help-26

[surreal mural]

can you construct a vector v such that $\langle x, v\rangle = |x_1| + ... + |x_d|$?

[patent trail]

on the left hand side you get |2| + |-3| + |5|

and on the right you get what you expect

can you construct a similar vector as my sign vector? for C

like @surreal mural said

[thorny flame]

rbit

[grand blade]

would v = (1, 1, ..., 1)?

[surreal mural]

not quite

that would just sum up all the elements, but not their absolute values

[bold urchin]

i’m too used to the real version of this kind of problems, what is |x_i|? like if x_i = x_ir + ix_ic then is |x_i| = x_ir^2+x_ic^2 or |x_i|=|x_ir|+|x_ic|?

[surreal mural]

$x_1 \cdot v_1 = |x_1|$

[thorny flame]

rbit

[patent trail]

no

you want a quantity such that when multiplied by x_i you get |x_i|

[grand blade]

is this to do with the sign function cos im not sure how that works here

[patent trail]

so |x_i|/x_i

[patent trail]

there’s no sign function in C

[surreal mural]

forget about sign functions, its pretty straightforward

[patent trail]

it was just an example for R

[patent trail]

this is x_i*/|x_i|

equivalently

x_i x_i*/|x_i| = |x_i|

as we want

[surreal mural]

you think so?

[patent trail]

yes

[surreal mural]

no

[patent trail]

x_i x_i* = |x_i|^2

x_i* being the conjugate

[surreal mural]

oh is that a conjugate?

[patent trail]

yes

[grand blade]

ok so v = x_i*/|x_i|

?

[patent trail]

well more or less

[surreal mural]

just be careful with zero

[patent trail]

yeah

[grand blade]

right so that gives me another vector to use C-S but how does that go into the inequality with sqrrt(d)???

[patent trail]

v = (x_1*/|x_1|, …, x_d*/|x_d|)

[surreal mural]

whats the norm of v?

[patent trail]

well try it

[grand blade]

hmmm ok

[patent trail]

notice that every coordinate of v has amplitude between 0 and 1

what happens when you sum d numbers between 0 and 1

it’s less than or equal to d

but you need to take care of zero coordinates too as @surreal mural said

[surreal mural]

I think this will actually equal d, because we need to take the absolute first

[patent trail]

yeah indeed

[topaz sinew]

@grand blade Has your question been resolved?

[grand blade]

is this it???

i mean i get the x_i cant rly =0 so idk what to do about that

@patent trail @surreal mural ???

[patent trail]

yes

except when a x_i is 0

but then it’s not very hard to fix the proof

[grand blade]

right like do i just mention that or am i meant to do something else

[patent trail]

just let the corresponding coordinate of v be 1

[surreal mural]

also need to take conjugates

[patent trail]

indeed

[grand blade]

oh so it could be -1 instead or 1

of*

[patent trail]

sure

for a 0 coordinate of x

[surreal mural]

and then yeah the part where you evaluate <x, v> could be more elaborate but idk

[patent trail]

but you need to use the conjugates in v

[grand blade]

so instead of saying v_2 = sqrt(d) its less than or equal to sqrt(d)

to deal with conjugates?

[grand blade]

ok

[surreal mural]

oh no that part is fine already

$\frac{x_i}{\abs{x_i}} = 1$

[thorny flame]

rbit

[surreal mural]

this one though is problematic

[grand blade]

right cos ig if x_i is negative then its -1 right

[surreal mural]

$\abs{\frac{x_i}{\abs{x_i}}} = 1$

i think this is what you actually meant

[thorny flame]

rbit

[grand blade]

oh yeh actually since im squaring them it wont matter right?

[surreal mural]

its complex numbers

[grand blade]

ummm so how does that affect what i have cos i havent done complex analysis yet so im pre dumb when it comes to complex numbers

[surreal mural]

shouldn't really affect what you already have, just put the absolute value bars and it will be fine (problem is with numbers like i, where i² = -1)

$\abs{\frac{\bar{x_i}}{\abs{x_i}}} = 1$

[thorny flame]

rbit

[surreal mural]

forgot about the conjugate

[grand blade]

oh ok then thanks

ummm and btw for the next inequality am i doing the same approach where i wanna create another vector to use C-S ???

[surreal mural]

oh no in this case I'd go with a similar argument to the first inequality

[grand blade]

did i do this right then cos idk how im using this for the last inequality

[surreal mural]

hmm first of all not sure what is meant by a monotone increasing series in this context

[grand blade]

like as in it is only increasing cos ur adding non-negative numbers

maybe not worded correctly but i think the idea is there tho

[surreal mural]

I'd start by choosing some k, such that $|x_k| = \max_{1 \leq i \leq d} |x_i|$

[thorny flame]

rbit

[surreal mural]

but I think from there your idea should be right

but this way its a bit easier to see

[grand blade]

yeh ok that makes sense

[surreal mural]

$|x_k|^2 \leq \sum_{i=1}^d |x_i|^2$

[thorny flame]

rbit

[grand blade]

yep thanks that does seem alot better

but anyways how could this method of working help with the 4 inequality

[surreal mural]

you can apply the same idea, choose a k and then just write out the inequality after squaring both sides and using a slightly different argument you can show that this also holds

[topaz sinew]

@grand blade Has your question been resolved?

