help-10

[scarlet gale]

Try with gf(x).

[dark steeple]

k

[scarlet gale]

@dark steeple What do you get for gf(x)?

[obtuse pebble]

@dark steeple Has your question been resolved?

[obtuse pebble]

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[crisp depot]

$(a + bi)^x = ((1 - \frac{|b|}{b})\pi + arccos(\frac{a}{\sqrt{a^2 + b^2}}) * \frac{|b|}{b})(a^2 + b^2)^{x/2} + (((1 - \frac{|b|}{b})\pi + arccos(\frac{a}{\sqrt{a^2 + b^2}}) * \frac{|b|}{b})(a^2 + b^2)^{x/2})i$

[crisp depot]

Is this accurate for when x is an integer?

And is it accurate for one solution of a non-integer power?

Since taking the nth root of a number creates n solutions

[warm shale]

Bman

[obtuse pebble]

@crisp depot Has your question been resolved?

[crisp depot]

<@&286206848099549185>

[obtuse pebble]

@crisp depot Has your question been resolved?

[obtuse pebble]

@crisp depot Has your question been resolved?

[obtuse pebble]

@crisp depot Has your question been resolved?

[crisp depot]

<@&286206848099549185>

[obtuse musk]

where is this coming from

[hybrid gull]

I have no idea where you got that

It looks quite complicated

[crisp depot]

I came up with it myself

With the help of a teacher,

[untold badge]

There are couple of things that bother me here

for example that only r depends on x

Let me check

[devout prawn]

How did you derive it?

[untold badge]

there is no way this is correct @crisp depot you're missing x somewhere at least

[warm shale]

Anna-Isabella

[obtuse pebble]

@crisp depot Has your question been resolved?

[obtuse pebble]

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[timid silo]

i need to solve this with Gaussian elimination

[royal basin]

and what is giving you trouble in doing that?

[timid silo]

wait a min

[royal basin]

@timid silo ?

[timid silo]

@royal basin

[royal basin]

oh, so you want your work to be verified

let's see

[timid silo]

i feel there's something wrong \

[royal basin]

yeah you screwed up the R1 <- R1 - R2

should have had [1, 5, -1 | 20]

[timid silo]

okay

is it right ?

[royal basin]

up until then you do not appear to have made any arithmetic errors

[timid silo]

is is the best solution or there's other ones better than this

[royal basin]

your solution is okay enough that i wouldn't worry about optimizing it

[timid silo]

idk i watch people always find easy solutions

i feel always mine are complicated and long

and sometimes i start in a wrong way

[royal basin]

you don't go into fractions

[timid silo]

no

[royal basin]

fractions are the most error prone thing in these

[timid silo]

wait i will do it

can you verify it for me

[royal basin]

sure

can you make it so you have one matrix per line?

[timid silo]

i don't understand

what do you mean

[royal basin]

i mean when you write on your paper

make it so that each matrix is on its own line

so that it's easier for me to verify

[timid silo]

ok

@royal basin

[royal basin]

??

you continued from work that i've already told you is erroneous

[timid silo]

no

aah

i'll do it again

should i start ?>

[royal basin]

okay now this seems correct

[timid silo]

@royal basincheck this

[royal basin]

your y is wrong

y is 23/7, not -23/7

[timid silo]

my bad

how can i avoid this errors

[royal basin]

practice

and attention to signs

[timid silo]

i need to start with fundamental

i guess

anyways thank you XDD

.close

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[frail turtle]

i don't understand what they've done to go from the second last step to the last step

[frail turtle]

could someone help me

[obtuse pebble]

@frail turtle Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[echo venture]

Uhmm hello
How do I answer this?

[obtuse pebble]

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[frigid oracle]

Let $S$ and $T$ be non empty subsets in $\bR$ with the following property: $s \le t, \forall s \in S and t \in T$

[warm shale]

dulg

dulg

[frigid oracle]

now im ngl, i have no idea how to do this

[royal basin]

try having one of your sets be a singleton

[frigid oracle]

but if my set is a singlton, then supS = supT implies S = T

[royal basin]

i said try having one of your sets be a singleton

not both.

[frigid oracle]

oh

[royal basin]

if they're both singletons then of course they'll be forced to coincide.

[frigid oracle]

but i still dont understand

let S be a singleton, and supS = supT

then wouldn't $S \cap T$ just be S?

[royal basin]

well if S is a singleton then yes it won't work

but not for the reason you tried to give

[warm shale]

dulg

[royal basin]

the condition (∀s ∈ S, t ∈ T)(s ≤ t) would have to fail

or i guess

well

SOMETHING would have to fail

however

try having T be the singleton instead

let T = {1}

[frigid oracle]

ok

supS = supT, and and so $ S \cap T$ would just yield 1, no?

oh wait

i see what you mean

i am wrong

the reasoning i gave for why having S as singleton is bcs s \le t lol

so it has to be T that is the singleton

let T = {1} and let S = {..., -1, 0, 1}

wait, but S intersection T is still 1

which is non-zero

argghhh

is this question telling me that no such non-zero sets exist given the conditions?

[royal basin]

no it isn't

try S = {1 - 1/n | n in N}

[frigid oracle]

oh wait

i read the question wrong

🤦

[frigid oracle]

oh

i see what you mean

supS = 1, but the intersection between T and S is empty because the set S never actually has 1 as its set, only numbers that approach it

i see

thanks

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[timid silo]

Hello,I am having trouble with factorization and looked up this method on YT but for some reason I keep getting the answer in the wrong sign

[lusty goblet]

Wait, factorization is completing the square, right?

[sage geode]

What I'd definitely do first before completing the square/factoring is dividing both sides by 4 so that the numbers are "nicer" to work this

[lusty goblet]

I'll do that too.

