help-10

[obtuse pebble]

@onyx gazelle Has your question been resolved?

[obtuse pebble]

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[inland thicket]

hello

my textbook show this

there is row operations very clearly at beggining

but at bottom i start to not understand what they did

howcome they remove i put in red i dont udnerstand does anyone know?

for making rang of matrix

[kind hawk]

I like how they stopped writing what operations they are doing

I assume this is a typo

[inland thicket]

clearly they are under the influence of chetnik lobby

not even experts know?

[viral blade]

the bottom left matrix i could believe could happen through row operations

[inland thicket]

yis

they explained that 1

[viral blade]

after that though idk how

[inland thicket]

woa

[viral blade]

maybe with column operations?

it should be possible then

[inland thicket]

[viral blade]

feels like if there's a way to accomplish it with row operations then there has to be a way with column operations as well but maybe not when the matrix is singular

row operations correspond to right multiplying by an elementary matrix

column operations would then correspond to left multiplying

so AX = B having a solution is then not the same as XA = B having a solution

[kind hawk]

well if you only want to find the rank of a matrix then doing both operations is fine

but really you should just stop when you have the matrix at the bottom left

[timid silo]

isnt that left multiplication? like having the elementary on the left?

[viral blade]

could be, i was working from memory and my linear algebra isn't that good

[timid silo]

but yeah column operations dont change rank and so these are valid steps, just unnecessary ones

maybe its there to show that there is 100% no way to reduce it even further so they bring it into this minimal form with only 1s on the diagonal, but it should be obvious that you can reduce it further

[viral blade]

yeah you're right

ah

because order goes right to left usually

er

idk

there's some way to use that fact to justify it but

[timid silo]

justify what?

[viral blade]

row operations being left multiplication

but i don't actually have an intuition for why they represent a transformation applied applied to a vector after the original matrix

[timid silo]

i guess just take the three elementary matrices and show their properties when left multiplying

[viral blade]

yea ik its showable

[obtuse pebble]

@inland thicket Has your question been resolved?

[obtuse pebble]

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[lone harbor]

i got as far as
Log3 (9t) - 2Log9 (12/t) = 2

but the 2 logs have a different base

[strange belfry]

they are each a power of 3 though

[lone harbor]

oh i see that

so should i square Log3 (9t)

[strange belfry]

also to remind of a log law: $log_{b^{p}}(x^{y}) = \frac{y}{p} log_b(x)$

[warm shale]

biggboy

[lone harbor]

we havent covered this though

[strange belfry]

you can derive it from the log law $log_b(a) = \frac{1}{log_a(b)}$

[warm shale]

biggboy

[strange belfry]

or just use the fact 9 = 3² and raise 9 to the power of "the equation" if that makes sense

[lone harbor]

oh ok

yea that makes sense

wait lemme try

got it

i actually used the log law for changing the base and setting them both to the power of 10

[strange belfry]

good work!

[lone harbor]

tnx

.close

[obtuse pebble]

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[timid silo]

I'm not sure where to even start with my question, let's see... Given two values; Item Quantity (0 - 100) and Item Price (0 - Any Positive Number) how can i determine a "value" that would put first in order the Items that have Lowest Quantity and Highest Price?

[drifting wraith]

well you can't assign a value like that

if it has to prioritize quantity

if you multiply price by 101 and subtract quantity it works but then price is more important

[timid silo]

Yea both Quantity and Price are important, but i'm only able to order by one of them. I expected to be able to obtain a value that would create a list with the most "rare" and "valuable" items ordered.

[high lily]

if you want to prioritise quantity, you could do something like

[timid silo]

Can you elaborate a bit further this for me? Why 101?

[high lily]

could try
(100 - Q) * (max(prices) or price cap or arbitrary large value) + price of item

[timid silo]

That looks interesting, does this formula have a name so i can study on the subject?

[high lily]

broadly this is weighting values

giving values a certain multiplier to increase importance when calculating a score

[timid silo]

Thank you!

[obtuse pebble]

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[static holly]

1/2 10 × v × √3/2 = 15√3
Right i've come up with this equation but i'm struggling on which steps to take first to rearrange it and get the correct answer of V because the website keeps rejecting all my annswers

Question is to find V then find W using this nuumber

[lost tree]

show ur work

oh wait

[static holly]

I did

at the top

[fierce lagoon]

You just divide by the coefficient of v

$\frac{1}{2}(10)(\sin(60^{\circ}))v \rightarrow \frac{5\sqrt{3}}{2}v$

[static holly]

But is it not possible to for example divide by each of the terms individually

for example

divide by root 3 over 2 first

simplify

then divide by 5

[fierce lagoon]

It's just like any fraction division

$15\sqrt 3 \div \frac{5\sqrt{3}}2 = 15\sqrt 3 \cdot \frac 2{5\sqrt{3}}$

[warm shale]

Umbraleviathan

[static holly]

how did you get 5 root 3 over 2?

[fierce lagoon]

Up here

I simplified the LHS of your equation

[static holly]

LHS?

[warm shale]

Umbraleviathan

[fierce lagoon]

Here I'll make it better

Left hand side

[static holly]

oh right

so that means the value of v is 5root 3 over 2?

[fierce lagoon]

No

I quite literally just simplified it

[static holly]

ah sorry

[fierce lagoon]

What you originally had is this:

$$\frac 12 (10)\left(\frac{\sqrt 3}2\right)v = 15\sqrt{3}$$

You need to simplify/condense the LHS:

$$\frac{5\sqrt 3}2 v = 15\sqrt{3}$$

[warm shale]

Umbraleviathan

[static holly]

wait yes

yes yes

sorry

I just thought you changed the RHS to that

I was desperately confused for one second

but I always have to simplify LHS first in this senario?