[grand blade]

@surreal mural i think this is correct and if so thanks alot for ur help

[surreal mural]

not sure if I can even follow this but I think you overcomplicate it a little

the most important step is to see that $d \cdot |x_k|^2 = \sum_{i=1}^d |x_k|^2$ which I think you did see? And from there showing that $\sum_{i=1}^d |k_i|^2 \leq \sum_{i=1}^d |x_k|^2$ just requires you to think about what it actually means for $|x_k|$ to be the maximum

[thorny flame]

rbit

[grand blade]

well ig that unless all the vectors have the same magnitude it be less than since ur adding the biggest magnitude d times

but doesnt my working still work?

[surreal mural]

can't really follow it, but if you don't make use of this maximum property somehow then probably not

oh hold on

yeah i think you did everything correctly

[grand blade]

well i guess i would go like x_1 + x_2 + ... + x_d <= x_k+x_k +...+x_k???

but if u think its good then imma leave it like that

[surreal mural]

only thing I would be careful about is when saying $|x_i| \leq \max |x_i|$

[thorny flame]

rbit

[surreal mural]

just notation wise

[grand blade]

yeh ok but rest is good right?

[surreal mural]

yeah

[grand blade]

ok thanks alot btw can i ask what resources did u use for linear algebra cos my lecture notes r rly superficial and skip alot of steps

[surreal mural]

just be careful with using the index i for multiple purposes at the same time

[topaz sinew]

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[surreal mural]

was reading a book but i think its not available in english

[grand blade]

rip 😦

ok thats fine thanks again tho i spent alot of time 😅

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@neon iron Has your question been resolved?

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[raven sparrow]

You can use monotone convergence here.
You would have to show that the sequence is (1) bounded (2) monotone.

Once you know it converges, then you can find the value by taking the limit on both sides of the recursive formula they give you and solving for the limit L.

[strong sable]

OMG Sorry, I sent wrong question here

My question should be

[raven sparrow]

Oh ok hahaha that's fine

So there's some mean value theorem type thing going on in here for sure

[strong sable]

yea but how🥲

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@strong sable Has your question been resolved?

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[chrome pulsar]

How equation for D' formed?
x' = x + x0, same for y, it is just translation

[chrome pulsar]

2nd order line

[topaz sinew]

@chrome pulsar Has your question been resolved?

[chrome pulsar]

what is that

[topaz sinew]

@chrome pulsar Has your question been resolved?

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[crisp glacier]

help

[versed elm]

Pls define a, b, c, d, and s

[crisp glacier]

sides of the cyclic quadrilateral opposite to the vertex

<@&286206848099549185>

<@&286206848099549185> ?

[sour bloom]

(1) consider area and Ptolemy

[crisp glacier]

okay

but

idk how to get sin beta

[sour bloom]

what have you tried so far

[crisp glacier]

well backtracking

sinA is BD/2R by sine rule

BD can be found in terms of sides

[sour bloom]

did you consider area of the cyclic quadrilateral?

[crisp glacier]

no

(s-a)(s-b)...

[sour bloom]

try finding ways to compute this area

(2) not sure what the tilde is

[crisp glacier]

i think area could work

[sour bloom]

(3) I'm thinking you can relate tan^2(beta/2) to some simpler function by using trigonometric identities

[crisp glacier]

yeah

tan^2(a/2)= (1-cos(a))/(1+cos(a))

[crisp glacier]

i think the wave mean this or that

as in a^2+c^2 or b^2+d^2

and thx i got the first one

[topaz sinew]

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[long fog]

Need help with the 2nd part

[neon iron]

what have you tried

[long fog]

cauchy's condensation criterion reduces it to proving sigma 2^(n/2^n)-1 converges

[snow nimbus]

The term in the sum roughly behaves like 1/√n. So which criteria would be natural to use here?

[cursive patrol]

rewriting n^(1/n) as e^(ln(n)/n) works

[cursive patrol]

?

[snow nimbus]

Sorry, I am blind. I read ^(1/2)

[cursive patrol]

oh fair

[long fog]

What after that tho

[cursive patrol]

series expansion

[long fog]

We're not allowed to use that yet

Havent been taught in our course yet

thats kinda the issue with a lot of these, but it makes sense

they're doing everything from scratch to make sure nobody is left behind

[cursive patrol]

ok. would have been useful information from the start

what are you allowed to use, then?

[long fog]

Sorry

[long fog]

1. basic definitions of limits and convergence/divergence
2. convergence criteria: comparison test and cauchy condensation test
3. some results eg. lim (1+1/n)^n= lim sigma 1/k! = e and some others

[topaz sinew]

@long fog Has your question been resolved?

[snow nimbus]

Sorry, if I make an unqualified suggestion again. This reminds me of my real analysis 1 course, in which we got a similar question. Is there any chance that you have the inequality n^(1/n) <= 1 + 1/(2sqrt(n)) hidden in the script?

*Back then we also had very limited tools and this was basically the only way we were able to solve this.

[topaz sinew]

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[long fog]

.reopen

[topaz sinew]

✅

[long fog]

yeah i was tryna find smth like n^1/n-1<1/n^p, 0<p<1

[snow nimbus]

Did you show the result limit n goes to infinity n^(1/n) = 1?

