help-42

[dull wagon]

what's your b here

[formal finch]

ohhhh cz 1 times 1 is 1

damnnnn

i meant 1^2 is jsu 1

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[glass vale]

Although not quite fully math (but taken from a discrete math textbook), I want to see if I am understanding the propositional logic properly.

My approach was forming the propositional statement:
The path on the left leads to the ruins if and only if you are a liar.
If the villager only tells the truth:

• T <-> F = F, will say no, implying it is the correct path
• F <-> F = T, will say yes, implying it is the wrong path
  If the villager lies:
• not(T <-> T) = F, will say no, implying it is the correct path
• not (T <-> F) = T, will say yes, implying it is the wrong path

To make this a question, would it simply work to say:
Would you say yes to the statement: "The path on the left leads to the ruins" if and only if you are a liar?

[calm coral]

@glass vale Has your question been resolved?

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@glass vale Has your question been resolved?

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@glass vale Has your question been resolved?

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[topaz yacht]

how do i prove q3?

as well as this page 😭

heloo?

[edgy jasper]

-0,4a(2a^(2)+3)(5-3a^(2)) help please

[calm coral]

@topaz yacht Has your question been resolved?

[calm coral]

@topaz yacht Has your question been resolved?

[harsh siren]

if you're trying to prove similar triangles, prove that two corresponding angles are the same

[topaz yacht]

okay

[calm coral]

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[shut hornet]

why does times 1/4 not get derivated out?

[brazen elbow]

$(kx)' = k(x')$

[potent lotus]

FungusDesu

[brazen elbow]

(k is a constant)

[shut hornet]

hmh wait so what if it was placed in side the parenthesis?

$(1/4*x^2-2ln etc.$

[potent lotus]

Katanapultti

[shut hornet]

oh wait in that way it's not the common multiplier tho

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[hoary acorn]

let <P(N), ⊆> be poset, does there exists A subset of P(N), s.t A doesnt have infimum?

Message #proofs-and-logic

[mortal orbit]

what do you think?

so A is an element of P(P(N)) for you right

[glass heart]

this might help
posts the solution

[mortal orbit]

in spoilers

the problem is you have to prove it

[glass heart]

well half of the problem is finding out whether its true or false in the first place

[mortal orbit]

kinda but I really had no other hint I could think of

like "try to find a formula for inf(A) if it does exist" also gives too much

[glass heart]

"play with some examples"

[calm coral]

@hoary acorn Has your question been resolved?

[hoary acorn]

guys

im sorry, for some reason i didnt get ping for your answers

@glass heart @mortal orbit

[glass heart]

well we didnt ping you

[hoary acorn]

Boyjee raised a concern to think about property

i thinked about generalizted intersection

which proved this :PPP

i proved that there is infinum for every A subset which is not empty

[glass heart]

👍

[hoary acorn]

This approach is correct? using the unary intersection of A?

[glass heart]

yes

[hoary acorn]

ty 😛

Have a blessed day

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[hoary acorn]

let A be subset of Z, non empty, proof there exists a in A s.t for all b in A, a<=b, it's kinda trivial, but is there any math tools to show that?

[vivid vine]

is it a drinker paradox?

[alpine stone]

Let A = Z

[upbeat venture]

except this isn't true, suppose A is the negative integers

are you given that A must be bounded?

[calm coral]

@hoary acorn Has your question been resolved?

[hoary acorn]

you are right

there is problm

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[remote mural]

anyone know how to solve this?

[vivid vine]

tf

is this

xDDDD

[tranquil pond]

that is just math nonsense

although actually it might be integrable

it's just written in an exceedingly obnoxious way and is clearlyt intended as a joke

[vivid vine]

-4/(3+e)

is the answer

dont ask

.

[tranquil pond]

yeah there's an awful loto f cancellation going on in there

[vivid vine]

you just substitute e_e = x

[tranquil pond]

the bounds are clearly just 0 to 1

[vivid vine]

and solve

almost everything simplifies

[tranquil pond]

i'm not motivated enough to try to sort out al the other tweakery

[vivid vine]

it took me 5 lines to solve xd

[remote mural]

yeah its not actually hard i think

3.1419526

its pi

.solved

[calm coral]

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[remote mural]

@vivid vine @tranquil pond

its pi

[tranquil pond]

yeah i believe you

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[flat pumice]

How many numbere are there below 200 which are not divisible by 3 or more prime numbers?

[crimson cliff]

what do you mean by more prime numbers?

[flat pumice]

Options are 168, 122,197,65

[crimson cliff]

isn’t 168 divisible by 3?

[dull wagon]

how is the question considering multiplicity

[vivid vine]

168 means the number of numbers that is the statement true for

[flat pumice]

The question is in Hindi, even i am not understanding it properly

Correct answer given is 168

[dull wagon]

e.g. for 8, is the factor of 2 only counted once (or three times)?

