help-49

[sage helm]

The formal wording gets to me sometimes 🫠

Okay thanks you two :)

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[last slate]

How can you represent a matrix on graph paper
As in can anyone represent [1 2 3] and [ 1 ]
2
3

[last slate]

Ughh

One is row and another is column matrix

[junior flower]

wdym “represent a matrix on graph paper”

[last slate]

What is geometric representation of a matrix

If it's just some numbers then what sense does it make?

[junior flower]

when you multiply a matrix and a vector you get a new vector

so you can interpret the matrix as something that transforms vectors to new ones

[last slate]

Like?

[junior flower]

it’s late and i’m on my phone soz don’t have the energy to type out a computation involving matrices

[oak nymph]

What about multiplication of matrices??

[last slate]

Sorry

[last slate]

Elaborate pls

[oak nymph]

Square matrix multiplication i can understand as a composition of transformations

But what about something like a 5x3 matrix multiplid by a 3x7?

Ah, sorry maybe i should take my questions to my own help channel

[last slate]

As in?
What are transformations?
I just started matrix

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

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[sage helm]

I've trouble articulating this

[sage helm]

Do I just say there exists no $x \in \set R$ such that $0 < x < 1/n$ for all $n \in \set N$

[grand pond]

Sadie Carnot (η = -1)

[sage helm]

For any x I'll give you a sufficiently large n

Isn't that kind of cheaty though since that's what I need to prove

[quiet hinge]

You are proving that the intersection of those sets is the empty set

[sage helm]

¯_(ツ)_/¯

I guess

[quiet hinge]

And you are proving it by showing that no real number could be in all of those sets because then such x would have to satisfy 0<x<1/n for all n in N

[sage helm]

Yeah well

I think internally I have strict rules for where I'd stop

[quiet hinge]

I mean if you want to prove 0<x<1/n for all n in N is impossible for any x, then choose n > 1/x by the archimedian property and you have found a counterexample

[sage helm]

Oh that makes sense actually

[sage helm]

I can just say this and tag on "due to the Archimedean property"

Okay thanks!

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[pearl void]

hi

[pearl void]

i was wondering if there is a better way to solve this than using ghe quotient rule

$\frac{d}{dx} \left( \frac{7x^3 + 27}{63x^7 - 28x^5 + 243x^4 - 108x^2} \right)$

[grand pond]

Alex

[cobalt swan]

Factorise maybe?

[cobalt sigil]

Yeah

[cobalt swan]

But ultimately u gotta use that u/v time tho

[cobalt sigil]

You can consider this function y

Apply ln y

And solve

[worthy wing]

,calc 243/27

[grand pond]

Result:

9

[worthy wing]

,calc 108/27

[grand pond]

Result:

4

[worthy wing]

,calc 63/7

[grand pond]

Result:

9

[worthy wing]

,calc 28/7

[grand pond]

Result:

4

[cobalt sigil]

This makes the work easier ,

The visualization easier

[worthy wing]

7x^3+27 is a factor

Of denominator

[cobalt sigil]

,w factorise 63x^7 - 28x^5 + 243x^4 -108x²

[grand pond]

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[pearl void]

what is that magic?

[cobalt sigil]

Wolfram magic

[cobalt swan]

Lol

[pearl void]

i mean he got oke factor doing some divisions

snyway thx

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[scenic urchin]

is the interval for the negatives values of sin 0 to - pi ?

[midnight plank]

Please don't occupy multiple help channels.

[winged wharf]

okay basically

what you said is technically right

but since sin is periodical function

[cold coyote]

well sin(x) is a periodic function so there are more intervals than that, but yeah one of the interval is that

[winged wharf]

there's a butch of intervals

but yeah 0 to -pi would give you every possible negative value of sin x when evaluated

,ask graph y = sin x

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[winged wharf]

like look even pi to 2pi gives you negative values

[winged wharf]

so yeah this is one of the intervals

[midnight plank]

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[dreamy lichen]

Try expressing the intersection of W1 and W2 as a set first

[bleak pier]

Set will be {x,y,z,t}?

[dreamy lichen]

it should be set of vectors

which vectors are in both in form (x, y, z, x, z) and form (0, x', y', z', t')?

[bleak pier]

Is this wrong sir?

[dreamy lichen]

Well its just a set of 4 variables

idek whats the meaning of that

[bleak pier]

(x,y,z,t)

[dreamy lichen]

well, this vector has just 4 entries

while both W1 and W2 have vectors with 5 entries

so W1 intersection W2 should have vectors with 5 entries too

[bleak pier]

Should I include 0?

[dreamy lichen]

E.g. one member of W1 intersection W2 could be
(0, 3, 4, 0, 8), because its both in the form (x, y, z, x, t) (for x = 0, y = 3, z = 4 and t = 8)
And also in the form (0, x, y, z, t), (for x = 3, y = 4, z = 0, t = 8)

And to determine the dimension, it would be a good idea to either find the general form of vector in W1 intersection W2, or find a basis

So if we have a vector of form (x, y, z, x, t), when can it also be written in the form (0, x', y', z', t') for some x', y', z' and t'?

Are you still here?