[topaz sinew]

@long fog Has your question been resolved?

[long fog]

Yea

[snow nimbus]

Did you use an upper bound to show it?

[long fog]

uhhh im confused as to how this is relevent to the qn

Sku

[snow nimbus]

I mean if you had this upper bound that I mentioned, the question would be immediately resolved (or any upper bound that is good enough)

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[trail sphinx]

one of the sat questions i was practicing with

[simple orchid]

ACT >>> SAT

jk

[trail sphinx]

i know that triangle GMN is isosceles and MH+NH is 3520 but idk where to go from there

[blissful spear]

Since MH and a NH are tangent to the circle, the angles GMH and GNH are both 90°. Which creates 2 right traingles.

[trail sphinx]

wait is that a property

[blissful spear]

Well a tangent to a circle will always be at a right angle with the radius of said circle

[trail sphinx]

oh wait yes

and then i would split the right angle in half to find the individual lengths right

[blissful spear]

(I hope I worded that correctly, english isn't my first language, so i apologize if something i say isn't comprehensible)

[trail sphinx]

don't worry it was perfect

ah wait

i think i got it now

2(168)+2(gh)=3856

wait that's wrong i think

i'll try it again

[blissful spear]

Yeah

[blissful spear]

It is

I also haven't fully figured out the sollution

Just wanted to chirp in with an insight that I thought you would find helpful

[trail sphinx]

it was helpful

i will try doing it again

im

stupid

@blissful spear it was just pythagorus theorem

1768^2 + 168^2 = (gh)^2

[blissful spear]

Are HM and HN the same length?

That's what I was trying to prove to myself this whole time

[trail sphinx]

if gm and gn are both equal

then h would be in the middle

i don't know the proof but intuitively that's how it should be

[blissful spear]

Yeah you're correct

[trail sphinx]

because if h wasn't in the middle, that would mean there was a tangent offset (therefore gm or gn weren't equal)

man

i completely forgot that the tangent of the point of a circle and the radius form a right angle

thank you so much though

[blissful spear]

No problem

@trail sphinx you should close the channel

[trail sphinx]

.close

[topaz sinew]

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[topaz sinew]

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[autumn plaza]

can i hab some help with diso

[neon iron]

factorize the top

turn the bottom into 1-cos^2

then factorize that and cancel factors

(this is all for LHS)

[neon iron]

use difference of perfect squares on denominator

[topaz sinew]

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[toxic aspen]

How do i determine the magnitude of a gradient vector?
For example [7,21,-14] ?

[simple orchid]

Have you learned about norms

[toxic aspen]

$\sqrt{7^2 + 21^2+(-14)^2}$

[thorny flame]

Merineth

[toxic aspen]

Would this be correct?

[modern rock]

dunno what a gradient vector is, but that is the formula for the magnitude of a vector

[neon iron]

agree ^

[toxic aspen]

Isn’t a gradient a vector?

.close

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[still sparrow]

Can someone explain part a and how dm/dt=2-.001m

[still sparrow]

I get that rate out is 2 l/t*m/200=m/100 or .001m

I don’t get have rate in is 2

Is it 2 because 1g of S/L hence as I have 2 l in the fore I have 2 grams of salt coming in

Giving me dm/dt=2-.001m

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[autumn plaza]

how i do dis

[autumn plaza]

guys... 😢😢

😭😭😭

<@&286206848099549185>

[fallow igloo]

Damn

[fallow igloo]

Have you compound it?

[autumn plaza]

whats compound

[fallow igloo]

Merge

Mix

[autumn plaza]

wut

[fallow igloo]

Can you explain where you’re stuck at?

Literally no idea?

[autumn plaza]

idk where to start

is it simplifying asinx+bcosx

[fallow igloo]

yes it is

[autumn plaza]

a = sqrt(28^2+45^2)

[fallow igloo]

Correct

[autumn plaza]

and then cos28/whatever that is x cos(pit/6 yada yada

[fallow igloo]

….

[autumn plaza]

...

[fallow igloo]

What’s yada?

[autumn plaza]

sorry im in bed because it kind of took a long time

wiat let me write it

[autumn plaza]

u get the point!

but then how can we make an equation for theta

[fallow igloo]

Notice angle compound formula

[autumn plaza]

sin(28/53) and cos45/53

oh..

[fallow igloo]

Set alpha or any other symbol

And mention sin alpha = 28/53 below

[autumn plaza]

so it’s not this

[fallow igloo]

Definitely not

[autumn plaza]

oh.

how do we do dis.

[fallow igloo]

28/53 and 45/53 are the results yield from trig function

[autumn plaza]

what trig function

[fallow igloo]

Sine and cosine

[fallow igloo]

👆

[autumn plaza]

I don’t get it

i think that cos theta = 28/53

and sin theta = 45/53

[fallow igloo]

Correct

[autumn plaza]

so now at

wat

[fallow igloo]

Then you should put theta instead of those figures

[autumn plaza]

why

[fallow igloo]

Why not?

[autumn plaza]

i thought we do liek dis

oh.

like dis

[fallow igloo]

yeah

[autumn plaza]

they didnt put theta there tho

[fallow igloo]

You should put theta in the parentheses of sine or cosine

[autumn plaza]

but they didnt put it in their parentheses 😢

[fallow igloo]

Are you referring to this?

