help-0

[timid oracle]

[queen imp]

sorry idk what to do

[earnest bay]

all good

[queen imp]

u should use one of the non-occupied help channels

[earnest bay]

I thought I was in this one?

[timid oracle]

oh sorry

[queen imp]

ya u are

[earnest bay]

oh ok

[queen imp]

it's okay

[earnest bay]

90<x<180

Once I get the angle I can use vector addition to solve for Vector shuttle

<@&286206848099549185>

[lone heart]

@earnest bay Has your question been resolved?

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@earnest bay Has your question been resolved?

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@earnest bay Has your question been resolved?

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@earnest bay Has your question been resolved?

[cinder sundial]

Is there anything that implies BD/CD=15/24

The solution says BD/CD=15/24 but I couldn’t figure out why

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[slow canopy]

I know the binomial theorem, but I'm a bit confused maybe by the wording here?

aren't a and b just 5x^2 and 2y^3 in the shown binomial? What does it mean when it gives the other values to a and b?

I'm working with this understanding

If I could even just get walked through the first couple here it would help a ton

<@&286206848099549185>

I expanded the binomial like this, still very much struggling since there are no instances where, for example, there's an x^6 * y^9 (since I need to find the coefficients of x^a * y^b and a=6, b=9)

maybe I expanded wrong?

[lone heart]

@slow canopy Has your question been resolved?

[noble sinew]

When k=3 you get the term a^3b^3=x^6y^9

[slow canopy]

does that mean the answer to a is 5^3, 2^3

or would the coefficient be 6C3 * 5^3 * 2^3

[noble sinew]

You also have 6 choose 3

[noble sinew]

Yes

[slow canopy]

thank you, that helps a lot

I think I forgot how to multiply exponents in my initial expansion

.close

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[zenith compass]

$\frac{\partial^{4n}}{\partial x^{2n} \partial y^{2n}}|_{(0,0)} \frac{1}{1+x^2+y^2}=(2n!)^2\binom{2n}{n}= \frac{(2n!)^3}{(n!)^2}$?

[ocean seal]

Trenton

[zenith compass]

I am not sure about if my computation is correct

Can anyone help me to have a look? Thanks

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@zenith compass Has your question been resolved?

[lone heart]

@zenith compass Has your question been resolved?

[civic jetty]

can anyone help me

[civic jetty]

whats square root of 123456789

[gray ingot]

use a calculator

and dont post questions in an occupied channel

[civic jetty]

uh ok

[left sphinx]

Hi, my calculator says
11111.1110605555554405416661433534692458
But that list of numbers can go long, but yea let's stop occupying an occupied channel

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[zenith compass]

.reopen

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✅

[zenith compass]

$\frac{\partial^{4n}}{\partial x^{2n} \partial y^{2n}}|_{(0,0)} \frac{1}{1+x^2+y^2}=(2n!)^2\binom{2n}{n}= \frac{(2n!)^3}{(n!)^2}$?

[ocean seal]

Trenton

[zenith compass]

Can anyone help me to see if this computation is correct?

[bitter vault]

It's not correct. The $(2n!)$ should be $(2n)!$. I responded to you in the MVC channel

[ocean seal]

1345631

[zenith compass]

Ok

.close

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[brave spoke]

hi can i have some help please so,
ive worked out that b = 0

but im unsure on A as i got a = 6 but according to the mark scheme a = 5.
can someone explain how to find A

[tardy rapids]

(2-3)/(a-2) = -1/3

Because it is the gradient, slope of the line,

The slope between any two points must also be the same

[brave spoke]

oh that makes sense so i would just do 3 + 2 = 5 which is A
because 2-3 / 5-2 = -1/3

[civic jetty]

deez nuts

[sudden hinge]

Another easy way is to construct the line equation and simply plug in the points:
$$y = -\frac{1}{3} x + c$$
Plug in the first point
$$\implies 3 = -\frac{2}{3} + c \implies c = \frac{11}{3}$$
Then you have the full equation and you can simply plug in the $x$- or $y$-value of the other two points.

[ocean seal]

nvx

[brave spoke]

i understand how you've plugged in the points but how would i find a and b this way

[ripe rain]

From the above conclusion we get $y=- \frac 1 3 x +\frac {11} 3$

[ocean seal]

Deep.

[ripe rain]

@brave spoke

[brave spoke]

yes that makes sense , would i just do the same with the other points?

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[shut parcel]

Can someone explain question b) for me? I don’t understand how to calculate the loss and gain

[shut parcel]

@alpine sable This spot already taken

[knotty oriole]

Yeah

[shut parcel]

Can someone explain question b) for me? I don’t understand how to calculate the loss and gain

It says that to calculate the loss in revenue, you would have to 40*20 = $800.
I know how to get an answer of 40 because 320-280 = 40
However, I don’t seem to understand where they’re getting 20 from?

It says that to calculate the gain in revenue, you would have to do 280*10 = $2800
280 because it’s the final price where 320 is the initial price…
10 because quantity increases by 10 each time

Not sure if this is right

<@&286206848099549185>

Someone help pls😭😭

<@&286206848099549185>

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[serene moon]

is this just

vector v has to be 0

therefore u*v (0 vector)=0?

[tawdry portal]

Why would v have to be 0?

[serene moon]

bc how else would magnitude of u+v vs u-v be the same? @tawdry portal

[tawdry portal]

u=0 for instance but it actually holds as long as u and v are perpendicular

Because of Pythagoras theorem

[serene moon]

but u=0 is still valid?

