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Not sure what to do
,rotate
let d be the total distance travelled. then, we have that 24 = a(1) + b(d-1). then we just solve for d
do you understand how i got this equation?
cool
Tyy
nwnw
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help
it seems like you rotate the shape clockwise then perform a reflection across the y axis
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can someone help me with 11b
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have you done sums of arithmetic sequences?
Yes
letting a_n = -4n -3, there are 2 formulae that we can use, (i) n/2(a_1 + a_n), and (ii)n/2(2a_1 + (n-1)d), where in this case n = 31
have you seen these formulae before?
Yes the first one
okay cool, so lets use that
we have that n = 31 (summing 31 terms), now we need to find a_1 and a_31
how should we do that
correct
Then we use a of 1 to find a of 31
we can just directly sub in 31 into the a_n equation
372?
what were your steps?
Used the equation
nah we have found a_1 and a_31
now we use the formula
the sum from n=1 to n=31 is 31/2(a_1+a_31), which is...
tyyy
nw
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Before i go any further, have i set this up correctly?
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The answer i want is the smallest number HIJ
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can someone help
!help
To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.
!status
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
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sin(x)/x is wrong, yes.
You're going to have to open another channel, this one's closing soon.
Ok
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How does D(ln5) become 0
So you dont need to know what the answer is?
What if its some infinite decimal number
it is some infinite decimal number.
But even as it is, it's a constant.
Like those infinite decimal places won't ever change.
So just e becomes 0 too
yeah.
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How do I determine the convergence/divergence of the following sum
I know that the sequence converges to 1
Can I directly affirm that the series is divergent?
,w sum from 1 to infinity (k(k+2))/(k+3)^2
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why is my answer here incorrect?
i took the antiderivative of each term and plugged in 5
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So 0 isn't a purely imaginary number right?
just checking
depends a bit on your definition of "purely imaginary"
I would generally count it because most stuff that is true for purely imaginary numbers should also be true for zero
except for the usual problems with zero like division
it's also part of the sets of real numbers, as well as set of integers, so it is also a real number, and an integer
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someone answer these for me i can't wait till the grade comes out 
wymmmm
Bruh y more than 1 question it’s too much for my brain to process 
We don't answer questions, we help you answer them yourself
i already did i just need to get em checked
its pre calc bruh cmon
plsssssssssssssssssss
Post your answers and we'll verify, then
alr alr
just tell me if its correct or not
the grade won't come out until Thursday and i cant wait that long
@last anvil Has your question been resolved?
Look right to me.
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Im supposed to calculate the angle cfd.
Aefg and adbc are squares
Im not sure what to do I'd appreciate your help.
I speak German and English
This is what I've thought of so far
i have been trying it for like 15min isnt there any other information?
@lilac matrix Has your question been resolved?
It's 2c
It says
We have 2 squares abcd and aefg
D is on the extension of side fg
And b is on the extension of ef
Right now we're at the topic of
circumferential angle theorem (I google translated that I'm not sure if it's the right term)
If that helps
interesting ques.
Let P be the intersection between AD and EF, can you say something about the triangles AEP and DFP?
The angle dpf and the angle EPA are the same?
they are simmilar triangles but what do you get from that?
I'm thinking
EPD and apf are also the same of course but I'm not sure about what else
What about the lines AE and FD?
They are parallel!
Yes I wrote that on the picture
What does that mean for the triangles?
I don't know the English word but the 2 other angles are the same size
Yes, so all angles are equal, that's called similar triangles
What about side lengths?
wow I figured it out finally.. damn this is a really good question
@lilac matrix try to focus on the quadrilateral BFDC
that's your hint
No
I don't know what that means can you describe it in words?
I think we did do something like that
But how do I know that all those points even are on the circle?
Or are they always gonna be on a circle?
ok
so the opposite of 2nd point also holds true
What do you mean by that
We have two very different ways of solving this
@lilac matrix which one do you want to go for?
which means when opposite angles of a quadrilateral sum up to 180 degree then the quadrilateral becomes cyclic or in other words all 4 points lie on a circle
Which ever one is easier to understand
Arguably his is easier if you memorize this cyclic quadrilateral thing
Am I allowed to take a break and ask for help later again?
channel gets closed due to inactivity I think
you can do that..but I rarely open discord
Ok thank you for all the help I'll try on my own and come back later if I don't understand
@past sand you can explain her the solution if you are present at that time
Sure
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Can someone explain to me how this was turned into the second line?
