help-26

[orchid geyser]

i dont really think i know what i am supposed to solve, if that makes sense

in the assignment

[stuck lintel]

you need to find for what s and t this is true

so just use vector operations

you'll get a system of two equations

which you can solve

[orchid geyser]

ohh okay, i got it now thanks!

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[fast drift]

need some help with sqrts, my teacher didnt explain it well and i need to learn how to write radical expressions as a quotient without sqrts in the denominator, as well as rationalizing some stuff

[empty sail]

Post the question you are stuck on and what you attempted

[fast drift]

alr

so this first oe

one

i got a little bit into it this is just simplifying radicals assuming x > o

sqrtx^121

so i did sqrtx^120x

= sqrtx^120 sqrtx

but im lost as to what i do next

[restive inlet]

consider why you split your radical and expressed it as
$$\sqrt{x^{120}}\sqrt{x}$$
then apply the appropriate radical/exponent laws and simplify further

[thorny flame]

ℝamonov

[fast drift]

so youd prime factor, right?

so itd be x^2^3 + 3 + 5

or really just x^2 + 2 + 2 + 3 + 5

[onyx shard]

can you tell me what is the sqrt of a^b ?

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[viral badge]

Anyone know how they derivate the function

It’s rule of chain but how

<@&286206848099549185>

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[celest glade]

what have you done so far

[neon iron]

tried to make some equations

[celest glade]

show us them

[neon iron]

P-2=3C-2

P-4=4C-4

[celest glade]

$3C-2$ reads as "(three times claire's age) two years ago"

[thorny flame]

illuminator3

[neon iron]

yep

so

then

[mild oasis]

Thats not what question says

[celest glade]

the text says "three times (claire's age two years ago)"

[neon iron]

2 years ago, P = 3C
4 years ago, P = 4C

[celest glade]

no

[mild oasis]

Theoretically yes but more like $P_{{2 years ago}}$ and $C_{{2 years ago}}$

[neon iron]

hmm

[celest glade]

.

[celest glade]

.

[celest glade]

.

[neon iron]

wtf

uh

[celest glade]

you have to put parenthesis

[mild oasis]

By what you wrote you just cancel from both sides and get
P = 3C
P = 4C
Only solution for this is P = 0 , C = 0

[neon iron]

true

thats why im am super confused

how do I write the equations?

[celest glade]

the first one should be written as

$$P-2=3\cdot (C-2)$$

[thorny flame]

illuminator3

[celest glade]

you have to figure out the second one on your own

[neon iron]

alright sweet

P - 4 = 4 (C-4)

?

[mild oasis]

Yes

Now figure out what the question asks

[celest glade]

no

pete was four times as old as claire

not the other way around

[neon iron]

c= 8

[mild oasis]

I mean that is what he wrote

[neon iron]

p = 20

[celest glade]

what he wrote is claire was four times as old as pete?

[mild oasis]

No

[celest glade]

wait

oh god my brain

my bad

[neon iron]

sweet

now how many years until 1:2

2:1

20 + x = (8 + x)/2

?

[mild oasis]

Almost

[neon iron]

4 more years

sweet thx

👏

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[mild oasis]

That is right answer just not right steps

[neon iron]

what steps

[mild oasis]

,w 20 + x = (8+x)/2

[thorny flame]

Results provided by WolframAlpha

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

Requested by Pluton#6301

[mild oasis]

Thats ur equation

[neon iron]

what steps

should it have been

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[mild oasis]

This is saying when will peter be twice younger than claire

[mild oasis]

Which is in -32 years

Even tho theoretically peter will still be older

[neon iron]

yes I understand

but how should I have writen the equation?

[mild oasis]

$$P + x = 2(C + x)$$

[thorny flame]

Pluton

[mild oasis]

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[neon iron]

this looks super easy but I have no idea how to do it

i-

2/3b = 3/4g

b+g=?

[celest glade]

you forgot an equation

[neon iron]

which one

[celest glade]

"an equal number of boys and girls passed the test"

[neon iron]

.

thats what this is

thats why I set it equal to each other

[celest glade]

my bad

[neon iron]

bruh

[celest glade]

I swear my brain is just not working properly right now lmao

[neon iron]

how should I work this problem?

[keen venture]

The number of boys who passed can't be a fraction. Can't have a half-boy pass

So the number of boys has to be a multiple of 3.

[neon iron]

yes

ok

and number of girls has to be a multible of 4

?

so multiple of 3 and 4

12

WOAH

so its 24

bruh

I forgot

about the multiples

thx

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[keen venture]

Very eager there. How many boys and girls is that?

Do they satisfy the equation?

[neon iron]

24

yesssir

[keen venture]

So how many boys?