What is the result of the equation after you do that?

(Ping me if you replied to me)

[timid silo]

do middle term factorization

you can do it like this -

4(x^2+10x+10)=-60 => x^2 +10x + 10 = -15 => x^2 + 10x + 25 = 0 => x ^ 2 + 5x + 5x + 25 = 0 => x (x+5) + 5(x+5) = 0 => (x+5)^2 = 0 => x = -5

This is a quadratic equation and there are more than 1 way to solve it

here you can learn about solving them -
if you wanna read -
https://www.mathsisfun.com/algebra/quadratic-equation.html
if you wanna learn through videos-
https://www.youtube.com/watch?v=qeByhTF8WEw

[obtuse pebble]

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[obtuse pebble]

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[crisp depot]

Sorry I see where I went wrong, I didn't write the full formula

[crisp depot]

Sorry about the mistake

This should be the formula

$(a + bi)^x = cos(((1 - \frac{|b|}{b})\pi + arccos(\frac{a}{\sqrt{a^2 + b^2}}) * \frac{|b|}{b}) * x)(a^2 + b^2)^{x/2} + (sin(((1 - \frac{|b|}{b})\pi + arccos(\frac{a}{\sqrt{a^2 + b^2}}) * \frac{|b|}{b}) * x)(a^2 + b^2)^{x/2})i$
Is this accurate for when x is an integer?
And is it accurate for one solution of a non-integer power?
Since taking the nth root of a number creates n solutions

[warm shale]

Bman

[crisp depot]

The rules say to share what I found

And here is a desmos I made of it

https://www.desmos.com/calculator/bbpxuid4gm

I tested it with to the second power

Sorry the mistake before

I am pretty sure this is the right formula

I am going from memory though

<@&286206848099549185>

[obtuse pebble]

@crisp depot Has your question been resolved?

[obtuse pebble]

@crisp depot Has your question been resolved?

[obtuse pebble]

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[crisp depot]

.reopen

[obtuse pebble]

✅

[obtuse pebble]

@crisp depot Has your question been resolved?

[crisp depot]

By noticing and then proving that when you multiply two complex numbers you take the sum of their angles from the positive x-axis and then multiply their magnitudes

[obtuse pebble]

@crisp depot Has your question been resolved?

[twilit knot]

It doesn't seem to work for c=3?

$$(a^2+b^2)^{\frac{x}{2}} \left( \cos{\left(x\theta \right)} + \sin{\left( x\theta \right)}i \right)$$

$$\theta = 2\arctan{\left( \frac{b}{a+\sqrt{a^2+b^2}} \right)}$$

This formula should work

@crisp depot

[warm shale]

Wheeler

Wheeler

[twilit knot]

It might be a good exercise to see if you can derive this

[crisp depot]

Okay I will do more testing

.clos

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[earnest shell]

How can i find dimension of M(n*n) symmetric matrix

[spiral maple]

find a basis for that space

[earnest shell]

how ?

[spiral maple]

That is literally the question

[twilit knot]

Or think in terms of degrees of freedom

How many entries are you allowed to choose arbitrarily

[earnest shell]

I saw formula in youtube n(n+1)/2 is that correct ?

[twilit knot]

Yes

[earnest shell]

I tried to do that but it will take a lot of time

[twilit knot]

Not really

[earnest shell]

thank you

[twilit knot]

If you think about it, whatever you choose for an entry in the upper triangular matrix, the corresponding entry in the lower triangular is determined

[earnest shell]

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[obtuse pebble]

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[earnest shell]

.close

[twilit knot]

And then you're free to choose the diagonals

So you can choose 1+2+...+n in total

[twilit knot]

Nw👍

[earnest shell]

I will try do that and use that formula to check my answer thank you

[obtuse pebble]

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[visual rampart]

Hey I was wondering if someone can help me understand these "ordinary differential equations" of a baseball's flight

[visual rampart]

https://user.eng.umd.edu/~nsw/chbe250/baseball-maxrange.pdf

what does dvxdt and dvydt mean? the change in velocity in x,y vector componets in respect to time?

and then dxdt and dydt would be the change in position?

[obtuse pebble]

@visual rampart Has your question been resolved?

[obtuse pebble]

@visual rampart Has your question been resolved?

[obtuse pebble]

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[plain otter]

Number of ways to arrange 7 blue and 6 green bottles, if exactly two green bottles are together, assume bottles are all alike except for the color

[plain otter]

I filled in the gaps of 7 blue bottles with 5 things, of which 1 was the gg pair, and others were g only, and i got 8C5 *5 i multiplied by 5 because the pair can be rearranged in each 5 times

I got 280 from here

But the answer given is diff

Please advise

[hallow zephyr]

8C5 seems to me that you are mixing the pair with other singular green bottles

At the start, just pick where the pair would fit in

@plain otter

[plain otter]

Yes

[hallow zephyr]

and then you can only fill the remaining slot with the other 4 green bottles

[plain otter]

So 8C1 7C4

Right?

[hallow zephyr]

sounds about right to me

[plain otter]

I got 70 from here

But the answer given is still diff

And i don't know why my method would be incorrect?

[hallow zephyr]

how'd you get 70 from 8C1 * 7C4

[plain otter]

Wait sorry

280

[plain otter]

Its still the same answer i got previously tho @hallow zephyr

[hallow zephyr]

yeaa I just noticed

hmm

[plain otter]

The answer given is 1080

It makes no sense at all

[hallow zephyr]

oh

7 blue bottles?

[plain otter]

Yes

[plain otter]

Yeah?

[hallow zephyr]

yea sorry i got no clue either
it seemed like 280 for me

[plain otter]

Oh ok, thanks anyways

[obtuse pebble]

@plain otter Has your question been resolved?

