help-0

[limpid spade]

what's the integral of 1?

[weak harbor]

X+C

o okay so im wrong

thanks!

.close

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[arctic wadi]

help

how to solve the problem a

[grand lagoon]

This channel is occupied

[arctic wadi]

ok

[median dirge]

@arctic wadi send your ques in a new channel

[arctic wadi]

which one？

in available？

[grand lagoon]

yep

[arctic wadi]

thanx

[lone heart]

@hazy goblet Has your question been resolved?

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[whole raft]

I need help my Teacher is ill and doesnt explain anything

[hazy goblet]

No

[drifting hull]

@whole raft Was verstehst du denn nicht? Anscheinend hast du ja zumindest einen Teil schon herausgefunden

[whole raft]

ne nicht wirklich war geraten

Mein problem ist ich weiß nicht was das in den Kästchen angibt

[drifting hull]

in I) II) und III) steht jeweils ein lineares Gleichungssystem

in $A_1$, $A_2$ und $A_3$ stehen die Gleichungssysteme, aus I), II) und III), nur sind sie hier umgeformt sodass jede Gleichung die Form y=... hat

[ocean seal]

Alexander42

[drifting hull]

Nun beschreibt eine Gleichung der Form y=sx+t ja eine Gerade. in $B_1$, $B_2$ und $B_3$ sind nun die Geraden eingezeichnet, die zu den Gleichungen in $A_1$, $A_2$ und $A_3$ gehören (jeweils beide Geraden von einem Gleichungssystem zusammen)

[ocean seal]

Alexander42

[drifting hull]

Und in $C_1$, $C_2$ und $C_3$ steht, wie viele Lösungen das Gleichungssystem jeweils hat.

[ocean seal]

Alexander42

[drifting hull]

Ist das soweit klar?

[lone heart]

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[frosty flume]

is this the correct way to present

I'm trying to say find A, B, C such that (the three equations below)

same for D, E, F and G, H, I

[lone heart]

@frosty flume Has your question been resolved?

[vocal hawk]

I don't understand your question

correct way to write it you mean?

if you mean how to write them

there's no correct way, they appear to be clear to me and that's the most important

but if you mean the most compact way, you could use matrices to express the three systems

[frosty flume]

yeah im trying to expand from the matrix

if that makes sense to the reader then it is alright

i just added the word find in front

was trying to see if there is a proper convention for writing something like that

[forest harness]

you could take out the words (the “find A,B,C such that” part) from the brackets but like mohad said it’s clear as long as you have the equations and the words there

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[timid hazel]

Where are you stuck? What have you tried?

[fossil sundial]

I'm stuck on finding value of x

I think I need 1 angle and 1 side

[dense lily]

.help

[lone heart]

Commands:
clopen: .close, .reopen
factoids: .tag
help: .help

Type .help <command name> for more info on a command.

[fossil sundial]

Lmao ur in wrong one

I think ABD is 53°

So BAD is 74?

Because ABD is the same as ADB

[sudden hinge]

notice that ABD and ACD are isosceles

[fossil sundial]

Idk how

[sudden hinge]

2 sides have the same length

[fossil sundial]

Is that one of the statements where a semi circle is split into 2 isoceles

[sudden hinge]

AB = AD = AC = radius

[fossil sundial]

How do u know the 2 sides have the same length

[sudden hinge]

because they are all radii?

the radius doesn't magically change

any line from the center point to a point on a circle's circumference has the same length, namely the radius

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[silk flame]

test the convergence

im pretty sure

its with xomparison test

what do we choose vn as/?

[ocean seal]

Navix

[lone heart]

@silk flame Has your question been resolved?

[silk flame]

i didnt understand

[noble sinew]

Start with saying if you are trying to find if the sequence or series converges

[silk flame]

mhm

we have to use a test tho

[noble sinew]

.

You haven’t posted a full question

So start with answering my question

[silk flame]

here

[noble sinew]

Convergence of sum or of series?

Like I said

[silk flame]

ohh sum

[noble sinew]

Gj

If you want to use comparison you need to make a guess if its convergent or divergent first

[silk flame]

wha

we dont do it like that

[noble sinew]

What?

That is how you use comparison

[silk flame]

we choose vn and then divide it with un

[noble sinew]

If you think its divergent you find another divergent series to compare to

And same for convergent

[noble sinew]

So again

[silk flame]

this is the question

[noble sinew]

Yes?

[silk flame]

😭

[noble sinew]

Hard to read what I am saying?

[silk flame]

i dont get u

[noble sinew]

Sounds like you don’t even understand comparison if you don’t understand me

[silk flame]

i do understand the way my teacher taught me

direct comparison

we never lrnt abt guessing whether the series is cgt or dgt

[noble sinew]

Do you agree its possible to show divergent and convergent using comparison

[silk flame]

yes

[noble sinew]

And to do that you need to compare with either another divergent or another convergent

To figure out which of the two

[silk flame]

if the condition satisfies

[noble sinew]

You need to have an idea if your series converges or diverges

How else are you gonna choose?

