And yes, 7200cm seems correct
#help-13
dusk finch
dusk finch
7200cm = 7200/100m
= 72m
gritty galleon
Yea, i wrote
```7200cm = 7200/100
= 72m```
Yeah, just add m as the unit in front of the fraction
peak dome
how to factorize this ?
cedar kilnBOT
how to factorize this ?
clear berry
how to factorize this ?
maybe substituting y=a/b might help
peak dome
i cant understand
peak dome
how did he factorised
6x^2 - 13x +6
to
c(x-2/3)(a/b-3/2) = 0
clear berry
either completing the square or direct application of quadratic formula
splitting the middle term seems more difficult to apply
peak dome
yes
clear berry
also for completing the square, consider dividing by 6 to make the quadratic coefficient 1
so, basically you're looking to factorise y^2 - 13y/6 + 1
cedar kilnBOT
Available help channel!
trim crescent
Hello, knowing the first line how can we deduce the second ?
trim crescent
Sh() is the hyperbolic sin function
dire geode
Usual methods for finding inverse function is to switch variables and solve for the other
trim crescent
Usual methods for finding inverse function is to switch variables and solve for ...
Switch variables what do you mean ?
trim crescent
I’m not trying to find the expression of sin-1 though i don’t think
dire geode
What do you think $sh^\inv$ means
wraith daggerBOT
riemann
trim crescent
What we did was change the x variable to sh(t) to help us integrate
dire geode
Well that's not obvious at all
trim crescent
Here s is x and u is t
Which equals the integral of 1
So we end up with this
And now i i’m a bit foggy as to what we’re doing @dire geode
dire geode
i too am foggy what you're doing
trim crescent
Haha
trim crescent
This all has to do with solving a non linear differential equation
But i’m stuck in the integral part of the problem i guess
dire geode
write out what you do know starting from the beginning instead of piecing stuff in random order
trim crescent
This is basically the beginning
I’m trying to solve this
@dire geode no you’re right we were just trying to find the inverse function
Thx man
Take care
.close
agile portal
What's the limit to + infinity of ln(x^(e^-x))
cedar kilnBOT
What's the limit to + infinity of ln(x^(e^-x))
crimson sedge
Plz help me
celest seal
do not ping mods for math help, and also dont barge into someone else's help channel
please read #❓how-to-get-help
crimson sedge
Mb
agile portal
It's occupied man..
upbeat dune
What's the limit to + infinity of ln(x^(e^-x))
(e^-x)(ln x)
Apply lopitals rule after bringing ln to the bottom
celest seal
l'hopital's rule isn't necessary here
agile portal
It will be undefined
celest seal
hm?
agile portal
Infinity decided by infinity
celest seal
i'm not sure how you're getting that
assuming you're applying l'hopital's rule, you should end up differentiating ln(x) in the numerator and e^x in the denominator
which gives 1/(xe^x)
agile portal
Man i can't use that
celest seal
then what did you do?
celest seal
alright, sure; then apply the continuity of the logarithm
$\lim_{x\to\infty} \ln\left(x^{e^{-x}}\right) = \ln \left(\lim_{x\to\infty} x^{e^{-x}}\right)$
wraith daggerBOT
Namington
celest seal
can you solve this?
celest seal
alright, sure; then apply the continuity of the logarithm
[explanation: continuous functions can be interchanged with limits, as i did with ln here]
agile portal
Man I can't cuz x will be infinity and infinity raised to 0 is not defined
agile portal
**Namington**
That was my question
celest seal
remember order of operations
higher exponents come first
so first e is raised to the -x
agile portal
It's infinity it's not a number?
celest seal
do you know what a limit is?
agile portal
Yes ofc
celest seal
you're not just "plugging in infinity", youre seeing what happens as your number approaches infinity
so the limit of x^0 as x approaches infinity is
agile portal
Yeah yeah
celest seal
the value approached by, say
1^0, 2^0, 3^0, 4^0, 5^0, ...
up as high as you want
youll note that these are all just 1
so $\lim_{x\to\infty}x^0 = 1$
wraith daggerBOT
Namington
celest seal
(in general, this is another case where order of operations applies; limits are functions, so we can simplify the contents of the function first)
(i.e. simplify x^0 before the limit)
(which gives the limit of 1 as x→∞, which is of course just 1)
now we take the natural logarithm of this
celest seal
.close
cedar kilnBOT
agile portal
??
cedar kilnBOT
Available help channel!
crimson sedge
how do I find this area
cedar kilnBOT
how do I find this area
digital cliff
its the sum of the areas of the 3 shapes
noble pumice
I don't even know where to start with this tbh
cedar kilnBOT
I don't even know where to start with this tbh
dire geode
is this a test?
cedar kilnBOT
@noble pumice Has your question been resolved?
noble pumice
<@&286206848099549185>
noble pumice
<@&286206848099549185>
noble pumice
anyone?
noble pumice
<@&286206848099549185>
cedar kilnBOT
@noble pumice Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
past bison
cedar kilnBOT
past bison
I'm having a really rough time with linear algebra (despite being pretty good at, and enjoying math) as I am not a fan of my professor. I don't even know where to get started with a question like this.