[plain otter]

.close

[obtuse pebble]

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[frigid oracle]

alright, im back again

[frigid oracle]

but i have no idea how to do this proof by contradiction

Let us define $S = \left{s: s \in \bR \right}$ and conversely, $-S = \left{-s: s \in S \right}$. There exists $t \in \bR$ such that $t \le s$ $\forall s \in S$ implies $t \le -sup(-S)$

[warm shale]

dulg

[frigid oracle]

My proof goes like this: let us assume that the negation is True. Specifically, assume $t > -sup(-S)$ if $t \le s$

Then, we can rewrite the statement as $-sup(-S) < t \le s$, which contradicts the result $-sup(-S) \le s$ because it imples $-sup(-S) < s$

but i am not sure if this is actually a good proof by contradiction, since i only referred to a previously proven result, but did not derive any contradiction with our contradiction hypothesis

[warm shale]

dulg

[frigid oracle]

any tips?

[royal basin]

$S = {s : s \in \bR}$? so $S = \bR$?

[warm shale]

Ann

[royal basin]

are you sure you did not mis-copy anything?

@frigid oracle your statement is false as written

[frigid oracle]

ah yeah

alright, S is a non-empty set of R that is bounded above, and therefore has a supremum

[frigid oracle]

bcs of the S part?

[royal basin]

so you want to write "S is a nonempty, bounded-above subset of R"

[frigid oracle]

which part do i need to rewrite?

[royal basin]

and yet you tried to give S a set-builder definition

and failed outright

[frigid oracle]

lol

[royal basin]

because the definition you wrote says "S = R"

[frigid oracle]

yeah i agree

[royal basin]

so S is a nonempty bounded-above subset of R

[frigid oracle]

i should've just stated the nonempty, bounded part

yeah

[royal basin]

and how exactly do we know sup(-S) exists?

(or depending on your prof, replace "exists" with "does not equal +∞")

[frigid oracle]

wait

lemme rewrite all this

[royal basin]

maybe we could skip all the bullshit and you could SCREENSHOT the question

[frigid oracle]

lol

i figured it'd be a good opportunity to practice my latex

:(

ok

my exercise asks me to prove equations 1 and 2

[royal basin]

so S is in fact bounded below, and not necessarily bounded above.

[frigid oracle]

yeah

i got mixed up with the axiom of completeness right before it

[royal basin]

do i take it that you have proved equation (1)?

[frigid oracle]

yes

but i am rather uncertaion of its proof

[royal basin]

it's literally equivalent to ``$s_0 \geq -s$ for all $-s \in -S$''

[warm shale]

Ann

[royal basin]

very little if anything to be uncertain of

[frigid oracle]

well, i tried to use proof by contradiction

and arrived at "the contradiction hypothesis contradicts the definition of sup(-S)"

alright, that assures me a bit more

but i do not know how to do (2)

i tried proof by contradiction, but i am not sure of the result

[royal basin]

t ≤ s for all s ∈ S

multiply that by -1

-t ≥ -s for all s ∈ S

[frigid oracle]

oh

[royal basin]

or equivalently, -t is greater than or equal to every element of -S

[frigid oracle]

mhm

[royal basin]

thereofre -t ≥ sup(-S)

that's it

no contradiction required

[frigid oracle]

what is this technique called?

[royal basin]

nothing

it doesn't have a name

[frigid oracle]

you just show it straight up?

alr, thanks @royal basin

much appreciate it

.clos

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[timid silo]

I need to show that n>= 1

[timid silo]

can I get an answer and an explanation?

[sage geode]

Have you tried induction?

[timid silo]

I’m not that familiar with induction

I was told to solve it with induction but I don’t really understand it

[west sierra]

Through induction you show that for n=1 it is true, then for n=k it is true and with help of equation you got from n=k you show it is true for n=k+1. If you can show all of this then by principle of induction you can tell it is true for all natural nos n

I might be wrong since I didn't study induction for long time

[timid silo]

I do not understand, you want me to solve n how?

[sage geode]

We aren't solving for n

We need to show that the equation is true for any n (in the natural numbers)

[west sierra]

Thou here it's asking only if it's true for n>=1

[timid silo]

I was meant to use sigma as well

[spiral maple]

yeah, just prove w/ induction

[sage geode]

What the method of induction states, briefly, is that

To prove that an equation is true for any natural number, you need to first show that the equation is true for the base case (most of the times it's n = 1, sometimes it can be n = 0, in this problem I guess going with n = 1 is fine). Then you need to show that if the statement is true for some natural number k, then it is true for k + 1 as well.

[timid silo]

this works for this problem right?

[sage geode]

Yes

Btw is this the first induction problem/exercise that you're encountering?

[timid silo]

no we have done induction before but I don’t really understand it

https://www.quora.com/How-do-I-prove-1-2-3-2-5-2-2n-1-2-n-4n-2-1-3-by-mathematical-induction
Is this correct?

looks like the method we did in class

[sage geode]

Yes, that is correct

[obtuse pebble]

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[weary flint]

Can someone explain to me how they split up 2s+3/(s+1)^2 ?

[wide drift]

they just factorized it

[unique solstice]

Looks like they made it 2s+2+1

[wide drift]

(s+1)^2=s^2+2s+1

[unique solstice]

Then split it into 2s+2 and 1

first goes to 2(s+1) and cancels out

[weary flint]

Aha I see

[wide drift]

oh i thought u talking abt different step
my bad

xd

[weary flint]

Okay I think I get it then, thank you for your help

.close

[obtuse pebble]

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[timid silo]

Is the expression "the sum of the square of x and the square of the opposite of x" equal to 0 or 2x^2?