[vivid vine]

your translation is fine here

[crimson cliff]

Ok, i understood the question

[flat pumice]

Please explain it to me. I wm still having headache what the problem is

[dull wagon]

lets consider a few numbers starting with 15

what's the prime factorisation of that

[flat pumice]

3×5

[dull wagon]

yes, note that there are only 2 different primes there

(less than 3) which will count towards what you want

compare that to
210

[flat pumice]

210 is out of our sample

[dull wagon]

whoops , ok. 182

[flat pumice]

2 ×7×13

3 primes here

[dull wagon]

which satisfies divisiblity by 3 or more primes, so this doesn't count towards what you want

[flat pumice]

I see

So i will count numbers which doesn't have 3 prime numbers divisible

[dull wagon]

3 or more

[flat pumice]

So i will pick numbers which are only divisible by one prime numbers

Like 2^3, 3^3

[dull wagon]

one or two

[flat pumice]

Yes like 2×5

2×5^2

There will be many numbers

How to count it quickly?

[dull wagon]

from the expected answer, only seems to be 32

[flat pumice]

Did you see English version?

[dull wagon]

yes

oh wait

i was thinking about counting the complement

products involving 3 or more different primes

and there should be 32 of those

[white atlas]

distinct primes?

[flat pumice]

Yes i count it is 168 lol

[flat pumice]

I guess no

If we substrate these numbers we get 168 answer

@dull wagon

If I substrate 199-31=168

How to count these 31 numbers

Is there any pattern?

@crimson cliff

[dull wagon]

brute force

considering bounds can make things more efficient

[flat pumice]

I tried to count but too much difficult

[calm coral]

@flat pumice Has your question been resolved?

[calm coral]

@flat pumice Has your question been resolved?

[flat pumice]

.close

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[vale flame]

im confused on how this is integration

[vale flame]

isnt this just diffrentiation

[leaden thunder]

integration relates very closely to differentiation

,tex .FTC1

[potent lotus]

riemann

[vale flame]

?

[vale flame]

idk im just confused on how there so much working involved for integration

when its just diffrentiated

[leaden thunder]

yea integration usually takes more work

[vale flame]

ok ig i just needed reassurance thx

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[vale flame]

where the 2u coming from

oh did they also diffrentiate left sidde

[leaden thunder]

implicit differentiation

[vale flame]

implicit

yh ok

thx

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[arctic grove]

I need help solving C

[arctic grove]

<@&286206848099549185>

[unborn gazelle]

could you solve the 3rd part of b)? you underlined that too

[arctic grove]

I don't need b anymore just c

I have no idea how t solve c

[unborn gazelle]

anyways if you did, you just have to subtitute the mileage for x into that equation

[arctic grove]

so y=-0.263 times 52,500 +38.435?

y=-0.263x+38.435 is the equation so I plug in 52,500 for x?

that feels wrong...

[unborn gazelle]

almost, but read the caution part as well.

you have to convert it kind of

[arctic grove]

the answer I got is 538.435

So I put 538.435 in the box?

is that corrct?

[unborn gazelle]

so, what did you plug for x?

[arctic grove]

I plug in 52,500 given to me in C

[unborn gazelle]

look at the table for the format of the miles

[arctic grove]

ok

[unborn gazelle]

you just have to convert the 52500 miles, and it should be good then

[arctic grove]

convert the 52500 miles to what?

[unborn gazelle]

the table doensn't simply contain milage but mileage * 1000, so what you plug in for x should be in a same format

[arctic grove]

ok so 52500* 1000=52500000

y=-0.263x+38.435 replace x with 52500000?

[unborn gazelle]

well, how would you read the first entry of mileage out loud? (the 28) what is it?

[arctic grove]

i read it with a Mileage of 280000, and Price of $30000

[unborn gazelle]

yes

[arctic grove]

yay!

now what?

y=-0.263*52500000+38.435= −13807461.565

[unborn gazelle]

not sure how to help without outright telling the answer :\, ill try explaning the equation. y = -0263x+38.435 means -> price(in thousand dollars) = -0.263(deprecation rate) * x(in thousand miles) + 38.435 thousand dollars for a brand new car

[arctic grove]

oh shoot...I need to do more...

[arctic grove]

Can you tell me step by step plz?

Step 1:y=-0.263x+38.435 replace x with 52500000 right?

[unborn gazelle]

then x would mean 525000000000 miles

but you have to make it so the value you plug in for x mean 52.5 thousand miles.

[arctic grove]

.close

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[scarlet saddle]

help?

[scarlet saddle]

i did

,calc 16/9 / 9

[potent lotus]

Result:

0.19753086419753

[scarlet saddle]

,calc 16/81

[potent lotus]

Result:

0.19753086419753

[old falcon]

start with (-1,9)

can you write the equation that you get from this?

[scarlet saddle]

wdym?

[old falcon]

it implies an equation right

[scarlet saddle]

yea

ohh

i see

[old falcon]

h(x)=y=ab^x

[scarlet saddle]

9=a(b)^-1

[old falcon]

yea

so

$9=\frac ab$

[potent lotus]

jan Niku

[old falcon]

pretty useful!

[scarlet saddle]

how?

[old falcon]

it means we can write 9b=a

[scarlet saddle]

how

how are u getting that

[old falcon]

you get this?

[scarlet saddle]

ohh yea

ohh so u multipllied by b?

on both sides?

[old falcon]

yup

[scarlet saddle]

why?