[fierce lagoon]

Well yeah

You wanna isolate v

You gotta gather up everything to make a coefficient of v

[static holly]

now I can just

divide the RHS by 5 root 3 over 2

[fierce lagoon]

Yes

[static holly]

15 root 3/ 1 * 2/5root3

so v is 15root3?

oh nvm it's 6

[fierce lagoon]

Yeah

[static holly]

then what do I use to work out w?

cosine rule?

[supple granite]

yeah

[static holly]

w^2 = 10^2 + 6^2 - 2 * 10 * 6 * root3/2

right so I got to

100 + 36 - 60root3

on the RHS

but the answer wants in the form a root b where both are integers and i'm not sure where to go next

[silk shell]

why are you using root3/2 for the value of cos60

[static holly]

oh wait

lol

just 1/2

Nah right so

I got to

w^2 = 100 + 36 - 120 * 1/2

but I still don't know how to get that in the form a root b where both are integers

[silk shell]

-120*1/2 would be -60

[static holly]

yes

120 + 36

no

100 + 36 - 60

[silk shell]

100 + 36 -60 would be 76

[static holly]

yup

then root 76

oh wait

I have to rationalise

I hate surds

I mean

simplify

[silk shell]

that’s what i was about to say lol

[static holly]

yeah

I haven't done surds in ages

I just keep forgetting the laws

but I'll be fine from here

we can close this help channel

[silk shell]

same, but practice makes perfect

[silk shell]

alr, if you need help you can dm me too idm

[static holly]

thanks boss!

[obtuse pebble]

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[torpid narwhal]

can any one help?

[high lily]

!status

[obtuse pebble]

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

[torpid narwhal]

3

[high lily]

show your work

[torpid narwhal]

i thought it was y=3x+10

[high lily]

work
not just final result

i.e. what did you do to get that equation

[torpid narwhal]

by looking at the graph

10 intercepts the y

and the line for every 1 goes 3

so gradient is 3

[viral blade]

you're counting grid lines

but look carefully at the axes

[high lily]

and the line for every 1 goes 3
exactly which part of the graph were you looking at

also ^, pay attention to the scale

[torpid narwhal]

OHHHHH

15x

y=15x+5

?

[high lily]

why's it +5 now

[torpid narwhal]

i mean 10

my bad

[high lily]

if you meant
y = 15x + 10
then yes, that would be correct

[torpid narwhal]

luv broski

come play me in chess

u free for a game?

[viral blade]

I'll play

[obtuse pebble]

@torpid narwhal Has your question been resolved?

[obtuse pebble]

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[indigo coral]

I am told to find f^-1(x) idk how to continue from here to get y on its own

[wise talon]

you are very close

like

consider 1/(1/y)

[indigo coral]

multiply by y ?

[wise talon]

no that would undo all your hard work

[indigo coral]

im confused what to do with this

change it to y^-1 ?

[wise talon]

we are taking a reciprocal

do you know what that is

[indigo coral]

nope

[wise talon]

ok

lets suppose we have a = b

now what is 1/b equal to

[indigo coral]

1/a

[wise talon]

good

can you see where im going

[indigo coral]

oh so

y= 1+x/2 ?

[alpine raven]

no lol

and you should solve for x

[wise talon]

these are not equal

[wise talon]

you said it correctly yourself

[indigo coral]

im confused

[wise talon]

alright

lets try something

lets say we have y = 293774x+92838483

now what is 1/y

[indigo coral]

1/293774x+1/928...

[alpine raven]

no

[wise talon]

ok

[indigo coral]

is that not what we did with a=b

[wise talon]

if we have f(x) = g(x)

then what is 1/f(x)

[indigo coral]

1/g(x)

[wise talon]

now

let f(x) = 1/y

g(x) = 1+ 2/x

what is 1/f(x)

[indigo coral]

1/(1/y) = 1+ 1/(2/x) ?

[alpine raven]

:/

[fading musk]

what

[wise talon]

ok lets try

this

$\frac{1}{(f(x))} = \frac{1}{(g(x))}$

[warm shale]

active mental mutilation liker

[wise talon]

now

try again

[indigo coral]

does f(x) just stay the same

[fading musk]

what even is f

[indigo coral]

ill post the question

[fading musk]

ik

ok

[indigo coral]

i got x=-6/5 for the first part btw

[fading musk]

you let y=f(x)

so $y = \frac{x}{x+2}$

ukm

[indigo coral]

y=x/x+2 ?

[warm shale]

DinVin95

[wise talon]

yeah they already got this far

we're trying to find the inverse now

[fading musk]

then you swap all y with x

because the range becomes the domain

[indigo coral]

x=y/y+2 ?

[fading musk]

when its inverse

[fading musk]

yes

then you solve for y

[indigo coral]

ok gimme a sec

uh i have xy+2x=y

[fading musk]

no

move it properly

make sure all y terms are on one side

[indigo coral]

2x = y-xy ?

[fading musk]

ye then take the y out and divide

[indigo coral]

2x/1-x = y

[fading musk]

put them in parantheae

parentheses

so you get $y = \frac{2x}{1-x}$ right?

[warm shale]

DinVin95

[indigo coral]

ye

[fading musk]

what is y?

