#help-27
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Wait I think I understand lol
sorry
I forgot that it can't be a negative when square rooting
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Given A and B are real, n x n matrices that satisfy AB = BA. Prove det(A^2 + B^2) >= 0
what have u tried ??
@pure stone Has your question been resolved?
@pure stone Has your question been resolved?
do you remember how matrix multiplication is done?
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Hello helpers
why is the output vector and input vector on the same plane?
its just for intuition right?
because 2d input vector and 2d output vector would be 4 dimensional
if you were to graph a function R^2 -> R^2 in the literal same way as we do R -> R then yes that would require 4 dimensions
here though visuals-wise the output vector is attached to its corresponding input
right so if I just took two different 2d planes, what would the resulting output vector look like
would it also start at 0,0 ?
or at the input vectors coordinates
Im trying to understand where the output vector actually begins and ends
because as you say its attached at the input
on its own 2d plane it would begin at (0,0) right?
"where a vector 'actually' begins and ends" is a moot question
it doesnt 'actually' begin/end anywhere at all
So its generally normal to be confusing
because I always get it wrong considering this way
whatever not worth digging into it
the visualization is good
thanks ann
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could someone help me with this problem?
@fair raft Has your question been resolved?
anyone?
@fair raft Has your question been resolved?
I think we could apply the definition of ρ-length on the equality given for ρ-length being invariant for mobius transformations of H
once we subtract both integrals, we shd get somethings like integral from a to b of (ρ(γ(σ(t)))∣γ′(σ(t))∣−ρ(σ(t)))∣σ′(t)∣ dt = 0
setting f(z)=ρ(γ(z))∣γ′(z)∣−ρ(z) should immediately give you the answer after applying the thing given in the hint
Correct me if I'm wrong tho, idk what those transformations and all mean 😭
@fair raft Has your question been resolved?
I am not gonna be too much of a helper but after no one responds for 15 minutes
I read it on the rules
it's safe to ping them at @ helpers
once
only once
okoko tysm
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Is this the right method
when price is 700 + 100 * 0, number of occupied rooms is 72 - 2 * 0
when price is 700 + 100 * 1, number of occupied rooms is 72 - 2 * 1
when price is 700 + 100 * 2, number of occupied rooms is 72 - 2 * 2
....
when price is 700 + 100 * n, number of occupied rooms is 72 - 2 * n
daily income is (700 + 100 * n) (72 - 2 * n)
and x = 700 + 100 * n
Wait...what
Hello?
@merry ivy Has your question been resolved?
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@merry ivy Has your question been resolved?
hello
I am taking algebra 2 this summer
should I get any textbooks
or rely on the normal curriculum work
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I need help learning elimination and substution
@unique crown Has your question been resolved?
which method do you want to use for this question first?
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Elimination, sorry
Had dnd on
so do you know what you need in both equations to use elimination?
Subtract
before you do, you need to make sure you have a term in each equation that will cancel itself out when added to or subtracted from the second equation.
that's why the whole process is called elimination - you're eliminating one variable from consideration.
as of now, though, using your question, there's no direct elimination possible. a naive subtraction here will still see both x and y.
to start using elimination in a situation like this, you should first match the coefficient of one variable in one of the two equations to its coefficient in the other one by multiplying.
in this case, the x terms are easy to match here.
hint: multiply the second equation by a number to get the x coefficient to the same as the one in the first.
can you do this step first?
pinging for attention as requested.
Appreciate it
yall know about quadratic equations like finding roots and vertex??
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it does.
you want the 5x to turn into 15x so that when you subtract the two equations, the 15x from the first equation and the 15x from the second cancel each other out.
What grade are u in 😭?
that's the whole idea behind matching coefficients: multiply one of the two equations by a specific number to get either the x/y-cofficient to match that of the other equation's, then add or subtract them to eliminate that variable.
I don't think we should be chatting in another OP's channel, but if it helps you somehow, whatever grade corresponds to age 15.
U are in grade 9
Nice
So 75x + 75y = 300
how did that happen?