[tawdry portal]

What do you mean?

[alpine sable]

hello

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[daring flume]

hi i need help -- i dont understand why cost=sint gets the points (pi/4, 5pi/4)

[tacit arch]

do you know the unit circle?

[daring flume]

OHHH

im so stupid

that makes sm sense lmao thank you

.close

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[white acorn]

can someone help me?

[fallen plinth]

Add the two equations together

[white acorn]

oh ok?

[fallen plinth]

Left side of first + left side of second = right side of first + right side of second

You'll get another equation

[white acorn]

2x-y-2x-2y?

[fallen plinth]

Yes

And since the 2x's cancel out, you'll have the value of y

Which then you can put into either of the 2 equations and solve for x

[fallen plinth]

That's only the left side tho

[white acorn]

i got -3y

[fallen plinth]

Ye for left side

That will be equal to sum of the right sides

[fallen plinth]

-3y = right side of first + right side of second

[white acorn]

-3y = -2x-2y

[fallen plinth]

Nonono the RIGHT sides

You already did the left side

[white acorn]

ok

[fallen plinth]

Whatdya get?

[white acorn]

x=y/2

it's wrong i think

[fallen plinth]

This is wrong yes

You did the left sides of the original equations and got -3y

[fallen plinth]

So now add up the right sides

[white acorn]

i got 3x=3?

[fallen plinth]

How did you get that?

[white acorn]

nvm i messed up

what do i plug in with -3x? the 2x-y or the -2x-2y

[fallen plinth]

Okokok let's take a step back

You're trying to find out what x and y are

So you add both equations together

To do this. You make another equation, where the left side is the sum of both left sides, and right side is sum of both right sides

$2x-y-2x-2y=-6-18$

[ocean seal]

Netimerus

[fallen plinth]

Does this make sense?

[white acorn]

yea

[fallen plinth]

Nah to plug smth somewhere, you need to know more exact values generally

[fallen plinth]

So rn you just simplify this

Tell me when you've done that

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@white acorn Has your question been resolved?

[white acorn]

i got -3y=-24

[vale sapphire]

looks good to me

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[pale kestrel]

<@&268886789983436800>

[junior blade]

[knotty spire]

At least some trolls are creative

[tacit arch]

eww

[knotty spire]

not this one

[sly mantle]

.close

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[gray heart]

how do i solve this?

[lone heart]

@gray heart Has your question been resolved?

[gray heart]

<@&286206848099549185>

[tight locust]

ratio of areas

[gray heart]

I don’t know how to find the area of triangle acd

[gray heart]

<@&286206848099549185>

[open folio]

Similar triangles?

[open folio]

Above

Break the triangles apart, identify corresponding figures and create a ratio

[gray heart]

but how do i find the area of the first triangle to find the ratio?

[open folio]

I just said, similar triangles, you need to find side AD

[gray heart]

is it 4?

[open folio]

Let me check

Yes

It is 4, now just solve for areas

[gray heart]

but why are the triangles similar?

how can i tell?

[open folio]

Some theorem, if you break apart a triangle like that you get similar figures

I forgot the details

Our geometry teacher showed us with pieces of paper

[gray heart]

well how do you personally know the triangles are similar?

.close

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[elder lark]

The question:

[elder lark]

Where I got to:

the answer is apparently infinity, but isn't sin(infinity) undefined?

[alpine sable]

∞ is undefined thus, sin(∞) and cos(∞) cannot have exact defined values

[elder lark]

when I had an answer that was just like cos(inf) it was undefined, but now that it's 1/2*inf - sin(inf) it's infinity? Why isn't it undefined? Is it just because 1/2 * infinity is so much larger?

[exotic glen]

sin(inf) and cos(inf) are something between -1 and 1 so inf minus something between -1 and 1 is still infinity

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[mellow vapor]

Can someone help me understand how to do error

[lone heart]

@mellow vapor Has your question been resolved?

[mellow vapor]

Nvm

[elder lark]

.close @mellow vapor

[mellow vapor]

.close

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[wild nest]

what does this mean?

[wild nest]

Help please

.close

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[cyan sapphire]

Hi, I'm trying to make a graph that looks like this, and I was curious if anyone knew a formula that would bring me to this

[cyan sapphire]

Kind of like it slowly spirals upward starting large at the bottom with a set radius, then at each ring having a smaller radius

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[cyan sapphire]

<@&286206848099549185>

[cursive badger]

parametrically it'd be very easy

z=t (you can modify the t to be 0.1t or 10t, it just governs how quickly it goes upwards, or you can make it negative to go downwards, or make it sint to make the spiral loop up and down on a given z interval, etc...)

x=f(t)cost (you can change the frequency of cost by making it cos2t or whatever, to make it spiral more per vertical z distance. same goes for the y)

y=f(t)sint

where f(t) is a decreasing function from where your t starts

such as 1/t, or 1-t, or (t+1)/(1+t^2), or 1/(lnt) or something

unless my brain has suffered a catastrophic failure, I do not believe that curves can be represented in three dimensions graphically

only surfaces

@cyan sapphire

[cyan sapphire]

I’m not understanding, what is Z in this case? I tried to graph this on desmos but it just looks like a circle

[cursive badger]

z is a third dimension

[shell widget]

Try geogebra.

[cursive badger]

use a 3d grapher like

https://c3d.libretexts.org/CalcPlot3D/index.html

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[foggy crypt]

How can I calculate the best outcome?