[indigo coral]

uh i dont have a value for it

[fading musk]

scrill up

scroll up

[indigo coral]

x/x+2

let them equal to each other ?

[fading musk]

no…

you already finished

[fading musk]

now swap back

[indigo coral]

what

swap back for what

[fading musk]

do you remember i swapped x and y

[indigo coral]

oh so x=2y/1-y

[fading musk]

thats because it turned into f^-1(x) as the domain and range is swapped

[fading musk]

no…

now you already get y=f^-1(x)

and so the question is basically asking for y

[indigo coral]

i have to state the domain now

[fading musk]

you already found f^-1(x)

[indigo coral]

y= 2y/1-y ?

[fading musk]

in what case is $\frac{2x}{1-x}$ not defined?

[warm shale]

DinVin95

[fading musk]

no…

[fading musk]

y=f^-1(x) here

[indigo coral]

right

[fading musk]

finding the domain

[indigo coral]

ye

[fading musk]

in what. case?

[indigo coral]

dont understand

[fading musk]

is 2/0 defined?

[indigo coral]

yes ?

[fading musk]

let me ask again

if so, what is the value of 2/0

[indigo coral]

i mean it doesnt make sense

[fading musk]

yup

can you give a value to 2/0?

[indigo coral]

no

[fading musk]

well so it is not defined when the denominator is 0

[fading musk]

recall from this

so when x=?, f^-1(x) is not defined

now fill in the “?”

[indigo coral]

so do i sub in the x=-6/5

from part i of the question

[fading musk]

no

tbh both questions are not related to each other

now you know that a/0 is not defined right

[indigo coral]

yes

but we dont know what x is

[fading musk]

so if you want to find the point where f^-1(x) is not defined

then make the denominator 0

so 1-x=0

[indigo coral]

so x=1

[fading musk]

thats right

now substitute x=1 into f^-1(x)

what do you get?

[indigo coral]

error since its not defined

[fading musk]

good job

now figure out this: is there any other points that f^-1(x) is not defined?

[indigo coral]

is that how you state a domain ?

[fading musk]

not yet

i am teaching you to find the domain

[indigo coral]

like any other x value ?

[fading musk]

tes

yes

[indigo coral]

other values besides 1 would work

[fading musk]

thats right

now you found the domain

and you need to state it

how would you use a notation to describe EVERY number

[indigo coral]

uh x is not equal to 1 ?

[fading musk]

no

do you know the real number symbol?

[indigo coral]

N ?

nvm its R xD

[fading musk]

yes

[indigo coral]

so x= R

[fading musk]

so the domain is real numbers except 1 right?

[indigo coral]

yes

[fading musk]

dont forget to exclude 1

so it is R{1}

wait

no

let me write it in latex

discord disabled

so the domain of f^-1(x) is $R \ {1}$

you use the \ to exclude

R \ {1}

[fading musk]

this is the domain

[indigo coral]

lets say x=1 works

what would i state then

[fading musk]

R

[indigo coral]

ah ok

[fading musk]

all teal nymbers

real

you can use union if u want

[indigo coral]

i see

thanks for the help

[fading musk]

(-inf,1)(1,+inf)

no

np

[indigo coral]

.close

[obtuse pebble]

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[obtuse pebble]

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[shadow lava]

Complimentary angles make trigonometry so much easier ☺️

[shadow lava]

My really any help needed, but just a statement

[timid yacht]

what?

[shadow lava]

The unit circle is designed for this

Pi/2 can be found at each 90 degrees

[shadow lava]

You can now switch between function and cofunction at your hearts whim

Incredibly powerful stuff imo

https://tenor.com/view/zelda-gif-19935744

https://tenor.com/view/goku-dragon-ball-level-up-anime-gif-gif-22760068

.close

[obtuse pebble]

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[lost tree]

,rotate

[charred topaz]

I am struggling to find the necessary values to solve this problem

Is there a possible substitution or something in my work that can be simplified further or? Yea not sure

[obtuse pebble]

@charred topaz Has your question been resolved?

[blissful bane]

Recall what $P(A|B\cap C)$ means

[charred topaz]

Given B and C what is the probability of A

Is that what you mean?

[warm shale]

CapitalW

[blissful bane]

What’s the definition of P(A|B)?

[charred topaz]

Given that Event b has occured what is the probability that event A will happen

[blissful bane]

P(A|B)= P(AB)/P(B) right?

[charred topaz]

yes

wait quick question

[blissful bane]

Similarly apply the definition to this

[blissful bane]

sure

[charred topaz]

I was using bayes theorem and the conditional porbability formula you just gave there

when do i know which to use?

I feel as though bayes is like the universal one where it applies to everywhere if you have the needed information whereas conditional not so sure

[blissful bane]

sorry brb

[charred topaz]

all good

tag me when your back 🙂

[blissful bane]

@charred topaz

Back, sorry my bad

[charred topaz]

All good

[blissful bane]

so let’s how we can tackle this problem

[charred topaz]

😄

[blissful bane]

and see how conditional probability is related to Baye’s theorem

[blissful bane]

So here we wanted to find $P(A|B \cap C)$

[charred topaz]

yea

[warm shale]

CapitalW

[blissful bane]

can you apply the definition of condition proabrbility there?

[charred topaz]

i did

did you see my work above too?

i can post it again if you want

[blissful bane]

$P(A|B \cap C) = \frac{P(ABC)}{P(B \cap C)}$

[warm shale]

CapitalW

[blissful bane]

Right ?

[charred topaz]

Yea

Is there a reason why it's not written A n B n C in the numerator?