crimson sedge
do you know the definition of the null space?
past bison
It is the solution to Ax=b where b is the zero vector
crimson sedge
right, so if y is in Nul(A), Ay=0
past bison
but besides that no. My professor does no type of graphical explanation or anything beyond just a definition like that so. A lot of the concepts are still very vague to me beyond just some definition
crimson sedge
if Ax =[28,-28,56], then what is A(x+y) equal to when y is in Nul(A)?
past bison
so then just constant multiples of that vector?
crimson sedge
so is x+y a solution to the system?
past bison
How would I know? Should that vector be a constant multiple of v=[1, 0, 2]?
crimson sedge
How would I know? Should that vector be a constant multiple of `v=[1, 0, 2]`?
no
you've found that A(x+y)=[28,-28,56]
the right looks awfully like the vector we want
past bison
What is the vector we want?
I mean it jumps out to me that it can be reduced to [1, -1, 2] if that means anything lol
past bison
What are the implications of y being in Nul(A)? Like I am aware that Ay=0, but I don't know what that means
crimson sedge
(Ax + Ay) = ([28, -28, 56] + [0])?
you did the calculation here
then you can pick anything in the span
crimson sedge
for instance, yes
(there's an infinite number of choices for y tho)
(anything in the span works)
past bison
I still have trouble understanding what Span is. Isn't it like the number of vectors that can be made by constant multiples of those vectors?
crimson sedge
past bison
So like Span([1,0,0],[0,1,0]) could be like [0.5,0,0] or [0.5,0.5,0] etc
crimson sedge
yup, thats right
past bison
Okay so...
This means that any arbitrary vector that can be created by a scalar multiple of [-2,-3,1] and [2,-2,-2] makes the matrix A equal to the zero vector?
crimson sedge
yes
past bison
Ah okay. So in order to find vectors v_1, v_2, and v_3 which are different from v, I need to...?
How do I know whether a vector is a solution or not a solution?
Well I guess I can't because A is arbitrary right?
I just know the Null space, some vector v which is assumed to be a solution, and that A times some vector x equals [28, -28, 56]
crimson sedge
we can set v_1 = v+y, where y is in Nul(A)
.
past bison
so [1,0,2] + [-2, -3, 1] = [1, -3, 3] ... is a solution?
v = [1,0,2], and y = [-2, -3, 1] is in Nul(A)
crimson sedge
so [1,0,2] + [-2, -3, 1] = [1, -3, 3] ... is a solution?
yes
past bison
... why is that also a solution?
crimson sedge
(Ax + Ay) = ([28, -28, 56] + [0])?
.
past bison
I apologize, I still just don't understand
What information do I extract from Ax = [28, -28, 56]?
crimson sedge
you know that A[1,0,2] = [28,-28,56]
past bison
And why is some arbitrary vector that is a solution, plus a vector in the Nul(A), also a solution
crimson sedge
and A[-2,-3,1] = [0,0,0]
crimson sedge
And why is some arbitrary vector that is a solution, plus a vector in the Nul(A)...
because of the calculation you did before
A(v + y) = Av + Ay = Av + 0 = Av
past bison
you know that A[1,0,2] = [28,-28,56]
I know that v = [1,0,2] is a solution, but why do I know that A times this vector v equals [28, -28, 56]?
crimson sedge
its given in the question
past bison
ohhhh
i misinterpereted the question
I thought
was just some random piece of information. not that x=v
Ah okay. So to summarize, I know that A[-2,-3,1] = 0 because that vector is in Nul(A)
and since A[1,0,2] is a known solution
then [1,0,2] plus anything in the Nul(A) must also be a solution because it's essentially adding 0
It make sense logistically. But I don't quite understand the implications of it.
crimson sedge
yup, thats all correct
crimson sedge
It make sense logistically. But I don't quite understand the implications of it.
here, the implications are that we have more than one solution (in fact, an infinite number of them)
past bison
crimson sedge
you havent added it to [1,0,2] yet
past bison
nvm
I just mistyped the first one, and for whatever reason they're all correct now, but If you wouldn't mind, I still have some questions
past bison
here, the implications are that we have more than one solution (in fact, an infi...
This I get. However I know that v=[1,0,2] is a solution, and adding 0 to that still gives a solution... right 5 + 0 is still 5... however the result we get from v + x, such that x is Nul(A) is not v... [1,0,2] + [-2,-3,1] = [-1,-3,3] ... [1,0,2] != [-1,-3,3] ... so I am confused as to why this is still a solution?
wait
lethal jackal
so if you have Av = b
and you have Ax = 0
then Av + Ax = b
but linearity of A says
Av + Ax = A(v+x)
so A(v+x) = b
rare slate
can yall help me
cedar kilnBOT
can yall help me
dire geode
can yall help me
cedar kilnBOT
@rare slate Has your question been resolved?
past bison
cedar kilnBOT
past bison
How do I go about finding this transformation matrix? Up until this point I have just been brute forcing the solutions as they have been trivial, but this one has been tricky
I have that [1,0] -> [.75, .75], [1,1] -> [1.5, 0], [1,0] -> [.75, .75], and [0,0] -> [0,0]
Im guessing I need to somehow turn this into a system of equations and solve... but Im not sure how
.close
ebon jungle
I think my brain is broken
cedar kilnBOT
I think my brain is broken
ebon jungle
And this question is the culprit
How can they say
y times x hat plus x times y hat
Like how do you even work with that??? I tried to draw a diagram because initially when I just used mk^2y as the horizontal component, my prof was like, y is a vertical component
So now I’m like ???? But then what’s the significance of x hat then if we’re only supposed to go off of the variable ?? Do I ignore it?
