help-27

[slow rampart]

wait i think ive figured it out

is this correct?

[wicked turtle]

that interface makes it really hard to read, but it looks right to me

assuming y is intended to be a function of x, as opposed to just a constant

[slow rampart]

okay okay! Thank you 🙂

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[rough veldt]

The drive motor of a particular CD player is controlled to rotate at a speed of 240 rpm when reading a track 5.5 centimeters from the center of the CD. The speed of the drive motor must vary so that the reading of the data occurs at a constant rate.
(d) Find a function S that gives the drive motor speed in rpm for any radius r in centimeters, where
2.3 ≤ r ≤ 5.9.
I dont know what to do

[narrow plank]

read this

[rough veldt]

no one wants to help me bruh

[sweet zinc]

what have you tried?

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[torpid ocean]

I did the geometric mean formula

so first term pi/4

i mean -pi/4

and then r would be -pi/4

n = 20

but how would i reduce it?

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[clever sorrel]

I want to understand this in step explanation

[mint flower]

Someone check help 10

[clever sorrel]

Class 10 ICSE

[plain sky]

did you try to expand both sides?

[clever sorrel]

Trigonometrical Identities

[clever sorrel]

No i need to prove that LHS = RHS

[plain sky]

what those 2 mean?

[clever sorrel]

Left hand side = right hand side

We mostly took LHS

I have solution but

[plain sky]

did you try to expand the most left side (1+sinA) /cosA with 1+sinA ? and the one left of it with cosA ?

[clever sorrel]

I want to know how that marked equation changed to 2sec A

[plain sky]

as you can see, 1+sinA can simplify themselfs out

and
1/cosA = secA

[clever sorrel]

Ah f me

Tysm

[plain sky]

np m8

[clever sorrel]

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[clever sorrel]

.Reopen

.reopen

[devout snow]

✅

[clever sorrel]

@plain sky

But it will be 2/cosA

So 2secA?

2 Sec A?

[main gull]

What's 1/cos equal to?

[clever sorrel]

Sec A

[clever sorrel]

Is this right?

[main gull]

Yes

[clever sorrel]

Ty both

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[main gull]

Because that's what the key says

[devout snow]

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[fickle jetty]

How did he get that span from the upper matrix

[orchid sierra]

please don't start a math question in the middle of a math problem.
Please post the entire problem statement.

presumably, they let alpha be the values they got, one at a time, to get the vectors

but you're not showing the matrix that simplifies to the one you are showing after substitution

[fickle jetty]

ummm this is the entire problem, and the bot said channel is available

[orchid sierra]

"still need" implies this is work in progress

where did this problem begin?

[fickle jetty]

this is the main question

[orchid sierra]

there we go

[fickle jetty]

answer sheet is 3 pages but doesn't explain jack shit so I just wanna know, how did they get 2 eigen vectors for a singular eigen value

in this case for eigenvalue = 1

[pulsar dock]

they just skipped all the calculation

[orchid sierra]

I noticed

I got some eigenvalue formulas but it looks like alpha CANNOT be 0, 1, 2

since then the eigenvalues would not be distinct

meaning un-diagonalizable

but the problem is for some reason assuming alpha is those values

(I am not the dude to do this problem)

[pulsar dock]

Just because the eigenvalues are not distinct does not mean the matrix is not diagonalizable.

Anyway, they basically took (A-I)x = 0 and solved that system. In parametric form you have (s,t,0) so two linearly independent solutions are (1,1,0) and (1,0,0). That's where the eigenvectors come from.

[orchid sierra]

*not distinct

[pulsar dock]

The hope is you can do it in your head, take A-I gives you a matrix of 0's everywhere except the 1 and the 2 in column 3. This has a RREF with 0's everywhere except a 1 in the bottom right corner. Then you see that x = s, y = t, z = 0 are your solutions

I'll edit thanks

[orchid sierra]

no I mean you got the theorem right, I am scratching my head because under the conditions for alpha, it looks like these eignvalues have repetitious pairs

I'll just watch though

sorry

[pulsar dock]

That's the point, for some alpha you need to look at the dimension of the eigenspace

I'm not going through these calculations, and this is more for you Disorganized, but the idea is that if two eigenvalues are the same, you take A-lambda I and the RREF needs a row of zeros for each repeated eigenvalue otherwise the solution to (A-lambda I)x = 0 won't have enough free variables to construct your eigenvectors.

[orchid sierra]

I see, thank you

[pulsar dock]

Take something like $\begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$ for example.

[woven radish]

Hexicle

[pulsar dock]

It has two eigenvalues, 1 and 1

[orchid sierra]

lol, I have that Brilliant page open rn

[pulsar dock]

It is not diagonalizable because the RREF of A-I is $\begin{bmatrix} 0 & 1\ 0 & 0\end{bmatrix}$

[woven radish]

Hexicle

[pulsar dock]

So you cannot construct the second linearly independent eigenvector

I don't know what Brilliant page you are talking about, it's the obvious first choice for an example after you've seen the Jordan Canonical Form.