[blissful bane]

Both are same

[charred topaz]

Okay just making sure because I see ABC everywhere but i just write as A n B n C

Hello you still there?😄

[blissful bane]

@charred topaz

My bad, my internet is a mess

Let’s get back,

[charred topaz]

Hello all good

[blissful bane]

So let’s try to find the numerator

[charred topaz]

Im eating lunch while doing this hehe

[blissful bane]

$P(A|B \cap C) = \frac{P(ABC)}{P(B \cap C)}$

[warm shale]

CapitalW

[blissful bane]

We need to use the information given to figure out ABC which intersection of all events

[charred topaz]

yup

[blissful bane]

Notice, It’s given P(C| AB)=1/2

So if we apply definition $P(C|AB) = P(CAB)/P(AB)$

[warm shale]

CapitalW

[charred topaz]

yup

[blissful bane]

1/2=P(ABC)/P(AB)

So if we can find P(AB), we can find P(ABC) right?

$\frac{1}{2}= \frac{P(ABC)}{P(AB)}$

[warm shale]

CapitalW

[charred topaz]

yup

[blissful bane]

So can you figure out P(AB) ?

Try applying the definition to each of the given

We have to essentially find P(ABC) which adds up the numerator

hint: Use definition to P(B|A)

Since it’s given $P(B|A)=1/4= \frac{P(BA)}{P(A)}$

[warm shale]

CapitalW

[charred topaz]

oh I just got oit

it

[blissful bane]

Since we already know P(A)=2/5

P(AB)= 2/5*1/4

Can you do the rest ?

[charred topaz]

I was going through my work and found my mistake hahaha, I had written P(C) in the denonminator instead of P(B)

I needed to know what B n C was but couldn't find it

because I wrote it wrong 😦

[blissful bane]

My bad, i didn’t checked that, i usually have a hard time looking through other peoples work

even mine, i cant find

[charred topaz]

All Good thanks for helping though

you made me look back at what you were doing and what I did

and then realized I had the wrong denominator

[blissful bane]

Now Baye’s Theorem

[charred topaz]

Oh yea that question

[blissful bane]

It’s just an exploitation of conditional probabilities

There are two ways you can write P(AB) in terms of conditional probabilities

ie P(AB)= P(B|A)* P(A)

Or P(AB)= P(A|B)* P(B)

Notice the difference

[charred topaz]

they are the same thing no?

P(BA) = P(B|A) * P(A) ?

[blissful bane]

Yes, they are same but in some context we might only have only information

[charred topaz]

all gotcha

so what the information is given is what we use

[blissful bane]

There is a classical example of a cancer related problem

[charred topaz]

Gotcha what is that exactly?

If you don't have time can you send like a link or something?

I'd like to read it

[blissful bane]

Sure, Baye’s theorem is sort of powerful, it’s good to understand it

[charred topaz]

yea rlly is if you get the information you need

[blissful bane]

https://www.eecs.qmul.ac.uk/~norman/papers/probability_puzzles/cancer.html

[charred topaz]

oh interesting

your classic java websites hehe love em

[blissful bane]

Here they discuss, the probability of having cancer give you are a smoker.

Yep yep

Try reading it,

[charred topaz]

i will thank you again 🙂

do you close this or i?

[blissful bane]

Sometimes it hard to get a specific data type of data so we can use it’s other conditional definition to make over

Have a good one

[charred topaz]

.close

[obtuse pebble]

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[ancient jacinth]

these equations should be the same yet they have different solutions, why?
and how do i solve the problem 3sin(2x+30)=tan(2x+30)
-180<x<180

[timid silo]

you will start with the principle that: -1<sinx<1
so: -1<tanx<1

and you countinu

[ancient jacinth]

ok

[ancient jacinth]

what does this do though

[ancient jacinth]

i mean tan(x) can be greater than 1 and less than -1

its bounds are infinity and -infinity

[timid silo]

yes but it is an equality with this principle you can use the fact that if sinx is bounded then tanx is also

[ancient jacinth]

hmm

i see now

[ancient jacinth]

why can i not divide by tan though

[timid silo]

tanx can be equal to 0

[ancient jacinth]

so then what can i do

cos(2x+30) cannot equal 0 then

[timid silo]

@ancient jacinth

[ancient jacinth]

hmm

hmm

[timid silo]

you understand ?

[ancient jacinth]

yes but i dont know what to do next

why are we doing this

[timid silo]

with this fact you can also solve sin2x+30

[ancient jacinth]

-3<sin(2x+30)/cos(2x+30)<3

[timid silo]

its a principe of trigonometry and trigonometric circle

[ancient jacinth]

where do we go from here

[timid silo]

you can continue from this

[ancient jacinth]

-3<sin(2x+30)/sin(2x-60)<3

?

[ancient jacinth]

i still dont see why im doing this

[timid silo]

no it is necessary to frame cos2x+30 because it is at the denominator and also sin2x+30

[ancient jacinth]

how does this help me solve the problem

cos(2x+30) is not equal to 0

so

[timid silo]

wait I will approach you to help you

[ancient jacinth]

what

[ancient jacinth]

how are they equal

its just greater than or less than right

[timid silo]

this should be like this : -31.249/2<x<-28.7/2

[ancient jacinth]

i dont understand

.close

[obtuse pebble]

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[earnest oriole]

Is anyone available to help me with a quiz?

[kind hawk]

we dont help you cheat

[deft sun]

just believe in yourself

[dark stirrup]

Your quest breaks TOS

[obtuse pebble]

@earnest oriole Has your question been resolved?