[old falcon]

to get 9b=a

because this is very useful

it means if we have another equation we can replace all the a with b

[scarlet saddle]

huh?

f what

ohh i see

9*b=a

i see

lol

[old falcon]

so write the other equation implied here

you have one more point

[scarlet saddle]

16/9=a(b)^3

[old falcon]

except, dont use a

[scarlet saddle]

y

[old falcon]

instead of a

[scarlet saddle]

9b

[old falcon]

we can

ya

[scarlet saddle]

16/9=9b(b)^3

[old falcon]

and look

you have an equation of one variable

these can be solved!

[scarlet saddle]

so 9b^4=16/9

divide by 9?

[old falcon]

sure

[scarlet saddle]

,calc 16/9 / 9

[potent lotus]

Result:

0.19753086419753

[old falcon]

no

keep the equation

[scarlet saddle]

solve for b

[old falcon]

you solve for b

dont start crunching numbers just yet

[scarlet saddle]

how?

[old falcon]

$b^4 = \frac{16}{27}$

[potent lotus]

jan Niku

[scarlet saddle]

nvrm i got it

its +- 2/3

which one do i use?

• 2/3?

wait no -2/3

since its decaying

[old falcon]

yea

beat me to it

[scarlet saddle]

wait my teacher used 2/3

[old falcon]

err wait

[scarlet saddle]

+2/3

[old falcon]

no

its the base

take the positive

[scarlet saddle]

how?

[old falcon]

sorry thought it was the exponent

[old falcon]

(-a)^x is actually not super well defined

[scarlet saddle]

wdym

[old falcon]

only positive numbers to the x are nice

[scarlet saddle]

always use the positive?

[old falcon]

negative numbers to the x are an advanced topic

[old falcon]

for a base to the x, yes

[scarlet saddle]

@old falcon yk what i couldve done

?

[old falcon]

huh?

[scarlet saddle]

i couldve done this

16/9 / 9

which is 16/81

and since -1 to 3 is four spaces, right?

i could do square root 16/81

which is 2/3

so i know from each whole number there is 2/3 change

right?

so y=a(2/3)^x

[old falcon]

id have to think about that

but if it works for you!

[scarlet saddle]

does it work?

[old falcon]

have to think

at the gym

brain off

[scarlet saddle]

lol no pains no gains

[calm coral]

@scarlet saddle Has your question been resolved?

[calm coral]

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[inland sinew]

the first sequence im havin a bit of trouble with

[calm coral]

@inland sinew Has your question been resolved?

[upbeat venture]

what have you done so far?

[calm coral]

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[latent zinc]

I want to show that this algorithm sorts all arrays correctly

[latent zinc]

I want to use induction

Not sure what the function for this would look like tbh

The base case is simple enough, as an array of [1] is already sorted

[calm coral]

@latent zinc Has your question been resolved?

[calm coral]

@latent zinc Has your question been resolved?

[atomic palm]

the induction is baked into the statement

assume the left and right halves are sorted by the mergesort call

then the merge operation will guarantee that the entire array is sorted

[latent zinc]

how would you write it for a paper?

[atomic palm]

Mergesort(A,i,i+1) is the base case

Mergesort(A,i,j) calls Mergesort(A,i,m) and Mergesort(A,m+1,j)

by the inductive hypothesis, both Mergesort(A,i,m) and Mergesort(A,m+1,j) are sorted

then show that using the merge algorithm, the array on [i:j] is also sorted

[calm coral]

@latent zinc Has your question been resolved?

[latent zinc]

not sure how to proceed

[latent zinc]

because they are previous iterations?

[latent zinc]

basically since we know that Mergesort(A,i,m) and Mergesort(A,m+1,j) are sorted, we can know that Mergesort(A,i,m), which returns a A[i,j] also will be sorted

[calm coral]

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[brave sluice]

If an assignment is worth 6.25%, the midterm is 25%. A student gets 108% on the assignment and 82% on the midterm. If the bonus mark from the assignment gets shifted to the midterm, what is the students score out of the 25% for the midterm?

[dull wagon]

what bonus mark

oh, wait, 108%?

[brave sluice]

yea

the bonus is 8%

[dull wagon]

what ahve you tried

[brave sluice]

is the answer 22.5%

[dull wagon]

how are you getting that

[brave sluice]

is it right?

[dull wagon]

i haven't done the calculations yet

[brave sluice]

so for the assigment the bonus is 8% you agree?

[dull wagon]

show me how you're getting it and i'll be able to check whether its valid

8% of the assignment marks, yes

[brave sluice]

ok, and the mark for the midterm is 82% so 82% + 8% = 90% for the midterm

good so far?

[dull wagon]

no

[brave sluice]

whats wrong

[dull wagon]

can't add percentages like that due to the different weighting of marks for each task

[brave sluice]

how else would i do it

[dull wagon]

for an intuitive way, consider having the whole course out of 100 points
with the assignment worth 6.25 points
and the midterm worth 25 points

how many bonus points do they have from the assignment

[brave sluice]

8?

[dull wagon]

no

numerical amount of points, not just the percentage

[brave sluice]

not sure

[dull wagon]

8% of the assignment marks

100% would get you the full 6.25 points

of is akin to multiplication

what is 8% of 6.25
i.e. what is 8% * 6.25

[brave sluice]

0.5

[dull wagon]

yes

now similarly how many points do they currently have for the midpoint (out of 25) before shifting?

what is 82% of 25points

[brave sluice]

20.5

?