[orchid sierra]

I washed out of grad school, only heard about Jordon Canonical Form there

I understand this though

thank you

[pulsar dock]

Without getting into the details, think of it as a way of getting something like a diagonalization when you don't have enough eigenvectors

[devout snow]

@fickle jetty Has your question been resolved?

[fickle jetty]

I think

I got it ish

thanks a lot, once again the solution explanations are f***ing trash

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[dawn plinth]

Can anyone figure this out

[pure anvil]

All the numbers have something in common except one of them

[keen rampart]

look at the last digits specifically

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[celest tide]

https://cdn.discordapp.com/attachments/931454414089228329/966261270259056640/766268632703434794.png

[celest tide]

https://cdn.discordapp.com/attachments/931454414089228329/966262002802630687/766268632703434794.png

[steel sage]

(2^y-2^x)(2^y+2^x)=4^y-4^x=2^y-2^x so y=x or y=log_2 (1-2^x)

[celest tide]

I understand the second solution but why y=x?

is it because if y=x, then (0)(2^x + 2^y) = 0?

ig so yeah thanks

.cox

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[woven warren]

Greetings. I cant figure out how i am supposed to solve the task no.4 i tried using the length of the stretches above me but that didn't work out and now im kinda stuck on this task

[woven warren]

The translation of the task is : You are given the ratio of 2 stretches AB and CD. Calculate the lang of the stretch CD

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[woven warren]

<@&286206848099549185>

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[wispy ginkgo]

.open

.question

.reopen

how do i open a help channel??

[wispy ginkgo]

oh

i need help on some questions

[winter patrol]

[wispy ginkgo]

"there are two tourist companies offering discounts on the same day, one of the companies offers 20% off if participants have more than 4 people, and the other offers $25000 off per person if more than 3 people. Group A wants to join the 20% off program and group B joins the $25000 program.

Q1: what is the formula of the question? (using x)

sorry for the late response was translating it from chinese

<@&286206848099549185>

anyone?

screw this then

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[grand pivot]

Can someone help me with L

[bold prism]

what do you need to do?

[grand pivot]

Inverse it

[bold prism]

what have you tried

[grand pivot]

This

I mean 1-3x

[bold prism]

seems correct to me

what's the problem?

[grand pivot]

Well my book answer key says its 3x-1/4x-2 , but technically my answer is the same, right?

[bold prism]

Yes what you have and the what the book says is the same

but the book one is cleaner

just multiply by -1 on the numerator and denominator

[grand pivot]

Oh okay thanks!

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[lost kettle]

pls help

[lost kettle]

It is known that f(x) = ln(2 + x).
a. Write down the 5 nonzero terms of the Taylor series around the point x=0
b. Use part (a) to find f(2.1) (4 decimal)
c. Find the maximum error of ln(2.1) using n = 4

[brisk prawn]

Show your work

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[trail igloo]

I need some help, I don't get how they did it at number 1 and I'm trying to break it down but I just can't get it

[trail igloo]

ah I see

nope, I can't understand it. I'm so sorry

yes

ahhh

I get it

and they did the same thing for 7 I assume

ah

for number 2 though 3^(1/3) is a bit confusing

ohhhhh

so then it's gonna be (6)^(1/3) x (3)^(1/3)?

or no

sorry brain fart moment lol

for q3, since it's a fraction, I assume things will go differently?

[whole cosmos]

not really. you still raise 2/5 to the 3rd power so you can have a common power

[trail igloo]

ah I see

I got the idea now, thanks a lot for the help!

for q5 how do I do it

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[grave iris]

No birds swim.
¬Ɐx(Bird(x) ˄ Swim(x))

[grave iris]

is this correct? seems wrong to me

because when I flip it I get:

ⱻx ¬(Bird(x) ˄ Swim(x))

[patent tartan]

Yes and that evaluates to "not(bird(x)) or not(swim(x))"

which doesn't exactly resembles the statement

[grave iris]

"There exist X, where X is not bird or X not swim" != No birds swim.

so therefore ¬Ɐx(Bird(x) ˄ Swim(x)) is wrong right

[patent tartan]

hmm wait a minute

[grave iris]

I have trouble understanding this since yesterday

sorry if im making it confusing

I flipped it based on this

[patent tartan]

It would be more intuitive to represent the statement using an implication and then convert that into lower-level logic symbols

So for example: "No birds swim":
Bird(x) => Not(Swim(x))

[grave iris]

yep I agree implication makes it easier to understand but the question im doing is asking this

if statement is not correct, provide the correct one

[patent tartan]

Oh it's only asking true/false

[grave iris]

yes

:)

[patent tartan]

if you rewrite the implication above

[patent tartan]

can you transform it into the representation given in the question?

