help-27

[restive river]

Can anyone help with 7c pls

[copper fossil]

think of it like this first, what is $P(Y_1 \leq 20 \text{ and } Y_2 \leq 10)$?

[woven radish]

MattDog_222

[restive river]

1 and 9/20

[copper fossil]

the length is not 20

the length is 19

the mean is 10.5

[restive river]

Why are we calculating the mean?

[copper fossil]

well i was just showing that the length isn't in the middle

its measure is 19

P(Y_2 < 10) = 9/19

[restive river]

Ohhhhh

Why isnt it 20 tho?

[copper fossil]

because that would be a set on [0,20]

$$P(X \leq x) = \frac{x-a}{b-a}$$

[woven radish]

MattDog_222

[restive river]

Ohhh alr I see what u did

So if its 9/19

If you draw twice

U just square it?

[copper fossil]

the point of P(X<20) was to show independence. And yes you would square it

its like whats the probability that you flip 2 coins and they're both heads

(1/2)*(1/2)

[restive river]

Ahhhh

[copper fossil]

in this case (9/19)(9/19)

[restive river]

Thanks I got it now

[copper fossil]

u might wannacheck a and b

a is 5/19

b is 15/19

[restive river]

Alr thanks

[copper fossil]

idk why the prob did it from [1,20]

like if it was meant to be confusing/particular

[devout snow]

@restive river Has your question been resolved?

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[restive river]

Can anyone help with 8b pls

[restive river]

I got 3/16

But it's wrong idk why

[copper fossil]

did u get 0.28, 0.22, 0.22, 0.28 for a?

[restive river]

Yeahh

[copper fossil]

i got that Pr = 0.2016

but i dont know for sure if im right

how did ug et 3/16?

[restive river]

Yeah that's right

[restive river]

I did a sample space diagram

[copper mango]

P(Total >=7)=P(Total=7)+P(Total=8)

[copper fossil]

^

[restive river]

Oh wait

[copper fossil]

which is $P(X_1 + X_2 \geq 7) = P(X_1+X_2 = 7) + P(X_1+X_2 = 8)$
and $P(X_1+X_2 = 7) = P(X_1 = 3 \text{ and } X_2 = 4) + P(X_1 = 4 \text{ and } X_2 = 3) = 2 P(X_1 = 3 \text{ and } X_2 = 4) $

[woven radish]

MattDog_222

[copper fossil]

and for 8 its x_1=4 and x_2=4

[restive river]

Why is stopping at 8 though?

[copper fossil]

because probability of scoring a 9 is 0

also I think i figured out that ur 3/16 is for a uniform/equal distribution

but the die is weighted

[copper mango]

how do you make anything more than 8 with adding 2 numbers from {1,2,3,4}?

[restive river]

True

[restive river]

Ohhh that's why

Then what is the probability?

[copper fossil]

compute $2 P(X_1 = 3 \text{ and } X_2 = 4) + P(X_1 = 4 \text{ and } X_2 = 4)$ with the weighted distribution from A

[woven radish]

MattDog_222

[copper fossil]

and the reason u can multiply the first 2 term by 2 is because theres 2 equal probable ways of getting 7 (3 then 4, or 4 then 3) and they're independent events

so using part A, what is the probability that you roll a 3 and then a 4

[restive river]

0.22 × 0.28?

[copper fossil]

yes

2(0.22*0.28) + 1(something)

[restive river]

0.28^2?

[copper fossil]

yea

and summing those gives the book answer

[devout snow]

@restive river Has your question been resolved?

[restive river]

Thanks

[devout snow]

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[restive river]

how is integration connected with the area of a fuction

[restive river]

integration is the area under a function

why

riemann sums

[copper mango]

definite integrals is literally defined to be it

Like... the definition of a definite integral is area

[restive river]

which stems from the riemann sum

do you have a proof?

i cant find why the area is equal to that

[copper mango]

proof of a definition doesn't exist

so no

You cant prove a definition

end of

[restive river]

It works by making slices of the function into small slices and adding them up

simple meaning ^

it would be rly helpful if u can help me find a pdf of the proof :3

[copper mango]

There's no proof that definite integrals give area

CAUSE IT'S A DEFINITION

[restive river]

its like asking for a proof of a limit

actually bad take ignore me

delta-epsilon definition

searching for why F(b)-F(a)=area of f between a and b if f(x)>=0

[gaunt quiver]

Because it's how it was defined to be

[copper mango]

You mean a proof of FTC2?