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@foggy crypt Has your question been resolved?

[placid zinc]

Can you explain the problem to some degree?

[foggy crypt]

Sure, sorry didnt see your message, Im trying to find out which combination of situations may end with the best outcome, but I dont know if there is any way to see If I can do better

This one was the best I got, I would spend 3 points on Sun making it 60 6 on Mon making it 40 and three on Wed making it 40 as well

[foggy crypt]

<@&286206848099549185>

[placid zinc]

Not really an explanation. What's a "combination"? What's a "situation"? What's a "best outcome"?

[foggy crypt]

Is it there on the first picture, The situations are made of 6 day values, and I need to use a combination of 5 situations. In which 40 is the lowest positive and 60 the best positive, above 60 is okay but it is redundant. One of the "must" situations has to be used at least.
An all days start with 19 points, you also have 12 spare points to allocate whenever convenient, but no more than 7 in the same day.
The best outcome would be to have as much days positive as possible

Is it too confusing? I tried to make it as clear as possible

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[pale reef]

i tried usong the chi rem theorem but uhhhhh im not too sure if my ans is right

[vale wigeon]

what is your answer?

[lone heart]

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[vale wigeon]

[pale reef]

88 (mod 195)

[vale wigeon]

let's see

,calc 2*88 mod 13

[ocean seal]

Result:

20

[vale wigeon]

er

well 20 mod 13 is 7 so close enough

,calc 88 mod 15

[ocean seal]

Result:

13

[vale wigeon]

@pale reef checks out

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[novel echo]

I can only conclude that P(A intersection B') = P(A)-P(A intersection B)

<@&286206848099549185>

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@novel echo Has your question been resolved?

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@novel echo Has your question been resolved?

[novel echo]

What the hell

.close

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[fresh sedge]

can someone go through this with me please

[random zinc]

yes just treat it like, you would do with an equation

[fresh sedge]

okay give me one moment

could y be 3?

[zenith lodge]

try y=3 and evaluate the LHS if you want to

[random zinc]

if you plug in 3 the equation is true, but at the end you have to get something like: y < some value. The equation is then true for alls y < some value. You are supposed to find this value

[gray isle]

have you ever solved equations before?

[fresh sedge]

.close

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[gray isle]

e.g if instead of an inequality you had the equation
3(y-4) = 6,
would you be able to solve that?

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[lost saddle]

Pls anyone can help

Mathematics discrete

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[lost saddle]

<@&286206848099549185>

🙏

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[lone heart]

@lost saddle Has your question been resolved?

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@lost saddle Has your question been resolved?

[tacit arch]

.close

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[astral solstice]

What means the notation $\mathrm{Id}_{V}-\pi: V \rightarrow V$

[ocean seal]

Simplex

[lone heart]

@astral solstice Has your question been resolved?

[limpid spade]

wut

[astral solstice]

Yea, and $\pi$ is a Projection

[ocean seal]

Simplex

[astral solstice]

And I have to prove, that $$\mathrm{Id}_{V}-\pi: V \rightarrow V$ is a projection

[ocean seal]

Simplex
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lone heart]

@astral solstice Has your question been resolved?

[lone heart]

@astral solstice Has your question been resolved?

[lone heart]

@astral solstice Has your question been resolved?

[slim spire]

@astral solstice these are linear maps I assume, linear maps can be represented as matrices and the difference of maps correspond to the difference of matrices

[ocean seal]

SkyTwX

[astral solstice]

Allright, thanks!

,close

.close

[lone heart]

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[next cliff]

Hey there!

[next cliff]

I'm looking to see if anyone can help convert $\frac{(n-1)!}{(n_1-1)!(n_2)!(n_3)!}$ + $\frac{(n-1)!}{(n_1)!(n_2-1)!(n_3)!}$ + $\frac{(n-1)!}{(n_1)!(n_2)!(n_3-1)!}$ into an equation w sigma

[ocean seal]

eswag

[lone heart]

@next cliff Has your question been resolved?

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[olive oar]

I'm making a presentation and am having troubles defining the core of a group?

[olive oar]

any tips?

[dense compass]

isn't the core the intersection of the conjugates of a subgroup H of G?

couldn't you just write the definition?

[olive oar]

yea

cause then i can use that to get a normal core right

[dense compass]

yeah the normal core is when H is the largest normal subgroup

[olive oar]

yea

[dense compass]

what is the objective of the presentation? I don't quite understand what you want to do?

[olive oar]

lemme show you

im defining the minimal faithful degree

cause i work with it later on

i just need to explain foundations before i get into it

[dense compass]

yeah, I mean those are the definitions

all seems to be fine

[olive oar]

k thanks

.close

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[zenith compass]

Let $n\in\mathbb{N}\setminus{0}$. Suppose that there exists some $z\in\mathbb{C}$ such that for any $u\in\mathbb{R}$, $u^2+n<|z|$. Then $n^2<-1$.

I am going to negate the aforementioned statement.

So I reformulate the statement as follows:

For any $n\in\mathbb{N}\setminus{0}$, if there exists some $z\in\mathbb{C}$ such that for any $u\in\mathbb{R}$, $u^2+n<|z|$, then $n^2<-1$.

Then I do the negation as follows:

There exists some $n\in\mathbb{N}\setminus{0}$ such that ( there exists some $z\in\mathbb{C}$ such that for any $u\in\mathbb{R}$, $u^2+n<|z|$ ) and $n^2\geq -1$.