[autumn plaza]

yere

[fallow igloo]

They transform trig(theta) into actual value

For example
Trig(theta) = a number

[autumn plaza]

oh

[fallow igloo]

You should put trig(theta) or number

[autumn plaza]

ok so if the number is not one on unit circle we put thet?

[fallow igloo]

Not trig(number)

[fallow igloo]

wdym

[autumn plaza]

OH

oh

ok

so how did they get theta

[fallow igloo]

They define it

The coefficient before sine and cosine stands for two side respectively

[autumn plaza]

so I can’t do this

its wrong tho

[fallow igloo]

Are you deciding to transform it into cosine or sine?

[autumn plaza]

it says form acos(bt-theta)

im trying to find theta

how do i find theta

[fallow igloo]

cosine stands for the coefficient of cosine in function h

[autumn plaza]

why cant i use tan inverse

[fallow igloo]

You can

But this is how I’m used to do it

[autumn plaza]

tan inverse is wrong why

oh nvm

..

i did it wrong,,

OK THX 🍮

NIGHT

[fallow igloo]

……

[autumn plaza]

...

NIGHT...

[fallow igloo]

.close

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[autumn plaza]

.close

[topaz sinew]

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[shell peak]

can someone help me with this problem

especially no. 5

[cunning kayak]

so what would be your approach

have u thought smthin abt it

[simple orchid]

Sounds like an engineering class seat arrangement

[cunning snow]

Show your work

[cunning snow]

[pearl trench]

bro

[shell peak]

so 8 seats so i would put 1 boy and 1 girl right away at first and last seat and then i have 4 girls and 2 boys left so ill add them and then it would be like

1 × 6 × 5 × 4 × 3 × 2 × 1 × 1

yes idk how pls help

[cunning kayak]

select 1 boy from 3 boys

[shell peak]

only 3 boys

[cunning kayak]

and 1 girl for 5

[shell peak]

yes so i already put him and her

so how do i do this probability

[cunning kayak]

5C1 * 3C1 * 6!

[shell peak]

oh yes the combination but we are rwquired to use fcp

and i really dont know how

like idk how to solve it using fcp

[cunning kayak]

FCP?

[cunning snow]

What is fcp

[shell peak]

fundamental counting principle

either u times it or add

[cunning snow]

Yea u can do it with that too

[cunning kayak]

3 × 6 × 5 × 4 × 3 × 2 × 1 × 5

3 boys can seat so 3

and any of 5 girls can seat so 5

[cunning snow]

Given that all of them are independent cases so we multiply

[cunning kayak]

same concept like you did for 6 5

[cunning kayak]

uh?

I have mult ;-;

[cunning snow]

If a boy sits on the seat 1 it doesn't affect the girls positioning on the last seat so the cases are independent

[cunning snow]

Uh?

[shell peak]

wait

why is the last 3 and 5?

and then 6 5 4?

[cunning kayak]

question says first must be boy

[shell peak]

because u still need to picj among the boys?

and there are 3?

[cunning kayak]

and last should be girl

[cunning kayak]

yes

[shell peak]

ohhh

[cunning kayak]

and hence we have fixed positions

[shell peak]

hey thank u so much

[cunning snow]

U didn't understand leave it

[shell peak]

because the 5 is the only one i didnt solve

btw vansh

[cunning kayak]

yes?

[shell peak]

u know that fcp can be multiplied or add, how do i distinguish when to multiply or add

[cunning kayak]

adding is generally for cases

[shell peak]

can u expound?

[cunning kayak]

yes

[shell peak]

pls

[cunning kayak]

if you are choosing outfit with 3 shirts and 2 pants, you multiply cuz each shirt can pair with each of pants
so, 3×2=6

For adding,
if you can take 3 different bus routes or 2 different train routes and for it you add them cuz choosing
one of them wouldn't involve the other
so, 3+2=5

adding is used for mutually exclusive events

whereas multiplication for independent where multiple things can happen at same time

[cunning kayak]

here only 1 event takes place at a time

[topaz sinew]

@shell peak Has your question been resolved?

[shell peak]

ohhh so u choose between taking a bus or a train so they are not dependent?

[cunning kayak]

yea, can also connect with real life

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[haughty wren]

I don't know if this image is very readable, so I linked the pdf. I would appreciate if someone looked at my workings and could tell me where I went wrong. The end result has an extra negative sign which I'm not sure how it got there. I know my workings might not be very clear, so feel very free to ask me to clarify things I've written 😅

[haughty wren]

Also I'm not sure if the thing I did where I just got rid of the sums on both sides is actually doable. It just felt right I guess?

[thorny remnant]

imtypo asking a question

[rigid cloak]

Fr

Bro is asking bout his working

On deriving the formula for the nth term of the Fibonacci sequence

In terms of golden ratio

[haughty wren]

Yeah, what's up with that

[rigid cloak]

Right?