[dull wagon]

yes

[brave sluice]

21 out of 25

[dull wagon]

yes

[brave sluice]

in terms of the 82% of the midterm, how would that change?

oh wait nvm

.close

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[dusty crater]

what distribution can part b be represented by? answer for part a is 0.574

[calm coral]

@dusty crater Has your question been resolved?

[calm coral]

@dusty crater Has your question been resolved?

[dusty crater]

<@&286206848099549185>

I think it's binomial

[lament pollen]

Whicj question

[dusty crater]

but for some reason, I don't get the right answer

[dusty crater]

part b

[lament pollen]

I'm still learning binomial

🙂

So will be able to help in some days maybe

Your age?

[dusty crater]

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[remote mural]

i know you use simultaneous equations

i just dont know what you do with the 28

[celest stream]

u write down relative initial velocity,acc

and use kinematics equations

[remote mural]

i havent done that

[remote mural]

ye

9.8 and -9.8

[celest stream]

like s = ut+1/2at^2

[celest stream]

u havent done this?

[remote mural]

yh i have

[celest stream]

have or havent

[remote mural]

i thought it was different to suvat

[celest stream]

ok

just find relative acc and initia vell

and put it in that eqn

[remote mural]

initial is 0 for the first right?

[celest stream]

yea

[remote mural]

do i use 28 as s?

no right?

[celest stream]

yea

and then u find time

[remote mural]

oh

ohh cause then i find the time they meet

and then i can use any of them to work out the distance

thx

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[vestal kite]

i dont get the proof for iii)

[vestal kite]

how is <gog'(v), v> = <g'(v), g'(v)>?

[upper sparrow]

How do you have g* defined?

[vestal kite]

@upper sparrow we know that f = gog*

g is an endomorphism

[upper sparrow]

But how do you get g*? As in, if I give you g, how do you describe g*?

[vestal kite]

i dont know the english word is it self adjungated?

[upper sparrow]

Sure generally you have that (g(v), w) = (v, g*(w)) at least

[vestal kite]

(g(v), v) = (v, g*(v))which theorem is that or where can i find that info .o

so self adjungated would be <v,F(w)> = <F(v), w> thats it isnt it?

[upper sparrow]

Yea, adjoint is the F*, if you're self adjoint then F = F*

[vestal kite]

ah okay so <gog'(v), v> is logical but i dont get how he gets this <g'(v), g'(v)>

where does the g'(v) on the right side come from and where did g go on the left side i dont get it rn @upper sparrow

[upper sparrow]

Because as per before for the adjoint you have (g(u), w) = (u, g*(w)) for any g

So set u = g*(v), w = v

[vestal kite]

oh wow

i just didnt see it

thank u very much!

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[coral moss]

is the answer pi/4 ?

[brazen elbow]

yes

,w integrate sqrt(sinx)/(sqrt(sinx)+sqrt(cosx)) from 0 to pi/2

[potent lotus]

Results provided by WolframAlpha

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

Requested by fungusdesu

[modern peak]

yep

[blazing coyote]

yup

[coral moss]

yay

thanks

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[remote mural]

I want some elaboration on the formula for finding the surface area and volume of a $n$-dimensional sphere. Those being:
\begin{align*}
S_{n-1}(R) &= \4{2\pi^{\ff n2}}{\m\Gamma{\ff n2}} R^{n-1} \
V_n(R) &= \4{\pi^2}{\m\Gamma{\ff n2 +1}} R^n
\end{align*}
with $R$ is the radius

[viscid silo]

wheeeee

[remote mural]

uhhh

ok there u go

Anyways, our professor worked through the derivation in class but id appreciate someone walking with me thru it

[remote mural]

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[remote mural]

(ill open later)

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[cobalt cave]

any idea?

[calm coral]

@cobalt cave Has your question been resolved?

[cobalt cave]

<@&286206848099549185>

[mossy gulch]

Is this some kind of pattern?

[cobalt cave]

yep

[potent lotus]

Roman_Garland

[cobalt cave]

it's 3

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[civic vortex]

i think it's the amount of 1's in the pattern

[cobalt cave]

.reopen

[cobalt cave]

that actually make sense

very smart thank you!

[calm coral]

@civic vortex Has your question been resolved?

[civic vortex]

i dont have a question wym

lmao

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[onyx quiver]

how do i do this?

[hasty fiber]

what's your working definition of "continuous at a point"?

[onyx quiver]

if a limit from both sides go to it ?

[hasty fiber]

what's "it"

[onyx quiver]

the point

[hasty fiber]

you'll need to be much more specific

[onyx quiver]

oh ok what is your definition

[hasty fiber]

that would give away the answer

[hasty fiber]

what is "the point"?

[onyx quiver]

😿

[onyx quiver]

The point ur trying to prove is continuous

The x value i mean ?

[hasty fiber]

alright that's more specific

it's incorrect but it's specific

[onyx quiver]

lol I tried

[hasty fiber]

you're indeed looking at a limit, and specifically the limit at that point

the definition of continuous at x=3 is $\lim_{x\to3} f(x) = ???$

[potent lotus]

Steakanator

[onyx quiver]

Oh yes

[hasty fiber]

you need to identify what goes in place of the ???

[onyx quiver]

So the y value the function meets at x =3 ?

