help-42

[simple musk]

Ker(f)=<(0,0,1,0)>

I repeated the wrong vector lmao

again

f(1,0,1,-1) = (-1,0,0,1)
f(0,1,0,0) = (0,0,1,0)
f(0,0,1,0) = 0
f(0,0,0,1) = (0,0,1,0)

Im(f)=<(-1,0,0,1),(0,0,1,0)>

Ker(f)=<(0,0,1,0)>

S =<(-1,0,0,1),(0,1,0,0),(0,0,1,0)>

Ker(f) ⊆ Im(f) ⊆ S

now we need to check if (1,0,1,-1) not in Im(f)

and if {(1,0,1,-1),(0,1,0,0),(0,0,1,0),(0,0,0,1)} is a basis of R4

I repeated the wrong vector

again

f(1,0,1,-1) = (-1,0,0,1)
f(0,1,0,0) = (0,0,0,1)
f(0,0,1,0) = (0,0,0,1)
f(0,0,0,1) = 0

Ker(f) = <(0,0,0,1)>

Im(f) = <(-1,0,0,1),(0,0,0,1)>

S =<(-1,0,0,1),(0,1,0,0),(0,0,1,0)>

Ker(f) ⊆ Im(f) ⊆ S

again bro

I repeated wrong vector

f(1,0,1,-1) = (-1,0,0,1)
f(0,1,0,0) = 0
f(0,0,1,0) = (0,1,0,0)
f(0,0,0,1) = (0,1,0,0)

ker(f) = <(0,1,0,0)>

im(f)=<(-1,0,0,1),(0,1,0,0)>

S =<(-1,0,0,1),(0,1,0,0),(0,0,1,0)>

Ker(f) ⊆ Im(f) ⊆ S

yes bro

now we need to check

(1,0,1,-1) not in Im(f)

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

i) 1 = -a
ii) 0 = b
iii) 1 = 0 + 0
iv) -1 = a

we did it

let's check if {(1,0,1,-1),(0,1,0,0),(0,0,1,0),(0,0,0,1)} it's a basis of R4

,w rank {{1,0,1,-1},{0,1,0,0},{0,0,1,0},{0,0,0,1}}

[potent lotus]

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[simple musk]

nice bro

another solved one

.solved

[calm coral]

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[simple musk]

let $\mathbb{H} = { x \in \mathbb{R}^4 \mid x_1 + x_2 -2x_4 = 0 }$ and $\mathbb{S} = \langle (1,0,1,0),(0,1,2,-1) \rangle$ Define a linear transformation $f : \mathbb{R}^4 \to \mathbb{R}^4$ such that $\operatorname{Im}(f) = \mathbb{H}$ and $\mathbb{S} \subseteq \operatorname{Ker}(f \circ f)$

[remote mural]

i wanna b this intelligent but am only 16 💔

[swift dragon]

intelligence doesnt come with age, knowledge does

(generally)

[remote mural]

very bad at maths and dk where to start

[potent lotus]

938c2cc0dcc05f2b68c4287040cfcf71

[swift dragon]

first step is to not intrude onto other's help channels

[simple musk]

x1 = -x2 + 2x4

[swift dragon]

!help or chat a bit in #discussion

[calm coral]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[simple musk]

<@&268886789983436800>

(x1,x2,x3,x4)=(-x2+2x4,x2,x3,x4)

x2(-1,1,0,0)+x3(0,0,1,0)+x4(2,0,0,1)

H = <(-1,1,0,0),(0,0,1,0),(2,0,0,1)>

f(v1)=w1
f(v2)=w2
f(v3)=w3
f(v4)=w4

im(f) = H

S ⊆ ker(fof)

this kernel of fof condition is tripping me up

that's all v such that

f(f(v))=0

ker(f) ⊆ ker(fof)

but ker fof is bigger than ker f

is it me or ker(fof) = ker(f)??

yeah i guess that was the trick

im(f)=H and S ⊆ ker(f)

f(v1)=w1
f(v2)=w2
f(v3)=w3
f(v4)=w4

f(v1)=(-1,1,0,0)
f(v2)=(0,0,1,0)
f(v3)=(2,0,0,1)
f(v4)=0

f(v1)=(-1,1,0,0)
f(v2)=(0,0,1,0)
f(v3)=0
f(v4)=0

S = <(1,0,1,0),(0,1,2,-1)>

f(v1)=(-1,1,0,0)
f(v2)=(0,0,1,0)
f(1,0,1,0)=0
f(0,1,2,-1)=0

ker(f(f(x)) all x st f(f(x))=0

ker f(x) all x such that f(x)=0

f(x) = a

f(a) = 0

stop tripping, ker fof = ker f

period

anyways

Im(f) = H

S ⊆ ker(f)

[calm coral]

@simple musk Has your question been resolved?

[simple musk]

I need some help proving ker(fof) = ker(f)

[calm coral]

@simple musk Has your question been resolved?

[calm coral]

@simple musk Has your question been resolved?

[calm coral]

@simple musk Has your question been resolved?

[simple musk]

f(0,1,0,0) = (-1,1,0,0)
f(0,0,1,0) = (0,0,1,0)
f(0,0,0,1) = (2,0,0,1)
f(1,0,1,0) = 0

[calm coral]

@simple musk Has your question been resolved?

[calm coral]

@simple musk Has your question been resolved?

[idle yarrow]

thats not true

[simple musk]

yeah

[idle yarrow]

do u know rank-nullity?

[simple musk]

only ker f ⊂ ker fof

[simple musk]

yeah, dim(im) + dim(ker) = dim(dom)

[idle yarrow]

yeah

so what should the nullity(f) be?

and whats nullity(f^2)?

f^2 = f compose f btw

[simple musk]

wait wait

nullity as in dim kernel, right

[simple musk]

dim ker = 1

[simple musk]

so?

