#help-26
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well the math is right
i just messed up with the hypotenuse
its -1 not 1
but the math is right hooray
What π
idk bro
Length of hypotenuse can't be negative π
quadrant 4 so its negative
what
π
Like it's a side of a traingle
Also
Nevermind that
I must mention
hows the answer neg
Sin is negative in the 4th quadrant
ah
yes because sin is y-xis
so because its sin its negative?
First of all
Which makes cos(-45Β°) = (-1/β2)/(-1)
since its in quadr 4
a is side length π
yes
Wait
no
i assumed it was 1 because its the radius
Seems correct
but the answer is negative so idk
Sin will be 1/β2/1
what
Opposite side / hypotenuse
Sin will be , (1/β2)/-1
Hypo is negative
this guy said it cant be negative tho
Isn't 1/β2 also negative π
I confused your method with a different one
oh
Proving it in general is easier for me
wait so its (-β1/2)/-1
Idk man
Sides of a triangle shouldnt be negative π
broo
WHAT OD U MEAN NO
Imaginary number
ahhhh
dnvbiuvweoiwfefhbovwnm
okay
how is the answer negative then
when does it turn negative
a^2 + a^2 = x^2
a= 1/β2 x
Sin(-45) = - ( 1/β2)(x) / x
sin (-45) = - 1/β2
thats the only way it would make sense
This is what I'd generally do
Maybe
oh my god bro im gonna kms
ok whatever next question
Isnt this just the y value of point A?
sqrt3/2
teacher doesnt put the answers on the slides
kinda dumb
Yep
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How would i know what x is when sin x is 3/5
General solution?
as y = sinx
just put values
you need to know inverse trignometry to understand it
as a = arc sin(3/5)
Result:
0.60181502315205
,w sin(37deg)
approximately, but yes
but note that for the question that you showed, getting the exact value of angle a is not necessary to answer the question
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ok so
you see the second deravtive graph
how does that show the maxima and minima
its just a straight line
actually first derivative graph determines point of max or min
second derivative graph is for concavity
look at the first derivative graph and see how the roots line up with the original graph's local min/max
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help me
i multiplied the recpricol of -1/20
There is multiple ways of doing it.
and then once i did that i got (x+75)^3/2 = -200
and then powered both sides by ^2/3
so i got x+75 = -200^2/3
but if you looked closely, the right side is -200
.
yeah
yea -200 to the poewr of 2/3
which can be written like sqrt(-200^3)
$sqrt(-200^3)$
there is no solutions from what i got
ok same
cuz i got -10 and 10i when i plugged x back into the equation
so that means no solutions correct?
cuz for me i got X = -75 Β± β40000
because you got (x+75)^3/2 so its not negative but left side is negative
yes
so then for this eqation: sqrt(5x+6) = 1+ sqrt(3x+3 or 2 i forgot)
i got 2 and -1
but when i plugged -1 back into the eqatuon i got 2 different decimal ansewrs
but when i plugged in 2 i got 4 on both sides
is it 3 or 2
yea wait its 3
i might have plugged in -1 wrong in the calculator tho cuz i was running out of time
it might have been -1 aswell
Incorrect
It's $\sqrt[3]{(-200)^2}$
sqrt
SWR
Step 1: Start with the equation
we doing sqrt(5x+6) = 1+ sqrt(3x+3 now?
yes
Five x plus six equals one plus the square root of three x plus three.
i got 2 and -1 but when i plugged in -1 it didnt give me the same answer on both sides
but i feel like i might have plugged it in wrong on the calculator
You forgot the minus
No. There is no plus or minus
Oh
400
You are right
I be mixing them up too 
But this still stands I'm quite sure
cuberoot 40000
34.19 is what i got
This is incorrect
but is it an irrational?
wait but in my math test am i suuposed to know if there are parenthesis when it says -200^2
You deduce this based on your own workings.
π π π π π
wait im so confused
Stop spamming please
ok
ok but how do i know if there are brackets if i get to -200^2 from evaluating
is that given or something
would either one of them give me 10
from what
cuz it was x -75 Β± cuberoot(40000) or is tehre a different number given if it was cuberoot(-40000_
The parentheses must be there. Because at one point, you alegraically got to $(x+75)^{3/2}=-200$. So when you raise both sides to power $2/3$, that exponent is applying to all of $-200$, which you represent through parentheses
SWR
Again. There is no $\pm$
oh i see
SWR
so if you subsituted X into the equation with 40000 cuberoot would it equal to 10
No
there is no solutions
No. You forgot the 75
cuz wehn i plugged x in i got -10 and 10i
misclick?
and the -75 i was about to add
Incorrect. There is one real (irrational) solution
what is it?