[dreamy lichen]

Even the 0 entry has to be included, (0, x, y ,z, t) isn't same as (x, y, z, t)

Oh you're offline :/

[bleak pier]

I am reading

[midnight plank]

@bleak pier Has your question been resolved?

[bleak pier]

What should be next?

(0,x,x,x,t)

Intersection vector

(0,x,y,z,t)=(x,y,z,x,t)

[bleak pier]

@dreamy lichen

Because x=y=z

[midnight plank]

@bleak pier Has your question been resolved?

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@bleak pier Has your question been resolved?

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[tidal turret]

calculate the area of the enclosed region between the x-axis and the graph of $\f(x) = \frac{6e^{3x}}{3+e^{3x}}$ for $0\leq x \leq5$

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[bold beacon]

Why don't you use an integral calculator?

[tidal turret]

so what you are saying that I need basically to calculate $\int_{0}^{5} \frac{6e^{3x}}{3 + e^{3x}}$

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

?

[shy hearth]

should be right

[bold beacon]

yep

[tidal turret]

okay

ok i didnt knew that, math isnt as fast for me

so I prolly need to do partial fraction decomposition

[shy hearth]

consider other techniques

[tidal turret]

ibp?

[bold beacon]

you can use an integral calculator. Full steps will be provide. Off the bat I can't see a way to perform partial fractions

[shy hearth]

u-sub

[tidal turret]

okay, u = 3 + e^(3x)

[shy hearth]

[tidal turret]

,, \frac{du}{dx} = 3e^{3x}

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

$\frac{dx}{du} = \frac{1}{3e^{3x}}$

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

$\int \frac{6e^{3x}}{3+u} \cdot \frac{1}{3e^{3x}} = \int \frac{2}{3+u} \cdot du$

[shy hearth]

shouldn't be 6u

oh you did e^(3x) = u

[last slate]

an integral calculator is not needed by any means here

[shy hearth]

this is easier

[last slate]

this function does have an elementary integral function

[shy hearth]

yep and e^3x/e^3x cancel out

then it's a simple integral

[tidal turret]

mmm

is this 2ln(3+u)

[shy hearth]

should just be du not du/dx

[shy hearth]

yep!

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

,,2\ln\left(3 + e^{3x}\right)

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[shy hearth]

• C

[tidal turret]

sure but we need to evaluate on 0 and 5 and subtract them

its definite not indefinite

[shy hearth]

oh right, it's a definite integral

my bad

[tidal turret]

,, 2\ln\left(3 + e^{15}\right) - 2\ln\left(3 + e^{0}\right)

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

,w 2ln(3+e^15) -2ln(3+1)

[grand pond]

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[tidal turret]

thank you for the help

lowkey thought partial fraction decomposition might help but as the other guy said, we got nothing to factor

whatever

thanks

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[night gyro]

is this correct

[night gyro]

i feel like im doin wrong

[edgy schooner]

What do you need to find?

[night gyro]

find the set of values of x

[edgy schooner]

Do you know the areas of the given figures?

[night gyro]

they didnt give us the area

area of rectangle is greater then triangle

aaa ruined the whole thing

[edgy schooner]

Oh this is a quadratic inequality then

[night gyro]

u sure

then i dont need to do question bc not in my topic list

[edgy schooner]

Yeah, if area of rectangle> area of triangle

(x-1)(3x-2) > 2x(x)/2

Solve this and you will get solution set of x

[night gyro]

kk got it boss is fine i dont need it 👍

[edgy schooner]

mb

[night gyro]

ty

ty

see yall

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[cobalt sigil]

Yes

Ok

We can battle them in the arena

No you have

So

We can choose any 3 of 9 in

9c3 ways

Now for 3 to be placed the probability is (0.01)³

So its 9c3 * (0.01)³

Here you wrote 4

Something 'between'

Here the question is different

C means combinations

ncr is defined as

n!/(r!×(n-r)!)

For 4 balls it would be 9c4 × (0.01)⁴

[midnight plank]

@last slate Has your question been resolved?

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[simple field]

the top face of the cylinder of mercury is not in contact with glass

[last slate]

You must only calculate the curved surface area of the cylinder and that is 2πrh

[simple field]

no, also the bottom face

[last slate]

Oh and add th area of the base too

[simple field]

yh

[last slate]

2πrh + πr^2

[simple field]

the surface area of a cylinder of height h and radius r is 2pi r h (the curved part) + 2 * pi r^2 (the top and bottom circle)

it doesnt matter

since oyu only need to know the surface area of the section that is contacting the mercury

the question is slightly vague in that sense

but the assumption to make is that the glass cyclinder is open at the top, so you can could actually pour the mercury in

and so the top face of the cylinder of mercury is in contact with the air

thus you dont include that top circle of surface area

gets rid of one of the circles

[last slate]

One circle's area is pi r squared

The top isn't covered, that is what we are assuming

R squared

[slender walrus]

don't conflate area with volume

[last slate]

See, let's get the basics right, the circumference of a circle is 2πr. A cylinder is formed when all these circles are stacked together, giving us some height. So the surface area of a cylindrical with no lids is the perimeter of one circle multiplied by the height. That is, 2πrh.

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[night gyro]

<@&286206848099549185>

[severe cloud]

!15m

[midnight plank]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[severe cloud]

What have you tried so far?