[haughty wren]

Yeah it's the Binet formula

[rigid cloak]

Yea

Let me see bro

Bro

I can't see the pic clearly

I'm on mobile

Pls can u send the pic sir

Separately

[haughty wren]

Yeah that's why I sent the PDF, since my workings are long vertically, taking a picture that's clear is not really possible. I could take multiple pictures of the different parts if you want

[rigid cloak]

Okok imma try

[haughty wren]

Here, go left to right, it should be a bit better

[rigid cloak]

Can u pls check the sum

Other side

I mean the right side

Left is correct

@haughty wren

I feel here might be the mistake

[haughty wren]

There I used the Geometric series formula

[thorny remnant]

Ntg much im just used to you answering questions

[rigid cloak]

,w sum (-a)^k from k=0 to n

[thorny flame]

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[rigid cloak]

Right?

[haughty wren]

n -> ∞ though

[rigid cloak]

n tends to infinity

Bro

We finding the nth Fibonacci term right?

[haughty wren]

Well yeah but I‘m using geometric series. 1/(1+x) = sum [n=0 to infty] x^n

But I just realized somwthing

[rigid cloak]

What

[haughty wren]

That only works for |x|<1 and phi is a bit above 1

[rigid cloak]

That's why

I am saying till n

But wait

We do not sum till n to get the nth Fibonacci term?

What?

Don't u think we sum till n?

Nth Fibonacci term is sum of n-1 terms right ?

[hallow depot]

no

[rigid cloak]

What

[hallow depot]

Nth Fibonacci term is sum of the last 2 fibonacci terms

[rigid cloak]

Fr

I'm getting mad I feel

[hallow depot]

$F_n=F_{n-1}+F_{n-2}$

[thorny flame]

everg

[rigid cloak]

Yea

Sry i am going mad

I'm sleep deprived i feel

[hallow depot]

I feel you

[rigid cloak]

@haughty wren sry just went mad right there

Sry for the inconvenience caused

[haughty wren]

All good

[rigid cloak]

F(x) working seems all good

Where is the mistake then

Wait he already realised that

Case for geometric series would not work as |x|<1 but phi is not

Is here where the mistake lies?

[haughty wren]

I think so. I‘d have to find a way around that

[rigid cloak]

What if we sum to n

Simply n

Can u try that once

Imma too tired do even geometric sum fr

,w sum (-a)^k from k=0 to n

[hallow depot]

oppa does fr mean "for real"?

[thorny flame]

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[rigid cloak]

Is this correct?

Why I don't feel so

Ya it is

Ok we see for the other one

,w sum (a)^(-k) from k=0 to n

[thorny flame]

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[rigid cloak]

phi^2-phi-1=0

[hallow depot]

yes it is correct

[rigid cloak]

phi^2=1+phi

[rigid cloak]

Ya imma use phi there

Wait let's see

First sum - second sum

By root 5

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[paper sage]

Hello

[paper sage]

I need help with this

can someone explain who did we go from eq 4 to eq 5

help 💀

Dam I am being ignored

[vestal sigil]

vtx is wrong

you made a mistake differentiating it

[paper sage]

can you help me out with it

[vestal sigil]

i found $\ v_{tx}=\frac{v , u_x , u}{2a} + \frac{v_x , u^2}{4a} - \frac{v_x , u_x}{2} - \frac{u_{xx} , v}{2}$

[thorny flame]

Emily

[vestal sigil]

just apply the functions composition rule and the product rule by the book, you should find what i found!

[paper sage]

Thank you so much

I'll do that rn

I appericiate you so much

[vestal sigil]

btw, im sure its correct, cuz after that i did

[paper sage]

yeah please do share

[vestal sigil]

vxt - vtx = 0

and i found the same result as them

[paper sage]

vxt value is correct right?

[paper sage]

can you like solve the entire equation, if it's not too much I am sorry

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[steep cliff]

can someone help with my thinking for such questions?

would the answer to the first question be n+1?

[primal patrol]

1. because you're not simplifying like terms, you should think that (a+b) appears n times and so there's essentially n holes and in each hole you can put either an a or b

so its 2 options for the first hole

2 for the second

[steep cliff]

hole?

[primal patrol]

.... 2 for the n^th hole

just using it for imagination

or ok let's start slow

[steep cliff]

ok

[primal patrol]

if n=1

[steep cliff]

yes

[primal patrol]

(a+b)

[steep cliff]

1 term

[primal patrol]

how many terms?

no 2

[steep cliff]

right

[primal patrol]

like each summand is one term

[steep cliff]

yea

[primal patrol]

ok n=2: (a+b) (a+b) = a^2 + ab + ba + b^2

[steep cliff]

and when n = 2

[primal patrol]

how many terms?

[steep cliff]

4

[primal patrol]

ok good

[steep cliff]

ab is one term yea

[primal patrol]

so what does it look like then?

yes

[steep cliff]

while a+b is a binomial

[steep cliff]

it doubled?

oh

it squared

[primal patrol]

no

for n=1

it was 2

its 2^n generally

[steep cliff]

for n=2 its 4

for n=3

its uh

a^3 + 3(a^2)b + 3a(b^2) + b^3

oh that's simplified though yeah

[primal patrol]

yeah don't simplify

think of it like this

there are n choices to make

in each choice

you have 2 options

[steep cliff]

a or b

[primal patrol]

either pick a or pick b

[steep cliff]

yes

[primal patrol]

so 2 x 2 x... x 2 = 2^n

[steep cliff]

OH SO FOR EACH VALUE N

I HAVE A VALUE A OR B

IS THAT WHY ITS 2^N

[primal patrol]

that's the definition of the exponential right

[steep cliff]

yess

[primal patrol]

if you multiply the same number n times

[steep cliff]

and then for question 2

[primal patrol]

you get that number to the power n

[steep cliff]

yeah

for q2 selecting either a or b from each of the n sets of brackets

[primal patrol]

a) well you have to pick n things, if b is picked r times, then a has to be picked how many times?