I multiplied 15 by the bottom equation
15 is overkill.
you want the 5x to turn into 15x, not multiply the whole thing by 15.
The 3 is negative though
we're eliminating x, not y. if you want to eliminate y, you'd have to scale both equations.
Would it be easier to use substitution?
for this one, using elimination kills the problem right away.
what is 3?
in that case, write the entire equation after you've multiplied by 3.
don't do anything else with it yet - just write out the equation as a whole.
15x + 15y = 60
right side?
there we go.
now, we see that we have a 15x in this new equation and a 15x in the first equation.
at this step, you can now subtract the two equations. the order doesn't matter.
(by the two equations, I mean the original first equation and the new second equation, by the way. don't use the original second equation for this.)
ok see, let me line up the two equations top to bottom.
15x - 3y = -30
15x + 15y = 60
oh, wait, never mind then. your 18y is correct, but not your right side.
a 60 and a 30 don't make an 80.
-30-60=90
after fixing this mistake, you're now on the right track to getting the value of y.
one minor tidbit though: if this was a free-response section, ensure you say which way you're doing the subtraction to result in this new equation.
So i plug 5 into y
why do that though?
I know y = 5 fits, but if you didn't know that, you're going to have to show how you got it.
you should make this step clear, in my opinion.
in this case, yes, you can solve for x now, but the question didn't really ask for the value of x.
Oh wait I realize that now
is there anything else you'd need help with?
Nope, thank you
8th
Im taking algebra 1 honors and ive been absent for 1 week and now we have a quiz 😭
so is there anything OP still wants to ask?
Not as of right now no
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i'm confused on how to get the vertex with -b/2a
which line in there is confusing
how do i get the y
f(3)
yes that's what f(3) means
sorry brain farting rn thank you 😭
all good
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hi im a little confused where did the -e^-x come from?
this is my teachers asnwer key
in the dy/dx?
if that's what you mean, then it's because the derivative of e^-x is -e^-x
and the reason we're getting that is because of the product rule
how does that work?
chain rule!
i know that the derivative of e^x is e^x

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✅ Original question: #help-27 message
okay so i fear im still confused
can you quickly walk me through this
its breaking me
let $f(x) = e^x$ and let $g(x) = -x$. then $e^{-x}$ is $f(g(x))$
higher!
what is it's derivative? the chain rule says that it's $f'(g(x))g'(x)$
higher!
okay got it
wait hold on
so why is f(x) e^x?
the -x isnt outside of the ^x, is it?
as in, why is f(x) e^x?
that is given in the question, the essence of doing f(g(x)) is putting the expression g(x) as the term for x
anothher example for constants would be f(3), this means you evaluate as x=3 in the equation, or if you do f(x^2) then you replace x with x^2, similarly this means that for composite f(g(x)) it just indicates that the x term in your f(x) now becomes whatever g(x) is
i see
i fear im still a little foggy but i have a basic understanding and i gotta keep pushing through
thank you!
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I don't get this. Why both of them are = to x.
Here is my evaluation,
For the first one, f-1(x)=x(domain), then f(x)=y(range)
For the second one, f(x)=y(range), then f-1(y)=x(domain)
do you mean the expression? or do you mean the range and domain
because correct me if im wrong, the doman and range of f operated on f^-1 and the method of f^-1 operated on f is diffferent
with regards to expression
you can think of f^-1 as an undo button, essentially if i have the result of f(x), doing f inverse takes the output of f(x) and gives you the original value
the undo button f inverse is similar to f^-1(y) =x and therefore i will explain it as such
since f(x) =y, then applying f^-1f(x) is essentially conducting f^-1(y) which returns x
for the first one, if i use f-1(x), i can get the x in the f(x) right? then, i plug the x into f, that just f(x), so it equals to y. That's what i think of.....
okay... let me think of an intuitive way to understand this yeah
My first thought is that x is a dummy variable.
