

[sharp crane]

exactly, its at least 5 colorless sides

[vagrant sand]

if we have 27 pieces laid out in a table the probability of selecting a piece within those 27 pieces that have AT LEAST 5 sides that are colorless would be 7/27

[sharp crane]

yes

[vagrant sand]

so thats the answer

[sharp crane]

no

[vagrant sand]

its 1/7

[sharp crane]

almost

how many out of the 7 remaining pieces have 1 side colored?

[vagrant sand]

6

[sharp crane]

so..

[vagrant sand]

oh so its 6/7

since 1 of them is colorless in all sides

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[maiden mural]

i am not able to solve this. i am not able to make it of the form dy/dx + Py = Q

[safe radish]

Please don't occupy multiple help channels.

[maiden mural]

alright, i closed my previous one.

[crimson field]

Divide by x and you shall see that it has homogenous coefficients. Use the substitution v = y/x.

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[vagrant sand]

4 coins probability of 2 heads probability of even heads

[vagrant sand]

wouldnt this be 1/2

<@&286206848099549185>

[plucky elk]

this is unclear. can you just show the original question

[vagrant sand]

that was how it got put in the internet

im assuming it says, if i flip 4 coins what's the probability of getting 2 heads? if i flip 4 coins what's the probability of even heads?

[plucky elk]

use binomial distribution

Let H = # heads in 4 coin flips

[vagrant sand]

oooppppsps

[plucky elk]

Then find P(H=2)

[vagrant sand]

fuck im stupid

the question was fucking unclear anyways

ive done this before

if its at least

you do (4c2) + (4c3) + (4c4) / 2^4

if its at most you do (4c0) + (4c1) + (4c2) / 2^4

if its exact you would use conditional probability

[plucky elk]

yea it's not clear if it's exactly 2 or at least 2

P(H=2) vs P(H>=2)

[vagrant sand]

i didnt even know it was asking that

[vagrant sand]

also less than as well

this is a bad question lemme do another one

[plucky elk]

How is it asking for P(X<2)

[vagrant sand]

i can flip 0 heads right?

or 1 head

[plucky elk]

i'm just asking why you think H=0 or H=1 is relevant here

[vagrant sand]

maybe it shouldnt be since we care about even heads right?

0 and 1 arent even heads

right?

[plucky elk]

oh 0 heads is even

but 2 heads is greater than 0

yea i have no idea what you're supposed to calculate.

[safe radish]

@vagrant sand Has your question been resolved?

[vagrant sand]

@plucky elk im confused now

why wouldnt 0 heads be relevent if its even heads?

[plucky elk]

yes it is you're right

[vagrant sand]

its just H = 1 not being relevent

since its not an even amount of heads

[plucky elk]

3 is also not even

[vagrant sand]

yeah

[vagrant sand]

vs P(H = 0)

[plucky elk]

Are you interpreting the question as P(H=0) + P(H=2) + P(H=4) ?

i think that's reasonable

[vagrant sand]

ye

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[topaz lintel]

why is the answer cos(x) and not |cos(x)|

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[lean otter]

hey how do I do this, but understand it and be able to replicate it evvverywhere

[lean otter]

i basically like, what do I do to that angle of 210, I know on positivve scalars the direction remains the same, and opposite vectors are 180 degrees but everytime i subtract I get weird numbers

do you jjust take 210 from 360? to get 40?

140**

150*

sorry doing head calculations

OOOOH

its like clocks right?

so stay relativvve within the 0-360 degrees

so do I add? And then restart the clock at 360?

add 210 to 360, then the difference/remainder will be my answer?

wait no

that doesnt make sense either

bruh this is like,, like one class I passed out n missed this someone give me the godcookie

https://tenor.com/view/somebody-help-kevin-hart-cold-as-balls-help-i-need-help-gif-18010251

https://tenor.com/view/this-is-us-rebecca-pearson-somebody-help-me-please-help-me-please-sad-gif-17710090

[noble notch]

Personally, I’d verify the magnitude of the resultant vector first if you haven’t already. Try find the components of vector v first, using the given info. Then perform the scalar multiplication and find the magnitude of the resultant vector. From there, you could use tangent or sine or cosine to get the direction angle depending on the quadrant the vector is in

Tho if you use tangent then that may just give you the reference angle so you’d have to do like 1 extra step and pay attention to the quadrant

[lean otter]

oh shoot thank you so much for the reply

does this help my reasoning a little bit more? Like how did he get 30

I can do the whole section of linear algebruh but i cant get this part

i have no idea how he got 30, should I google sine cosines and tangents?