[neon iron]

12

wait

[keen venture]

So 12 boys, 8 passed the test
And 12 girls, 9 passed the test

[neon iron]

17

oke

[viscid thistle]

3x4 = 12

no, 12 in total

you just multiply the denominator

[keen venture]

Make sure to check your answer

[viscid thistle]

oh wait ,

equal number of guys and girls

right s

so that is 12

nvm

[viscid thistle]

yeah

[keen venture]

Except no

[viscid thistle]

wdym

12 is divisible by 4 and 3

it’s the lcm of 3 and 4

or gcd

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[viscid thistle]

however you want to call it

[viscid thistle]

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[keen venture]

@neon iron feel free to open a new channel if you want a correct answer.

[neon iron]

was it not 17

.-.

[keen venture]

Oh wait haha okay I see what you did

[viscid thistle]

17

?

[keen venture]

Sorry I misunderstood. Good job you got it.

[neon iron]

thx

[viscid thistle]

12 each is 24

and ye, you’re right,

it’s 24

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[indigo haven]

for a hypothesis test how do we decide which type of tail test we do?

[indigo haven]

<@&286206848099549185>

[wide berry]

Depends on the situation

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[tawdry chasm]

Anyone know what this means?

The little formula at the bottom

The guide says it but it doesn't make much sense

Ah never mind, followed it and just fucked up my order the first time

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[lethal grove]

can someone explain why this isnt correct?

[sturdy oracle]

Oh boy

This is gonna be a moment

[lethal grove]

oh wait

[sturdy oracle]

Wait did you forget an n

[lethal grove]

no I think I just screwed up negative signs

[sturdy oracle]

Ah

[lethal grove]

this is what it looks in geometric form

[sturdy oracle]

LMAO

Yeah that happens

[lethal grove]

oh I pulled the 3 out of no where

[sturdy oracle]

Well

[vale jacinth]

Because it’s not -1 but (-1)^n

[sturdy oracle]

Hold on

[vale jacinth]

You don’t have ^n

[lethal grove]

i also cant pull out 2/3 right

[vale jacinth]

And it’s 1 above not -1

[sturdy oracle]

This isn't right I don't think

[lethal grove]

I turned the -1 into -3+2

[vale jacinth]

$\frac{1}{1-(-\frac{2(x-2)}{3})}$

[thorny flame]

Cogwheels of the mind

[vale jacinth]

$=\sum_{k=0}^{infinity}(-\frac{2(x-2)}{3})^{k}$

[thorny flame]

Cogwheels of the mind

[vale jacinth]

So it’s not -1 but (-1)^n

[lethal grove]

thats what I did

if I divide by -3 wouldnt I get -1 on the top

it should be -3-(-2x+2)

[vale jacinth]

It’s wrong

Your last =

[lethal grove]

it would be 3/-3-(-2x-2)

[vale jacinth]

Yeah

Which makes no sense

Because you want to expand it near 2 not -2

[lethal grove]

it is centered at 2

so dont I want (x-2)

[vale jacinth]

Yeah

But you write x+2

Wrote

[lethal grove]

this is correct tho right

[vale jacinth]

You simply replace x with x-2+2 at the beginning

[vale jacinth]

Correct but doesn’t make sense

[lethal grove]

how would you do it? I was just taught to do it this way

[vale jacinth]

I told you

[vale jacinth]

This

2x-1=2((x-2)+2)-1=2(x-2)+3

[lethal grove]

so then I get 1/1-[(-2(x-2))/3] right

[vale jacinth]

Yeah

[lethal grove]

how would you simplify it then?

pull (-2)^n * (x-2)^n/3^n?

[vale jacinth]

1/(1-t)=Σt^n

Yeah

[lethal grove]

can you explain why this doesnt make sense?

[vale jacinth]

Told you

You are not expanding it near -2

Why would x+2 help

[lethal grove]

oh wait because if I factor out a -2 I would get x+2

ok that makes sense, ty

[vale jacinth]

Np

[lethal grove]

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[covert cedar]

Can I get some help with the Ford-Fulkerson algorithm?

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[vernal vale]

So, I have a question that is to find the score pivot for a students t distribution

[vernal vale]

the whole problem:
During a spectacular thunderstorm, the distance from a given flagpole to 15 lightning strikes is measured. Statistics $\overline{X},s$ are computed, and the sample is found to be approximately normal. The unknown parameter in question is $\mu$.

[thorny flame]

jan Niku (Shuri for Honorable)

[vernal vale]

State the appropriate pivot with it's distribution, to find a score interval for unknown parameter

So clearly this is student t

can we just use that $s \to \sigma$ along $n$ or something?