[simple musk]

thats >= 2

[idle yarrow]

Yes

So ker(f) is a proper subset of ker(f^2) then

[simple musk]

yeah

ker f ⊂ ker fof

[idle yarrow]

U can actually determine the exact dimension of ker(f^2)

Can u see why?

[simple musk]

how ?

[simple musk]

bro, please elaboratee

[idle yarrow]

well do u know of the identity dimker(ST)<=dimkerS+dimkerT?

[simple musk]

https://math.stackexchange.com/q/1965966/1383337

[simple musk]

is something I just realized is true, though Is still an inequality

[idle yarrow]

but what does it say for ur current problem?

[simple musk]

for our purpouses
dim(ker(fof)) <= dim(ker(f)) + dim(ker(f))

and dim(ker(f)) = 1

so dim(ker(fof)) <= 2

[simple musk]

and we also know S ⊂ ker fof
dim(S) <= dim(ker(fof))
2 <= dim(ker(fof))
dim(ker(fof)) >= 2

hahah

and since we know dim(ker(fof)) >= 2 and dim(ker(fof)) <= 2

that must be only possible if dim(ker(fof))=2

[simple musk]

so?

[idle yarrow]

can u say anything more about ker(f^2) and S?

[simple musk]

they are equal

since their dimensions are equal they describe the same subspace

or maybe not, just give me a sec to think

S ⊂ ker fof

but they are same dimension

they must be equal

but unless ker fof ⊂ S we will never know

[idle yarrow]

well what is the definition of the dimension of a vector space?

[simple musk]

the length of the basis, so?

[idle yarrow]

as it says, any two bases of a finite-dimensional vector space are of the same length

moreover, if i have a basis and a linearly independent set with the same length, what can u say about the linearly independent set?

[simple musk]

what do you want me to say? they are equal to each other?

[idle yarrow]

so that proves this

[idle yarrow]

actually no

its that the set is also a basis

thats what proves ur claim

[simple musk]

okay

[idle yarrow]

so u know ker(f^2)=S

[simple musk]

S = <(1,0,1,0),(0,1,2,-1)> = ker(fof)

[idle yarrow]

and ur idea so far has to find a basis and define the elements f maps this basis to

[simple musk]

yeah exactly

[simple musk]

any ideas?

[idle yarrow]

u can construct a basis one vector at a time

u probably first want to describe a basis of ker(f) right?

[simple musk]

well

[simple musk]

which basis vector I send to 0

S = ker fof

what about ker f?

[idle yarrow]

again, whats its dimension?

[simple musk]

dim ker f = 1

[simple musk]

what about it

[idle yarrow]

so any basis of ker f should be just 1 vector

but what properties does this vector have to satisfy?

it cant be any arbitrary vector, right?

[simple musk]

well

[simple musk]

since Ker(fof)=S

and ker(f) ⊂ ker(fof)

then ker(f) ⊂ S

[simple musk]

should be in S

[simple musk]

problem is, which vector in S to map to 0

[idle yarrow]

well we dont need to decide that yet

lets just call it v1

so ker f = span(v1)

[simple musk]

v1 ∈ S

[idle yarrow]

now i want another vector

call it v2

[simple musk]

why not the angle notation <

[idle yarrow]

idk i thought u used axler

but wtv

i can use angle notation

can u think of something v1,v2 should span?

like what subspace makes sense next

[simple musk]

I am unsure

wdym?

where did v2 came from

and why we want a second vector?

[idle yarrow]

cuz in the end u want a basis v1,v2,v3,v4 of R^4

and u want to know f(v1) to f(v4)

rn u know f(v1)=0

[simple musk]

no

[idle yarrow]

and u have subspaces kerf, S=kerf^2, H of R^4

currently kerf=<v1>

dimS = 2

and dimH = 3

also kerf is a subset of S

if i have an element of S\kerf, what should f map it to?

[simple musk]

S\kerf?

[idle yarrow]

S minus kerf

the set of vectors in S but not in kerf

[simple musk]

to H

[idle yarrow]

thats true

also,

f^2 maps it to 0 right?

[simple musk]

the vectors in S\kerf?

the vectors of S\kerf are mapped to 0 under fof

[idle yarrow]

yes

[simple musk]

why did u brought up fof

[idle yarrow]

so if v2 is in S\kerf, u know f(v2) is in imf=H, but also, f(f(v2))=0

[simple musk]

I thought we were over with fof

[idle yarrow]

so what else do u know about f(v2)?

[simple musk]

f(v2) ∈ Ker(f) = <v1>

[idle yarrow]

great

so maybe lets just say f(v2)=v1

since i can scale v1 arbitrarily

but then f(v2) is in H intersection ker(f), so what new condition does v1 have to satisfy

before we only said v1 is nonzero and in ker(f)

[simple musk]

v1 ∈ H ∩ ker(f)

v1 ∈ Im(f) ∩ Ker(f)

[idle yarrow]

and also H intersection S

what can u say about H intersection S in ur original problem?

[simple musk]

S = ker(fof)

[idle yarrow]

ie what does it equal

[simple musk]

why

[idle yarrow]

how does kerf relate to S?