[grave iris]

let me try

you mean this right: Bird(x) => Not(Swim(x))

[patent tartan]

For all x: Bird(x) => Not(Swim(x))

[grave iris]

try to fit into ¬Ɐx(Bird(x) ˄ Swim(x))

Ok gimme a while I'm slow

Starting with:
Bird(x) => Not(Swim(x))

Implication elimination:
Bird(x) V Swim(x)

Therefore, back to the question:
For all x, Bird(x) V Swim(x)
Ɐx(Bird(x) V Swim(x))

@patent tartan

oh wait I'm wrong

[patent tartan]

You need to negate the left one

[grave iris]

yes let me post again sorry

Starting with:
Bird(x) => Not(Swim(x))

Implication elimination:
Not(Bird(x)) V Swim(x)

Therefore, back to the question:
For all x, Not(Bird(x)) V Swim(x)
Ɐx(¬Bird(x) V Swim(x))

[patent tartan]

Second one stays the same

so keep the Not(Swim(x))

[grave iris]

🤦

ok

[patent tartan]

Not(Bird(x)) OR Not(Swim(x))

[grave iris]

brain is slow

ill change

Starting with:
Bird(x) => Not(Swim(x))

Implication elimination:
Not(Bird(x)) V Not(Swim(x))

Therefore, back to the question:
For all x, Not(Bird(x)) V Not(Swim(x))
Ɐx(¬Bird(x) V ¬Swim(x))

...should I simplify further?

[patent tartan]

Now you could use DeMorgan

[grave iris]

oke

:D

[patent tartan]

to "pull out" the NOT

[grave iris]

ok imma do it

[patent tartan]

not(a) or not(b) = not(a and b)

[grave iris]

Starting with:
Ɐx(¬Bird(x) V ¬Swim(x))

De Morgan:
Ɐx¬(Bird(x) ^ Swim(x))

[patent tartan]

Yes

so now the only thing that is different

[grave iris]

oh!

[patent tartan]

is that in the original question the "not" was before the "for all" quantifier

[grave iris]

yep: ¬Ɐx(Bird(x) ˄ Swim(x))

[patent tartan]

are those two equivalent?

[grave iris]

hmm

lemme think about this

¬Ɐx(Bird(x) ˄ Swim(x)) <--- original statement
Ɐx¬(Bird(x) ˄ Swim(x))

does not look equivalent to me

[patent tartan]

Think about a simpler example

Say you have:
For all x: Not(Breathes(x))

Which means, well, whatever you pick, it doesn't breathe

Now, Not(For all x: Breathes(x))

Basically means, there exists at least one x such that it does not breathe

Which is not the same thing as saying it MUST be true for ALL of them

[patent tartan]

So yes, they are not the same

[grave iris]

Ɐx¬(Bird(x) ˄ Swim(x))
means: For all x, whatever x I pick is not a bird that can swim

¬Ɐx(Bird(x) ˄ Swim(x)) <--- original statement
means: There exists at least one x such that x is a bird that can swim

[grave iris]

up until this point is very helpful thank you

[grave iris]

Did I get the correct meaning for both statements?

[patent tartan]

First one yes, second one means: "There exists at least one x such that it either isn't a bird or does not swim"

[grave iris]

oh yeah because of De Morgan

[patent tartan]

Yep

[grave iris]

awesome

!!!!

thank you :)

https://cdn.discordapp.com/emojis/933036572444602419.gif?size=44&quality=lossless

[patent tartan]

glad i could help :)

[grave iris]

very helpful, thank you! have a good day :P

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[restive river]

Hey

[devout snow]

Please don't occupy multiple help channels.

[restive river]

I need help with these!!!

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[restive river]

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[charred jackal]

could someone help me with this question?

[devout snow]

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[compact vine]

I have a math question, but don't have a picture, I'll do my best to describe it. In a triangle ABC, The angle bisectors of A and B cross in a Point N. Prove angle ANB = 90 degrees + 1/2 * angle C

[restive river]

need to know which answer is corrot

[compact vine]

I already claimed this channel

[compact vine]

You have to post it again in an available channnel

[restive river]

ahh alr

[devout snow]

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[compact vine]

.reopen

[devout snow]

✅

[compact vine]

<@&286206848099549185>

Can someone answer my question?

How do I prove angle ANB = 90 degrees + 1/2 * angle C

[restive river]

ANB=180-0.5angleA-0.5angleB
angleA+angleB+angleC=180
angleA+angleB=180-angleC
so ANB=180-0.5angleA-0.5angleB=180-0.5(angleA+angleB)=180-0.5(180-angleC)=180-90+angleC=90+angleC

@compact vine

[compact vine]

thank you

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[restive river]

i know this sounds easy, but i missed a lot of calculus classes and now im lost, but what do we do incase of absolutes

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.close

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[warped pasture]

$x^3-2ax^2+bx+12
is dividable by
x^2-x-12$
$$a=?, b=?$$

[woven radish]

Tanjiha

[lost gazelle]

??

[narrow plank]

Long division?

[lost gazelle]

what do you mean

or divisible

$x^3-2ax^2+bx+12
is divisible by
x^2-x-12$
$$a=?, b=?$$

[steel sage]

Just f(-3)=0=f(4)

You have two linear equations of a and b solve them

[warped pasture]

There is no remaining

[warped pasture]

Any easier way?

[steel sage]

That’s easy enough

[devout snow]

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[hushed pier]

Can someone explain why they have
S=0, R=1 then Q(t) is 0
but right after that has the row where
S=0, R=1 then Q(t) is 1??