But then if y times x hat is the vertical component… do I take velocity as 0 ???? Since we have u times x hat as our horizontal velocity and apparently unit vectors mean nothing
I feel like I was happier before I saw this question
cedar kilnBOT
@ebon jungle Has your question been resolved?
sullen saffron
It seems to me that you have x'' = k^2 y and y'' = k^2 x. If you assume that y is a function of x then y''(x) = k^2 x. Integrate twice to get y(x) = 1/6 k^2 x^3 + c_1x + c_2. Let x be the parameter and then you have the parametric equations.
ebon jungle
x” here you mean to say the second derivative of?
The thing is, to determine those constants, I’ll have to use the initial conditions
So I need to know which equation is my horizontal component and which is my vertical
sullen saffron
y(0) = 0 will give that c_2 = 0. Since it starts there we can say c_1 = 0 in some way I think.
ebon jungle
Also the parameter that prof is referring to here is t
She wants x and y as functions of t
sullen saffron
The parameter can be anything you want. Set x = t.
ebon jungle
Hmm
sullen saffron
Remember there can be many ways to parameterise the same curve.
ebon jungle
y(0) = 0 will give that c_2 = 0. Since it starts there we can say c_1 = 0 in som...
I need to sub in the velocity at some point
sullen saffron
Yeah. x' = 0 and y' = 0 at the origin since I think we are to assume it starts at rest.
ebon jungle
It seems to me that you have x'' = k^2 y and y'' = k^2 x. If you assume that y i...
Let me try using your logic here to see if I can figure this out
sullen saffron
I'll check my book in the meantime for these systems of DE.
ebon jungle
Thank you
This was in my dynamics course
In the projectiles section
If that helps
sullen saffron
Ok. I've gone to my book and it seems that we should actually have in the end four arbitrary constants in the general solution. I assume x(t) and y(t) would have two each.
ebon jungle
Yea
But we gotta find those
Using our initial conditions
So I’ve found this piece in my notes and I wonder if it’s useful some how
sullen saffron
I'm not sure how much to present as my book tends to use less familiar methods.
ebon jungle
sullen saffron
Basically I get to y^(4) - k^4 y = 0.
ebon jungle
Like could I do this somehow
sullen saffron
This can be solved quite easily and substituted into the y of the equation containing x''.
ebon jungle
Which bit are you referring to ?
sullen saffron
I'll write my work up but the methods might not be familiar.
ebon jungle
I’ll have a look anyway
sullen saffron
Just note D^2 is the operator to mean second derivative, e.g., D^2y = y'', etc.
ebon jungle
Gotchu
sullen saffron
ebon jungle
Hmm I see what you’ve done
Using substitution is a smart way to go about it
I don’t think that exact method is the way to go about it since the question is worth 15 marks and it must somehow relate to what we’ve studied
But I’ll use that logic as well
sullen saffron
It might work out the way I described before though.
ebon jungle
I’ll write up a few methods and see which makes the most sense
ebon jungle
It might work out the way I described before though.
Yea
sullen saffron
I'm not sure though.
ebon jungle
That seems like the most logical one
I’m jus struggling to incorporate the velocity
Because if I split them into a x” and y”
Then I have to stick with du/dt for the horizontal
Which goes to dx/dt once I have it as a function of u(t)
So I can’t have a y on the other side
Which leads me to think substitution might be the thing to do
sullen saffron
Once you've got x(t) and y(t) in terms of the general solution you can then calculate x'(t) and y'(t) and use x'(0) = y'(0) = 0 along side x(0) = y(0) = 0 to potentially eliminate all the constants.
ebon jungle
Both methods you have suggested have been helpful
sullen saffron
You could perform a similar calculation to determine what x(t) is too.
ebon jungle
Hmmm
I asked a few of my masters friends and apparently they had the same question
And none of them got it right
Like back when they were studying under this prof
sullen saffron
Oh. I think this will get you there.
I'm going to work the problem out due to curiosity now.
ebon jungle
Hahaha
That’s the only reason I wanna get it right as wel
At this point I just wanna know what’s happening
sullen saffron
Lol. I get to D^4x - k^4 x = 0. They will have the same general solution.
sullen saffron
Oh. I didn't see the initial condition.
ebon jungle
This is consistent with the kind of stuff we’ve been doing in class too
But yea this solution I’ve sent u was marked wrong so 🤷♀️
sullen saffron
I think I've pretty much got the path to the solution.
I'm probably not going to check what the other students did.
