help-33

[brave spire]

Rad i think

[drifting nimbus]

in rad of course

[green sky]

Alrigh then

[rustic rain]

sry then why is it 10(2)

[green sky]

Which one?

[rustic rain]

q6c

[green sky]

It contains chord AB

[fathom ridge]

,calc 0.5(3.14 - 2.3)

[elfin berry]

Result:

0.42

[rustic rain]

btw what's a chord

[green sky]

Google it

[rustic rain]

so the answer for q6c is wrong?

10(2.3)+18.3

[twilit grove]

I would think it is. 2*10 to me looks like they're including AO and OB

[rustic rain]

so it's this?

[twilit grove]

if 18.3 is the length of chord AB, then yes that would've been my answer

[rustic rain]

alr thx

[marsh citrus]

@rustic rain Has your question been resolved?

[marsh citrus]

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[torpid basin]

I'm trying to figure out if I need to use sin, cos, or tan and how I'd type that into a TI-89

[gilded ivy]

elevation?

oh wait nvm

[torpid basin]

The angle of elevation

[gilded ivy]

yeah lol

i didnt see that

14 foot ladder 12 foot wall so base is (sqrt 52) foot

uh from there you uhhhh

wait cant you juse use trig

[gilded ivy]

ohhhh

you can use any i think

[torpid basin]

Okay

It worked

Thanks!

[gilded ivy]

tan(x) = 12/(root 52)
. = 59 degress right

[gilded ivy]

what was it

[torpid basin]

59

I have another question

I'm lost on what to do with this one

@gilded ivy

[gilded ivy]

hm?

just gimme a second my 12:30 am english skills are dying

im still trying to understand the question

[torpid basin]

No worries

[gilded ivy]

wait so shes on a building roof

40 meters away from airport

[torpid basin]

Yeah

[gilded ivy]

but its 400 meters tall

[torpid basin]

Yeah

[gilded ivy]

is 24 degrees up or down

[torpid basin]

Up

[gilded ivy]

and the 24 degrees is off the ground 24 degrees?

[torpid basin]

24° from the top of the building

[gilded ivy]

oh i see

wait the plane is coming towards her

[torpid basin]

Yeah

[gilded ivy]

so is she seeing it like this

[torpid basin]

Yeah she is

[gilded ivy]

alright

im gonna be honest this is way harder than before but i think i might be able to get it

[torpid basin]

No worries, I've got plenty of attempts

[gilded ivy]

💀

[gilded ivy]

i dont really have any idea how to get any measurements sooo

maybe ping helpers

[torpid basin]

Okay

<@&286206848099549185>

[torpid basin]

Since we're trying to find the planes altitude could we ignore the 400m and just work with 40m away and 24° up and add the 400m onto the answer?

[gilded ivy]

thats the thing i really dont know if you can apply those measurements

you can try

i mean you got a hefty amount of attempts left

[torpid basin]

Alright let's do that

Would I need sin, cos, or tan for this?

[gilded ivy]

you got the angle

you got the adjacent

the 40m

uhh

yeah youre missing a measurement thats the problem

idk how to get either hypotenuse or opposite

i mean if i had the opposite i might as well have the answer

[torpid basin]

Let's move to another answer and I'll come back to this one later today

[gilded ivy]

yeah ok

[torpid basin]

Alright here's the next one

[gilded ivy]

so trying to get hypotenuse

ok i wish i was better at trig but i dont think i could get this one etiher

i havent really been taught trig i just learn stuff off the internet

[torpid basin]

I got it, I had to type in 2500/tan(25)

[gilded ivy]

yeah sorry im not the best at trig

i know how to apply but thats about it

any rules and stuff is a no go for me

[torpid basin]

No problem, I'm not either

[gilded ivy]

i managed to clear 2 other help tickets in the meantime 🤣

[torpid basin]

Lol nice

Alright next one

[gilded ivy]

WTF ITS 1 AM

eh its fine

[gilded ivy]

so from the botoom of the slide to the bottom of the pool is 57 ft

we know the depression is 45 degrees

do we know any lengths of the slide whatsoever?

wait what the

[torpid basin]

45 feet on the slide

[gilded ivy]

am i reading it wrong or are there no measurements

oh

wait

but it says 45 ft to the pool from the slide

doesnt that mean off the slide

[torpid basin]

I think it's a straight slide

So essentially yeah 45 feet to the pool

[gilded ivy]

but do we have slide measurements

it just sayd 45 degrees

i dont see any slide measurements

[harsh rain]

hello

[gilded ivy]

striker can help

its 1am for me i need to go

[harsh rain]

is it approximately 45 feet

[torpid basin]

Alright, thanks for the help!