[fathom prairie]

I’ll help

Oh

[mortal anchor]

Yes

[severe cloud]

You summoned 3 people lmao.

[shy tendon]

lmaoo

[night gyro]

is the rule the third or the last option

[mortal anchor]

Yep u summoned 3

[fathom prairie]

Last

[mortal anchor]

Yes it is

[night gyro]

ty tht wht i was thinking aswell

[mortal anchor]

Np

[fathom prairie]

Reason why is because of 0 then 6 but it’s for b it’s 0 then 12 so there is a gap

[fathom prairie]

No

Probs

[mortal anchor]

Was gonna say that bro

[fathom prairie]

Lol

[mortal anchor]

I'm jk

[fathom prairie]

Hi jk I’m Midget

[night gyro]

is the last option yh

[mortal anchor]

😡

[fathom prairie]

Lol jk

[mortal anchor]

😡

[night gyro]

ok ty

[mortal anchor]

I'm jokin

[mortal anchor]

Np 👨🏿‍🦲

[night gyro]

.close

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[fathom prairie]

Hi jokin how are you

[mortal anchor]

😡😡😡🤬🤬🤬🤬

[mortal anchor]

Buddy I don't think u wanna mess with the demons inside me

[steel dirge]

I'm actively praying you're not an emo. Anyways, you guys need help?

[mortal anchor]

I am

[mortal anchor]

I'm jk

I'm not

[steel dirge]

[midnight plank]

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[true knot]

heyy guys, im at first year undergraduate physics course and i need help with linear algebra

[true knot]

it says "Given U and W the following subspaces of R4, Find the dimension and a base of U, W and U intersection W"

i've solved other exercises about base and dimension before, but the vectors were always given, i dont have any idea of what to do with "a, b, c, d"

[fossil knot]

Try to form a basis for each subspace

Since the vectors in U are independent of A, (1,0,0,0) is in the basis

[true knot]

thanks! I'm gonna try it! Soon i'll be back with the results

[midnight plank]

@true knot Has your question been resolved?

[true knot]

i isolated d = -b -c

Then expressed the vectors like this:

(a, b, c, -b-c) =>
a(1,0,0,0) + b(0,1,0,0) + c(0,0,1,0) + b(0,0,0,-1) + c(0,0,0,-1) =>
a(1,0,0,0) + b(0,1,0,-1) + c(0,0,1,-1)

And got to the conclusion that (1,0,0,0), (0,1,0,-1), (0,0,1,-1) can be expressed as a basis. Is it correct?

[midnight plank]

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[eternal musk]

the top is where I am now

the bottom is the answer

how do I rearrange it into that

[atomic jolt]

can we have this in latex?

[midnight plank]

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[eternal musk]

I did not delete it but yes

sure

[atomic jolt]

you need to open a new channel

[prime hornet]

^

[eternal musk]

.close

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[night gyro]

someone help bru

[nova iron]

Top one

[night gyro]

how

and why

[fathom tendon]

you give out the solution not the answer

do you know how to get the mean of something

[night gyro]

yh u add everything and devide by how much there is

[fathom tendon]

ok

so how much candy is in bag 1

[night gyro]

n

[fathom tendon]

bag 2

[night gyro]

n+5

2n-3 on the third bag

4n+2/3?

[fathom tendon]

me: That seems good

now you solve for n

[night gyro]

ngl

i got it thanks boss

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[last slate]

What am I doing wrong can’t get lnsiny

From coty

[visual tiger]

yep, also if u = cos(y), don't understand how -sin^2 becomes (1-u)^2 on the denominator

[midnight plank]

@last slate Has your question been resolved?

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[maiden river]

is there an easy way to prove this graph has no hamiltonian cycle?

[midnight plank]

@maiden river Has your question been resolved?

[polar mortar]

I'd look at what happens after the cycle visits the triangles on the side. Since the cycle needs to go through 12 - 21 - 32 and 14 - 25 - 34 You can look at the necessary path to connect those two segments in a cycle.
it should reduce to showing a few symmetric cases are impossible (unless I'm missing something)

[midnight plank]

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[worthy spruce]

could anyone help me with this question, I can get the uniqueness part of it, however, just solving that there exists a h and k st g = hk is giving me some greif. Any help would be great, thanks.

[worthy spruce]

I tried just trying to show that the image of gh^(-1) is in the kernel for some h, which seems like the right direction, but I don't know how to proceed.

because wouldn't that mean that h = g? which is not necessarily the case

[tribal temple]

the homomorphism maps from G to H, and g is an element of G

[worthy spruce]

yeah sorry, I dont follow

I guess we can then prove that H=G?

[tribal temple]

if $\phi : G \to H$, and $g\in G$, is there a statement you can then make from those two? (something very "obvious"!)

[grand pond]

@tribal temple

[worthy spruce]

oh, it pretty much follows directly from the 1st isomorphism theorem? is that what you were thinking?

[tribal temple]

Well, not quite what I meant-

[tribal temple]

I was looking for you to notice that $\phi(g) \in H$

[grand pond]

@tribal temple

[worthy spruce]

omg

haha

yeah, thanks a ton. that makes sense

cheers

.close

[midnight plank]

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[marsh ferry]

how to do this?

