help-49

[steady river]

for 1 @real marlin

[real marlin]

yeah

[steady river]

x can be

[real marlin]

bro i need work to show

[steady river]

uh..

sending...

x^2+14x−38=0

i made p = x

[real marlin]

is that allowed?

sorry

[steady river]

well u write it as p

x looks good to me so yeah...

[lyric charm]

if op finds it more convenient, why not

[steady river]

now add 49 to both sides

[lyric charm]

uh wait

wait

!nosols

[midnight plank]

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

[lyric charm]

you should not do ppls problems for them!

why fast?

[steady river]

probably school's load

[real marlin]

bro i have exam in 3 days i dont know any of this

[lyric charm]

please do not call me "bro".

[steady river]

not bro
respect

their identity

[real marlin]

oh

[lyric charm]

her*

[steady river]

check pronouns before talking

[real marlin]

yeah i just use that word but ok

[lyric charm]

https://www.youtube.com/watch?v=-KWsS2FZVTA&list=PLzr_MlWyOdDt87OAJCKpZtxUmYv8r1dCV

here is a playlist that you should watch

and/or go to khan academy

[steady river]

@lyric charm do i help or hop onto another section

[real marlin]

oh wow thanks. do u think i can learn this in 3 days

its most of algebra 1

[lyric charm]

dunno

you can cram some of it in for sure! but i would doubt that you can get from "zero algebra skills" to "good at algebra 1" in 3 days

you will have to study now and after the exam imo

[lyric charm]

should probably look for another section -- but remember not to do ppls work for them still

[real marlin]

yeah ill be back on this server in three days lets just say

thank you for the answers that was fast

[steady river]

||irony spotted||

[real marlin]

what

[steady river]

leave it

.

.close

!close

nvm

[lyric charm]

you can only close your own channel

helpfuls and above have permission to close anybody's

[steady river]

ok

im new

thx

[real marlin]

alright thanks ill be back

how do u close

[steady river]

hope you crash the exam..

[real marlin]

🙂

[steady river]

.close probably

[real marlin]

.close

[midnight plank]

Channel closed

Closed by @real marlin

Use .reopen if this was a mistake.

[severe grail]

what part are u confused in

nvm

[lyric charm]

too late

[midnight plank]

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[frosty sable]

how do i do c

[pastel garden]

just solve the inequality and check the interval against the zeroes you find

[night hawk]

Its a similar logic to A and B (assuming you did those)

[lyric charm]

have you been able to do a and b?

[night hawk]

Can you explain where youre struggling?

!status

[midnight plank]

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

[frosty sable]

This is my work for a and b

[lyric charm]

,rccw

[lyric charm]

mmm no good

[pastel garden]

im seeing a problem here

[lyric charm]

you cannot go from ab < 0 to any combination of a<0 and b<0

also unclear wtf your final answer to either one is

[frosty sable]

Im so fucking stupid

[pastel garden]

Think on what Ann said for a few minutes, maybe try redoing the problems, and if you don't feel confident in that endeavor, we're here

[lyric charm]

do not beat yourself up like this please

i will say that your factorization for a is correct

and likewise for b

[frosty sable]

Was I supposed to do something like this

[lyric charm]

this is good but a bit overkill

[pastel garden]

The solution for the zeroes of problem A appears to be right, explanation works for me.

[frosty sable]

It’s just how I was taught it

I just learned this today

[pastel garden]

do you still have a question about problem C and do you think you can solve problem B? because those problems don't add any difficulty compared to this problem A

[frosty sable]

I just cant figure out the factoring for c

[gaunt nimbus]

how abt completing the square?

👀

(i do not fw quadratic formula)

[pastel garden]

idk why I thought -6+2=-2 yeah you'll have to complete the square

[frosty sable]

I got it

.close

[midnight plank]

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[lament gulch]

Hi

[gaunt nimbus]

hello shinji

[lament gulch]

How do i do "Sum of 126 multiples of 7

Do i just do the normal formula?

[gaunt nimbus]

like

7 + 14 + 21 + ...?

[lyric charm]

"the normal formula" could mean a million things

[lyric charm]

24??? 💔

[twilit field]

*21

[gaunt nimbus]

shi

[lyric charm]

did you miss a word there?

did you mean "sum of the first 126 multiples of 7"?

[lament gulch]

Oh yea

My bad

[gaunt nimbus]

i am so braindead after acid-base equilibrium 🥀

[gaunt nimbus]

this right?

[lyric charm]

and what does "normal formula" mean to you?

[lament gulch]

Uh wait

Sn=n/2[2(a1)+(n-1)d]

[lyric charm]

well you are summing an AP, yes

[lament gulch]

The N should be 126?

Both d and a1 should be 7?

[lyric charm]

yes there are 126 terms

yes

[lament gulch]

So do i just gotta do the formula?

[lyric charm]

if you can't function without an obligation then yes

[lament gulch]

I didn't understand what you meant

[lyric charm]

otherwise there is an alternative: you could realize your sum is equal to 7(1+2+3+...+126) and then recall how to find the sum of the first n natural numbers

[lament gulch]

Uh

Our lesson is still at Arithmetic series

[lyric charm]

if you feel like you need someone telling you you "have to" do something then i will say so

[lament gulch]

Uh

I was just not sure tbh

The question was different from the 1st anyways

So yea

Lemme finish it first

[gaunt nimbus]

what 🥀

[lyric charm]

you CAN apply the formula for the sum of an AP

and you have all the data for it correct

im just saying there shouldn't be a "have to" attached to it

[lament gulch]

Sorry

Uh

Last question

[ebon timber]

hi

[lament gulch]

Is it the same for this one

Sum of the first 63 number of the series 19,15,11

[gaunt nimbus]

ap decreasing by 4

so it shoudl apply

[lament gulch]

Is ap like common difference?

