#help-0
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sills
i understand that there has to be more to it, right?
because i was talking to someone else and they had shown me this
I've only gotten to applications of derivatives, haven't even done integrals yet đż
Good luck
do uk what an integral is ?
yes
what is it ?
area under a graph
ya
definite integral will give in b/w the specified co-ordinates
indefinite integral will be the general area formula ( it is like an expression )
not like that wait let me share some pic for better understanding
ok
see this green is definite integral b/w point a and b
indefinite integral is the entire area's expression
understood ?
yes
cool
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I am just not sure what it is asking or how to do it
I had a problem on my review exactly like this
Did you understand what it was asking and how to do it?
We are just learning this stuff and I am not understanding ngl
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help
is this in a test?
tri c
I believe it is saying that you reflect the triangle over the line x=-1
(which is not what I had in that previous image)
hm
so x=-1 is the new mirror?
only for triangle c
i know that
how are you able to draw that quickly
Triangle C is just defined as Triangle B but reflected over the line x=-1
ah
too many years of using illustrator
it's worded strangely, but that's what it means by "reflected in"
how about triangle d
I'm trying to remember if in matrix form y is above or below x
the matrix
[-3
-4]
could either be (-3,-4) or (-4,-3) depending on whichever it is
-3 is for x
-4 is for y
ty
or at least that's usually howit works
matricies are weird, but I just checked and yup, thankfully they're normal here
so x for the top?
so you're translating triangle a by (-3,-4)
yup
so it should look something like this
so move triangle a from1,1 to -3,-4?
a translation of vector (-3,-4) would mean moving the point (1,1) 3 across and 4 down
basically
oh
you're taking the original point (1,1) and adding (-3,-4) to it
(1+-3,1+-4) -> (-2,-3)
That's how adding vectors work
You add the components together
makes sense
ok
okay so say you have a grid
when scaling in relation to zero, you are pretending that shape's corner is centered at zero
90 degree angle?
so for example
we're pretending that this triangle is centered at 0,0
even though it's not
how would this apply to a square?
well, the square is a tool for us to think about the scaling
because the triangle is "centered" at 0 now
we have to pretend make sure it scales properly
here's what it looks like scaled
The square is half the size it was
but if we're looking at the triangle, we'll see it's also moved
Does that make sense?
ah
I'm quite certain this is what it's asking for
so, you basically scale it but it also shift in the direction of the center
because of that invisible box
if it was a scale factor of 2, it would get further away
It rings a bell, one moment
ok
and no, I can't solve it XD
did you know that you can use negative numbers?
if you do there are 3 loops
The Collatz Conjecture is the simplest math problem no one can solve â it is easy enough for almost anyone to understand but notoriously difficult to solve. This video is sponsored by Brilliant. The first 200 people to sign up via https://brilliant.org/veritasium get 20% off a yearly subscription.
Special thanks to Prof. Alex Kontorovich for in...
found out through this guy
Fantastic channel
yea
engaging format too
do you think it will be like negative numbers?
like at first it wasn't accepted and now it's basic knowledge?
No idea
Sorry kinda busy
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i got this wrong can someone help
What did you use for U sub?
yea i used lnx for u
should i try this, i have only 2 attempts so i can only have 1 more attempt
or should i just do the decimal
no idea what the program wants
i would assume an exact answer, which you already had
so sure, go for it
dang i got it wrong
thats so weird hmm ill email my teacher about it
thanks tho guys
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Please help with negative polar coordinates
@cunning raven Has your question been resolved?
hello?
wait I have another question
Where did I do my work wrong? @tough hatch
not exactly but i need the answer right now because i need to hand in in 10 min
again, this is not a test
hw
I'll ask why it works later
The only reason I'm cramming rn is I took too long trying to understand the previous ones
you saw me post hw and never asking for answers straight away
this is the only instance, and it's because I'm running against the clock
@tough hatch
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It looks like to see more clearly that its a perfect square in $\mathbb{C}$
glittersparkles
$1-4+4i = 1 +4i +4i^2 = (1+2i)^2$
glittersparkles
@gusty yew but what kind of way did they find it?
Probably by the same concept as completing the square.
But instead of x you have i and i^2
So -8+6i=-8+9-9+6i=1+6i+9i^2=(1+3i)^2
This will not work for any imaginary number
(I think)
Yeah it only works if for some a,b real numbers x+iy =a^2-b^2+2abi
So the real part needs to be a difference of squares
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Hi, so I have $\phi : Z_{30} \to Z_5$ be a homomorphism, where $\phi (3) = 4$, how do I find a formula for $\phi (x)$?