[steep cliff]

that means i would have n number of a's

[steep cliff]

i have no clue

[primal patrol]

you have n decisions

b was picked in exactly r of those decisions

so how many decisions remain

[steep cliff]

n-r ?

[primal patrol]

yes

[steep cliff]

TYSM

in how many ways could b be chosen r times from the n sets of brackets?

would that be

n/r

[primal patrol]

part b is something you've likely seen : nCr

no

n choose r

[steep cliff]

i have not seen that

b!/n

[primal patrol]

actually it's kinda questionable here if its n choose r or multiply that with r!

I think the latter

b! what?

no

[steep cliff]

i meant r

over n

r!*

[primal patrol]

n choose r is n!/(r! (n-r)!)

[steep cliff]

ah i've never seen that before

in how many ways could bananas be chosen r times from the n sets of brackets

[primal patrol]

hmm strange they shouldn't be askng you then

[primal patrol]

yeah

[steep cliff]

im going ahead cause im not in school for 3 weeks so i need to

[primal patrol]

that's going to be n!/(n-r)!

[steep cliff]

be ahead of syllabus

[primal patrol]

ok

[steep cliff]

oh

[primal patrol]

for c) the answer is n!/(r! (n-r)!)

[steep cliff]

binomial coefficent?

[primal patrol]

yeah

[steep cliff]

ohhh

thank you so much man

i understand ths

.solved

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[clever sleet]

i got n = 54, answer key says n = 9?

[clever sleet]

my step by step was pretty straight forward cause it seems like an easy equation??
(log base 2 of n) - (log base 2 of 1)/(6) = log base 2 of 54
(log base 2 of n) - (0)/(6) = log base 2 of 54
(log base 2 of n) - 0 = log base 2 of 54
(log base 2 of n) = log base 2 of 54
n = 54

i'm just like... where did they get 9...

divide both sides by 6 or smth??

[old saffron]

I am sorry if this doesn't help much, but if n = 54, then the equality wouldn't make any sense since log_2(54) - (log_2(1))/6 = log_2(54) isn't equal

unless (log_2(1))/6 = 0

and it does

wait

yeah i am confused too now

[clever sleet]

...

[thorn violet]

log_2(1)=0 because 2 to the 0th power is 1.

[old saffron]

nah bro

[clever sleet]

i'm assuming the answer key is wrong (hopefully)

[old saffron]

you are right

they are tripping

[distant mesa]

The answer key seems to be wrong

[old saffron]

yeah

[distant mesa]

Ya

[old saffron]

sorry for not helping, you are totally right

[thorn violet]

You’re all good.

Try taking 2 to the power of each side of this equation for an easier simplification to see what’s going on.

[old saffron]

no way that is n = 9

[clever sleet]

my professor needs to renew his degree or smth

[thorn violet]

Lol everyone makes mistakes 🙂

[clever sleet]

this is the 3rd wrong answer on the answer key like ?? bro

whatevs

ty guys

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[trim wharf]

how can i solve this

the rate of change/slope is (5/-8)

all of these answers have the same slope since both x and y are multiplied by the same value for each one, which when divided by each other equal to 1

[opal vault]

what does a unit vector verify?

[trim wharf]

the magnitude is 1

ahh

i think i see

square both and then see which one is 1?

or sorry, add and then square root?

i forgot how it was done

[acoustic pecan]

youd be better off finding the magnitude of -8i+5j

[trim wharf]

64 + 25 = 89, square root of that

[acoustic pecan]

so if you want the magnitude to be 1, what would you do to -8i+5j to get the unit vector in the same direction

[trim wharf]

add 9i and subtract 4j??

[acoustic pecan]

no that wouldnt be parallel

think simpler

[acoustic pecan]

you only want to change the magnitude

[onyx summit]

v/|v|

[trim wharf]

uhhh

i dont know?

how can i only change the magnitude without changing the value of a and b

[acoustic pecan]

if you multiply/divide a vector by some constant, its magnitude changes by the same factor (the abs value of it)

[trim wharf]

yeah but then the value of i and j change as well

[acoustic pecan]

well yeah, they have to otherwise its just the same vector?

[trim wharf]

thats what im saying??

[acoustic pecan]

i and j are just unit vectors in their own right

[trim wharf]

so how can i change the vectors magnitude without changing anything else

[acoustic pecan]

you cant

no one ever said that

[trim wharf]

.

[acoustic pecan]

yeah, i mean you dont want to change the direction

?

[trim wharf]

ah

that makes sense

[acoustic pecan]

^

[onyx summit]

its the last option

[trim wharf]

so what do i need to divide the square root of 89 by to get it equal to one, is that what i should be asking myself?