[noble notch]

Well, if you visualise or draw a 210 degree angle, that lands you in Q3. Since the opposite direction cuts diagonally across into Q1 and that’s a distance of 180 degrees rotated clockwise (negative) direction then they just subtracted 180 from 210 tbh

To get 30 degrees

If that makes sense

[lean otter]

OH MY GOD, THANK YOU, YOU CLICKED IT so basically whenever I run into negative scalar parameters,, that change the direction of the original vector, just subtract 180 from its original vector? (Say the original vector is 170 instead of 210, it would be 350 (assuming my arithmetic is above a 3rd graders))

bbut since 210 is 30 degrees above 180, I get sent to the end of the clock again, passing 0 to get 30 degrees (ik its starting to breakdown bbut like, my expressives are awful when it comes tto logic)

[noble notch]

Well, if your angle was 170 then in the opposite direction it’d be 10 or something, right? 170 is in Q2 so you’d end up in Q1 if you went the other way. I think.

Sorry, if I make a mistake, I’m running off of literally no sleep lol

You could also think of it like if your angle was 170 and you subtracted that from 180 degrees (since this is the angle formed in relation to the x-axis) then you’d get 10 degrees

It basically depends on what quadrant your initial angle is in

[lean otter]

oh dog legit I got no idea what im doing dont worry, so im assuming the direction has nothing to do with the angle (the commpuuter people told me that but im a dumbass) so I only need to subtract n shit when its a negativve, as positive vectors always go in the same direction, so basically stop overthinking it, just do the subtraction when it calls for it. I think im on like precalculus shit or whatever if that helps

[noble notch]

No worries, I just got through precalculus + trig over summer so I get it lol

[lean otter]

Like quadrants are Q1 is that obv graph paper we all saw in school, but Q2, 3, and 4 were those othersides of the graph paper.

[noble notch]

Ah okay here this might help

[lean otter]

Should I start there? Or fuck do I have to go back all the way to geometry? Or should I jjust stay in linear algebruh and headbang until I find an answer and move on? Iont wanna be stuck til 40 but I also dont wanna like waste my time in a way ykno, be out here studying basic arithmetic for hours lol

[lean otter]

ahhh holy fuck

its like algorithmic, so if i see plus minus im going to be in quadrant 4, if i see minus plus im going to bbe in quuadrant 2

[noble notch]

Pretty much. You can tell which quadrant something is in either by the angle if it’s given, or by the vector’s components (think of it as the x,y coordinates)

Q1 is anything between 0-90 degrees, Q2 is 90-180, Q3 is 180-270, Q4 is 270-360

So if you saw a vector with these components <-3, 4>, you’d know the vector was in Q2

Or if the direction angle was 120 or something, for example

[noble notch]

Idk man tbh, I’m never taking linalg so idk LOL I only know vectors from doing basic stuff w/ em in trig

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[spark dagger]

can some one help me with this one

[spark dagger]

so a(t) = -32 ?

v(t) = -32t + C

p(t) = -16t^2+Ct+C2

[shell tide]

you know that v(0) = 10

and that p(0) = 400

so with the initial values you can determine C and C2

[spark dagger]

yes thank you

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[remote burrow]

I need help finding the property used

[eager ravine]

they just subtracted 7 from both sides

so what property is that

[remote burrow]

Subtraction property of equality?

[final halo]

You have your own channel

Don't invade someone elses

[remote burrow]

.close

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[lyric harbor]

Im new to proofs can someone help with me errors

[solar hazel]

what's the theorem?

[lyric harbor]

Suppose that integers 𝑚 and 𝑛 are perfect squares. Then the product 2𝑚 × 8𝑛 is also a perfect square.

[solar hazel]

then this is not valid

[lyric harbor]

yeah I know I just dont know where its wrong

what are the errors

[solar hazel]

you can't just let m and n be particular numbers

you can say something like...

let m = a^2 and n = b^2, where a and b are some integers

[lyric harbor]

ok is that the only thing wrong with it though

[solar hazel]

not sure how to answer that, the whole proof was based on letting m = 36 and n = 25

which makes it all wrong lol

[lyric harbor]

ok well thanks

.close

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[lean otter]

A ladder leans against a wall. The foot of the ladder is 5 feet from the base of the wall. The ladder reaches a height of 12 feet on the wall. Find to the length of the ladder.