Can anyone help me to have a look that whether I made them correct or not?

[ocean seal]

Trenton

[slim spire]

For all ... there exists ... such that P is negated by
There exists ... such that for all ... then not P

[lone heart]

@zenith compass Has your question been resolved?

[zenith compass]

There exists some $n\in\mathbb{N}\setminus{0}$ such that ( there exists $z\in\mathbb{C}$ such that $\boxed{for any}$ $u\in\mathbb{R}$, $u^2+n<|z|$ ) and $n^2\geq -1$.

[ocean seal]

Trenton

[zenith compass]

Do you mean this?

[slim spire]

There exists some z in C becomes for all z in C

[zenith compass]

Do you mean this?

There exists some $n\in\mathbb{N}\setminus{0}$ such that ( for any $z\in\mathbb{C}$ and $u\in\mathbb{R}$, $u^2+n<|z|$ ) and $n^2\geq -1$.

[ocean seal]

Trenton

[slim spire]

Now you are saying that this works for any u

But it should be there exists some u such that ...

You dont need to think a lot.
For all x in X, there exists y in Y such that for all z in Z blablabla
Becomes
There exists x in x such that for all y in Y there is some z in Z with NOT blablabla

All the quantifiers are inversed, and your proposition negated
I hope this is clear 😅

[lone heart]

@zenith compass Has your question been resolved?

[zenith compass]

There exists some $n\in\mathbb{N}\setminus{0}$ such that ( for any $z\in\mathbb{C}$ there exists some $u\in\mathbb{R}$ such that $u^2+n<|z|$ ) and $n^2\geq -1$.

[ocean seal]

Trenton

[zenith compass]

Lol I think I used the equivalent form before rather than negating it
Hence I mess up everything

[slim spire]

Im kinda getting lost too lol

Can you send again the original one? XD

[zenith compass]

Ok

Let $n\in\mathbb{N}\setminus{0}$. Suppose that there exists some $z\in\mathbb{C}$ such that for any $u\in\mathbb{R}$, $u^2+n<|z|$. Then $n^2<-1$.

[ocean seal]

Trenton

[zenith compass]

With your help, I get:

There exists some $n\in\mathbb{N}\setminus{0}$ such that ( for any $z\in\mathbb{C}$ there exists some $u\in\mathbb{R}$ such that $u^2+n<|z|$ ) and $n^2\geq -1$.

[ocean seal]

Trenton

[zenith compass]

Originally, I made a mistake so I write:

There exists some $n\in\mathbb{N}\setminus{0}$ such that ( there exists some $z\in\mathbb{C}$ such that for any $u\in\mathbb{R}$, $u^2+n<|z|$ ) and $n^2\geq -1$.

[ocean seal]

Trenton

[slim spire]

Oooh my

The rule imverse everything works but here you need to be cautious

You have:
There exists some n such that P => Q

So it becomes:
For all n, then not P=>Q

Since P=>Q is the same as not P or Q, then not P=>Q is the same as P and not Q

Just rrassemble the pieces together

[zenith compass]

Oh it is the de Morgan

For any $A$, if there exists $B$ such that for any $C$, $D$, then $E$.

$A=n\in\mathbb{N}\setminus{0}$

$B=$ some $z\in\mathbb{C}$

$C= u\in\mathbb{R}$

$D= u^2+n<|z|$

$E= n^2<-1$

[ocean seal]

Trenton

[zenith compass]

I can further write it as:

For any $A$, if $F$, then $E$.

$A=n\in\mathbb{N}\setminus{0}$

$B=$ some $z\in\mathbb{C}$

$C= u\in\mathbb{R}$

$D= u^2+n<|z|$

$E= n^2<-1$

$F=$ there exists $B$ such that for any $C$, $D$

[ocean seal]

Trenton

[slim spire]

👍 👍 👍 💯

[zenith compass]

So the negation is:
There exists $A$, $F$ and (not $E$).

[ocean seal]

Trenton

SkyTwX

[slim spire]

Everything you said is perfect

[zenith compass]

Ok so the negation will become:

There exists $n\in\mathbb{N}\setminus{0}$ such that (there exists some $z\in\mathbb{C}$ such that for any $u\in\mathbb{R}$, $u^2+n<|z|$) and $n^2\geq -1$.

[ocean seal]

Trenton

[slim spire]

PERFECT

@zenith compass 🎉

[zenith compass]

Thank you so much!!

[slim spire]

Np♡

[zenith compass]

.close

[lone heart]

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[lone heart]

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[cinder sundial]

.close

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[spark sky]

Hello guys

[spark sky]

I just have a question if I did this correctly

[lone heart]

@spark sky Has your question been resolved?

[spark sky]

<@&286206848099549185> any feedback?

[tacit arch]

what's T?

[spark sky]

Transformation

[tacit arch]

What's the equation for it

[spark sky]

T(x)=0

[tacit arch]

that doesn't define T

That defines the kernel of T

[spark sky]

This is not a linear transformation tho

It didn’t pass the test

[tacit arch]

jesus all i'm asking is write down something like

$T = ?$

[ocean seal]

riemann

[spark sky]

I dunno lol

What you mean by that

T(u)+T(v)

=

T(u+v)

T(cu) = cT(u)

@tacit arch

That’s the definition of linear transformation

Tx = Ax?

:/

I’m looking at the class examples

[tacit arch]

How do you do the problem if you're not given T?