[restive river]

whats that

[gaunt quiver]

oh my bad, this is actually provable

[copper mango]

$\int_a^bf(x)\dd{x}=F(b)-F(a)$

[woven radish]

Mosh

[restive river]

yes

[copper mango]

is this what you're asking about?

this is fundamental theorem of calculus

[restive river]

yes why the fk is this equal to fkng area so random

[copper mango]

there's your keywords

[restive river]

how do you guys find proofs that are legit and 100% correct
I dont wanna find a proof that some dude invented and might be half wrong or smtg

[gaunt quiver]

wikipedia for one

[restive river]

if you learned it in your math class its most likely right

im not given a proof

wikipedia is good i check it atm

thanks

.close

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[echo breach]

I want to design an algorithm for profit re-distribution in a corporation inversely weighted against salary. So 1million dollars spread over 20 people, distributed inversely against their starting amount. i.e. the people paid the least get paid more until they reach the next, and then it slows until they reach the next etc etc what would the math look like?

[devout snow]

@echo breach Has your question been resolved?

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@echo breach Has your question been resolved?

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[crystal yarrow]

do i consider the cases where
x is even and y is even,
x is even and y is odd,
x is odd and y is even,
x is odd and y is odd?

[torn vessel]

nope, well you could... but there's a theorem you can cite

have you been taugh Pidgeon Hole Principle?

[crystal yarrow]

yes

[torn vessel]

ok, what do you know about even - even or odd - odd?

[crystal yarrow]

idk

[torn vessel]

the sum or difference of two even number or two odd numbers is always even

[crystal yarrow]

oh ya

i didnt understand how to read ur problem sorry

[torn vessel]

Since ever positive integer is either even or odd, and you're picking 3 distinct positive integers, what does pigeon hole principle say?

[crystal yarrow]

if m objects are sorted into n boxes, there is a box > (m-1)/m objects

[torn vessel]

Right, so we have 3 numbers sorted into two boxes (even and odd), so one of them has more than (3-1)/2 = 1 object. So two of them must be the same (even or odd)

[crystal yarrow]

oh i see

thank you so much!

.close

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[crystal yarrow]

the objects are the 4 integers and there are 3 boxes

but idk what the 3 boxes represent

using the pigeonhole principle

[torn vessel]

It exaclyt the same as the last problem, just 3 boxes now, 3k, 3k+1, or 3k+2

[crystal yarrow]

ok

but if there are 2 integers in the box 3k+1, how do i show that the 2 integers are divisible by 3

@torn vessel

ohhh

i see

so u have ((3k+1) - (3j+1))/3

[torn vessel]

yup

[crystal yarrow]

the 1s cancel out

so u end up with k-j

lit

[torn vessel]

which is basically what happened with the other one

even/odds can be written 2k and 2k+1

See #❓how-to-get-help

so two odds is (2k+1) - (2j+1) = 2(k-j) is even

[crystal yarrow]

o ok thanks for clarifying

not sure how to approach this

the last part

[devout snow]

@crystal yarrow Has your question been resolved?

[crystal yarrow]

<@&286206848099549185>

[devout snow]

@crystal yarrow Has your question been resolved?

[slender mirage]

Did you try the previous 2 parts?

If you did, you'd have an idea as to what's going on +_+

1. Given the fixed positive integer 3, and the set T of 4 distinct positive integers. You have a set N of the possible remainders when the elements of T is divided by 3.

Since this set N contains 4 elements, but the remainders can only be {0, 1, 2} distinct at most .. by pigeonhole principle there must be at least two elements from T that leave the same remainder when divided by "3"

Hence, there always exists two distinct elements x and y of T such that 3 divides x - y.

[devout snow]

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[hot forum]

@jolly lily

[devout snow]

@hot forum Has your question been resolved?

[hot forum]

<@&286206848099549185>

[devout snow]

@hot forum Has your question been resolved?

[vast rain]

what have you tried?

[hot forum]

2/3

basically im multiplying 2/3 to get the amount of ingredient requried

@vast rain

[hot forum]

<@&286206848099549185>

[devout snow]

@hot forum Has your question been resolved?

[pseudo basin]

@hot forum do you still need help with this

[hot forum]

@pseudo basin no im good now

.close

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[restive river]

ive been making the shape

[restive river]

but im not sure how to fill in the circled area

[placid rover]

Think, it must be connected

Think about which squares must be connected to each other

Which must be disconnected

For example, the picture on the right

Tells you the square underneath 22 must be connected to it right?

[restive river]

?

[placid rover]

Ok, and now those 3 new dots you drew

The right picture tells you where the cut must be

The cuts will correspond in both pictures

[restive river]

so i was right?

[placid rover]

Right?

I haven't seen you make new cuts

[restive river]

ah

@placid rover

[placid rover]

Some ok, some guessing

[placid rover]

You drew 4 new lines ok?

And I think some follow, but some are just guesses.

Can you explain me your thought process for drawing them

[restive river]

i didnt guess at all

[placid rover]

huh???

Well I don't understand how you are drawing that

(and that clearly isn't the correct answer)

Also I just realised an error in my earlier picture (minor)

[restive river]

why not?