[crisp depot]

If what you mean by opposite is negative then the answer is 2x^2

$x^2 + (-1 * x)^2$ = $x^2 + -1^2 * x^2$

[warm shale]

Bman

[crisp depot]

Since $-1^2$ is equal to 1

[warm shale]

Bman

[spiral maple]

-1^2 isnt 1

(-1)^2 is 1

[crisp depot]

Yes sorry,

Since $(-1)^2 = 1$

[warm shale]

Bman

[crisp depot]

Then $x^2 + 1 * x^2$ = $x^2 + x^2$ or $2x^2$

[warm shale]

Bman

[alpine raven]

the opposite of x is -x
so it's x² + (-x)² which is 2x²

[crisp depot]

Therefore it is equal to $2x^2$

[warm shale]

Bman

[crisp depot]

@timid silo

[timid silo]

oh

[alpine raven]

if it was the opposite of the square of x, then it would have been 0

[timid silo]

im stupid

i thought opposite meant reciprocal

[crisp depot]

Everyone makes mistake

[crisp depot]

I didn't know what you meant by opposite

[alpine raven]

opposite is the symmetric of a number in the group (G, +), if im not saying shit

[crisp depot]

That would be $x^2 + x^{-2}$

[warm shale]

Bman

[timid silo]

yeah, makes sense

i was just overthinking

.close

[obtuse pebble]

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[rocky canyon]

I'm pretty lost on this question..

[rocky canyon]

Im trying to make it into a first order differential equation

[hot sonnet]

Hello, find how much salt enters the the tank in lb/min, and how much salt is drained in lb/min too

[rocky canyon]

where does the function come in tho?

[hot sonnet]

then you can get a ODE in the form
dm/dt = (rate in) - (rate out), where m is the mass of salt

[hot sonnet]

what do you mean? 🤔

[rocky canyon]

like m(t)

since I thought that's usually how these go

[storm lynx]

So you have y(t) being the amount of salt and t being time in minutes.
Based on the givens you can write out the way y changes over time.
y'=(rate in) - (rate out)
You can now solve to get y.
Now you can simply find the value of y at t=10

[hot sonnet]

Try finding the rate in and the rate out, then you will see or you can ask here

[rocky canyon]

wait amount of salt like what would be in lb/gal?

[hot sonnet]

no, just lb

[rocky canyon]

for the total rate I just got 9.8 lb/min

[hot sonnet]

mmm what do you get for rate in?

[rocky canyon]

10 lb/min

[hot sonnet]

ok, what about rate out? Remember that you need the concentration of salt (lb/gal) of the mix that is being drained so you can get the rate out

[rocky canyon]

OH WAIT ok ok

so y(0) = 4

which is why I need to use y(t) for the rate out?

to find*

y(t) x 5?

or I mean

y(t)/100 x 5?

[hot sonnet]

yes 👍🏽

[rocky canyon]

Okok Ill try that rn

the amount I got was unreasonably high, Im not really sure where I went wrong

it may have been while evaluating the differential equation

[hot sonnet]

mmm C is not 196

[rocky canyon]

but when I do ye^t/20=200e^t/20 + C at t=0, I get 196

even though it shouldnt be

oh wait a sec

I cant do that method

omg

since there's no t on the other side of the equation

[hot sonnet]

look carefully 👀

[rocky canyon]

.

AHahaf

[dim root]

i love diff equations great. the general rule is that linear changes in salt will be a constant if ds/dt

exponential changes will be x*a/b if equation =ds/dt

because when we integrate them, we will get a linear equation on the constant

and an exponential one on the other thing where x*a

[rocky canyon]

hmmm interesting

[dim root]

so, at t=o the salt is 4lbs, thats our starting value for 100 gal of water. we get in water where there is 2 lbs/gallon at a rate of 5 gal/min, but the volume stays the same as we are losing 5 gal a minute

[rocky canyon]

ahhh

[dim root]

let me just have a think of how you set these up again 😅 one sec

[rocky canyon]

alr np

[dim root]

@rocky canyon alright here

just had to make sure it was correct

so we get that ds/dt which is the difference in salt (ds)

per thethe difference in time (dt)

'is equal to'

take the linear forms first

at every moment, we constantly get 2lbs/100gal of salt

plus an additional 5 gal water

i hope that is correct

[rocky canyon]

wait why 2b/100gal?

lb*

[dim root]

oh thats the amount of salt

so our salt concentration is lbs/100gal water

its just something i decided

[rocky canyon]

ohhh

[dim root]

its easy to work with thats all

[rocky canyon]

alr ok I thought u were using my problem as an example, sorry

[dim root]

i am

lets go through it actually

so at t=0 the salt is 4 lbs salt in 100 gal of water

so our salt concentration is 4lbs/100gal of salt at t=0

[rocky canyon]

yup

[dim root]

lets leave that alone for now though, that will be made later

at every moment of time

we get 2 lbs of salt in one gallon

[rocky canyon]

mhm

[dim root]

sorry i read it wrong, at every minute we get 5 gallons of 2 lbs salt pr gallon

[rocky canyon]

np

[dim root]

i hate text questions as im dyslextic af

[rocky canyon]

i get that, esp with the big block of useless text

[dim root]

but we also lose 5 gallons of water every minute too

ok great, so lets just say for ever t one minute passes

lets take the linear forms first

for every minute we get 2 * 5lbs of salt

and 1* 5 gallons of water

no clue what an lbs or a gallon is but lets keep going 😄

[rocky canyon]

lolll

[dim root]

we also lose 5 gallons a minute

so our expression will look like this

ds/dt = 2*5+5*1-5*1-the tricky part

i wont bother with latex but you see whats there

[rocky canyon]

yeah

[dim root]

we get the amount of salt which is 2*5

and 5-5 is just 0 so we can cancel that out

ds/dt=2*5-tricky part

the only thing thats not linear in the expression is the tricky part

for every moment t we lose 5 gallons of a concentration of some amount salt

so we will lose s*1

thats one part of salt

the current concentration of salt

it changes with the current concentration of salt

[rocky canyon]

makes sense

[dim root]

i assume you know the basic rules of integration

[rocky canyon]

indeed

[dim root]

integration of linear diff equations i mean

so i will skip that part

we get that s(t)=c*e^(-t)+2

the constant will be calibrated to the first bit of info we got

we know that t(0)=4lbs/100 gallon

since we are dealing with amount of salt in the question not concentration we can just set 4=ce^0+2

and there we have it! c=2

[rocky canyon]

wow!