The x in first line belongs to the range, while the x in the 2nd line belongs to the domain.
if you doesn't feel safe about both using x, you can rewrite the x's in the first line into y's
so, they are not the same thing? 😨
The problem is, when I apply to the inverse+composite function, i feel weird....
btw i'll come back later after my class
Let's pick an example
Let f:A→B, be bijective, hence inverse of f exists, lets call the inverse function g.
suppose x in A and y in B, and f(x)=y
Since f is a bjiection, g(f(x))=g(y)=x
and f(g(y))=y
But what if A=B?
then x in A also means x in B, therefore we can use x (in A)
to say g(f(x))=x
and we can use x (in B)
to say f(g(x))=x
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Hey guys, I have a question about how i start with matthematics I am now in 11t grade and my knowledge of math is pretty poor: fuctions, basic logarithm's ect... But i want to learn way more math and i really want to learn and understand everything well but i searched about how i should start with math online and there where so many different ways and fields that i came to the conclusion that i really just need someone that has some decent expirience to talk with them about it, so if anyone wants to start a little conversation with me about the most efficient ways on starting math im all ears! (sorry for bad english its not my main language).
I'd say that the first move is to find a syllabus to structure your learning from this point forward
one sample syllabus I like to use is in the Bookshelves of Math LibreTexts, which covers you all the way to abstract algebra adequately
then, instead of focusing entirely on just reading, always do practice problems! you are only as good in math as the problems you understand and can solve, so consider spending some time trying out some problems after each topic
also, don't hesitate to ask when stuck! it would be nice if you tried the problems at least as much as you could first, but if you're stuck, someone could give you a tiny hint that could be enough for you to make progress.
and sometimes, while waiting for an answer (or even when voicing out the question for your helpers/teachers/tutors) you hit upon an idea you've never thought of
the last tip I can offer is to understand why formulas are what they are (i.e.: understand where they come from, why they work and how they apply)
Hello, I have searched on the internet for math libre texts and i have found the website wich looks good to me, i just don't understand what you mean with finding my sullabus?
https://math.libretexts.org/Bookshelves
if you go to this part of the site specifically, notice that each field is laid in order
the fields start at Arithmetic, then Pre-Algebra, then Algebra, Geometry, Precalc/Trig, Calculus, etc.
up to and including calculus, this is more or less the suggested study order for most people
after that, you can sort of jump around depending on what field you plan to major in
Ah ok its like a road map for the basics?
yes!
And do you suggest that i learn all of them?
not necessarily, but if you are going into pure math, you will see a lot of them
Alright thank you so much you are wonderfull
I finially have some good structure now
Thanks
No that was all for now, i will close this channel
aight, see you around, and welcome to the server!
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Hey i am now learning my math trough math libretexts but as i am going trough the excercises i just can't seem to find any solutions to the excercises so i can't know 100% if my answer is right or not. Is it just like that or does the Solutionpage stand anywhere that i can't seem to find?
the channel is closed btw. if you want to ask your question, please grab another channel
Oh ok
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prove that x=2 is the only solution
what are your steps so far? where have you reached?
I found that a=0 and x=2 is a solution
what is a here?
its not mentioned in the problem statemenet
the full statement is find a then find all solutions (all x)
yes, so where do you get stuck when you need to prove x = 2 is the only solution? try to see if the same logic works
you've already found cbrt(3x^2-11) + sqrt(2x-3) + 4 - x^2 - x <= 0
so a^2 <= 0 and so a = 0
yes
when does the equality in AM-GM happen
when the two terms are equal
when the terms are equal right?
so use it to prove that the equality only happens when x = 2
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shouldnt this function have a jump discontinuity at x = pi/2
floor of sin(90)=1 is 1 right?
yes, it's just a removable jump discont and desmos cant really show these well
if i am asked to plot it on paper i will show the discontinuity right
yes
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Pls help me I'm stuck here i wanted someone to explain me
can you express 2.25 as a fraction first & foremost?