[restive river]

hmm

im confused with what is wrong

[placid rover]

https://i.imgur.com/eurqKIp.png

This is correct, 19 shouldn't have been there

Well SUPPOSE this is true

How on earth does that make an 8x8 grid

That bottom cut doesn't even connect to the top

Let me tell you now --- this is already wrong

This cannot be deduced

@restive river let's go through this step by step

https://i.imgur.com/h7Pqx7A.png

first line you drew

Do you know why you drew it (line on the right of (1))

[restive river]

yes

[placid rover]

https://i.imgur.com/z3YKcVq.png

Because it transfers over from the right ok?

In the same way you can transfer 2 more lines but not more

Do you follow so far?

You drew an extra line after this

But where does it come from???

[placid rover]

So the green line on the left is new

and it corresponds to what I marked on the right

You get why?

[restive river]

yes

[restive river]

is this good?

[placid rover]

well yes cus thats what we originally had

(I'm ignoring the red dots, cus they don't help atm)

I am just talking about the cuts

[restive river]

i dont know if i know how to continue

[placid rover]

You have this

We have drawn everything we could on the left ok?

You are stuck on the left

but you can go to the right

There is new information to be gained

[main gull]

I'm curious, what was the original question because it seems like you've been stuck on this for hours

[placid rover]

Take the 7x10 grid with a 1x6 hole in the middle.

Split it into 2 pieces (along grid lines) that can be joined into a 8x8 grid

[restive river]

go to the right?

[placid rover]

Look at the right grid.

You can now draw new lines there.

[restive river]

ah

[main gull]

Restriction is it has to be 2 pieces?

[placid rover]

yes.

https://i.imgur.com/HaHaC4p.png

[restive river]

i dont know what i should draw tho

[placid rover]

Ok good, very good

Now you go back to the left...

Because you just exhausted the information

[restive river]

?

[placid rover]

yes...

keep going

huh wait

yh ok

That isn't a cut there lol

https://i.imgur.com/Ddofs37.png

You can't cut along the edge of the hole

You have already hit it 😄

[main gull]

This exercise feels like it's a maze solving algorithm

[placid rover]

yh it aint maths 🤦

just lateral thinking/whatever

@restive river now you hit the hole, you can do some stuff on the right

[restive river]

no i havent

[placid rover]

In fact, you can do a lot of stuff on the right

Look, this thing is the hole

Your scissors just hit it

[placid rover]

uhhhhhhhhhh

🤔 1 sec

I don't agree uhhh somethings gone wrong, thinking.

@restive river

https://i.imgur.com/E3iJ49Z.png

Ok, look, this is what we have so far

[placid rover]

I don't think this extra extension follows

Can you see why?

If you assume 27-28-29 are connected, you can do this at most.

This problem aint easy to explain 😅

[restive river]

ok

[restive river]

why not?

also, i got to go to math class in 9 minutes

[placid rover]

It doesn't if you look carefully

You are halfway to the answer btw

You can draw that extra line because of the hole...

[restive river]

hmm

[placid rover]

I find it hard to explain but hmmmmmmmm

[restive river]

i dont see it

[placid rover]

A lot of this is just 'smart' guess work

😓

Maybe someone else can put it better into words

But basically, this was my thought process

In trying to make these cuts

[placid rover]

Next is the issue of trying to fill the hole of that 7x10 grid

[restive river]

i dont know how to do thi

oof

its ok if i dont finish this question

honestly, i dont really have the energy to finish

[placid rover]

Rather than click the pieces that way

you can also click them this way

[restive river]

wait what

[placid rover]

That is basically how you will fill the hole

(what I'm seeing in my mind right now)

[restive river]

i cant even follow

[placid rover]

urgh

[restive river]

its ok

we dont have to finish

[placid rover]

Ok

[restive river]

i cant just submit what i have

[placid rover]

lets stick with the original

https://i.imgur.com/9vv8K92.png

Can you see how we must click the pieces together like this

[restive river]

which is pretty much nothing

We start with 70 squares, and since we take out 6, now the area is 64. This means that if we were to make a square, it would be an 8x8. 
    To make this square, we know that we are going to have two pieces that form an 8x8 and two pieces that form a 7x10. These two are interchangeable, so instead of cutting the 7x10 into two pieces to make the square, we can cut the 8x8 to make a 7x10 rectangle.    

[restive river]

not really

[placid rover]

Imagine what the top arrow is doing

it will be dragging the right piece

up 1, left 2

But nvm ok, some of what I'm saying isn't terribly obvious

[restive river]

^

i need a break

[placid rover]

yh sure

rip

[restive river]

i got 2 mins before my math class

[placid rover]

@slender mirage hi

[slender mirage]

why'd we start with where we did and not an edge?

[restive river]

isnt it always

[placid rover]

https://i.imgur.com/HaHaC4p.png

The logic is we can go forwards or backwards

If we go from 8x8

We are making 2 cuts on 2 edges

ok?