much simpler than what I did tbh

[dim root]

wait i fucked up somewhere

one sec heh

@rocky canyon sorry, the equation is obviously ds/dt=2/100-s/100

[rocky canyon]

oh yeah

[dim root]

so we get

yeah sorry about that..

[rocky canyon]

don't worry about it

[dim root]

so at t=0 we get 4/100=ce^(-t/100)+2

and we solve like that

you may graph it in wolfram or desmos

notice the consentration never quite reaches 2 lbs pr gallon

but it gets damn close, its a horizontal asymptote

[rocky canyon]

isn't s(t) just 4 and not 4/100?

[dim root]

its late...

[rocky canyon]

lol it's ok you helped enough

I appreciate your time

[dim root]

Thanks a bunch and good luck with diff equations

[rocky canyon]

thanks 🙂

[dim root]

its imo the only useful math we do

[rocky canyon]

wow that's actually really motivating

to do well on this unit at least

[dim root]

yeah! did a lot of this in chemistry

titration and that sort

[rocky canyon]

ohhh

.close

[obtuse pebble]

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[restive torrent]

Im pretty sure (x-2)^2 turns into (x-2)(x-2), and there is (x+2) left so (x-2)(x-2)(x+2) = o

all of this: (x-2)(x-2)(x+2)

you factor it out

[upbeat island]

what is a "two-term expansion"?

like taylor series?

[restive torrent]

it means a Binomial Expansion

I think its asking you for 2 Binomials

i am not really sure

ok, tell me the solution and ill try to reverse solve it

What math is this?!

that's a math

[upbeat island]

i've taken many years of calculus and this is new to me lmao

[restive torrent]

?

Lol, im in math 2. Im just a freshman in hs. I just love to help people. you know....

i just joined as a helper

oh ok

is this like pre calc or calc?

oh ok

<@&286206848099549185> miles needs help with this question. He knows the answer but doesn't know how he got it. Has been waiting for 20 mins. I'm a helper but I'm not sure!

@silent flame someone will help you in a Lil bit. sr for not helping.

[obtuse pebble]

@silent flame Has your question been resolved?

[obtuse pebble]

@silent flame Has your question been resolved?

[obtuse pebble]

@silent flame Has your question been resolved?

[warm shale]

Miles

[royal basin]

guess $y_p = (Ax^2 + Bx + C)\sin(x)$ by the looks of it? you have order 1 resonance going on

[warm shale]

Ann

[royal basin]

should work

A, B and C will be in terms of λ obviously

@silent flame

[obtuse pebble]

@silent flame Has your question been resolved?

[obtuse pebble]

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[sturdy rune]

Hello I have to solve this but we dont get the answers..
i have to say whether its correct or not correct and I was hoping someone could check if I'm correct

I tried both with a example, I think both are correct?

1. Arcsin(sin(-pi/2)) = Arcsin (-1) = -pi/2
2. This one is not correct because sin(arcsin(-1)) = sin (-pi/2) = -1

[sage geode]

x has to be inbetween 1 and -1 sothat arcsin(x) exists

[sturdy rune]

so they both wrong then?

[sage geode]

No, just the 2nd one

Just write that -1 <= x <= 1

[sturdy rune]

the first sin(x) always ends up between -1 and 1

[sage geode]

Yes

[sturdy rune]

But is it then equaal to x?

wouldnt it be x + kpi in some cases?

[sage geode]

The 1st one is correct

[sage geode]

You'd dothat if you were solving an equation

[sturdy rune]

but it is an equation right?

[sage geode]

With an unknown

[sturdy rune]

mhm alright

and if x is [-1, 1] in the 2nd, then sin(arcsinx)) will always be = x?

[sage geode]

Yes

Otherwise arcsinx wouldn't exist in the first place

[sturdy rune]

Allright okay

and imagine

arcsin²(sinx) , would that = x² then?
and sin²(arcsin²x) = x²

Given the limitations of x for arcsinx?

[sage geode]

By arcsin^2(x) you mean arcsin(arcsinx) or (arcsinx)^2?

[sturdy rune]

(arcsin(x))²

[sage geode]

The first one would be true

Not sure about the other one

[sturdy rune]

allrigt thanks

.close

[obtuse pebble]

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[timid silo]

@timid silo could u help again

The other channel is closed

And why would be use t=0 and s=1.5km if it moves away from it?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

at 0th second the parachute is deployed which is 1.5km, second sentence

[timid silo]

Is it travelling away from the launch site or towards it?

away

Ahhhh

Then how would I find b

I tried using the velocity

After 25s V=0

i was thinking of using 800m/s as v then 800 = 2ct + b

[timid silo]

What would t be?

[obtuse pebble]

@timid silo Has your question been resolved?

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[sand pilot]

My school book says the following: let d1 and d2 be parallel lines, and let P a plane containing d1 and P' a plane containing d2, both P and P' intersect at a line called Q, therefore Q is parallel to d1 and d2.
I not really sure if that statement is true because the line that intersects with both planes could be not parallel to any of d1 or d2 making this statement false

[shut sundial]

I'll call Q lowercase for consistency. Let's assume that q isn't parallel to d1 (we want to lead this to a contradiction). Then, since q and d1 are in the same plane P, q intersects d1 in a point X. Now X is on q, so X is on P'.