[torpid basin]

It has a constant slope

Of 45°

[marsh citrus]

@torpid basin Has your question been resolved?

[marsh citrus]

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[azure trench]

stuck on b

[lucid zenith]

what's the circle's radius

[azure trench]

root 18

[lucid zenith]

so draw a circle

with radius root 18

[smoky terrace]

3sqrt2

[azure trench]

lmao

[lucid zenith]

and draw a chord of length 6

[azure trench]

yh nvm

😭

thanks

.close

[marsh citrus]

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[thorny kayak]

need help with part a

[thorny kayak]

it’s a past paper for revision, not an exam

[real creek]

The first thing that I would try is t formulae

Since they are easier to work with than the trig functions

[thorny kayak]

i haven’t done t formulae

the mark scheme wants me to use trig rules but i don’t understand the process

like tan = sin/cos

[real creek]

Can you send a link to the MS?

[thorny kayak]

yeah

question 13a

oops wait hang on

there you go

[real creek]

thanks

I'm just doing it on paper then will send photo

[thorny kayak]

ok

[real creek]

I tried to explain my thought process whilst doing it

If you want more explanation just lmk. The hardest thing was spotting that you can multply but the cos()/cos() and then factorise out 1+sin()

@thorny kayak Do you understand it or want me to explain further?

[thorny kayak]

thank you so much i understand it now

your thought processes really helped

[real creek]

np if that's all you can close the channel 🙂 (.close)

[thorny kayak]

.close

[marsh citrus]

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[final plover]

Hiw do I slove this

[final plover]

The triangle is isocseles

We are looking for the radius of the inner and outher circle

And AO is the radius of the outer

Brb ill explain in a min

.close

[marsh citrus]

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[wary bluff]

How would I convert (x^2)/(8+x^3) to a mclaurin form

[wary bluff]

I started wit findin the first 3 derivatives but that gets so messy

apparently theres another way

but idk

[main idol]

derivatives are not necessary

do you know the maclaurin series for
$\frac{1}{1+y}$

[elfin berry]

riemann

[wary bluff]

wait I can convert it into a geometric series form?

[main idol]

yes

[wary bluff]

wait wont it be

1/1-x?

so

$\frac{\frac{x^2}{8}}{1 - (- \frac{x^3}{8})}$

[elfin berry]

Calc II Victim

[main idol]

yea that's right

[wary bluff]

$\sum_{n=1}^{\infty} \frac{x^2}{8} (- \frac{x^3}{8})^{n-1}$ ?

[elfin berry]

Calc II Victim

[main idol]

simplify and you should be done!

[main idol]

you can try plotting the polynomial just for a few terms (maybe 5?) in desmos and compare with the original function

[marsh citrus]

@wary bluff Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[jagged turret]

Hey. Calculus problem. Can you help me solve this? Idk how to start

[dire spire]

optimisation

draw the box first

label sides

come up with a function

find the derivative

then the minimum inside your domain

[jagged turret]

volume of 8 cubic feet?

what am i suppose to do with that lol

@dire spire ! saad

[dire spire]

what is the width?

[jagged turret]

2x

[full orchid]

this question is weird

[dire spire]

no, “twice as long as it is wide”

[dire spire]

width is x

what is the length?

you’re not helping

[full orchid]

sorry lol

[jagged turret]

tbh man

i didnt really go over this

in class

shits hard

gonna need extra help

on this one

twice as wide, i think of "2x"

[dire spire]

read the question, write down what you’re given first

[jagged turret]

8 cubic feet = volume

idk how to convert that

[dire spire]

volume is 8, good

now

width?

“twice as long as it is wide”

[jagged turret]

16?

[dire spire]

no

not double of volume

[jagged turret]

help me make use

of volume

[dire spire]

you’ll get there

[jagged turret]

how is volume

found

whats the formula

[dire spire]

you need to know what to use first

width is x

length is 2x

height is y

volume = lwh

can you isolate y?

[jagged turret]

i labeled

all the sides

you listed

[dire spire]

good, now you can isolate a variable using the volume formula

[jagged turret]

4x+2x+4y

isolate y?

[dire spire]

how’d you come up with that

[jagged turret]

4y=-4x-2x

divide by 4

y=x-1/2x

[dire spire]

huh

4y?

volume = length * width * height

[jagged turret]

2x times x times y

good?

[dire spire]

yup

[jagged turret]

=8

[dire spire]

yup

[jagged turret]

isolate 8?

[dire spire]

isolate y

[jagged turret]

4x^2=y?