That leaves 2 edges untouched

[slender mirage]

Yeah I saw how you're making the cuts

[placid rover]

Therefore we need to preserve 2 edges of 8

on the original

That's how we know how to start

(that gives 2 possibilities, but you can dismiss one of them)

[restive river]

im so fucking annoyed with myself

[slender mirage]

I mean .. where you ended at "18" you could've started with "1" and moved left till you got to "8" move up and it's the same tracing +_+

[placid rover]

That is true

but uh

[slender mirage]

that tracing though... I'm not sure many people could come up with this this spontaneously

👀 Shuri's Shuri afterall

[placid rover]

Just trying to translate thought process into a diagram

No, it's not obvious - I already know the answer, so I am trying to justify it without cheating

[restive river]

its never obvious

[slender mirage]

(ヘ･_･)ヘ┳━┳

[placid rover]

Like Ansh said though, the solution.... uh probably has rotational symmetry

At least, this is a 'good guess'

for what it might look like

So you could try this and see if it works 🤔

@restive river

[slender mirage]

btw crabbo.. start with a new set of grids: 7 x 10 with 1 x 6 hole... and the 8 x 8 on the right...

Now, where your "18" is right now, mark "1" with red. mark correspondingly "1" red at the bottom right edge of your
8 x 8 grid.

You wanna start from here, and move to the left till you touch a wall on either of the two grids! At the same time, start from the top right corner of both these grids and mark this corner cell as "1" with green. start moving right and try to steer clear of overlapping with the red marks. One way is stopping each time you reach a wall with one colour and changing colours to check on how the other is doing

A couple rough works will give you some insights as to what the conditions is for green and red marks to not overlap! [ you have to keep every mark adjacent and not diagonal or randomly placed lol ]

[restive river]

hmmm

ok

i will look into it

[devout snow]

@restive river Has your question been resolved?

[slender mirage]

So far so good lmao

[placid rover]

this is an interesting approach

[slender mirage]

Nope.. I've locked it out

there's no way to fix that 2 x 6 grid in that 2 x 4 space lol

[placid rover]

Well you're trying to draw that line without lifting your pen

idk why?

[slender mirage]

Not really.

[placid rover]

Not sure precisely what the idea is then

this surely follows.

(or maybe not idk)

[slender mirage]

and the 2 x 2 that remains? 👀

[placid rover]

it belongs to the right piece

[slender mirage]

whaaa- how will you add them together then >_<

[placid rover]

It cannot belong to the left, clearly

[slender mirage]

Aren't you a block too high? (@_@;)

[slender mirage]

wait- that's allowed ?

[placid rover]

yh this aint the right answer

no its wrong.

[slender mirage]

thought so lmao

[placid rover]

some wrong assumptions there

but anyways.

🤷‍♂️

[slender mirage]

Not really assuming anything so far. Also, remove that second vertical line frm the left lol

[placid rover]

that's yours 💤

[slender mirage]

That's all I got 💤

[devout snow]

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[slender mirage]

Done lmao

[placid rover]

Im kinda done for today smh

but yes I see the idea

[slender mirage]

👀 no worries. I figured it. Ty

[placid rover]

wait

[slender mirage]

Hmm?

[placid rover]

did u use this method after knowing the answer

or before

[slender mirage]

I don't even know the answer yet lol

[placid rover]

????????????????

[slender mirage]

I just think this is one right way

[placid rover]

wtf u just got it

[slender mirage]

Yeah I started from the top right and bottom left edges simultaneously

[placid rover]

wym u dont know the answer yet smh

[slender mirage]

And changed edges every time I reached a wall

[placid rover]

u just did it 😒

[slender mirage]

Welll... the inspiration came from your dots and lines tho 👀

.close

[devout snow]

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[slender mirage]

.reopen

sadge

@restive river 👀

[devout snow]

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[bold locust]

I'm taking a course on basic group theory and I'm tasked to proof that within a group for some elements A, X and Y that if $AX = AY$ then $ X = Y$. My proof is as follows: $AX = AY \implies A^{-1}AX = A^{-1}AY \implies (A^{-1}*A)*X = (A^{-1}A)Y \implies EX = EY$, thus $X = Y$. (E is the identity element). I'm just wondering if that first step is even allowed, "multiplying" by the inverse on both sides.

[woven radish]

Tiessie

[storm crater]

The proof is good. 🙂

And yes that is allowed. If two things are equal they are also gonna be equal if you do the same operation with them.

the same reason why x = y => x + 3 = y + 3 or more generally for any mapping f you have x = y => f(x) = f(y)

[bold locust]

I thought so, just wasnt sure if I was allowed to put A-1 in front

I'm assuming I may also put it at the end as long as I do it on both sides

[storm crater]

Yes, the mapping in this case is just f(g) = A^-1 g

yes

[bold locust]

thanks for your time :)

.close

[devout snow]

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[vestal prairie]

could someone help me with this question?

[devout snow]

@vestal prairie Has your question been resolved?

[devout snow]

@vestal prairie Has your question been resolved?

[restive river]

idk do @ helpers

without the space

[ivory kayak]

try #groups-rings-fields or #advanced-number-theory

[vestal prairie]

<@&286206848099549185>

[devout snow]

@vestal prairie Has your question been resolved?