Atehortua
I know that $\phi(x)^n = \phi( \underbrace{x + x + \cdots + x}_{n\text{-times}})$, so we can use that on $\phi(3)^n$, but since 3 divides 30 I can't get $\phi(1)$ this way
Atehortua
or any element that generates $Z_{30}$ for that matter
Atehortua
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Use exponent rules
Same thing lol
You can distribute the exponent
You don't add exponents, you multiply when an exponent is to an exponent
More simplifying
So can I just add the power
Seems so
2b(c-a) is not 2bc-a
No
2bc-2ab
Oh
So
Somehow I ended up like this
@pine kettle
Can I do
2c(a-b)
Cuz in both side they have 2c
Why couldn't you?
! BASUäšá´Źá´ł
$x$^2c(a-b)
```Compilation error:```! Missing $ inserted.
<inserted text>
$
l.55 $x$^
2c(a-b)
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.```
2c(a-b)
Yeah
Distribute
Ok,
It will help you see like terms
So what's the point
Is that the final answer?
I don't think so as my books says the answer should be "1"
That means you didn't distribute or do exponent rules correctly
Where was I wrong
Man,
Sry dude
I ended up messing up
And can you tell md
How do I use this bot @ocean seal
And thanks for the help I got the answer
I appreciate it
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Why is this not allowed in 28?
I can multiply the numerator and denominator by the same constant without issues right?
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I don't know if questions about genetic algorithms are appropriate to ask here since It partially falls under theoretical computer science, but I thought I should give it a go anyway.
I am currently working through Schemas in genetic algorithms and came across the following question:
"How many possible schemas are there of length 'm' binary strings".
Intuitively my answer would be "2 to the power of m", however, this answer is wrong. Instead, it would be "3 to the power of m", but I don't entirely understand why that is the case. Would anyone be able to help me? Thanks.
@neon gull this isnât your channel, delete your message and go to an unoccupied channel
Sorry
what's a schema?
A schema is a set of bit strings that can be described by a template made up of 1, 0 and *, where * stands for âwild cardâ ( 1 or 0).
hm
so then wouldn't it be kinda obvious that there are 3^n such templates
each template is a string over {0,1,*}
and different templates clearly describe different schemas
Lmao, that does make sense. I guess I was confused by the wording of the explanation that was given on the worksheet for the question
"There are m positions in the schema, and each position can take one of 3 possible values: 0, 1 or m. So the total number of possible choices is 3 to the power m."
0, 1 or *
So would I be right in thinking that m is just basically * ?
whoever wrote that answer key for some reason replaced the wildcard symbol with m
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it's Sam
Use this and tell me what you get on both sides
Ok
Now remove log from both sides
What do u get
Now write 9 and 27 in terms of power of 3
Good
Now get the powers of 3 on both sides
it's Sam
Ok so now both base are same
So the power needs to be equal
What do we get
Good
There's other way too
By keeping log
it's Sam
I feel the presence of papa bread
\begin{aligned} (x+2) \log 9 &= x \log 27 \ (x+2) \log 3² &= x \log 3³ \ 2(x+2) \log 3 &= 3x \log 3 \ 2(x+2)&=3x \end{aligned}$
it's Sam
\begin{aligned} (x+2) \log 9 &= x \log 27 \\ (x+2) \log 3² &= x \log 3³ \\ 2(x+2) \log 3 &= 3x \log 3 \\ 2(x+2)&=3x \end{aligned}$
```Compilation error:```! Package amsmath Error: \begin{aligned} allowed only in math mode.
See the amsmath package documentation for explanation.
Type H <return> for immediate help.
...
l.55 \begin{aligned}
(x+2) \log 9 &= x \log 27 \\ (x+2) \log 3² &= x \log 3...
You're in trouble here. Try typing <return> to proceed.
If that doesn't work, type X <return> to quit.```
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Hey, how to solve this? Thanks
(To prove or disprove the statement)
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Shouldnât it be i-(whatever)j + (sjzhusb)k
Because the determinant
<@&286206848099549185>
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Find the centroid of the region bounded by the given curves: $y=3x+5, y=0, x=-1, x=2$
How would I do this when there's two x-values? I've tried doing the integral where a = 0 and b = 2, but it doesen't give me right answer
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hi, could someone help me understand this worksheet?
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I have a Google sheet that I am trying to figure out the following.