[distant moth]

going into exams with 1 percent knowledge and 99 percent faith

[obtuse pebble]

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[spice chasm]

|x+3|²-4|x+3|-5=0

[spice chasm]

Solve x

[timid silo]

try treating |x+3| as another variable, say t

[spice chasm]

Wait Is wolfram allowed

[timid silo]

what do you get?

[spice chasm]

t²-4t-5=0

[timid silo]

almost

there you go

[spice chasm]

so (t+1)(t-5)=0 so t=-1 and t=5

[timid silo]

yup

now what do you think is the next step?

[spice chasm]

System of equations

[timid silo]

not really, we just have two different answers for x

[spice chasm]

1. t=|x+3|
2. t=-1

[timid silo]

yes, thats one of the solutions

[spice chasm]

oh it’s transitive property

[timid silo]

other one is with t=5

[spice chasm]

t=-1=5=|x+3|

[timid silo]

well be careful when you write that because you just wrote 5=-1

with t=-1; we get that |x+3|=-1; if we remove absolute value we get x+3=±(-1)

[spice chasm]

but doesn’t t equal both

[timid silo]

yes but in different instances

[spice chasm]

Oh so it’s 2 cases

[timid silo]

yup! and for every case of t if im not mistaken we have two cases for x, giving four solutions for x

[spice chasm]

|x+3|=-1 which i don’t think has any real solutions and |x+3|=5 which results x=2 or x=-8

wait

,w |x+3|=-1

[warm shale]

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[timid silo]

ah of course that one doesnt have solutions

[spice chasm]

Shouldn’t it have complex solutions

[spice chasm]

like log(-1)

[timid silo]

hmmm

[spice chasm]

i factored this into (|x+3|+1)(|x+3|-5)=0

[timid silo]

we can take it as $sqrt((x+3)^2)=-1$

how do i use latex

[spice chasm]

Idk xD

[timid silo]

we can take it as $\sqrt{(x+3)^2}=-1$

[warm shale]

operandi8827

[spice chasm]

oh yea

[timid silo]

sooo

[spice chasm]

you put \ whenever there is a symbol or function

[timid silo]

yea it probably has got something complex

yes

[spice chasm]

wait nvm √ can’t be negative

,w solve for complex roots of √(x)<0

[warm shale]

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[spice chasm]

ok

,w |x+3|²-4|x+3|-5=0

[warm shale]

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[obtuse pebble]

@spice chasm Has your question been resolved?

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[vestal dome]

Hello there I got a question about "Topological Space" :
I have to show that in the metric space ( E, d ) applies : M1 = [0, 1/2) is a "open ball".

I am not sure how to work with the definition :

[warm canopy]

In E, think about what an open ball of radius r centred at zero looks like

[vestal dome]

@warm canopy (Is it okay that I am pinging you?)
So:
U_r (0) = |0 -y| < 1
(something like that?)

[obtuse pebble]

@vestal dome Has your question been resolved?

[kind hawk]

that can't be correct as written cause on the left side is a set and on the right side is not

what you mean is U_r(0) = {y in [0,1]: |0-y|<r}

so which y do satisfy that

[vestal dome]

@kind hawk
But my E=[0,1) so it should be " y in [0,1) "
But this isn't the end of the proof or am I wrong?

[kind hawk]

yeah sorry typo

U_r(0) = {y in [0,1): |0-y|<r}

point still stands. which y are in that set

which y satisfy |0-y|<r

[vestal dome]

the biggest r we can use should be r=1/2, because our M1 = [0,1/2]

IF this is rigth then y< 1/2 :/

[kind hawk]

lets ignore M1 for a moment

maybe you are thinking too complicated rn

0<r<1

can you write the set {y in [0,1): |0-y|<r} differently

easier

[vestal dome]

y < r

[kind hawk]

that's an inequality, not a set

[vestal dome]

I am sorry, give me a moment to find the german "word" for it

[kind hawk]

dann machen wir das halt auf deutsch

[vestal dome]

Danke

Was genau meinst du mit "set" ?

[kind hawk]

naja menge halt

U_r(0) ist eine menge

nach definition die menge {y in [0,1): |0-y|<r}

aber diese menge kann man einfacher schreiben

[kind hawk]

das ist ja schon der richtige gedanke, ist halt einfach nur keine menge

[vestal dome]

Ich bezweifele, dass es richtig ist aber wäre es vielleicht die Menge E ?
( Da |0-y| = y und y aus [0,1) )

[kind hawk]

ne es ist nicht E

es ist ein interval

welches

[vestal dome]

Ein anderes Intervall außer [0;1[ fällt mir nicht ein.

(Was genau sagt diese Menge aus? Ist das die Menge an Punkten die im Radius r sind? Wenn ja, wie stellen wir r fest ?)

[kind hawk]

ich will jetzt ne antwort hören die von r abhängt

[vestal dome]

(0, 0+r)
Es ergibt aber keinen Sinn

Nur ein paar Verständnis Fragen:
Die Menge U_r(x) kann man so sehen, als wäre es ein Kreis mit dem Mittelpunkt auf x und dem Radius r. Somit würde unser Kreis dann bei "x_1= x-r " starten und bei "x-2 = x+r" enden. Kann man sich U_r so vorstellen?

y müsste dann innerhalb von (x-r, x+r) sein, damit er innerhalb des Kreises ist

[kind hawk]

so in etwa. aber hier sind wir nur in der menge E unterwegs und da gibt es keine zahlen unter 0

[vestal dome]

Aber weil unser Mittelpunk x dem Randpunkt unseres Intervalls entspricht, also 0, kann es ja keinen x-r geben

[kind hawk]

wären wir in R, dann wäre das interval (-r, r)

hier sind wir aber nur in E, darum ist das interval [0,r)

[vestal dome]

Intervall des Kreises wäre (-r,r) groß, oder?