[devout snow]

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[steep fjord]

What did I do wrong?

[steep fjord]

The answer is not 1/2.

[deft hemlock]

distributed the negative wrong

[steep fjord]

[deft hemlock]

,w -(1+sqrt(5))/(1-5) + (1-sqrt(5))/(1-5)

[woven radish]

Results provided by WolframAlpha

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

Requested by Yottachad#6969

[deft hemlock]

okay we have:

$-\frac{1+\sqrt{5}}{1-5} + \frac{1-\sqrt{5}}{1-5}$

[woven radish]

M.E.G. Yottachad

[deft hemlock]

commute the addition first

$\frac{1-\sqrt{5}}{1-5} -\frac{1+\sqrt{5}}{1-5}$

[woven radish]

M.E.G. Yottachad

[deft hemlock]

then distribute the negative

[steep fjord]

oh

[deft hemlock]

$\frac{1-\sqrt{5}}{1-5} +\frac{-1-\sqrt{5}}{1-5}$

[woven radish]

M.E.G. Yottachad

[deft hemlock]

so do you understand?

[steep fjord]

uh

what happened to the 1

on the left

[deft hemlock]

i didn't consider it

cuz you didn't manipulate the 1 at all

(so i just wrote out what u did wrong)

u could just add one to this and get the same thing

[steep fjord]

oh

[steep fjord]

is this step necessary

[deft hemlock]

wdym

you HAVE to distribute the negative

you can't just make the 1 negative

cuz that would be wrong

[steep fjord]

no i mean

just leave it as

• 1 + sqrt(5)

[deft hemlock]

??????

[steep fjord]

like it is here

[deft hemlock]

well technically yes

but isn't the whole point to simplify it

?

[steep fjord]

uh

yeah

ohhh

i always do that step mentally

like

i don't even think about it

[deft hemlock]

well this time you forgot the distribute the negative

so just make sure u do and ur all good

[steep fjord]

wouldn't it still be just
1 - 1 = 0

• sqrt(5) - sqrt(5) = -2sqrt(5)

?

[steep fjord]

i didn't even do step one

[deft hemlock]

yes thats right

b/c you forgot to distribute

[deft hemlock]

you don't have to

[steep fjord]

i didn't do this

uh

[deft hemlock]

pretty much you did this

[steep fjord]

oh

[deft hemlock]

$-\frac{a+b}{c} \neq \frac{-a + b}{c}$

and this is clearly wrong

so:

[woven radish]

M.E.G. Yottachad

[steep fjord]

i considered the fraction on the left as positive and ignored the - sign at the very left because i thought the result would then simply be turned negative by it

lmao

[deft hemlock]

what is $-(1+\sqrt{5}) + (1-\sqrt{5})$

[woven radish]

M.E.G. Yottachad

[deft hemlock]

?

[steep fjord]

uh

-2 sqrt(5)

[deft hemlock]

correct

that's all you did wrong

note that its not 0

[steep fjord]

lol i had my doubts about that

[steep fjord]

yeah

oh so

[steep fjord]

this isn't really necessary right?

it's just a bit confusing if you don't do it?

to me at least

[deft hemlock]

yep

[steep fjord]

ohhh

[deft hemlock]

i had to look at it for a while lmao

[steep fjord]

alright

[deft hemlock]

for a second there i thouhgt u just didn't see the answer

cuz it looked right

(like last time)

[steep fjord]

bruh...

wait you mean the picture i sent looked right?

💀

[deft hemlock]

yep

[steep fjord]

bruh

lol

[deft hemlock]

but we see now that it was wrong

[steep fjord]

yes

thank you very much

[deft hemlock]

wait b4 u go

https://media.discordapp.net/attachments/488084340807565332/939870267876966400/unknown.png?width=510&height=676

did u like the old honorable color

or the new one"

server owner changed it for ppl who use light mode (who uses light mode?)

[steep fjord]

honestly i prefer the new one

[deft hemlock]

fr?

[steep fjord]

it fits the rest of the color palette better

yeah the old one was far too pale

[deft hemlock]

good point

[steep fjord]

almost looked like white

that's why it was hard to see too

i suppose this is not what you wanted to hear lmao

[deft hemlock]

but like

who does this lmaoo

[steep fjord]

my eyes-

[deft hemlock]

light mode is painful

[steep fjord]

i prefer it on dark mode btw

[deft hemlock]

i do like the new owner color

[steep fjord]

obviously on light mode too

also uh

maybe my opinion is invalid because i've used light mode before 💀

[deft hemlock]

still valid

[steep fjord]

not saying im biased im just stupid because i've used it

[deft hemlock]

cuz u came to the dark side

[steep fjord]

like i said

i think it just looks nice on my phone at full brightness under sunlight

like

really nice

[deft hemlock]

it does look nice on phone + full brightness

(so light mode)

lol

[steep fjord]

lol

[deft hemlock]

anyways if you have any more questions feel free to ask

if not you can use .close to close channel

[steep fjord]

alright

[deft hemlock]

cya

[steep fjord]

i'm gonna close it

thank you

[deft hemlock]

you're welcome

[steep fjord]

.close

[devout snow]

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[visual minnow]

any1 who can help me out w/ this? ty in advance

[devout snow]

@visual minnow Has your question been resolved?