Yea it would be 225/100 i.e. umm 9/4
Ohh i seee , I see something like loga-logb
that's half of one law
We have log2 and log3 , 9/4 is square of 3 and 2 right
Sorry my bad
Yes
ok, do you have paper on you
Thank you tho you ignited my brain cells 😭
I'm thankful
I gotta have the approach developed
Its 2 lg(3) not 3lg(3), also im pretry sure you mixed up the values for lg(2) and lg(3)
Ah yeah , marks deducted like this in exam 😭
Ty for correction @coarse flume
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if LHL and RHL limits both are +infinity then can I say that the limit exists at that point
and the limit is +infinity
strictly speaking, no
<@&268886789983436800> troll.
whether +∞ counts as "existing" is kinda author-dependent
Limits taking infinite values are a lot trickier than limits taking finite values
who is my author
is this dude even alive
didn't he die in like 05
What Ann means is that you can define a notion of limit existing if f goes to infinity, it depends on if the text you're reading uses that
im js solving this problem
(at -3)
Personally I'd say that's infinite but it depends on your curriculum
do i say limit of -3 infinit
so infinity ehh
this thing doesn't have a universal answer
not undefined
it depends on what curriculum you're following, go read your textbook to find out
this will be explained somewhere
i thought math was supposed to be objective
it is objective in the sense that once you fix your definitions everything has exactly one answer
but fixing those definitions is subjective
I think that in this case the answer is +∞
suppose im doing a continuity question and i get both side limits equal to +infinity
then what should i conclude abt the limit
does it exist or not
Because both left-hand and right-hand limits exist and are +∞
reading is an important life skill 
what would YOU do
read his textbook 💀
We would all read the textbook
I would check which one will not make me lose marks by reading my textbook
waaat
okay let me open the pdf
ok here
damn chartbit is here
you didn't need to go straight to the answers, usually your textbook will explain the content to you in the pages before the problems, you should read those pages when you're stuck
its easier to read the answers
i just learnt
ok on it
1 sec
it says does not exist
but why did it write +inf in the answers then
just write dne
we write +infinity for these limits because it is more informative than just saying "DNE"
even if "DNE" is technically true it's less helpful
which one is more helpful in solving continuity questions
well you should just get into the habit of writing 'infinity' where it applies and then just remember that also implies it DNE
Which is basically what the book said. It's just a convenience to convey the f grows without bound
are undef & dne the same
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Why is x^-1 = 1/x^1
x^3 = x * x * x
x^2 = x * x (we divided by x)
x^1 = x (we divided by x)
x^0 = 1 (we divided by x)
x^-1 = 1/x (we divided by x)
it's defined this way because it preserves many nice properties and patterns
that exponent property being...
$x^{a+b}=x^{a}x^{b}$
🫎 Moosey 🫎
and it works nicely with the other exponent properties as well
$x^{(ab)}=\left(x^{{(a)\right)}^{b}}$
also why is x^0 = 1?
🫎 Moosey 🫎
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well, again, it works nicely with your exponent properties
you know x^(a+b)=x^a x^b
if you have x^2 * x^3 then that's equal to x^5
2^2 * 2^3=2^5
2^0*2^1=2^1, it only makes sense for 2^0 to be 1 in this circumstance
but we don't have to add 1^1 to 1^0...
well yeah because 1^x is just going to be 1
its the same reason why we define square root as x^(1/2) as well
an nth root as x^(1/n), because when we take the n-th power, we get back to x (atleast for odd powers)
(x^n)^1/n=x (for even n its |x|, because even root functions don't account for negative signs)
#help-27 message you could read this conversation here
,calc (4a)^2
The following error occured while calculating:
Error: Undefined symbol a
,w (4a)^2
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@stiff karma Has your question been resolved?
what further questions are you working on that helpers didn't already answer
@stiff karma Has your question been resolved?
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I have to test this equation for x-axis, y-axis, and origin symmetry. x = 2y^4 - 9y^2
First I was doing y-axis
I know that you gotta substitute x for -x
But now I am confused
Would I make the negative x back to positive?
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its pretty easy to see if you try to achieve the LHS by expanding the RHS in this
@royal laurel Has your question been resolved?
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Am I crazy or does this not have a gcf
I think what you need to do is to find lowest power of variables and coefficients that appear in all terms
i.e. 2x^2 + 4x= (2 * x * x) + (2 * 2 * x).