[thorny flame]

jan Niku (Shuri for Honorable)

[vernal vale]

<@&286206848099549185>

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[strong birch]

i got a question

[strong birch]

let me say it

[prime hollow]

do u have any guesses or ideas so far?

[strong birch]

yes I understand it lol.

i actually saw a post of this

A it should be

[prime hollow]

a makes sense

[strong birch]

everyone knows this

actually, the post said that it was B

that was what the people on the post said

eh

[prime hollow]

I am not sure what in general you are asking. it sounds like u understand.

[strong birch]

yes but I am saying that the people in the forum said it was B

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[hazy skiff]

having troubles understanding how am I going to do this with the sum formula

[next helm]

what type of sum is this

arithmetic or geometric

[hazy skiff]

arithmetic i believe

[next helm]

good

what is the common ratio

[hazy skiff]

-5

@next helm

[next helm]

yes

what is the formula for sum of an arithmetic sequence $\sum_{i=0}^n a_i$ where $a_i = a_{i-1}+d$

[thorny flame]

Invictus

[hazy skiff]

that looks like recursive formula

[next helm]

im asking you what the formula is

im not giving a formula, just quantifying it

[hazy skiff]

would it be sum arithmetic sequence formula

[next helm]

yeah

the formula for the sum is $\frac{a_0+a_n}{2} \cdot n$

[thorny flame]

Invictus

[abstract wadi]

you're got a_0 already, now figure a_n.

[hazy skiff]

so ther is 2 step to solving this?

[abstract wadi]

You can make it one too.

Then again, it's alright if you do it in two.

[hazy skiff]

this is my first time doing it kinda confused

[abstract wadi]

How do you figure a_n?

[hazy skiff]

um........

[next helm]

so the question says that there are 22 terms

the first term is 6 and the common difference is -5

how do we get to term 22 from term 1?

we subtract 5 21 times

so the last term is 6 - 5 * 21 = -99

[hazy skiff]

ohhhhh

i got it now

[next helm]

great

[hazy skiff]

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[wraith ice]

Im trying to get a formula to add an increasing number to itself over steps. So 10, then 10+10, then 10+10+10 for any number of steps. I'm pretty sure this is possible I just can't figure out the formula

[wraith ice]

Also obviously not just for 10

[leaden tusk]

what have you tried

[wraith ice]

Google searching honestly? Math isn't my forte I mostly just need it for a program im working on

Everything I try to find though results in either adding increasing sequential numbers or excel stuff

[leaden tusk]

well... what is another way to write repeated addition?

a more compact way?

[wraith ice]

Multiplication right?

[leaden tusk]

yep

[wraith ice]

Am I stupid what did I miss I thought I thought about that

So like... for 10 steps using 25

How would I multiply that?

[leaden tusk]

as in, 25 is your multiple and you're doing (25+25+25) etc

?

10 times?

[wraith ice]

Yeah but the thing is it's increasing

[leaden tusk]

indeed it is

[wraith ice]

Oh im dumb

Okay yeah I was massively over thinking this

[leaden tusk]

happens to the best of us lol

[wraith ice]

Thank you for that

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[graceful elk]

Where did the 0.8 come from?

[graceful elk]

Nvm 1-0.20

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[twilit smelt]

Would like to have some guidance on this question

[twilit smelt]

alongside with this hint

<@&286206848099549185>

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[brazen burrow]

<@&286206848099549185>

[alpine mist]

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[quartz jay]

so you have the expression 3(5+75)/3 + 45x+20=10
and you want to find out what 'x' is right?

all good, what you need to do is simplify the expression down

[prime hollow]

No, u need to distribute

No need to factor in this case right now because u are simplifying some whole numbers

[alpine mist]

just to be clear, you have $\frac{3(5+75)}{3}+45x+20=10$?

[thorny flame]

Zybikron

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[grim rover]

Hi, i just wanted to double-check

[grim rover]

This proof uses additive notation : https://math.stackexchange.com/a/2182688

Why not use composition of functions as the operation in the group?

<@&286206848099549185>

[worthy storm]

in that particular example, homomorphisms don't have inverses in general

[grim rover]

oh wait

that might make sense

since |G| != |H|, there's no bijection --> inverse operation may not be well-defined?

[worthy storm]

there's also no identity if G and H are distinct, and G isn't embedded in H

even if G and H are the same, there are homomophisms that are not invertible, for example the trivial homomorphism that maps all elements of G to the identity element of H

[grim rover]

Ah ok ,that makes sense

[worthy storm]

on the other hand if S is any set then the collection of all bijections of S to itself forms a group under function composition

(the symmetric group on S)

[grim rover]

yup, that i've got down! Thanks!