It makes you try to emulate them and you make weird errors when you read theirs first.
ebon jungle
Makes sense
Here’s the only two comments that the prof gave btw
Oh
I think I realised what the other students did wrong
sullen saffron
What's that?
ebon jungle
You can’t express x” as v dv/dt
It has to be like
Hold on let m write it out
Let’s assume at u is horizontal velocity and v is vertical velocity
That’s why their methods don’t work
Because they’re differentiating vertical velocity with respect to horizontal velocity
sullen saffron
I've not read their workings at all.
It seems impossible to eliminate all the arbitrary constants.
ebon jungle
Basically they’ve equated acceleration to v dv/dy
sullen saffron
I'm going to check my answer idea again carefully.
ebon jungle
Okay
ebon jungle
It seems impossible to eliminate all the arbitrary constants.
Which makes me think the prof is just testing how our minds approach this
sullen saffron
I'm pretty sure it's easy with careful consideration. I could be wrong though, I'll see now.
ebon jungle
I’m gonna jus go talk to her in the morning
It’s 1:25am, I’ll @ you tomorrow if you want to satiate your curiosity
sullen saffron
Ok them. I should be done one way or another by then.
sullen saffron
Glad to help. 🙂
shrewd blaze
I’m gonna jus go talk to her in the morning
Close it if you dont have any questions
cedar kilnBOT
@ebon jungle Has your question been resolved?
cedar kilnBOT
Someone help please
hoary spoke
Please
Anyone
abcd is a 4 digit number divisible by 5
bcd is divisible by 9
acd is divisible by 11
abd is divisible by 7
Where a,b,c and d are digits of that number
hoary spoke
bcd is divisible by 9
bcd is 3 digit
hoary spoke
acd is divisible by 11
Same here
hoary spoke
abd is divisible by 7
And here
Anyone?
Someone please
This question is cracking my head since last 3 days
cedar kilnBOT
@hoary spoke Has your question been resolved?
hoary spoke
Help me
cedar kilnBOT
@hoary spoke Has your question been resolved?
cyan wren
Help me
Alright, go through it step by step
abcd is divisible by 5. What does that tell you about the last digit?
cedar kilnBOT
half dagger
Where do I get started in finding the equation of an ellipse that passes through two points: (-2, 4) and (3, 1)?
upbeat gorge
@half dagger what is the formula for an ellipse
upbeat gorge
lol
half dagger
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
olive stag
Derivative of ln(2x)
cedar kilnBOT
Derivative of ln(2x)
tropic oxide
that sounds like the derivative of log mixed in with chain rule. but you are correct so far.
sonic stag
why tho
sonic stag
how do you know if a polynomial can be simplified or not
or if they result in a polynomial
hang on nvm i got it
.close
karmic epoch
help
livid hound
what have you tried?
karmic epoch
nothing. those were my last 2 questions.
livid hound
try something
cedar kilnBOT
@karmic epoch Has your question been resolved?
gritty galleon
try something
I couldn't complete my explanation but i told him how to calc the area of first two shapes (indirectly)
cedar kilnBOT
cedar kilnBOT
Available help channel!
ashen field
cedar kilnBOT
ashen field
can anyone help with this ? do i need to convert it to the closed form first?
and if so is the complimentary function just the fraction and there is no particular solution
or is the particular solution +3/2
mighty drift
usually, don't do more than what is asked of you
especially when you don't know if it's even possible
mighty drift
have you studied sequences of the form u(n+1) = f(un) before ?
ashen field
it’s what i’m studying at the minute so i thought it might be related
mighty drift
it's easy to prove un >= 0
it isn't hard to show that f(sqrt(3)) = sqrt(3)
then it is sufficient to show f is increasing on [0, sqrt(3)]
ashen field
hmm thank you i’ll think about it
halcyon spire
I need help solving these + theres more
cedar kilnBOT
I need help solving these + theres more
halcyon spire
idk
one of my friends gave me that answer
The only one i done was the one in pen
cerulean sail
Well maybe try it again 
how would you find x if your friend didn't tell you the answer?
halcyon spire
idk how to start tho like idk any steps
thats why i wanna know how to solve it
Is there an app or calculator i could use to solve these that would tell me the answer and steps to it
jaunty pumice
cedar kilnBOT
@halcyon spire Has your question been resolved?
trim crescent
Hello, how would you solve this non linear differential equation ? Number 4
cerulean sail
is it not simpler than that? you could separate the variables
From $1 + x y' = e^{y}$, you can change it to $x y' = e^{y} - 1$ and hence $\frac{1}{e^{y} - 1} y' = \frac{e^{-y}}{1 - e^{-y}} y' = \frac{1}{x}$
wraith daggerBOT
@cerulean sail
cerulean sail
Note that you get another $\frac{f'(y)}{f(y)}$ again
wraith daggerBOT
@cerulean sail
cerulean sail
Mostly experience tbh 
I'm not usually that switched on 
cerulean sail
two in a day on many occasions erm, I mean
trim crescent
I’ll try to take it on from here, will open an other ticket if i get stuck
Thank you so much man ❤️
cerulean sail
Yep sure thing 
you'll be fine from there I'm sure 
trim crescent
Can i maybe add you just in case ? 😇
cerulean sail
Sure thing
cedar kilnBOT
cedar kilnBOT
Available help channel!