[tiny drum]

induction?

[glacial slate]

you can also get it from differentiating the binomial theorem (I mean the one which says sum from k=0 to n of of nCk = 2^n)

[midnight plank]

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[bleak pier]

so here are my equivalence class
1-{11 22 33 44}
2-{11 22 33 44 12 21}

[misty gorge]

equivalence classes should be subsets of S

in this case

an equivalence class consists of the things that are equivalent to each other

[bleak pier]

like?

example pls

[bleak pier]

is this wrong?

[misty gorge]

like {1,2} is an equivalence class, for example

because (1,2) is in the relation, and there isn’t anything that says “1 is equivalent to 3” in R

the ordered pairs are sort of saying “these elements are equivalent in R’s eyes” when R is an equivalence relation

so we can ask the questions like “in R’s eyes, what is 3 equivalent to?”

[bleak pier]

3 is related to which?

[misty gorge]

yeah we can ask a question like that

[bleak pier]

33
44
two equivalence class

but here 1 is related to itself and 2

and same prolbem with 2

[misty gorge]

we want that though

like that’s fine for us

to say 1 is related to itself. in fact, all elements are related to themselves when we have an equivalence relation like this.

can i ask you a question that sounds different but is actually the same as this?

[bleak pier]

yes you can ask

[misty gorge]

if i have the set {a,b,c,d}, and a=b, and b=c, then how many different objects are there in my set?

[bleak pier]

2?

[misty gorge]

yeah

that’s right

[bleak pier]

a a a d

[misty gorge]

yep

[bleak pier]

so here two different cases in my question

related to iteself

and related to vice versa

1 2
21

[misty gorge]

so i’m going to say that the problem you’re doing is the same thing. there are actually more than two. replace “related to” with “equal to” in your mind for a second

so the relation R says “ 1=1, 1=2, 2=1, 3=3, 4=4”

now how many different objects are there?

[bleak pier]

3?

[misty gorge]

yep

and an equivalence class is just like “treat all the related things as equal”

[bleak pier]

so we are asking a simple question to the relation R

which is defined on the set A

how many different objects in it

[misty gorge]

yeah, and R will answer differently from us, because it thinks some of the objects we see as different are actually the same.

[bleak pier]

yes

[misty gorge]

so we will say “there are 4 objects in S” but R will tell us “i only see 3 equivalence classes”

because R thinks 1 is the same as 2

[bleak pier]

[1]= 11
[2]= 22?

[misty gorge]

[1] is all the things R thinks is the same as 1

[bleak pier]

i see

so [1] will be 11 22?

or 12 21?

[misty gorge]

no need to use double numbers here, we just want the things that R sees as 1, which will be in S. so R thinks 1 is the same as 1, and same thing for 2 (R thinks 2=1), so [1] = {1,2}

[bleak pier]

i see

[misty gorge]

so let’s try the same thing for [2]

[bleak pier]

[2]={21}?

[misty gorge]

yep, and notice that’s the same set as [1]

[bleak pier]

yes

so it will be one case or two?

[misty gorge]

okay great. what about [3]?

[bleak pier]

i meant total 4 classes?

[misty gorge]

just one. it’s the same equivalence class

[bleak pier]

or just 3?

[misty gorge]

in total you get 3, we won’t count it if we get the same set as we’ve already counted

[bleak pier]

[1]={1,2}
[2]={1,2}
[3]={3}
[4]={4}

[1]=[2]

[misty gorge]

don’t use double numbers

dont use double numbers in the [1] or [2] case either. we want to be precise with what we’re saying, so we want a subset of S. this means we really have to say “what is the collection of things in S that are the same as 1?” “what is the collection of things in S that are the same as 2?”

that will be a set, so we have to use the curly brackets, and separate things with commas clearly

closer now

[misty gorge]

remember this!

okay it gets even better, the only thing left to change is [1] and [2] to include everything

okay great

so those are the equivalence classes

[bleak pier]

is this correct now?

[misty gorge]

there are 3 different ones total

yes it is

[bleak pier]

so we can do it directely by asking that questioin

[misty gorge]

yep

[bleak pier]

so what if they ask me to count total classes?

is there any sort heck?

[misty gorge]

so you do this process, and then count the number of different sets that you get

for small sets it’s easy. as you get practice with it you might develop some tricks

but this is really exciting stuff, this is really foundational to a lot of mathematics

[bleak pier]

i see

interesting

suppose we have a set {a,b,c}

[misty gorge]

yep

[bleak pier]

we will have total 2^3=8 relations

[misty gorge]

wait uh, this is not how you count them in general

[bleak pier]

and classes will be

[misty gorge]

what do you mean exactly by relations here

we haven’t defined a relation

[bleak pier]

ohh i wrote it wrong

8 are subsets

[misty gorge]

there are 8 subsets yeah

but there are no classes because there is no relation to ask what the classes are

[bleak pier]

in the subsets ab ba same?