[hasty ice]

n mod m = k
nx mod m = kx mod m

i think

[visual minnow]

uh

since n mod 3 = 2 --> n = 3k+2

i end up with like 5n = 5(3k+2) = 15k+10 =
but then i get lost at the last part

[hasty ice]

15k mod 3 = 0

10 mod 3 = 1

[visual minnow]

oh so itd be 1?

[hasty ice]

i think so

[visual minnow]

alright thanks

[devout snow]

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[shadow valve]

I lost where my channel was

[devout snow]

Please don't occupy multiple help channels.

[shadow valve]

.close

[devout snow]

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[ivory kayak]

it hasn't been even close to 15 minutes, what a bamboozle

its fine, im sure someone will come around to help

definitely do, it is much more likely someone will help you if you do

(and the first thing we ask is always what have you done so far anyway)

[devout snow]

@dawn breach Has your question been resolved?

[devout snow]

@dawn breach Has your question been resolved?

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[devout snow]

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[restive river]

Hey I am having trouble figuring out the cardinality of this set? I assume it would be 3 because there are 3 variable, is that correct?

[copper mango]

not quite, they want to know how many sets you can make given a choice in those elements

Hint: An upper bound will be 2^3, so you can easily list out all 8

[restive river]

oh so basically making all possible sets with these elements, in this case 3 sets of 2, is that the correct thinking? so 8 like you said

.close

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[restive river]

hello

[restive river]

for #4 i keep getting to this part

-(1/2)y^-2 = (5/3)x^3 + c

but i do not know how to simplify from there

[graceful cosmos]

Multiply both sides by 2
Multiply out the negative
Take a power of -1/2 to both sides

[restive river]

oh so a negative square root?

wait

square root on the bottom?

so 1/ the rest

[graceful cosmos]

Wording is odd haha. More like an inverse square root

[restive river]

isnt that a square root under 1

[graceful cosmos]

Ye

[restive river]

so like x^-1/2

yea

is just 1/sqrtX

damn i had no idea you could undo it like that

[graceful cosmos]

Easier to think of "take the power and multiply it by -1/2" to get rid of it

[restive river]

smart, wrote that down

thanks!!!

[graceful cosmos]

But of course you have to do that to the other side as well, and the power doesn't split over +

[restive river]

what do you mean split over + ?

like turn the negative to positive?

[graceful cosmos]

Doing the rest, you'll be up to:
y^(-2) = -(10/3)x³ - 2c

[restive river]

wait -2c? i was told you dont multiply the -2 to c in this case and just leave it

since we don't know what c was

my teacher bullied that into me

anyway i think i got it now

ty

.close

[devout snow]

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[tawny coral]

how do it will i just substitute the given values for the function then evaluate it?

[shy wing]

yes

use limit properties

[tawny coral]

do i still need to make a table of values like this?

or i just evaluate it example in letter a then the answer would only be 10?

[shy wing]

you already have the limit values

[tawny coral]

ok thanks

.close

[devout snow]

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[devout snow]

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[plush pendant]

Determine when the following quadratic equation is increasing and decreasing:
$h(x)=4(x-1)(x+6)$

[woven radish]

Zizi_Sunee

[plush pendant]

I never understood what this meant 😅

[deft hemlock]

well first take the derivative, set it equal to 0

then make a sign chart

[plush pendant]

How do I do that? 😳

[deft hemlock]

what class is this for

alg 2?

[plush pendant]

Yes

[deft hemlock]

okay my b

notice first that the quadratic is going up

[plush pendant]

No worries!

[deft hemlock]

then think about the parent quadratic

its decreasing to the left, and increasing to the right correct?

[plush pendant]

yes

[deft hemlock]

so when i shift it

those values shift as well

(so left of the VERTEX, and right of the vertex)

so do you know what the vertex of that quadratic is?

[plush pendant]

(-2.5, -49)

[deft hemlock]

now which do we care about

[plush pendant]

technically -5/2

[deft hemlock]

the x value or the y-value

when talking about increasing/decreasing on parts of its domain

[plush pendant]

the x value?

[deft hemlock]

correct

cuz domain involves x values

not y values

[plush pendant]

right

[deft hemlock]

so based on this

can u figure out where it would be decreasing/increasing

[plush pendant]

increasing on (-5/2 , infinity)?

[deft hemlock]

yep

so where would it be decreasing

[plush pendant]

and decreasing on (-infinity , -5/2)

[deft hemlock]

correct

[plush pendant]

oh, ok, tysm!