[spark sky]

I did this way @tacit arch

This was the first part of the question

T is usually not given in these questions

🧐

@tacit arch left the building?

[tacit arch]

these two vectors are not linearly independent

the kernel can be described by just one of them

[spark sky]

Oh

So it doesn’t matter that x2=-x3

Or x3=-x2

I mean I guess x3=-x2 is basically x2=-x3

When you think about it

[tacit arch]

🧠

[spark sky]

Better now? @tacit arch

Also do you usually need T in this questions for real?

He never ever gives us T

[tacit arch]

looks right

[tacit arch]

It tells you in the question

[spark sky]

Ohhhh

Like that

I thought you wanted T as a matrix or something

[lone heart]

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[lone skiff]

h

[lone skiff]

Hello guys is there a way to find the solution here with using the calculator?

to try every option or somtthing

if anyone know please tell me

help

how to solve question like this please?

[elder grail]

which topic do u have this under?

[lone skiff]

systems of linear equations

[elder grail]

like have u learnt either matrices/ basic 3D geometry?

[lone skiff]

I know about matrices and determinants

[spark sky]

I think this is a plug in question?

You are gonna plug a b c and for the equations

[elder grail]

not really u cd do it with determinants

[lone skiff]

What do you mean?

how to plug?

[spark sky]

It tells you what a is

[lone skiff]

I knew something that the system will be consistent if I write the 3 equations on the calculator and gets me result

so I put the system on the calculator, but the problem is I don't know the values of a ,b, c

[spark sky]

It tells you what they are

[lone skiff]

so in the choice (a)

-2a+3b+c=0

[spark sky]

a is x1 + 2x2

[lone skiff]

a=1
b=1
c=-1
is these numbers valid

[spark sky]

What is this question under

Is this matrices

[lone skiff]

systems of linear equations

[lone skiff]

I what is this : |

[spark sky]

What you happen if you plugged in first equation into second equation

[lone skiff]

x1=-2x2 - x3

[spark sky]

It says a is equal to x1 + 2x2 + x3

[lone skiff]

we get x1 and put it in the second equation

[spark sky]

What would happen if you plugged a b and c into -2a+3b+c

Do you get 0

I dunno what you are doing

[lone skiff]

I don't understand what you mean with plugged

[spark sky]

I would do it differently but then Km not sure if that’s correct

[lone skiff]

what is plugged

like this or what?

[lone skiff]

yes this is the right answer, I want to know how did you get it

@spark skyare you there?

[spark sky]

Wait what

That’s the right answer?

[lone skiff]

yes (a) is the right answer

[spark sky]

Hmmm

Hmmm

[lone skiff]

how to solve this?

[spark sky]

Like this maybe

I dunno

[elder grail]

ok u know how to solve a linear eq in 2 variables ryt?

[lone skiff]

yes

with the calculator

or i can do it with pen/paper but it takes time

[elder grail]

ok ur allowed calculator?

[lone skiff]

Yes

[elder grail]

ok tho u won't really need it

ok so first try getting the eqs into eqs with 2 variables (basically eliminate 1 variable)

try doing that n tell me if u can

[lone skiff]

ok

[elder grail]

u got something?

[lone skiff]

gimme minute

look from the first eq I got x1

now we can put it in eq2 and eq3

is this what you meant?

[elder grail]

not really

just notice that (2 times eq 1) -eq 2 eliminates x1

tell me if u get this

[elder grail]

this isn't wrong tho but i feel might complicate things

[lone skiff]

Look

Some friend solved it for me

omg do I have to do all these steps to get the solution ?

can't I get the solution faster?

[elder grail]

they're doing it in a similar manner

[elder grail]

there is a faster way but it depends on whether u've learnt planes

tho u can solve without getting the concept cuz the formula is just determinants

[lone skiff]

ah

look I have a question here

I sent it again so we can see it

[spark sky]

I got the answer

[lone skiff]

the solution is a right

[elder grail]

yeah ik

yes

[spark sky]

It’s a

You need to row reduce the equation

[lone skiff]

so if I put random values from my mind
like
a=1,b=1,c=-3
this will make the equation
-2a+3b+c=0

right?

[spark sky]

It’s a linear algebra question

[elder grail]

that won't help in general

[spark sky]

Don’t you know matrices?

This is a linear algebra question

[elder grail]

yeah much faster with that

[spark sky]

You need to reduce row the matrix

And then make sure that there won’t be a pivot on the right side of the matrix

To be consistent

[lone skiff]

why , isn'y (a=1,b=1,c=-3)
will make the equation valid?

so I can solve the matrix with these valid

cuz it will be consistent

right?

on the calculator

so I put this matrix with(a=1,b=1,c=-3)
and if I get value, so it is the right answer

if I don't , so it's not the right answer

the problem is, i get wrong answer in all cases

[spark sky]

Where do you get those values

[lone skiff]

idk why

[lone skiff]

from my mind

[spark sky]

You can’t randomly choose values

[lone skiff]

and it's equal to zero in this equation

[spark sky]

This matrix looks like it has indefinite answers as long as the last row is 0=0

[lone skiff]

aaah

[elder grail]

there's a "none of these" option so u might not want to do that

[lone skiff]

so calculators get's math error if the matrix has indefinite answers too?

[spark sky]

Don’t you know matrices?

[lone skiff]

and indefinite is still consistent right?