idk crap about this subject but it seems like
the 3R(0) corresponds to X(1)^2 + X(2)^2 + X(3)^2 (as they mentioned in a, E(X(t)^2) = 3 = R(0) for any t)
2R_X(-1) corresponds to 2X(1)X(2) (delta in time of -1)
2R_X(-2) corresponds to 2X(1)X(3) (delta in time of -2)
2R_X(-1) again corresponds to 2X(2)X(3) (delta in time of -1)
but don't ask me what it means, I just tried to pattern match the start with the end
@opaque galleon
wait I think it didn't say for any t
they said at time t
the only thing it can mean is that the formula works for any time
ohh okok
@opaque galleon Has your question been resolved?
yeah you right
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a semigroup with an identity is a monoid right
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T
Problem 69
idk what it’s looking for or idk what “determine the numbers so that their product is the greatest” means
is it looking for the x intercepts ?
ultimately you want the vertex
the average of the x-intercept/roots tell you the location
ok
I did it right then ok
I thibk
was it when it said “their product is the greatest” = looking for vertex ?
after getting a quadratic after expressing the product in terms of a single variable, pretty much
yeh
y = -f (x)
we flip over the x axis
nvm got it
y=1/2f(x)
it getting wider ?
or smaler
nvm
got it
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Help
nerd
?
joke
Ok
i am gonaa live sirver
If I was a nerd I wouldn’t be asking for help lol
Anyway I do need help
<@&286206848099549185>
@hidden sparrow Has your question been resolved?
<@&286206848099549185>
lol I’m not getting help anytime soon it seems
<@&286206848099549185>
what is your problem with this exercise?
I don’t know how to solve it
And my approach is incorrect
So I’m lost
No help….
<@&286206848099549185>
Are you familiar with what the modulus/argument are for a complex number and what an Argand diagram is?
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can someone please explain how 1 - 1/e approx equals to -1/e
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if you miltiply by 0 you can just get rid of that whole term
0n = 0
am i missing something?
so it would just be:
1/2n^2 + 1/2n + 1/8 i believe
oops
didnt see the original question
it does, 1/8 just means remainder was 1 when you divided by 8
although idk how to feel about the 1/2 in the answer, i think since they are both odd it can be written to add the two 0.5s
0n + 8 = 8
4n^2/8 = 1/2n^2
4n/8 = 1/2n
1/8 = 1/8
basically just dividing by 8
and multiplying by 8 to see the remainder of 1
½n² + ½n + ⅛
8 / (4n^2 + 4n + 1)
4n + 0 + 0
4n + 0
1
and since we have half an odd number plus half another there must be some way to put them together so that we aren't halving anything
yep
so we have
½(4n² + 4n + 1) + ½(2n + 1) + ⅛
which is
2n² + 2n + ½ + n + ½ + ⅛
combine like terms
2n² + 3n + 1 + ⅛
still ⅛ which is a remainder 1 after division by 8
R1*
^
(1 * 8) + 1 is 9
because remember n is our odd number so n = (2n + 1) (yeah should probably change variable names lol)
actually we could san n = 2p + 1
so then we have p^2 / 8 = 2p² + 3p + 1 + ⅛ when p is odd
hang on
no i think im wrong actually
actually now im just confused
:|
ok, lets say n is our odd number. n = (2p + 1).
then we square n and get 4p² + 4p + 1
we divide by 8 and get ½p² + ½p + ⅛
ok, so we have ½p² + ½p + ⅛
½p² + ½p must necessarily be a whole number because if you half two whole numbers and add them together the 0.5 parts will add back together into a whole 1. this means that the answer is (some whole number) + ⅛ which proves that if you divide an odd number squared by 8 you get a remainder of 1.
im just not sure how to put it in figures
yw :)
hmm?
we still haven't figured out the first one
it depends if an explanation via text is accepted as opposed to figures
but it probably is
so i guess you could say we have
its basically remainders again, your proving that a^2 mod 10 cant be 3
because if a number ends in 3 then when you divide it by 10 the remainder will be 3
for example 13 / 10 = 1 remainder 3
yeah same idea
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all paths which go down by a row at each step are shortest paths, so it would start with 3 options, then in all cases have 3 options again. so its 3^<how many rows are there - 1>
what isnt
there are 3 options, but also there are 5 options and all kinds of options
oh man i didnt see that lol
yeah its more complicated than that
i dont understand what numbers your putting in, but this isnt my help channel and i should be doing school, so bye for now
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In need of help.
close this channel
.close
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/end
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is it true that $v(t)=v(0)+\int_0^t a(s)ds$
Jash
but wfa says its false
,w does f(5)=f(0)+integrate from 0 to 5 of f'(x)
Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online
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forgot how to do these
you multiply the frequencies by the midpoints, and sum them
then divide by the total frequency
i believe anyway, its been a long time
how do i find the midpoint 🥹
midpoint of 20 to 25 is 22.5 for example
oh
you can just find the average of the boundary numbers
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Hi can someone prove this formula to help me understand it better for my exam?