[devout snow]

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[hushed pier]

https://cdn.discordapp.com/attachments/903481106819612714/966942625548623942/unknown.png

[hushed pier]

Can someone help me w/ this? idk where to start

i read my textbook for jk flipflops and d flipflops so i know what it does but not sure what the truth table should look like

do i basically do:
J=x, K=y
then do truth table for x,y,A
then put A and Y' into the XOR gate which is then equal to B?

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[craggy quiver]

i dont quite understand the question

so normally oh wait this is taylor not maclaurin

this shouldn't be on my spec

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[craggy quiver]

i dont understand this question

at all i think?

[inner plinth]

which

[craggy quiver]

the first part

how do you show that thing

do you just do

(x + iy)^3 = 1, work out W, then sub it in??:?

thats all im thinking rn

[inner plinth]

you can do that but that's gonna take time

try looking at the equation which gives the cube roots of unity

[craggy quiver]

i mean theres the

cos(ntheta/3) + isin(ntheta/3) thing ig?

or is it 2theta i forgot

[inner plinth]

no I meant

what does it mean to be a cube root of something

[craggy quiver]

that thing ^3 = that number

[inner plinth]

yea, so the cube roots of unity are just values which satisfy

$x^3=1$

[woven radish]

R31_

[inner plinth]

right?

[craggy quiver]

yh

i can do the z = e^(2kpi/3) thing if thats what u mean?

if i sub that in for n = 1 will that work?

[inner plinth]

so what can you do with it

its a polynomial

[craggy quiver]

i mean the obvious thing is cube root but idk what that would achieve

cos(0) + isin(0) = 1??

[inner plinth]

try subtracting 1 from both sides

[craggy quiver]

x^3 - 1 = 0, try and factorize it somehow?

[inner plinth]

yea, do you know difference of cubes

[craggy quiver]

briefly come across it many times but never learnt it

[inner plinth]

yep

[craggy quiver]

(x-1) (x^2 -x -1)

oh and solve x^2 -x -1 gives the other roots

[inner plinth]

+x+1

[craggy quiver]

oh yes sorry

then just sub solutions in?

[inner plinth]

so you know $\omega$ isnt $1$

[woven radish]

R31_

[inner plinth]

which means it will satisfy what

[craggy quiver]

the equation

[inner plinth]

yes

so substitute $\omega$ in $x^2+x+1=0$

[woven radish]

R31_

[shy quest]

another way u could think about this is using the Geometric Progression sum

and the fact that omega^3=1

[craggy quiver]

why does it not work

if i let w = e^(2i pi/3)

[shy quest]

it does

[craggy quiver]

oh is it just my calculator does not support me writing e^ipi

oh yh calc give math error

[inner plinth]

,w (e^(2ipi/3))^2+e^(2ipi/3)+1

[woven radish]

Results provided by WolframAlpha

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

Requested by R31_#9293

[craggy quiver]

yh turning it into cos isine thing makes it work

geometric progression sum?

[shy quest]

yeah

[craggy quiver]

how?

[shy quest]

u know what that is?

[inner plinth]

yea, basically the formula comes from factoring

since $x^{n+1}-1=(x-1)(x^n+x^{n-1}+\dots +1)$

[woven radish]

R31_

[inner plinth]

so $x^n+x^{n-1}+\dots +1=\frac{x^{n+1}-1}{x-1}$

[woven radish]

R31_

[shy quest]

and in our case n=2

[inner plinth]

yes

[shy quest]

@craggy quiver makes sense?

[craggy quiver]

i think so

[shy quest]

u know how u get the sum of a geometric progression?

[craggy quiver]

yh just the formula

[shy quest]

then don't use this

use the approach suggested earlier

[craggy quiver]

oh ok, ill have a think about what you have said just anyway but yh might be a bit above me lol.

thanks

[shy quest]

sure once u learn this it'll be really simple

[devout snow]

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[torpid night]

This is just a quick question but

[torpid night]

What is that number that's circled in red?? does that look like 0*10^3 to you guys

[craggy quiver]

work backwards

sub in the delta t value at the bottom

into the equation at the top

and see what it comes out as

deffo not 0... cant have 0 in denominator

[winged hazel]

its an 8

[torpid night]

This is for physics 1 class btw

[winged hazel]

just bad hand writing

[torpid night]

weird and bad handwriting

[winged hazel]

Yeah, not sure if you meant what the constant\varible is, that i dont know, but the number is 8000, if thats what you meant

[torpid night]

and even then, this is the "solution" to one of my problems with different numbers but it doesnt explain where 8000 comes from

nvm im dumb

[rotund heron]

its 8

[torpid night]

Yeah I just realized that the question gives you a (number)kiloWatts, the example had 8kW, and I got 6.89, i was just supposed to plug it in

I hate to bring in my physics work in here, but I ran into another issue;

nvm me asking helped me realize again

But either way guys, thanks for the help on the arithmetic

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[craggy quiver]

i dont know what it means by find the greatest value of |z-w|

here is the mark scheme

im guessing its something to do with the dotted line?

idk

oh i just scrolled down on the mark scheme and

here is the bit about z (i still dont understand it)

but i also dont understand what that dotted line is

2 questions then

[devout snow]

@craggy quiver Has your question been resolved?

