help-49

[latent elk]

No

A force doesnt need to be 0 at t=0

It could be any number

[jolly roost]

but then how would it be at rest

[latent elk]

Cause before t =0, there's no force being applied

So its at rest

Then at t =0, we apply a force

So its no longer at rest

I think u messing up F and V

Function of V need to be continuous, Function of F doesnt

[jolly roost]

i see, but what if it said starts at (0,0)

[latent elk]

So t = 0, F= 0?

[jolly roost]

yea

[latent elk]

Then solve like this

Since we know t = 0 => F =0 and t = 4 => F = 4, and F changes linearly

[jolly roost]

but hows that right considering that the impulse is mv which is 1(4) = 4, but i only got an impulse of 2

[latent elk]

This the graph must be a triangle

[latent elk]

Cause u made a wrong assumption

The initial problem clearly state that the force is steady, AKA F = const

And not at t = 0, F = 0

Thats a wrong assumption

At t = 0, F = 4 because, we know at t = 4, F = 4, and F is const!

[jolly roost]

ok imma change up the question

A smart cart is initially at rest and then is influenced by a increasing linear force. The cart has a mass of 1 kg. At 0 seconds, the force is at 0 N. At 4 seconds, the velocity of the cart is 4 m/s. What is the impulse?

now?

[latent elk]

Then the impulse is 2Ns

Using your graph

[jolly roost]

but why cant we just use I = mv

and do 1(4)

[latent elk]

The a is not const then

Sorry

a

not v

[jolly roost]

hmm i see

so if acceleration is not constant, then we can't use I = mv?

[latent elk]

F is variable thus a is also changing according to F

[latent elk]

I = m * deltaV

[jolly roost]

is deltaV not just vf - vi

and vi is 0, so deltaV = vf

[latent elk]

Well

Its just then u need to calculate dv over that 4s

which is not straight forward

At t = 0

F =0

F = ma

f = m * dv/dt

We might need to use jerk

Since its dF = m * da/dt

dF = m * d(dv/dt)/dt

[jolly roost]

but is the answer still 2Ns?

[latent elk]

Yep

Using F-t graph now is more fissable then doing integral at this point

[jolly roost]

but i still dont get why we cant simply use m(vf - vi) = 1(4 -0) = 4

is it because there is a change in acceleration?

[latent elk]

Yep

[latent elk]

this is wrong btw

If F is linear

Then a is linear

But v will be not a linear

So over large time period (4s) we cant use v(at 4) - v(at 0)

[jolly roost]

i see

so what would vi be in that case then

[midnight plank]

@jolly roost Has your question been resolved?

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[native ruin]

How do you figure out this problem? I just need help on what to find to get the answer.

I am thinking that I need to find M, then find that section of K, which then I add 8 to. Would that be correct?

[fallow scarab]

did you try using similar triangles

[fickle harness]

thts the trick

k=12

[vale zephyr]

i think this is wrong

[fickle harness]

use 20/15=x/8

sorry 10.667

can counter check

this question is solved

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[terse gulch]

So with logarithms, we have (b^e = N ) = (log b of N = e) ... I know that logs with negative bases are undefined, but then why can we easily see that log (-4) of (-64) = 3 ? It's not undefined if I can calculate it, no ? Or am I missing something ?

[terse gulch]

Initial problem was this :

[runic hamlet]

you dont usually want to only evaluate very specific inputs

you are often much more interested in all of the inputs

and for negative basis and negative inputs you get huge problems

[surreal moon]

(in real numbers)

[runic hamlet]

yes you can define individual exceptions but those are very much exceptions

[surreal moon]

btw you can have negative logs if you deal with complex numbers. But the exact workings are pretty .. uhh... complex.

[terse gulch]

so we just apply the blanket rule that log with negative bases are undefined because of the times it's impossible to calculate them, like for log-4 of -16 or any other of the same type ?

[burnt flame]

well logs of negatives work in specific cases, but again, they're specific

[surreal moon]

We leave negative logs as undefined because, for any $n<0$, $n^x$ will not be real for most $x$

[terse gulch]

yeah, I'm currently only working with real numbers, I think this is precalc for now

[grand pond]

SWR

[burnt flame]

like (-2)^2 is real but (-2)^(1/2) isnt

[terse gulch]

ah gotcha, so really it's to avoid the complexity in real number systems... this feels like something the teacher is going to go "remember when I said that?... well it's not 100% true"

but thanks, that makes it somewhat better because I know the rule, I was wondering why we applied even for "solvable" cases. but what SWR said sounds good to me

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[pseudo karma]

Convinced my book is wrong, ive come up with 439 every which way.

[grim vector]

You have to evaluate ?

[dawn dagger]

have you tried a calculator

[pseudo karma]

Book giving answer as 443

I have no idea how to insert this on a phone calculator I havent done math in 20 years.

[dawn dagger]

,calc 2 + 21^2

[grand pond]

Result:

443

[grim vector]

2+5 = 7

7*3

21

21²

441

[pseudo karma]

How is 6-8 = 2

[dawn dagger]

abs value

[grim vector]

Abs

[keen herald]

Absolute value of 6-8

[dawn dagger]

train abs

[pseudo karma]

Omfg lmao

No wonder, thank you.

[grim vector]

Nw

[pseudo karma]

Yeah thats right strait bars down are Absv

Thank again.

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[calm fiber]

How would i approach this problem?

[midnight plank]

@calm fiber Has your question been resolved?

[calm fiber]

<@&286206848099549185>

[fervent ember]

use the given informations. "the diagonals of any rhombus are perpendicular bisectors of ech other" - what does this mean to you?

[midnight plank]

@calm fiber Has your question been resolved?

[calm fiber]

that every side must be the same?