So thanks

.close

[midnight plank]

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[lyric charm]

oh yeah that's that one riddle

[viral dagger]

fff trick gets us that f(3n)=f(f(f(n)))=3f(n)

[lyric charm]

,w factorize 1992

[grand pond]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by ann_dec

[viral dagger]

gimme a sec

f:z->z and f strictly increasing implies f(x)>=x
n=1 gets f(f(1))=3
if f(1)=1, then that implies f(f(1))=f(1)=3 which is impossible so f(x)=/=x, that implies f(x)>x
if f(1)>=3 then f(f(1))>=f(3)=3, but we determined that f(x)>x, so impossible
this f(1)=2

f(f(1))=f(2)=3 so f(2)=3

f(f(2))=f(3)=6
so f(3)=6

[golden comet]

Ohhh I think I have seen this somewhere on the internet

[viral dagger]

f(f(3))=f(6)=9 so f(6)=9

since 6=f(3)<f(4)<f(5)<f(6)=9 then that implies f(4)=7 and f(5)=8

you can repeat this for every multiple of 3, but im not sure how to write it exactly

[runic hamlet]

when in doubt, induction

[golden comet]

yeah something like that, but like that is going to take forever to reach 1992

[viral dagger]

yeah that was i wanted to use but im not sure how to really do it, lemme write it first

claim that f(n)=n+9 forall n>=9
by strong induction, assume it holds for all n up to k, base case is true, then
f(k)=k+9 <=> f(3k)=3k+27
f(k+1)=k+10 <=> f(3k+3)=3k+30
since 3k+27=f(3k)<f(3k+1)<f(3k+2)<f(3k+3)=3k+30, which implies f(3k+1)=3k+28 and f(3k+2)=3k+29

but how do you make sure it covers all x>=3

ehh wait actually why is f(3k)=3k+9

is it really the case

[lyric charm]

your claim is false

[viral dagger]

yeah i noticed that

[lyric charm]

1 -> 2 -> 3 -> 6 -> 9 -> 18 -> ...

[viral dagger]

whats the correct clain then

[viral dagger]

oh yikes

so i dont think this will work then

f(1)=2
f(2)=3
f(3)=6
f(6)=9
f(9)=18
f(12)=21
f(15)=24
f(18)=27
f(21)=36
f(24)=45
f(27)=54

ohh the pattern is further up

so f(n)=n+9 forall n>=9?

or am i doing concluding something too fast

[runic hamlet]

I would compute at least two-three more values. we just saw that two values that followed the n+3 pattern wasnt enough

[viral dagger]

awh

we dont know f(7)

what if you just do this borderline manually

[lyric charm]

what are f(4) and f(5)

[runic hamlet]

and what is f(27)

[viral dagger]

7 and 8 respectively

oh

f(7) is 12, f(8) is 15

pattern seems quite vague

[viral dagger]

f(27)=54

my current idea atm is just finding values manually till your close to the 1992 but idk if thatl work

call me crazy but

$f(x)=x+3^{\floor{\log_{3}(x)}}$

[grand pond]

skissue.in.a.teacup

[viral dagger]

or maybe i am

[runic hamlet]

clearly doesnt work for f(7) and f(8)

compute more values and hope that a pattern will pop out

[viral dagger]

im trying to figure out when it "jumps"

it jumps at 3, 9, 21 and im not noticing anything atm

it seems its linear up to some numbers then changes equation

[runic hamlet]

given how the number 3 is important, maybe ternary helps

[viral dagger]

ternary?

[runic hamlet]

base 3

[viral dagger]

o

[vivid yoke]

Since it's mapping Z+ into Z+ and it's increasing we have f(4)=7, f(5)=8 , f(10)=19 , f(11)=21,... It seem like f(3n+1)=f(3n)+1 and f(3n+2)=f(3n)+2

It's just an idea idk if it will help anything

[viral dagger]

is this the pattern?

how do i write a piecewise function in latex

[raven zenith]

$$|x| = \begin{cases}
x & \text{if } x \geq 0 \
-x & \text{if } x < 0
\end{cases}$$

[grand pond]

astraea 💫

[viral dagger]

honestly i cant have an idea on how to write this explicitly

like its recursive is it not

it doesent matteranyway since were not finding f

on proving it..

ohh wait i see it

.close thanks guys!

[midnight plank]

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[midnight plank]

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[viral dagger]

the line y=5x-10 meets x^2-4x+y^2-32=0 at P and Q, find the distance between them

[viral dagger]

,rccw

[viral dagger]

where did i go wrong

[hard shard]

youre missing a squared on the 4*68/26

[viral dagger]

why the square?

[hard shard]

oh wait yeah

why do you think its wrong? should the answer be an int

[viral dagger]

my friend says the answer is 12

and my teacher checkmarked his answer

[hard shard]

the circle has a radius of 6

so that sounds wrong

oh nvm it does go through the center

[viral dagger]

oh wait does it?