[kind hawk]

"kreise" in R sind intervalle der form (x-r, x+r) wie du gesagt hast

wir sind hier aber nicht in R, wir sind nur in E und darum läuft die welt anders

weil es halt einfach keine zahlen unter 0 gibt

darum können die auch nicht im kreis drin sein

[vestal dome]

Also ist der Intervall des Kreises [0,r)
und r können wir jetzt beliebig wählen solange es nicht r>= 1 ist

[kind hawk]

yes

wenn r>=1 dann ist es halt einfach nur [0,1)=E

[vestal dome]

Zum Beispiel könnte ich jetzt sagen dass:
-M_1 = [0, 0.3)
-M_2 = [0, 0.9)
-M_3 = [0, 0,5)
Alle diese Mengen sind offene Mengen in E.
Es würde sich weiterhin um eine offene Menge handeln, auch wenn wir nicht bei 0 starten oder?
Zum Beispiel: M_4=(0.3 , 0.7) oder M_5=[0.3, 0.5)

[kind hawk]

also die ersten paar sind offen, ja

M_5 ist nicht offen

weil es halt zahlen unter 0.3 gibt

[vestal dome]

Ah, ok!
Vielen Dank für die Hilfe!

Noch mal zu der Aufgabe: Um zu zeigen dass M_1 =[0, 1/2) offen ist, muss ich schreiben:

U_r(0) ={y in E: |0-y|<r} => Damit ist die Menge [0,r) definiert, wobei r aus E genommen werden kann und nicht r>=1 sein darf

[kind hawk]

r darf alles sein

bzw sollte natürlich positiv sein

für r>=1 ist die menge halt nicht [0,r) sondern nur [0,1)

aber trotzdem darfst du r so wählen

[vestal dome]

Aber [0,2) wäre doch ungültig, weil unser E nur bis 1) geht

[kind hawk]

du darfst trotzdem r=2 wählen. als menge kommt dann nur [0,1) raus aber das hindert dich nicht daran r=2 zu wählen

[vestal dome]

OH !
ok, danke nochmal den Hinweis!

Nochmal vielen vielen Dank für die Hilfe

[kind hawk]

gern

[vestal dome]

was studierst du bzw. hast du studiert, wenn ich fragen darf?

[kind hawk]

mathe

[vestal dome]

Welches Semester?

[kind hawk]

ich bin fertig mit dem studium

[vestal dome]

Oh cool!
Ich habe gerade mit dem Studium angefangen (erster Semester) 😅
Hättest du vielleicht paar Tipps (vill. auch Bücher), wie ich meinen "Mathe-Verständnis" verbessern könnte?

[kind hawk]

also die ersten zwei-drei semester sind das schwierigste weil man da halt lernen muss mathematisch zu denken

danach wirds "einfacher" weil man zumindest weiß wie man an zeug herangehen kann

darum am anfangen durchhalten

viel üben

immer schön fleißig die hausaufgaben machen

[vestal dome]

Bin dabei und versuche so gut es geht mitzukommen.
Es ist halt ein komisches Gefühl, wenn man sieht, dass die Kommilitonen den ganzen Stoff besser verstehen und viel schneller vorankommen als einem selbst.

Naja, ich will dich jetzt auch nicht mehr stören und auch nicht diesen Channel "unnötig" besetzen.

Nochmals vielen Dank für die Hilfe

[kind hawk]

ja sowas passiert halt. nicht zuviel vergleichen, dann fühlst du dich nur schlecht

arbeitest du alleine?

oder hast du dir ein paar andere gesucht

[vestal dome]

Ich habe paar gute Kommilitonen, die mir immer helfen, wenn ich Hilfe benötige. Aber die ganze Zeit sie zu fragen, fühlt sich so an, als würde man Ihnen die ganze Zeit auf die Nerven gehen. (vor allem wenn viele andere diese Personen um Hilfe bitten)

[kind hawk]

ich sag mal so, solange sie nicht genervt wirken ist alles gut

manche leute sind halt einfach nett

und helfen gerne

[vestal dome]

Das ist sehr faszinierend. Die versuchen immer jeden zu helfen, wenn sie gefragten. Und ich bin auch sehr glücklich solche Personen an meiner Seite zu haben.
Ich finde es auch sehr cool, wie schnell man auf dieser Server Hilfe / Antworten bekommen. ich kann mich nicht genug dafür bedanken 😄

[kind hawk]

wobei das auch sehr stark auf die frage und auf den tag ankommt

manche leute haben hier kein glück mit fragen

aber so ist das halt wenn ein server auf freiwilligen passiert

🤷

[vestal dome]

Mal schauen, ob meine Mathekenntnisse auch irgendwann ausreichen um anderen zu helfen 😄

[kind hawk]

also wenn du mathe studierst dann könntest du ez mit schulstoff helfen

quadratische gleichungen oder so gibts hier genug

aber naja, solltest erstmal gucken dass du deine zeit benutzt um deinen stuff zu verstehen

[vestal dome]

Ich könnte es mal versuchen.
Ich studiere IT-Sicherheit und habe als Mathe Module : Analysis I und LADS I

[vestal dome]

Das stimmt 😄

[kind hawk]

was ist LADS

linear algebra ?