[noble sinew]

Your idea might be wrong but you still need to start somewhere

[silk flame]

the other series vn we usually calculate 1/the highest power of n in the denominator-highest power in the numerator

to get the other series

to compare woth

[noble sinew]

Why not do the same here then?

[silk flame]

the highest poweris n in both cases

3

[noble sinew]

Yes?

[silk flame]

so itll be n^0?

[noble sinew]

How did you get that

[noble sinew]

Surely you meant smth like 1/(n^(3/3)) here

[silk flame]

so itll be 1/n?

[noble sinew]

That is what you said (at least meant) is it not?

[silk flame]

yea lemme try solving

and then ill tell u

its dgt?

[noble sinew]

yes

[silk flame]

okee

thnku smm

.close

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[wide plaza]

i need help finding what to put in this ratio

[rotund plank]

1/18

[wide plaza]

oh ok

that makes sense actually

cause 18pi is circumference

1 is basically x

because thats the arc length

thx @rotund plank

.close

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[pallid sail]

Hello!

[limpid spade]

jsuew

hhi

[pallid sail]

,, Let $G = { (a,b) | a, b \in \R, a \ne 0} $ and operation ⊕ defined like this $$(a, b) ⊕ (c, d) = (ac, ad + b)$$. Prove that (G, ⊕) is a magma.

[limpid spade]

what

[pallid sail]

rip i suck at latex

Let

Prove that (G, ⊕) is a magma (groupoid - however you call these things)

[ocean seal]