Now P' is a plane containing both X and d2, so since d1 and d2 are parallel and d1 contains X, P' also contains all of d1. But this means that d1 is a line which is also in the intersection of P and P', and d1 is not the same line as q. Therefore P and P' intersect in more than just one line. This contradicts the original problem statement (that P and P' intersect in just one line), hence the last assumption we made (that q isn't parallel to d1) is false. We conclude that q is parallel to d1 after all.

The same argument but with d1 and d2 swapped also shows that q is parallel to d2.

[obtuse pebble]

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[frigid oracle]

Show that $s \le t$ if and only if $s* \subseteq t*$

[warm shale]

dulg

[frigid oracle]

I have little experience with these types of proofs, so what do i do?

my idea is that i show that $s* \subseteq t*$ implies $\left{r \in \bQ: r <s \right} \subseteq \left{r \in \bQ: r < t \right}$

[warm shale]

dulg

[frigid oracle]

which shows that sup(S) <= sup(T) since both of their infimums are -\infty

which means s <= t

is my idea correct?

[slim leaf]

what is s*?

[frigid oracle]

yeah sorry, i was a bit all over the place with the notation for the sets

s* or S = {r \in Q; r < s} and t* or T = {r \in Q: r < t}

s* was the notation my book used for the dedekind cut were s \in R

[slim leaf]

which way are u trying to prove?

[frigid oracle]

how the "if" statement leads to the "then" statement

that is, if s* \subseteq t*, then s \le t

[slim leaf]

Ok, we have that S is a subset of T. For any r < s, we also have r < t.

You know s is the supremum + t is some upper bound.

you should be able to conclude the proof.

If you can't understand the first statement, let me clarify. Since S is a subset of T, any element in S is also an element of T.

[frigid oracle]

ah ok

i see

i can say that t is some upper bound of the set S, and all upper bounds are greater than or equal to the least upper bound

and this is if and only if S \subseteq T

[slim leaf]

yeah

[frigid oracle]

is my idea correct?

haha, thanks

.close

[slim leaf]

np

[obtuse pebble]

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[haughty totem]

sup y'all!

quick question, is my graph already correct?

thank you in advance ✨

[high lily]

no

[haughty totem]

please elaborate

[high lily]

your graph isn't correct

[haughty totem]

why?

[high lily]

do you know what a parent function is?
what do you think you graphed and why did you choose to graph that function

[haughty totem]

i just inserted that equation on a Mathway calculator

that is why i'm asking if the calc's answer is already correct or not

[high lily]

ew, mathway, but what exactly are you putting into mathway
and what exactly is it pumping out

(and also you shouldn't be using mathway or any calc for something like this)

[mental cloak]

Shouldn t this just be a straight line which is parallel to the x-axis? since the function hits always c for every value of x?

[haughty totem]

oh, is it just that?

oh well

[topaz bridge]

for some reason if you ask mathway to graph f(x)=c, it instead graphs f(c)=c

[obtuse pebble]

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[obtuse pebble]

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[wintry edge]

hi! need help with the proof for this because i am really stumped

[wintry edge]

is this true? and if it is, how do i prove it? really sorry

[obtuse pebble]

@wintry edge Has your question been resolved?

[wintry edge]

<@&286206848099549185>

[timid silo]

This is rather trivial.

[wintry edge]

How so? Sorry

also thank you so much for replying @timid silo

[timid silo]

No no why say thank you haha

[wintry edge]

Sorry haha

if i may ask, why is the statement true? i can't seem to understand it sorry

[timid silo]

Was figuring out something, but was false.

[spiral maple]

Not true

[restive acorn]

Consider

f(x) = -x

f'(x) = -1

f''(x) = 0

[wintry edge]

oh right, thank you so much! i was a little confused because the exercise asked us to prove it

thank you!

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[cunning zealot]

Can someone run me through the stages of this? Wildly stuck.

[spiral maple]

apply the substitution

[cunning zealot]

(e^3x)/u

[spiral maple]

apply the substitution fully

[cunning zealot]

where else am I going

[spiral maple]

need to make e^(3x) into a function of u.

[cunning zealot]

would it be (u-1)^3?

[spiral maple]

oh wait, let me rephrase

$e^{3x}=e^x\cdot e^{2x}$

[warm shale]

Mosh

[spiral maple]

so you need to write e^(2x) as a function of u

[cunning zealot]

is it just (u-1)^2 or am I doing something wrong

[spiral maple]

yep, it's exactly (u-1)^2

so $\int\frac{e^{3x}}{1+e^x}\dd{x}=\int\frac{e^{2x}}{1+e^x}e^x\dd{x}=\int\frac{(u-1)^2}{u}\dd{u}$

[warm shale]

Mosh

[cunning zealot]

how did the e^x and the dx at the end turn into du?

[timid silo]

u substitution

with u = 1 + e^x

[spiral maple]

$u=1+e^x\implies \dd{u}=e^x\dd{x}$

[warm shale]

Mosh

[cunning zealot]

ah okay thank you I think I know where I'm going now

.close

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[sterile spire]

Pretty sure this is correct just a bit rusty at math so I’m a lil confused why when we find f(x)>0 we take the negative interval and the other way around for f(x)<0

[obtuse pebble]

@sterile spire Has your question been resolved?

[sterile spire]

<@&286206848099549185>

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[wanton lark]

hello

[wanton lark]

i have a question if you have a right angle triangle

with these dimensions

and you rotate it around the shorter leg (doesn’t say by how much)

and they are asking what geometric shape is created and what are the dimensions of said shape

i thought it could be isosceles triangle or a circle

but i am not sure

[simple siren]

By shorter leg you mean side with length 3?