[simple musk]

well, ker(f) ⊂ ker(fof)

and ker(fof) = S

so ker(f) ⊂ S

v1 ∈ H ∩ S

[simple musk]

idk, is prolly dim 1

[simple musk]

we would need to find it

[simple musk]

H ∩ S = <v1>

[idle yarrow]

well numerically, using the vectors given in the question

[simple musk]

how

so I find the intersection between S and H or no?

you just gotta write S as a generic point in R4 and make it satisfy equation in H to find the constraint and put the constraint back into the generic point of S

s in S, a,b in R

s = a(1,0,1,0) + b(0,1,2,-1)
s = (a,b,a+2b,-b)

H : x1 + x2 -2x4 = 0

a + b -2(-b) = 0

a+b+2b=0

a+3b=0

a = -3b

s = (-3b,b,2b-3b,-b)

H ∩ S = <(-3,1,-1,-1)>

(-3,1,-1,-1) = -3(1,0,1,0) + (0,1,2,-1)

(-3,1,-1,-1) = (-3,0,-3,0) + (0,1,2,-1)

[simple musk]

so?

[idle yarrow]

M8 u gotta start thinking for urself

U have 2 vectors v1,v2 for which u know their image from f, ie f(v1)=0, f(v2)=v1
And u know v1 is in <(3,1,-1,-1)>

U just need to find v3, v4

The last condition u need to satisfy is for H=im f

So think about that

And figure it out

[simple musk]

v3 = e1, v4 = e2

you can just use standard basis vectors

as for the image

f(3,1,-1,-1) = 0
f(h3) = (3,1,-1,-1)
f(e1) = h1
f(e2) = h2

where H = <h1,h2,h3>

[simple musk]

I am trying, is just hard exercise bro

[simple musk]

solved in #linear-algebra

.solved

[calm coral]

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[manic pewter]

hlep please

there is not asolution

online

I ended up getting 3/91

but there is not answer choice for that

I baseically did 6 choose 2 * 4 choose 2 and divided that by 2 and then divided all of that by 15 choose 4

is there a mistake in my thinking

the 15 choose 4 is essentially choosing 4 chords from the 15

and 6 choose 2 is choosing 2 of the points of the 6

and then another 2 for the 4 choose 2

so then we can just find those two choords, and the rest will connect themselves. as in we find two, and the other two are decided

and then divide by 2 for repetition

what is wrong with my answer

is it just simply a computational error?

help would be much appreciated

thank you

[remote mural]

this is where you are making a mistake

you see to account for repitition or indistinguishable chord sets...it has to be divided by 3!

Let me explain u with small examples

Say you have 4 people A,B,C,D and you have to create 2 teams with 2 members each ...you go on with (4C2*2C2) / 2! ....we divided the whole by 2! because whether we select A,B with 4C2 or we select C,D with 4C2 , both results in same distribution of teams...teams here are indistinguishable to each other

Similarly say now you have A,B,C,D,E,F 6 people and you have to create 3 teams of 2 members each...you will do (6C2*4C2) /3! ....we divide by 3! here to account for 3 indistinguishable teams here

NOW another case which is more valid and directly links to the Q is if out of 6 people, we need to create 2 teams of 2 members each...last 2 would be left behind...this still doesn't change the answer as we want all 3 groups (2 teams and the group of 2 people left out) to be indistinguishable so we still do the same thing i.e (6C2*4C2)/3!

[manic pewter]

huh

thats weird

interesting

so how might I go about thinking about this in the future

and how can I avoid mistakes like these

thank you very much for the helpful response, and taking the time to write that by the way

[remote mural]

Just remember when, why, and how we account for indistinguishability.

so now you know which option is correct?

[calm coral]

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[meager quarry]

anyone?

[frosty spruce]

are there something missing in those boxes or it's just an indefinite integral?

[meager quarry]

its an indefinite integral

@frosty spruce

[frosty spruce]

I just tried weierstrass substitution and it failed very quickly

oof

[meager quarry]

never heard that lol

[frosty spruce]

https://en.wikipedia.org/wiki/Tangent_half-angle_substitution

[meager quarry]

oh , havent taught us this yet

maybe in college

[frosty spruce]

you should have met this before, since this integral is very high-level

[meager quarry]

nope , never , this is a 12th grade integral

[frosty spruce]

no way

[meager quarry]

it is haha

and there are endless questions like these , in 12th grade

[frosty spruce]

but they shouldn't be that hard

i used Wolfram Alpha to solve it and it gave me a monstrosity

[meager quarry]

never heard of it either lmao

[meager quarry]

oh hell naw , the answer isnt even close

the answer is 2(tanx)^1/2

[frosty spruce]

Then it's probably the problem's fault

[meager quarry]

nope , alr checked it

[frosty spruce]

that looks so wrong

💀

[meager quarry]

it is what is given in the book

[frosty spruce]

if you differentiate it it doesnt even come close

[meager quarry]

have you heard of a method , where we take a part and put it equal to a variable , then differentiate it , and then try to make its differential in the integral along with the value we chose to put in the variable

[frosty spruce]

I hope this makes enough sense to you

the only way that is the answer is if the problem doesn't have the minus sign

[meager quarry]

idk bruh , i'll ask the teacher once

[frosty spruce]

k

[meager quarry]

i have another one tho if you can

[frosty spruce]

though, just in case, i'll solve the one without minus sign if you need

[meager quarry]

no no its fine , i'll cross check it with the teacher

[frosty spruce]

alr

[frosty spruce]

Hint: sin(2x)=2sin(x)cos(x), then let u = cos(x), du = -sin(x)dx

[meager quarry]

oh damn , i was doing it right then

thanks , i tend to not beleive myself sometimes lmao

[frosty spruce]

np

[meager quarry]

.close

[calm coral]

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[calm coral]

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[honest lotus]

Need help with precalculus analytic geometry

[ivory saddle]

Did you solve for where the intersection is

[honest lotus]

Nope

I have 0 clue on what to do

[ivory saddle]

Well you have 2 straight line equations that intersect each other at some point (x,y)

You want to solve the equation of a third like that passes through that point

And said 3rd line has a gradient of 4

So you want to make eq1=eq2

In order to see where that intersection is

[honest lotus]

okay hold on let me try sumn

okay didnt work out

how do I find the point of intersection between the two equations

[honest lotus]

I had an idea with this

but do I need to convert the equation to y=mx+b form?