Аdriаn Sheрhаrd

\begin{gather*}
Let $G =  \{ (a,b) | a, b \in  \R, a \ne 0\} $ and operation ⊕ defined like this $$(a, b) ⊕ (c, d) = (ac, ad + b)$$. Prove that (G, ⊕) is a magma.
\end{gather*}
```Compilation error:```! Missing } inserted.
<inserted text> 
                }
l.57 \end{gather*}
                  
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.```

[pallid sail]

`` Let $G = { (a,b) | a, b \in \mathbb{R}, a \ne 0} $ and operation ⊕ defined like this $$(a, b) ⊕ (c, d) = (ac, ad + b)$$ Prove that (G, ⊕) is a magma/groupoid

There we go

[ocean seal]

Аdriаn Sheрhаrd

[lone heart]

@pallid sail Has your question been resolved?

[pallid sail]

<@&286206848099549185>

[vague coral]

you show that G has a neutral element and that oplus is associative

[pallid sail]

Ty

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[pearl nimbus]

how do you get 225?

[pearl nimbus]

@hlepers

@mental coyote

<@&286206848099549185>

[placid zinc]

See rules on helper pings

You have an answer key. I'm not certain what you may be missing.

[pearl nimbus]

i want to know how do you get 225?

[placid zinc]

,calc 5000*0.045

[ocean seal]

Result:

225

[placid zinc]

Interest is $225 per annum

[pearl nimbus]

any other ways

possible dividing by 100?

i did that before but forgot why i did it

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[lone heart]

@pearl nimbus Has your question been resolved?

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[plush fox]

Hey I just need an explanation on how to work my way through this problem, the sheet says to evaluate this problem.

[lone heart]

@plush fox Has your question been resolved?

[plush fox]

<@&286206848099549185>

[lone heart]

@plush fox Has your question been resolved?

[calm perch]

use the log power rule

followed by the log addition/multiplication rule

[plush fox]

.close

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[subtle heath]

yes, only a is correct because it is given that y has open lower limit 0 and closed upper limit 2. No other condition is given so we have to assume all real numbers are a part of Y

[short plank]

tqvm bro 🙂

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[runic lava]

can someone help?? is it possible that the value of the ice cream will be 0?

or even -1?

[marsh sonnet]

-1?

[runic lava]

is that allowed?

[wary stream]

No, because you can't have negative people surveyed

[runic lava]

seee

[marsh sonnet]

it always be a positive number

[runic lava]

but the thing is 6 people only ate ice cream but the sum of all persons in the ice cream is 7

how can i solve this? :((

[marsh sonnet]

c. U=25
6+9+6=21
does not love any = 25-21 = 4

[runic lava]

thankss!! but im talking about letter b

[wary stream]

Don't solve the problem for the person

[marsh sonnet]

oh sory

ice cream only is 0

[runic lava]

i know the answer is 4..

[runic lava]

but wouldn't be that be a problem? since they're 7?

[wary stream]

Even if you already know the answer, people here, who help are suppose to teach the people asking/lead them to the correct answer, not do the work for them

[marsh sonnet]

ice cream and burger are 4
ice cream and hotdog are 4

[runic lava]

oooohh so its the problem thats wrong

thanks .76!

.close

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[marsh sonnet]

see the 3,1,3 came from burger and hotdogs

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[rigid flower]

.close

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[runic lava]

Hello! Is there another way to get the subset of Let 𝐵={𝑥:𝑥 is even number less than 15}? Give 2 subsets of B using the roster method.
My first subset is B: {2,4,6,8,10,12,14}

[alpine sable]

What is the "roster method"

That is a valid subset

{2} is a valid subset too

same with {0}

theres actually quite a lot of subsets

[runic lava]

its listing the numbers

[alpine sable]

Oh okay

A subset of some set A is just a set in which every element in the set is an element of A

So alongside your example

you could have {2}, {0}

I think a good way to get subsets would first be to write B out as its elements

but its probably fine in this case

[runic lava]

oohhh B= {2}{,4},{6},{8},{10},{12},{14} acceptable??

[alpine sable]

All of those individually would work too

{2} works

{2, 4} works

{4} works

does that answer your question?

[runic lava]

yupp! thank you so much!

.close

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[shadow raptor]

So um this is like 7th grade geometry

[shadow raptor]

I have to find m(angle) CBD

[oblique spire]

Find CBD

?

[shadow raptor]

Yea

[oblique spire]

Well here is a hint

CBE is 90 degress

And DBE is 59

Pay some focus on that info

[shadow raptor]

I have that already I just dont get like what it is fully saying

[oblique spire]

U don't get the picture?

[shadow raptor]

Because the missing angle is 31

Well we just learned this

and I missed a day of school

[oblique spire]

It's indeed 31 degress

Think of

When I removed DBE

What are se left with

[shadow raptor]

with 31*

[oblique spire]

The DBE angle

[oblique spire]

Or the missing angle?

[shadow raptor]

Yea

[oblique spire]

Then what are u Steuggling with

[shadow raptor]

Ohhh I think im starting to get it

[oblique spire]

[shadow raptor]

It is simple

I was overthinking the hell out of it

[oblique spire]

Rip

[shadow raptor]

Well thanks

.close

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[lone heart]

@rigid flower Has your question been resolved?

[rigid flower]

.close

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[inner current]