rich swift
Hello
rich swift
I need help with these 2 questions
@uncut veldt
<@&286206848099549185>
Bruh
Hello
<@&286206848099549185>
tribal dagger
avg nikalna hai?
rich swift
what
cedar cypress
If it takes 3 hours for Joshua to make 1 table, how much does he do in 1 hour?
tribal dagger
for 2nd maybe 3.5 =x+5/2
cedar cypress
Yes and what about the other guy?
rich swift
1/7
cedar cypress
Great
Now let's assume the number of hours taken to be h, and we know how much each of them can do in 1 hour, and we know that the sum of both their works should be 1 table
Can you form an equation out of that?
cedar cypress
Ok lemme ask you this
If I eat 1 apple in 1 hour, how many do I eat in 5 hours
Assuming constant rate
rich swift
5
cedar cypress
Of eating apples
Yes
Now if I eat 1/3 of an apple in 1 hour
How many can I eat in 'h' hours?
rich swift
1/3h
cedar cypress
h/3, yes
Now if he makes 1/3 of a table in 1 hour
How much table does he make in h hours
rich swift
h/3
rich swift
h/7
cedar cypress
And what are they building in total in h hours?
rich swift
Idk
rich swift
h/3+h/7
cedar cypress
They require to make 1 full table
So the portion made by the first one+ the portion made by the second one
What should that equal to?
rich swift
1
rich swift
h/3+h/7=1
plucky dagger
Explain it please
cedar kilnBOT
Explain it please
cedar kilnBOT
cedar kilnBOT
Available help channel!
crisp socket
cedar kilnBOT
crisp socket
how can i calculate this
jaunty shoal
.close
cedar kilnBOT
Available help channel!
crimson sedge
cedar kilnBOT
crimson sedge
I got 60 but they say it's 70?
this is my working:
30/30+x = 1/3
x=60
the "given they play badminton" should be the first row on the table
since no tennis is missing I set it as x
so out of the people who play badminton which is 30+x
only 30 play
and that probability is 1/3
so x should be 60
where is my mistake
this is from our first lesson in probability
cedar kilnBOT
@crimson sedge Has your question been resolved?
cedar kilnBOT
@crimson sedge Has your question been resolved?
fair geyser
yes
x is 60
that's the no tennis cell
the question asks for the other cell, for which you go 200−30−40−60
lost chasm
cedar kilnBOT
muted bear
yo @/riemann help us out
lost chasm
solid juniper
sullying your own post 
vast pike
what a troll
cedar kilnBOT
cedar kilnBOT
Available help channel!
crimson sedge
Yo
dire geode
I Love esting meat
feelings go in #discussion
vast pike
yo
dire geode
.close
gentle remnant
Let f(x) = $e^x$ for $0 \le x \le ln2$ and f(x) = 0 elsewhere.
Find $x_{0.50}$ and $x_{0.90}$
wraith daggerBOT
BlewiiQ
gentle remnant
I did this:
$e^x = 0.5$
$x= -0.6931$
wraith daggerBOT
BlewiiQ
gentle remnant
Which is just not true, because p(x) only equals e^x for 0 <= x <= ln2, and -0.6931 is less than 0 obviously.
crystal raptor
Feels like theres a ton of context you're hiding here. whats p(x)? what does $x_{0.50}$ mean?
wraith daggerBOT
ΣAC
dire geode
That's literally the question verbatim.
nobody knows what $x_{0.50}$ means
wraith daggerBOT
riemann
dire geode
$F(x_{0.90}) = 0.9$ solve for $x_{0.90}$
wraith daggerBOT
riemann
dire geode
Let f(x) = $e^x$ for $0 \le x \le ln2$ and f(x) = 0 elsewhere.
Find $x_{0.50}$ ...
do the same for $x_{0.50}$
wraith daggerBOT
riemann
gentle remnant
that is what i did.
i got x= -0.6931. which must be false. so i did something wrong and my professor offers zero help. ever.
dire geode
that is what i did.
no. you found $f(x_{0.90})$. that's not the same
wraith daggerBOT
riemann
dire geode
Use the formula that relates F(x) to f(x).
wraith daggerBOT
BlewiiQ
gentle remnant
but is not the integral of e^x e^x?
and so the answer would still be $\e^x=0.5\ x=ln(0.5)\ x=-0.6931\$ right?
wraith daggerBOT
BlewiiQ
dire geode
Your F is wrong
dire geode
Use the formula that relates F(x) to f(x).
F(x) is a definite integral
go find it in your book
gentle remnant
This is why i hate this professor. They don't even teach the course. Some random videos of somebody writing down examples with very little work shown leads to extreme confusion for me. And when asking them to explain where I'm going wrong I get no response ever. At least you told me to go google it (no book for the course, just crap examples). The professor doesn't even do that.
dire geode
wild misconception that you learn all you need from lectures
gentle remnant
there are no lectures. The videos are of some (not the professor) guy working through an example in 30 seconds to a minute.
My entire course has consisted of me relying on this discord and google to figure out what anything is.