[fervent ember]

no.

[calm fiber]

why not?

each triangle has the two sides

and using pythagorean theorem each side should be the same

[fervent ember]

what does it mean for the diagonals?

[calm fiber]

they are at right angles? they have been cut into two equal halves?

[fervent ember]

yes. use this information.

[calm fiber]

would i first try to find the mid point in this case?

or the center

and then the distance from A or C to the center

[fervent ember]

you have A and C given, so it is clear to determine the point of intersection of the two diagonals.

[calm fiber]

yeah ik that i can do

but my main issue is where to go from there

i can also find the equation of the perpendicular line which consitutues of b and d

[fervent ember]

you started with the question "how would i approach tis problem?".

[calm fiber]

hmm ok

but the thing im genuinely stuck on how to proceed 😭

[fervent ember]

did you determine the point of intersection?

[calm fiber]

yes

(3,2)

[fervent ember]

and now determine a perpendicular vector.

[calm fiber]

the equation of the perpendicular line?

is that what you mean?

yeah i did that too

[fervent ember]

so whats the remaining question?

[calm fiber]

a) Determine possible coordinates for B and D 😭

[fervent ember]

any point on this line?

[calm fiber]

would i say any points in (said line)?

oh

oh my gosh

bruhhhhhhhhhh

thank you thank you

so turns out i basically solved this question

but i thought it had to be specific

😭

this is genuinely frustrating

[midnight plank]

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[maiden forge]

open

[maiden forge]

im doing this problem

ive done this bit so far

im stuck on the top right corner

im trying to simplify the root 6

and got root 3 x root 2

i dont know if ur supposed to get two of those prime numbers

[silent elbow]

you already have a root 3 in the denominator and the you can get a root 2 from the 14

[maiden forge]

i thought u can only do that to numbers in a surd

14 is like a normal number its not in the surd bracket thing

[silent elbow]

I've never heard of a surd in math before. What are you refering to?

[maiden forge]

gg

u dont know surds??

oh wait

u might call it something else to be fair

[silent elbow]

I probably do, but maybe they're called something different were you are from

[tired vapor]

surd is some weird way to refer to square roots

[maiden forge]

yeah

[maiden forge]

this i guess

[dawn dagger]

are you surd?

[maiden forge]

@versed summit

[tired vapor]

yeah i am surd about it

[maiden forge]

nvm

im gonna ping helpers

[silent elbow]

I'm working on it, just slow in latex

[versed summit]

You pinged?

[tired vapor]

essentially with your question, just simplify the square roots by factorizing. for example $\sqrt{28}=\sqrt{7\cdot 4}=2\sqrt{7}$

[grand pond]

fish

[versed summit]

Ok what've you done so far?

[maiden forge]

did this so far

i think i simplified the top part correctly

[tired vapor]

try getting both in terms of $\sqrt{3}$, so you can make it one term (your top part is correct)

[grand pond]

fish

[versed summit]

You have!

Very good

[silent elbow]

$14\sqrt{6} \cdot 5\sqrt{3} = 7 \sqrt{2} \sqrt{2} \sqrt{3} \sqrt{2} \cdot 5 \sqrt{3} = 7\sqrt{2} \cdot 2 \cdot 3 \cdot 5$

[versed summit]

Alright what do you notice now?

[silent elbow]

That's for the denominator

[versed summit]

Maybe something in the numerator wink wink?

[maiden forge]

that looks so cofnusing

[silent elbow]

Yeah agreed

[maiden forge]

ive tried simplifying this part of the numerator

but i got two prime numbers and dont know what to do now

[grand pond]

Gruttan

[versed summit]

The numerator is the one on top not the the one below

[maiden forge]

oh yeah i did that bit

[silent elbow]

Maybe that's sligtly better

[versed summit]

Hold that thought. This is very useful

[versed summit]

You can simplify it further?

[maiden forge]

heres what i did so far for the numerator but i know you can do 30 x 10 x 7

so 2100 on numerator

[versed summit]

Good, so let's focus on the denominator now

[maiden forge]

ok

[maiden forge]

yea sort of but not really

it just looks confusing in text format if you get what i mean

[versed summit]

Now you got 5root3 multiplied by 14 times root3 times root2

What do you see?

[silent elbow]

It's not in text format tho, but I think I get what you mean

[versed summit]

Also Idk latex yet, pls help me @silent elbow

[maiden forge]

the root 3

[versed summit]

Yes, so that multiplies to become 3

[maiden forge]

wait where did u get that from

[versed summit]

You said that 14root6 can be written as 14 times root3 times root2

[maiden forge]

oh yeah nevermind

yeah sorry about that

[versed summit]

Here

[versed summit]

Np

You should continue from here

[maiden forge]

wait so we can do that?

[versed summit]

Why not?

What law do you think we're breaking?

[maiden forge]

i didnt know u could have two primes when from when you simplify it

[versed summit]

We can have as many as we like. Are you familiar with prime factorization?

[maiden forge]

yes

[versed summit]

Because this is just that but with the square root sign on top

[maiden forge]

as in i didnt know u could simplify it and leave it as this

because isnt it basically root 6

[silent elbow]

Sure, I didn't know anything either a couple hours ago and I just searched up this tutorial on yt. https://youtu.be/ydOTMQC7np0?si=Nx9iN1ALJVE4N6yP

You can skip all the non-math stuff and skip straight to 36 minutes

[versed summit]

But we multiply 5root 3 to it

[versed summit]

It is

[versed summit]

Thanks mate

[versed summit]

Yep this is it

[maiden forge]

so the root 3 x root 3 gives us 3 on its own

[versed summit]

Yes

[maiden forge]

so we do like 5 multiplied by 3 to get 2100/14 root 2 x 15

?