why does that one part get turned into ]1/2, +infinity] if it cant be bigger than 1/2 or it would get zero?

[calm grove]

Do you know wavy curve method?

[calm grove]

No if x∈ (1/5 , 1/2) then
(1-5x)(1-2x) < 0

[inner current]

why?

i mean i get that it cant be exactly 1/2 or 1/5

but why does it need to be bigger than 1/2 instead of smaller?

[calm grove]

2x5(1/5 - x)(1/2-x)

Put x<1/2 or x>1/5

1/5-x <0
And 1/2-x>0
So overall it shall be <0

[inner current]

i dont want it to be <0

[calm grove]

Look if x>1/2
Then (1-5x)<0 , (1-2x)<0
And thus their product shall be > 0

And if x<1/5
(1-5x)>0 , (1-2x)>0
And thus their product shall be >0

we have to find solutions (values of x) for which the overall inequality is >0

Which is satisfied by x<1/5 , x>1/2

[inner current]

oh

[calm grove]

I am kinda sure that you are little confused rn

Ok let's try this way

[inner current]

you mean like how -*- = + ?

[calm grove]

Yes

[inner current]

yeah but in that its +*-

which would be -

[calm grove]

@@

@inner current
I have (1-5x) and (1-2x)right?

[inner current]

yes

[calm grove]

Can I write 1-5x as 5{(1/5) - x) ?

And 1-2x as 2{(1/2) - x} ?

@inner current

[inner current]

yes

you can

[calm grove]

So now focus on (1/5 - x) and (1/2-x) part ok?

[inner current]

yea

[calm grove]

Suppose I take x<1/5
Then with this inequality 1/5-x>0

Just shifted the x to the right side of the inequality

You agree?

[inner current]

i agree

[calm grove]

Ok good

Now since x<1/5 => x<1/2

Because 1/2 is greater than 1/5

If x is smaller than 1/5 then of course it's smaller than 1/2

You agree?

[inner current]

i agree

[calm grove]

Nice

[inner current]

is that it?

[calm grove]

So since x<1/2
I can write (1/2 -x) > 0

[inner current]

yea

[calm grove]

We have already shown (1/5-x) >0 here

[calm grove]

And here (1/2-x) >0

Both are positive terms

So their products is also >0

You agree?

[inner current]

yes

[calm grove]

.

[calm grove]

.

Thus x<1/5 is a solution

[inner current]

yes

oh

alright

[calm grove]

And now the x>1/2 part

Ok here we go , since x>1/2
So x-1/2 > 0
Implies 1/2 - x< 0

And since x>1/2 then x must be greater than 1/5

x>1/5
x-1/5>0
1/5-x<0

So product of two negative numbers > 0

[inner current]

huh

[calm grove]

You didnt get it?

[inner current]

i mean

i do get it

but its kind of a long difficult way to calculate that

[calm grove]

Yes that's why the wavy curve method is introduced

For these situations

So that you don't have to think so much

Saves time

[inner current]

oh that

yeah i know that

we call it differrent

[calm grove]

Here you just wanted explanation of why solution is (-inf, 1/5) U (1/2, inf)
That's why I gave you an explanation

[inner current]

but i just looked it up and i use it

[calm grove]

Oh

Yes use it on this inequality

[inner current]

ok 1 sec

[inner current]

RIGHT

this makes sense

thanks

.close

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[ocean seal]

Aero

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[unique garnet]

Hi, i have a question about mathematics. the linear formula is y=ax+b right. can somebody verivy this for me

[placid creek]

yes

[unique garnet]

yayy

thx

[placid creek]

b is the y-intercept
a is the slope of the linear function
y is the point's vertical coordinate
x is the point's horizontal coordinate

[unique garnet]

ah thx

.close.

[alpine sable]

yes the equation of a straight line can be modelled like that

[alpine sable]

.close

you added an extra .

[unique garnet]

AH

.close

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[rich basin]

(F)

[rich basin]

When consider when x = 0, i get u = +- 1

Which value of u do i take?

[severe sluice]

why do you need to consider that?

[rich basin]

To be put in the limits

Replacing 3 and 0

I am to replace the u values and as such I would also have to replace the integral limits

[severe sluice]

oh i was just confused bc you said "when considering x=0"

[rich basin]

What do you think why it is that u = 1

And not also u = -1 ?

[lone heart]

@rich basin Has your question been resolved?

[solemn grove]

the sign doesn’t really matter
0.5u^2-ln|u|
so it doesn’t matter

@rich basin

real translation is 1, but both give you the exact same answer

[lone heart]

@rich basin Has your question been resolved?

[solemn grove]

f????????

[supple tundra]

You keep taking positive roots. It should be $\int_{0}^{\pm \frac{\pi}{4}} \bigg|\frac{\sin{\theta}}{\cos{\theta}}\bigg|$

[ocean seal]

azeem321

[rich basin]

But why take positive root?

[supple tundra]

If you take positive $\frac{\pi}{4}$, then it will be $\int_0^{\frac{\pi}{4}} \frac{\sin{\theta}}{\cos{\theta}}$. However, if you take negative, it will be $\int_0^{-\frac{\pi}{4}} \frac{-\sin{\theta}}{\cos{\theta}} = \int_{-\frac{\pi}{4}}^0 \frac{\sin{\theta}}{\cos{\theta}}$

[rich basin]

Why take -sin(theta) ?

[ocean seal]

azeem321

[rich basin]

I know you are trying to flip it

[supple tundra]

When you integrate from $(0,-\frac{\pi}{4})$

$\sin{\theta}$ is negative

Remember $|\sin{\theta}|$ gives the positive value

[rich basin]

But isn’t this not finding the area but the integral

[supple tundra]

Whoops. I calculated your integral wrong

let me do it again 1sec

[supple tundra]

and both integrals give the same value

The point is, it doesn't matter whether you take positive or negative as long as you consider the signs given from modulus function

[rich basin]

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[robust phoenix]

is it 22 ?

[lone heart]

@robust phoenix Has your question been resolved?

[karmic rapids]

show what you've done so far please

[robust phoenix]