[wanton lark]

yes

i thought it could be something like this

[simple siren]

I suppose that's how it is

[wanton lark]

but which one

the triangle or circle

[simple siren]

See the triangle is formed if you kinda flip

If you rotate it it won't even come to that point

[wanton lark]

yes i get that

[simple siren]

And it would still be a circle with radius 4

[simple siren]

Yeah

[wanton lark]

i get that if you rotate it it would look like this but tbh idk anymore

[simple siren]

Well rotate it 1 degree each time and mark the outer most point

You could see it makes a circle

[uncut robin]

I vote circle too

[simple siren]

But that's just if the thing keeps rotating or is meant to be duplicated in every rotation

[uncut robin]

just because it doesn't tell you by how much you would need to rotate it

[simple siren]

Also, I'm confused is it the line that changes or the whole triangle on that point?

[wanton lark]

kk I trust yall and the radius will be a Pythagorean theorem

[simple siren]

No ig

It would be it if it's moving from that axis

If it's moving from the right angle side it's just the longer length

Just try to visualize it in your mind it's easier

[uncut robin]

the question is just not clear

[simple siren]

Yep that's where the hypotenuse is radius

[simple siren]

Just needs labeling on points

[uncut robin]

yeah

[simple siren]

I just would like to ask was I good? Like in explaining it?

[wanton lark]

So around this point?

i guess thats it

well thanks

how do i close this?

[mild ocean]

it’s .close

[wanton lark]

kk

[wanton lark]

Yeah you did great

[obtuse pebble]

@wanton lark Has your question been resolved?

[obtuse pebble]

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[mild ocean]

it could possibly help to rewrite it as 4*(1000a+100b+10c+d) = 1000d+100c+10b+a

[wanton dagger]

Ok

[mild ocean]

so 4000a+400b+40c+4d = 1000d+100c+10b+a

[uncut robin]

Ok

[mild ocean]

maybe that isn’t the best idea actually

because you would get conflicting results

[wanton dagger]

Oh

Ok

[mild ocean]

stuff like 4a = d and 4d = a

[wanton dagger]

So what should I do

[mild ocean]

not sure, that was my only idea

[wanton dagger]

Ok

Should I ask a helper?

[wanton dagger]

Yes

[astral ivy]

Hm

[mild ocean]

actually i think i might have been on the right track

[wanton dagger]

Oh

[astral ivy]

Yeah I think you were

But the problem is that you can’t do 4a = d because 4a might be bigger than 9

So it’s not really a digit

And then you’d have to take it mod 10 and it would get very messy

There might be an easier solution

[uncut robin]

a can't be bigger than 2

otherwise the number on the right side will have 5 digits

[astral ivy]

Yeah since 2500*4 = 10,000 we know abcd < 2500

[mild ocean]

4000a+400b+40c+4d = 1000d+100c+10b+a, so 3999a+390b+40c+4d = 1000d+100c, so 3999a+390b = 996d+60c

[astral ivy]

So a = 1 or 2 (can’t be 0 because then not 4 digit number)

[mild ocean]

hmm

i actually don’t have any other ideas other than just guessing and checking

[uncut robin]

I mean A has to be 2 because it's a multiple of 4

so just work with that

if a is 2 then d is 8

[astral ivy]

Smart

Wait couldn’t d be 3 @uncut robin

[uncut robin]

so substitute a=2 and d=8 here

[uncut robin]

no

[astral ivy]

Why

[uncut robin]

because a would be d/3

but that would make A a fraction

[astral ivy]

Why

[uncut robin]

wdym why, the digits can't be fractions

[astral ivy]

Like

I was thinking

4*d = a = 2 (mod 10)

So d could be 3

[uncut robin]

idk how that works

[astral ivy]

No like why would a = d/3

Where’d the 3 come from

[uncut robin]

from nowhere

that's why it can't be three

[astral ivy]

What lol

[uncut robin]

4a = d
you said d can be 3
4a = 3
a = 3/4

not an integer

A is either 2 or 2

[astral ivy]

What if b >= 3? Then multiplying it by 4 would carry over to a and mess it up right

[uncut robin]

yeah

they are just digits

can't be 12

because that would be two digits

nothing can be carried over

[astral ivy]

Let’s say abcd = 2311. Then 4*abcd = 9244, and the first digit is not 4*a = 8

[uncut robin]

no

the number could be

2xx8

you would have to isolate b and c from the equation quantum sent earlier

[uncut robin]

give me a sec

yeah the number is just 2178

[astral ivy]

Nah I’m just saying that a=2 doesn’t imply d is only 8

Jeez dude lol

[uncut robin]

it does

really

[wanton dagger]

Any solutions?

[astral ivy]

But why exactly

[astral ivy]

2178 is abcd

[uncut robin]

because

[wanton dagger]

Okay let me check

[uncut robin]

idk if I'm able to explain

it just is like this lol

[astral ivy]

This might sound dumb or pretentious

But it’s a good exercise to force yourself to explain

Also, I just don’t understand it too lol

[uncut robin]

ok so you have to numbers

abcd and dcba

you know that 4(abcd) = dcba

[wanton dagger]

Yes

[uncut robin]

and both numbers have to be four digits

and each digit has to be a digit aka a number less than 10 and bigger than 0

[wanton dagger]

Yup

[uncut robin]

now, if we said A was 3

[astral ivy]

Well a and d can’t be 0, but b and c can right

[uncut robin]

they could yes

but not in the solution

so, anyway, if A was 3, you had that 4A=D

[wanton dagger]

Ok

[uncut robin]

where a and d are the fourth digit

so if A was 3 then D would be 12

[wanton dagger]

Yes

[uncut robin]

meaning that the other number would be 5 digits long

[astral ivy]

Ok is this the explanation for why a = 2:

abcd *4 must be a four digit number, so it must be less than 2500. Thus a < 3. And since a can’t be 0 and it must be even (since dcba is a multiple of 4), then a can only equal 2.