[versed summit]

Better!

We can do 14*5=70

70*3=210

That's better because we can directly cancel the 2100

[maiden forge]

oh ok

[versed summit]

There would be a root 2 in the denominator though

[maiden forge]

so that would be better than the way i said?

[versed summit]

It would mathematically be the exact same

[maiden forge]

its the same but this way is a bit longer?

[versed summit]

Bingo

[maiden forge]

ah ok

[versed summit]

If you think like this you can start solving these in your head and become a pro

Now all you got to do is get the final solution

[maiden forge]

finding out answer now

5 root 2?

[versed summit]

Done

[maiden forge]

oh ok cool

thank you so much

[versed summit]

Good that you didn't make a single calculation error here

[silent elbow]

ggs guys😎

[versed summit]

Yep we got em boys

[maiden forge]

oh wow

yeah ur explanation actually helped out a lot

i got a different type now

[versed summit]

I'm glad

[maiden forge]

adding them

[versed summit]

Ok what do you think?

[maiden forge]

this looks easier than the previous ones

you just add the normal numbers outside the surd?

[silent elbow]

I would suggest trying it on your own for a minute or two, and then if you get stuck ask for help

[versed summit]

Yes

[maiden forge]

so in this case 7 root 3?

[versed summit]

5+3=8

So 8root3

Alright if that's it, I gotta go now, I have to sleep

Gn y'all

[maiden forge]

this is how im gonna lose marks on the test 😭

[maiden forge]

thank you so much for your help and have a good night

[versed summit]

Thanks

[versed summit]

Yep keep double checking dw

[silent elbow]

My biggest advice for this kind of mistake is write every step down very neatly, good luck!

[maiden forge]

yeah you're right

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[stark harbor]

i need help

[stark harbor]

im not sure how to do scenario 1 question 6

and scenario idk how to do scenario 2 ques 4,5

.close

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.reopen

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✅

[stark harbor]

hi

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@stark harbor Has your question been resolved?

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[fresh abyss]

Why wouldn't choose 1 be optimal solution? You will always be part of the winning pair and receive payouts?

[rose trout]

No. If the chosen numbers are 0.1, 0.7, 0.8, 1 for instance, the middle pair wins.

[fresh abyss]

Ic, so the problem is about maximizing how big the number is without deviating too far away from 0.5?

[midnight plank]

@fresh abyss Has your question been resolved?

[fresh abyss]

<@&286206848099549185>

[midnight plank]

@fresh abyss Has your question been resolved?

[fresh abyss]

@gray widget Please anyone?

I just need a small hint

[knotty vigil]

What about 0.5?

[fresh abyss]

Why is it 0.5?

[eager galleon]

you have to set up some integrals and find the expected value of (X+A) - (B+C) with A,B,C being uniform random variables in [0,1] and X being the number you pick and then maximize its value

unfortunately i don't have the brainpower right now to explain better but i hope it works as a hint

[fresh abyss]

Would you split this up into different cases for different values of X?

Bc I’m unsure how to take into considerations the value of X relative to the three others (since that affects which pairing is made)

[eager galleon]

i'd presume yes, assuming A is the partner number, case 1 being A >= B, C (or B,C >=A) and case 2 being B >= A >= C (can replace B and C)

i think you have to multiply by some constant somewhere because of the different possible orderings of A,B,C

take it with a grain of salt, i still haven't worked out a solution

[fresh abyss]

The thing I don’t logically understand with this problem is how there can be an optimal answer. Isn’t the best value of X entirely dependent on what 3 numbers were drawn? Since it’s uniformly distributed, each number is equally likely, so if there is a high variance among the other numbers, you want to go for a middle value, but if there’s is small variance (all 3 numbers are around 0.2, say) then you go for a hig h value

[eager galleon]

it's uniform distribution, so the variance is known to be 1/12

var(A - B - C) = 3 * 1/12

[fresh abyss]

Why did you subtract the variance of the three?

[eager galleon]

from here

[midnight plank]

@fresh abyss Has your question been resolved?

[midnight plank]

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[still cove]

can someone explain why can we write Y1 = -2X1 +X2
AND Y2 = X1 +X2

like why they didnt give any explaination for that

[fallow scarab]

Have you taken linear algebra

[midnight plank]

@still cove Has your question been resolved?

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[ember sage]

how do u determine the power of x's n the answer

[heady plume]

Starting with the right.

Rightmost: Remainder : 55
Next: x^0 [constant term] : 27
Next : x^1 : 11
Next : x^2 : 4
Next : x^3 : 2

[ember sage]

is it always like that?

[heady plume]

if you start with right and move in that order, yes

[ember sage]

ok thanks

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[ancient sonnet]

hey hey, quick question. is this the right series of formulas for determining a riemann sum?

[sharp coral]

that is mostly correct for calculating a left- or right- riemann sum with equal-length subintervals

but you need to be consistent with the sum variables (the variable should be i, and the upper bound should be n)

and a left- riemann sum goes from i = 0 to n-1, whereas a right- riemann sum goes from i = 1 to n

[ancient sonnet]

oo o ok ty, ill fix that then

very much appreciated!!

.close

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[twilit field]

I'm trying to integrate this by converting it into polar coordinates

[twilit field]

Getting really strange limits

I'm getting $\int_{0}^{\pi} \int_{0}^{\sin(\theta)}d(r)d(\theta)$

[grand pond]

What a wonderful world!

[surreal moon]

How?

[twilit field]

What's wrong

[visual tiger]

I can see a few things

first of all why isn't it rdrdtheta

second of all how did you get the bounds

[twilit field]

Why should it be rdrrdtheta

[visual tiger]

dxdy = rdrdtheta

[twilit field]

oh, it;s the jacobian

[visual tiger]

and can you explain the bounds

[twilit field]

Hmm

Oh right

okay

I see

The line r=1 is at at a constant distance from the origin

r=1

[visual tiger]

the line?