I did 5*24

3*19

3*28

and found the difference

I just dk

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[lone heart]

@spare sierra Has your question been resolved?

[tardy rapids]

You can write (2n+2)! as

(2n+2)•(2n+1)•(2n)!

[tardy rapids]

We can split it into two limits,

$\lim_{n\to \infty} (-1) \frac{6n+1}{6n+7} \cdot \lim_{n\to \infty} \frac{x^6}{(2n+2)(2n+1)}$

[ocean seal]

5lttrz

[tardy rapids]

Left limit approches -1

Right limit approches 0

So the series converges

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[lone heart]

@boreal idol Has your question been resolved?

[boreal idol]

.close

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[lavish ether]

suppose $G$ acts on $X$ transitively and faithfully and that for some $x\in X$ we have that $G_x\trianglelefteq G$. show that $G_x={e}$ (where $e$ is the identity element of $G$)

[ocean seal]

Beous

[lavish ether]

i already have that $gG_xg^{-1}=G_{gx}$ for any $g\in G$ and $x\in X$

[ocean seal]

Beous

[lavish ether]

i tried using this on the equivalent definition of a normal subgroup where for all $g\in G$ we have that $gG_xg^{-1}\subseteq G_x$

[ocean seal]

Beous

[lavish ether]

but i'm getting nowhere

[lone heart]

@lavish ether Has your question been resolved?

[lavish ether]

<@&286206848099549185>

[oak perch]

What is G_x? {g: gx=x}?

If so, That’s trivial, because G_gx=g(G_x)g^-1=G_x since G_x is normal

So any g from G_x, g fix any element

Since G faithfully acts on X, g must be e

[lavish ether]

yeah G_x is the stabilizer of G wrt x

[oak perch]

Okay so it’s done

Anything else?

[lavish ether]

oh i think i get it

thanks

.close

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[silk flame]

using partial diff

[meager snow]

do u know what the jacobian is?

[silk flame]

yea

we use that?

[meager snow]

yeah

well not really because in this case you only have 1 function

you can just differentiate in each variable and equate the both to 0

[silk flame]

yea

i did that

thats where im stuck

dof/dox=2x-3y+2

dof/doy=-3x+2y

are these correct

@meager snow

[oak perch]

Let them be zero

Solve this linear equation

[silk flame]

ar ethey correct?

the eqs

[oak perch]

They are

[silk flame]

idk what im doing wrong im not getting it

[oak perch]

Read your textbook

[silk flame]

could u help in solving the eq

[meager snow]

so, when you look for a critical point you want the derivatives to be 0

[silk flame]

i get that

[meager snow]

not just put both equations equal to 0 and solve the system of eq

[silk flame]

where im stuck at

is

solving the eq

i think im messing up the signs or smth

[oak perch]

After you obtain this solution $(a,b)$ of this system of linear equations
You obtain a matrix $\begin{pmatrix}\frac{\partial ^{2} f}{\partial ^{2} x}(a,b)& \frac{\partial ^{2} f}{\partial y \partial x}(a,b)\ \frac{\partial ^{2} f}{\partial x \partial y}(a,b)& \frac{\partial ^{2} f}{\partial ^{2} y}(a,b)\end{pmatrix}$

[ocean seal]

Cogwheels of the mind

[oak perch]

Check that it’s positive definite (a_11 and determinant being positive)

[silk flame]

yea ik that tooo

the values if x and y

[oak perch]

Then you can conclude that f(a,b) is minimal

[silk flame]

from solving the eq''

i need to find the points tho

[oak perch]

?

[silk flame]

x and y

[oak perch]

You can’t solve that linear equations?

High school students can do that

[silk flame]

well idk where im going wrongsdjk.df

IK

[oak perch]

?

Those equations are correct

Why you said you did something wrong?

[silk flame]

OH WAIT I THINK I GOT IT

smthng to do woth signs

i was confused

is y -6/5

and x -12/15?!@!

[meager snow]

x=4/5 and y= 6/5

[silk flame]

yea i got where i went wrong

thanku

.close

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[humble ferry]

So im trying to solve

[humble ferry]

$$\frac{dy}{dt}=\frac{22.5-y}{50}$$

[ocean seal]

Nope

[humble ferry]

So i tried separating it so that i get

$$50\int{\frac{dy}{22.5}}=\int{dt}$$

[ocean seal]

Nope

[humble ferry]

so i then have

$$50Ln\lvert 22.5-y \rvert + c_1 = t+c_2 $$

[ocean seal]

Nope

[humble ferry]

rearranging i get

$$y=e^{\frac{t+c}{50}}+22.5$$

[ocean seal]

Nope

[humble ferry]

however i have that when t=0 y=0, so ive made a cunt of something, anyone able to see what?

[meager snow]

pretty sure you messed up a minus sign

[limpid spade]

e should be -

[meager snow]

yeah

[humble ferry]

Im not going to lie im looking at my working and i dont see where i have but if e is negative it means i can actually solve it so we'll go with it

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[humble ferry]

.close

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[fierce olive]

Oh here my math need to solve.

[fierce olive]

There is a homework for last week of school

<@&286206848099549185>

[tacit arch]

read this

[fierce olive]

Yes

can you help me how to do it

Like tutorial

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@fierce olive Has your question been resolved?

[tacit arch]

,rotate

[tacit arch]

do you know the formula for volumes of cones and spheres?

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[dawn osprey]

at first sight i thought of Lagrange, but no idea what to do with that info

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@dawn osprey Has your question been resolved?

[dawn osprey]

<@&286206848099549185>

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@dawn osprey Has your question been resolved?

[oak perch]

I don’t think it’s true. Let f(x)=2sqrt(x), then it asks you to find 0<a<b such that b/sqrt(a)<1/sqrt(b), so sqrt(b)<sqrt(a), impossible