If I send a message to the professor, his response is always "You're wrong, if you want the correct answer, feel free to contact me." But when I ask for the correct answer or better yet, an explanation of how to get the correct answer, I never get a response back.
.close
tired inlet
tired inlet
I have to create a trigonometric graph to model tides in Makara beach and another graph to model the tides in Lyall beach
I'm currently working on the graph for Makara beach
I got the amplitude to be 0.25 and the vertical shift to be 0.85
cedar kilnBOT
@tired inlet Has your question been resolved?
mild burrow
Can someone help out with yhis I've tried so hard but couldn't get it
mild burrow
X belongs to the interval [0;+inf[ btw
dire geode
Can someone help out with yhis I've tried so hard but couldn't get it
can you type it
your handwriting is hard to read
dire geode
did you try using taylor's remainder theorem?
mild burrow
We didn't study it
So I don't think I'm allowed to use it anyways
I tried to go on from -1<cos(x)<1 but didn't land me anywhere
waxen crater
Boston is 5 hours behind London.
A flight from Boston to London left at 19:35 and took 6 hours and 40 minutes. What time did it arrive in London?
waxen crater
is it asking for the time zone in London or what time it arrived in London(Boston time)
waxen crater
alr ty
cedar kilnBOT
@waxen crater Has your question been resolved?
lost kindle
what is the difference between congruence class and remainder class in mod arithmetic?
lost kindle
Like 3 and 5 belong in the same remainder class modulo 2
jaunty mural
if thats how u use it i see no difference in terms of practicality
perhaps its technical
jaunty mural
what about congruence class
like technically
well nah actually i would consider the 2 to be synonymous
as far as im reading the context
the classes are the same
===
Each class is the set numbers which are multiples of n apart when ur in mod n
Yeah, same meaning different word.
jaunty mural
itd best to stick to just one
more people use 'congruence class'
or even better
'equivalence class' perhaps
depends on context possibly
lost kindle
Isn't congruence class a subset of equivalence class?
Like it's equivalence plus some properties that it adheres to
which is the same remainders
jaunty mural
I think its a specific type of equivalence class perhaps???
Im not entirely sure on the details
I feel like its completely semantic tbh
I use the 2 interchangeably
lost kindle
Oh ok
jaunty mural
Because technically you can declare all things inside your equivalence class to be 'congruent'
and then its a congruence class
its just a matter of whether it makes 'sense' to do so
Congruence has the connotation of some kind of 'equality'
But that might not make sense for certain equivalences.
zenith sail
I think equivalence class is just a more general term that describes this kind of relationship across areas of math
crystal raptor
congruence classes are specific types of equivalence class yes
they play well with the ring addition
jaunty mural
congruence classes are specific types of equivalence class yes
Whats the difference
i must be missing something then
crystal raptor
for modular arithmetic? there is no difference
because the equiv relation is a cong relation
jaunty mural
for the cases where there is a difference
crystal raptor
oh stuff like if a ~ b and a' ~ b' then a + a' ~ b + b'
congruence relations respect the operations on your object
jaunty mural
oh ok understood, mb.
lost kindle
Ok so the main objective is I'm trying to explain why modular inverses are "unique" despite multiple values can satisfy ax=1(mod k), where x is inverse. Do I explain it because we consider equivalence class or congruence class to be unique?
jaunty mural
you are considering set equality
lost kindle
This doesn't sound right
lost kindle
you are considering set equality
set equality? What's that?
crystal raptor
right, all those values will lie in the same class
jaunty mural
equality of sets. or 'classes'
crystal raptor
so the modular inverse is unique
lost kindle
Wait so is the best way of explaining it is because x represents the set of all values in congruent class or equivalence class?
I'm just struggling with wording it properly
jaunty mural
I find it best to use notation that distinguishes between the classes and the elements
So one common one is [a]_n := {a + kn : k in Z}
crystal raptor
Wait so is the best way of explaining it is because x represents the set of all ...
yeah thats actually the right terminology, a member of an equivalence class is a representative of that class
jaunty mural
When working in modular arithmetic, you are defining new operations that work on the sets themselves
It so happens these operations follow 'what you would expect' when working with the individual elements
(and thats what it means for a congruence class to be a congruence class as pointed out above - it respects operations on the algebraic structure)
jaunty mural
Wait so is the best way of explaining it is because x represents the set of all ...
in the usual notation 'x' really represents the class x is in
its meant to denote [x]_n
lost kindle
It so happens these operations follow 'what you would expect' when working with ...
wait sorry I got a little lost here
lost kindle
what does [x]_n mean
lost kindle
Oh ok
jaunty mural
ill omit the subscript since its clear what mod we're in
[2] = {2 + 5k : k in Z}
[3] = {3 + 5k : k in Z}
When you normally write:
,,2\times3\equiv1\pmod5
wraith daggerBOT
jaunty mural
you are formally writing
lost kindle
6=1+5k, for some k?
lost kindle
oh
jaunty mural
A multiplication operation has been defined on sets now.
[a][b] := {xy : x in [a], y in [b]}
Thats how multiplication is defined, first of all.