[twilit field]

Sorry, I mean the curve

so $\int_{0}^{\pi} \int_{0}^{1} r dr d(\theta)$

[grand pond]

What a wonderful world!

[sudden yacht]

This is the area of semicircle 😉 , so that's just π•1²/2

[twilit field]

yeah

so π/2

[twilit field]

I know, I'm just learning polar substitutions now, so wanted to use that

[sudden yacht]

Ahn alright sorry

[visual tiger]

so kinda different from your first bounds

this one works now

[twilit field]

Got it

thanks

This feels really sus

I'm getting

$\int_{0}^{\pi}. \int_{-1}^{1} r^3 dr d(\theta)$

[grand pond]

What a wonderful world!

[sudden yacht]

Two things:

• why from 0 to π?
• r can't be negative

[twilit field]

I see

okay

I think I'll have to split it into 2 regions then

[sudden yacht]

If you draw that region (in cartesian coordinates) what shape do you get?

[twilit field]

A circle

[sudden yacht]

Yep, so theta will go from 0 to 2π

[sudden yacht]

There's no need

[twilit field]

hmm and r

r goes from 0 to 1, right

so $\int_{0}^{2 \pi} \int_{0}^{1} r^3 dr d(\theta)$

right

[grand pond]

What a wonderful world!

[twilit field]

so I'm getting $4 \pi ^4$

[grand pond]

What a wonderful world!

[twilit field]

which feels sus

[twilit field]

Is this right

huh

it's π/4

what am I doing wrong

[twilit field]

Whit, won't this just be 2π/4 = π/2

That's wrong, is it not?

nvm

Got it

[sudden yacht]

Sorry, I've just seen the messages rn

Glad you worked it out 💪

[heady plume]

:| the bounds for theta is [0, π)

[sudden yacht]

Mmh I don't think so

[sudden yacht]

At least looking at these cartesian bounds

[twilit field]

I feel like the bounds would be identical here

$\int_{0}^{2 \pi} \int_{0}^{1} \ln( 1+ r^2) dr dt$

[grand pond]

What a wonderful world!

[twilit field]

Of course integrating this is a pain, but that's a different story

Thanks

[twilit field]

Rather messy

I know it's a standard result

but still

[heady plume]

rln(1 + r²) - 2r + 2arctan r

[twilit field]

oops

I forgot the Jacobian

Yeah, in that case pretty easy

[heady plume]

0.5[(1 + r²)ln (1 + r²) - r² - 1]

[twilit field]

Yeah

thanks

This is much more interesting

Here theta would vary from 0 to π/4 I think?

and r from 0 to 2

so $\int_{0}^{\pi/4} \int_{0}^{2} sin(\theta) + cos(\theta)$

[grand pond]

What a wonderful world!

[twilit field]

Or is that wrobng

anyway, I have a class now, will get back to this afte that

thanks

.close

[midnight plank]

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[midnight plank]

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[worthy plover]

how would you write this as a function?
'A taxi costs 6$ for the first 3 miles, with every mile after the third costing an extra $.20'

I got as far as y=.20x +6 but I don't know how to reflect the part that the taxi only costs $6 for the first 3 miles

[teal sluice]

y=.20(x-3)+6

for x>3

[worthy plover]

oh okay, thanks for the help for sure :]

.close

[midnight plank]

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[dense harness]

<@&268886789983436800>

[midnight plank]

Channel closed

Channel closed due to the original message being deleted.
If you did not intend to do this, please open a new help channel,
as this action is irreversible, and this channel may abruptly lock.

[frail heron]

.close

[midnight plank]

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[tidal turret]

Let $f: \mathbb{R}^4 \to \mathbb{R}^4$ be the linear transformation defined by
[
f(\mathbf{x}) = (x_1 - x_3, x_3 + x_4, x_1 + 2x_2 + x_4, -x_1 - x_4)
]
and let
[
\mathbb{H} = { \mathbf{x} \in \mathbb{R}^4 \mid x_1 - x_4 = 0 }.
]
Then, $\operatorname{Ker}(f \circ f) \cap f^{-1}(\mathbb{H})$ is equal to

Select one:

\begin{enumerate}[label=\alph*)]
\item $\langle (1,0,0,0); (0,0,1,-1) \rangle$
\item $\langle (1,0,-1,1) \rangle$
\item ${0}$
\item $\langle (1,0,1,-1) \rangle$
\end{enumerate}

[grand pond]

938c2cc0dcc05f2b68c4287040cfcf71

[brisk iris]

Start in ker H

What’s ker H

[zealous schooner]

What is ker H?

H isn't a linear transformation

[tidal turret]

bro I am tripping hard as balls

[tidal turret]

still has a kernel bro even if it's a subspace of R4

[tidal turret]

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

[tidal turret]

like just evaluate (x1,x2,x3,x4) = (x1-x3,x3+x4,x1+2x2+x4,-x1-x4) bro then find the kernel of that

I got it i got it bro

I got the idea

I got it

let me draw it

[brisk iris]

I’m sorry

[tidal turret]

we still need the preimage of H under f

[brisk iris]

My nephews are screaming 🙀.
I meant what’s f-1(H). There’s no kerH

[tidal turret]

wdym there is no ker H?

wdym?