[spark sky]

I mean you can’t solve this without linear algebra

And looking at your friends answer

You must have learned linear algebra

[lone skiff]

yea so i have to do it in the hard way 😐

I was just seeking an easy way with the calculator

since I'm allowed to use it

[spark sky]

Yeah there’s no easy way to this question

[elder grail]

there is actually

[spark sky]

You can enter the matrix into calculator

[lone skiff]

say it

[lone skiff]

and?

[spark sky]

And you can see what you get

Row reduce it

[elder grail]

so u need to just make a determinant

ok wait

[spark sky]

Wait determinant?

[elder grail]

i'll see if i can look it up for explanation

[spark sky]

Why would you need to find determinant

[lone skiff]

o

[elder grail]

the reason's related to planes so

3d geometry

[spark sky]

We are not trying to prove if this is a consistent matrix

[elder grail]

cross products

[spark sky]

We are trying to make sure this is a consistent matrix

[elder grail]

yess

so the determinant equates to 0

[lone skiff]

Yes

I got that too

[spark sky]

Finding determinant will only help if this is a consistent or inconsistent matrix

And this is a consistent matrix

So it shouldn’t be zero

If you say the determinant is 0

[elder grail]

ok i'm not sure what's the name but u replace one set of coefficients with the constants n get the determinant to be 0 for consistency

[elder grail]

lol this isn't but the set of eq is with a specific condition

[spark sky]

Yeah

[lone skiff]

replace it with what?

[spark sky]

That’s why finding determinant has no use

[elder grail]

https://www.cliffsnotes.com/study-guides/algebra/algebra-ii/linear-equations-in-three-variables/linear-equations-solutions-using-determinants-with-three-variables

just check this once

[spark sky]

And how will that help

With this question

You are finding the determinant of the matrix

[elder grail]

so in our case we have D=0

for consistancy we need Dx,Dy n Dz 0 too

*as per the notations in the link

[spark sky]

Lol

You are just complicating it for him

[lone skiff]

🥲

[spark sky]

It’s harder

Dx Dy

[elder grail]

just find one

u'll see it satisfies option (a)

[lone skiff]

I think this is the shortest road right?

[spark sky]

Why do you have to go into derivatives

@elder grail

[elder grail]

that's not derivative 😂

[spark sky]

Oh nvm

It’s not

[elder grail]

just the same determinant with a,b,c replacing x's coefficients

[spark sky]

Just a confusing notation

[elder grail]

so it's a 2 step thing

yeah true

[spark sky]

This took me 3 minutes to solve

It’s not a hard question at all

[elder grail]

yeah not too much work

[spark sky]

You need to know how to REF

And what consistent matrix is

[elder grail]

@lone skiff ditch the determinant approach if it feels complex

[spark sky]

An augmented matrix is consistent if there’s no pivot after the equal sign

There you go

[lone skiff]

I'm curious how do you like math like this.. I really hate it and study it in order to pass the exam

[spark sky]

Linear Algebra is fun

[elder grail]

ok that's prolly for a different channel XD

[lone skiff]

Alright thanks anyway for answering 🤍

[spark sky]

Stuff I do

[lone skiff]

.close

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[restive shadow]

smallest value of n for which 100n^2 < 2^n holds true

[tacit arch]

binary search

[restive shadow]

yea that can be done. But how can I know the value mathematically

if we apply log on both sides we get,
2 log 10 base 2 + 2 log n base 2 < n

[tacit arch]

Unsolvable

at least in terms of elementary functions

[restive shadow]

(Although we can try to guess the value by trial and error) can't we get the value mathematcally

[restive shadow]

oh ok

[tacit arch]

feel free to try https://en.wikipedia.org/wiki/Lambert_W_function

[restive shadow]

so we have to do it by trial and errr?

[tacit arch]

.

[restive shadow]

alright

.close

[lone heart]

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[restive shadow]

wait. how do i free the channel

[elder grail]

automatic ig

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[magic tartan]

can someone explain this

[lone heart]

@magic tartan Has your question been resolved?

[alpine sable]

Do u have the figure

[magic tartan]

no this is all i have

[alpine sable]

I could draw a rough figure and I have done the proof

Excuse me for the bad handwriting

Basically equal chords are equidistant. And they form 90 degrees. Adjacent sides are equal. This is possible if it's a square or kite(Right angle Euclidean kite)

So in either case XG=YG

Hence triangle XGY is isoceles hence opposite angles are equal

And since in both kite and square the diagonals bisect equal angles hence XGY=YGX angle

[magic tartan]

hmm

yeah I get it now

thx

[lone heart]

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[alpine sable]

f(x) = 0.3 * sin(x + 1) + ln(0.3x + 1)

The graph of this function in the given interval [2;20] represents the cross section of a racing track. Each of the 3 maxima represents a hill, with the 3rd one being the goal. Now at that goal, a camera is placed so the middle hill can be viewed. How do I figure out how high the camera has to be put up so it can see the middle hill and not be blocked by the hill it's on?

Obviously I can't just use a function that crosses both the middle hill and the goal and then take the height difference between the two functions at the goal; then it would also intersect with the hill ("the ground would block it's view"). So It'd need to be a function that goes through the middle hill and another point P, but how do I go about finding that?

[alpine sable]

this is what I meant by the floor blocking it

but it would need to be something like this

[lone heart]

@alpine sable Has your question been resolved?

[alpine sable]

Would any of the <@&286206848099549185> perhaps have any idea?

[lone heart]

@alpine sable Has your question been resolved?

[elder grail]

so idk if i cd help much, but where exactly is the cam placed?

like only on the maxima (perpendicular to that point) or anywhere on hill 3?