[flat gorge]

the equation for that line would be (x + 6y - 4) + k(3x - 4y + 2) = 0, make slope = 4

put slope = 4, and sub value of k

[honest lotus]

okok

[flat gorge]

so next time you might wanna save time on evaluating the intersection point and doing y = mx + b stuff

[honest lotus]

aight

.close

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[tulip herald]

any ideas on what to do with this

[manic pewter]

yes, thank you very much

just to clarify, we would do this if we had 3 left over right

so if it was 7 choose 2

and 5 choose 2

then would we still divide by 3!

or is it just when the amount being chosen is equal

[tulip herald]

.close

[calm coral]

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[manic pewter]

A square is tiled by 24 congruent rectangles. points A and B are the midpoints of a pair of opposite sides of the square on each side of the line AB, four rectangles are chosen at random and colored black. The square is then folded over line AB. compute the probabilty that exactly one pair of black rectangles is coincident.

[manic pewter]

I need help verifying my answer to this

I got 24/2475

@remote mural

I tried to use your idea

where you divide by 3!

except for this problem I tried dividing by 4 fact

so

idk

could anyone please check and lmk if the answer is correct

thanks

[gilded atlas]

Do we have the rectangles configurations?

[manic pewter]

nope, thats all

they are just arranged inside it

[manic pewter]

I mean they are symmetric

from what I imagine

12 on one side

and 12 on the other

I dont quite think that it would be split down the middle

or weirdly

[gilded atlas]

we still have to prove it tho

[manic pewter]

its split down the axis of symmetry of the square

and they are all congruent

[remote mural]

Hey

I believe it should be 20/483

favourable ways of selecting = 24* 1* 22 * 20
Total ways= 24* 23* 22* 21
P(exactly 1 pair coincide) = favourable/total =20/483

[calm coral]

@manic pewter Has your question been resolved?

[gilded atlas]

first case: 3 on one side, 1 on the other

• pick 3 rectangles on one side: 12C3
• pick 1 corresponding rectangle on the other side: 3
• we can do this in 2 ways (3 on the right & 1 on the left, or vice versa): 2
  => total for this case: 12C3 * 3 * 2
  second case: 2 on each side
• total pairs (both matching and not): (12C2)^2
• pick 2 on one side: 12C2
• pick 2 on the other side that don’t correspond to the first 2: 10C2
  => subtract matching cases: (12C2)^2 - 12C2 * 10C2
  now, the probability: (12C3*3*2 + (12C2)^2 - 12C2 * 10C2) / 24C4

So... 41/161

[remote mural]

hmm, i think there are 4 rect on each side of the line?

or did i misread

[remote mural]

for your latex you should use #latex-testing or #latex-help

[gilded atlas]

Oh... exactly 1 match... hang on

[remote mural]

oh ok, ty

[remote mural]

yeah!

$\frac{\binom{12}{1} \times \binom{11}{3} \times \binom{8}{3}}{\left(\binom{12}{4}\right)^2}$

[potent lotus]

raydohad

[remote mural]

is this the answer?

you can model this questions like this also i think
consider you have two bags with numbers 1 to 12
you take 4 numbers from eachbag
probability that exactly 1 number is same

i think

[gilded atlas]

first case: 3 on one side, 1 on the other

• pick 3 rectangles on one side: 12C3
• pick 1 corresponding rectangle on the other side: 3
• we can do this in 2 ways (3 on the right & 1 on the left, or vice versa): 2
  => total for this case: 12C3 * 3 * 2
  second case: 2 on each side
• total pairs (both matching and not): (12C2)^2
• pick 2 on one side: 12C2
• pick 1 on the other side that don’t correspond to the first 2: 10C2
• pick 1 corresponding rectangle on the other side: 2
  => total for this case: 12C2 * 10C2 * 2
  now, the probability: (12C3*3*2 + 12C2 * 10C2 * 2) / 24C4

[remote mural]

is this not the case? im confused by your answer @gilded atlas

[remote mural]

there are in total 4 rect

oh

[calm coral]

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[remote mural]

n each side of the line AB, four rectangles are chosen at random and colored black
i think it means 4 on one side 4 on other?

[calm coral]

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[molten sphinx]

What exactly am I doing here?

[indigo grove]

You input x in g(x) then you input the the value of g(x) in f(x)

[molten sphinx]

I just need to do it for one of the values in each collumn?

[indigo grove]

Yep you take g(3) then f(g(3))

[molten sphinx]

I got it

.close

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[floral turtle]

yo

[floral turtle]

How do u solve this?

I need help with b and b

Top and bottom

I'd really appreciate it

[glad parrot]

For the first one

Consider 12 = 3*4 and use log rules

[glad parrot]

The first rule here to be exact, thx yobo

[floral turtle]

Huh?

I don't understand

[glad parrot]

Write 12 as 3*4

[floral turtle]

Hmm alr

One sec

[glad parrot]

And then use the first rule in the image that yoboi linked

[floral turtle]

I have a headache already

[glad parrot]

No 3*4

Not 3⁴

[floral turtle]

Alr

Now what?

@glad parrot

<@&286206848099549185>

Anyone?

I need help for 2 questions asap

My exam is on Monday and I got so much to study for

[regal bronze]

Use the expansion property (the first one) and re write the log(3*4)

[floral turtle]

Idk how to do that

What's x and y?