You then prove that it turns out
[a][b] = [ab]
And the choice of representatives a and b do not matter
Thats the formal way modular arithmetic is defined
And a similar thing is done for all other operations
wraith daggerBOT
jaunty mural
So this. Is actually
[2][3] = [1]
Hence [2] and [3] are multiplicative inverses, once you have shown [1] is the multiplicative identity
lost kindle
so since [2] and [3] = 2+5k and 3+5k respectively, [2] and [3] represents the set of all possible values that satisfy this inverse?
jaunty mural
yes
jaunty mural
yes
lost kindle
which is correct?
jaunty mural
the values in the set are congruent to each other
lost kindle
Ok
jaunty mural
both - congruence implies equivalence
The key point is that any representative of [2] multiplied by any representative [3] will result in a representative of [1]
jaunty mural
Ok so the main objective is I'm trying to explain why modular inverses are "uniq...
to answer this
lost kindle
So because the values in the set are congruent, we consider each values to be "equivalent", and the set itself as unique?
jaunty mural
and also any solution x to
[2][x] = [1] must lie in [3]
jaunty mural
So because the values in the set are congruent, we consider each values to be "e...
forget about using both congruent and equivalent - just use one. Congruent
lost kindle
Ok
jaunty mural
basically yes
jaunty mural
and also any solution x to
[2][x] = [1] must lie in [3]
it would be an issue say
if x could lie in [3] or [4]
or if not everything in [3] worked
then the inverse would be poorly defined
lost kindle
if x could lie in [3] or [4]
then that would mean inverses aren't unique anymore, right?
jaunty mural
yes
jaunty mural
or if not everything in [3] worked
in this case, inverses wouldnt even lie in a class
lost kindle
or if not everything in [3] worked
and this would mean inverses don't satisfy all the values?
jaunty mural
and this would mean inverses don't satisfy all the values?
u would fail to say 'the solution is x = 3 mod 5'
because thats not true
lost kindle
oh because mod is a congruence relation, and it must satisfy for all values in this congruent class?
for it be consider a "mod" inverse
jaunty mural
The other way round really 😅 to be pedantic
I said lets avoid using both words
but to be precise
You can prove mod is an equivalence relation fairly easily (reflexive, symmetric, transitive)
What makes it a congruence relation also
Is that operations such as inverse are 'well defined' and 'preserved' under the relation
Addition, multiplication, additive inverse, multiplicative inverse.
So this exercise is part of the steps required to prove 'mod' is a congruence relation.
lost kindle
Addition, multiplication, additive inverse, multiplicative inverse.
Wait so you saying, all these operations must be proven to show that mod is a congruence relation?
jaunty mural
yes.
lost kindle
Ok got it
jaunty mural
in particular, I think its a congruence relation in the ring structure
jaunty mural
Addition, multiplication, additive inverse, multiplicative inverse.
perhaps not all for a ring, cant remember
(it may be only 2 need to be shown, and the others are then already necessarily true, cant remember)
lost kindle
Wait ok, so to recap basically to answer why modular inverses are unique, I need to show that [x]_n*[y]_n=[1]_n. To show that I need to show that [x][y]=[xy]. And in doing so, I proved that the set of all values in [x] and in [y] are congruent, meaning the set is unique?
jaunty mural
Im not sure what your exercise expects from you
is this 1st year uni
or highschool
lost kindle
no I'm in high school
jaunty mural
k, in which case its a lot simpler
lost kindle
and I'm basically writing a paper, and one of the goals is to show that modular inverses are unique
jaunty mural
The key point is that any representative of [2] multiplied by any representative...
and explaining this idea will suffice
But its certainly worth understanding the underlying workings early on
jaunty mural
and I'm basically writing a paper, and one of the goals is to show that modular ...
oh a paper not an exercise
Yeah in which case the full explanation would be worth it
jaunty mural
It is worth you checking the definition of a ring
The integers under the usual addition and multiplication are a ring structure
And equivalence classes modulo n also satisfy a ring structure under the 'expected' addition and multiplication operations
[a]+[b] := {x+y : x in [a], y in [b]}
[a][b] := {xy : x in [a], y in [b]}
A multiplicative inverse for some x in a ring is defined to be any y which satisfied
xy = yx = 1
where 1 is the multiplicative identity
You need to show [1] is the multiplicative identity
And then what was said above
cedar kilnBOT
@lost kindle Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
crimson sedge
$-\sqrt[3]{x^9}$
cedar kilnBOT
$-\sqrt[3]{x^9}$
wraith daggerBOT
okokok
humble karma
Remember the n-th root of an expression is the same as taking the expression to the power 1/n
crimson sedge
Remember the n-th root of an expression is the same as taking the expression to ...
wut
humble karma
For instance $\sqrt[2]{x} = x^{\frac{1}{2}}$
wraith daggerBOT
Azyrashacorki
crimson sedge
$\sqrt[3]{x} = x^{\frac{1}{3}}$
wraith daggerBOT
okokok
crimson sedge
is this right
humble karma
Yes! In general, $\sqrt[n]{x} = x^{\frac{1}{n}}$
crimson sedge
ok so where do go from therer
humble karma
Whoops
wraith daggerBOT
Azyrashacorki
humble karma
Well if we take another example $\sqrt[5]{x^4} = (x^4)^{\frac{1}{5}}$
wraith daggerBOT
Azyrashacorki
humble karma
Can you apply some law of exponents then?