[lean otter]

a. What geometric figure can be drawn from the problem?
b. What theorem comes out to your mind to solve the problem?
c. Draw a diagram showing the wall leaning against the wall and make the necessary labels and mark x as the unknown.
d. What is now our working equation?
e. Final answer: the ladder is ___ feet tall.

a. triangle
b. pythagorean theorem
c.
d. a²+b²=c²
5²+12²=c²
25+144=c²
169=c²
13=c.
e. 13

Is it correct guys?

c.

[tranquil sentinel]

The ladder reaches a height of 12 ft, it's not 12 ft long, so x should be 12 and 12 should be x in this case

The rest is correct

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

okay thank you

[tranquil sentinel]

Yw

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[vital dirge]

not sure where to begin

[vital dirge]

I dont think I can even factor the denominators because i dont want to bring in complex numbers

perhaps I should multiply everything by both denominators?

nevermind!

I got it

.close

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[narrow nova]

why the tangent axis is drawn at x=1 on the unit circle

[dusty sable]

It’s thales theorem

Sin(x)/cos(x) = tan(x)/1

[narrow nova]

So according to that 1/ root { Sin^2x / Cos^2x + 1 } equals to x/y, right?

[spice grove]

Because you still need a right angled triangle, the line x = 1 is perpendicular to OA

[dusty sable]

What is y and x in your question ?

[safe radish]

@narrow nova Has your question been resolved?

[narrow nova]

found it

[safe radish]

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[safe radish]

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[opaque sorrel]

hi can someone tell me how to identify the pattern here?

im not quite sure how to continue

[steel lark]

can you replace T(n-2) with T(n-3)+T(n-4) at the 6th line

[opaque sorrel]

changed it with this

hopefully this makes more sense

[opaque sorrel]

idk i think the way i did previously was too lengthy

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@opaque sorrel Has your question been resolved?

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[opaque sorrel]

.reopen

[safe radish]

✅

[feral linden]

Use induction to show that $T(n)=\sum_{k=0}^{r} \binom{r}{k}T(n-r-k)$

[flat frigate]

Cogwheels of the mind

[opaque sorrel]

wait how do I do it with big O notation thou

[feral linden]

No idea

And doesn’t seem relevant

[opaque sorrel]

how did u derive the formula

[feral linden]

Just observed that pattern

[opaque sorrel]

erm, how? did u write the answers for various input n or smth

[feral linden]

Just prove it, by induction on r

It holds for r, prove it holds for r+1

[opaque sorrel]

okay lemme try later

thanks

[feral linden]

It’s just

Replace the expression for r, of each T(n-r-k) with T(n-r-k-1)+T(n-r-k-2), then immediately follows

[opaque sorrel]

oh right

so induction is assume x prove x + 1 right

something like that

[feral linden]

Yeah

[opaque sorrel]

so here u replaced x with x - 1 and x - 2 (defined by the qn)

[feral linden]

No I replace each term T(n-r-k) with T(n-r-k-1)+T(n-r-k-2)

Then in new sum, you will see that coefficients of T(n-r-1-k) equals C(r,k)+C(r,k+1)=C(r+1,k+1)

[opaque sorrel]

Okay think i see it now

Lemme try

[safe radish]

@opaque sorrel Has your question been resolved?

[safe radish]

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[trim condor]

does this look correct to you guys?

[slow fern]

Yes

[trim condor]

thanks

how do i solve this question?

[tall bough]

Look at the graph of a normal distribution

[trim condor]

it looks to me that its the part with maximum frequency

so i assume b is correct? unless it also applies to a which in that case the answer would be Both

[tall bough]

It's b

[trim condor]

gotcha

just confirming this is d?

after looking at the uniform distribution

im gonna stick with rectangle

is this correct?

[lean otter]

Yes it is, according to the “68-95-97 rule”

[trim condor]

i was also confused about this but my friend said she thinks that its c

is she right

it makes sense after reading it

[lean otter]

What’s the y-axis of the normal distribution

[trim condor]

the dependant variable?