I am giving 50% to all my sales team of net and or profit of sale depending on the product. I figured it out my issue here is that some of my items cost is more then the sale, For instance line 19 price is $3 I am charging $4 for the product. In this case I want to make sure i am not losing money. So I need to do a few steps that i completed in the G,H,I,J col. But i cant figure it out a way to get the percentage amount at the end of the day between 50% Profit and lets say 48% profit. The reasoning i need to figure this out. I have a backend system that i can enter the percentage of net ONLY. This sheet will tell me what i need to put in the backend at the end of the day.
I want to show the percent they are getting for each year.
Year 1,2,3 the amount they will be getting in percentage.
Col L and M can be used for any math purposes
row 5 and 6 are different then the rest
Another screenshot
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What are you supposed to do?
oh sorry
use the gradient formula
$m = \frac{y_2-y_1}{x_2-x_1}$
ShitakeMush
Sorry my internet is being weird - point-slope form:
y - y1 = m(x-x1) -> m is given as 1
Let's put it in point-slope form and see what happens:
y - y1 = 1 (x-17)
y - y1 = x - 17
y = y1+x-17
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How do I solve this problems correctly and even if I do, how do I figure out the value of the other sides?
What have you tried?
I already figured out the first problem, by diving 9/15 and multiplying the answer of that by 9, getting 5.4
I'm just out of ideas in the second problem though.
I tried doing the same thing, but I don't know how I can get 7.86.
what if you knew XW?
I do not know how to determine the other sides besides the two already mentioned.
I'd start with two different expressions for cos(ZXW)
what are they?
leave XW just like that, no numbers yet
put these expressions for cos(left lower angle) equal
express XW
knowing it you can find a cosine too, and using cosine and XW you can also get WY
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what have you tried?
I havenât tried much yet just because i didnât know where to start so i opted for some other practice problems first đ
Not sure i follow the set up
not a math guy. use to be one, but forgot it during my older years. I can use the guidance on how to do it
@waxen adder let's go to #help-14
Ok
Sorry i didnt know who was communicating with who
Telocall Profit is the company profit
Your profit is the sales persons profit
income is the overall profit
Our cost...
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can someone please help
@frozen matrix Has your question been resolved?
just... uh, separate this into parts and do integrals over each part?
I did for likf
3 times
used calculator
still says its wrong
the big part is 6859/6
left part 59/6
right part 104/3
its just not accepting
and I have no idea what im doing wrong
left?
ugh im not sure
Kagna Gang Advocate Adavocowana
, w$-\int_{-10}^{-9} 90-x^2+x \dd x$
oh
I found it 59/6 I think
but the right part
I did integral of 100-x2 from 10 to 12
minus 12
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anyone know how i might go about verifying this?
ive tried
changing sec2x
to
1/cos^2x-sin^2x
and 1/2cos^2x-1
but
idk how to bring the things to the numerator
Start from the right side
Turn cot into 1/tan
You will then get a pythagorean identity
oh ok lmao
ill try
and then do i just work backwords
to prove it fromt he left
But yes
lemme give it a try
Do work backwards after
So you know how to go from left to right
It is good practice
right
hmm
i ended up at
cos(2x)/sin^2xcos^2x
idk what happened
i just tried a bunch of things
hereâs my working
i tried to make a cos2x
because of how 1/cos2x= sec2x
but idk how to get rid of this sin^2xcos^2x
@pine kettle
I'm going to try and get Notability on my personal iPad so I can work this out
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A small jar contains 41 dimes and quarters. The value of coins in the jar is 6.65. How many quarters are in the jar?
A small jar contains 41 dimes and quarters. The value of coins in the jar is 6.65. How many quarters are in the jar?
How can I set this equation up for elimination?
have you written down the equations themselves yet?
before you do any algebraic shit you need something to do said shit on
@boreal sedge
yeah yeah i know and I cant figure out what the equations are
okay so you should have begun with that
rather than jumping straight to pure algebraic techniques
i have 6.65=10x+25y so far
what's x and what's y?
x is the dimes and y is the quarters
x is the number of dimes, and y is the number of quarters
okay then your equation as written is wrong
ah
the left-hand side is (presumably) in dollars while the right-hand side is in cents
so what's up with that?
that's more like it.
so that's the equation describing the total value of your coins.
there's one more equation to be written down.
right and it includes the amount of coins
I thought it was just 41=x+y but surely not
why do you think it isn't 41=x+y?
mainly cause my brother said it was wrong
and is your brother 100% infallible?