[dawn osprey]

yeah seems that f(x)= -x is a counterexample too

weird problem tho

.close

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[peak gorge]

How do you solve cos^2 + cos = 0

[knotty spire]

Do you mean $\cos^2 x + \cos x = 0$?

[ocean seal]

Remavas

[buoyant kayak]

just like how you’d solve any quadratic

[knotty spire]

Product equal to 0 rule

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[lunar ledge]

Yes?

Does this require group theory? or are you trolling?

[limpid spade]

he is trolling my friend

[lunar ledge]

Ahhhh

[limpid spade]

there is no clear legit mathematical pattern

its 8

[lunar ledge]

Well yes, I do see that each of the variables are not a set of eachother

[limpid spade]

-_-

[sly mantle]

.close

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[thorn kindle]

What am I missing here?
dA = r dθ dr

[thorn kindle]

Am I supposed to assume that r equals the average of the two radii

The context is polar integration if that helps

Using the fact that R = Δr + r

I have further reduced this to

Δθ Δr r Δr/2

Which is even more confusing

[lone heart]

@thorn kindle Has your question been resolved?

[tacit arch]

product rule?

$A = r\theta$. $dA = r d\theta + \theta dr$

[ocean seal]

riemann

[thorn kindle]

this is for polar integration

dA = d(theta) dr r

the dA is supposed to represent that sector of an annulus as i drew in the diagram

[tacit arch]

oh then it's $dA=$ A (outer sector) - A (inner sector)

[ocean seal]

riemann

[thorn kindle]

yes

but then i get this expression in terms of R and r.

when the formula i'm given is just dA = d(theta) dr r

am i just supposed to consider the limiting case as d(theta) goes to 0? giving me a factor of 2r/2 = r?

[tacit arch]

this looks right

where's the original problem?

[thorn kindle]

https://math.libretexts.org/Courses/Mount_Royal_University/MATH_2200%3A_Calculus_for_Scientists_II/9%3A__Multiple_Integration/9.3%3A_Double_Integrals_in_Polar_Coordinates

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[thorn kindle]

i just don't understand how the formula for dA is gotten. i don't see why you have to make the specific assumption that dA represents the trapezoidal region

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[vapid flame]

I need help with 2c

my working out was:
n = 10 * 0.95¹³ = 5.133, which means there's more than half the initial amount of insulin left. The answer given in my textbook is "Yes, about 44% of the initial insulin is still present". but idk how i'm supposed to get that

<@&286206848099549185>

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@vapid flame Has your question been resolved?

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[reef field]

help 1+1

[reef field]

.close

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[polar cloud]

if i have two matrices AB = XY

[polar cloud]

four matrices*

how do i find A?

i know you can seperate inverses if B was on the left, can you similarly do that on the right?

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[jade mantle]

Is this even solvable? I don't even know what to do anymore With which method should I approach this?

[raw fractal]

My first instinct is to solve for x in the first equation, so x = 1 - iy, then substitute that into the second equation to find y.

[void niche]

this works. its two equations in two variables, you can solve this using typical systems of equations methods

substitution, elimination, gauss-jordan elimination, whatever

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@jade mantle Has your question been resolved?

[jade mantle]

I think it worked, let me verify. One question, would taking the is out be a problem? Maybe it was that...

Got X= 3/2 and Y= i/2, the first equation is right with those, but the second one wrong

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[tranquil kestrel]

may i know is it true that only option b is right?

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[limpid stone]

3f(x)+f(1/x)=2/x, show that f(-x)=-f(x)

[limpid stone]

for these types of questions

would i try to sub in x for some sort of value in order to get f(x)?

or is there a better way to do it

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[tacit arch]

the only way to solve it for all x is to be as general as possible, so no specific value of x is enough

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[alpine sable]

$3f(x)+f(\frac{1}{x})=\frac{2}{x} \text{ show that } f(-x)=-f(x)$

[ocean seal]

aspwil

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[errant dagger]

Im struggling to understand how to do ii

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[errant dagger]

<@&286206848099549185>

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@errant dagger Has your question been resolved?

[gusty gorge]

So the question asks you to assume that the Moon undergoes uniform circular motion. Therefore, the centripetal acceleration is given by a = v^2/r

[vale wigeon]

no that's part (i)

part (ii) is just applying the inverse-square law

and ensuring it matches up

[errant dagger]

idk how to do that

the inverse square law of what

[vale wigeon]

gravity, what else

accel is proportional to distance^-2

[errant dagger]

whats the proportionality constant thoughb

I understand the concept

not really the equation / solution

[vale wigeon]

$a = kr^{-2}$

[ocean seal]

Ann

[vale wigeon]

when $r = r_E$ (radius of earth) you have $a = g$ and you need to find the value of $a$ when $r = r_E+r_M$

[ocean seal]

Ann

[errant dagger]

HM

hm*

answer doesnt agree

i got a = (g(r_e)^2) / (r_e + r_m)^2