[grim swallow]

Try graphing these on desmos.
R^2 is isomorphic to C with clear conversion maps between them

[devout snow]

@craggy quiver Has your question been resolved?

[craggy quiver]

i dont understand this

[devout snow]

@craggy quiver Has your question been resolved?

[leaden aurora]

1 sec

[leaden aurora]

Ok. I've read through the problem. But I don't understand what you're confused about

Firstly, do you understand why that length is the largest distance?

or the greatest value of |z-w|

[craggy quiver]

i dont understand what |z-w| is

[leaden aurora]

|z-w| is the distance between z and w

But

[craggy quiver]

z is the origin in this case, right?

[leaden aurora]

we know z can be any value in the shaded part

[craggy quiver]

oh yes alright

[leaden aurora]

The condition given in (ii) states that $|z|\leq 4$ and $-\frac{\pi}{2} \leq \arg{z} \leq \frac{\pi}{3}$

hm latex broke?

Seems latex is down according to #math-discussion

I am sorry

So z can be an value in the shaded region because of the condition given in (ii)

But we want the one which gives the largest distance from w

[craggy quiver]

it seems its clear from looking at the diagram, that the bottom is gonna be the furthest distance. is there any more mathematical approach though?

[leaden aurora]

hmm i was dreading that question

lemme think about it

well actually it should be obvious

imagine rotating the red line clock wise

We know that it will reach a maximum when it has been rotated so the entire line is a diameter

[craggy quiver]

ah yes truetrue

i think that satisfies me, thanks for the help as always 🙂

[leaden aurora]

cool and getting the length you can do right? just trig

[craggy quiver]

yh i got that

[leaden aurora]

cool np

[craggy quiver]

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[devout snow]

@naive umbra Has your question been resolved?

[devout snow]

@naive umbra Has your question been resolved?

[tiny dagger]

<@&286206848099549185> someone help this man

[carmine sand]

Sample size (n) =1500

Sample proportion (p) = 0.39

z value at 95% confidence interval =1.96

I forgot how to find xD

0.365<p<0.415, is there answers like this

[naive umbra]

ye there is

[naive umbra]

how did you get this?

[naive umbra]

how about the margin of error?

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[urban bear]

how can I find all the possible solutions to this so I can get the roster form of this set?

[devout snow]

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[shadow bolt]

does anyone know how to show that an function has rotational symmetry?

[shadow bolt]

we are told

which is x = -a

and that

would it be suitable to substitute p = -a

q= 2a^3 - ab + c

and see if this satisfies?

<@&286206848099549185>

[devout snow]

@shadow bolt Has your question been resolved?

[hybrid snow]

Find the second derivative

[shadow bolt]

I did

6x + 6a (and for inflection) =0

[hybrid snow]

Yeah so what would x be

Well you have it above I saw

x = -a

Which is p

[hybrid snow]

I think so

[shadow bolt]

alright, thanks

[shadow bolt]

when I did it before I got a = 2

would this be meaningful

or likely just an error

[hybrid snow]

Yeah you can solve for b

And then eventually solve for c

I guess finding maxes and mins

Can help you solve for b

[shadow bolt]

so once I found b and c

[shadow bolt]

would you show the max/min are equally spaced or something?

[hybrid snow]

Well you can find Q

And then plug it into that 180 degree funny spinny equation

[shadow bolt]

oh yeah

thanks very much

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[timber pebble]

want to give this a go in questions channels before i go to dedicated

anybody give some guidance on where to even start with this one?

a hint is provided to combine the formulas for the coefficients

idk if that helps

its not immediately evident to me that it would be

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[timber pebble]

I have I think a silly question

[timber pebble]

I'm trying to convince myself that $\int _{-\pi} ^{\pi} i\sin \qty( z \sin \theta - n \theta ) ) \dd \theta = 0$

[woven radish]

jan Niku (Shuri for Honorable)

[timber pebble]

i get even and odd is a thing its just

i dont necessarily believe it here

there should be some circumstance where we can complicate the inside of a sine function enough to make that property not hold, right?

i mean we definitely can

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[strange heath]

I dont get this

[native lake]

its asking the rate of depreciation

[strange heath]

I tried doing y1=18500(1-0.87)^x

ya ik that

but idk how to find the amount

[native lake]

p sure its 87%

18500 is initial value

n is number of years

0.87 is the depreciation

to convert 0.87 to a percentage

87%

no wait

after 1 year

it would be 18500 x 0.87

it would be 87% of its original value

so that means its depreciated by 13%

does that make sense?

[strange heath]

why did u do 18500 x 0.87

and to get my final amount is it just 18500(1-0.87)^1

@native lake

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[shadow bolt]

hey

[shadow bolt]

I don't know if I'm allowed to do this

but please can you help me again?