probably not
did he explain in any way why he thinks you're wrong, or did he just say it outright and act hostile to you over it?
well yes, it's your brother who is in the wrong here
yes
now you can solve this system of equations in whatever way you like
unless you have been explicitly instructed to use some particular method in which case follow said instructions
the bottom?
the x+y?
why not the top, and why not by 20? that way you get rid of all those decimals.
i mean, if you want to make your own life unnecessarily difficult and the numbers unnecessarily big, then by all means feel free to do so.
haha preferably not
okay fair
so i multiply by 100 on the top to make them into whole numbers
well that doesn't really help me much
uhh
if what you're doing doesn't help you much, then why not heed some of my suggestions?
you multiply the top equation by -10 on both sides, and get what?
it sounds like you're not listening to my suggestions at all
the way you're trying to do something completely different in what i can only describe as flopping around like a fish washed up on shore
do you want to solve the problem or do you want to keep floundering?
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could someone please try (a)
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can someone help me with this qus
Did you try anything
add em and remember x=r cos(theta)
y=r sin(theta)
nope dont know what to
what do u mean?
how does that eliminate theta
Eliminate usibg cos(theta) =x/r
why make that substitution tho
@fast zenith Has your question been resolved?
that's the formula for going from polar to cartesian or vice verse
oh i havent learnt that
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Let $R$ is a reflexive and transitive relation on a set A. Prove that relation $R \cap R^{-1}$ is a relation of equivalence on $A$. Denote $\mathcal{S}$ partition of set $A$ induced by relation $R \cap R^{-1}$ and we define on a partition S following relation:
$$
\preceq={(X, Y) \in \mathcal{S} \times \mathcal{S} ;(\exists x \in X)(\exists y \in Y) x R y}
$$
Prove that $\preceq$ is a partial order on a set $\mathcal{S}$.
Note.\Relation $R$ is reflexive and transitive on a set $\mathbb{Z}$, but it is not a partial order. This task basically describes general manual, how it is possible to get partial order from this kind of relation. In this case it means that numbers $a,-a$, which impair asymmetry, we will consider them as same and will will use them.
Michal
I have already proved that R intersection R^-1 is and equivalence relation, but i dont know how to solve second part of the task.
which part of the definition are you having trouble verifying?
$$
\preceq={(X, Y) \in \mathcal{S} \times \mathcal{S} ;(\exists x \in X)(\exists y \in Y) x R y}
$$
Michal
Prove that this is a partial order.
which part of the definition of a partial order are you having trouble verifying?
refl, antisym or trans?
basically all
prove that for all $X, Y, Z \in \mathcal S$:
\begin{itemize}
\item $X \preceq X$
\item $X \preceq Y, Y \preceq X \implies X = Y$
\item $X \preceq Y, Y \preceq Z \implies X \preceq Z$
\end{itemize}
Kanga Gang Annihilator Ann
i think it's just a matter of writing out the definitions
ok I will try
maybe it will help if you first show \preceq can be defined in a different but equivalent way, with both \exists replaced with \foralls
REFLEXIVE:
$\forall$ X: [X$\in S \implies (X,X)\in \preceq$]
Michal
do you insist on writing $(X,X) \in \preceq$ and not $X \preceq X$?
Kanga Gang Annihilator Ann
you are supposed to show that your preceq satisfies it
take an arbitrary equivalence class of R cap R^-1, call it X. is it true that there exist x and x' in X such that x R x'?
REFLEXIVE:
$\forall$ X: [X$\in S \implies (X,X)\in \preceq$]
\X is equivalence class of $R\cap R^{-1} \implies R\cap R^{-1}$ is a relation of equivalence, hence for $x\in R\cap R^{-1}$ holds that $xRx$ .
Michal
ok, im going to try other properties
@misty bobcat Has your question been resolved?
it's antisymmetric not asymmetric
yes, typo
and this is not correct as written
for transitivity:
from X preceq Y you get x in X and y in Y such that xRy and from Y preceq Z you get y' in Y (different than y) and z in Z such that y'Rz
however because y and y' are in the same eq class of R cap R^-1 you get y R y'
thus x R y R y' R z
x R y, and y R y', and y' R z
Hmm that's weird
And asymmetry is ok ?
no
you should start by taking two arbitrary elements x and y in X and Y respectively and showing that x R y & y R x
But that's what I did. Or not ?
no, you just asserted it outright
you have from X preceq Y that there exist x' and y' with x' R y'
and you have from Y preceq X that there exist x'' and y'' with y'' R x''
x, x' and x'' are all in the same equivalence class
as are y, y' and y''
x R x' R y' R y
and y R y'' R x'' R x
and that well give you your goal
Not
maybe you could send a picture of the exercise as it appears in your book?