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[grim rover]

Ok, I need help with linear Characters

[grim rover]

http://sporadic.stanford.edu/bump/group/gind2_2.html

Specifically this question

The only thing I know is that G is isomorphic to G*

<@&286206848099549185>

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[narrow torrent]

Do you understand why?

[silver valve]

no not really

[narrow torrent]

Or need I explain?

Ok.

[silver valve]

do i need to take extreme values of sinx and cosx?

[narrow torrent]

sin and cosine are bounded below by -1 and above by 1

[silver valve]

nvm ill let you explain

[silver valve]

yep

[narrow torrent]

Agreed?

[silver valve]

yes

[neon iron]

sin^2(x)cos^2(x) is bound in (0, 1/4)

[neon iron]

when sinx equals 1, cosx will be zero so their product will be zero

in that case

[narrow torrent]

Yes agreed

I see my error

[silver valve]

hold on yeah

i got (0, 1/4) as well

[neon iron]

use this indentity to generalise f(x) , sin^2(x)cos^2(x) = sin^2(2x)/4

[silver valve]

sin^2 x(1 - sin^2 x)
-(sin^2 x - 1/2)^2 - 1/4

[silver valve]

never knew this indentity

[neon iron]

sin2x = 2sinx*cosx, square this one and you will get that

[silver valve]

i didnt know this either lol

anyways i got $\frac{\sin^{2}{2x}}{2(2 - \sin^{2}{2x})} - 1$

[thorny flame]

kinglacto

[silver valve]

so i think our problem comes down to solving the range of $\frac{\sin^{2}{2x}}{2 - \sin^{2}{2x}}$

[thorny flame]

kinglacto

[silver valve]

can i substitute y = 2x? or ?

hmm no nvm

[fluid phoenix]

no

[fluid phoenix]

no

[silver valve]

welp

[silver valve]

is this correct atleast?

[neon iron]

no that's not correct f(x) =$\frac{3\sin^{2}{2x}-4}{4 - 2\sin^{2}{2x}}$

[thorny flame]

Anonymous67

[silver valve]

hmm ok ill redo the working

but, to not waste your time, how would i proceed after this?

[neon iron]

assume sin^2(2x) to be t and write an equation for t in terms of f(x), and it will be an inequality problem as 0<=t<=1

[silver valve]

ive solved such inequalities but completely forgotten- hold on -

no wait

im confused on how to solve the ineqaulity

so $\frac{3t - 4}{4 - 2t}$ where $t \in [0, 1]$

[thorny flame]

kinglacto

[neon iron]

yes then rearrange it

for ex., y = x+4 => x= y-4

so that you get an equation like t= ...

[silver valve]

ohh so y here is f(x)?

[neon iron]

yes.

[silver valve]

$t = \frac{4 + 4 × f(x)}{3 + 2 × f(x)}$

[neon iron]

this is wrong..

[thorny flame]

kinglacto

[neon iron]

yeah thats right, now apply 0<=t<=1

[silver valve]

hmm it has two f(x)

seems hard

[neon iron]

do you know how to solve these inequalities ?

[silver valve]

i am not sure I really understood this

[neon iron]

remember what we supposed t to be

[silver valve]

i got [-1, -1/2]

i think my working isnt the most efficient tho

how would you have solved it?

i first proved that 3 + 2f(x) is always positive

then cross multiplied across

to get $0 ≤ 4 + 4 × f(x) ≤ 3 + 2 × f(x)$

[thorny flame]

kinglacto

[silver valve]

and split it and solved

[silver valve]

anyway to avoid this?

[silver valve]

yes i re-did it and got this

[neon iron]

take 2 cases first would be t >=0, $\frac{4 + 4 × f(x)}{3 + 2 × f(x)}\geq{0}$

[thorny flame]

Anonymous67

[silver valve]

isnt it always greater than or equal to 0?

ohh wait this proves that the denominator should always always be +ve

[neon iron]

and a fraction $\frac{x}{y}$ is greater than or equal to 0 if both denominator and nominator are postive or if both are negative.

[thorny flame]

Anonymous67

[silver valve]

yeah and if they are both negative, the -1 factor cancels out?

[neon iron]

yes.

[silver valve]

and so it reduces to this?

[neon iron]

no, it will reduce to ${4 + 4 × f(x)}\geq0$ and ${3 + 2 × f(x)}\geq0$

[thorny flame]

Anonymous67

[silver valve]

so f(x) ≥ -1 from the first
and f(x) ≥ -1/2?

that doesnt sound right

ik for a fact that the answer is [-1, -1/2] since thats the TB answer

[neon iron]

yeah then take those values of f(x) which satisfy both equations.

[silver valve]

f(x) ≥ -1/2?