[alpine sable]

perpendicular to the maximum

and its supposed to cast exactly onto the maxima of the middle hill

[elder grail]

like a sort of tower ryt?

[alpine sable]

yep

let me make a sketch real quick

[elder grail]

okie

[alpine sable]

something like this

and if d is the "height" of the cam

what is the smallest d possible

so that the line is not blocked

[elder grail]

oh ok i get ur q

so u cd probably try finding the eq of the tangent to the curve passing through the 2nd hill's maxima n tangential at some point on the 3rd hill

n find the intersection of that with the line perpendicular to the 3rd maxima?

@alpine sable

[alpine sable]

but is a maximums tangent not a constant?

[elder grail]

no we're not taking that

[alpine sable]

or am I not understanding you

[elder grail]

so the tangent just passes throught the maxima, not tangential

*2nd maxima

it is tangential at a diff point on hill 3

[worthy mica]

yeah newbienoob is correct i think

[alpine sable]

ohhh yeah I believe Iunderstand now

but how would I go about finding that point ?

is it just trial and error ?

[elder grail]

no like take a general point

say x1,y1

find the general eq of a tangent through that general point

n then substitute the 2nd hill's maxima in that eq n see what value of x1 satifies it

[elder grail]

yeah will prolly look something like this

[worthy mica]

To phrase newbienoob's method another way: you want to find a general tangent line to the curve on the third hill, and see which one of those passes through the second maximum as well

[elder grail]

yes

[alpine sable]

I'm a bit confused as to what you mean by "general tangent line"

[elder grail]

ok so imagine u have a point x1,y1 satisfying this eq

(y1 is just 0.3sin(x1).....)

[alpine sable]

this point would be the 2nd hill ?

[elder grail]

no this cd be any point in general

on the curve

[alpine sable]

but it has to be on the curve

okee

[elder grail]

so our target is to first find the tangent's eq at this point

(so basically differentiate)

[elder grail]

yupp

so our 'x1' is the 'x' here (just a convenient notation)

now plug in the point of the 2nd maxima into this eq

waitt no

so this is the slope u get ryt?

[alpine sable]

if i plug in the x of the 2nd maxima i get zero

or wait

do you mean

[alpine sable]

by this

replacing y and x with the coordinates of the 2nd maxima ?

[elder grail]

nono forget that for a moment

so this is what u get after differentiating?

[alpine sable]

yes this is the first derivative

[elder grail]

ok so this is your slope at point (x1,y1)

[alpine sable]

x1, y1 was any point on the original function

[elder grail]

now do uk the point-slope form of a line?

[elder grail]

yes

[alpine sable]

haven't heard of that until now

but i see the formula on the net

y - y1 = m (x - x1))

[elder grail]

yes

so m is

[elder grail]

this

so this is the eq of a line passing through (x1,y1) n with slope m

follow this much?

[alpine sable]

like this ?

[elder grail]

yess

[alpine sable]

please continue speaking, I feel I'm beginning to understand

[elder grail]

now plug in the maxima point for the variables x n y

[alpine sable]

x or x1?

[elder grail]

x

x1 is what we want to find

[alpine sable]

oh rigt

[elder grail]

so what's the 2nd hill's maxima?

[elder grail]

yes plug these into x,y

n tell me what u get for x1

*u may need calc cuz it's not solvable manually ig

[alpine sable]

but there are two unknowns in there still

that not a problem?

[elder grail]

oh wait the 0.3x n cos(x+1)

they need to be x1 not x

cuz the tangent is at x1

so slope depends on x1

[worthy mica]

Just want to point out this is the third hill's maximum! not the second hill

[elder grail]

$y-y_{1}=(\frac{0.3}{0.3x_{1}+1}+0.3cos(x_{1}+1))(x-x_{1})$

[ocean seal]

newbienoob

[worthy mica]

Oh ignore what I said sorry

[elder grail]

this should be ur eq

[alpine sable]

yeah the first ones outside the domain

[alpine sable]

yesss and I put in these for y and x

https://www.wolframalpha.com/input?i2d=true&i=1.904+-+y+%3D+\(40)Divide[0.3%2C0.3x%2B1]+%2B+0.3+*+cos\(40)x%2B1\(41)\(41)+\(40)13.338+-x\(41)

[elder grail]

so here x1,y1 is the req point for tangency

and x,y will be the maxima

[elder grail]

here we already know y

[alpine sable]

oh oh oh

i know that

that is the x of the 3rd hill!

and then we can solve for y

[elder grail]

yes

so y is just f(x)

so try plugging that too

[elder grail]

f(x) u mean?

[alpine sable]

oh wait no i missread something

[alpine sable]

so I have this

[elder grail]

just read this again once

tell me if all good till here

[alpine sable]

okay let me try to say how I understand it

i had

y -y1 = m(x - x1)

I put in the 1st derivative for m

and the coordinates of the 2nd maxima for x and y

x1 and y1 are the coordinates for the point I want it to be tangent to

[elder grail]

yes perfect

now y1 u can find

in terms of x1

u understand this?

[alpine sable]

that means i put in an x1 and get a y1? 😅

[elder grail]

so u had this ryt

[alpine sable]

yes

[elder grail]

x1,y1 satisfies this eq

[alpine sable]

wait

i think

i get it

x1 is the x of the 3rd maxima

then I get a y

i now have two points

and through these goes a function

[elder grail]

not maxima, it's just on the 3rd hill

[elder grail]

yes this is correct

x1 is unknown, but the function connects x1 n y1

or explicitly, y1 = 0.3*sin(x1+1)+ln(0.3x1+1)

makes sense now?