I got 2, 3, and 4

3 numbers

2 letters

Which one is x and which one is y

Wtf is this shit

[regal bronze]

x=3 y=4

2 is the base of your log
Its "a" in the page

[floral turtle]

Alr

What now

[regal bronze]

You have -log3 in the question
These two cancel each other out

[regal bronze]

And use the third property for the log4

[floral turtle]

?

What r u talking about

-log 3?

[regal bronze]

Here

[wheat owl]

Log2(12)-log2(3) = log2(12/3)=log2(4)=log2(2^2)=2log2(2)=2

[floral turtle]

So if it's a minus Infront

[wheat owl]

3log2(2)=3

[floral turtle]

It crosses out?

[regal bronze]

Why

[wheat owl]

So we have 3+2

5

[floral turtle]

U guys are making me even more confused

I need one person explaining at a time

Idk who to listen to now

[wheat owl]

Basic rules

[regal bronze]

Yes you have
3log2 + log3 + log4 -log3

[wheat owl]

Just like log2(3)+log2(4)=log2(3*4)

[floral turtle]

So were left with logbase 2 4

[regal bronze]

Dont give the answer this fast

[regal bronze]

The 3rd property in the page
That says
Logbasea x^n =n logbasea x

[floral turtle]

I give up

[regal bronze]

And 4=2²

[floral turtle]

Fuck this exam

[regal bronze]

Lol

[floral turtle]

Ima skip this section

Doesn't make sense

I'm gonna lose it

[wheat owl]

Good job

[floral turtle]

Can someone write it out cuz wtf

Or we can get someone who actually knows how to do this

And knows how to explain SIMPLY

[wheat owl]

Go watch a video about logarithms properties

And stop yelling

[regal bronze]

[wheat owl]

In less than 10 mins you'll be able to solve this

[floral turtle]

Nah cuz they don't do the same question

Skip this I'll figure it out later

Can u help me out with a different question should be easier

[wheat owl]

Well you gotta learn how to solve it
Not just memorise the question

[floral turtle]

b)

Bottom one

[wheat owl]

Lemme show you

[floral turtle]

I'll watch a video in a bit

[floral turtle]

The bottom one?

[floral turtle]

Wtf is that

[wheat owl]

The solution

[floral turtle]

Huh

Same question?

Where did those extra numbers cone from

[wheat owl]

Yes
The bottom one

The powers

Trust me

[floral turtle]

Shit just made my day worse

I'm skipping this question as well

Ima do the easy ones

[wheat owl]

Go watch a video about the logarithms properties...

[floral turtle]

Couldn't care less

[wheat owl]

It would be like 10 mins

Explaining what happens

It's nothing complicated fr

[floral turtle]

Not allowed any formulas either

[floral turtle]

The vid doesn't explain that question

[wheat owl]

https://youtu.be/Jtv9Lnf7Zw8?si=Xw_JGxXjEjGzi0D-

[floral turtle]

With the sqaure roots or whatever

Ok forget this

[wheat owl]

Srt of x equals x^1/2

[floral turtle]

Let's do something that's not logs

[wheat owl]

Ok...

[floral turtle]

I'll be able to choose the difficulty level on the exam

So let's do 7

[wheat owl]

Okay

P = 610,000 + 610,000*0.026t

[floral turtle]

Yep

[wheat owl]

That's a

In b

You'll let the function p equal 1 million

And solve for t

In c you'll let the function p equal 2*the population given

[floral turtle]

That's a?

Thats it?

[wheat owl]

Yeah that's the function?

The equation that represents the population

[floral turtle]

Alr

Ok lemme try out b

Yea nvm

Idk

Do I plug in 19 as t?

Nvm idk wtf I'm doing

@wheat owl

[wheat owl]

Hey

In b

It needs the time the population will reach 1 million

[floral turtle]

How am I supposed to know

[wheat owl]

And we have an equation that represents the population right?

[floral turtle]

If it doesn't tell me when

[wheat owl]

Stay with me and calm

[floral turtle]

Yes

Alr

Just finished my English exam that's why I'm like this mb

[wheat owl]

So then
If we need to find the time it takes for the population to reach a million we need to make the equation that represents the population equal 1 million
And get T which is the time taken

Since as it said
T represent the time in years

[floral turtle]

That's what I did

What do I plug in as t tho?

[wheat owl]

You don't plug a value as t

[floral turtle]

Oh

[wheat owl]

You make
610,000 +610,000*0.026T=1,000,000

T is like x

[floral turtle]

Ohhh

[wheat owl]

And you solve for it

[floral turtle]

Ok one sec

How 💀

Alr one second

[wheat owl]

okay...

Ping me

[floral turtle]

Do I divide by 610000?

@wheat owl

[wheat owl]

Look

The equation is

610,000 +610,000*0.026T=1,000,000

[floral turtle]

Oh

[wheat owl]

You have to subtract 610,000 from both sides

Then divide

By what's multiplied by t

Which is 610,000 * 0.026

[floral turtle]

Wym subtract?

Like I'm confused

Subtract 610k from 1m as well?

[wheat owl]

Kush
Do you know basic math? 😭

How old are you

[floral turtle]

Nah but like

There's 2 610ks

Which one do I get rid of?

English exam fucked me up this is from 5 lessons ago

I did this back in October I don't remember

[wheat owl]

Idk how to simplify this more
To solve for x
You make X only in one side and the other numbers in the other side

Here we have a number +T times a number = another number

So first we need t times a number in one side

[floral turtle]

Just write it out

[wheat owl]

So we will subtract the first number from both sides

[floral turtle]

I can't learn like this tbh

[floral turtle]

Ok there

U could've said this

[wheat owl]

Whatever you say kush

[floral turtle]

Now we divide by 610k

6.393

610k ÷ 390k

Yo @wheat owl

Is this right?