[vestal dome]

jup

[kind hawk]

und das DS?

[vestal dome]

Lineare Algebra und diskrete Strukturen

[kind hawk]

ah

[vestal dome]

Ich muss jetzt leider gehen. Habe morgen noch Mathe-Lernwochenende und muss darum schon um 7:30 aufstehen

[kind hawk]

dann ab ins bett mit dir

gott 730 sonntag

[vestal dome]

Danke nochmal und schonen Abend noch

[kind hawk]

gute nacht

[vestal dome]

Ja, ich muss mit der Bahn 1:30h zur Uni fahren

[kind hawk]

uff

[vestal dome]

Langsam hat man sich daran gewöhnt 😄

[kind hawk]

keine wohnung gefunden in der stadt?

[vestal dome]

War am Anfangen auf der Suche, doch dann habe ich mich umentschieden. Bin bei den Eltern geblieben. Erspart viele kosten und auch "Zeit"

[kind hawk]

naja gut dafür musst du jeden tag 3h bahn fahren

[vestal dome]

😬

Aber hat auch viele Vorteile 😄

[kind hawk]

aber naja gut ich kenne natürlich deine situation nicht

[vestal dome]

Die 3h kann man aber auch meistens sinnvoll nutzen

[kind hawk]

ok wenn du das kannst

ich finde bahn kommt immer sehr drauf an

[vestal dome]

Ich kriege halt immer einen Sitzplatz und habe meisten meine Ruhe

Man kann schon gut während der Fahrt mit den Hausaufgaben anfangen

So ich muss jetzt wirklich los, ich wünsche dir noch eine schöne Nacht !

[kind hawk]

jo

gute nacht

[vestal dome]

Gute Nacht
(Wie schließt man diesen Channel? )

[kind hawk]

.close

[obtuse pebble]

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[obtuse pebble]

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[final matrix]

if I have the bounds of an integral described as 1 <= y <= 2, but i'm revolving around an X axis (and thus, want my bounds in terms of x), how can I translate the y bounds to x bounds?

[final matrix]

or am I thinking of this the wrong way?

I already solved for y

where y is y = 4x-1/2

[fierce lagoon]

Use shell method

Instead of washer

[final matrix]

this is the surface area of a curve

i.e. arc length revolved around an axis, not a solid of revolution

.close

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[drowsy carbon]

Hey I am struggling with solving for x for this question

[drowsy carbon]

I know I have to use a rule but Idk what rule

[high lily]

instead of separating the logs combine them
with essentially the same laws(s)

[drowsy carbon]

Ok let me try

[obtuse pebble]

@drowsy carbon Has your question been resolved?

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[foggy ore]

how do you solve integrals that go from 0 to x?

[foggy ore]

like this one

or is it just a normal integral?

[static beacon]

Yoo would

Integrate normally

But when it comes to minusing the limits

Instead of a number you would just have x

[foggy ore]

ah okay, thanks!

!close

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[strange yoke]

I really dont understand how to do this

d/dx of x ^ sinx

[solar trellis]

Write it as exp(sinx.logx)

[strange yoke]

What is exp?

[solar trellis]

e^

[wise talon]

e^sinx * logx

[strange yoke]

Why am i doing e

Why cant I do one of the ones above

[wise talon]

because you know how to differentiate it

the power rule only applies for

x^n where n is an integer

sinx is a function

[strange yoke]

Ok so I can do

I dont understand how to rewrite it

I know that e^ln of something just gives us that something

[wise talon]

it was given to you

how to rewrite it

just scroll ip

up*

[strange yoke]

That doesnt make sense to me

I dont get why it'd be rewritten like that

[solar trellis]

Do you agree the x^sinx = e^(log(x^sinx))?

[strange yoke]

Why log not ln

[wise talon]

log is lm

n

here

usually when log is used its used to mean natural logarithm ln

[solar trellis]

You could use either

By log I mean base e

[strange yoke]

Like this?

[solar trellis]

Yes

Now use ln(x^a) = a*ln(x)

[strange yoke]

But I dont have an ln outside

[solar trellis]

Wdym

[strange yoke]

There has to be an ln so that I can do that

I dont have an ln

Like i nfront of the e

[solar trellis]

There's an ln here

Doesn't matter where it appears

[strange yoke]

Do i do the rule inside the exponent

[solar trellis]

look it's e to the power (something)

You can do whatever you want to (something)

If a = b then e^a = e^b

You can rewrite the exponent however u want

[strange yoke]

Is it this

[solar trellis]

Yes

Now chain rule

[strange yoke]

Is one of the things just e^x

I dont know how to do the chain rule here

[obtuse pebble]

@strange yoke Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[wise talon]

by the way try uhh #multivariable-calculus

if no one responds

not yet let me see if any helpers are available

I only know single variable analysis so

cant help you much

[noble kindle]

npnp, i was thinking maybe this can be turned into a single variable analysis

[wise talon]

lets see your work

[noble kindle]

using g(t) = sint/t

[wise talon]

,w plot sinx/x

[warm shale]

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[wise talon]

lemme see if g is continuous

oh yup it is

[brisk matrix]

not at x = 0

[wise talon]

not at 0 tho

dang beat me to it

[brisk matrix]

problem specifies (x,y) =/= (0,0) tho

so we good

and takes care of x=0 and y=0

[wise talon]

oh thats the equivalent of defining g to be 1 at x=0

which actually goes to show sinc is cont on all R

[noble kindle]

i've only written down this so far. I was trying for some substitution to turn g(x,y) into something like that, then use the epsilon_1 in the pic

[wise talon]

set xy=t

[noble kindle]

right

[wise talon]

thats actually it lmao

[noble kindle]

what my delta?