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

still has a kernel bro even if it's a subspace of R4

not that the kernel of H is relevant to the Que, though

[brisk iris]

Kernel applies to a linear mapping , not a subspace

[tidal turret]

Kernel = Nullspace

[brisk iris]

No

[tidal turret]

is same thing, wdym the term is restricted to linear maps

anyways is not like I care, in spanish is "Nulo" for both linear maps and subspaces

in English maybe they invent more words

[brisk iris]

By definition,

Let’s not debate this. It’s not relevant to your problem

[tidal turret]

yeah, let's continue let me draw my idea

[brisk iris]

Now, what’s f-1(H)

[tidal turret]

why start with the preimage

can we start with the kernel, null or w/e is called of the linear composition

[brisk iris]

Ok

[tidal turret]

[brisk iris]

Write f as a matrix.

[tidal turret]

you are multiplying the matrix twice and then finding the nullspace of the resultant matrix?

[brisk iris]

yup

[tidal turret]

you know what is am doing?

fuck my English and my keyboard sucks ass

[tidal turret]

like i evaluate standards basis vectors under f?

and the outputs i place as columns in a matrix bro?

mmm fuck

[tidal turret]

damn bro, I am alone in my house, id wish I had a screaming nephew, or someone to talk to

[brisk iris]

I'm sorry I just can't focus, too much noise. I'll be back in 10 to 15

[brisk iris]

Waiiit

I have a much faster way

Can you compute det(f)

[tidal turret]

det([f]_EE)

[brisk iris]

Yeah

[tidal turret]

what about it

[brisk iris]

Can you compute it

[tidal turret]

,w det {{1,0,-1,0},{0,0,1,1},{1,2,0,1},{-1,0,0,-1}}

[grand pond]

Results provided by WolframAlpha

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

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

we have kernel of f with dim >= 1

,w rank {{1,0,-1,0},{0,0,1,1},{1,2,0,1},{-1,0,0,-1}}

kernel of f has dimension 1

[tidal turret]

bro

do not leave after saying that

[brisk iris]

Sorry I thought it was not 0 give me a moment

[tidal turret]

,w Nullspace[{{1,0,-1,0},{0,0,1,1},{1,2,0,1},{-1,0,0,-1}} * {{1,0,-1,0},{0,0,1,1},{1,2,0,1},{-1,0,0,-1}}]

[grand pond]

Results provided by WolframAlpha

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

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

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

[tidal turret]

exactly as mines

my way or the highway

but atleast we verified ker fof is correct

[midnight plank]

@tidal turret Has your question been resolved?

[brisk iris]

Hey, still there ?

[tidal turret]

ye

[brisk iris]

Ok, take a look at this

Your solution is ker(f²)

@tidal turret since you computed ker(f²)

You got your answer

[tidal turret]

?

[tidal turret]

sorry i was in FaceTime with my sister

[tidal turret]

how do I find the preimage of H under f

[brisk iris]

You don’t

You don’t need it

[tidal turret]

bro in this case finding the inverse of A doesn't work because det(f)=0 remember

. . .

[brisk iris]

Im not talking about this inverse of A

Its the pre image of H by A

[tidal turret]

yeah

[tidal turret]

C is canonical?

why not use E

[brisk iris]

Yes any basis will do

[tidal turret]

kerA is a subspace of ker A^2

[tidal turret]

you computed kernel of (f)?

broo

[brisk iris]

Yes

[tidal turret]

this seems ok

but you skipped the steps of finding ker(f) bro

[brisk iris]

Cause you can do that in a calculator right ?

[tidal turret]

sure but in the exam bro

[tidal turret]

like this bro, I fixed some things

[brisk iris]

In the exam you’re not allowed to use a calculator ?

[tidal turret]

only scientific

but finding ker f is simple

[brisk iris]

The calculations you have to do here is finding both kEr f and kEr f^2

But if you want to be time effective without a calculator, that’s not the way to go.

[tidal turret]

is so confusing bro

[brisk iris]

What is ?

There is a different (faster way) when you don't have a calculator

[brisk iris]

If you look closely, you'll notice that answers b,c,d are included in answer a.

[tidal turret]

ker fof is not necessary bro

or yes?

no man, ker fof entirely contains ker f

ker(f) n H is a stronger implication

I mean

I am not even sure anymore

I messed up

I messed up

I messed up

[brisk iris]

Look

I love reasoning, much more than calculations. So I did a lot of reasoning, and they I do calculations when I have to

However

If you had to do this during an exam. You should do it differently

[tidal turret]

however what

[brisk iris]

You have to look at the answers, and test them

Do you want me to write how I would do that in an exam ?

[tidal turret]

usually exam is not multiple choice thoo

but i get your point

[brisk iris]

You start by testing answer a, because you notice that all other answers are included in a

If a) fails (which it won't), you move on to (b) or (d).

If both fail, it has to be (c)

[tidal turret]

,w kernel {{1,0,-1,0},{0,0,1,1},{1,2,0,1},{-1,0,0,-1}}

[grand pond]

Results provided by WolframAlpha

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

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[tidal turret]

ker(A) is not contained in H

any vector in ker(f) does not satisfy the equation of the plane H bro

ahh is so confusing i think I messed up again

fuck my life

I think I get it tho

[brisk iris]

Wait

Noise again

[tidal turret]

x in ker fof iff f(x) in ker f
x in f^-1(H) iff f(x) in H

f(x) in H n Ker f
but H n Ker(f) ={0} only trivial intersection

ohh, I'm stupid

let me explain with drawing

[tidal turret]

I got it bro

it wasn't hard per se, it was just confusing

but naming f(x) = y clear things up

look if ker(f)=<(-1,0,-1,1)>

then f(-1,0,-1,1)=0, so if I create an alias for f(x), call it, f(x)=y
where x is (-1,0,-1,1) and y = 0

we want f(x) in H and f(x) in ker f
0 is in H and in ker f
but particularly
f(f(-1,0,-1,1))=0
so (-1,0,-1,1) in ker fof

[brisk iris]

Wait. You’re right but then the solution would be just ker f

[tidal turret]

solution is just ker f according to answers

[brisk iris]

Yeah

Ok

Do you see how to fix my picture ?