Ah you are right 
really?
I forgot the square root nonsense
wait the problem is on the test i wrote x = -75 Β± cuberoot(-40000)
Step 1: Start with the equation
Five x plus six equals one plus the square root of three x plus three.
Step 2: Isolate the square root
Move one to the left side by subtracting it from both sides:
Five x plus six minus one equals the square root of three x plus three.
This simplifies to:
Five x plus five equals the square root of three x plus three.
i showed work though
Algebraically, you would arrive at $x=\sqrt[3]{40000}-75$, but this would not work. You would end up with $-10=10$
Step 3: Square both sides to remove the square root
Square both sides of the equation:
The left side becomes the square of five x plus five.
The right side becomes three x plus three.
Step 4: Expand the square
The square of five x plus five equals twenty-five x squared plus fifty x plus twenty-five.
So the equation becomes:
Twenty-five x squared plus fifty x plus twenty-five equals three x plus three.
Step 5: Move all terms to one side
Subtract three x and three from both sides:
Twenty-five x squared plus fifty x minus three x plus twenty-five minus three equals zero.
This simplifies to:
Twenty-five x squared plus forty-seven x plus twenty-two equals zero.
Step 6: Solve the quadratic equation
Use the quadratic formula to solve for x. The formula is:
x equals negative b plus or minus the square root of b squared minus four a c divided by two a.
Here, a is twenty-five, b is forty-seven, and c is twenty-two.
SWR
oh shit wrong question
Consider a simpler problem: $\sqrt{x}=-1$. If you square both sides, you get $x=1$, but that clearly is not a solution.
SWR
nah this is on my top 10 dumbest moments
Me too
ok what about the second question
Half awake on mobile one eye open. Not at my best
i got 2 and -1 but i said -1 was an extranueos solution
This second question?
?
Start with
negative one over twenty times the quantity x plus seventy five to the power of three over two equals ten.
Step one. Eliminate the fraction.
Multiply both sides by negative twenty to cancel out negative one over twenty.
x plus seventy five to the power of three over two equals ten times negative twenty.
This simplifies to
x plus seventy five to the power of three over two equals negative two hundred.
Step two. Check the left side.
The power three over two means taking the square root first and then cubing the result. The square root of any number is always zero or positive. This means x plus seventy five to the power of three over two cannot equal a negative number.
Final answer. There are no real solutions to this equation.
this is what i did
you gotta eliminate the sqrt first
.
$\sqrt(5x+6)\ = 1+ \sqrt(3x+3)\$
```Compilation error:```! Missing $ inserted.
<inserted text>
$
l.50 \end{document}
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.```
Never seen someone here write math in all words like this 
from ELA
$\sqrt{5x+6}=1+\sqrt{3x+3}$
SWR
i think imma go now
Why is -1 extraneous?
cuz when i plugged it in the calc i got on the left side a 3.1irrational and the other side i got a 3.4 irrational
but i think i may have plugged it in wrong in the calc
ig not
it not
granted when i was checking the answers the class had ended like 5 seconds ago and i had to give the paper
Yes. You may have
nooo
what was wrong
for the first question i providede i said x = -75 +- cuberoot(-40000)
use your time to double check, heck, even triple
when it was no soliution
bru
and the second one i said x = 2 and not x = -1
and then i dont know if i needed to rationalize the denominator for one of them
but its ok.
cuz i had to find the inverse of 3x^8 = y
there is always another day
Hard doubt
cuz the next question was asking to see if they were inverese and it was f(x) = 2x^5 and g(x) = 5throot(x/2)
and the fifth root was not rationalized
and also is the rationalized verison of 8throot(x/3) 8throot(3x)/3
?
quick question, whos your fav mathmatician
Nobody
welp
Maybe Sal khan
yeahhhhh
lol
$\sqrt[8]{\frac x3}=\frac{\sqrt[8]{3^7x}}3$
SWR
guys
i might have read the question wrong
my freinds are sayig it was 60 - (1/20) * (x+75)^3/2 = 10
and they got 25
i swear to GOD on my paper there was no 06
60
it must have gotten cutten off in the printing no way

@rocky crane Has your question been resolved?
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Am I correct in thinking that what the author means by "the same" is really "proportional"?
Nope, the same is correct
By "the same linear combination", I guess the author just means that the coefficients that multiply $\mathbf{x_1}$ and $\mathbf{x_2}$ are the same as the coefficients that multiply $f(\mathbf{x_1})$ and $f(\mathbf{x_2})$, respectively.
ucheo
omg i love that font
whats the font pls?