@dim quest Has your question been resolved?
@dim quest Has your question been resolved?
Better go to physics at #old-network
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is there a method for finding the eular path or is a lot of it just trial and error
I know you have to start and end with an odd node
which in this case is Z and Y
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help
Have you made any progress so far? Or is there anything from the original problem you don’t understand?
Maybe you could start by understanding what it means to be in Thom class (&Thom Space) and the normal bundle before you start the problem
Okay
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trying to expand this log expression, not sure where I'm going wrong:
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can someone explain how the ^15 turned into those?
maths 1 but yea it includes it
you can either convert it to the e^(i theta) form, or you can use this equation
sec lemme type it
You should rewrite it into polar form like bunny suggested
or use the binomial theorem if you hate yourself
$(a+bi)^n = \sqrt{a^2+b^2}^n(\cos(n\theta) + i \sin(n\theta))$
did i get this first try?
and of course $\tan(\theta) = \frac ba$
should be square root (then to the n)
lemme check that i wrote this out correctly though, see if it matches the work above
yeah i think this works out
@fathom flicker have you checked?
but like either way the "easier" way is surely just use polar form unless you want to memorize this

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I’m not familiar with that form off the top of my head
Seems plausible though
oh i meant to ping the question asker, my bad
im pretty sure it directly derives from the polar form method anyways
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is that graph a rational graph?
@worldly plover Has your question been resolved?
what is a rational graph?
I mean I just wanted to take one of those lines into account
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can anyone help me with a few questions on my pre-test? i have an exam coming up tomorrow and i really need to understand all this cuz im not confident at all
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
I don't know where to begin.
try to bring it to a simpler form
then you substitute (2x + pi/3) with u
and it turns into
sin(u) = 1
ok so since sine is a periodic function (meaning it repeats) there could be multiple solutions for example the solution is x = 90 deg but x = 360 + 90 could also be a solution
but after you get it into the form that i said, then u can be 90 deg, 350 deg, 610 deg etc, then you solve like 2 or 3 equations like 2x + pi/3 = pi/2 and check if x is in the [0, 3pi] range
i am trying to understand
gimme a sec
im still kinda confused
how do i find how many solutions
there are
you try it until x gets out of the range thats given to you
if you still dont understand we can go on a call
so should i call you?
yeah
okay
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Did I do this right?
wait isnt the decks corner where the two lines intersect?
Ah?
y = x-2 intersects y at -2 and x at 2
you made it intersect x at 1
also the other one y = -2x +1 is wrongly drawn
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Ohhh
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Why did he get two different answers using synthetic division versus polynomial divisions
Omg name
Nvm
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Hia
Short question
How do you know just because PX = 1/3 of PR, that the rest are 1/3 each
.
Also if you wanted to calculate XY would it just be -1/3a +1/3b
Becuase 'apparently' its split into 3
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How do i solve this
What steps do i do first
Y=mx+c
Also product of slopes of this one and perpendicular line is -1
Will it be 3y = 2x + 6?
Oh
How do i know if it’s -1
Divide both sides by3
yes
For perpendicular line...y=m2 x + c
Oh
Slope of perpendicular line
2/3 (m2)=-1
Ye
Ye
Ok
-3/2
So for like two..y=(-3/2)x+c
Ye
Will 3/2 be positive then
Flip sign?
like it’s 2/3
Why flip sign
No need
-3/2
ye
.