[hard shard]

yes

center (2,0)

[viral dagger]

damn

[hard shard]

sillying

[vivid yoke]

yeah it's 12, I did it and got 12

[viral dagger]

the first thing i did was to write it in its square form but i didnt check it passing trough the center :(

[hard shard]

but lets figure out why this doesnt work

oh yeah your minus turned into a plus

[vivid yoke]

Oh yeah, at second last step

[viral dagger]

ohh

fuck

alr ty

.solved

[midnight plank]

Channel closed

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[viral dagger]

i thought it was the -b/a so i got rid of the -

[midnight plank]

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[grave kernel]

need an explanation

i got till the upper bound of A part

after that

idk

and whats sup A

[lyric charm]

sup(A) = supremum of A, a.k.a. the least upper bound of A

@grave kernel is the red underline the first part that's unclear to you?

[grave kernel]

nope, after the upper bound of A, i dont get it

i started with a small problem now im in a huge rabbit hole

[grave kernel]

okay now

what is least upper bound 😭

[lyric charm]

out of all the possible upper bounds of A, the lowest one.

[grave kernel]

okay

right

makes sense

[lyric charm]

the completeness axiom says this is always a thing assuming at least one bound exists.

1. Let A = {nx : n ∈ N}.
2. If (a) were false, then y would be an upper bound of A.
3. But then A has a least upper bound in A.
4. Put α = sup(A).
5. Since x > 0, we have α - x < α,
6. and α - x is not an upper bound of A.
7. Hence α-x < mx for some positive integer m.
8. But then α < (m+1)x ∈ A,
9. which is impossible, since α is an upper bound of A.

can you name the earliest step in this breakdown that is unclear to you? @grave kernel

[grave kernel]

3rd one

[lyric charm]

axiom of completeness

if A has an upper bound at all, then it also has a least upper bound

[grave kernel]

oooooooooooooo

right

okay now alpha

is that

and alpha - x is not an upper bound cuz alpha already is the least upper bound

now

7th one

hows that coming

[lyric charm]

α - x isn't an upper bound of A, therefore there's some point in A that beats it

(and by "beats" i mean "is greater than")

[grave kernel]

oh wait yeah, i forgot A was the nx set

or mx

[lyric charm]

it's the set of all positive integer multiples of x

[grave kernel]

yeah

okay

now

the rest

yeah

i got it

damn

thank you so much

thanks a lot

.close

[midnight plank]

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[midnight plank]

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[chilly gazelle]

need help

[exotic pelican]

what happens if you plug 1 in?

[lyric charm]

!status

[midnight plank]

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

[alpine crag]

Don't they all have 3^n in them?

[civic gazelle]

just factor out 3^n

[midnight plank]

@chilly gazelle Has your question been resolved?

[chilly gazelle]

got it thank you

[silk geode]

let 3^n as Y

and solve it like quadratic

thats what we usually does

[gaunt nimbus]

,calc 3 + 1 - 9

[grand pond]

Result:

-5

[gaunt nimbus]

this is it

[chilly gazelle]

Got it thank you everyone

[civic gazelle]

?

[gaunt nimbus]

!solve

hwo do i use it

[desert siren]

!done

[midnight plank]

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

[desert siren]

[midnight plank]

@chilly gazelle Has your question been resolved?

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[chilly gazelle]

.close

.close

[midnight plank]

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[rare maple]

prove that the set {1,2,3,4,5.....15} cannot be divided into 2 disjoint sets following these conditions. The first set contains 2 elements and the product of the 2 numbers must equal the sum of the numbers in the other set

[rare maple]

i have proved it using mods but there were 4 cases to deal with

so i am looking for a better method

[lyric charm]

,calc 15*16/2 - 3

[grand pond]

Result:

117

[lyric charm]

can say immediately the product of our two numbers must be at most this much

,calc 15*16/2 - 15 - 14

[grand pond]

Result:

91

[lyric charm]

and at least this much

[civic gazelle]

just find integer solutions for xy + x + y = 120

[lyric charm]

meaning that we can't have both of our numbers be smaller than 11

[lyric charm]

how does this happen

oh like x and y are our chosen numbers

got it

[wary valve]

Hmm

[civic gazelle]

add 1 to both sides and you can factor the LHS

[wary valve]

SFFT 🥶

[rare maple]

So xy=120 -x -y

since xy were chosen

(X+1)(y+1)=121

so x and y have to be same

which isnt possible

is this correct

[civic gazelle]

yep

[rare maple]

Thanks for the help again

This was so much better than my mod bash

.solved

[midnight plank]

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[unkempt skiff]

im so confused what did I do wrong

[unkempt skiff]

ok wait I changed it

it’s still not right tho…

[hushed mauve]

the first function is wrong i believe

if i understand it correctly, the first function and your choice for its continuation are discontinuous

try evaluating the first function at x = 1

[vast loom]

Do you not understand how to proceed to get the answer? or are you struggling to do the math?

[unkempt skiff]

like 1-cos pi x right

I got 0

nvm

[vast loom]

yes, so not 1

[unkempt skiff]

Wait yea

[vast loom]

lol

[hushed mauve]

hang on

[vast loom]

but thats not correct

its not 0

[unkempt skiff]

oh

[hushed mauve]

1 - cos(xpi) for x = 1 is 1 - cos(pi)

[unkempt skiff]

what

[hushed mauve]

how did you get 0

[vast loom]

do it again

[unkempt skiff]

yea

[hushed mauve]

cos(pi) is?

[unkempt skiff]

oh yea

[vast loom]

cos pi = ?

[unkempt skiff]

Oops.