You and me both. I love it, too. I specifially thought... "I love this font, I wonder which one it is". I tried to find the name, but couldn't. However, the font I'm using for my own notes (by coincident) is either very similar or the same.
I can get you the name of this font I'm using...
Font: Cambria
I'm also using "Cambria Math" for the math symbols.
i remember i have an alt in this server with that latex font, let me check
dont ask me why
Hmm, you're saying the same a and b multiply both x and f(x), but he's talking about the function itself.
Not the expression before the function is resolved.
Or maybe you mean the multiplication by the constant inside the function?
i think its eulervm font
Thanks for the name. Am gonna save it π
A linear combination of two vectors $\mathbf{x_1}$,$\mathbf{x_2} \in V$ is an expression of the form $a\mathbf{x_1} + b\mathbf{x_2}$, where $a$ and $b$ are elements of the field over which $V$ is defined. So a linear combination can be thought of as an operation uniquely determined by the coefficients $a$ and $b$. If $f$ is a linear transformation, then $f(a\mathbf{x_1} + b\mathbf{x_2}) = af(\mathbf{x_1}) + bf(\mathbf{x_2})$. In this sense, the linear combination determined by $a$ and $b$ remains unchanged under $f$.
ucheo
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welp after searching in google it seems to be nazanin
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How do i find the greatest common divisor and least common multiple of (xΒ³-3xΒ²+4x-12) and (xβ΄-16)?
I just don't know what to do with x
it GCD x^+4?
Factor both terms first
I know how to factor but I don't know what exactly i should do with the x, where should i put it?
Th x represents the variable and stays within the polynomial during calculations. To find the GCD and LCM, factor both polynomials, identify common factors for the GCD, and combine all unique factors for the LCM.
Show your factors
are you sure you factored correctly?
Soo for the 1st one i should tuen it into (x-3)(xΒ²+4) and the 2nd one into (xΒ²-4)(xΒ²+4)?
Ohhh okayyy
Is there anyone willing to help me on a large amount of geometry π i canβt understand it at all and itβs the only class i have left
From (xΒ²-4)(xΒ²+4) to (x-2)(x+2)(xΒ²+4)?
ill do a portion. claim your own channel
Dw i have time
Ok thank you
you can make the first one (x-2)(x-2)
LCM=(xΒ²+4)(x+2)(x-2)(x-3)?
Oohhh okay thanks
Wait no then wouldnβt you be left with -4x
yes
? lemme see
(x-2) (x-2)
x * x = x^2
-2 * x = -2x
-2 * x = -2x
-2 * -2 = 4
x^2-4x+4
So it can just stay as (xΒ²+4)
?
Yeah I mean can u think of any other way to simplify
No I can't
Thanks for helping me on this one
I also don't know what exactly to do for another exercise tho
GCD(a,b)=48
a+b=576 is another exercise I don't understand
Lemme fix it rq
You should be fine but also you could also do (x^2+4)^2
Wait show what u mean
Thatβs when signs alternate
Well, discord aint
Soooo what do i do with this thingy
What is the context
Is this a linear algebra question ?
What values can a and b be
be back in 5 min
No idea π
Oh I was thinking when you originally wrote D (a,b) = 48 that it was the distance between 2 points
Yes i accidentally did that
So what is GCD mean ?
I meant to write the greatest common divisor
Like GCD(6,2)=2 if im not mistaken
Itβs saying a and b are both multiples of 48
And you just have to find what two values that are multiples of 48 = 576
Wait
I got that wrong
Or wait it may still apply
Yeah show just in case Iβm misunderstanding
Yeah itβs pretty much just the explanation I gave
This one
Umm lemme try
a= 48x, b=48y
a+b=576=
=48x+48y=
=48(x+y)
576Γ·48=12
x+y=12
Then i find the values?
x could be 1, y could be 11
Anddd
X could be 5 and y 7 i think
So then a=48Γ1 , b=48Γ11 or a=48Γ5, b=48Γ7
I think?
my brain aint braining this
She got convincing results
Since both numbers have 48 as GCD, you can just write them as 48x and 48y, and you have the info of a+b=576
Equate, 576=48x+48y
Factor 48 and divide 576 by it
aww man i hate algebra 2
Are you to get specific values for them or not
oh wait this is algebra 1
I mean if i multiply the x and y that i get with 48 it matches the answer
Any expression for 12 will be the answer
Even negative numbers
Smth like -18 for a and 30 for b will fit
I maybe forgot to say it says that they have to be natural numbers
My question is, is that ok or do you need specific numbers
You can then exclude
For a and b
Like does a need to be a specific value
I think that they have to be composite numbers
Idk how it correctly translates into English
Does the question say smth like: find βall possible valuesβ of a and b, or find βthe valuesβ of a and b
Non prime?