[dire spire]

4/x^2*

[jagged turret]

m

bb

mb

4/x^2=y

[dire spire]

now define the surface area function

[jagged turret]

huh

[dire spire]

you're minimising material needed

[jagged turret]

i have lwh now

[dire spire]

S.A. = 2lw + 2lh + 2hw

[dire spire]

thats volume, i.e. the space inside the container

you need to find the function that tells you the minimum amount of material needed to build the box

i.e. surface area of the box

[jagged turret]

ok 4x^2 + 16/x + 8/x

i get it

[dire spire]

now

derivative of the function

do yk how to take the derivative?

[jagged turret]

yes but

i simplified

it to

4x^3+24/x

is this accurate

before i take derivative

[dire spire]

4x^3?

[jagged turret]

i combined

the 4x^2

with 24/x

by adding

them together

gave left an extra x

[dire spire]

oh wait nvm

yes

got it

take the derivative of that now

wait no

4x^2 + 16/x + 8/x
= 4x^2 + 24/x

[jagged turret]

ia dded

em together

but

ok

8x-(24/x^2)

[dire spire]

bruh

[jagged turret]

thats my surf area?

[dire spire]

use brackets then

its confusing

i thought it was 4x^2 + (24/x)

[jagged turret]

[ 8x - [(24)/(x^2)] ]

then u said to find dy/dx

[dire spire]

that looks correct

now set that = 0 and solve for x

[jagged turret]

1.442

none of em lol

1.6 is closest

@dire spire

[dire spire]

wait

[jagged turret]

l0l

something is wrong with our problem

before we took dy/dx

🔶 🔶 🔶

@dire spire

<@&286206848099549185>

[dire spire]

wait

relax

[jagged turret]

the aids kind

jk. its calculus

we're stuck

@still temple

@rugged cobalt pls baby girl ❤️

halp

[jagged turret]

here

@flat raft @stoic saddle @stark trail lets get the team on this lol

if ya'll free idk

[remote heron]

it says its open on top

so the top part isn't part of surface area

so
S.A. = lw + 2lh + 2hw

[jagged turret]

wait so

we did sa wrong?

lw not 2lw?

1.817

still wrong? lol

@remote heron @dire spire

.close

[marsh citrus]

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[short kelp]

This is finding the inverses, I’m confused on what I’ll do next after x=4/y - 2

[remote heron]

you solve for what y is

[short kelp]

+2 on each side right?

[remote heron]

yeah, try to isolate it

[short kelp]

Ok I have x+2=4/y

[remote heron]

so then you would multiple both sides by y to get it out of the denominator

[short kelp]

Oh ok how would that look?

X+2*y=4

[remote heron]

remember it multiplies everything on the left not just the 2 so you would use parenthesis to show
(x+2)*y=4

[short kelp]

Oh right

Divide by (x+2) to get y on that side right?

[remote heron]

yup

[short kelp]

Ok thank you!

[remote heron]

$y=4/[x+2]$

[elfin berry]

Alexandria!

[short kelp]

I have it it’s fine

.close

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[marsh citrus]

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[still temple]

Help?

[still temple]

Can someone explain how to do this to me.
All I know is that AB=CD which is =6
and AC is 10
to 10^2-6^2=B^2
100-36=64
root 64=8
so BC and AD=8
This is all i can figure out bymyself.

[cobalt sentinel]

Sat go crazy

[still temple]

yeah

[cobalt sentinel]

Maybe set up proportions for similar triangles

Have u tried that

[still temple]

wdym?

[cobalt sentinel]

Do yk what similar triangles are?

[still temple]

yeah triangles that are similar in shape but not in measurement i think

[somber flame]

Consider the area of △ADC

[still temple]

1/2 bh right?

1/2 (8)(6)=1/2 48=24

Hello?

[cobalt sentinel]

Oh I thought sed was helping you

[still temple]

he came and left

[cobalt sentinel]

Ok so going back to this, similar triangles have the same angles

[still temple]

okay

[cobalt sentinel]

Consider triangles ADC and DEC

They are similar

[still temple]

yeah

[cobalt sentinel]

Ok so now, we can set up proportions relating their side lengths

[still temple]

AB=DC

and AD=BC

BC is 8

so AD is also 8

[cobalt sentinel]

$\frac {AC}{DC} = \frac {DC}{CE}$

[elfin berry]

Stephen

[still temple]

?

im confused

[cobalt sentinel]

What are u confused abt

[still temple]

why does AC|DC equals to DC|CE

[cobalt sentinel]

Because triangle ACD and triangle DCE is similar, we can set up proportions with corresponding side lengths

AC and DC are sides of triangle 1

DC and CE are sides of triangle 2

AC and DC are both hypotenuses

[still temple]

so AC|AB would also be similar?

to AC|DC and DC|CE?

[cobalt sentinel]

Yes, that’s correct

[still temple]

okay I understand

so would it be 10|6 = 6|x and we cross mulitpy?