Same thing for c I just multiply 610k by 2 and add that instead of a million

Yo u there?

@wheat owl

This exam bouta be light

.close

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[remote mural]

What do I do with the 0?

[swift laurel]

what about it?

[remote mural]

Like

Do i put ot in the f(x) thingy

[swift laurel]

every root should appear in the factored form of the polynomial

[flat gorge]

T_T is it a mark the incorrect option question?

[remote mural]

😭 um idk

I'm doing this based off of the notes the teacher made is do

Us*

[flat gorge]

why you marked wrong on b, c is the y - axis perhaps upside down

[remote mural]

Wakt

Huh

[solid goblet]

wait.? wont the leading coefficient be negative?

[remote mural]

No?

This is the notes our teacher told us to write

So if the x int is -4 in the equation it becomes +4??

[remote mural]

I'm cooked

[solid goblet]

the graph is continuously decreasing for large values of x so that's only possible when the leading coefficient is negative?

[solid goblet]

this what im talking about^

[flat gorge]

2) As x -> ± inf, your fn goes -inf but x^4 is dominating, so leading coeff should be -ve```

refer to Cid's figure

[remote mural]

Ope

I see

[flat gorge]

Also, for the fourth part, it's just factor theorem

[remote mural]

Factor all of x int?

[flat gorge]

hmmm?

[remote mural]

Idk.. I haven't understood this at all in class 😭

[solid goblet]

yeah you can even try to desmos the graph with k<0 it should be negative

[remote mural]

Okok

.close

I'll try again later I'm getting out of school rn

[calm coral]

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[flat gorge]

Baii

[remote mural]

Baii

[calm coral]

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[devout meteor]

i know that this is the formula but i dont know how to find any of the variables

[tidal grotto]

try drawing a picture of the scenario :)

https://tutorial.math.lamar.edu/classes/calci/volumewithrings.aspx this also may help

(this has some worked through examples)

[devout meteor]

i have an example on my notes but it doesnt have 3 functions

but thx

[tidal grotto]

oh those aren't 3 functions

well two of them are

x=4 is a vertical line

y=0 is a horizontal line (function)

and y=x^2 is a function

[devout meteor]

y=x^2; x=4; y=0

[tidal grotto]

yes

x=4 isn't a function with respect to x

as i said, drawing a picture will help to figure out what you need. (it may even be simpler than you think)

[tidal grotto]

yes

and you're rotation this region around the x-axis and getting a volume

[devout meteor]

okay so why does it say to use the washer method if there isnt a radius on the inside? because we r rotating on the x axis and the function touches the x axis

oh it says disk OR washer method

do the disk method here works best

[tidal grotto]

it would seem so

[devout meteor]

so like this right?

okay i got it thx

.close

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[indigo ginkgo]

can i get some help with this exercise?

[jagged raptor]

could you tell me a bit about what you've already tried or what parts are confusing you?

[indigo ginkgo]

yep yep, so i know that b) e) and f) are wrong

but the thing is that i dont know how to prove a) c) or d)

i really dont understand how i can approach the problem of is det B equal or diff from 0

[indigo ginkgo]

any ideas?

[jagged raptor]

you're right b, e and f are indeed false when det(a) = 0

let's quick recap what it means for det(a) = 0

imagine a matrix as representing a transformation of space. The determinant tells us how much matrix scales areas (in 2D) or volumes (in 3D), and so on

If det(A) = 0, it means this transformation squishes space down in some direction, essentially collapsing dimensions

this has some important consequences for things like inverses and solutions to equations

[indigo ginkgo]

i dont need to bend space time to know that a matrix is invertible only if det A diff 0

its pretty basic actually

[gilded abyss]

How do I solve this using the graph?

[indigo ginkgo]

not the right channel buddy

[jagged raptor]

my apologies if i was going too basic there

[indigo ginkgo]

no no

u were actually saying some really complicated stuff

that i didnt quite understand

i dont know how a matrix is squishing space into a direction

but, does it have relevance regarding this exercise?

[jagged raptor]

hmm

squishing space it's a way to interpret what a determinant of zero means geometrically, but it's not directly needed to solve this problem

[jagged raptor]

not really in terms of directly proving the answers

it's more background understanding

for this exercise, we really need to use the properties of determinants and matrix operations more directly, like you already know about invertibily and determinants

forget about "squishing space" and focus on what you know about determinants and matrix multiplication

[indigo ginkgo]

right

so how do i solve this

[jagged raptor]

a) There exists B ∈ M_n(R), with B ≠ O_n, such that AB ≠ O_n

in simpler words, this is asking "" can we find some matrix B which is not the zero matrix, such that when we multiply A by B, the result is also not the zero matrix?

think about the identity matrix

[indigo ginkgo]

i just checked with 2 x 2 matrixes and yes it works

which would mean that c) is wrong

but how is d) wrong then

why would det B have to be 0

[jagged raptor]

if det(B) ≠ 0. If det(B) is not zero, it means that matrix B is invertible

do you remember what "invertible" means for a matrix?

[indigo ginkgo]

what does it mean?

i actually got no clue

[jagged raptor]

think about like this: with regular numbers, "invertible" is kind of like sayinh you can "undo" multiplication

[indigo ginkgo]

am i undoing a matrix?

i really dont understand what ure trying to explain to me

[jagged raptor]

let me do a example

[indigo ginkgo]

are you trying to explain to me that B*B^-1=In?