It's just about that R is reflexive and transitive on set Z. And elements a and -a impair asymmetry.
It's not in English language...
I translated it
ok well something is clearly getting lost in translation! because i have no idea where Z even came in until now!
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it's about infinte geometric series
and basically my thinking is 1/2 = 0.5 and the formula for a infinte geometric series is u sub 1 divided by 1 - r
so it would be 0.5 divided by 1-0.5
which is 1
but the answer key says it's 1/3
<@&286206848099549185> sry for pinging a bit earlier then i'm supposed to but i really need to go to sleep
what's your r?
i'm fairly sure it's 0.5 right? cause it bounces 0.5 to 0.25 to 0.125
it would affect the signs of your summands
i do not know what a summand is
term being added
the r you used and what you calculated would be what would happen if the flea only jumps right
so then r is -0.5?
yes
ohhhhhhh
wait yeah that makes sense when i put it on a number line
thank you for the hlp
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Hey, how to prove that the determinant of a skew-symmetric matrix of size n is 0 if n is odd? Thanks
look at the charpoly tbh
characteristic polynomial
@peak iron Has your question been resolved?
$p_A(\lambda) = \det(A - \lambda I) \ = \det(-A^T - \lambda I) \ (-1)^n \det(A^T + \lambda I) \ = -\det(A + \lambda I) \ = -p_A(-\lambda)$
Kanga Gang Annihilator Ann
so the charpoly as a function is odd
ahh
Thanks a lot!
I'm unsure about smth, how do you formally conclude, from the fact that the charpoly is an odd function, that determinant is 0 if n is odd? đ I kinda see why but I don't know how to say it formally
I see, thank you!
p_A(0) = det(A)
I didn't see this formulated like this in my book, is this a consequence of something?
if u plug in lambda = 0 in the equation p_a (lambda) = det(A - lambda I), you get p_a (0) = det(A - OI) = det(A)
thanks đ
Thank Ann(again) đ
Dbout
0I, not OI
@peak iron Has your question been resolved?
Ann I got another question xD
we want least-squares solution, but they say: z(1) has to be an exact value, z(1) = 1.5 instead of "normal" least-squares
and they ask: "what does it change to the problem? what are the parameters to be determined?"
hm.
well i suppose this lets is express one of the coefficients (alpha, beta, gamma) in terms of the other two
and then do what looks like ordinary least squares
how would you do that?
whoops sorry I forgot to desactivate the ping
alpha * f(1) + beta * g(1) + gamma * h(1) = 1.5
isolate one of the greek letters as you normally would
just basic algebruh
thanks đ that's true
Isn't taking determinant of A and -A^T enough here
what do you mean?
and btw last question:
how to find this with QR?
I mean starting from det(A) = det(-A^T)
(theta_2 just means parameters)
you mean -transpose
Is saying det(A)=det(A^T) true?
det(A) = det(-A^T) = (-1)âż det(A^T)
So first check if what I asked is true then you can continue from here
ahh I think you're right
lemme check
(-1)âż becomes -1 as n is odd
yes it is correct
so if -detA = detA then detA=0?
What do u think?
Argument?
nah forget
thanks!
What's the meaning of argument?
like your proof
nah np đ
^
No y_i on it
z(x) = y
Ok so you must've tried something by now
yeah
first how do you write it as Ax=b?
I have an idea but unsure
First you have to get answer for this question then use it in the question you sent now
yeah but I mean how to use QR for least squares in general?
it's Sam
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What do you mean?
nvm
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howâd we get this?
,tex
(a - bi)(c + di) \\
&= ac + adi - bci + bd \\
&= (ac + bd) + (ad - bc)i
citrusmunch
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
this channel is in use already
,tex
(a + bi)(c - di) \\
&= ac - adi + bci + bd \\
&= (ac + bd) + (bc - ad)i
citrusmunch
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)

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is this correct?