[alpine sable]

not quite yet but I feel like it might when we're done

[alpine sable]

so i put this for y1 and then solve for x1?

[elder grail]

ok so for now subst that for y1 in the eq

yess

[alpine sable]

aaaa

curse this phone

[elder grail]

,rotate

[alpine sable]

sec

oh tahts beast lol

[elder grail]

lol

perfecto

[alpine sable]

yeah it got a bit longer than I anticipated

[elder grail]

now find x1

wolf it

[elder grail]

expected

cuz there r many values

so our req value will be the one closest to, but greater than 1.904 ryt?

cuz it needs to be on hill 3

n 1.904 is on hill 2

[alpine sable]

yeah that makes sense

x ≈ 3.74562487400014

that's the closest

[elder grail]

yes

wait a min

[alpine sable]

sooo now I have x1 and x2?

[elder grail]

-_-

[alpine sable]

oh god

no

what happened

[elder grail]

u subst x n y opp

whops

n i missed it too -_-

issok not much of a prob just reverse 1.904 n 13.338

waitt

damn i'm going nuts

it's correct

-_-

[alpine sable]

ahaha i feel you man

[elder grail]

y such a pathetic function?

[alpine sable]

but issok we almost done ryt

i have no clue

[elder grail]

yeah

ok so now we got x1 n we can subst it back in the original eq

y-y1=m(x-x1)

this

this is our line

[alpine sable]

but wait is that now what I just used to get x1 with ?

[elder grail]

n x1 will be 13.383 (closest to 13.338, not 1.904)

[elder grail]

yes now we get the general eq

[alpine sable]

okay so I just take this

and put x1 in?

so that

y-y1=m(x- 13.383)

[elder grail]

hold on a bit

[alpine sable]

yeah no p

[elder grail]

ok so looks like the x1 is 19.3933

plotted n checked

the others intersect before itself

[alpine sable]

the comfort in hearing that

[elder grail]

lol

n a close up on hill 3

u see the 2nd point?

that's the maxima

shd've realized this q better off done numerically/graphically

[alpine sable]

yah so now I only need to figure out the diff between the two

[elder grail]

not really

plot x=(the x value for maxima)

see where they intersect

n then find diff

basically the green gap u see

[alpine sable]

oooh

[elder grail]

i'll be honest i've never done worse calculations even with a calc at hand

[alpine sable]

bruh my mind is fucking blank

[elder grail]

-_-

[alpine sable]

yeah I like

[elder grail]

lol

[alpine sable]

holy shit

i appreciate your work

[elder grail]

so see one thing u cd do

[alpine sable]

but ifeel like for 11th grade this is not quite the intended solution

xd

[elder grail]

u cd prolly dm me later sometime (cd be anytime)

we'll do it when our brain's ain't drained up XD

[alpine sable]

yeah

[elder grail]

yeah this is pure numerical method

n 0.3 is the most random value one cd find

this is an assignment?

[alpine sable]

homework

[elder grail]

damn

[alpine sable]

its like a worksheet

we started looking at trig functions in analysis last week

and now teach is ick and we gotta do these at home

[elder grail]

smh

[alpine sable]

like the other stuff is 100% legible and stuff but wtf is this

[elder grail]

ok btw we might have to open up this channel

[alpine sable]

yeah

[elder grail]

switch to dm?

[alpine sable]

ill send an fr and come back l8r

yep

[elder grail]

okie

[alpine sable]

.close

[lone heart]

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[upbeat gorge]

It’s not a trapezoidal prism though

[lone heart]

@thick lynx Has your question been resolved?

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Send your question here to claim the channel.

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

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

[near hollow]

Got this problem from my teacher for our homework due on Monday, but its so hard, please someone help. "What is the probability of an n by n binary matrix being invertible, assuming all n by n binary matrices are equally likely"

[lone heart]

@near hollow Has your question been resolved?

[near hollow]

not sure how to describe it in terms of fields but its the matrix space of n by n matrices where all the elements are either 1 or 0

[round mantle]

Use central noodle theorem?!

[alpine sable]

what is that 😆

[round mantle]

You can convert a matrix to either udon noodle or Chinese noodle soup

[alpine sable]

wow, I never knew about that

[near hollow]

yeah i guess you would if u added matrices together

ye but how do u find the probability they are linearly independent?

so we need to count the number of different ways of forming a linearly independent set of n n-dimensional vectors

Hmmm so im thinking of starting with the basis vectors and then adding them to each other in different ways to find all possible linear independent basis of R^n

and then we know there are n! ways of putting them vectors into the matrix

and also there are 2^(n^2) different matrices

but problem is adding the basis vectors doesn't ensure linearly independent right?

like if we add them all to each once we get the 1n matrix

is there some rule that im missing so that i can add these basis vectors together and still produce all linearly independent n dim vector sets without double counting and creating a vector set that isnt linearly independent

we do because of reordering

you could have e1, e2,e3,...en or e1+e2, e1+e3... en-1+en ext. there are too many to count cuz all upper and lower triangular matrices are invertible if the diagonal is full of 1s

but the number of combinations for the first vector depends on the other vectors aswell

there are 2^n

ye

what's the answer then?

hmm idk man

for 2 by 2 you get 3/8