[hasty fiber]

f(3) would be a more ideal answer

[onyx quiver]

Oh ok

So the yvalue is 5 ?

[hasty fiber]

no

[onyx quiver]

Oh wth

[hasty fiber]

your original definition of f has (x-3) in the denominator

it also has (x-3) in the numerator

so f(3) = 0/0

[onyx quiver]

So indeterminate

So it doesnt exist ?

[hasty fiber]

indeterminate is a term used only in limits

it's undefined at x=3

[onyx quiver]

Ohh ok

So its discontinuous

[hasty fiber]

yes

[onyx quiver]

Wait im right about the hole being at (3,5)

[hasty fiber]

you are

the point x=3 is a removable discontinuity

[onyx quiver]

Yes

i see now

Tysm

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[onyx quiver]

.reopen

[calm coral]

✅

[onyx quiver]

Ok nvm imconfused

Wouldnt it be lim x>3 f(x) = 5

So if the limit equals 5 then what is all that bs I said about to prove a point is continuous you look at if the limit exists

[hasty fiber]

it's not enough for the limit to exist

it has to exist and equal the value of the function at that point

[onyx quiver]

So if a limit has a hole then its not a limit

[hasty fiber]

what

[onyx quiver]

Wat

Nvm im just gonna stop confusing myself

[hasty fiber]

you're confusing me too

what do you mean "if a limit has a hole"?

[onyx quiver]

Since thers a hole at (3,5) is that the reason why lim x>3 f(x) does not exist

[hasty fiber]

who said the limit doesn't exist?

[onyx quiver]

Oh ok it does

I thought the limit can't exist for it to be discontinuous

[hasty fiber]

this is precisely why I specified "removable discontinuity"

this type of discontinuity is present when the limit exists but the function is undefined

[onyx quiver]

Oh ok

The reason why F(3) doesnt exist is because x cant equal 3

Im a genius

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[unkempt kite]

Hiiii

[unkempt kite]

working on counting rn

im trying to understand when to determine it is best to use the Formula of distributing n identical objects in r distinct boxes (n + (r-1) / n) , or some other counting formula

came across this problem, i could draw it out mayb
but id rather do it mathematicaly

The formula for permutations of n distinct objects taken r at a time is this i think

[calm coral]

@unkempt kite Has your question been resolved?

[upbeat venture]

personally I'd find the total number of possible exercises, then subtract off the amount where you do exactly 2 in a row, then subtract off the amount where you do exactly 3 in a row

and that would be pretty easy I think

[unkempt kite]

that makes sense

so there isnt a distinct formula i should be using

total number of possib exericses is just 7! = 7 x 6 x 5 x 4 x3x2x1 = 5040

Exactly 2 in a row= 6! × 2!

then xactly 3 in a row= 6! × 3!

5040 - 1440 - 4320

wait no

somethings off none of the options have that sum

[split basin]

Well you see when you remove exactly two in a row you are also removing 3 in a row

Because out of those combinations there is also a combination where the three line up

[unkempt kite]

ah ok so i can just omitt the subtraction of the 3rd row

5040 - 1440 = 3600

[unkempt kite]

hmm doesnt seem like its an option

[split basin]

Don't you need to minus two together 3 times?

Also i think it minuses all three together 3 times but we need only once so we add 2 times

[unkempt kite]

sorry wym minuses 3 times and add 2 times

[upbeat venture]

that's why I said exactly lol, that usually implies not more or less than

so do the cases where there are only 2 in a row, not 3, then do the cases where there are 3 in a row

then subtract from the total number of possible exercise routines

[unkempt kite]

was that not this here?

[unkempt kite]

1440

[unkempt kite]

720

7! = 5040

[unkempt kite]

subtracting from the total number of exericse routines

which is 5040

wouldnt that be 5040 - (1440+720)?

[upbeat venture]

it is, I was reiterating lol

i just did what I said and one of the answers is what I got so

[unkempt kite]

i tried doing what you said here, but the options aren't one of them, perhaps i misdid smth?

[upbeat venture]

how did you get those numbers in the parentheses

[unkempt kite]

case for 2 in a row

case for 3 in a row

i think i mightve miscalculated tha

[upbeat venture]

okay sure, walk me through what you did for this

a helpful tip is to first assume everything is indistinguishable, then make it distinguishable

it's much less confusing that way

[unkempt kite]

i treated the leg exercises as a single unit

since they should stay together

[upbeat venture]

okay, how many possible ways did you get for exactly 2 units to be in a row

that's the first question

[unkempt kite]

i got 1440

by doing 6! x 2!

but that clearly wasnt right 😔

[upbeat venture]

okay I see what you were doing I think

but this actually includes scenarios where there are 3 in a row

and it also doesn't account for order

lemme give a picture to help visualize

these are the possible layouts for 2 exercises in a row, and the number on the right represents the number of possible orderings with EXACTLY 2 in a row, not including 3

do you see where I got 4 for the first row?

[unkempt kite]

is it because 3 of the exercises are legs

so 7 - 3

[upbeat venture]

each blue circle represents a leg exercise, in no particular order

so in the first row/layout, there are only 4 possible places to put the third exercise

otherwise there are 3 in a row

[unkempt kite]

ohhh ok

[upbeat venture]

I did the same thing for each row

hopefully you see where I'm going with this

if we focus on the first row, there are 4 places you can put leg exercises. after that, you need to make everything distinguishable, so multiply 4 by 3!*4!

you can do something similar for the other case

[unkempt kite]

so for the second row

id have 3 by 3!×4!