[hybrid snow]

I am in constant suffering

Uh sure

If I have time

I have like uh

3 minutes lol

[shadow bolt]

ok cool, please just tell me when you need to go (thanks)

[hybrid snow]

Aight what's the problem/question

[shadow bolt]

same as yesterday

I'm struggle to find b and c

the derivative is 3x^2 + 12x + c = 0

and inflection at (-2, -16 + 2b + c)

but I can't seem to find anymore information to use

[hybrid snow]

Inflection happens when y" = 0

So you can solve for c in terms of b

[shadow bolt]

wait, how?

[hybrid snow]

Oh wait hold on

That's the point of inflection?

[shadow bolt]

yup

[shadow bolt]

this is

[hybrid snow]

Aight I gtg but I think you plug those into the 180° thing

I don't think you need to find actual numbers

[shadow bolt]

ok, thank you for your time 🙂

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[restive river]

hey

[restive river]

how would i go about doing these types of questions?

i tried rearranging the formula 1/2 ab sinc but got an answer really far away

[patent tartan]

For both?

[restive river]

mainly the first one

if i figure out how to do the first then im hoping i can do second on my own

[patent tartan]

If you solve sin(x) = 1200 / (28 * 52) you should get two solutions: (1) x = 55.51 and (2) x = 124 (approximately) on the interval [0, 2pi]

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[restive river]

how do i find the height

[main gull]

Pythagorean theorem

[restive river]

i tried but i dont know the base

[main gull]

Oh, wait, similar triangles

You can apply that

Create a proportion

[restive river]

could you walk me through that please

[main gull]

What side relates to the 10cm?

[restive river]

8cm?

[main gull]

Yes

Because notice, how the 10cm is the hypotenuse of the bigger triangle, 8cm is the hypotenuse of the medium size one

[restive river]

yes

[main gull]

What about h?

[restive river]

h it relates to 8 csm aswell?

[main gull]

No

[restive river]

oh

[main gull]

h is the base of the medium triangle, what is the base of the bigger one?

[restive river]

6cm

[main gull]

Yes

Now, if you recall proportions are in the form of $\frac{a}{b} = \frac{c}{d}$

[woven radish]

dldh06

[restive river]

ohh

ok so

does that mean

10/6 = 10/h?

[main gull]

Almost, I think you made a typo

[restive river]

oh

i dont know were i went wrong sorry

[main gull]

Why are there two 10s?

[restive river]

6/h

its that right

[main gull]

No

[restive river]

oh

im sorry i never learned similar triangles i just transfered to a new school

[main gull]

I asked what side relates to 10cm, you said 8cm, where is 8cm?

[restive river]

oh

so

8/h

[main gull]

Yes so what is the proper equation?

[restive river]

10/6 =8/h

[main gull]

Now solve for h

[restive river]

ok

h = 4.8

[main gull]

If you calculated that right, then sure

[restive river]

i crossed multiplied then devided 48 by 10

its that how you do it

i think thats how you do it

thankyou for the help

i found it out

have a nice day :D

[devout snow]

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[restive river]

can i please get help with this question

[devout snow]

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[restive river]

.close

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[tacit needle]

I'm new to linear algebra, I only know a handful of things, however I'm solving a complex problem where a certain techinque seems seems to be very helpful.

I found a 2d matrix that describes a linear transformations. I know how to convert a vector from normal coordinates to the new ones. What's a method to get the vector from the transformed coordinates to normal coordinate system?

[ivory tapir]

$v_1 A = v_2$, $v_2 A^{-1} = v_1 A A^{-1} = v_1$

[woven radish]

KurtDee

[tacit needle]

I'm sorry, I don't really understand what's going on

[ivory tapir]

Can you define your new coordinate system

You have a linear transformation

A

You convert your vector using $xA$ or $Ax$ whatever

[woven radish]

KurtDee

[ivory tapir]

traditionally $Ax$ now that I think about it

[woven radish]

KurtDee

[ivory tapir]

so $A \in \mathbb R ^{2 \times 2}, x \in \mathbb R ^2$

[woven radish]

KurtDee

[ivory tapir]

You get some $y \in \mathbb R^2$, $Ax = y$

[woven radish]

KurtDee

[ivory tapir]

If you want to undo the transformation, you need to find $A^{-1}$ such that $A^{-1}A = I_2$ where $I_2$ is the identity matrix

[woven radish]

KurtDee

[ivory tapir]

such that given $Ax=y, A^{-1}A x = A^{-1}y \implies I_2 x = A^{-1}y \implies x = A^{-1}y$

[woven radish]

KurtDee

[tacit needle]

Oh, I see now

[ivory tapir]

So if $A^{-1}$ doesn't exist, you cant undo your transformation (for a lot of reasons)

[tacit needle]

How do I find the $A^{-1}$?