[cobalt sentinel]

Yes

[still temple]

10x=36

x=3.6

[cobalt sentinel]

Yes

[still temple]

OHHHHHHHHHHHHHHH.

It's actually super easy

Thanks for hleping

helping*

[cobalt sentinel]

Np

[still temple]

your a W teacher otherwise than the wait time. 👍

.close

[marsh citrus]

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[cobalt sentinel]

Lol

[marsh citrus]

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[marble glacier]

https://gyazo.com/408cdec64f053408b0e3336c145f569c

[marble glacier]

bro Ma's life these expressions are equivalent

calculators lying

all they did is factor an a out the bracket

oops

[cobalt sentinel]

What abt the minus sign

[marble glacier]

yea i must've typed it in the calc wrong

[cobalt sentinel]

[marble glacier]

.close

[marsh citrus]

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[lean falcon]

am i right with D?

[marsh citrus]

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

[desert socket]

yes

It is 44*sin72°/sin27°

approx 92

[lean falcon]

tysm same thing i thought

[lean falcon]

am i right with C here?

jus got a few more

[marsh citrus]

@lean falcon Has your question been resolved?

[wind sun]

seems right

[marsh citrus]

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[azure trench]

For P(A U B complement) the red part is what you’d include?

Even though xy is part of B?

[sharp harness]

U is (i assume) union, anything that's either of those counts

xy isn't in the complement of B, but it is in A

[azure trench]

Oh I see

[sharp harness]

the only thing that doesn't get included is anything that isn't in A, and also isn't in B complement (so it is in B), which is just the y-xy bit

[azure trench]

Ohhhh

Ok thanks

.close

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[marsh citrus]

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[pastel nacelle]

Hello, Can you help me to solve this problem please, i cant find a primitive of (x^2)*sqrt(1-x^2)

[nova totem]

You need to change to polar coordinates

[pastel nacelle]

like x=cos(u)?

[nova totem]

Sure yeah

Well no

You need two parameters

You need to parameterize the region not the boundary curve

[pastel nacelle]

ok thanks

.close

[marsh citrus]

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[void finch]

Anyone here who owns a ti-84 calculator? Mine show inaccurate results. Anybody here who knows a fix?

[marsh citrus]

@void finch Has your question been resolved?

[void finch]

<@&286206848099549185>

[vocal cloak]

Yes

[void finch]

Whats wrong with my calculator lmao

Tried to reset it. Didnt work

[vocal cloak]

Replace the battery 🔋

Open the screw on the back of the calculator

[void finch]

The battery has nothing to do with how the calculator functions outside of the battery being dead which isnt the case

[vocal cloak]

Insert a USB drive

[void finch]

It's fully charged,

[void finch]

Obvs it has something to do with the software. The hardware does what it's supposed to do.

[bright jay]

Inaccurate how?

[void finch]

X^2-4X+4=0
X=2 and not what the calculator shows (X=1.99999)

[bright jay]

That technically rounds to 2, but you probably need tighter bounds

[void finch]

Zooming in doesnt help.

Ik it rounds of to 2. But it should be accurate. This is a simple function where i know the answer should be rounded off. This is not the case with more advanced functions and makes me mess up

[bright jay]

The calculator is probably doing some interpolation method to find the roots, if it upsets you that much talk to their support team

[void finch]

Ofc it upsets me. I have to redo my exam because my calculator is off. Ofc I am upset

[void finch]

Where do I contact them?

[bright jay]

Look it up

You can google ti support numbers

[void finch]

K

.close

[marsh citrus]

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[torpid owl]

I don't know where to start with this problem. Assistance needed.

[torpid owl]

Here is the previous question :

[stoic saddle]

that's actually extremely helpful

bc now all you need to do is calculate f(pi/6), f(pi/3) etc. based on the formula you answered with here

[torpid owl]

is the previous question right?

and would I plug in pi/6 in the sin part?

@stoic saddle When I plug in pi with the equations I just get 20

so it's not right

hello?

[stoic saddle]

show work?

yes the prev question is correct

[torpid owl]

I did 20+10sin(pi/6)

[stoic saddle]

yes...? and what does that give you?

[torpid owl]

20.09138...

20 regardless

[stoic saddle]

no

first off your calculator is in degree mode

[torpid owl]

oh...

[stoic saddle]

second you should know sin(pi/6) without a calculator

[torpid owl]

you regard me too highly..

[stoic saddle]

or sin(30°)...