[tidal turret]

how

[brisk iris]

Wait

[tidal turret]

yeah exactly bro

🤓

[brisk iris]

Sorry I really struggled to focus today

[tidal turret]

no worries mate, I enjoy math when I struggle

[brisk iris]

Too much screaming at home

[tidal turret]

take care of your nephew bro

time is the only thing that doesn't go back

I appreciate the help I couldn't have done it without u bro, it wasn't hard, it was tricky

[brisk iris]

I do. He screams of excitement while playing Fortnite, and I can’t leave the room cause Fortnite should be supervised

[tidal turret]

🤭 😂

seems like fun

I will close this, thanks everyone

.solved

[midnight plank]

Channel closed

Closed by @tidal turret

Use .reopen if this was a mistake.

[midnight plank]

Available help channel!

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>.
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[oblique gyro]

okay so how do i do this again can i just cancel the cosxcosy

[white gate]

I think you think too much

[oblique gyro]

well can you help me think less

[white gate]

Use the cos(x-y) and cos(x+y) formula

[opal garden]

try expanding cos(x-y) and cos(x+y)

[oblique gyro]

i did

but now what

[noble hatch]

what?

where

[white gate]

Where

[chilly adder]

the direction is kind of correct if im following it right

[oblique gyro]

it’s cosxcosy+sinxsiny
/
cosxcosy-sinxsiny

[oblique gyro]

on my paper

[white gate]

Divide numerator and denominator by cosxcosy

And that's it

[chilly adder]

$\frac{1+\tan x\tan y}{1-\tan x\tan y}=\frac{1+\frac{\sin x}{\cos x}\frac{\sin y}{\cos y}}{1-\frac{\sin x}{\cos x}\frac{\sin y}{\cos y}}=\frac{1+\frac{\sin x}{\cos x}\frac{\sin y}{\cos y}}{1-\frac{\sin x}{\cos x}\frac{\sin y}{\cos y}}\times\frac{\cos x\cos y}{\cos x\cos y}$

bruh

[quiet parcel]

bro had patience to write all that

mad respect

[chilly adder]

ok yeah

[grand pond]

@chilly adder

[chilly adder]

this what i meant

this is "reverse-expanding" the angle sum identities if that makes sense

[noble hatch]

@oblique gyro have you solved any of your hw questions yourself? cause ive seen your previous ticket and it might be better for you to work on some easier problems

[oblique gyro]

what..?

i’m confused is this shade like

[noble hatch]

no not really

[chilly adder]

nothing is shady

unless

[white gate]

bull's shade!

[noble hatch]

tl:dr dont bite off more than you can chew

[white gate]

tl;dr start easy

[chilly adder]

tl;dr bruh moment

[opal garden]

tl;dr hi

[oblique gyro]

idk what u being shady for saying start easy as if i haven’t already started easy? i wouldn’t be working on these had i not done easier things before

[chilly adder]

well then

is your question answered

[oblique gyro]

like i’m confused on what this mentality of struggling on a problem equating to dropping the problem itself

[chilly adder]

i continued your direction

#help-49 message

here

[opal garden]

I feel like you stick with one approach too long. Try to quit when one method stops working and try another

[noble hatch]

im not admonishing asking for help, its just that after finishing one question you asked for help on another not 5 minutes later

not trying to be hurtful here btw

[white gate]

no no no, its fine, its just that you overthink what could've been ended in 2 lines of reasoning

you should practice more about testing simple approach first then going for more complicated approaches when it fails

[noble hatch]

just work up in difficulty a bit slower

[chilly adder]

this

sometimes occam's razor works

[opal garden]

we belive in you Jayiee!

[oblique gyro]

yes because the chat was still open and i already had a specific question about cancelling a property

[chilly adder]

different questions regardless

[oblique gyro]

and mind you idk why @chilly adder decided to jump in to throw shade

like let’s keep it civil

[chilly adder]

what did i do?

[white gate]

Shade

[chilly adder]

im not even white wdym

i think i answered your question

[opal garden]

Hey Jayiee! What do you need?

Can you speak clearly?

[white gate]

!topic

[midnight plank]

Please read the channel description before posting, and stay on topic.

[chilly adder]

going back to this

yes your approach of writing tan into sin/cos is valid

if you continue it this way

but what we recommend here is to go with the approaches that come to your mind the first time

sometimes the most intuitive approaches outshine some complicated manipulation

especially in these

[white gate]

Wtf why are there so many bots recently

Today i met like 3 of them here

[formal blade]

This is very condescending. If you don't have anything relevant to the problem to say you should probably not chime in with things like this.

[chilly adder]

lmao damn

to be fair

[opal garden]

I don't really get what he's aiming at. Just looking for clarity in what he means, as I feel like what he's saying is scatterd. Might have been worded a bit harshly, but I had no negative intent

[chilly adder]

the one who asked the question started accusing people out of the blue

[formal blade]

If they aren't asking multiple questions or breaking any obvious rules there is nothing wrong with opening channels back to back like this. Sometimes people want more support.

[fathom onyx]

"Could you phrase that a little more clearly?" would be a nicer way of asking that question perhaps

[white gate]

Can we just stop here please

[chilly adder]

right

[oblique gyro]

Let’s not redirect this at me when i asked a simple question for yall to go off track about how I shouldn’t do the problem if it’s too hard

[formal blade]

The overall implication that they can't communicate clearly because you can't understand them is still just unproductive compared to you asking them to rephrase something specific

[chilly adder]

off-track?