[deft hemlock]

alright

use .close to close channel

[plush pendant]

.close

[devout snow]

Channel closed

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[devout snow]

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[deft mist]

How do I prove the big Oh of the following code, I believe the count of basic operations is 3n and that the big Oh is O(n)

[devout snow]

@deft mist Has your question been resolved?

[supple knot]

can you express the results one iteration at a time?

e.g. -1, ...

[deft mist]

i dont know

[supple knot]

Oh you're talking the run time of the loop?

yes it's O(n)

[deft mist]

how do i prove it mathematically?

[supple knot]

this is it?

3n = O(n)

are you asking how to prove that?

[deft mist]

yes

[supple knot]

Example 2 here:
http://www.cs.utsa.edu/~bylander/cs3233/big-oh.pdf

use the definition on page 1

[deft mist]

am i understanding this right?

forgot to remove the 3n+7

and make it only 3n

[supple knot]

yea

[deft mist]

.close

[devout snow]

Channel closed

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[deft mist]

thank you

[devout snow]

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[small yew]

Can anyone help me with this? Sorry for the erasures

[devout snow]

@small yew Has your question been resolved?

[small yew]

<@&286206848099549185>

[main gull]

Sum of angles in a triangle equals 180

[small yew]

Yeah

[main gull]

Did you apply that?

[small yew]

I mean do i have to equate something?

[main gull]

180

[small yew]

I cant apply it yet since it has variables

I can't find the x

I have to equate something but i dont see it

[main gull]

That's how you find it

You equate it to 180

Sum of all angles in a triangle equals 180

So sum up the 3 angles given and equate it to 180

[small yew]

Ohh i get ya

But dont i have to find angle g and o

[main gull]

You have to find x first

Wait, I see what you mean now

[small yew]

Answer was 30 so i substitute it right?

[main gull]

Yeah, should be it, because similar triangle, I think

[small yew]

Wait im confused between m<G and <G

Can u help me differentiate it?

[main gull]

m<G is notation for measure of angle G

[small yew]

Can you help me point it out on the diagram?

@main gull

[main gull]

It’s angle G

[small yew]

So is m<G = <G?

[main gull]

m is just shorthanding the word measure

So m<G means measure if angle G or measure of <G

All means the same

[small yew]

Ohh thank you!

.close

[devout snow]

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[somber frost]

where do i start?

ive used cosine law

but that didnt work

[solid trail]

ah

classic problem

https://youtu.be/zUuLgtxKI_4

here’s a similar problem

[somber frost]

this talks about area though, im trying to find the side lengths, but im not sure how to orient the triangle to find that. Any suggestions?

[solid trail]

you can find the area

by finding the side

it’s an equilateral triangle

[solid trail]

vice versa

[somber frost]

so if i find the area

i can find the side lengths?

[solid trail]

A = √3/4 s² is the precise formula

s = side

so yea

[somber frost]

okay i will try this out

[devout snow]

@somber frost Has your question been resolved?

[devout snow]

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[limpid lodge]

can a homie help me with this problem

[exotic stump]

simply evaluate the left hand ones

[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

[unkempt lance]

how can i not use cosine law here

but here i can

sorry for the ping

[haughty torrent]

??? why ping me?

[unkempt lance]

sorry

[haughty torrent]

well, you have two sides and the angle between them, cosine law makes sense

[unkempt lance]

but it doesnt work

i dont get the vr

[haughty torrent]

what cosine law did you do?

[unkempt lance]

c^2=a^2+b^2-2abcosC

[haughty torrent]

i know

what values did you put in

[unkempt lance]

250 for a

50 for b

[haughty torrent]

there's no reason why it shouldn't work

[unkempt lance]

(250)^2+(50)^2-2(250)(50)Cos65

i get 233.3121

but my teacher gets 266

[haughty torrent]

where did 65 come from?

[unkempt lance]

angle

[haughty torrent]

you need to use the angle opposite the edge you want to find

not the angle adjacent

[unkempt lance]

oh so i have to know that angle

to find vr

not 65

[haughty torrent]

yes

[unkempt lance]

OOOH ok

one last question

how did he get 8 degrees here?

the 8 was supposed to be there

but when u form the triangle

he put it bottom any reason why

[devout snow]

@unkempt lance Has your question been resolved?

[restive river]

Well you would move counter-clockwise from north to be 8º west of north as it's mutually perpendicular to north. It would be the same as you said if it was mentioned 8º north of west, implying that he was moving west earlier.

[devout snow]

@unkempt lance Has your question been resolved?

[devout snow]

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[drowsy falcon]

What is the probablity of getting sum of 10 with 3 dice,
I am getting 45/216 as I tried to solve it this way

$x\le 6$

$x_1 = 6 - x$

$y \le 6$

$y_1 = 6 - y$

$z \le 6$

$z_1 = 6 - z$

$x + y + z = 10$

$6 - x_1 + 6 - y_1 + 6 - z_1 = 10$

$x_1 + y_1 + z_1 = 8$

so $10C_2$ which is 45

[woven radish]

cosh²(x) - sinh²(x) = 1

[devout snow]

@drowsy falcon Has your question been resolved?