[torpid owl]

why do you have this image saved on your computer? 😆

👍

regardless thank you @stoic saddle

.close

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[next shale]

$S= log_a(bc) + log_b(ca) +log_c(ab)$ a, b, c be real numbers greater than 1

[elfin berry]

Nomad_InSearchfor_

[next shale]

Find the smallest possible value of S

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[thorny onyx]

how do I find the vertical and horizontal asymptotes?

will I be using the derivative?

[bitter dawn]

No

It's not necessary

For vertical you need to find the points where you have an improperty in the division

For the horizontal, compute the limits in + and - infinity

[thorny onyx]

what does that mean?

[bitter dawn]

Which one?

[thorny onyx]

all of it

[bitter dawn]

A vertical asymptote is where the function diverges

A horizontal asymptote is when it has a finite limit in either + or -infinity

[thorny onyx]

so how would I set that up to calculate that?

[bitter dawn]

Have you done limits ?

[thorny onyx]

not for a while

actually nvm

how do I find the intervals were f is increasing/decreasing?

.close

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[sharp harness]

do you have a question...?

[winter meteor]

What is wrong about this?

[sharp harness]

the step $0^0 = 00^{-1}$ isn't valid because $0^{-1}$ doesn't exist

[elfin berry]

bee

[winter meteor]

[sharp harness]

(depending on the definition of exponentiation 0^0 might also not exist but that's not particularly relevant in this context)

only numbers that aren't 0 have reciprocals

[winter meteor]

so i have to prove 1/0 equals something?

[sharp harness]

"1/0" is not a valid expression

if you want you could try to prove that there exists an x such that 0x = 1

but you're not going to get anywhere because 0x is always 0 which isn't equal to 1

that's "why" there's no 1/0, you can't simultaneously be able to divide by 0 and have multiplication by zero always give zero

[winter meteor]

surely

[sharp harness]

?

surely what

[winter meteor]

Surely this is too much for me to think about

Thank you for your help!

.solve

.solved

.close

[marsh citrus]

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[winter meteor]

🐢

[marsh citrus]

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[glad depot]

hi! so i'm a little confused about the quadratic imaginary number fields whose integer rings are UFDs

[glad depot]

its supposed to be d \in {-1,-2,-3,-7,-11,-19,-43,-67,-163} right?

but somehow in my lecture they seem to be saying something else?

what am i missing here?

[marsh citrus]

@glad depot Has your question been resolved?

[glad depot]

ah i think i can ping now <@&286206848099549185>

[marsh citrus]

@glad depot Has your question been resolved?

[marsh citrus]

@glad depot Has your question been resolved?

[marsh citrus]

@glad depot Has your question been resolved?

[glad depot]

k im just gonna close it

.close

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[fathom prairie]

Can someone calculate the total of males and females in this graph?

[long arch]

yeah you can

[delicate maple]

You.. add the values?

[marsh citrus]

@fathom prairie Has your question been resolved?

[marsh citrus]

@fathom prairie Has your question been resolved?

[marsh citrus]

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[lean monolith]

This is a problem in Micheal Artin's book.

Do I show that for A, B in the set of M, AB^-1 is in M

?

Before that I need to show that A is also in GL(n, R) before proceeding

[green sky]

For what part are you saying ?

[lean monolith]

subgroup

[green sky]

Is this for inverse?

[green sky]

No thats not what you have to do

[lean monolith]

oh...

then?

[lean monolith]

I mean isn't there a theorem that if H is subgroup of G then for x,y in H, xy^-1 is also in H

[green sky]

Your group is $GL_n(R)$ and $H$ is the set of matrix of form $M$, you have to show $H$ is subgroup of $GL_n(R)$

[elfin berry]

fäf

[lean monolith]

yes

[green sky]

First of all x,y here aren't A,B

H is group of matrix of form M

[lean monolith]

I thought those were group elements...

that's what I was taught

that's true for any general group

[green sky]

Those are block matrices in M

Not M

[lean monolith]

yeah

[green sky]

You have some $M = \begin{bmatrix} A&B\0&D\end{bmatrix}$ and $N = \begin{bmatrix}E&F\0&I\end{bmatrix}$ in $H$ and you have to show their closure

[elfin berry]

fäf

[green sky]

Do you think we will get MN such that it's A11 and A22 belong to GL_r(R) and GL_(n-r)(R) respectively

[lean monolith]

I actually meant this from the start

if now N is invertible and you show MN^-1 belongs in H

[green sky]

Sorry

But

[lean monolith]

then H is a subgroup of GL(n, R)

[green sky]

You wrote A,B in set of M

[lean monolith]

oof

my bad

[green sky]

Find MN

You know how block matrix multiplication

I'll explain you the answer, we can go step by step

[lean monolith]

ok wait

oh it's just the multiplication of blocks

MN would also be of block form

[ AE K ]
[ 0 DI ]