1–1

1—1

[vast loom]

no

[unkempt skiff]

2

[vast loom]

ye

[hushed mauve]

uh no

[unkempt skiff]

/2

[hushed mauve]

nonono

[vast loom]

exactly

[unkempt skiff]

2/2

1

[hushed mauve]

wait, wait

from the top

cos(pi) is what again?

[vast loom]

no 1 is correct

[unkempt skiff]

-1

[vast loom]

cos pi is -1

[hushed mauve]

oh right correct

yea proceed

[unkempt skiff]

okay

2/2

1

[hushed mauve]

oh then i might have screwed myself over

[unkempt skiff]

Is B not right

[hushed mauve]

ignore my very first message

[vast loom]

check out the third one

[unkempt skiff]

Oh

Nwwww

okay

uhh

[vast loom]

the limit when x comes from <1

the <1 is important

[unkempt skiff]

is it -inf

[vast loom]

no

[unkempt skiff]

😭

[vast loom]

|x-1| / x-1

you answer is almost correct

[unkempt skiff]

rlly

[vast loom]

its almost 1

[unkempt skiff]

-1.

help

[vast loom]

yes

[unkempt skiff]

ong

[vast loom]

because top is positive and bottom negative

[unkempt skiff]

😭😭😭😭😭😭😭

ya okay thx

wait so

[vast loom]

so first 3 limits are good now

[unkempt skiff]

It’s not C then

?

[vast loom]

I havent looked at the letters

[unkempt skiff]

oh

[vast loom]

lets get the limits correct first

then the letters

[unkempt skiff]

okaay

ty

[vast loom]

so the last one

how did you do it

why 2

[unkempt skiff]

I did

dif of cubes

[vast loom]

im not saying right or wrong, just walk me through the process

[unkempt skiff]

for top

to cancel out x-1

so then it’s

[vast loom]

yes

[unkempt skiff]

(x^2+x+1)-1

3-1

2

[vast loom]

ok yep

so all limits should be good

now, the letters

first one

[unkempt skiff]

okay

factor

I got

3

uh

[vast loom]

?

[unkempt skiff]

is that eight….

Right

[vast loom]

eight

wym

[unkempt skiff]

tight*

RIGHT

[vast loom]

oh type lol

[unkempt skiff]

typos

mb

[vast loom]

wym 3

like C?

[unkempt skiff]

No like

in order for the function to be continuous at 1 LHL must = RHL must = f(1) right

[vast loom]

yep

[unkempt skiff]

so I was evaluating

limit of A

[vast loom]

ok

and it is?

[unkempt skiff]

3?

[vast loom]

how did you get that

[unkempt skiff]

I factored

(X-1)(x-3)

Canceled out denominator

[vast loom]

and if x -> 1?

[unkempt skiff]

Nvm.

[vast loom]

lol

[unkempt skiff]

OMG

SORY

I rlly

[vast loom]

what

[unkempt skiff]

can’t do math rn

LMFAOO

so embarrassing

[vast loom]

nah you're doing good tbh

[unkempt skiff]

🙇🙇🙇🙇

[vast loom]

just few dumb mistakes, but its not like you're bad at math

i do the same all the time

[unkempt skiff]

ty 😭😭

[vast loom]

so limit is?

[unkempt skiff]

-2

[vast loom]

yes so you can put it with the right one

[unkempt skiff]

yes so number 2

[vast loom]

since all the limits are different, there's only one answer

[unkempt skiff]

that’s right rifhr

[vast loom]

now B

should be not too hard lol

[unkempt skiff]

yea lol

[vast loom]

so thats ok

[unkempt skiff]

1. B

2. A

[vast loom]

yep

now C?

[unkempt skiff]

okay

C

Is

1 sec

[vast loom]

take your time

[unkempt skiff]

am I tripping

Or

Des

[vast loom]

maybe

[unkempt skiff]

Does

F(x) become 1

[unkempt skiff]

lol

[vast loom]

no its not 1

same as before, x>1

[unkempt skiff]

yea doesn’t

[vast loom]

🧠

[unkempt skiff]

Absolute value become (1-x)

/ 1-x

[vast loom]

-(1-x)

[unkempt skiff]

??

why

[vast loom]

because x > 1

so (1-x) <0

[unkempt skiff]

wait what

OH

[hushed mauve]

if x is greater than 1, 1-x < 0, but because of the absolute value, we just multiply by -1 to make it positive again

[vast loom]

can you send the problem again, the original message disappeared i dont know why

[unkempt skiff]

oh

[vast loom]

so you understood the limit of C?

[unkempt skiff]

wait so we just treat it as absolute balue first and sub in to see if we get engine or positive?

Negative

[unkempt skiff]

The

Ur

Ye

[hushed mauve]

@vast loom here it is

[vast loom]

if you have |x| where x <0 then it becomes -x

[vast loom]

thank you

[unkempt skiff]

ohh

i was taught so poorly 😔

[vast loom]

because you want the result to be positive

[unkempt skiff]

ouhhhhh

makes sense ty both

[vast loom]

so -x is minus times minus if x is negative

[unkempt skiff]

ok so

[vast loom]

and last D

[unkempt skiff]

ya

wow

[unkempt skiff]

2

[vast loom]

yep

so its ggs

[unkempt skiff]

yay

tysmmmmmm

[vast loom]

:)

[unkempt skiff]

Omg

[vast loom]

you did everything, i corrected small mishaps

good work

[unkempt skiff]

tyty, appreciate both of yall 🙇🙇🙇

[hushed mauve]

i barely did anything hahah

all you and him

[midnight plank]

@unkempt skiff Has your question been resolved?