[indigo ginkgo]

sure

go ahead

[jagged raptor]

if you multiply a number by 2, you can "undo" it by multiplying by its inverse, which is 1/2 (or dividing by 2). So, 2 and 1/2 are inverses of each other for multiplication because 2 * (1/2) = 1. And also (1/2) * 2 = 1. The number '1' is like the "identity" for multiplication

[jagged raptor]

yeah

[indigo ginkgo]

buddy

i know some stuff about matrices

i know all the theory there is to learn

i just need help understanding this exercise

[jagged raptor]

oh

mb

[indigo ginkgo]

not all of it, we werent taught eigen values and stuff

but i dont think its that hard

anyway

so, any idea on how to actually show why d) is wrong?

because the det B diff 0 is throwing me off

[jagged raptor]

If det(B) ≠ 0 and AB = O_n, then A must be O_n. But we know A is not O_n. So, such a B cannot exiist

[indigo ginkgo]

i cannot give an example that would work is B diff 0, so its obiously wrong, but how would u prove it

[indigo ginkgo]

why would it have to be On

what does the determinant of B have to do with whether or not A is On

[jagged raptor]

wait a minute

im gonna draw

[indigo ginkgo]

this looks like its really complex

[indigo ginkgo]

omg

i havent even thought of this

ure amazing

thank you

what about c) tho

how would you prove that c) is wrong

oh

you can just give an example there

nvm

wow

thx

i got more matrix exercises

if you feel like u can do more, tell me

[jagged raptor]

sure, just send i will help

[indigo ginkgo]

okk wait one sec to write down the solution for this one u sent first

[jagged raptor]

np

[indigo ginkgo]

ill just type the next one for u

so

let D be a determinant different than 0; D e M3(R) and its elements can be 1 and -1. Then:
a) D=1
b) D=-1
c) |D|=1
d) |D|=2
e) |D|=4
f)D=2

[indigo ginkgo]

are you still here?

[jagged raptor]

yep

[indigo ginkgo]

i think i got it actually

theres just 4 possible determinants

unless u just multiply the determinants by -1

bc then ud have 8

[jagged raptor]

it's not quite that you multiply the determinant by -1 to get more values

the determinant itself can be positive or negative

[indigo ginkgo]

yea i guess

[jagged raptor]

for a 3x3 matrix with ±1 entries, the determinant will always be an integer

so, the possible determinants are numbers like 0, ±4, ±8, ±12, and so on multiples of 4

[indigo ginkgo]

why is it by 4

also, it can only be 4

ever

[jagged raptor]

since we are given that det(D) ≠ 0, the smallest possible non-zero value for the magnitude or absolute value of the determint

is?

[indigo ginkgo]

its -4

idk where u got + - 12

and + - 8

[jagged raptor]

when i said "multiples of 4," I meant exactly that - any number you get by multiplying 4 by an integer (positive, negativem or zero)

[indigo ginkgo]

anyway, the answer was indeed e so i got it

[jagged raptor]

yep

[indigo ginkgo]

i got anotherrrrrrrrrr

wanna try?

[quick talon]

Hello everyone !

[indigo ginkgo]

hi

.close

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[near agate]

im not sure what this is asking for

[near agate]

https://i.imgur.com/l7Sjr7h.png

[fleet verge]

This is asking you to prove that ( f\bigl(f^{-1}(D)\bigr) \subseteq D ) and then give an example where the inclusion is strict.

[potent lotus]

Samuel

[near agate]

I dont see how the first given function and sets help with that?

im just not even sure where to start

we did nothing like this in class but its assigned for homework

[quick talon]

When you don't know where to start. Start writing what you have and see where it leads.
So you know D included in S what does that mean

[near agate]

D is a subset of S

[quick talon]

Ok, how would you write that mathematically

[near agate]

every element in D is also in S

[quick talon]

Ok good. So let x belongs to f(f-1(D))

[near agate]

then x belongs to f(f-1(S)) ?

[quick talon]

This is what you need to fill

[near agate]

so f(y) is an element of D

then x is an element of D

[quick talon]

Why would f(y) be an element of D ??

[near agate]

i dont know

i dont get how the first function from A -> B helps

[quick talon]

It's just a definition

When you dont know how to prove a causality , just write the definition of your starting point and explore where it goes (while keeping in mind the result you want to get

[near agate]

how did you get that y is an element of f-1(D)

[quick talon]

Because this

if x belongs to f(H). That means x = f(something), with something belongs to H

in this case H is f^-1 (D)

[near agate]

i see

thank you

[quick talon]

You're welcome, feel free to ask if you have more linear algebra questions

[near agate]

oh this is linear? its for my real analysis class

[near agate]

so f: A -> B and Y subset of B were never used?

[quick talon]

The whole first sentence is the definition of reverse image.

it's just to tell you (remind you) of what is the reverse image

and then they apply it to X --> S

[near agate]

ah

wait how did we get that x = z

[quick talon]

Because we showed that f(y) = x & f(y) = z

so x = z

[calm coral]

@near agate Has your question been resolved?

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[fallen crag]

How do i factor the denom here

[pallid halo]

could try the rational root theorem

or maybe get lucky and notice that x=-1 is a root

[fallen crag]

https://tenor.com/view/icb-i-can't-believe-shaking-my-head-duck-cigarette-gif-8035362469672454220

i havent touched that since precalc

[pallid halo]

oh, well it says that the only possible rational roots are 1, -1, 3, -3

and then if you try each of those, you see that only -1 works

[fallen crag]

i appreciate you jumping into it but i forgot how to apply rrt

so i needa figure that out first lol

[pallid halo]

but that's enough to get a quadratic factor

[pallid halo]

and then you can use the quadratic formula to get the other two roots

[ancient grotto]

yeah, thats exactly how you do it

[fallen crag]

i dont get it

[pallid halo]

your a0 is 3

[ancient grotto]

let me see if latex will let me list it

[pallid halo]

and your a_n is 1

[fallen crag]

so 3/x^3?