Now (6,-1) lies on both lines
We substitute
They don't know it...im helping derive intuition
do you know slope point form of a line
Yes
Of course u can derive standard results
no thats just slope intercept
It's almost over
ah right
Wait looks like u do
i replace y and x
Ye i learned but i forgot
But i renember some of it
you could think of it as rise over run for this line is ofc -3/2
What’s the easiest way to do
assume a point x,y
I’m confused what are we talking about
Why is there extra "y"...u already substituted value
Then what is it supposed to be
What
thats the line
Oh
DW it's over almost
Ok
-1 = -3/2(6) + b
?
is this rifht
then
Or what is the right equarion
what
i just told you
Find b
-1=-3/2x +c
Oh
so y=-3/2x+8
Which equarion is easier
Over
Percy or this one
Percy
Oh
slope point form is a derived result
Whag is derived
Oh
Let me redo but
Let me try to do the steps
So first i change the equation to
Let me try and i will see if i get 8
That's better
This algebra video tutorial provides a basic introduction into point-slope form. It explains how to write a linear equation given a point and the slope. It also explains how to write a linear equation given two points. Finally, it discusses how to graph a linear equation in point-slope form.
Access Full-Length Premium Videos: ...
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Lighting...
U need a better picture?
Yea or send the question
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<@&286206848099549185>
is it possible to get some help or?
been waiting for over an hour
Hello. Notice that it's 1/2 -1 = -1/2, so the exponent in x and y is -1/2
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oh Okay but What will i do further down to solve the question?
you can isolate lambda in one eq and substitute in the other one
also use 3000-x-5y = 0
for example, x = 3000-5y
How do i isolate the lambda?
you have
50/sqrtx - lambda = 0
and
50/sqrty - 5lambda = 0
you can isolate lambda in one of these 2 eqs
and replace in the other one
oh okay and then use the 3000-x-5y after
yes
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As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
sorry, plus u wont get better if i give u the answer
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help me please
Stay in your original channel please
You closed it after i sent my message. Also, you should generally stay in the channel you first opened instead making a new one just because no one came to help you quickly enough.
idgaf
People are less likely to help you if you abuse the system or act hostile
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I'm trying to understand this part of linearizing.
I'm not sure what is happening here
why they make that substitution y = p+u ?
what's that u ?
its from this page
https://ocw.mit.edu/courses/18-03sc-differential-equations-fall-2011/resources/mit18_03scf11_s11_1text/
I think you need to give more context yh
yh ok
dont think i can help out myself
ok so i understabnd something just reading the screenshot
it's a recap of diff equations studied so far in the course basically, in the form of a conversation
The idea is to center at your equilibrium
so, redo the axes and make the equilibrium the origin
thats the effect y = p + u has
p is your new origin
u will be the relative change from that origin
Don't know if that made any sense
mmmm kinda but still forming the image in my head
p would be the value at equilibrium basically
cause i don't care what happens in the transient
right?
If you were to compute this, a sensible thing would be to 'center' the limit at 0
that is making the substitution u = x - 1
I think the same idea is being used here
mm ok
x = 1 + u
is another way to write this
and thats just like the y = p + u here
i'm not used to think this way but i understand
i mean never had to recenter to compute the limit
but ok
not sure why it would make more sense
i think i understand the parallel now.
why would be better to center it?
Whenever you focus on a random point
isnt it better to make it the new origin
In this particular case, try and you will see it makes things a bit easier
This is a bit of intuition that makes sense to me though, having encountered the same idea in many places.
Topology, algebra, analysis etc etc
im trying yes
If you only care about the change relative to a point
then that point is essentially your origin
If you only write down the change
then it is.
why u = x-1 ? that 1 is from the limit or from the numerator?
i'm not sure i understand the processing of ideas here
you see that function and you see ... something, the makes you think , i can shift this to origin!
what do you see?
cause even if i have just u in the numerator the denominator becomes messier.. how am i shifting this?
this is super interesting i didn't find it explained so far...
could you tell me what should i look for to study this or make practice over this? so i don't keep spamming the server
I think you're overly worrying about the change of variables and missing the point that is interesting to begin with.
You have some function y(1-y/p) = y- y^2/p and as y tends to p. This tends to 0 as y tends to p, but that doesn't really give you any insight into what's going on near p.
If your limit was near 0 instead, you could use an approximation by dropping the highest power of the polynomial, so any change of variables that accomplishes that would achieve the same goal
From y(1-y/p) you could just use u = 1-y/p so that y = 1-u/p. All I did was choose a linear function of y that was easy to substitute into the equation, then you have (1-u/p)u and as y tends to p u tends 0 so that you can drop leading terms.
The whole point was the strategy of changing the equation to study a polynomial near 0 so you can drop a leading term to simplify. The substitution itself wasn't particularly magical
thanks for taking the time!
i don't understand what you mean by
"that doesn't really give you any insight into what's going on near p."
what sort of insight you mean?