answer is this

[velvet pelican]

im pretty sure they assumed $r_E + r_M \approx r_M$

[ocean seal]

Frosty

[velvet pelican]

since r_E << r_M

@errant dagger ^

tbh if you use your thing and plug in the values, you should get the same answer or like +/- 1dp

[errant dagger]

thats a bit annoying 😦 but ig i get that thank you

wrong one nvm

idk how to solve those

and also the answers say for [2,1] its ca + db which is not what i got

i got ca + dc

[velvet pelican]

looking at the 1st and 4th equation, what can you deduce?

[errant dagger]

a^2 = d^2

[velvet pelican]

yea

but you can now simplify that to 2 cases

which can be done separately

[gusty gorge]

you just have to find one

so maybe you could set a = -d or something

[errant dagger]

this isnt infinite solutions though is it?

its 4 eq with 4 variables

[velvet pelican]

wait nvm i thought it was find all LOL

[velvet pelican]

it likely is

[gusty gorge]

a=-d does not work lmfao

[velvet pelican]

wait does it give b=0 or smth

[gusty gorge]

actually, maybe it does

it basically makes the two center equations dead

a^2 = -bc

maybe set a = 1, b = 1, c = -1, d = -1

[velvet pelican]

mhm works

[gusty gorge]

,w [[1, 1], [-1, -1]]^2

[ocean seal]

Results provided by WolframAlpha

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Requested by Saccharine#3780

[errant dagger]

what??

How did you do that

[gusty gorge]

we start from a^2 = d^2

[errant dagger]

yeeey

[gusty gorge]

since we only have to find one matrix that works, I'm just going to assume that a = -d

this zeroes out the middle equations, so we don't have to worry about them

and the only constraint we have to meet is a^2 = -bc

and since I don't like calculating square roots, I just chose a = 1

[errant dagger]

so you just arbitarily pickjed a value for a?

[gusty gorge]

yeah

[errant dagger]

so bc = -1 and

oh bc = -1

from eq 1 and 4

so since b and c can really be anything and still works for the two middle equations

you just have to find solutions for these two?

like theyre independtant from eq 2 and 3 essentially?

[gusty gorge]

If you set a = -d, then equations 2 and 3 are satisfied by any choice of b and c

[errant dagger]

yeee

so then arbitarily pick either b or c to be -1

[gusty gorge]

you could pick b = 2, c = -1/2 or whatever too

[errant dagger]

so like b = -1 c = 1

yeeee

okie i get that tysm :)_

how do i prove this one?

A,B are equal size square matrix

[gusty gorge]

I guess the lousy way is to show it by showing ABv = (-A)(-B)v for all v using the summation notation stuff

[lean spire]

use the fact that aTv = Tav

where a is a scalar, v is a vector, and T is a linear map

ie T is a matrix

[lone heart]

@errant dagger Has your question been resolved?

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[errant dagger]

I said i is false and ii is also false

but ii is true and i dont know why

I dont understand the inverse rules when multiplying and shit

[velvet pelican]

i think the rule was like $(AB)^{-1} = B^{-1} A^{-1}$

[ocean seal]

Frosty

[velvet pelican]

@errant dagger

[lone heart]

@errant dagger Has your question been resolved?

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[tawny fable]

Hi there

[tawny fable]

I have some more questions in sets theory

@vale wigeon Sorry for not answering yesterday! It would be great if you could explain the function we talked about again 🙂

[vale wigeon]

what function

[tawny fable]

And for the other question:

Let A in R be a countable set.
We define B = {x - y | x,y in A}

a. prove that it is possible that B = Q
b. Show that it is not possible that B = [1,pi]

[vale wigeon]

...

one at a time please

i don't remember which function you were talking about

[tawny fable]

Ok yeah sorry

I have to give an example of a bijective funtion from the set of functions f : R --> {0,1} s.t. for all x in R\Q f vanishes, to the set of functions g: Q --> {0,1}

[vale wigeon]

\ not /

[tawny fable]

yup

[vale wigeon]

okay

[tawny fable]

So you said something there and I wansn't to see that

[vale wigeon]

so i called the first set S and the second T

and then i said:

the bijection from $S$ to $T$ is given by $f \mapsto f\big|_{\bQ}$

[ocean seal]

Ann

[vale wigeon]

are you able to see this image now or are you still having technical issues that i have no control of

[tawny fable]

what is the meaning of f --> f|_Q

yeah yeah I do

thanks

[vale wigeon]

$f\big|_{\bQ}$ is the restriction of $f$ to $\bQ$

[ocean seal]

Ann

[tawny fable]

When you say that do you mean that you change the input or the output?

[vale wigeon]

no

i mean what anyone else means when they talk about the restriction of a function to a subset of its domain

[tawny fable]

I think we might use some other words to say that in Hebrew

So let me google it for a sec

sorry

oh ok

you create a function with a narrower domain

[vale wigeon]

yes that's what restriction means

[tawny fable]

So you get a function as an input right?

[vale wigeon]

yes

we are creating a bijection between S and T, which is a function whose inputs are themselves functions from a certain class

[tawny fable]

I think I get it

and we don't actually care about the fact that the function vanishes for every x in R\Q right?
We just deal with it's domain

[vale wigeon]

well no, we do care

the fact that we're only considering functions that vanish outside Q is what makes this map one to one

[tawny fable]

Ummm

Ok let me understand that

So for example

we define h: S --> T s.t. h(f) = f_Q

right?

now f_Q : Q --> {0,1} ?

[vale wigeon]

f|_Q but yes

[tawny fable]

Ok we haven't done that yet in the course