[neon iron]

yes it would have been if you calculated second part correctly. f(x)>= -2/3

[silver valve]

holy shit yeah -

so f(x) ≥ -1

now we consider the case where both are negative?

hmm wait no

so we got lower bound

[neon iron]

no, f(x)>=-2/3

[silver valve]

oh yeah its greater than -1

umm

desmos says [-1, -1/2] as well

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[silver valve]

what does this even mean?

[topaz sinew]

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[fallow vigil]

ig it means that this eq holds for all x values

[silver valve]

it actually doesnt, the answer is 0

but how do we arrive at that?

[fallow vigil]

prolly we need to equate the constant term n the coefficients of the terms with variables to 0

[silver valve]

hmm yeah that makes sense

[fallow vigil]

cuz if we assume this has a solution then this eq should hold for all x values

[silver valve]

and there exists no such common value

[fallow vigil]

yess

[silver valve]

got it thanks!

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[spiral briar]

Is this correct or incorrect

[neon iron]

@spiral briar correct

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[neon iron]

<@&286206848099549185>

[ruby acorn]

Read this

[vital relic]

close this since u have #help-6

.close

[neon iron]

ok

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[mossy nest]

how to calculate this ?

[clear kindle]

Can you see why [ \sum_{k = 0}^{\infty} \binom{n}{2k}\
= \sum_{k = 0}^{n} \binom{n-1}{k} ]?

[thorny flame]

Vigne

[clear kindle]

If so, this is a well known result, and so I'll leave it to you like this.

[mossy nest]

i cant see it

[clear kindle]

Imagine a row in Pascal's triangle.

You know that each term in a row is produced by summing the two terms above it in the previous row.

Now think about every second term in the n-th row, and the two terms in the (n-1)-th row that sum to it.

If you do this for every other term in the n-th row, you should be able to see that it covers all the terms in the (n-1)-th row.

A visualization might help.

Thus, the sum of every other term in the n-th row is equal to the sum of the (n-1)-th row.

[mossy nest]

thanks, i will try it

[clear kindle]

Note: I added that 0

[neon iron]

Or apply Pascal's recursive formula

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[mossy nest]

i will check it

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[wooden violet]

Pls help wtf do I do

[gentle valley]

Is there an equation for l

Slope of j is 6/11

@wooden violet

[wooden violet]

How

Why

[wooden violet]

No there is not

OMG

THWRE IS

HOLT SHIT

[gentle valley]

The slope of a line is y/x where (x, y) is any point on the line

[wooden violet]

THIS QUESTION IS APART OF A PREVIOJS ONE I DIDNT EVEN NOTICE

I see

The equation for L is 5+y-2x=0

[gentle valley]

You can solve j and l for x and y. These x and y values represent the point where the lines intersect

You will end up with 2 points through witch the line l passes

Then you just find the equation for l

[wooden violet]

Ok what?

[gentle valley]

Solve j and l for x and y

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[mortal wolf]

hello

[mortal wolf]

I believe 15 is 1

though this more specullaton

and i think it will be the same for 16

but if someone could explain why or why not

[solid wing]

If Z = {E} I think they are d-separated

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@mortal wolf Has your question been resolved?

[mortal wolf]

I'm gonna be honest

I'm not sure what Z i

is

for 16

A is blocked

so that blocks one

and then C to be then blocked at E

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[upbeat yoke]

A 30 item survey gives the following points for responses A, B, C, and D: A = 5, B = 3, C = -2, and D = -4. A person’s score is found by totaling the points for all responses. Jason gave 8 A responses, 6 B responses, 12 C responses and 4 D responses. Find Jason’s score for the survey

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[upbeat yoke]

<@&286206848099549185>

[pale breach]

(8 x 5) + (6 x 3) + (12 x -2) + (4 x -4) = 18

[upbeat yoke]

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[agile tide]

how do i describe the transformation of these?

[analog herald]

vertical stretch since the coefficient is greater than one

[agile tide]

so i can say when there a 2 before x it will stretch inwards compared to y=x^2?

[analog herald]

ye

[agile tide]

thank uu so muchh

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[graceful elk]

Could I get some help with b?

[graceful elk]

Does complement fall under this?

[gilded anchor]

Complement of one, not both if that's what you're asking

[graceful elk]

Yup

so basically

1-0.6?

Since the quiz is the complement

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[frosty nacelle]

Why is it 6x^2 instead of just 6x

[empty sail]

Square foot

[frosty nacelle]

Huh

If we draw a diagram it’s a square cube right?