[opal garden]

sure, as I said. Worded it poorly, but that's what I meant

[chilly adder]

i suppose i answered your question already here: #help-49 message

it is you who pressed on

[formal blade]

You don't need to pester them so much about their problem choice.

[white gate]

oh boy.

[formal blade]

You can say it once and let them choose to take or leave your advice

[chilly adder]

lmao.

[oblique gyro]

Pressed on what? My last message was very much minutes ago

[formal blade]

No need for this to be drawn out

[fathom onyx]

On track - Does what Shirihan said help you @oblique gyro ?

[formal blade]

How about ya'll move on to the problem at hand rather than pedagogical meta-advice?

[opal garden]

!topic

[midnight plank]

Please read the channel description before posting, and stay on topic.

[oblique gyro]

Well what he said where because after he posted the image there was still a lot of rambling clearly off topic

[oblique gyro]

Like this was posted 20 minute ago

[formal blade]

Okay good, lets get back on track.

Might help to repost the question down here.

[fathom onyx]

!repeat

er

[formal blade]

And maybe where you left off with what you understand too ofc

[chilly adder]

rambling where?

[fathom onyx]

is there a bot command to repeat the question?

[formal blade]

Drop it

[chilly adder]

okay

[chilly adder]

i ask this question again

[oblique gyro]

So that’s before you sent the image, let’s clock what you were doing in the 20 minutes between then and now

anywayy

[formal blade]

You also drop it

[noble hatch]

question here

[chilly adder]

well

[white gate]

Answer:

• Expand numerator and denominator of LHS
• Divide numerator and denominator by cosxcosy

[chilly adder]

answering the exact question asked

yes you can cancel cosx cosy

assuming that gives you the desired result

[white gate]

Note:

• The idea of dividing cos to obtain tan doesn't only appear in this problem, it is more common than you think.
• Simple extra homework as an example:
  Given tan(x) = 3/4. Calculate A = [sin(x) + cos(x)]/[3sin(x) + 2cos(x)]

In fact, this idea is so common that it can expand all the way to very advanced problems you can meet in the future (particularly calculus)

But that's a different topic for now

[midnight plank]

@oblique gyro Has your question been resolved?

Channel closed

Closed by @oblique gyro

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[midnight plank]

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• Show your work, and if possible, explain where you are stuck.
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[umbral sparrow]

am i right with C here?

[rocky storm]

explain your thought process

[dusty portal]

!show

[midnight plank]

Show your work, and if possible, explain where you are stuck.

[umbral sparrow]

i would but its on paper and i dont have a phone (ik its embarrasing)

but i got C

[dusty portal]

What did you do, though?

[umbral sparrow]

i used the law of cosine

[dusty portal]

Okay.

[umbral sparrow]

so would i be correct with 81.325 here

<@&286206848099549185>

[dusty portal]

What is this for?

[round parcel]

Can you type out your work? You asked this before in another channel with answer D that time (<#help-42 message>).

If you want to check it yourself, you can also use the law of cosines again and fill in all the values this time (including 81.325) and see if the sides equal each other.

If you do that, please say something like 6^2 = 5^2 + 3^2 - 2(5)(3) cos(93.82 degrees) or something.

That way, we can check to see if you have the right idea.

[midnight plank]

@umbral sparrow Has your question been resolved?

[midnight plank]

Channel closed

Closed due to timeout

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[wind inlet]

I have to find the non-permissible values for x. This is what I got, but in the answer key, it's π/2 + 2πn, nEI. Why's that..?

[fallow scarab]

show the answer key

[wind inlet]

11b

[fallow scarab]

probably typo then unless the question asks for something else

[wind inlet]

Oh shoot, I just checked, and the teacher made some corrections in the answer key. I got a picture, but didn't correct it on the paper

It is + πn, ye

Thanks!!!!

[kindred dagger]

What does $n \in \mathbb I$ mean?

[grand pond]

King Leo

[kindred dagger]

Like, what is the set I

[wind inlet]

N can be any integer

.close

[midnight plank]

Channel closed

Closed by @wind inlet

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[midnight plank]

Available help channel!

Send your question here to claim the channel.

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[last slate]

Let A and B be subspaces of X,then A $\cup$ B might not be a sub space of X.

[last slate]

Why is this the case?

nvm

.solved

[midnight plank]

Channel closed

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[midnight plank]

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[last slate]

True or false

(k)

Oen min

$c_1(x-1) +c_2(x-1)^2+c_3(x-1)^3 = b_1x^3+b_2x^2+b_3x^2+b_4x+b_5$

[last slate]

And when I expanded the left side

I got

$c_3x^3+(c_2-3c_3)x^2+(c_1-2c_2+3c_3)x+(c_2-c_1-c_3)$

[last slate]

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[midnight plank]

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[fierce canyon]

when x=-1 y=2.5 when x=0 y=4 when x=1 y=7 when x=2 y=13 find the relationship between x and y

[dusty portal]

Huh?

I suppose find the first, second, and third differences between the pairs?

[fierce canyon]

their differecnes double but im not sure how to use it

srry i wasnt on discord

[dusty portal]

Okay.

So we have the first differences as 1.5, 3, and 6. Correct?

[fierce canyon]

yeah

[fierce canyon]

how do we use this to find the relationship?

[dusty portal]

Take the second and third differences.

[fierce canyon]

3 and 6?

[dusty portal]

[fierce canyon]

oh wait do u mean like 1.5 and 3 and then 1.5?

[dusty portal]

Yes.

Now that I see it, I don't know what to do with this information.

[fierce canyon]

o

<@&286206848099549185> ?