[upbeat venture]

yep

do you see where the 3!*4! comes from btw

[unkempt kite]

is it the arrangement with the remaining exercises after placing the leg ones

like the number of ways to arrange the exercises within the leftover slots after placing the leg exercises

[upbeat venture]

you might be right and I'm not understanding, but what this does is determine where the leg exercises can go, it doesn't determine which ones are where. so right now, all leg exercises are indistinguishable, and so are the leftover ones. so to make the leg exercises dist, we have 3! ways, and to make the others dist, we have 4! ways

[unkempt kite]

oh ok

we had some formulas for these types of qs

in our labs

like here for example

i was thinking of trying to to do something similar

for the problem now

here the hall must have 2 walls

rn, 2 leg exercises cant go back to back

so ( n choose n - k) formula is what i was thinking could work

[unkempt kite]

but then again the n and k value is what im trying to determine too, if i were to proceed with this

[upbeat venture]

this isn't the same situation tho since a non-leg exercise can be between 2 leg exercises

so you can't use that formula sadly

the argument I made is essentially that same argument actually

[upbeat venture]

just a little different bc of this

[unkempt kite]

ohhh

so we pretty much need to make these leg exercises distuingshable

3! ways, and 4! ways

[upbeat venture]

yep pretty much

[calm coral]

@unkempt kite Has your question been resolved?

[unkempt kite]

did you get 864

[calm coral]

@unkempt kite Has your question been resolved?

[unkempt kite]

oh

.close

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[unkempt kite]

!close

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[remote mural]

🫡 Relearning linear algebra after forgetting everything

[remote mural]

What equation must be satisfied for the augmented matrix to correspond to a consistent​ system? Select the correct choice below and fill in the answer box to complete your choice.

k+2g + h = ???

I have no idea how to figure this out

wait

i misread the matrix

Am i doing this right or naw

[kindred estuary]

No R3 + 2R2 doesn't equal that

[remote mural]

should've bee 2R_1

[kindred estuary]

Even with R1 that math is wrong

[remote mural]

[kindred estuary]

What's 2 * R1?

[remote mural]

21

backwards

[kindred estuary]

What

[remote mural]

bro idk

Thank you for your help. You've helped me realize i'm hopeless

[kindred estuary]

Then you should review elimination

https://www.youtube.com/watch?v=eDb6iugi6Uk&ab_channel=TheOrganicChemistryTutor

[remote mural]

I know what elimination is, i dont need to review

just addition and subtraction mistakes

TY TY

.close

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[kindred estuary]

No it wasn't

If you don't recall how to do 2 * R1, it's more than just addition and subtraction mistakes

[calm coral]

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[main musk]

What should I do here

[main musk]

The book factors the LHS like so but idk how they do that

isnt $iz\cdot z=|z|^2$?

[potent lotus]

The Great D

[main musk]

ahh nvm its $z*$

[potent lotus]

The Great D

[pallid halo]

z times z conjugate is |z|^2

i assume z^* means conjugate here

[main musk]

yes yes I see

makes sense yes

thanks

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[tulip basin]

Where can I go from here for this inductive proof?

[remote mural]

yeah everything you written so far its correct

[tulip basin]

Thanks. I always kinda get stuck at this point with these inequality proofs that have a summation

[eager citrus]

try factoring out a 1/2 out of the last k+1 terms in the summation

[outer wave]

I wouldn't directly write that the sum is smaller than 2. (I'll define a:=sum 2^(k+1) and b:=sum 2^k ), now you can calculate that 1/a + 1/b < (here you use the assumption) 2 + 1/b.
Then you simply calculate both sides with -1/b and you get what had to be shown

[calm coral]

@tulip basin Has your question been resolved?

[tulip basin]

I still dont understand. Are these sums right? And i don't know how you can directly use the assumption to say 1/a < 2

[outer wave]

The idea is, that if a < b,
then a + c < b + c
and if you subtract the c, you get a < b
It's so easy that people overthink it, but you are actually done,
so for you case sum(1/2^i) + sum(1/2^(k+1) < 2 + 1/sum(1/2^(k+1), if you subtract the last part for the right inequality, you get your assumption and that's the proof

[calm coral]

@tulip basin Has your question been resolved?

[calm coral]

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[remote mural]

I found the slope of the tangent line to be 3

which lines up

but what you do from there is what I don't get

[cursive coral]

Do you know the point-slope formula for a line?

[remote mural]

Y2-Y1/X2-X1?

or is it f(x+h)-f(x)/h?

[real briar]

Y - Y2 = m ( X - X2 )

[remote mural]

how do you derive this

not familiar with it

[real briar]

Sorry?

[remote mural]

how do you get this formula

[random adder]

$$y_1 = mx_1 + b$$
$$y_2 = mx_2 + b$$

[potent lotus]

casework

[random adder]

You subtract those 2

[remote mural]

like in a literal sense?

[real briar]

You differentiate the given equation to find the slope

Now do you know a point where the tangent goes through?

[random adder]

What do you mean?