[green sky]

Jes

[lean monolith]

AE and DI are from before

[green sky]

Ues

Jes

[lean monolith]

K is something that comes

[green sky]

Yeah leave K

[lean monolith]

but if we do this we have to show everything else

like

Associative

Identity

[green sky]

So let's first make clear the definition of H

When does some block matrix M belong to H

When the A11 and A22 belong to GL_r(R) and GL_(r-n)(R) respectively

[green sky]

No need

[lean monolith]

oh ok

[lean monolith]

yes

[green sky]

We have to show this

So

AE must be in GL_r(R)

We know the closure of GL_r(R)

[lean monolith]

that's becuase GL(r, R) is a group itself

yes

oh we done

lol

[green sky]

Yeah so AE belongs to that group

[lean monolith]

similarly DI

[green sky]

Same goes for DI

That means MN is in H

[lean monolith]

yes

[green sky]

1st part done

Now you know how to find inverse of upper triangular block matrix?

[lean monolith]

no sadly

[green sky]

What will be inverse of M

That's a bit sad

Yeah

Won't lie

[lean monolith]

🥲

[green sky]

Well can you guess

[lean monolith]

some kind of upper triangular block again?

inverses of upper triangulars in case of 2x2 is upper triangular again so

that's my guess

[green sky]

Yeah but a bit different

[lean monolith]

oh

[green sky]

Well let me just find something rather than typing

[lean monolith]

don't tell me it's lower triangular

[green sky]

The diagonal become A^(-1) and B^(-1) and that's what matters for our question

[lean monolith]

yeah

thought so

[green sky]

No its not

[lean monolith]

oh

[green sky]

The third entry changes tho

It doesn't matter in this question too so we can go ahed

https://math.stackexchange.com/questions/2316555/the-inverse-of-a-block-upper-triangular-matrix

Here's something if you need info

So going ahed we gotta show that A^(-1) is in GL_r(R) and B^(-1) is in GL_(n-r)(R) for M^(-1) to be in H

Since GL_r(R) and GL_(n-r)(R) forms group it is obvious

So yeah M^(-1) does belong to H

I hope everything is clear to this point

[lean monolith]

oh now I got it

that's what I just got trying to do it myself

[green sky]

Nice

So this is proof of H being subgroup of ...

[lean monolith]

K is basically AY + XI

so if you just put the new stuff, you get what's there in the link you shared

[green sky]

Yeah but we can write it as K since it doesn't play any role in showing something belongs to H

[green sky]

Yes

[lean monolith]

i know, I just did the entire calc so I was pointing out

very messy stuff, reminded me of why I hated doing calculations using matrices while doing Linear Algebra, Linear Functionals and Transformations were my best friends as those were functions

but thank you very much for putting up with me

[green sky]

[lean monolith]

[green sky]

Well the question is remaining

[lean monolith]

oh yes, after that

uh... M^-1 exists

now what do we show

oh shit,

unique identity?

[green sky]

We showed M^(-1) belongs to H already

H is a subgroup

So now we have to solve for homomorphism

[elfin berry]

fäf

[lean monolith]

yes

this is easy to show

[green sky]

So we have to show f(mn)=f(m)f(n)

Yeah it is

[lean monolith]

we just did the multiplication thing

[green sky]

Yeah and what was its first entry?

[lean monolith]

AE

so we have it

[green sky]

Phi(M) phi(N)

[lean monolith]

basically the group multiplication of the first entries

yes

[green sky]

Now we have to find kernel

[lean monolith]

kernel is also easy I think

[green sky]

You can show that $M=I_n$ gives us identity matrix $I_r$ for homomorphism

[lean monolith]

just find the block matrices with the identity of GL(r, R)

[elfin berry]

fäf

[lean monolith]

yes

[green sky]

No this is for homo

Not kernel

[green sky]

Yeah it is

[lean monolith]

wait, we just showed for two arbitraty M, N in H that phi(MN) = phi(M)phi(N)

aren't we done

....oh...

[elfin berry]

fäf

[lean monolith]

yeah I meant that already

[green sky]

We have to show identity maps on identity

[lean monolith]

oh.

fudge

I completely forgot about that

hmm so wait

[green sky]

Set of such M is kernel

We have to find M such that phi(M)=I, which implies A=I

[lean monolith]

yes

[green sky]

You're welcome

[lean monolith]

You helped a lot

I tend to be annoying so

Thank you again

[green sky]

[lean monolith]

[green sky]

You weren't

[lean monolith]

okay now I'm closing this channel

hope we meet again later on

[green sky]

I'm pretty bad at group theory

[lean monolith]

same

[green sky]

I hate it

Do you?