<@&286206848099549185> pls

[raven abyss]

can you state all info given

[vast ginkgo]

look at the pattern in the differences

[fierce canyon]

thats what i was given

[vast ginkgo]

what happens to the difference for each increase in x

[fierce canyon]

it doubles righgt

[vast ginkgo]

yes

[ancient plank]

I've plugged it into desmos and found that it's cubic polynomial

[vast ginkgo]

what expression doubles each time x increases by 1

[fierce canyon]

wdym by that?

[vast ginkgo]

like what type of function can double its value every 1 x

[novel flax]

the successive difference of the terms are in geometric progression

1.5, 3, 6...

[fierce canyon]

would it be like 2^n

[vast ginkgo]

hint it's an exponential equation

[vast ginkgo]

yes

that's the factor for every x

[novel flax]

so its general term would be given by

[vast ginkgo]

now what's the difference at x = 0 (to x = 1)

[fierce canyon]

3

[vast ginkgo]

that would be considered your base change/increase

yes

now you can get 3(2)^x

for each increase

at x = 0 this will be 3

what do you have to add to get y = 4

[fierce canyon]

1

[vast ginkgo]

yep

that's your equation

[fierce canyon]

but why multiply by 3?

[vast ginkgo]

because 3 is the base change

3(2)^0 = 3(1)
3(2)^1 = 3(2) = 6
3(2)^-1 = 3/2 = 1.5

basically the value it should be at zero

it's just used to scale the exponential function

to the correct values

[fierce canyon]

mb i had to do smt, so basically ur saying that the base for 2^n was 1 and now that its 3 we multiply it by that to get 3,6,12....

[vast ginkgo]

i guess better to say initial value

the y intercept of an exponential function is always the coefficient a in a*(b^x)

a in this case is 3 because we start with a difference of 3

as x increases, this doubles

as x decreases this halves

because of 2^x

the difference still follows the points you gave above
from x = 0 to 1
3*2^1 - 3*2^0 = 3*2 - 3*1 = 6 - 3 = 3
from x = 1 to 2
3*2^2 - 3*2^1 = 3*4 - 3*2 = 12 - 6 = 6

(the +1 would cancel out since you're subtracting anyway)

[fierce canyon]

o think i understand it now tysm

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[wind inlet]

I'm trying to understand how I solved his question. I have no clue how the + cos 2x in the denominator changed to smth I'm not even able to understand where I got from💀

[white gate]

cos2x doesnt even change

the actual thing that changes is cos4x

[wind inlet]

Right... But I'm not sure what I did in the third step?

In the denominator

[white gate]

it's factorization

if you imagine in your head that u = cos 2x then the denominator becomes 2u^2 + u - 1

[wind inlet]

I think I get it

Thank you!!!!

.close

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[gaunt jetty]

,, \sum_{k=1}^n 2^k \ne 2^n

am i misunderstanding something in what you are writing?

[grand pond]

Aero

[gaunt jetty]

You are erasing the sum entirely

and resolving it to just 2^n

how did you go from orange to green here

[midnight plank]

@teal rover Has your question been resolved?

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[wet grail]

hi do people in here know how to do boolean?

[kindred dagger]

I personally dont, but lots of people do. Post your question and someone will help

[wet grail]

X'Y' + Y'Z + XZ + XY + YZ'

[rose trout]

What is the task?

[kindred dagger]

@wet grail

[wet grail]

simplify it

i'm stumped by the x'y' + xy

[rose trout]

Well first off Y'Z + XZ + XY have redundancy

The term Y'Z is redundant, so you can drop it.

Then, you can do it again with the last three terms.

[wet grail]

so then would the answer be XZ?

[rose trout]

No.
You can go from
X'Y' + Y'Z + XZ + XY + YZ'
to
X'Y' + XZ + XY + YZ'

Then use redundancy on the last three terms to drop another term (XY).

I don't think it simplifies to less than 3 terms

At least not with just OR, NOT and AND operators.

[wet grail]

so would yz' be the redundant one

and then i keep going till i have one left?

sorry my teacher hasnt taught me this yet

[rose trout]

You won't have one left

[wet grail]

oh

i didnt read that

so it would just be X'Y' + XZ + YZ'

[rose trout]

Yeah I don't think you can do much better afaik

[wet grail]

A’BC + AB’C + ABC’ + ABC

Can you help answer this one as well

[rose trout]

You can kind of factor things out a bit

For instance, the last two term can be "factored" as AB(C + C')

[wet grail]

can you explain to me why

cuz I see it as A'BC + ABC

A'B(C) + AB(C)

but i dont know how it becomes AB(C + C')

[rose trout]

ABC’ + ABC = AB(C' + C) = AB

You can see it as having the form XY + XY'.

This always simplifies to just X

Since it's telling you that the value of Y "doesn't matter"

[wet grail]

Y + Y' = 1?

how about A'BC + AB'C

Would it be C(A'B) + C(AB')

[rose trout]

You can see it like that, but it doesn't simplify as much

[wet grail]

so what would i do next after i get ab from AB(C' + C)

[rose trout]

You can factor again with the second term

So

A'BC + AB'C + AB
A'BC + A(B'C + B)

Then what do you think B'C + B is ?

[wet grail]

C

[rose trout]

Hmm not exactly

You can try and do a mini truth table for it

If B=C=0, then B'C + B = 0
If B=1, C=0, then B'C + B = 1
If B=0, C=1, then B'C + B = 1
If B=C=1, then B'C + B = 1
What does the expression behave like then?

[wet grail]

or function

would it be B + C

[rose trout]

Yes

So that means that you get A'BC + A(B+C) = A'BC + AB + AC

Then what do you get if you factor B out of the first two terms?

[wet grail]

B(A’C + A + AC)?

B(A’C + A) + AC

[rose trout]

Yes

And then what does the brackets become

[rose trout]

As per

[wet grail]

A + C

BA + BC + AC