wraith daggerBOT
okokok
humble karma
The negative sign just hangs on outside $-\sqrt[5]{x^4} = -(x^4)^{\frac{1}{5}} = -x^{\frac{4}{5}}$
wraith daggerBOT
Azyrashacorki
humble karma
so $(x^9)^\frac{1}{3}$
That works, but also remember that $(x^a)^{b} = x^{a\cdot b}$.
wraith daggerBOT
Azyrashacorki
humble karma
Now you can simplify it further
crimson sedge
so $x^{9*\frac{1}{3}}$
wraith daggerBOT
okokok
crimson sedge
so that was easy
humble karma
Yes!
crimson sedge
zamn
humble karma
So essentially in the exercise, you just change the root to an exponent and apply the exponent rule.
crimson sedge
so is it $x^3$
humble karma
The minus sign just multiplies to whole thing
wraith daggerBOT
okokok
humble karma
Yes that would be the simplification they asked. $-\sqrt[3]{x^9} = -x^3$
wraith daggerBOT
Azyrashacorki
crimson sedge
**Azyrashacorki**
so it would be negative ?
humble karma
Yes it would be negative
crimson sedge
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
crimson sedge
How would I solve this?
cedar kilnBOT
How would I solve this?
crimson sedge
dx is 3/4 cos?
copper crest
dx is 3/4 cos?
Yessir
crimson sedge
I've replaced x and dx and simplified it to (9/16 sin^2)/sqrt(9 - 9/16 sin^2)
not sure what to do from here
crimson sedge
Show your work
buoyant latch
Where’s your dx in the first line
What are you integrating with respect to
There’s a lot wrong with your notation that makes it confusing to see what you’re doing
And I’m sure it’s confusing for yourself as well
crimson sedge
crimson sedge
oops
buoyant latch
That’s how the dx turns into dθ
Also what happened to all your limits on the integral
There aren’t even equal signs in your working out
crimson sedge
There aren’t even equal signs in your working out
buoyant latch
Ok now why have you subbed in x = 3/4 sin θ but somehow the limits didn’t change
cedar kilnBOT
@crimson sedge Has your question been resolved?
neat wave
cedar kilnBOT
neat wave
can anyone help me understand why is there a in numerator and 1 in the denominator
gilded elm
gonna need more context for this one
dire geode
show the whole page
neat wave
gilded elm
So basically A_n is just the nth term?
neat wave
its the arithmetic mean of nth terms togather
atleast thats what i assume
so i assume numerator is supposed to be all the values of A_n terms added togather
and den is the nth number of terms
but based on that this seems weird
ANYONE HELP? PLS
amoeba is the powerhouse of the cell
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
mossy warren
cedar kilnBOT
mossy warren
I'm kind of stuck on this ngl, could someone guide me through it?
crimson sedge
How does the (1) become positive after extracting X outside of the ()
crimson sedge
Because there is also a minus sign in front of the Bx
Thanks
cedar kilnBOT
Available help channel!
cold light
help with these question.
1+ i = (1+ r/m)^m where i = 0.1025 and r = 0.1. Now find m.
vagrant elbow
Replace i with 0.1025 and r with 0.1
cold light
Replace i with 0.1025 and r with 0.1
I know the answer but I can't do the working.
answer is m= 2 guess work but how do I do it properly.
1.1025 = (1+ 0.1/m)^m
cedar kilnBOT
@cold light Has your question been resolved?
twin mulch
Let ABC be a triangle. Point K belongs to side BC such that KC = 1/2 KB, I belongs to segment AK such that AI = 1/3 IK. BI intersects AC at E. Calculate AE/AC using similar triangles.
twin mulch
What have you tried
i tried to solve it but can't :(
twin mulch
And what is point E?
BI intersects AC at E
cedar cypress
BI intersects AC at E
Yeah I read it now
twin mulch
Yeah I read it now
Can u help me:?(
cedar cypress
Yes sure gimme a second
twin mulch
Yes sure gimme a second
ty so much <3
livid hound
i tried to solve it but can't :(
show what you've done
twin mulch
show what you've done
I drew the figure and recorded as many proportions as I could
worldly marsh
Maybe you can try vectors
twin mulch
Maybe you can try vectors
i'm grade 8 and don't know what is vectors :(
worldly marsh
sorry
twin mulch
Yes sure gimme a second
are you okey:?
worldly marsh
I solved it with similar triangles
cedar kilnBOT
@twin mulch Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
steel thistle
i dont understand what does the 3rd step does
cedar kilnBOT
i dont understand what does the 3rd step does
jaunty mural
"when r u dividing by 0"
steel thistle
what is the purpose tho
livid hound
there's an additional restriction that you can't have values that would result in division by 0
jaunty mural
well u cant divide by 0
livid hound
any such values would be excluded from the solution set
steel thistle
but why isnt the solution start from - 3
livid hound
it also gives a critical location where signs of expressions change