[midnight plank]

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[midnight plank]

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[lament citrus]

If we know a Boolean formula is satisfiable first hand, whats the way to find any one satisfying assignment

[main current]

If you can simplify to a sum of product form, you're good

Just take any one of the terms, set the values such that each factor is true, and that's a satisfying input

[lament citrus]

Converting into sum of product form is not trivial

[main current]

Problem is, even this isn't always easy to do

[lament citrus]

Yeah

[fresh sparrow]

what's the context for this? what scales are we working with here?

[lament citrus]

So like does the knowledge of a particular formula being satisfiable, give any hint into a not-so-convolutedly solving for any one satisfying assignment?

[fresh sparrow]

i may be mistaken but I'm pretty sure you can't beat regular SAT by a lot

[lament citrus]

? Can you elaborate

[fresh sparrow]

SAT is the complexity class for boolean satisfiability, which checks whether a formula is satisfiable or not

[lament citrus]

Basically like, if you have enough patience to convert into dnf, then you may as well not know of a formula being satisfiable or not, but if we know beforehand that a formula is satisfiable, will we be able to easily find one satisfying assignment

[fresh sparrow]

hmm

[lament citrus]

<@&286206848099549185>

[zealous heart]

Dppl algorithme ?

[lament citrus]

We already have an oracle for sat

We just need a way to find a satisfying assignment

Given that we already know the formula is satisfiable

(the other case is trivial, because if not satisfiable ,no satisfying assignment possible)

[midnight plank]

@lament citrus Has your question been resolved?

[civic lynx]

unless im missing something, you can either bruteforce it via truth table, or consider the subformulas and try to make each one true

[lament citrus]

We need the subformulas to be simultaneously true

Assuming we are in CNF, we need all clauses simultaneously true

[civic lynx]

yeah, so considering the original sentence, you can start with some options for a subformula, then narrow it down

e.g. considering $(P \wedge (P \implies Q)$, consider the subformula $P \implies Q$. Either
\begin{itemize}
\item $e(P) = e(Q) = \textbf{T}$, or
\item $e(P) = \textbf{F}$ and Q is whatever
\end{itemize}
but in the second case, then $P \wedge (P \implies Q)$ would be false, so we can narrow it down to the first case, and get something that satisfies the entire sentence

[grand pond]

haseeb

[civic lynx]

so you are satisfying the entire sentence, but you break it down into the smallest subformula, then keep checking as you go up

[lament citrus]

I see what you're saying yes

Which makes me realise I made a crucial flaw in posing the question

And now that I realise the flaw, I realise the answer too

[lament citrus]

Thanks for this, but one question I have js

Is

[civic lynx]

lol i love when that happens

[lament citrus]

Isn't this justa slight modification over just brute force checking F or T

For every variable

We are doing brute force but also taking out impossibilities

[civic lynx]

basically, yes. i think there are some algos for checking with CNF, but they basically do the same thing (check each variable by clause and assign it a value one at a time)

but you are skipping some lines of the truth table based on what the sentence looks like

[lament citrus]

Also the question I should have asked was, if we knew a way to instantly know if any formula is satisfiable or not, can we easily find satisfy ing assignment and I think the answer to thta would be to set variables one by one to T or F, check the oracle, if unsat, flip, and move to next variable

[lament citrus]

Makes sense

[lament citrus]

And if sat, don't flip and move tk next variable

Anyway thanks a lot!

.close

[midnight plank]

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[civic lynx]

hmm im not sure what this means, since every variable would need some kind of valuation initially, so that you can actually evaluate it

but if it works then

[lament citrus]

The assumption is that we have an access to an oracle for finding out if any given formula is satisfiable or not

[civic lynx]

ohhh ok i see what you're doing

sub in e(P) = T for some P, check this "new sentence" for SAT, and continue

but im not convinced that you'd get a satisfying assignment at the end

[lament citrus]

Why not?

[midnight plank]

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[mystic salmon]

Approximately 10% of adults in the United States are left handed. Out of a random sample of 500 adults, what is the probability that 12% or more will be left handed?

[mystic salmon]

i'm genuinely so confused

i have no idea on how to start

i learned the central limit theorem and other basic things about sampling distributions from ap stats

but no matter how much i reread the notes and even the solution (this is a practice problem) i can't understand

[lament citrus]

First of all you need to identify the sort of distribution the left handed people will follow

What sort of distribution do you think it'll be

[mystic salmon]

hmm

i think it would be a binomial distribution

[lament citrus]

Why so

[mystic salmon]

because you could identify the amount of left handed people as 'sucesses'

and also i don't see how the central limit theorem fits into this because that relates to means of several sampling distributions being approximately Normal

[lament citrus]

Correct

[mystic salmon]

one thing i don't get how it fits in though

[lament citrus]

What does clt say

[mystic salmon]

is specifying that the sample has 500 adults

ohhh

since the sample size is greater than 20-30 the sampling distribution is approximately Normal

[lament citrus]

Correct, and which theorem tells you that

[mystic salmon]

clt

so then we can just do

hold on

how would i arrive at the standard deviation

if i'm using normCdf on a calculator

oh

sqrt((np)(1-p))

so it would end up being normCdf(lower = 60, upper = 500, mean = 50, standard deviation = 6.7082), right?

i checked the solution and it matches

thanks a lot

.close

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[mystic salmon]

Approximately 10% of adults in the United States are left handed. Out of a random sample of 500 adults, what is the probability that 12% or more will be left handed?