It's not obvious what the behavior of the differential equation looks like near p
so in the end i would get this
and you say "as y tends to p u tends 0 so that you can drop leading terms."
why y tends to p?
ok i think i see, its for the substitution you made
u = (1-y/p)
i think i see what's happening here now, can you confirm @remote skiff ?
i'll look up for linearization i think that's just a topic that has not been explained yet
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D is middle point of AC, and i need to find DEBC area
is 12 the length of AD or AC
Ac
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find the area of ADE and subtract it from the area of ABC
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is my proof correct? the instructions were:
looks good but you carry (k+1)/(k+2) the entire way through
you dont have that equality from the start, so only write it at the bottom when you show it
otherwise it seems like you're assuming its true, which is what you're supposed to show
I don’t understand the line where there's (1/k+1) on the left, where are the parentheses ?
Ah, ok, sweet thank you! I feel great right now because I haven't been able to understand induction for the last couple of days
And how do you go from that line to the next ?
I was just showing that I was factoring out (1/k+1), I don't think it was necessary but it was part of my workflow to keep it straight in my mind
I repeat this question jic
(k/1) + (1/1) = (k+1)/1
(k+1)/1 * (1/k+2) = (k+1)/(k+2)
Did you forget order of operations?
And this would also imply the 1/(k+1) you factored out disappeared
yes 🤦♂️
Ahhh I see what you mean now
The way to do it : put everything on the same denominator
my factored term is my +1, not the (1/k+2), correct?
Ill write it with parentheses so we all agree
$\frac{1}{k+1}\Big(k + \frac{1}{k+2}\Big)$
rafilou2003
This is your 2nd to last line rewritten with parentheses
Now try to put everything on the same denominator
$\frac{k}{1} + \frac{1}{k+2} = \frac{k^2+3}{k+2}$
You didn't do it properly
Am I dumb am I failing algebra rn
yeah
$k = \frac{k(k+2)}{k+2}$
k(k+2) = k^2 + 2k
rafilou2003
And then you add it to the other fraction
damn I am forgetting all kinds of things
Ok so can you edit this to get a correct statement?
$\frac{k(k+2)}{k+2} \cdot \frac{1}{k+1} = \frac{k(k+2)}{(k+2)(k+1)}$
ronin
$\frac{k(k+2)}{k+2} \cdot \frac{1}{k+1} = \frac{k(k+1)+1}{(k+2)(k+1)}$
ronin
Why did it become a multiplication instead of addition?
I dont know 😭
Complete this statement: $\frac{k(k+2)}{k+2} + \frac{1}{k+2} = ...$
rafilou2003
k(k+2)+1/(k+2)
Yes, can you expand the numerator?
Yes, does this seem familiar?
No to be honest
look at your quadratic identities
Try to factor the numerator is what I'm hinting at
(k+1)(k+1)
Yes
then (1/k+1)[(k+1)(k+1)/(k+2)] becomes k+1/k+2
Yes
Wow my algebra is rusty, thank you so much
ngl I was like "why so much rigor when can I start handwaving" 😭
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Does anyone know anything about the college classes: Topics math and Contemporary math
What is your question bro
Possibly, though you'll probably have much better luck if you have specific questions related to stuff within those classes... 
I have a video on my watchlist from computerphile something about origin of bytes
@warm canopy 
I don't think I have anything useful that I could say myself 
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i have no clue how to do this
thales ?
we can use the fact that T and W are midpoints of the legs
meaning, the increment in length from SR to TW is the same increment as from TW to UV, afaik
here we can see that on the right, the addition green length is twice the additional blue length
same works for the left side
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Mb😭I got the option of either taking College algebra/pre calc or topics math/contemporary math
I want to know which is simpler
If it’s easier than the above
I have no clue what the classes consist of
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Might be a better question for the discussion #math-discussion #discussion channels, these channels are more for like specific homework flavoured questions tbh
ok ty
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Sorry for the pic in spanish. I've been trying with a group of friends to solve this, however we just get constantly stuck. There's no intersection between S and v or H and v, and we need to somehow get a 3 vector subspace to transform into a 2 vector subspace, while having that second subspace transform into a single vector.
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What really bothers me is that you cannot do something like SintH to Sintv
<@&286206848099549185>
<@&286206848099549185>
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help
how awful that they make you write this using text
it doesnt really seem like youve used enough parens
not that this will make it more legible
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