You could see that only 2x is common in both the terms
I have no idea how to do that
it does have a gcf
there's no constant term
nah
so it's not about the numbers 3, 8, 5, 44 (their gcd = 1, of course)
but it's about...
why are you showing us a completely different part of the question
You could write:
3x^4 = (3x^3) * x
8x^3 = (8x^2) * x
5x^2 = (5x^1) * x
44x = (44) * x
What is common in all the terms?
,rotate
I gave up on the other one
did you have a look at this?
So I’m trhing the other one but the answer doesn’t seem rigjt
Yea but I’ve never done anything like that before so I don’t understand it at all
can you factorise $x^2 + x$ for example?
south
Yes
I understand that part but I don’t know what to do with the • X outside the ()
But you need to solve the equation
it implies one of the roots is 0
yes, so that implies $x (3x^3 - 8x^2 - 5x - 44)$
south
and you can factor inside the parenthsis
But there’s four of them so now you just have to x^4
But then your just where you started?
wdym?
Do you have to factor out the gcf or can you just solve it by doing long devisition twice
you need to factor the gcf
There’s four •X
the gcf is x
You just gotta find what you took outside(which is common in both the terms), that is your gcf.
As stated here, only x can be taken out
I don’t know how to do long deviation with an extra x on the outside …
you just leave it there
can you long divide (3x^3 - 8x^2 - 5x - 44) by (x - 4)?
Yes
to make a simpler example, imagine I have $x(x^2 - 3x + 2) = 0$
I factor $x^2 - 3x + 2 = (x - 1)(x - 2)$
subbing that back into the equation, $x (x - 1)(x - 2) = 0$
south
I kinda understanding factoring out X but I don’t know how to do long devision with just the value X on the outside
after you long divide
you get $x(x - 4)(\text{some quadratic}) = 0$
south
But why don’t we divide the X?
cause x can be 0
so dividing by 0 would be undefined
do you understand this?
No I understand us getting (X-4)( quadrant ) if we where just using 3x^3-8x^2-5x-44
I'm asking if you understand this or not
I know you know how (polynomial) long division works
I mean techinally only polynomial long division I can’t do normal
okay, what don't you understand about this?
How is it equal to zero and how the X value is there and was not in the long deviation
to find the roots of a polynomial, you set it equal to 0
do you agree?
I'm giving you a new question which is to find the roots of $x(x^2 - 3x + 2)$
south
Are the roots the zeros?
yes
I don’t do that then I use the box or quadratic formula
yes, I know you don't need polynomial long division here
do you understand how $x(x^2 - 3x + 2) = 0$ becomes $x(x - 1)(x - 2) = 0$ then?
south
But it says to use long divisiton
,rcw
your problem is why there's an x on the outside, right?
My teacher sped through directions
I can help
read through all the previous conversation then
Yea usually I would just divide it was by a factor of (X+4
okay
but if you divide by $x$ on both sides, you get $(x - 1)(x - 2) = 0$
@mellow willow what are u stuck at
south
Wdym on both sides
you've lost a solution: you've lost x = 0
Wait
(X-0)?
Noo
I’m saying is that a zero?
0 is a zero, correct
@fossil locust
X=1
X=2
So could I just do long devision twice one time with (X+4) and then with (X-0)
💔
to divide by x, you don't need long division haha
you just divide every single term by x
I just don’t know that way
but then yes, you then divide that cubic with (x + 4)
okay, let's start with something simple
@mellow willow write the work that u are stuck at
$\frac{x^2 + 2x}{x} = \frac{x^2}{x} + \frac{2x}{x} = \cdots$?
south
you're being disruptive
I don’t really understand
I've chatted with them for half an hour and figured out what they don't understand
im with obama on this one
it's like $\frac{30 + 50}{10} = \frac{30}{10} + \frac{50}{10}$
south
agreed
Wouldent it be just the same thing
yes, they're equal
when we split the fraction, those two things are equal
I just don’t get how this relates
so now can you answer what this becomes?
we're talking about how you can divide an entire polynomial by x without needing long division
I don’t really understand how to decide by just X tho
$\frac{2x}{x}$ is just dividing 2x by x
Mirror
surely you know what 2x / x is
yes, 2/1 is 2
$\frac{2 \cancel{x}}{\cancel{x}}$
south
oh yeah and when you cancel, that becomes (2 * 1) / (1)
(South, if I'm not being helpful feel free to tell me to stop!)