I solved this question by noticing that since n(p) and n(1-p) are both larger than 10 and n is larger than 30 we can assume that the sampling distribution is approximately Normal. To find the mean, I multiplied (0.12)(500) to get 60. To get the standard deviation, I did sqrt(np(1-p)) to get ~6.7082. With this, I performed a normCdf function on my calculator with the parameters lower=60, upper=500, mean = 50, stdev = 6.7082. This let me to get the same answer as is provided. However, the solution is different. They used mu sub p-hat = p to get the mean and sigma sub p-hat = sqrt((p(1-p))/n) to get the standard deviation. They then performed normalCdf(lower = 0.12, upper = 1, mu = 0.1, sigma = 0.0134), and got about the same answer as me. This confused me, as they used different formulas.

For the next question, when I tried doing it my way, my mean and standard deviation was wrong:
Candidate A and B are running for a seat in the state senate. The election results showed 75% voted for candidate A and 25% voted for candidate B. To confirm the results, recommend a
sample size that could be taken which would result in an approximately normal sampling distribution. State the mean and standard deviation you would expect to see in these samples.

What are the distinctions between:
mu sub x = np
sigma sub x = sqrt(np(1-p))
mu sub p-hat = p
sigma sub p-hat = sqrt((p(1-p))/n)

and how would I know when to use which? Furthermore, why do both methods lead to almost the same solution*?

*normCdf(lower = 60, upper = 500, mean = 50, sigma = 6.7082) is about 0.068018 while normCdf(lower = 0.12, upper = 1, mean = 0.1, sigma = 0.0134) is about 0.067779

Apologies if this question got a little wordy, but I've been really confused about this topic for the past few days

[mystic salmon]

i realize now this is not concise

if there's a better place to ask this then please redirect me

[steel crest]

i would ask one question at a time, you have two

[mystic salmon]

i see

What are the distinctions between:
mu sub x = np
sigma sub x = sqrt(np(1-p))
mu sub p-hat = p
sigma sub p-hat = sqrt((p(1-p))/n)

and how would I know when to use which? Furthermore, why do both methods lead to almost the same solution*?

*normCdf(lower = 60, upper = 500, mean = 50, sigma = 6.7082) is about 0.068018 while normCdf(lower = 0.12, upper = 1, mean = 0.1, sigma = 0.0134) is about 0.067779

disregarding the candidate scenario entirely, would this be a more reasonable question to ask?

i don't know how to ask just 1 question about this because both of my questions are dependent on each other in this instance i feel

[steel crest]

when approximating a parameter, mu_subscript indicates the expected value of the estimation

[mystic salmon]

so would that be an approximation of the mean?

[steel crest]

so I wouldnt expect to see mu_x, you sure you saw that? I would think it's more likely to be

$\mu_{\overline X}, \mu_{\mathbb {E}[X]}$

[grand pond]

gfauxpas

[steel crest]

something like that

but

[steel crest]

let's say not

ah

I dont think that's good notation

[mystic salmon]

i'm struggling to understand the difference between these 2 kinds of formulas

[steel crest]

they should use capital letters

for random variables

$\mu_X$ is then simply the expected value for the R.V. $X$

[grand pond]

gfauxpas

[mystic salmon]

if that's the case then what's the difference between mu sub X and mu sub p-hat?

[steel crest]

it's stupid to use a lowercase x because that symbol isn't introduced beforehand, so it's random, might as well use Greek letters or emojis

[mystic salmon]

the way it's presented on the formula sheet doesn't make it seem like there's much of a difference

[steel crest]

give me one moment while i see what conventions are used online

I don't like your worksheet

it's making questionable choices about presentation of the symbols

but I'll tell you what i think the sheet means

[mystic salmon]

i can't do much about it sadly since this is the only sheet i can have out during proctored exams

so i have to get used to it

[steel crest]

1. $\mu_x$ and $\sigma_x$ surely should be $\mu_X$ and $\sigma_X$, there's no possible justification for using a lowercase x

[grand pond]

gfauxpas

[steel crest]

and it's telling you: these are the parameters of the variable X. X has mean blah and st.dev. blah

[mystic salmon]

yeah that part makes sense

[steel crest]

X has been introduced in the first line, but "x" has not been introduced at all

[mystic salmon]

if X is some binomial distribution

[steel crest]

now, when they say $\mu_{\hat p}$ and $\sigma_{\hat p}$

[grand pond]

gfauxpas

[steel crest]

they seem to be implicitly introducing a sampling process

[mystic salmon]

yes

this unit is all about sampling distributions

[steel crest]

yes but the sheet you showed me doesn't introduce the sampling until the line "if X-bar is the mean..."

it's defining the horse before the carriage

[mystic salmon]

this is taken from an old ap stats formula sheet

[steel crest]

also $\overline{x}$ is bad notation, because the RV's name is $X$

[grand pond]

gfauxpas

[steel crest]

anyway, what I think the sheet means is

if you were to treat this X as a sample of a larger distribution

then you are estimating p-hat with your sample

the average estimate of p-hat will be p

[mystic salmon]

i don't really understand this part

if X is a distribution in itself

you're saying you're treating the distribution as a sample of an even larger distribution?