And there’s only one too and one bottom

And 4 sides

[empty sail]

Square foot is area

[frosty nacelle]

Oh

[empty sail]

Don't ask in an occupied channel

[plain acorn]

oh mb

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[vernal bluff]

hey, im having a bit of trouble with this question. I'm not sure exactly how to approach it- like im not even sure what the final answer should look lik

i was originally thinking for pt1 it was either 1/100 or 1/i given that i is 100-(number of drinks so far)

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[vernal bluff]

<@&286206848099549185> pls

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[grave cloak]

so i dont know how to solve this problem; but theres a similar question with like some dude marrying brides

basically the question is like "if there are 100 girls lined up and the person needs to select like the "best", but after he passes one he can never go back, where is the optimal place to stop/optimal girl to date or smth like that"

the answer to the problem was place #(100/e)

idk that might give some insight @vernal bluff, this might have smth to do with e (when u talk about part b of this problem)

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[lusty ferry]

Need help with a convergence problem. Not sure where to start other than comparison test

[lusty ferry]

Believe I need to isolate highest terms?

[crude anchor]

do you know if it converges or not?

[lusty ferry]

No, trying to figure that out

[crude anchor]

it looks kinda like a harmonic series based on its highest terms so i assume it diverges

but i could be wrong, thats justa guess

[lusty ferry]

How do I go about?

[crude anchor]

i think comparison test is a good approach

[lusty ferry]

It looks like it fits but Im not sure how to start

I think 5n^2 / 7n^3 but doesn't seem correct

[crude anchor]

lemme think about it for a bit

[lusty ferry]

5 / 7n because the n's cancel out?

[crude anchor]

ok i got it

it def diverges

and you can show it throught he limit comparison test

with the harmonic sequence $1/n$ as your test sequence

[thorny flame]

llspacebarll

[crude anchor]

you compute $\lim_{n\to\infty}\frac{5n^3+n^2}{7n^3+20}=\lim_{n\to\infty}\frac{5+\frac{1}{n}}{7+\frac{20}{n^3}}=\frac{5}{7}$

[thorny flame]

llspacebarll

[crude anchor]

and since 5/7 is positive finite the LCT tells us that our series has the same convergence as the harmonic series

which is divergent

[lusty ferry]

Your powers are off

[crude anchor]

nope

[lusty ferry]

5n^2

?

[crude anchor]

i divided by 1/n

the harmonic series

this is what i used

[lusty ferry]

Why is 5n^2 changed to 5n^3 in the beginning of the lim?

oh!

🥴

[crude anchor]

yeah sorry, i made it a bit confusing

skipped a step

[lusty ferry]

Which step?

The division by 1/n?

[crude anchor]

yeah

[lusty ferry]

?

[crude anchor]

yeah

[lusty ferry]

Thanks

Alternating series?

Or not, because everything is positive

Ill start a new inquiry.

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[shrewd anvil]

Can someone help me get started on how to evaluate this limit?

[sweet shard]

surely you can simplify it a little bit

[shrewd anvil]

I can? Sorry, I'm very unfamiliar with factorials 😓

[wide berry]

Write n! = n(n-1)(n-2)...1, and similarly for (2n)!. Then simplify

[shrewd anvil]

Does the square on n! not make it so I can't cancel anything?

[mellow venture]

$\frac{((n)(n-1)(n-2)....(1))^2}{((2n)(2n-2)(2n-4)*...(2))}$

[thorny flame]

Ugu'yaränikeri'u

[shrewd anvil]

I don't see any simplification possible there...Am I dumb?

[gentle valley]

Try dividing the denominator by 2

[shrewd anvil]

OH

[mellow venture]

isn't answer infinity

(n!)^2 outgrows (2n)!

[shrewd anvil]

It seems like it, since I've simplified it to 2n! now

[mellow venture]

,w (10!)^2 and 20!

[thorny flame]

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[mellow venture]

weirddd

by that logic answer would be 0 I think

[gentle valley]

[shrewd anvil]

Did I simplify wrong? Yo, I suck at this

[mellow venture]

u will get (1/2)*(n!)

if u multiply by (1/2) to nullify the multiplication by 2

,w limit of (n!)^2/(2n)! n to infinity

[thorny flame]

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[mellow venture]

☠️

,w limit of (1/2)*(n!) n to infinity

[thorny flame]

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[shrewd anvil]

Oh derp

I see my mistake

I multiplied by 2/2

And then didn't cancel the 2 lmfao

Ty ty

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[coarse jasper]

Hell

[coarse jasper]

Hello*

I had a question about this

I'm here to confirm my answer

So first you multiply

I think this is how

(I'm putting this in a calculator so I don't have to write it out)

That would equal

Then u cancel the 2

and get 5/17 - 3/17i

Correct?

[neon iron]

Hm

[gentle valley]

Correct

[neon iron]

wait lemme check

ah someone already did

[coarse jasper]

Ok

so it should be amounted as

Right?