I never take the gcf out in general which is why it’s even more confusing
Like I cant remember me doing that ever
Can I back us up a step?
@mellow willow are u in grade 9?
Isn’t quadratic equation for grade 9
See I’m not good at math bht some how get A’s
dude you are not helping
I likely forget it or learned it in middle school which basically means I did not learn it because of Covid
Ahh
Allie, backing us up a step: $f(x) = x({3x}^{3}-{8x}^{2}-5x-44)$ is what we're starting with
Mirror
so if f(x) = 0 what does that mean?
please don't take any conversations unrelated to the helpee's problem into help channels
I don’t really know
X intercept?
yes, that's related. if i were to tell you that $a \cross b = 0$ can you deduce anything about a and/or b?
Mirror
Right! That's the idea here.
$x({3x}^{3}-{8x}^{2}-5x-44) = 0$ means that $x = 0$ or ${3x}^{3}-{8x}^{2}-5x-44 = 0$, are you with me so far?
Mirror
Uh
it looks long and complicated but it's just our a x b example from before
No I think it’s just not showing up for me
(Think x = a and The Big Scary Function = b, does that help?)
bc it doesn't matter what we multiply by once we multiply by zero
But I just don’t know how to do long divisiton witu x(the quadratic )
We don't have to! Let's break it up.
We know that if x = 0 the function equals zero, so let's set that off to the side and remember it later.
Now our big function is ${3x}^{3}-{8x}^{2}-5x-44$ and we're looking for when that function is zero.
Mirror
We also know that x=4 is an intercept, so what does that mean?
Wait so do we just divide the quadratic and not the x
ofc! you're doing fantastic too, btw! sometimes just needs another way to word
Possibly. Can you be more specific?
(What are we dividing by what?)
There we go! Can you tell me why you're dividing by $x-4$?
Mirror
Cus it’s a zero
(It's the right move here, by the way: just want to make sure you understand why!)
Perfect.
Next step here is the long division, which (as far as I understand) you can do?
So do that and report back if you encounter some more trouble!
(Broader picture, by the way:
${3x}^{4}-{8x}^{3}-{5x}^{2}-44x) = 0$ if $x = 0$ OR $({3x}^{3}-{8x}^{2}-5x-44) = 0$ \newline
$({3x}^{3}-{8x}^{2}-5x-44) = 0$ IF ($ x-4 = 0$ OR <some other function> = 0)
Mirror
I’m not sure how to continue because I need to write smth witu i but I don’t know how since it’s not a perfect square
this is pre-cal?
Let me double-check that long division
Yea
Long division is correct 
,w (3x^3-8x^2-5x-44)/(x-4) quotient and remainder
I haven’t done get before
the Wolfram i mean
For c would I leave it as is
Ohh
I'm inclined to agree that we leave it as is since there are no real solutions to $3x^2 + 4x + 11 = 0$
Mirror
,w sqrt(-116)
Do you have any other questions, Allie?
I know I have to do the quadratic formula since it’s a step and it’s a little similar to a class problem but it’s not a perfect square
yes, in fact you can check the discriminant of the quadratic
,calc (4)^2 - 4(3)(11)
Result:
-116
< 0, so no real roots
They did this above!
Yea but I need to write it with i some how
ah, of course cause your class wants the complex zeroes
then yes you need the quadratic formula
^
you can substitute $\sqrt{-116}$ into $i\sqrt{116}$ and keep crunching!