[lean otter]

Like what is its interpretation

[trim condor]

im not sure what you mean

[lean otter]

The y-axis represents the probability

[trim condor]

oh thats what you mean

[lean otter]

So based on the bell curve shape of the normal distribution, values near the center have a higher probability

Or are more likely

[trim condor]

oh yea since the middle is higher for normal distribution

and since y is probability the center is more likely to occur

I have like 2 more questions left, is this one correct?

[lean otter]

No that’s wrong

I’d look at the definition of the normal distribution

[trim condor]

okay just a sec

oh i see

the answer is true

it is not applied for drv

[lean otter]

Correct

[trim condor]

thankyou for the help!

im going to get my sister to help me with the other hmwk questions

cya :))

.close

[safe radish]

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[severe pebble]

A straight line with negative slope passes through (8;2) such that it cuts the coordinates axes at points P and Q. Find the equation of the line if OP+OQ is minimum where O is the origin

[severe pebble]

so since the slope is negative I know the intercepts are postive

because it passes through (8;2)

but that's about all I know

[keen comet]

What's OP + OQ

[severe pebble]

wdym

read the question

[inner tapir]

is it the sum of the lengths of those segments?

[severe pebble]

It is

yeah

oh that's what he was asking

[inner tapir]

ye

[severe pebble]

it's the length of op and oq

their sum

[keen comet]

So how do you minimise a function?

[severe pebble]

differentiate, don't have function though

[keen comet]

ok then find function

[severe pebble]

and that's why I'm here

[keen comet]

Well do you have any equations for lines that pass through (8,2)?

[severe pebble]

y = mx + b
2 = 8m + b
b = 2-8m
b is also the y intercept

0 = mx + 2- 8m
x = (8m - 2)/m ?
Now I just sum?

[keen comet]

What's the interpretation when you do both of those operations?

[severe pebble]

b is OQ

Which I have in terms of m now

[keen comet]

You already have OP

Just identify it

[severe pebble]

and I can write the x intercept in terms of m too

[keen comet]

Yes

You have everything you need

[lean otter]

it would be easier if the equation would be in terms OP and OQ without the slope

[severe pebble]

let me do it the other way, and then get back to this

[lean otter]

ok i was late

[keen comet]

Actually I don't think this is easier

[severe pebble]

Basically

[keen comet]

Though I think that leans more on geometrical knowledge

[severe pebble]

I have to do

(8m-2)/m + 2 - 8m

assuming m is the slope

I have to minimise this, yeah?

[keen comet]

Try it

[lean otter]

we can find a constraint and then if a+b has to be minimum then a=b

[severe pebble]

is m -4

[keen comet]

no

[severe pebble]

-1/2

[keen comet]

yes

[severe pebble]

That's it?

Thank you very much!

[keen comet]

Try it for variable positive quadrant point (a, b)

It will change your answer into letters

Hopefully you get more intuition

[severe pebble]

what does that mean?

Try what

[keen comet]

Instead of (8,2) consider any positive variables a, b at point (a,b)

Like (0.5, 0.5)

[severe pebble]

yeah a and b, got you

let me work it out and share my work

a and b are both positive?

[lean otter]

yes

[severe pebble]

and m is negative?

or wait nevermind m could be anything let's say

[keen comet]

m has to be negative or you won't be guaranteed axis intercepts

[lean otter]

m has to negative or it wouldn't cut the coordinate axes

[severe pebble]

oh wait a and b are positive, so yeah

Sorry

[keen comet]

||#bots message||
If you need to check your answer, you can compare to this

[severe pebble]

b/a = m^2

[keen comet]

Seems to be missing a negative sign

[severe pebble]

why negative sign, a and b are positive

[keen comet]

oh

Well yes m^2 satisfies that equation

[severe pebble]

i mean you told me that bruh

[severe pebble]

perfect, i got it now

[keen comet]

But you can write m after writing out the m^2 lol

[severe pebble]

m = -sqrt(b/a)

[keen comet]

nice

very nice

[severe pebble]

thank you

[keen comet]

So now you know the minimizer for arbitrary place at top right

[severe pebble]

I sure do

good day

.close

[safe radish]

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[zenith glen]

I need help with this please, for both questions

[zenith glen]

For the first one, should I find FB and use Pythagoras theorem there?