[random adder]

Subtract one from the other

[remote mural]

how would the substraction process go

[random adder]

You get the formula

[remote mural]

ok I get y1-y2

[random adder]

$$(y_2) - (y_1) = (mx_2 + b) - (mx_1 + b)$$

[remote mural]

where does the b go

[potent lotus]

casework

[random adder]

B cancels out as you can see

[remote mural]

okok

perfect

but doesn't mx

aswell

oh wait no

[random adder]

No

[remote mural]

it doesn't

[random adder]

m is factored

[remote mural]

x2-x1

[real briar]

You take the m as a common factor

[remote mural]

$$y_2 - y_1 = mx_2 - mx_1 $$

[potent lotus]

wyldinwilliam

[remote mural]

oh i see

[real briar]

dy/dx = 2x - 3
So f'(3) = 6 - 3 = 3 = m

[remote mural]

$$y_2 - y_1 = m(x_2 - x_1 )$$

[potent lotus]

wyldinwilliam

[remote mural]

ah there we go

okay then, so we have 3 for m

[random adder]

Now just plug x_1 and y_1

[remote mural]

$$y_2 - 3 = 3(x_2 - 1)$$

[potent lotus]

wyldinwilliam

[real briar]

There you go

[remote mural]

but why x1 y1

is that because that is the initial

x and y

$$y_2 - 3 = 3x_2 - 3$$

[potent lotus]

wyldinwilliam

[remote mural]

wait mbv

actually no what

do the 3's cancel

$$y_2 = 3x_2 $$

[potent lotus]

wyldinwilliam

[random adder]

Yea its good

If the slope is correct

[cursive coral]

Hold on you plugged in x and y backwards

[random adder]

It doesnt matter

If you plug it for x_1 , y_1 or x_2 , y_2

[remote mural]

is the answer ^^

not sure how i'm off

[real briar]

No they used (1,3) instead of (3,1)

[cursive coral]

No i mean the point (1, 3) was plugged instead of (3, 1)

Yes that

[remote mural]

ah okay

$$y_2 - y_1 = m(x_2 - x_1 )$$

[random adder]

Oh i see

[potent lotus]

wyldinwilliam

[random adder]

Yea sorry didnt see that

[remote mural]

$$y_2 - 1 = 3(x_2 - 3 )$$

[potent lotus]

wyldinwilliam

[remote mural]

$$y_2 - 1 = 3x_2 - 9 $$

[potent lotus]

wyldinwilliam

[remote mural]

$$y_2 = 3x_2 - 8 $$

[potent lotus]

wyldinwilliam

[remote mural]

$$0 = 3x_2 -y_2 - 8 $$

[potent lotus]

wyldinwilliam

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[remote mural]

I'm so fucking loss

[remote mural]

How do i keep messing up?

[old falcon]

this process is really tedious and error prone

its normal to make a lot of mistakes

im not sure i understand the specific question here

[remote mural]

oooh

[remote mural]

I fixed the mistake

[remote mural]

Basically I just had to make sure 0 = 0 otherwise the system is inconsistent or smth like that

TY for help again

.close

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[old falcon]

https://tenor.com/view/funny-animals-thirsty-kitten-drink-up-ice-water-gif-10692620

[remote mural]

Homework assignment 1 out of 11 complete 🫡

[calm coral]

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[remote mural]

im not sure what im doing wrong unless the equations i used were wrong so ill attach those and my work

[remote mural]

and i know sin and cos are negative in quadrant 3, and tan is positive

the answers i uploaded also evaluate to the same value as what i get when directly plugging the value of cos theta into the formulas so i dont think theres an error in my work. idk what the issue is

[glacial nacelle]

This is correct

However, we are not finding them for theta, we are finding them for theta/2

We know that pi < theta < 3pi/2, so pi/2 < theta/2 < 3pi/4

this range is entirely in quadrant 2

so sin is positive, cosine is negative, and tan is negative

that should help

[remote mural]

hmm ok! i'll see what i can do with that, thank you.

great! i just changed the signs and it was able to accept the answers. I appreciate the help and wasn't even thinking about applying the division by 2 to the given domain

this discord server is honestly so much more helpful than trying to ask chatgpt and getting some made up explanation

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[supple marsh]

The great mathematician Countalot discovered the Incredible numbers: he does not know yet if they are finite or infinite, but he made the following conjecture:

If they are infaite, then at least one of them hes 8 distinct prime factors.

One of his students, Countagain, showed that Conntalot's conjecture is false.

Therefore he proved that:

A. if Incredible numbers are finite, then none of them has 8 distinct prime factors

B. if Incredible numbers are finite, then all of them have 8 distinct prime factors

C. Incredible numbers are infinite and all of them have 8 distinct prime factors

D. Incredible numbers are infinite and none of them has 8 distinct prime factors

E. Incredible numbers are infinite

[supple marsh]

I ll go with D

It seems like the exact opposite what the argument states

[trail ingot]

yea D is right

[supple marsh]

Are you sure

[trail ingot]

yes

negation of (P implies Q) is (P and not Q)

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[rocky flower]

hi, i was wondering if there is any way of factorising x^3+x+1 over complex numbers

[pearl skiff]

since x^3+x+1 has 3 root in C,
if you can find the 3 roots,e.g. a,b,c
then you can write it like
(x-a)(x-b)(x-c)

so it's more like a cubic question

[rocky flower]

wait, i think i found it using wolfram

thanks for the help!