Ohh ok
Mirror
that is technically correct but i think you can clean it up a lil more
My teacher did say in clas it’s fine to not simplify but how would I?
since you're at a stage where the answer is complete i will just tell you!
since $4 \cross 29 = 116$, $\sqrt{116} = 2\sqrt{29}$
Mirror
Ohhh kk
One more question for another problem I’m trying to do the box but I don’t know if 7 would be positive or negative
It'll be negative in both. A great way to approach is to work from the innermost set of parentheses outward
so before you even start distributing turn $(x-[7+3i])$ into $(x-3i-7)$
Mirror
This doesn’t seem right
,rccw
(one thing I will say is that for this, I would personally be tempted to write what we're expanding as ((x - 7) - 3i)((x - 7) + 3i), from which it becomes a tiny bit easier as you can difference of two squares to begin with)
Really which one?
I’m not supposed to have I left I’m very confused
Ohhh I see
@mellow willow Has your question been resolved?
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Are there any nice propeties for fin-dim vector spaces that neatly carry into the inf -dim case
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Here for example, is it x y^-2?
…
well, what more do you want?
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Hi can someone just tell me what the question means by sum of the perimeters of region a,b,c and d
,rccw
Like does it means the entire square (basically combine a,b,c and d and the perimeter of that is 152) or is it that like if we took it apart individually perimeter a + b + c + d = 152
perimeter of A + perimeter of B ... = 152
id assume its double counting edges due to how its phrased
Oh alright ty
If it meant perimeter of the whole thing then it would probably just state that
Rather than give of each region
Yea i got kidda confused cus if it was like one by one
Double edges would be counted
😭
Anyways ty
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\begin{gather*}
2 a^2 + \sqrt{b} = 8 \
a^3 + ab - b = -34
\end{gather*}
Thomas
you want to solve for a, b right?
yeah
what have you tried so far?
Are we given anything else
not rlly
For example are they integers or smth
,w a^3 + a(8-2a^2)^2 - (8-2a^2)^2 = -34
ok so there's one good solution
the other roots are all very ugly, yes
It's still a quintic though
so...now what?
where did you get this problem from?
from a friend's book from their extracurricular class
this is a terrible problem cause you can only solve it through guessing
.
you should ask your friend what the intended solution is
i think it's a, b are real numbers
I mean, like what chapter this question comes from
and if there's a worked solution
what grade is this?
Advanced grade 9
if you know that a must be a rational number, then the rational theorem gives the possible values of a as:
plusminus (factors of 30)/(factors of 4)
that's still too many possibilities to check
I think this is a terrible question and you should skip it
okay
alright, if you're done for now, type .close
.close
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I'm just learning this now sort of, as I understand it, D) would be DNE, but what do I put for C??? the coords for the darker point? both?
When I first saw c i assumed it was asking what d is asking
but since it's not idk what it's asking
i don't know what to do about the hole
it's asking what the value is at x = -3
the darker point means its defined at x=-3 right?
isnt that just a removable discontinuity
therefore we have that it is discontinuous but still retains a limti
not at -3
that's a jump isn't it
but i guess i have the same question about m
or k i mean
since it has 2 holes at g(4)
Huh
how
my ass is not getting above 80 in this class 💀
isnt that point a removable discontuity bro
yes bro it is
no?
Why
oh wait
im regarded
sorry i meant to say
that is a jump discontuity
so the limit from both sides approaches diff values
so we have discontinuity and no limit for that point
did you read my question
yes
Bro i just wrote for part D
i was asking about part c
its just gonna be the y value of the dark point
I meant for part C
because the function is defined there
isnt it a jump disconuity?
and what about k
it is discontuityies
we dont care about limits or continuity for just evaluating the function at a certain x value
we just care about what the value is
if it is discontinuous how can it have a value?
same deal
does discontinuous mean not defined?
It looks like there's a value above?
at y=4?
if it is not defined at the point then how can it have a value?