[woven radish]

Frogieder

KurtDee

[ivory tapir]

The easiest way is using an augmented matrix

I'll try to explain the reasoning why it works

So you can set up a matrix $\begin{bmatrix} a_{11} & a_{21} & | & 1 & 0 \ a_{21} & a_{22} & | & 0 & 1 \end{bmatrix}$

[woven radish]

KurtDee

[ivory tapir]

And reduce it to $\begin{bmatrix} 1 & 0 & | & b_{11} & b_{21} \ 0 & 1 & | & b_{21} & b_{22} \end{bmatrix}$

[woven radish]

KurtDee

[ivory tapir]

With B being your $A^{-1}$

[woven radish]

KurtDee

[ivory tapir]

The reasoning is that every row operation of a matrix can be considered a matrix transformation in itself

I can prove that if you dont understand

so lets say you have some list of transformations

$A * C_1 * ... * C_N = I_2$

$C_1* ... * C_n$ is effectively $A^{-1}$

[woven radish]

KurtDee

KurtDee

[ivory tapir]

Does this make sense

[tacit needle]

I'm sorry, I don't know what does the bar in the middle of those bigger matrices mean

I'm still getting into it, this field of maths is mostly new to me

[ivory tapir]

The bar is just symbolic

doesnt mean anything

https://www.mathsisfun.com/algebra/matrix-inverse.html

A separate resource can explain it better than me

[tacit needle]

Oh, I see now

Thank you so much

.close

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[dusk mica]

the mean mass of 12 people is 80 kg. two of the people have a mass of 75kg. find the mean mass of the remaining 10. the answer i got is 81. is that right...

[dusk mica]

edited my message

[timber pebble]

well how could you check?

[dusk mica]

um idk

i just removed the mass of the two people

from the previouis total (960)

and then divided by 10

instead of 12

[timber pebble]

so 2 people have mass of 75kg

[dusk mica]

ya

oh

[timber pebble]

and 10 people have an average mass of 81 kg each

[dusk mica]

ohhh

i could check by

[timber pebble]

whats their total mass?

[dusk mica]

the fact that it increased by 1

and the mass decreased by 5

0.5 x 2 = 1

[timber pebble]

👀

[dusk mica]

ok so its correct ( i think) right

[timber pebble]

i dont understand what youve just said

[dusk mica]

um

[timber pebble]

but your answer is correct

[dusk mica]

perfect

thx

[timber pebble]

was just gonna say

[dusk mica]

.close

[devout snow]

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[timber pebble]

check the mean is still the same

275 + 1080

etc

[dusk mica]

thanks for da help

[devout snow]

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[cosmic quest]

hello

[cosmic quest]

so to preface this question the last time i did anything with matrixes was 2006.

im given that

\begin{matrix}
0&1 \
2&0
\end{matrix}

[steel sage]

Use two \ to create a new row

[woven radish]

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

[wooden veldt]

\\

[cosmic quest]

2^.5 root 2 whatever you wanna call it or write it

because matrix A^2 is

\begin{matrix}
2&0\
0&2
\end{matrix}

[woven radish]

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

[cosmic quest]

so A^2=2*(1)

so how do i reverse engineer this and take say root 3 or root5 or 3+root 7 is some matrix?

[wooden veldt]

Sorry your question is a bit unclear, are you asking how to find "roots" of matrices?

[steel sage]

$A=\begin{pmatrix}
+/- \sqrt{2}&0\
0&+/- \sqrt{2}
\end{pmatrix}$ or $\begin{pmatrix}
a&b\
\frac{2-a^{2}}{b}&-a\end{pmatrix}$

[woven radish]

Cogwheels of the mind

[cosmic quest]

ya lets make it more concrete

find some matrix B such that
B^2=3 +7^(1/2)

[wooden veldt]

Psst \pm

[steel sage]

I thought he was asking find A such that A^2=2I

Anyway, method is the same, just directly calculate it

$A=\begin{pmatrix}
+/- \sqrt{3+\sqrt{7}}&0\
0&+/- \sqrt{3+\sqrt{7}}
\end{pmatrix}$ or $\begin{pmatrix}
a&b\
\frac{3+\sqrt{7}-a^{2}}{b}&-a\end{pmatrix}$

[cosmic quest]

oh is it both? ok.... hmm any recources to show me why that is?

[woven radish]

Cogwheels of the mind

[steel sage]

Like I said, directly calculate it

[wooden veldt]

@steel sage use \pm instead of +/- :)

[steel sage]

Set A=(a,b;c,d) solve a,b,c,d

Thanks

$A=\begin{pmatrix}
\pm \sqrt{3+\sqrt{7}}&0\
0&\pm \sqrt{3+\sqrt{7}}
\end{pmatrix}$ or $\begin{pmatrix}
a&b\
\frac{3+\sqrt{7}-a^{2}}{b}&-a\end{pmatrix}$

[woven radish]

Cogwheels of the mind

[cosmic quest]

riiiiiight ok that kinda makes sense

brb lol

since matrix multiplication takes an eon to do by hand

ok im only partway through analysing this but i can see why either c and b(in the matrix[a,b;c,d]) are zero or a=-d

ok cool i think i like the rightmost one better but i do understand it thanks for the help folks

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[hybrid snow]

Well draw a diagram

You always gotta draw some visual

[devout snow]

@vestal aspen Has your question been resolved?