[misty gorge]

yes, but make sure that you use the right notation for it, because it’ll be a pain down the line to correct yourself

{a,b} and {b,a} are the same

[bleak pier]

r={aa bb cc } reflexive relation

sorry for brackets

[a]=a
[b]=b
[c]=c

3 classes

[misty gorge]

R = {(a,a),(b,b),(c,c)}

[bleak pier]

for this relation

[misty gorge]

yes

with brackets of course

[bleak pier]

yes yes

i guess this concept of classes is very important in maths

it sounds amazing

[misty gorge]

yes they’re everywhere

they come up in geometry a lot, also in algebra

in analysis even

real numbers secretly use equivalence classes a TON

you can define equivalence relations on really really big sets, and end up with a small set of equivalence classes, and then do interesting math to that

for example

if i take the natural numbers and define an equivalence relation
{(0,0),(1,1),…
(0,2),(2,0),(0,4)(4,0)(2,4)(4,2)(0,6),…
(1,3)(3,1)(1,5)(5,1)(3,5)(5,3),…
}

where i just have all the even numbers related and all the odd numbers related

then the equivalence classes are

[0] = {all even numbers}
[1] = {all odd numbers}

which is a cool example of an equivalence relation on the natural numbers

[bleak pier]

it is like partition?

i meant it sort out relations

[misty gorge]

yes

in fact equivalence relations are secretly the same as partitions

every equivalence relation gives you a partition (into “things that are equal in R’s eyes”) and every partition gives you an equivalence relation ((a,b) is in R if they’re in the same set of the partition)

[bleak pier]

so this equivalence class and partition same?

i meant in this example equivalence class and relation are same?

[midnight plank]

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[last slate]

Multiplying a × a^-1 isn't giving me identity matrix

[sharp coral]

show your work on that

[last slate]

Oh shit shit shit

Oh I was multiplying corresponding terms

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[viral dagger]

is there like a compilation of properties or like formulas for the relationship between triangles and circles, along eith triangles and their special lines

[viral dagger]

i mean like fmproperties of incircle, excircle, properties of angle bisector, ect

i keep running into questions that use properties i didnt even know exists and i just want to try to learn em all

[edgy schooner]

i guess the Mathematics textbook is a good compilation...

[viral dagger]

wdym

[last slate]

sin^2 = cos^2 - 1, u = root 1 - u ^2

[edgy schooner]

I mean that that there are some textbooks which have these relations and formulas

[viral dagger]

well i dont have the textbooks that you are talking about ig

[last slate]

I have some textbooks

I'll look it up, what do you want?

[viral dagger]

im not even sure which exists unfortunately

[last slate]

Right, holdon

Oh god

,rotate

🗿

[viral dagger]

wth

thanks ig

[midnight plank]

@viral dagger Has your question been resolved?

[viral dagger]

what are the formulas for excircles of a triangle?

[viral dagger]

how do i find the circumradius of a triangle

is it actually this

oh nvm

why the fuck is it made like that and not with a semiperimeter

[midnight plank]

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[viral dagger]

.close

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[slow swift]

steal

[tulip wigeon]

tldr

[slow swift]

hold on

[pearl hull]

Steal wat

[rain wasp]

my man there isnt a shortage of channels, no one is competing

[slow swift]

sorry, just tried to be funny

anyway

i dont know how to get the overall success rate

[rain wasp]

show "example 2.3.9"

[slow swift]

sorry?

oh gotcha

[rain wasp]

the text mentions of example 2.3.9

[junior flower]

real

[slow swift]

wait hold on example 2.3.9 isnt really relevant, my question is pretty simple: how should i calculate "overall success rate"? im supposed to be comparing a legitimate medical test, and a bogus test that just says whoever takes it doesnt have the disease

[rain wasp]

slayla

[rain wasp]

sure

[junior flower]

fungusdesu

[slow swift]

for the legitimate test, i can get p(diseased|testedpositive)

but i cant do that for the bogus one

since no one ever tests positive

[slow swift]

sorry, wdym?

[rain wasp]

no nothing, im just acknowledging you

[slow swift]

oh okay

but yea, my idea of getting the overall success rate would be to use p(d|t) and p(d^c|t^c), but i can only calculate one of those for the bogus test

is there anything u guys want me to elaborate more on

would it be better if i made this a post instead?

[midnight plank]

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[slow swift]

.close

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[gusty vigil]

for the first bit do i just it prove it by inserting the values in the gp equation? also how do i deduce them as a function of n? especilly for u_n

[gusty vigil]

,racw

,rcw

[gusty vigil]

Working

[tiny drum]

you can try just plugging in the formula for u_n into the formula for v_n, see what you get

[gusty vigil]

how do i do that for u_n when idk what the r value is

[tiny drum]

just try it

[gusty vigil]

okay

i didnt understand

[tiny drum]

so you are given u_{n+1} = 3u_n - 4

[gusty vigil]

yes

[tiny drum]

but you want u_n = ? instead of u_{n+1}

[gusty vigil]

yes

[tiny drum]

so you just change the index to get u_n = ?

then substitute that into the formula given for v_n

[gusty vigil]

yes that is how i got all of this

[gusty vigil]

what im asking is how do i deduce both of them as functions of n

[tiny drum]

you were finding particular values of v_n. I am saying not to do that, and find another formula for v_n

[gusty vigil]

oh

[tiny drum]

oh, your original post asked about the first part

[gusty vigil]

yeah mb i got the first part, now the second

Now I need to make n the subject right

[tiny drum]

should be raised to the nth power, not n - 1 (so you can get v_0), right

[gusty vigil]

but isnt the formula for gp that

.