[lean monolith]

oof

i don't exactly hate it but it's cumbersome

[green sky]

Yeah hopefully

[lean monolith]

i like analysis and topology more

[green sky]

So you are in my line just a bit behind

Tell me if you change your mind anyday

[lean monolith]

yes I'm just a 2nd Year Math Major, preparing for the Finals

will do

[green sky]

Cool you are doing some harder questions then

[lean monolith]

oh really?

I thought this was the standard 😭

[green sky]

I started group theory basics in my third year

[lean monolith]

damn

I guess it's India

[green sky]

Yeah but my 3 year ago self won't find it this easy

[green sky]

My brown color gave it away

[lean monolith]

same I'm suffering alongside my whole class, I'm just in slightly better condition that others

[green sky]

As expected

[lean monolith]

oh no I meant I'm in India

you are also?

[green sky]

Yeah

[lean monolith]

niceee

Where did you complete your Masters from?

[green sky]

Wait ill dm you

[lean monolith]

okay wait

okay

.close

[marsh citrus]

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[green sky]

Check

[green sky]

.close

[marsh citrus]

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[lilac pine]

How should I correctly set up the ranges for the double integral?

If I’m not mistaken it should be the red regions, right?

[marsh citrus]

@lilac pine Has your question been resolved?

[lofty gyro]

looks good to me

what ranges did you set up for the integral? let's see if it's correct

[lilac pine]

That's what I need help to do, I'm not really sure how to do it given these regions

[lofty gyro]

i see

since the question requires to be in the order of dxdy

we search for available x such that for every possible y, it's valid

so, let's look at y first

the range for y must be constants in this case

so, it'll be -1 to 1

now, take a look at x

it can be a constant or in terms of y

we pick something middle-ish to check

let's say the slice of y=0.5

the red region spans from x=0.5 to 1

let's pick another slice, let's say y=0.25

the red region spans from x=0.25 to 1

so, for y is positive, the range will be y to 1

now it's your turn for the part when y is negative

[lilac pine]

So for y is negative: let's say -0,5, then the red region spans from x=0,5 to 1.

and for y=-0,25, then the red region spans from x=0,25 to 1

[marsh citrus]

@lilac pine Has your question been resolved?

[lilac pine]

So for y is negative, the range will also be y to 1?

[marsh citrus]

@lilac pine Has your question been resolved?

[quiet anvil]

@lilac pine It will help other users check your work if you typeset your solution using latex. Is:

$$\int_{x=0}^{x=1} \int_{y=-x}^{y=x} f(x) \dd{y} \dd{x}$$

what you meant?

[elfin berry]

OmnipotentEntity

[quiet anvil]

(This is the most natural way I can think of to set up this integral, but I believe you were trying a slightly different way where you integrate over y on the outside and x on the inside, in that case this integral would seem to be:)

$$\int_{y=-1}^{y=1} \int_{x=|y|}^{x=1} f(x) \dd{x} \dd{y}$$

[elfin berry]

OmnipotentEntity

[lilac pine]

I didn't have a solution, but if I would, then I'll give latex a try.

[quiet anvil]

well, does either of the two solutions I posted help you?

[lilac pine]

I'm not really sure, I don't know how to set up the integral ranges, so I don't know what to do with the solutions. @lofty gyro showed me that for y is positive, the range will be y to 1, but I don't know what to do with that yet.

[marsh citrus]

@lilac pine Has your question been resolved?

[marsh citrus]

@lilac pine Has your question been resolved?

[marsh citrus]

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[untold gust]

How would I do this problem

[vestal forge]

what have you tried?

[marsh citrus]

@untold gust Has your question been resolved?

[untold gust]

idk where to start

[vestal forge]

you can start by just adding up the fractions, but instead of just adding them all together, start by grouping them first

after that, you'll notice that you can simplify them using the given fact

[desert socket]

Multiply numerator and denominator of individual fractions by such power of w that all fractions have w^5 in numerator

[untold gust]

.close

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[proud ice]

No ocupes multiples canales

Si necesitas ayuda, solo pregunta en uno canal y espera

[marsh citrus]

@young jungle Has your question been resolved?

[marsh citrus]

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[full orchid]

I don't know where to start with this

[marsh citrus]

@full orchid Has your question been resolved?

[full orchid]

<@&286206848099549185> i still don't know where to start

[idle lotus]

what does "M" mean here?

[marsh citrus]

@full orchid Has your question been resolved?

[full orchid]

no clue

i haven't seen it anywhere before this question

[marsh citrus]

@full orchid Has your question been resolved?

[proven flicker]

theres a formula for the error on tylor series

[marsh citrus]

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[molten mantle]

.

Hi there!