It is defined
that's the point
okok so if it's not a hole then i put that answer right
on the test it'll be clearer
this is like bad scanning from a textbook lol
you just put whatever the value of the function is
whats confusing you?
does discontuity mean that it can still have a value at the point?
yes
just means its not continuous
okay thanks
how do i find the limit at 2 from the graph
unless i "estimate" at 4 but i feel like we haven't estimated anything in a long time
wait i'm confused
so u cant evaluate the limit
just look at what the y value goes towards as x gets close to 2
hs you are not helpful
do i estimate 4

probably
they wouldn't make it 3.99 or somethjng
that's dumb
is that the answer for all of these
okok
thanks so much
i feel like we've been doin a lot of equations and not a lot of pulling from graphs so the graphs are trippin me up
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Can someone check my working for biii)? \
let $P(x) = x^3 - 6x^2 + 9x - 15 = 0$; therefore $P(x+1) = x^3 - 3x^2 - 11$ thus $e_3(\alpha - 1, \beta - 1, \gamma - 1) = 11$
i cant find the answer sheet anywhere 
woomy
,w expand (x+1)^3 - 6(x+1)^2 + 9(x+1) - 15
yes
the logic is fine aswell?
yeah because the roots of P(x+1) will be alpha-1, beta-1, gamma -1
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A 5 x 7 rectanglebis partitioned into 1x1 squares. Then two opposite 1x1 corners squares are removed resilting in the figure. How many squares with integer lenght sides can be found using the line segments in the resulting figure
what's your logic for this
33 1 by 1 squares cus (5 x 7) - 2
Then 6 2 by 2 squares
Then 2 3 by 3 squares
i think there might be more than 6 fr
there are more ways to position a 2×2 square in there.
let me reproduce the grid, one moment
these are the possible locations of the southwest corner of a 2×2 square within the grid
if a 2×2 square fits in the grid, its SW corner has to be one of the green tiles.
this is the same but for 3×3 squares.
and this is for 4×4.
indeed. the question did not ask for the max number of non-overlapping 2×2 squares
just how many positions a 2×2 square could go in so that it's contained inside the grid
ok, let's go back: do you understand the diagram with the green color
No 😭
if a 2×2 square fits in the grid, its SW corner has to be one of the green tiles.
southwest as in bottom left.
Wait lemme try to process what u are saying 💀💀💀💀
i was wondering what u meant but then i realized that u were talking about the bottom left corner or southwest piece of the 2x2,3x3 and 4x4
like the positions they can occupy
yes thats in fact exactly what im talking about
Okok i understand it now
No 😭😭 i kept trying to think what u mean by bottom left
Then realise oh
When u said i thought like u meant the 2 x 2 squares can only fit onto thr green part
💀💀
Then i got confused on why the unshaded part cant be used
ok there was no need to take that offensively
i BOLDED the words "southwest corner".
chill
ok yeah it's whatever.
😭🙏
some of us are dumb and it takes time to process
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just a sec
✅ Original question: #help-27 message
@pseudo basin can u explain how to use combinitions to do this
How would u even use combinations here
💀
no like if i learnt that then i can do it for even larger ones
like 2025*2029 and 4x4 is removed
from corners
and i just got denied 
Lol
Anyways wait how do u use combinations in such questions
Im genuinely curious
i have seen some for like a rectangle
but never with pieces removed
so i wanna know
you know how to do it for complete rectangles I take it?
uh ic
for a full rectangle the limiting factor is whichever of width and height is smaller
or... ok not quite
i guess you kinda can make a formula for a rectangle of size m by n with m <= n but i honestly do not feel like writing it all out
here
I'll js explain the steps, simplification if upto you, alr?
ite
so to find total number of squares in a mxn grid
oh wai
we take cases finding total number of ixi squares such that 0 < i < min(m,n) + 1
@plucky marten @carmine veldt get your own channel
ah alr mb
itw
this used to be @compact hawk's channel
so yall need to take a new one in one of your names
.closw
.close
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What am I doing wrong? Everything I read tells me this is how to do it but I keep getting the wrong answer
I feel like I am losing my mind and am over this
I've tried this 8 times
I keep looking for info on what I can be doing right, but all the resources avalible say this is the right way to do it
<@&286206848099549185>
pretty sure that's abt torque balancing
but you can't really do it without knowing the weight of the diver
From what I understand, you get that by subtracting the two values but that might be the part that I don't get
@snow shore Has your question been resolved?
How should I find the weight
@snow shore Has your question been resolved?
@snow shore Has your question been resolved?