nice

Can you explain please again why is it one to one and onto?

and how does it relate to the definition of the functions in S?

[vale wigeon]

i feel like i will have to repeat the same thing like ten times before you get it

and i don't have the energy to do that right now

two functions in S that get mapped to the same function in T are equal on Q due to being mapped to the same function in T, and are equal on R\Q due to them both vanishing on R\Q by definition of S

that's for our map being one to one

to see that it is onto, observe that any function on Q can be extended back into a function on R by filling in its values at irrational points with 0

[tawny fable]

got it

great

thanks!

You think you'll be able to help me with this one as well?

Let A in R be a countable set.
We define B = {x - y | x,y in A}

a. prove that it is possible that B = Q
b. Show that it is not possible that B = [1,pi]

[tawny fable]

Sorry about that. It's just the first time I see that restriction thing

idk if it's a good way to prove that it is possible to say that if A is Q for example

Then A is countable

and. then x - y in Are in Q

and actually we get B = Q

?

Or do I need to show a onto function from B to Q in order to say that?

<@&286206848099549185>

@vale wigeon You're out huh?

.close

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[maiden dawn]

im stuck proving EOHD is cyclic ;-;

[lone heart]

@maiden dawn Has your question been resolved?

[maiden dawn]

<@&286206848099549185>

[maiden dawn]

nvm i solved it

damn

.close

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[subtle ether]

Can someone help me?

[knotty spire]

what have you attempted?

[subtle ether]

I have tried to subject xy from e

as in

for e to the power xy

I subjected xy to xy = -infinity

I don't know what to do here

[knotty spire]

have you tried just substituting in the expressions for x and y

[subtle ether]

?

t^2 and t^-1 ?

[slender gull]

Yes.

[pine portal]

yo

[subtle ether]

I got t = -infinity from that

[pine portal]

this questiob will take 1 sec

can I say that the y axis of alpha is 3,5??

from the picture?

[slender gull]

@pine portal #❓how-to-get-help

This channel is unfortunately occupied by someone else.

[slender gull]

How? show your work.

[knotty spire]

@subtle ether what do we have to find here?

$\frac{\dd z}{\dd t}$

[ocean seal]

Remavas

[lone heart]

@subtle ether Has your question been resolved?

[lone heart]

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[alpine sable]

how do i do this?

[gray isle]

start by drawing a diagram

[lone heart]

@alpine sable Has your question been resolved?

[agile prawn]

if u want to figure it out, use a coordinate plane first

then draw the diagram

im pretty sure

[alpine sable]

Keys to solve it after drawing a diagram:
Find a 90° angle
Draw possible lines between the points (0,0), A and B
Find their gradients
Rule: angle between a line and x-axis is the inverse of the tan() of its gradient.

[abstract wigeon]

^Expanding on this, You can find a right triangle between O,B and point on x-axis (3/5, 0) (vertical line along B) as the other vertex.

[lone heart]

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[glass lichen]

How old are you if you don't mind me asking

But also read #❓how-to-get-help

[calm grove]

Send the question and don't mind any emotional damage please

[glass lichen]

<@&268886789983436800>

[tame falcon]

Thanks

.close

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[remote berry]

Can any1 provide any advice on solving this

[lone heart]

@remote berry Has your question been resolved?

[abstract wigeon]

You can multiple each joint probability with the corresponding # buses + # cars to get the overall expected value

For example, starting from the top row:
0.025 *(0+0) + 0.015 * (0+1) + 0.010 *(0+2) + .....

If you need expected values for certain cases (for example, only where buses =1) then you can look for that row/column (column for buses =1)

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@remote berry Has your question been resolved?

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[tight thunder]

guys im doing derivations, can someone explain why everything -1 in exponent

this is task we solved in school on class

[rose raven]

Power rule

Times by the power, drop the power by 1

[tight thunder]

i cant understand

like f'(x)

[rose raven]

So 3x^3 goes to 3*3 x^2

[tight thunder]

man 😄

like i will die

my brain not competitive

to understand this

[noble sinew]

its just a rule; just accept it

[tight thunder]

like

i just want to get a glimpse

is that it?

like derivations are just

-1 on exponent

[noble sinew]

no that is only for derivative of something on the form x^n

[tight thunder]

so there is like alot of rules?

that i have to memorise

[noble sinew]

sure? Or just check your notes

[tight thunder]

like im just trying to understand whats it all bout

you get me

[rose raven]

You can prove it from first principles

If you want

[tight thunder]

man

[rose raven]

It will get to the same end result

[noble sinew]

he obv doesn't have the math skills to prove smth

[tight thunder]

yes

[rose raven]

That’s quite rude

[tight thunder]

this red circled thing

confuses me extra

so i transfer -1 into exponent

replacing it by 0

i would bang my head against wall rn if i understood why that is done

[noble sinew]

and accurate, saying you can prove this when you can tell he likely hasn't done a single proof in his life

is just pointless

[rose raven]

I can show you the proof from first principles, but it's basically a short hand way of doing it