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I know how to do a question like this when the differential is = say t^2 - t + 2 or something
But Iâm not sure how to do it when it = sin
Feel free to ping
@peak coral Has your question been resolved?
hi luke, i dont wanna seem rude but ill try to be honest...what u posted doesnt feel like an interesting problem at all. it looks like one of those trivial easy exercises you can solve only by understanding the question and applying basic formulas.
i really hope you don't take this in a bad way. but i just wanted to tell the truth
at some point, it might be time for you to work and stop relying on people on discord servers. we won't be here to help you with your 1st grade questions in exams for example. you will have to count on yourself, but unfortunately that is not smth you're particularly used to.
this is not meant to offend you at all. sorry if it did, just tell me and i'll delete it
@raven rivet if your message is somehow serious, iâm in the help channels all the time and iâve never seen this person, so i have no clue what youâre talking about
@peak coral Has your question been resolved?
i do have a clue
Ahahahahaha sounds rather passive aggressive bud, this is the first time Iâve asked a question in maybe a month
If you have such a fine grasp of differentials do you care to explain or point me in the direction of how itâs done?
Iâve literally explained I understand how to do do these types of questions and the exchange of real number t variables to trigonometry has thrown me off but label it as â1st grade exam questionsâ all you want
<@&268886789983436800> is this really the type of people that represent this sub?
this is not acceptable
And thanks @raw shard for actually pointing me to something I can use to better my knowledge
Apologies for their rudeness. We try not to encourage people like this on the server; thanks for letting us know about them.
No worries lol, couldnât tell if it was a troll or not considering he seems to believe I rely heavily on this discord to get my degree
Donât know if you know how to solve them or just found the correct notes for it @raw shard but Iâve had a go at it now, I applied the quadratic formula and got -2+-2j which I applied to the homogenous formula and got
e^-2t(c1cos2t+c2sin2t)
somehow i didnât even think to ping the mods, oops
Which leads me to believe itâs the second option and the value of a1 is -2
No problem was just taken a back a bit when I saw that response hahaha
iâll try it myself really quick if you have a few minutes
same lol
still convinced that person is a well written troll or something
i have to do something really quick, so it will be probably an extra 2 minutes
iâll ping you when iâm done
No problem at all, thanks a bunch
@peak coral iâm asking because it looks a bit weird, but does that say 1*sin(3t) on the right?
because there seems to just be a space between the 1 and the sin(3t)
Yeah i think its just how the software enters it
I just used sin(3t) I donât think the 1 is there for any particular reason
so far i have something in terms of sine and cosine, so iâm not sure if iâm doing it wrong or something else
because i did solve the homogeneous version of the DE
Ah ok, well my final answer was the one that included sine and cosine so hopefully thatâs the right way of doing it?
Thatâs what I finished with if itâs any help
Says by solving the homogenous equation however I didnât think there was one as itâs got to be nonhomogenous as itâs using imaginary numbers no?
Iâll send the full question incase itâs any help
ok
The bits with blue are just me guessing answers Iâve not finalised anything yet and am going to change them
Iâd assume the first can only be B and the next question where itâs asking for alpha 1 and 2 Iâd leave alpha 2 empty and alpha 1 would be -2
yeah the answers they put are something from a DE with real solutions
i guess letâs just focus on the stuff we know we can do
Ah, so thereâs real solutions for this?
Iâm confused then hahaha
,w x^2+4x+8 = 0
Yeah thatâs what we both got I assume ?
Alright, and from that and our nonhomogenous identity id assume itâs the second option? Even tho itâs not exactly the same
the sine isnât being multiplied by t
ugh this question terrible, at least the answers are
although i assume A and B are numbers
Yeah I did wonder why A didnât have a multiple of t however none of the other answers make sense?
well itâs up to you at this point lol
Typical university questions
since your guess is as good as mine, and itâs your work
Well are you familiar with solving particular solutions at all?
As I can show you what Iâve done for that even tho my handwriting is awful
Itâs everything to the right of the line if itâs even readable
iâve done it before, i donât exactly remember though so i have a page open to help
so i can do it
Cheers
@peak coral i got $\frac{-1}{145} \cdot (12\cos(3t)+\sin(3t))$ for the particular solution
quantum
Oh, I mustâve done something wrong then?
@scarlet cargo #âhow-to-get-help
yeah you did, i even put it in wolfram alpha and mine was correct
<@&268886789983436800>
Lord almighty thatâs some question
@scarlet cargo dont troll
ok their historys all trolling
How did u manage to put it into wolfram to get an answer?
ban is good too
Iâm not good using the bot tbh lol
this looks very cursed
Ah you guys do it in this American way with yâ instead of d^2x lol
Thatâs what it was I was confused at the difference
we do it the same
Oh?
just different notations
Oh right I see