[neon iron]

yeah

[coarse jasper]

Alr

Thanks

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[prime hollow]

Well, what is a maximum and a minimum?

[topaz sinew]

@dusty orbit Has your question been resolved?

[dusty orbit]

max is the higher point marked and min is the lower

[prime hollow]

So, on the graph where is the highest and lowest point?

[dusty orbit]

(-2,2) high and (1,-4) low

[prime hollow]

Yes

[dusty orbit]

yeah?

i do not believe i can put that on both though.

<@&286206848099549185>

[prime hollow]

What do u mean?

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[floral solar]

It’s only asking for minimum

[dusty orbit]

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[vernal vale]

I have a silly question

[vernal vale]

Find the first 3 terms in the maclaurin expansion of $\frac{\cos z }{ 1-z}$

[thorny flame]

jan Niku (Shuri for Honorable)

[vernal vale]

so i have an incredibly strong feeling this is wrong but

cant i just multiply the two series

as long as i restrict the domain

like it should just be

$\sum z^n \sum \frac{ (-1)^n z^{2n} }{ (2n) ! }$

[thorny flame]

jan Niku (Shuri for Honorable)

[vernal vale]

as long as i restrict |z|<1

this gives me an incorrect first 3 terms, i think

[solemn wigeon]

well you have to make sure you multiply the polynomials correctly

for example

$(x + x^2 + x^3)(x + 2x^2 + 3x^3)$

[thorny flame]

Brontochad (CV for honorable)

[solemn wigeon]

would be

like the coefficient of x^3 would have multiple combinations

e.g. 2x^2 and x

x and x^2

[vernal vale]

i dont understand

are you saying that the product of series wont work

[solemn wigeon]

i mean

why wouldnt it

finnite number (convergent series) times another cnovergent series would just be convergent

[vernal vale]

im a crazy or does it click forward in powers of 3n on z

im pretty sure the first 3 terms should have a z in it

[solemn wigeon]

this is what i was saying about the polynomial multiply thing

you have to write (part) of the series out

and then make sure that you are doing all the combinations to get all the coefficients

[vernal vale]

but we will certainly get a power of 1

in that method

where the series will not produce this

[solemn wigeon]

wdym

[vernal vale]

👀

the expansion should have a z^1 term

the product of the series does not have a z^1 term

[solemn wigeon]

$(1 + z + z^2 + \dots)(1 - \frac{z^2}{2!} + \frac{z^4}{4!} + \dots)$

[vernal vale]

sure

[thorny flame]

Brontochad (CV for honorable)

[vernal vale]

this has a z term

and a z^2 term

[solemn wigeon]

well you get

1 + z + z^2 - z^2/2!

[vernal vale]

okay

now compare the product of series

different

[solemn wigeon]

wait

did you think you could just remove the second series symbol

(if so, you cant)

[vernal vale]

i dont understand

[solemn wigeon]

$\sum a_n \sum b_n \neq \sum a_nb_n$

[thorny flame]

Brontochad (CV for honorable)

[solemn wigeon]

cuz that's not how series works

[vernal vale]

oh, yea i didnt think was true but then my professor said something like this in class

i was like this makes no sense

[solemn wigeon]

yeah hold on

i have a book about this

[vernal vale]

no this makes sense

im glad i was right

[solemn wigeon]

therse a bunch of theorem

[vernal vale]

this is why you need to multiply terms

[solemn wigeon]

yes

cuz polynomail

same reason why (a+b)(a+b) is not a^2 + b^2

[vernal vale]

kk

thanks

.close

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[solemn wigeon]

youre welcome

[vernal vale]

oh

🙇

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[agile mulch]

I know that torque = force * radius
so im guessing the force for the whole gear is 16.33 because 50Ncm / 3
then do i divide by each tooth?????

[agile mulch]

<@&286206848099549185>

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[undone fog]

Hey, I have a small question; if two balls are thrown, how do I find the minimum speed at which the first ball must be thrown so that it collides with the second ball before hitting the ground?

Ball A: Dropped from 10m
Ball B: Thrown upwards from 2m

What I have is that acceleration for both is $-g = -9.81 ms^{-2}$, and thus getting the integral of ball A and B for velocity, we get

$\overrightarrow{v_{A}} = -gt + V_{0} = -gt$ because $V_{0} = 0$ for A
$\overrightarrow{v_{B}} = -gt + V_{0}$, and we don't know $V_{0}$ for B.

Thus, for displacement,

$\overrightarrow{x_{A}} = -\frac{1}{2}gt^{2} + 10$ and
$\overrightarrow{x_{B}} = -\frac{1}{2}gt^{2} + V_{0}t + 2$

Setting the displacements equal to each other, I get $V_{0_{B}} = \frac{8}{t}$, but I have no clue where to go from there.