[steel crest]

sampling from a distribution always creates its own distribution called the sampling distribution

[steel crest]

yes

[mystic salmon]

hmm

[steel crest]

and in fact this can iterate

you can be sampling from a sampling distribution

likewise you're also using the samples to estimate the standard deviation of the underlying distribution that you're sampling from

the average estimation for the std. will be sigma_p-hat

[mystic salmon]

why does mu sub p-hat equal p though? does it mean that the means of the sample proportions will be the proportion of the overall ratio of successes to total items in the sample

[steel crest]

that needs to be proven, yes. i forgot the proof offhand

it's a theorem, not trivial

https://math.stackexchange.com/questions/226237/expected-value-of-a-binomial-distribution

[mystic salmon]

so mu sub p-hat would be focusing more on the probability of picking something from the sample?

[steel crest]

uh

it's the best estimate of the underlying distribution's rate, given the information you collect by sampling

is how I look at it

[mystic salmon]

if these 4 formulas are identifying different things, then how did me and the question writer arrive at the same answer when I used the formulas with sub X and they used the formulas with sub p-hat?

or should i ask that in another help channel

[steel crest]

well that's not really a math question, that's a horoscope question

why is it that I tripped this morning but didnt fall

[mystic salmon]

i think i understand a little bit better now

thanks for the help

[steel crest]

I'm glad!

[mystic salmon]

.close

[midnight plank]

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[molten bay]

So Order of n of a mean a^n=e

and i know that O(a^p)

Gcd(p,n)=1

So O(a^p) is n

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[last slate]

Thats no math

[ember sage]

Well

Its got some

,rcw

[ember sage]

Ples

[midnight plank]

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[analog carbon]

i think its 100

81x-1 = 1/2(360-100x)

im js not sure

[civic gazelle]

it's 100 but your reasoning is incorrect

[analog carbon]

oh alr

what should it be

[civic gazelle]

connect the center O of the circle to P and R

you'll get a quadrilateral with angles (81x - 1) + (100x) + 90 + 90 = 360

[analog carbon]

ohh alr ty

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[versed wasp]

A bit off topic but it's kinda math

Why is this placement of 9 wrong in this sudoku game?

[golden comet]

why did you think the 9 is there?

[lyric charm]

who's saying it's wrong though

there doesn't seem to be anything on the board rn that it's contradicting with

[golden comet]

well it is highlighted red, which probably means it is wrong

[bronze wigeon]

is that the only spot on that column where a 9 fits?

[golden comet]

nope, it can go to r9c3

[versed wasp]

Yeah

I think because both 7 and 9 are possible and there is no certainty of a single number yet. We can't say for sure?

[bronze wigeon]

You can try solving the other parts of the puzzle first

[golden comet]

I think you just guessed there too early

[bronze wigeon]

or now that you know where 9 isn’t

[versed wasp]

Yeah this is it

Thanks

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[soft glade]

So I'm pretty far into this question but I don't know what to do next. I know that $\triangle CDE \cong \triangle CBA$ by the cyclic quadrilateral property or whatever, and because of that, I've managed to deduce that $\frac{2\sqrt{2}}{3}CA=CB$. What should I do next?

[grand pond]

747244351179980930

[vital lava]

Hello guys, who have an idea to answer this

[livid python]

!occupied

[midnight plank]

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

[livid python]

Plus you got your channel

[vital lava]

Im new to discord and dont know how it work, my bad

[soft glade]

all good

[civic gazelle]

hmm

An observation I see herr is that AE and BD are altitudes of the big triangle

[soft glade]

oh yeah

[civic gazelle]

so we get AE/BD = AC/BC

[soft glade]

I noticed that those were right angles by thales but I didn't think of them being altitudes

[civic gazelle]

then you might confine those into the right triangles inside the circle

set up equations

idk if it'll work or we just end up with what we started

[soft glade]

how do we derive this?

what property tells us this?

[civic gazelle]

1/2 AE.BC = 1/2 AC.BD = area of the big triangle

[soft glade]

ohh

I see

[civic gazelle]

haven't tried it on paper tho

[hidden turret]

OK here's a thought: if BD is an altitude of triangle ABC, then by Pythagoras AD^2 + DB^2 = 30^2, and DC^2 + DB^2 = CB^2

but you can express both DC and CB in terms of AD

[soft glade]

that's a good point

[hidden turret]

So you should end up with two equations in just DB and AD

I am hoping that they are not just the same equation...?

I think (?) AD = 6√(5)

But I did that pretty quickly so plz confirm

[soft glade]

the system I have is $AD^2+DB^2=900$ and $(2AD)^2+DB^2=(\sqrt{8}AD)^2$

[grand pond]

747244351179980930

[soft glade]

so far

let me further simplify

[hidden turret]

yes agreed

[soft glade]

so therefore

$DB=2AD$

[grand pond]

747244351179980930

[soft glade]

right?

[hidden turret]

I got same

[soft glade]

30 60 90 0:

[civic gazelle]

just came back

oh wow

[hidden turret]

I don't think it's a 30-60-90

[civic gazelle]

nvm it's not 30 60 90

[hidden turret]

because DB is not a hypotenuse

[soft glade]

good point

sad

oh well

[civic gazelle]

but we can get those sides nonetheless

[soft glade]

I'll just substitute $DB=2AD$ into the second equation

[grand pond]

747244351179980930

[hidden turret]

Yah then once you have all the sides you could use Heron ig

hella messy tho

[soft glade]

ad=$sqrt{180}$ I think?