I need to use Picard's iteration method on the following ODE: x' = 2(x + t) x(0) = 5

I have to use the iteration twice, so I need the function of x2, and Im given the t = 1, so finally I have to calculate x2(1). The solution is 26.67 but I cant seem to get it right.

[molten mantle]

I got that x1(t) would be equal to 5 + t^2 + 10t

and if Im right x2(t) would be 5 + integrate[0, t] (t + 5 + t^2 + 10t)

so far I got this, but it isnt right, cuz if I take it as x2(1), it equals to 15.833..

Could someone explain where Im missing something, or whats the problem?

[marsh citrus]

@molten mantle Has your question been resolved?

[marsh citrus]

@molten mantle Has your question been resolved?

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[placid oak]

[placid oak]

I found, $$A=\begin{pmatrix} 1 & 2 \ 4 & -10 \end{pmatrix} \text{ and } B=\begin{pmatrix} -1.5 & 6 \ 2.5 & -4 \end{pmatrix} $$

[placid oak]

Tr(A) != Tr(B), so A cannot be similar to B.

But, the answer key states that A is similar to B.

Can I confirm this as a discrepancy?

[marsh citrus]

@placid oak Has your question been resolved?

[marsh citrus]

@placid oak Has your question been resolved?

[marsh citrus]

@placid oak Has your question been resolved?

[pallid frigate]

Yes, you may. You can take a look at this stack exchange post to figure out precisely how: https://math.stackexchange.com/a/413419/825751

[placid oak]

.close

[marsh citrus]

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[wary void]

Is there any way to show the relationship between the angle theta and the motion of the tent / the moment against the base without using drag coefficients and stuff? Im doing AS-Level physics atm

[wary void]

Like how can making the angle shallower make the tent more stable

or if thats even the correct approach when describing this

Its making me depressed

[marsh citrus]

@wary void Has your question been resolved?

[wary void]

are you a mong

[knotty trellis]

this channel is taken. Dont spam

[wary void]

.close

[marsh citrus]

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[proven holly]

can anybody help me, please?

[marsh citrus]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[proven holly]

1. I don't know where to begin

[marsh citrus]

@proven holly Has your question been resolved?

[proven holly]

<@&286206848099549185>

[frosty briar]

Im not sure how to do 1 but I think the idea is that the spanning vectors of that plan are orthogonal to the normal vector

like you have (x1,y1,z1) \cdot (1,-2,3) = 0

this gives you 1 equation

then (x2,y2,z2) \cdot (1,-2,3) = 0

this gives another equation

and the third one comes from the fact that (2,1,-1) is on the plane

actually you only need 2 I believe

[sleek pivot]

So for the first problem you have this formula $\vec{r}\cdot\vec{n}=\vec{a}\cdot\vec{n}$

[elfin berry]

Parzival

[sleek pivot]

Where n will be the normal vector and a is the position vector of that point

@proven holly

[frosty briar]

if you dont mind me asking, what is r and a here

[sleek pivot]

$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$

[elfin berry]

Parzival

[frosty briar]

so one of the spanning vectors?

[sleek pivot]

Yes

[frosty briar]

I see, thanks

[sleek pivot]

anytime

[frosty briar]

So both our approaches are similar?

I guess mine is a little more time consuming tho lol

[sleek pivot]

Um IDK this is what I was taught

[frosty briar]

my school was teaching this shit but I was too bored in that class so I never learned it

[sleek pivot]

LOL

.close

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

@vital oriole Has your question been resolved?

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@vital oriole Has your question been resolved?

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[ebon schooner]

Okay so I know how to solve this question, but I'm confused with my answer

so thats the first part to get the value of BC

and by using the law of sines

we get

sin35/5.2=sinC/9 (if we were solving for C in this case)

and that would simplify to

and then by adding 83.1 to 35 and subtracting from 180

you get 61.9

but if I solved for B first

and when you subtract 35 and 50.5 from 180

you get 94.5

so depending on how I do it I get two different values for C

I have no idea why I get two different answers and I've checked over my work as well

<@&286206848099549185>

[amber ruin]

It might be because of approximation error, the calculations seem correct

[ebon schooner]

calculator approximation error? but can it be that big considering 83.086 is nowhere near 94.5?

[amber ruin]

if you use sine law on C and B none of the answers are correct it seems

$\frac{\sin B}{\sin C} = \frac{7}{9}$

[elfin berry]

ta

[ebon schooner]

so both answers are correct?

how about the ambiguous case? can it be applied here?

[amber ruin]

actually yes

when you are using law of sines it doesn't take the 3rd side into account

[ebon schooner]

but can't ambiguous case only be applied to SSA pairings?

[amber ruin]

whatever the case both of the answers are wrong just plug them into this formula