[trim wharf]

bruh

im trying to learn

[onyx summit]

mb

got excited

[trim wharf]

-_-

[acoustic pecan]

you would divide the vector itself by its magnitude

though its been spoiled

[trim wharf]

i mean i was getting there

thats what i was thinking of since thats the only way to get one in this cases

alr then its seems to be D

[opal vault]

think about v = -8i+5j

[trim wharf]

thanks for your help

[opal vault]

yes great

[trim wharf]

lol

[opal vault]

since v = |v|*unit vector

to get the unit vector

you gotta divide by magnitude

[trim wharf]

i should write this one down

thanks guys

👍

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[trim wharf]

Since A * I = A, I multiplied both sides by the inverse of A to get I = 1. If I put all the values of the matrix as 1, would the matrix equal onw?

[trim wharf]

nope i got it

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[trim wharf]

Ignore parts a and b

for part c, I got T * T as well

I put T into the calculator and cubed it to get the 3rd year

nvm screw this

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[amber star]

The next question discusses the importance of definability. Let M be the set of names. Prove that M is a finite definition if and only if the following condition holds: there exists a finite series of atomic verses pi1 , . . . , pin and a truth function v:AP→{F,T}, so that all names f :{F,T}n →{F,T}
v ∈ M ⇐⇒ f(v(pi1),v(pi2),...,v(pin)) = T.

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[neon iron]

Question about partial fraction decomposition, how would I decompose this?

[neon iron]

The leading power of the numerator and denominator is the same

[cursive patrol]

when the leading power of the numerator is greater than or equal to the leading power of the denominator, you first divide the polynomials until this is no longer true

[loud oasis]

start by performing polynomial long division, then write it in the form (quotient) + (remainder)/(denominator)

[neon iron]

wow ordinary differential equations really does draw from everything ive learned

algebra precalc calc 1 calc 2 linear algebra and some calc iii

Got it, thanks!

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[strong sable]

Let ( k ) be any non-zero constant, and the function ( f(x) ) satisfies ( f'' - kf' - f = 0 ) on the interval ([a, b)), and ( f(a) = f(b) = 0 ). The correct statement among the following is:

(A) ( f(x) ) is monotonically decreasing first and then monotonically increasing on ([a, b]).\
(B) ( f(x) ) is monotonically increasing first and then monotonically decreasing on ([a, b]).\
(C) ( f(x) ) is identically zero on ([a, b)).\
(D) ( f(x) ) has exactly one zero within ((a, b)).\

[thorny flame]

riyobi

[topaz sinew]

@strong sable Has your question been resolved?

[topaz sinew]

@strong sable Has your question been resolved?

[strong sable]

Any ideas are welcome!!

[old saffron]

hi

[old saffron]

are you there

[wet harbor]

You can try solving the differential equation.

It's constant coefficient. Shouldn't be that hard

It's also homogeneous which makes it even easier

[strong sable]

How?

i may lack some knowledge of this

Any keywords?

[wet harbor]

Auxiliary equation

[strong sable]

r^2 - kr - 1 = 0
(r - (k/2) - √(k^2/4 + 1))(r - (k/2) + √(k^2/4 + 1)) = 0
r1 = (k/2) + √(k^2/4 + 1)
r2 = (k/2) - √(k^2/4 + 1)
f(x) = c1e^(r1x) + c2e^(r2x)
This way?

[wet harbor]

this works

i was checking if there will ever be complex roots but turns out there isn't

you can now calculate c1 and c2 by substituting the boundary conditions

at a and b

[strong sable]

They're all 0

What does this indicate

[wet harbor]

that C is true

[strong sable]

Why?

[wet harbor]

why is C true or why does it indicate that C is true ?

it indicates C is true because if it's zero everywhere then it's zero on the given interval

[strong sable]

Yes I misread

So basically this q is an ode question?

[wet harbor]

there could be another tricky solution that we don't see

but I honestly don't see it at the top of my head

[strong sable]

This is already tricky enough

[wet harbor]

fun fact that's the standard solution xD

everyone can do that

tricky solutions usually don't follow a standard method

you figure something out just by looking and make a solution out of it

like maybe for example turn this question into a matrix problem and figure out that the matrix is invertible

therefore it has only one solution which is the zero vector

thus leading to C

there could be many ways you can approach this

[strong sable]

It's an ode question but it's placed in exercises of single var calculus...

[wet harbor]

i mean, the world is small, math is all connected

by the way I also just realized the differential equation is only valid on the interval

so it could have any other value outside of it, it doesn't follow the ODE except there

[strong sable]

Then what should i do

[wet harbor]

nothing, the solution still works, because it only asks for the value in the interval

[strong sable]

I see

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[lusty lagoon]

Hi need a little help with part c, i can't seem to expand it for some reason

[rocky lagoon]

In part (c), you're asked to show that the sequence vn = un - l, where l is the solution to the equation l = al + b, is a geometric sequence. By substituting the recurrence relation un + 1 = aun + b into the expression for vn + 1, and using the relationship l = al + b, we find that vn + 1 = avn. This confirms that (vn)n => 0 is indeed a geometric sequence with common ratio a

Also, translate the question, you'll get more help

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[lusty lagoon]

Sorry I had to go somewhere, thank you so much for your help the answer was right there in plain sight. I really appreciate it

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[neon iron]

i need help for my subsusituion grade 10 math work its due in 1 hr and i have a test i just need a bit of gudiance thats all