[thorny flame]

beanbeanjuice

[undone fog]

Do I do; $v = -gt + V_{0} = -gt + \frac{8}{t}$?

[thorny flame]

beanbeanjuice

[undone fog]

however, it's difficult, because I can't solve that because I don't have time

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@undone fog Has your question been resolved?

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[frosty nacelle]

i dont understand how its A = (w+4)(h+2)

ik that A = hw

but

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@frosty nacelle Has your question been resolved?

[supple badge]

You are working out the width of the garden and the borders

Which includes the green bit

On both the left and right side of the garden, there is a 2m border

So 4m additional to w in the width = (w+4)

Height has 1m up and down, (h+2)

Area = bh = (w+4)(h+2)

[frosty nacelle]

wouldnt for width itd just be 2 and for h itd be 4?

[supple badge]

No, you have to account for each side

[frosty nacelle]

uhj

hold on

wouldnt width be

the top and the bottom

and h would be left and right side?

[supple badge]

No

[frosty nacelle]

its labled as that in the picture

[supple badge]

Starting from the left of the image, you go 2 meters across, then w metres, then another 2

So the width is w+4

[frosty nacelle]

OHHHH

I see it now

omg

i was confused

ahh

alright

thank toy

[frosty nacelle]

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[ancient ivy]

Hello
When setting up the integral for the surface area of a revolution your variables must be in terms of the axis of rotation
For example, to revolve around the x-axis you must be in terms of x.
However for a specific homework problem I am asked to setup the integral for both the x and the y axis. Yet for the box where you would enter the equation for the x-axis, there is a dy at the end, implying that it should be in terms of y. Furthermore, solving for y and getting the equation in terms of x seems impossible with the equation given.
Perhaps I am just misunderstanding something

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@ancient ivy Has your question been resolved?

[ancient ivy]

<@&286206848099549185>

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@ancient ivy Has your question been resolved?

[hollow hedge]

It's not really true that when you are doing a revolution around an axis, the integral that evaluates the volume/surface of the solid created has to be in terms of that axis

for example, when evaluating the volume of a curve f(x) from x=0 to say, x=4 around the y-axis, the integral does not have to be in terms of y

it can be in terms of x too, and this integral can be gained through what is know as the shell's method

the shell's method exists because of those naughty curves that cant be isolated in terms of y

and similarly for surface of solids

but a bit easier to remember

im not sure if you are familiar with these integrals, but the surface area of a solid when revolving a curve around the x-axis can be evaluate as the integral of 2pi * yds

and ds is different depending on what is given

ds=sqrt(1+y')dx if y is given in terms of x, the bounds of integration is also in terms of x of course

and ds=sqrt(1+x')dy if x is given in terms of y, same is said about the bounds of integration

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[normal marsh]

I need help with 62

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@normal marsh Has your question been resolved?

[normal marsh]

<@&286206848099549185>

[floral rampart]

what'd you try to do?

[normal marsh]

I wrote out the general form of a polynomial and tried substituting the conjugate to no avail

[floral rampart]

so you used binomial theorem etc.

[normal marsh]

no just the general form of a polynomial

I have no idea how to approach this

[floral rampart]

so you have $$a_0 + a_1\overline{z_1} + ... + a_n\overline{z_n} = \overline{a_0 + a_1z_1 + ... + a_nz^n}$$

[thorny flame]

ohNoiAmHere

[normal marsh]

okay

how do we prove it?

[floral rampart]

im thinking gimme a sec

so far what ive tried is mimicking the proof of the conjugate root thm

ah i see

$$\overline{z+w} = \overline{z} + \overline{w}$$

[thorny flame]

ohNoiAmHere

[floral rampart]

by definition of complex conjugation

you can just apply that n times

also, proving this isnt that hard

let z = a + bi, w = c + di -> conjugate(z + w) = (a+c) - (b+d)i, conjugate(z) + conjugate(w) = a - bi + c - di = (a+c) - (b+d)i

[normal marsh]

aight

oh I get it now

thanks

.close

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[oak perch]

Help for ai) when I tried to solve it normally I got a different answer when I put it in the calc cal someone explain why?

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[carmine junco]

A small block of mass m slides down a rough plane and climb at the angle of 45° to the horizontal. If the block starts from rest and has velocity v after traveling a distance x down the plane, find the exact expression for the work done against friction in terms of m, g, x, and v.

[carmine junco]

I've worked this much

But im a bit confused about the question, if it just asks for the work done against friction, isn't it just 5✓2 N?

Where does the m, g, x, and v come to play

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[carmine junco]

.close

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