[wise talon]

exercise for you

hey it's not too bad

[obtuse pebble]

@noble kindle Has your question been resolved?

[noble kindle]

<@&286206848099549185>

.close

[obtuse pebble]

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[unkempt sparrow]

How does sin20 simplify into sin(pi/3)?

[haughty swan]

uh its pi/9

[unkempt sparrow]

I'm not following

[haughty swan]

degrees to radians

its a measure of the arc length

conversion factor is pi/180

[unkempt sparrow]

is pi/3 20?

[haughty swan]

no its pi/9

[haughty swan]

pi/3 is 60 degrees

[unkempt sparrow]

so this is wrong then?

[haughty swan]

oh

[timber fox]

that is not 20 im afraid

[haughty swan]

you mean 2\theta

well that is a difference

[unkempt sparrow]

I'm confused on the step where it says simplify the sine function in the second term. It looks like it simplifies sin(20) to sin(pi/3)

[haughty swan]

$f(\theta) = 2\sin\theta + \sin(2*\theta)$

[warm shale]

rishi e

[haughty swan]

they just plugged in

$\theta = \pi/6$

[warm shale]

rishi e

[haughty swan]

2 time pi/6

2 times pi/2*3

pi/3

[unkempt sparrow]

how do you get from sin(20) to sin(2 * theta)?

oh is it sin 2 * theta and not sin 20

[haughty swan]

its not a 0 its a theta

[unkempt sparrow]

did I read it wrong

[haughty swan]

yeah

[unkempt sparrow]

thats what i get for not wearing my glasses

thank you so much lol

[haughty swan]

lol nah you good

[unkempt sparrow]

.close

[obtuse pebble]

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[brisk matrix]

lol #help-14

similar names too

[obtuse pebble]

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[obtuse pebble]

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[subtle cargo]

I've found that 10=FE×GD using the angle bisector theorem and then used the fact that the sum of two sides of a triangle must be larger than the third side always and narrowed it down to A and B but I can't solve it after that

[obtuse pebble]

@subtle cargo Has your question been resolved?

[subtle cargo]

<@&286206848099549185>

[obsidian isle]

are you familiar with the law of sines

[subtle cargo]

Yes

Sina/a=sinb/b

How would that help tho, we have no angles given to us

[obsidian isle]

simple

just solve this system

[subtle cargo]

Ah I see, thank you

.close

[obtuse pebble]

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[hallow quiver]

Assume you are a family who has $100,000 in a Bank account. You have two housing options: to take a bank loan and buy a property or continue to rent an apartment.

[hallow quiver]

Needed help with part c)

My f rent (t) eqn =

and f buy (t) eqn

[obtuse pebble]

@hallow quiver Has your question been resolved?

[ionic dock]

Hey guys I need a help with this

[hallow quiver]

im aware that I have to take derivative of f_rent (t) - f_buy (t)

but not sure how to solve that

<@&286206848099549185>

[obtuse pebble]

@hallow quiver Has your question been resolved?

[hallow quiver]

<@&286206848099549185>

[limber quartz]

@hallow quiver
f_rent - f_buy is convex if its second derivative is positive on the interval and concave if it is negative on the interval.

yeah, bad times.

this part is easy

differentiation is a linear operator

so the difference of the functions' second derivatives is the same as the second derivative of the whole thing

...not that either is easier or harder

please show your answers for a and b

[hallow quiver]

so if the second derivative is negative it would mean the function f_rent-f_buy is concave?

[limber quartz]

yeah

[hallow quiver]

here thats for a

[limber quartz]

this is in your notes

[hallow quiver]

thats my eqn for b

[limber quartz]

oh, they got red and tiny for some reason

you gotta rewrite that

[hallow quiver]

oh I can write it out again

[limber quartz]

formatting is terrible, not because it is red or small

you messed it up a litte

if you don't want to use latex, use parentheses () and karots ^

[hallow quiver]

f_rent (t) = 100,00 (1+ 2/100) ^t - (19,000 t) for 0 ≤t≤10

f_buy(t) = 500,000 (1+2/100)^t - (65,000 t)

for 0 ≤ t ≤ 10

[limber quartz]

ok, so assuming those are correct, you need to differentiate up to the second derivatives and do a second derivative test on the difference

$\frac{d}{dx} a^x = a^x \ln{(a)}$

[warm shale]

Disorganized

[hallow quiver]

I did something like this

[limber quartz]

where did you get 48,000? should be 46

in fact, you can divide out 1000 from all of this for now

the problem statement says to assume quantities of 1000 in the function

oh hang on a sec

[hallow quiver]

oh sorry yes 46000

[limber quartz]

$f(t) = -400(1.02)^t + 46t \
f'(t) = -400(1.02)^t \ln{(1.02)} + 46$

[warm shale]

Disorganized

[limber quartz]

do second derivative

[hallow quiver]

ok

[limber quartz]

||$f''(t) = -400(\ln{(1.02)})^2 (1.02)^t$||

[warm shale]

Disorganized

[limber quartz]

this thing.... is never 0

however, its behavior isn't a mystery

it's negative. Always.

[hallow quiver]

yep so

this is how I tried solving it