[wanton dagger]

Ok

[uncut robin]

yes

exactly

[astral ivy]

Cool

So now why does that necessarily imply that d = 8

[uncut robin]

because the fourth digit of the second number has to be 4 times the fourth digit of the first number

where abcd is the first number

and dcba the second

[astral ivy]

Does it? I just showed that if abcd = 2311, then abcd*4 = 9244. The first digit, 9, is not four times 2

[wanton dagger]

So how did you get the answer I’m still confused lol

[astral ivy]

Because thunder7 is cracked

[wanton dagger]

Lol

.close

[obtuse pebble]

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

Any one can solve this problem hah

[proper glade]

2th

[astral ivy]

twoth

tooth

[astral ivy]

Do you know how to find the volume of a box?

[proper glade]

lwh

[astral ivy]

Yeah

This metal piece is made of two boxes — if you can find the volume of each box, you have the volume of the whole thing, right?

[proper glade]

ok ok

2225+83

but i dont know the bottom

fe value

[astral ivy]

Well if you cut the metal from C across horizontally, you have your two boxes

The first box has volume 2*5*22

What about the second box? (Hint, remember that it extends under the first box. You have to account for that.)

@proper glade

[proper glade]

832+8 ah i got it

[astral ivy]

(8+2)*3*22

[proper glade]

we should add 2+8 aha for ok i got it thanks

[astral ivy]

Ok what’s ur final answer

Should be 880

[proper glade]

880 cm**3

your is wrong

[astral ivy]

Yeah cm^3

[proper glade]

you missed cm**3

ok bye

[astral ivy]

Cya

[obtuse pebble]

@proper glade Has your question been resolved?

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[lucid flume]

Can anyone tell is this method right?

[hushed moat]

Interesting method

[sage geode]

Seems to be correct

I'd still check the answer by plugging in just in case if one of the coefficients must have a certain value

[hushed moat]

I think you have to put y back into the original equation

[hot sonnet]

I think it's correct too

[hushed moat]

Since you're only meant to have 2 constants

[hot sonnet]

Just don't forget to plug in y in the original equation so you can solve for one of the constants k

It's a second order differential equation, so you should get only 2 constants of integrations as unknowns

[lucid flume]

Now how will I go back from it?

$$ y'' -2y' + 1 = 0 $$

[warm shale]

Opshorah

[hot sonnet]

You have to get y'' and y' in terms of k1 k2 and k3, not k

For example

$y''=4k_1e^{2x}$

[warm shale]

leonardogtf

[hot sonnet]

Do the same for y', you should get k2 = 1 when you plug in y'' and y' in the original equation

[mild ocean]

well

[mild ocean]

this is certainly a unique method

darn i don’t remember if i pinged you or not, sorry if i did since i’m not helping

[supple quest]

lol, u good?

$$ k_2 = \frac{1}{2} $$

[warm shale]

👑Overlord Prince Khan😈

[hot sonnet]

Well I might be wrong, thanks for giving the right answer

Anyway, the important thing is that you can get a value for k2

[supple quest]

yeah

[hot sonnet]

Yes, k2 = 1/2, I hadn't given it a try yet, just tried to eyeball the answer but I forgot the 2 😅

[supple quest]

BTW, @lucid flume your method is somewhat unorthodox but indeed not wrong

[lucid flume]

uh thanks 😅

[supple quest]

But in order to make your method consistent and yield the actual answer you have to find the value of k2

then it will be perfect

[lucid flume]

oh ok

thanks for the help

Can you help me out in my homework? @supple quest

[supple quest]

Well, I can only help you. You have to do it on your own

Close it if you're done

[lucid flume]

thanks 💓 🥰

.close

[obtuse pebble]

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[cedar bluff]

In a linear transformation, do we map coordinates (representation of a vector with respect to a certain basis) or do we map vectors?

[junior inlet]

uhhhhhh
i think you can view linear transformations as mapping the basis vectors

so all the other vectors (represented by a linear combination of those basis vectors) will change as well

[cedar bluff]

In this question, if I want to map 1 + x + x^2 + x^3 to an output, am I mapping (1, x, x^2, x^3)^T to an output or am I mapping (1, 1, 1, 1)^T?

[junior inlet]

i think that's not a vector

1, x, x^2 and x^3 are bases

[cedar bluff]

yes but I have to find the matrix of linear transformation.

[junior inlet]

yeah?

just find out what those vectors should map to

and bingo

you got the rowsno, that depends of your matrix

[cedar bluff]

My answer is actually $$A = \begin{pmatrix} 1 & 3 & 4 & 8 \ 0 & 0 & 8 & 0 \ 0 & 0 & 0 & 21 \end{pmatrix}_{3\times 4}$$

[warm shale]

Aarjav

[cedar bluff]

And answer according to one of my teacher in college is

$$A = \begin{pmatrix} 1 & 0 & 0 & 0 \ 3 & 0 & 0 & 0 \ 4 & 8 & 0 & 0\ 8 & 0 & 21 & 0 \end{pmatrix}_{4\times 4}$$

[warm shale]

Aarjav

[royal basin]

it should be a 3 by 4 matrix

[cedar bluff]

That is exactly what I think. But when I mentioned this today to that teacher, she said that it is a way of representing.

[royal basin]

"it is a way of representing"? were those her exact words?

[cedar bluff]

ya kind of

[royal basin]

...