[tiny drum]

yeah I guess so

[gusty vigil]

then how would i make n the subject

[tiny drum]

you don't

[gusty vigil]

then how do i the second part-

[tiny drum]

you've already found v_n as a function of n

[gusty vigil]

oh-

does that mean i dont need to make n the subject

[tiny drum]

no, why would you

[gusty vigil]

i thought making a function of n means making n the subject

[tiny drum]

no

if y = mx + b, y is a function of x

or if y = ax^2 + bx + c, y is a function of x

[gusty vigil]

ooooh

[tiny drum]

etc.

[gusty vigil]

okay

i still couldnt find u_n though

[tiny drum]

yes you can. you are given another formula for v_n. use that

[gusty vigil]

OHHH

okay tysm

.close

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[twilit field]

Find the number of ways in which 10 balls can be selscted from 10 identical green balls, 10 indentical blue balls and 9 identical red balls

[twilit field]

so I essentially have to find x+y+z=10

such that $x\leq 10$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

$y \leq 10 $

[edgy schooner]

combinatorics

[twilit field]

$z \leq 9$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

other than brute force

how would I do it

i.e listing out the possilble solutions are permuting them

[junior flower]

generating functions is an option but kinda need to know some special stuff to do it like that without a computer

[edgy schooner]

does the order matter?

[junior flower]

doable by hand though

[twilit field]

yeah, won't work here

all objects aren't distinct

[edgy schooner]

Do you mean that each ball isn't unique?

[twilit field]

yes

[twilit field]

any other way?

[junior flower]

there are "formulas" for problems like these that you can get out of doing generating function stuff

[twilit field]

ah

probably beyond me for now

[junior flower]

idr what they are off the top of my head

i would just have to do generating function stuff if i had to do a problem like that

[twilit field]

I did it by listing out the solutions when it came in my exam

😭

thanks

.coose

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[junior flower]

https://math.stackexchange.com/questions/203835/enumerating-number-of-solutions-to-an-equation there are stackexchange posts like this you could look through

but generating functions are super helpful for problems like this (and combinatorics in general)

[novel herald]

[midnight plank]

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[silent bramble]

for b) how can i put it to integrate y first ?

[dawn dagger]

hmm maybe try first to sketch the region

[silent bramble]

yeah that s also a problem not that good in sketching these, if there s a way to do it only by looking at the limits ...

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[tired moth]

how is the common ratio of this gp 21/20? the calculation behind it?

[midnight plank]

@tired moth Has your question been resolved?

[tired moth]

<@&286206848099549185>

[fleet niche]

according to einsteins theory of the north, you can just multiply the axes

and you get the ratio

[tired moth]

yes, im getting 21/10?

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[lone hemlock]

hi how could i solve $\frac{\pi^2}{3}+4\sum_{n\geq1}^{}\frac{(-1)^n}{n^2}\cos{(3n)}$

[grand pond]

PA

[visual tiger]

hey, if you consider cos(3n) as the real part of something...

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[hardy shuttle]

can someone help me figure out what to do with the last part

i figured out everything else, i just dont get how the last part is 8

where does it want me to plug 1 into?

I plugged it into d(x), its not 8 and I plugged it into D'(x) it's not 8

[tiny drum]

f(1)

[hardy shuttle]

thats not 8 tho

[tiny drum]

that is d(x), not f(x)

[hardy shuttle]

oh

i see

thank you haha

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[ornate finch]

imagine a coinflip game where the players have to bet or dont bet if it will land on heads(you cannot bet on tails)(however there are strategies i.e. following the hand movement that allow some people to predict better than 50 50) say when person A bets on heads he is correct 80% of the time while person B's heads bet is correct 70% of the time. person C does not have any skill in following hands but his advantage is that he knows the winrates of person A and B and is able to see if they are betting or not. what would be the best strategy for person C and what would his winrate be?

[ornate finch]

i would guess the best strategy would be to only bet when both A and B are betting, cause A has an 80% winrate and the fact a 70% winrate is also betting, should make me even more confident, but does that increase the 80% even higher for person C?

[carmine sigil]

This is a problem about conditional probability.

You find P(heads | A and B bet heads)

[ornate finch]

alright, so lets assume P(heads) is 0.5

[carmine sigil]

Unfortunately, the problem is a little bit underspecified

[ornate finch]

ye i was asking chatgpt and he said 56% chance for person C while in my mind clearly it is at least 80 if not higher, so im probably not giving enough details in the question, what information did i skip?

[carmine sigil]

In order for you to do better than 80% winrate, it cannot be the case that B only bets when A does.

Please don't ask ChatGPT

ChatGPT is horrible at math

[ornate finch]

ohh haha

[ornate finch]

can you elaborate?

[carmine sigil]

If we assume that whether A or B bets is independent of the other, we can solve this question

Otherwise, if A and B are dependent, then we would need to know how.

[ornate finch]

ok so A and B are independant, they both use their own unique strategy of i.e. following hand movements or checking the direction wind is blowing

[carmine sigil]

Do you know the formula for conditional probability?