[grand pond]

747244351179980930

[soft glade]

oops

bad latex

hey

that's the result you got earlier

[hidden turret]

good for me lol

[soft glade]

and then there's just heron's

I guess

might be easier if I like prime factorize everything when doing it

let me get the side lengths first

ok so I got $CA=18\sqrt{5}$, $CB=12\sqrt{10}$, and obviously $AB=30$

[grand pond]

747244351179980930

[hidden turret]

Yes I got same

Does not look that conducive to heron though

[soft glade]

agreed

I mean tbf, though unused, it is an aime problem

so theoretically it should take time

there's probably a better way though

[hidden turret]

Are you looking for solution in closed form?

[hidden turret]

yes most likely

[soft glade]

closed-form?

[hidden turret]

as in, not a decimal approximation

Normally I would say this is that time to break out your calculator

[soft glade]

yes

[hidden turret]

but then you will get decimal approximation

[soft glade]

well it's an aime problem so it has to have a solution from 0-999

unless the rules are different because of it being unused

so it has to simplify to something nice even if the process isn't pretty

[hidden turret]

OK, I see now, you don't have to use Heron

AC = 18√5

DB = 12√5

Area of triangle is 1/2 their product since DB is an altitude

[soft glade]

thank god you saw that

I was genuinely about to start using heron's

alright

hey and the sqrt(5)s cancel

or not cancel

whatever

$18125$

[grand pond]

747244351179980930

[soft glade]

60*18

60*9 oops

forgot to divide

540

540!!

[hidden turret]

See you have to be lazy and unwilling to work like me otherwise you will not look for shortcuts hard enough 😄

[soft glade]

thanks @civic gazelle @hidden turret

[civic gazelle]

no problem

[hidden turret]

happy to help

[soft glade]

!close

oops

[civic gazelle]

.close

[soft glade]

wrong prefix

.close

[midnight plank]

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[midnight plank]

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[neat silo]

Find the value of 5xy - 3zy if (x + 2y + 4z)^2 >= (5 + 4y^2)(x^2 + 1 + 4z^2)

[neat silo]

I dont know where to begin here

[hoary pulsar]

I'd start with playing with the terms, maybe expanding or moving things from one side of the inequality to the other

[hushed mauve]

i'd say expand everything out first, find your xy and zy

!occupied

[midnight plank]

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[hoary pulsar]

Is this for a specific class? There could be a topic specific method

[neat silo]

Yea im thinking smth like that

im trying to make them somewhat factorable >= 0

like smth + smth + smth >= 0

i got this

[silk geode]

Just remove brackets

Oh nvm you couldn't

What about just tryna factor out all the X,Y and Z

And leaving only numbers

[neat silo]

thats what im trying to do

[midnight plank]

@neat silo Has your question been resolved?

[hoary pulsar]

Okay I found an answer using a very non-algebraic method

I tried plugging the inequality into desmos, and after fighting with it for a while I managed to nail down a point that satisfies the inequality

idk if it's the only one, but there is at least a answer

[neat silo]

ss it please

I got like x = z and x = 1/2y

[midnight plank]

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[hoary pulsar]

x = y = z = 1/sqrt(2)

[midnight plank]

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[lyric charm]

1071.20/(6.15%) is the principal

so what are you talking about?

@worldly pine

[worldly pine]

why in this question we multiplied with 100 then and not in the earlier question...
(both questions are slightly different)

[midnight plank]

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[midnight plank]

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[runic hamlet]

$\frac{1071.20}{6.5%}\cdot 10%\cdot 5$ ?

[grand pond]

Denascite

[worldly pine]

8240

[lyric charm]

idk about anyone else but i have no idea wtf you're asking

anymore

[worldly pine]

I accidentally deleted the pinned message..so..

did you get why 1071.20/6.5 is multiplied with 100

[lyric charm]

they ask for principal directly

[worldly pine]

i know only this way..i get why they did this here..
but

[worldly pine]

in this..it will also be multiplied with 100.. isn't it?

@lyric charm

[lyric charm]

does it matter?

look, if you want to explicitly find the principal and you feel that you have to multiply by 100 for that

you can do it

nobody is forbidding you

i am not your mom, i cannot stop you from multiplication by 100 like it's something illegal or dishonorable

[worldly pine]

yes you are not..
i just had a doubt

will the answer be 8240 x 100 =824000

.close

[midnight plank]

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[midnight plank]

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[rugged vine]

if a simple graph has some odd vertices, how do i show that there are 2 vertices with an even number of common neighbours?

[midnight plank]

@rugged vine Has your question been resolved?

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[molten bay]

Is it 5?

x^8=-1

Then i applied

.close

[midnight plank]

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[desert siren]

Yes good application of .close

[midnight plank]

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[obsidian cliff]

may sound dumb but for parts c,d,e are we allowed to assume that the inverse exists (and thus f is bijective)

[runic hamlet]

its not the inverse

its the preimage

that always exists

[midnight plank]

@obsidian cliff Has your question been resolved?

[obsidian cliff]

oh

wait what

oh fuck me time to redo all those exercises

[obsidian cliff]

whats the difference

[runic hamlet]

well for starters the inverse doesnt have to exist....

the inverse is still a function

the preimage of a set is a set

[obsidian cliff]

right

ok

my bad

sorry i forgor that $f^{-1}$ does not necessarily mean the inverse

[grand pond]

lifelong dumbass

[obsidian cliff]

im just going to close the channel until i get

some more progress first

.close