[neon iron]

can we hop on vc

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[neon iron]

r u stupid

ok wtv

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[dull maple]

can som1 help me?Im stuck here and im really confused on how to continue and i think i did most of the calculations wrong .. the result of the expression should be 3a^4b

[wary tulip]

slay

[neon iron]

slay

[restive inlet]

you have notation issues, inappropriate use of ()

those () i maked with those red x shouldn't be there
you're missing () around the 6a^2 in blue

[dull maple]

yes im confused on how to use those🥲

[restive inlet]

for that component, you should have
ab multiplied to whatever 5a^2 + a^2 simplifies to
i.e. you should have ab * 6a^2 or equivalent
as opposed to the ab + 6a^2 you had

with expressions adjacent to grouping symbols, multiplication is implied

[dull maple]

okay

[restive inlet]

what you had there: $5a^2\red{(\black{+a^2})}$ represents the product of $5a^2$ and $\gray{+}a^2$ which isn't what you want

[thorny flame]

ℝαμΩℕωⅤ

[dull maple]

i think i understand now

[restive inlet]

try continue from
$$\br{ -\frac 32 a^3b + ab(6a^2)}\br{-\frac 23 a} + 6a^4b$$

[thorny flame]

ℝαμΩℕωⅤ

[dull maple]

yes i writed this now too

i think now it should be fine

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[placid drum]

did i do 11 and 13 right?

[placid drum]

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[lyric grove]

I was on Chapter 1 of Calculus by James Stewart and I'm currently doing this question. How do I approach part c?

[lyric grove]

like how do i find the "least squares regression line"?

[craggy haven]

least squares is a pretty complicated thing. Are you intended to use computer software for this question? Is that what that green symbol means?

[lyric grove]

i guess it does

so i just use a software and give the points as inputs

?

[wary tulip]

hi hayley :waves:

[lyric grove]

and get the equation?

[craggy haven]

yea

[lyric grove]

hmmm

thankyou!

[lucid junco]

slayla

[wary tulip]

water beam

[lucid junco]

slayla

hello

[wary tulip]

water beam

hello

[lucid junco]

https://tenor.com/view/anime-bye-bye-maki-wave-soredemo-ayumu-wa-yosetekuru-when-will-ayumu-make-his-move-gif-26195508

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[midnight sinew]

I'm not too sure what to do with (b)

I don't understand what it means by angle to z-axis

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[eternal pike]

3 darts are thrown (equal probability of landing anywhere on the dart board). What's the probability that they all land on a same half of the dart board? I am not sure how to start this

[gloomy lance]

Let's take that dart boards is divided in 2 equal part

As we know there is 3 darts there is equal possibility so we can say that 2×2×2 so it will be 8

Let's think that all dart land on 1 st half or 2nd half that means we have 2 possibility so we can do 8/2 which is 1/4 and 25% chance

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[odd flower]

so for this

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[odd flower]

.reopen

just think abou the total combinations of the dart landing

1st dart has 2 choices where to land

2nd also

and 3rd also

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[lusty cedar]

How do I they know f of 4 is 3

[thorny flame]

Result:

2550

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[lusty cedar]

why is subbing in 4 to the first deravative equal to 3

i get stationary point is three

but still

oh they sub it into the integral

I don't get it though still

oh ffs cause y = f of x

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✅

[lusty cedar]

whats the method for part c here?

[hidden zinc]

So first u can convert tan into sin/cos and then ull get a quadratic in sin

[lusty cedar]

yh I get all that

its just part c Im lost on

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[long lake]

choose 2 negative numbers and 2 positive numbers from 14 numbers (total 14= 6 positive and 8 negative). multiply those 4 numbers, and we need a positive number. find probability of it being positive

[long lake]

my doubt

if I do (8C2 × 6C2)/14C4 then it gives correct answer

but if I do 8C2/14C2 × 6C2/12C2 then it's wrong answer

can someone tell why is second approach wrong

[cunning snow]

U mean multiply those 4 numbers?

[long lake]

yes

[cunning snow]

Why does the other condition exist if we multiply those 2 negative no anyway get a positive no

[long lake]

I don't understand

which condition

second approach?

[cunning snow]

Multiply those 4 numbers and we need a positive no

[long lake]

its a bigger question, and I'm asking for only one part of it

[cunning snow]

What's your question?

I'm confused

[long lake]

from 6 positive and 8 negative numbers, 4 numbers are choosen and multiplied. what is probability that resultant is a positive number

I don't need answer for whole question. only the case where we choose 2 negative and 2 positive

other two cases would be all 4 negative, and all 4 positive numbers

<@&286206848099549185>

[cunning snow]

Ok you can choose two numbers negative no by 6c2 and two positive by 8c2 and now we need to multiply since the cases are independent
6c2 × 8c2

[long lake]

just tell me why this method would be wrong @stuck hearth

[stuck hearth]

This

[long lake]

what's wrong with choosing those numbers one by one

[stuck hearth]

Over counting

In your wrong answer

You count ab cd and cd ab as different combinations

But they’re the same

[long lake]

if i choose 2 negative numbers out of 14 numbers then it is 14C2. then 12 numbers remain. and I choose 2 from it, 12C2

[long lake]

?

[stuck hearth]

If I give you a,b,c,d

What is the combination of 4c4