[drowsy falcon]

<@&286206848099549185>

the answer is 27/216

[somber trail]

@drowsy falcon i think u have to make sample space two dice thrown

And then work ur way up

By considering the sun of the two dices can't be 10 or greater than 10

[drowsy falcon]

like 9 then 8 then 7?

[somber trail]

And also their sum should be greater or equal to 6

I mean u know abt sample space right?

[drowsy falcon]

yes

[somber trail]

So make sample space for two dices thrown

And than make conditions there

[drowsy falcon]

like this right?

[somber trail]

Wait a sec

Yeah that method is wrong

Sry

[drowsy falcon]

which

@somber trail

[somber trail]

The one i told

[drowsy falcon]

ok

[somber trail]

@drowsy falcon

I got it

I just did one condition wrong

That sum of two dices should be greater than or equal to 4

So now it's coming 27/216

[drowsy falcon]

so i have to count?

theres no mathematical way?

[somber trail]

Yeah I made sample space for two dices thrown

I think this is easy

But sure there's a mathematical way that's just connected to it

[drowsy falcon]

i love to see the mathematical way though

[somber trail]

@drowsy falcon k

[drowsy falcon]

i mean appreciate your effort but

[somber trail]

@drowsy falcon nah it's ok maths way is more fun

@drowsy falconhey I've a doubt u used n+r-1cr-1 right?

[drowsy falcon]

yes

[somber trail]

But it's for non negative

Like X1 y1 z1 can be zero

In that case

[drowsy falcon]

then 8C2

[somber trail]

I think so it's n-1cr-1

[drowsy falcon]

which is 28

[drowsy falcon]

sorry this

which is 36

[somber trail]

Yeah ik

But im saying the formula n+r-1cr-1

Is off numbers that can be zero

Like x+y+z=n

Here if x y z can be zero than we use that

It's for non positive

[drowsy falcon]

yes if x, y ,z can't be zero we use n-1Cr-1

[somber trail]

Yea for positve

[drowsy falcon]

still 9 off

[somber trail]

Yeah ik

That's what I'm figuring

[drowsy falcon]

hey but
$x \le 6$, $0\le6-x$ and $x_1 = 6-x$

[somber trail]

Ok?

[drowsy falcon]

oh wait no its other way around

[somber trail]

I mean im sticking to it x is numbers on 1st dice y is no on 2nd dice so 4=<x+y<=9

[woven radish]

cosh²(x) - sinh²(x) = 1

[drowsy falcon]

yes

[somber trail]

@somber trail isnt that Correct?

@drowsy falcon yeah

[drowsy falcon]

we need to get possible ways of getting x+y = 4 and then subtract it from possilbe ways to get x+y = 9

[somber trail]

K ..

[drowsy falcon]

wait

[somber trail]

@drowsy falcon also i think the x+y+z=10 that has answer has 10-1c2 that's 36 but here we said x,y,z are+

So we are also counting the cases where they are like 7 or 8

We need to subtract those cases from 36

[drowsy falcon]

not 8 because we exlcuded zero same with 7

if one would be 7 then other need to be 0 and then other 1

[somber trail]

Nah I'm saying in the original question

It said sum up to 10

So the 36 method is correct

[drowsy falcon]

oh

[somber trail]

Cuz u get 36 then u subtract those cases that are having x,y,z greater then 6

[drowsy falcon]

wait i got it

no wait

yes i did @somber trail

the number of ways getting 7 will be 3! and number of ways of getting 8 will be 3!/2!

which is = 9

36-9 = 27

.close

[devout snow]

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Remember:
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[true oyster]

how to find anti derivative of 3^x

[placid rover]

convert 3 to a power of e

[late fern]

soo...

it becomes e^ln3^x?

[devout snow]

@true oyster Has your question been resolved?

[devout snow]

Channel closed

Closed due to timeout

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[restive river]

can anyone help with part bii pls

[restive river]

shouldnt it be at (theta - 77.3) = 180?

[slender mirage]

[devout snow]

@restive river Has your question been resolved?

[restive river]

<@&286206848099549185> ?

[devout snow]

@restive river Has your question been resolved?

[restive river]

.close

[devout snow]

Channel closed

Closed by @glass beacon

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[subtle fiber]

<@&286206848099549185>

[tiny carbon]

#❓how-to-get-help

[subtle fiber]

H

I would really appreciate, if anyone help me solve this problem

[tiny carbon]

Express things in terms of sin and cos?

[subtle fiber]

Hold a min

Next ?

[subtle fiber]

I already tried it , but i didn't get it

[tiny carbon]

Sin(2theta) is that what that is?

In the denominator on the left side

[subtle fiber]

Reciprocal of cosec

[tiny carbon]

Yes I know

[subtle fiber]

Yeah ?