P(X | A) = ...

[ornate finch]

i had a probability course in uni, but its long ago and i dont remember how to do stuff anymore haha

but i will recognice what the syntax means at least

[ornate finch]

so this is the chance X happens, given the A happened right? or was it reverse

[carmine sigil]

The first

[ornate finch]

ok

[carmine sigil]

P(X given A) = P(X and A)/P(A)

[ornate finch]

ahh yes i recognise that one

[carmine sigil]

Unfortunately, we don't have any way of specifying P(A) that's just information we don't have

However, we can forge ahead anyway

P(X | A and B) = P(X and A and B) / P(A and B)

And due to independence P(A and B) = P(A)P(B)

[ornate finch]

0.8 * 0.7 = 0.56

[carmine sigil]

I'm not sure this has a single answer

No, P(A) is the probability that A bets on any given throw

Which isn't specified

[ornate finch]

ahhh

[carmine sigil]

What if A only bets when they're really really confident, like 1 in every 20 throws?

[ornate finch]

what would be the issue with that?

[carmine sigil]

Might change how the two probabilities interact.

[ornate finch]

ahhh

[carmine sigil]

Did you get this problem from a problem set

Or did you make it up?

[ornate finch]

ye that is indeed what im confused about, i might not be able to know the answer then if i do not know both of their amount of bets?

[ornate finch]

i made it up, as a substitude for a more complex situation

[carmine sigil]

Ok, so imagine a given sequence of heads and tails, a random sequence where each entry is independent of the previous and following.

HTTHTHHHHTHTTHTTHTTHTHTTH

So A chooses 10 and gets 8 of them correct on average

B chooses 10 and gets 7 correct on average.

But there's an underlying sequence

That is 50/50 on average

So the more densely A and B choose, the fewer "free" heads there are, meaning that A and B become more dependent on each other

And the less densely they choose the more "free" heads there are, so they become less dependent.

[ornate finch]

ahhhh

that makes sense yes

[carmine sigil]

The fact that there's an underlying truth here makes this problem not as trivial as you had hoped

What is your original problem?

The one you are attempting to solve via analogy

[midnight plank]

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[tidal turret]

find $k \in \mathbb{R}$ such that $\lim_{n \to \infty} \left(1 + \frac{k}{n^2} + \frac{4}{n^3} \right)^{6n^2} = 4$

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[midnight plank]

@tidal turret Has your question been resolved?

[tidal turret]

I got $k = \frac{\ln(4)}{6}$

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

is it true?

[midnight plank]

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[novel herald]

every semicircle has radius 3 and is centered at the middle of an edge

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[ruby imp]

How can you tell this is absolutely convergent? I know if the sum of the absolute value of the n^th term converges, then the original series converges absolutely. So do i do the ratio test on this series but ignore the (-1)^n?

[sharp coral]

well the ratio test only says whether a series is absolutely convergent (not conditionally), although i don't think that would be the easiest to apply here

i would instead take the absolute value (which just gets rid of the (-1)^n), then use something like the limit comparison test with a p-series (which is usually a good strategy for rational functions)

[ruby imp]

so I should be comparing it to 1/n^2?

which converges

and its smaller than 1/n^2 right

[sharp coral]

that seems to be true, although i think it would be easier to establish the limit comparison

[midnight plank]

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[hardy shuttle]

If I want to foil this out, $x(11-2)(14-2x)$ do I only distirbute the x to the first paranthesis

[grand pond]

dingypine

[subtle zinc]

yes

[hardy shuttle]

thank u

.close

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[worthy wing]

No?

.reopen

[midnight plank]

✅

[hardy shuttle]

?

[worthy wing]

You have to distribute everything

After the x

[hardy shuttle]

like this?

is this what u mean?

[worthy wing]

No

You do first with 11-2x

Then u do both parenthesis

[hardy shuttle]

aha i see

thank you

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[bleak pier]

Is this question wrong?

[surreal moon]

How can the question be wrong?

[bleak pier]

How to select mode class. It is 2 times 6

@surreal moon

[midnight plank]

@bleak pier Has your question been resolved?

[bleak pier]

So how @surreal moon to select it?

Could you help me in first step?

Because for mode i have to choose the mode class then i can apply the formula

[bleak pier]

<@&286206848099549185>

[cobalt sigil]

Two modal classes !

[bleak pier]

Yes

[cobalt sigil]

Do you have the answer ?

[bleak pier]

Naa bro

[cobalt sigil]

without any extra data , both should be correct answers

Its multimodal

[bleak pier]

Have you solved out such problem before?

[cobalt sigil]

No , mode was excluded from our syllabus

is it a MCQ?

[bleak pier]

Yes

[cobalt sigil]

Then choose the correct option

Are both of them in options ?

[midnight plank]

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[scenic urchin]

anyone could help me out on the
Theorem Corollary of the Intermediate Value Theorem

[scenic urchin]

idk if it translates this way

[midnight plank]

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.close

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[rich barn]

Can anyone show me how to do this

[midnight plank]

Please don't occupy multiple help channels.

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[bleak pier]

what is the difference between identity relation and reflexive relation?