[vestal aspen]

Oh yeah I've done that

But I'm not sure if it's done correctly

[hybrid snow]

Let's see it

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[scenic solstice]

Is this true or false

[devout snow]

@scenic solstice Has your question been resolved?

[sweet zinc]

try to solve one side.

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[stray lark]

Can someone explain why for a) Negative, why it isn't closed brackets?

[stray lark]

I'm referencing this definition

[graceful cosmos]

This definition is wrong

[stray lark]

o

[graceful cosmos]

But most calc classes won't see the difference

[stray lark]

I can't tell

I'm p sure that my class

had a debate

on why this was right

so tbh I don't really know

[graceful cosmos]

So for example, it wouldn't be negative on [0, 1.25] because the slope is not negative at 0 or 1.25 haha

Another good case is √x, where the function has a positive slope on (0, inf) but has no derivative at all at x = 0

[stray lark]

oh

ohhh

so technically

it's not increasing on those points

[graceful cosmos]

f'(0) = 0
The function is not increasing or decreasing there.

[stray lark]

right

my teacher said something along the lines of

functions don't increase at a point

but between an interval

[graceful cosmos]

Weird thing to say

[stray lark]

I see your point though

like it's not increasing there

since f'(0) is 0

[graceful cosmos]

But you know they're the one grading you so do whatever you can for those grades

[stray lark]

Yeah I will

But thanks for the reasoning

[graceful cosmos]

In general, behavior on (a,b) says nothing about x = a, or x = b.

[stray lark]

it makes sense why they used open brackets

[stray lark]

right just like 0 and 1.25

[graceful cosmos]

The corollary is weird.

[stray lark]

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@stray lark Has your question been resolved?

[devout snow]

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[final matrix]

I do not understand when to make groupings. i.e: (x + 28) + (4y - 2).

I know how to SOLVE with groupings, but I do not know WHEN to make groupings. I need help on that

[brisk prawn]

What are you solving?

[final matrix]

I am just confused on when to make groupings

It is just something that popped up in my head, and I get paranoid if a question about groupings comes up

[torn vessel]

groupings of what?

[final matrix]

Like parenthesis

(x + 2) + x = 3/4

[novel goblet]

Well, grouping is to group like terms together

[torn vessel]

thats.... wildly general

[novel goblet]

In this example, the like terms are the numbers, ie 28 and -2

So, this can be grouped as x + 4y + 28 + (-2)

[final matrix]

Ohh

[novel goblet]

For this example, the like terms are x. So you can group as x + x + 2 = 3/4
2x + 2 = 3/4

[final matrix]

So basically, groupings are there to simplify things?

[novel goblet]

yes

indeed

[final matrix]

Oh I see

Alrighty

Thank you

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[novel goblet]

welcome

[devout snow]

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[craggy quiver]

why do you reject -

wait i can prolly guess it tbf

cosh y graph is like U

so inverse needs to be one to one

so cut it in half

.clos

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[craggy quiver]

i keeep rubber ducking

[devout snow]

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[devout snow]

@olive swallow Has your question been resolved?

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@olive swallow Has your question been resolved?

[devout snow]

@olive swallow Has your question been resolved?

[olive swallow]

.helpers

<@&286206848099549185>

[supple trench]

just ask wolfram alpha for self-check

,w d/dx cot(x)

[woven radish]

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[supple trench]

,w domain of -(csc(x))^2

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[supple knot]

@olive swallow this is your answer. just write csc in terms of sine and you'll see why

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[tropic ledge]

How do i solve b?

[tropic ledge]

I can solve part a i get that: (P: T - 2mg = 2a) and (Q: 5mg - T = 5ma)

I know for part b I need to find acceleration but how?

[devout snow]

@tropic ledge Has your question been resolved?

[magic mauve]

https://discord.gg/physics

[tropic ledge]

<@&286206848099549185>

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@tropic ledge Has your question been resolved?

[tropic ledge]

Nvm i was being such an idiot

.close

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[odd peak]

I feel like I made a mistake somewhere

[wooden veldt]

What is the question?

[odd peak]

In the first picture

I have to first find the second derivative of (e^x)/(4+e^x)

Then set the numerator to zero

,w second derivative of (e^x)/(4+e^x)

[woven radish]

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[odd peak]

Oof

,w first derivative of (e^x)/(4+e^x)

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[odd peak]

Ok well at least I got that right

Here's my second attempt at that second derivative

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[odd peak]

<@&286206848099549185>

[devout snow]

@odd peak Has your question been resolved?

[odd peak]

<@&286206848099549185>

[supple knot]

use use the result of your first derivative

[odd peak]

But it's asking for cupped up and down

So we need a second derivative

[supple knot]

the second derivative is the derivative of the first derivative.

[devout snow]

@odd peak Has your question been resolved?

[odd peak]

Yes but is my second derivative correct?

[pulsar dock]

I don't understand why you would use wolfram alpha to confirm your calculation of the first derivative, but wait over 3 hours for someone to check your second derivative.

[odd peak]

Because Wolfram gave a derivative that I don't know if it is equivalent to the one I found