.close

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[fierce compass]

median

[fierce compass]

intuitively speaking, why is the formula for the median = (n + 1) / 2 ?

[inner delta]

What's n here?

[fierce compass]

number of values in the dataset

[inner delta]

Oh that median

I thought the triangle one

[swift laurel]

that's not true for all data sets... where did you see this formula?

[fierce compass]

well it's true for datasets w/odd number of inputs

[fierce compass]

it's not that deep, just a highschool formula

i tend to sometimes forget it and would like to know the reason behind the +1 to the n in there

[swift laurel]

it's not true, though. the median of a data set is just the middle value when you sort the data points by size. if there are two middle points you take the mean of the two of them

[covert scroll]

Fr

.occupied

!occupied

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[fierce compass]

well u use that formula to find that position

._.

$\Large M_{odd} = \dfrac{ n + 1}{2}^{th}$

[potent lotus]

pl_ans_me

[fierce compass]

$\large M_{even} = \dfrac{\dfrac{ n + 1}{2}^{th} + \dfrac{n}{2}^{th}}{2}$

[potent lotus]

pl_ans_me

[fierce compass]

^th

[gray pebble]

ur overthinkingi t

[fiery sun]

I really want to ask the most simple quest but im not trying to get roasted. I cant for the life of me figure out how to do this simple mathematical question

[swift laurel]

!occupied

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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).

[fiery sun]

oops

[fierce compass]

hey, @swift laurel , any opinion would be nice. Don't hesitate, even if it's not too accurate or smth

[swift laurel]

there's a sort of disconnect between when we mean "halfway through a dataset", and "half of a number". take a 7-point set. The "halfway mark", if we start counting from 1, is really halfway between 1 and 7, but half of 7 is really halfway between 0 and 7

we can compensate for that by adding an extra 1/2, since the difference between those notions of "halfway" are only different by 1/2

[calm coral]

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[fringe hamlet]

$\sum_{n=1}^{\infty} \left( \frac{n}{n^2+1} - \sqrt{\frac{n^2}{n^4+1}} \right)$

[potent lotus]

alee

[fringe hamlet]

how can I understand if after a certain index "n" this difference becomes positive, I thought of making the limit of n-> +infinity of this thing, but I don't know if it's correct

[thick pond]

What's your goal with the series in the end?

Do you want to show first term > second term for sufficiently large n?

[tranquil pond]

if you can show that the individual terms are (past some point) positive and increasing then the series necessarily diverges

because if the sequence is positive and increasing then the limit of the sequence is not 0, and thus the related series divergees by the divergence test

[fringe hamlet]

see if the series converges or diverges

[thick pond]

test if the terms go to 0 or not

Start with that

[fringe hamlet]

ok

[thick pond]

lim(n to inf), etc.

[fringe hamlet]

lim n->+infinity a_n=0

is 0

[thick pond]

so if it is 0, then you'll need a diff test

[fringe hamlet]

but shouldn't I first see if the series is on positive terms?

[thick pond]

Well, if lim (n -> inf) != 0, then the series div.

Otherwise, we need other info

[fringe hamlet]

Can I see what the series is asymptotic to?

[thick pond]

I don't think it's really asymptotic to anything

[fringe hamlet]

-1/n^3

so it's a series with negative terms?

if -1/n^3 is right???

[thick pond]

What is -1 / n^3

[tranquil pond]

if yiou can squeeze it with 1/n^3 then you can show it converges because sum of 1/n^3 obviously converges

[fringe hamlet]

what the series is asymptotic to

-1/n^3

if it is asymptotic at -1/n^3 then the series converges

but the question was whether before doing these things, I had to and how to see if the series was on positive terms????

[tranquil pond]

just use the absolute value. if sum |a_n| converges, so does sum a_n

[fringe hamlet]

ok

can i send a photo?

[tranquil pond]

yes, help channels have no embed restrictions

[fringe hamlet]

I did both the asymptotic comparison and the limit in the same steps

Is correct ?

[tranquil pond]

i won't guarantee it but let me point out that if you can show that |a_n| < 1/n^2 for sufficiently large n, then that gives you convergence, and i think you have that

[fringe hamlet]

so what should I add in my exercise?

my question was: "how can I understand if after a certain index "n" this difference becomes positive, I thought of making the limit of n-> +infinity of this thing, but I don't know if it's correct"

[tranquil pond]

i don't think that is a legitimate test of whether a series converges

[fringe hamlet]

how can I do

[calm coral]

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[hushed tartan]

i drew the coordinate system at O (the blue lines) so that the x-axis is parallel to AC to have alternate interior angles, the question is find the coordinates of A,B and C the work is easy but i feel that i have something wrong idk

[hushed tartan]

let the origin be O, what i did is that i found the angles OAC and OCA using law of cosines then i used the fact that each of these angles are alternate interior angles with the angles on the x-axis respectively

so A(7.3cos(OAC),7.3sin(OAC)) and C(5.1cos(OCA),5.1sin(OCA))( i have angle and radius so i can use these to convert to rectangular coordinates as above)

is this work correct

or am i missing something

[calm coral]

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