[surreal moon]

For any set $X$, a relation $R$ is just a subset of $X\times X$. The relation $R$ is said to be reflexive if, for all $x\in X$, $(x, x)\in R$. A relation $I$ is an identity relation of $X$ if, for all $x, y\in X$, $(x, y)\in I$ if and only $x=y$.

[grand pond]

SWR

[surreal moon]

In short, every identity relation is reflexive, but not every reflexive relation is an identity

[midnight plank]

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[primal nest]

In a triangle ABC, AB = 9, BC = 3 and AC = 7. The points M on
AB and N on BC are located such that MN //AC and AM =BN .
What is the value of MN ? I calculated all the angels but im stuck here.

[midnight plank]

@primal nest Has your question been resolved?

[bleak pier]

Can you tell me an example of it?

[midnight plank]

@primal nest Has your question been resolved?

[midnight plank]

@primal nest Has your question been resolved?

[primal nest]

can anyone help?

[last slate]

The two triangles are similar by AAA property, right? Do you think we can use that?

[last slate]

The answer is 5.25 , I think

[primal nest]

how did you come to that?

[last slate]

Do you know the answer?

So I can double check?

[primal nest]

no i do not

[last slate]

Do uk the concept of similarity is triangles?

[primal nest]

yeah

[last slate]

Look closely, triangle BMN and BAC are similar

Can you see it?

Angle B is common

[primal nest]

yes

[last slate]

Angle N and C are corresponding angles

BNM and BCA,I mean

[primal nest]

yes N and C are the same

[last slate]

And BMN and BAC are equal because the parallel lines are interesting the same line, so the angle at which they intersect would be the same

So since all corresponding angles are equal, by AAA property we can conclude that they are similar

With me so far?

Since the triangles are similar, the ratios of corresponding sides are equal

[primal nest]

yes i think im on to something

[last slate]

This is what you should do after that Ig

@primal nest did you understand?

(9/4)/3 = 9/4 x1/3 btw

If you think I made a mistake or if you didn't understand something, let me know

[primal nest]

i got it thank you very much

[last slate]

No problem!

[primal nest]

.close

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[calm cypress]

not sure how i would go by this

the first 1 i get

it would be b

but the 2nd one is it d?

[round garnet]

d or f

cuz longest side corresponds to largest angle

[calm cypress]

yea thought so

and a is right?

[round garnet]

yea

[calm cypress]

alr 1 more

is this just regular cos and tan etc

[round garnet]

a for a

f for b

[calm cypress]

howd you get that tho

[round garnet]

cause the longest side has the largest angle

smallest side has smallest angle

and angle in between has the angle in the middle

so if u use that logic in reverse

[last slate]

Nooo the longest side is opposite to the largest angle

[round garnet]

yea

thats what i said

[calm cypress]

so for a the greatest angle is b

[last slate]

In a right angle triangle, 90 degree angle is always the largest angle

[calm cypress]

and least is a

[last slate]

Yes, not a

[calm cypress]

uhhh ok

[round garnet]

i found smallest angle mb i was doing part 2

yea so basically just know that side lengths correspond directly to angles

if the side length is the longest the angle opposite is the largest and angle'

applies to every side

[calm cypress]

ok so d is longest

f is smallest

[round garnet]

yea

[calm cypress]

ok perfect

have a good one

.close

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[calm cypress]

.reopen

[midnight plank]

✅

[calm cypress]

i got another question

for this can i just move the x to the sin and multiply sin30x12

and how would i solve this

cross multiply?

then isolate?

[glossy jacinth]

Yes. 12*5/3 = x => 60/3 = x => x = 20

[glossy jacinth]

sin 30 = 1/2

[calm cypress]

but can i use the other method

[glossy jacinth]

=> 12 / 2 = x => x = 6

[calm cypress]

x= sin30x12

[glossy jacinth]

Yes, this is correct. Just put in the value for sin30.

[calm cypress]

my calc does that automaticlly

thanks though

.close

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[sharp wagon]

hey im stupid, could i have some help with this?

[sharp wagon]

yeah i dont rememebr how to find it tho :\

yeah

aight so 261

so then what do i do?

i dont know ;-;

dividing ?

squaring?

like about 16

number divided by that number

[grand pond]

faiyrose

faiyrose

faiyrose

Result:

16.155494421404

[sharp wagon]

yeah i think i get it, thanks.

[midnight plank]

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[pearl star]

just the answer please im in a roblox game

[oak ether]

okay

ill try

[pearl star]

sure

[oak ether]

not sure

[pearl star]

bro

[still nacelle]

!nosols

[midnight plank]

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

[pearl star]

oh

okay ill just work with you then

[tawny hedge]

hopefully u win

[pearl star]

thanks 🙏

[still nacelle]

ok, do you know any properties of determinant that can be used in this question?

[bright gust]

dayum boi

[pearl star]

i think so yes

[still nacelle]

ok which ones do you know

[pearl star]

u multiply 3 with x and

6 with 4

[oak ether]

ik

[pearl star]

for a

[pearl star]

buddy stop typing here

[oak ether]

i asked copiolate

look

[tawny hedge]

💀

[still nacelle]

!nogpt

[midnight plank]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[worthy kestrel]

🍞

[pearl star]

bro

😭

[tawny hedge]

aint no way bro asked copilot