[ancient grotto]

a_0 and a_n are coefficients here

[fallen crag]

ah

the coeff for x^3 is just 1

so 3

and i need to do that for all positive and negatives?

[pallid halo]

yep so your only possible rational roots are the factors of 3

(including the negative factors)

[ancient grotto]

$\pm\qty[\frac{3,1}{1}]$

[potent lotus]

;(

[ancient grotto]

oh my god it worked

my premable is fucked too

[fallen crag]

hold on

so after i do 3/(coeff of x^3) i move on to 3/(coeff of -x^2)?

and so on

[ancient grotto]

yup

[fallen crag]

is that what it's saying

[ancient grotto]

nonnonnono

[ancient grotto]

lookie here!

[fallen crag]

okay so we need to check if 3, -3, 1, and -1 are factors?

not factors, rational roots mb

meaning i do f(3), f(-3), etc

[pallid halo]

yep

see which one(s) give you 0

[fallen crag]

okay but what if i have coefficients

yea its convenient that its one here

but what if it were 4x^3 - 2x^2 + 5x + 3 for example

+-3/4, +- 3/2, +- 3/5?

[pallid halo]

then you have to consider 1, 1/2, 1/4, 3, 3/2, 3/4 and also their negatives

[fallen crag]

wdym

how did you get those numbers is what im asking

[pallid halo]

the theorem quickly becomes less useful as the coefficients get bigger

[pallid halo]

the factors of 4 (the leading coefficient) are 1,2,4 and their negatives

the factors of 3 (the constant term) are 1,3 and their negatives

[fallen crag]

ohhhh

so you only take into account the leading coefficient

[ancient grotto]

and constant, yeah

[pallid halo]

yep the coefficients of the highest power and the constant term are the only ones that play a role here

[fallen crag]

okay so -1 is a zero

meaning what

[pallid halo]

meaning x+1 is a factor

[fallen crag]

one of my factors will be (x+1)?

[pallid halo]

yep

now you can divide by x+1 to find the rest

[fallen crag]

so rational root theorem then synthetic division

and then i can set up my problem

[pallid halo]

pretty much

[fallen crag]

damn so calculus really is like 90% algebra and the 10% is knowing how to set up the equatione

[pallid halo]

you'll still need to know how to integrate the partial fractions

but yea a fair amount of the work is finding the partial fractions to begin with

[calm coral]

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[safe oriole]

is there a way of doing this without brute force?

isn't hard checking all possibilities tho, it's only 16

[ancient grotto]

4 member:1member, 3member:2member

4 member is guaranteed to have it i believe

[amber bolt]

it's easy if two numbers and their difference is possibly 2 numebrs total

then set with 2 doesn;t have 1 and doesn't have 4
which means it has 3, 3 can't go with 1 and 4

then 5 can't go anywhere

the other intepretation idk

oh but {1,2}{3,4,5} would work if it can't be 2 numbers

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[tough star]

lets say a projectile is thrown perfectly horizontally, if there is a fan that is added in the direction the projectile is thrown, does that increase air time and horizontal range?

[void umbra]

I mean ideally it would increase horizontal range but not air time ig

[silver tapir]

yes i think so

[tough star]

yeah thats what i was thinking as well, but how come is it that if a faster projectile is lauched then it stays in the air the same time? like if a bullet is launched perfectly horizontally it woluld have more air time then if a ball is launched horizontally

[prime skiff]

air time depends on vertical component of motion

so if the projectile is thrown perfectly horizontally

the fan will also only act in the horizontal direction (since you said its in the direction the projectile is thrown)

[tough star]

yea

[prime skiff]

so

range will change

but the time of flight remains the same

[tough star]

but what abt this?

for air time

[prime skiff]

because you aren't changing the vertical component

[prime skiff]

all objects will take

[tough star]

rlly

[prime skiff]

the same amount of time to fall

if theyre launched at the same velocity

sorry wait are you saying

[tough star]

but here its not launched at the same velocity

cuz the fan adds to the velocity, no?

[prime skiff]

right but the vertical velocity is the same

but its thrown perfectly horizontally

so the vertical velocity is 0

right

in your scenario the fan will act the same direction as the projectile

[void umbra]

It travels a further distance in the same time

[tough star]

so lets say 2 balls thrown perfectly horizontally, no y initial.

a ball travelling at 100 m/s and a ball launched at 1 m/s will hit the ground at the same time?

[prime skiff]

yes

i will draw up what it looks like for you

[flat gorge]

it's not an ideal case T_T The fan does not affect the air resistance surrounding the bullet (which is already less significant than the ball, thanks to it's mass).

[void umbra]

Yea it's only ideally

[tough star]

i mean it makes sense in the sense that only vertical velocity affects air time, but logically wouldn't it make sense for a ball to hit the ground after a bullet

[void umbra]

You can go outside and test it yourself

With like any ball ig

Won't be exact but you'll get the idea

[prime skiff]

do you assume this because of the difference in mass?

if so that's not the case even though it may seem like it

objects regardless of their mass will accelerate downwards at the same rate

[tough star]

hmm i see

[flat gorge]

cited by Principles of Physics

[prime skiff]

in practice though im ignoring other factors which may affect air time

[flat gorge]

If you're so interested, you can skim through 6-2 of Principles of Physics. The pdf is available online

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[plain bone]

Seems like you got it right

Or are you having trouble understanding why 0 is right