Like those who don't share a common divisor apart from 1
Like 1 and 11
And 5 and 7
Ohhh
They only divide by 1 and themselves
But
1 is not prime
1 is just excluded
So you only get 5 and 7
Because it only has 1 divisor?
Ohhh okay
Anytime
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WAIT

Are the prime numbers values of a and b or x and y
Should be x and y
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I keep getting C but its supposed to be E?
I use a u substitution
where u is x-3
and du is 1
so its the integral from -1 to 1 of 1/u^2
which i then get as finding from -1 to 1 -1/u
which is just -1 - (-1) or 0
in the interval [2,4] there is a point where the function is not defined
x=3
Then you must calculate the integral from 2 to 3, and from 3 to 4
oh so if you take the definite integral of a function which contains an undifiend point, is it always N/A
*DNE
probably
did you change the bounds of integration?
yes you did, sorry
yeah i did
yeah DNE
wait ur right
theres an asymptote that goes to infinity
thank you!
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how to find the gcd (n, 209)
those three informations are given , forget about the french
i know that the the $gcd(n,209) \ | \ n$ and $gcd(n,209) \ | \ 209$
<rajel />
does this mean n ~ 4 mod 19
yes
so final one says n ~ 137 mod 209
that means n don't divide 209
so we have gcd < 209
idk this but from n ~ 137 mod 209 you get n = 346 which satisfies the first two conditions too
how did you know
oh make sens
appears that the first two eqs lead to the mod 209 one
209 = 11 x 19 if that helps
so the gcd is bigger than 209
so gcd can only be 1, 11, 19
how does that follow?
first two eliminates gcd being 11 or 19
so I think it easily equals 1
ah my bad got confused a bit
gcd is the greatest common divisor
so like it must divide both n and 209
does a number bigger than 209 divides 209?
no
so we can say gcd<209
given that the gcd can only be {1,19,11} , after verifying it seem to be only 1
because none of the other values divides n
takin 19 , which have remainder of 4 so it doesnt divide , and takin 11 it has a remainder of 5 non of them have a remainder of 0
yep
I'm 90% sure that it's gotta do something with the Chinese remainder theorem
ok thx
i'll search for it thx too
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do not ping us for help or when there's nothing going on please
The helpers are volunteers
yes, he's not obligated to help you at all 
They don't have an obligation to help you within some time frame
Depends on a lot of things how fast this happens
It is just random internet strangers after all
unrealistic to expect this given that people here are volunteers, please be empathetic
use a calc?
Why not just check on a calculator?
you have access to the internet
You have an internet connection
Nice job, admitting this is a ban-evasion account 
You can use the command ,calc
Yeah fair lmao
Anyways, stop trolling please.
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Let $a_n$ be a sequence such that ${a_n : n \in \mathbb{N}} \subseteq [a, b]$. Let $\epsilon > 0$, a sequence $a_n$ is epsilon-dense if for each $x \in [a, b]$ and for each $N \in \mathbb{N}$ there exists an $n \ge N$ such that $|a_n - x| < \epsilon$
jason2D
so if i understood that correctly, that just means that every number in the interval is a sub limit right?
this channel is occupied
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but this is impossible isnt it?
sounds right
I don't think so since this is for one specific epsilon
no
someone else will be able to surely
well the questions continues to say "given a sequence a_n that is epsilon-dense for any epsilon > 0..."
but how can every number in an interval be a sublimit?
the cardinality of an interval over R is larger than the cardinality of N
this is not true of an epsilon-dense sequence
what does it mean then
here's a crude drawing of a (b-a)/4 dense sequence for instance
but the sequence i have is epsilon dense for any epsilon
it means that for any point "x" in the interval, you can always find arbitrarily large n such that a_n is at a distance < epsilon from x
okay that makes more sense then
how so
I didn't read that bit lol
because then what you can do is build a subsequence $x_{n_k}$ such that $|x-x_{n_k}|<1/k$ for each $k\in \bN$
derivada.schwarziana
but thats just all of the rationals
by picking $\varepsilon=1$ to get $x_{n_1}$, $\varepsilon=1/2$ to get $x_{n_2}$ and so on
derivada.schwarziana
not necessarily, another example of a sequence that's epsilon-dense for any epsilon is all rationals that can be written as $a/2^m$ for some $a\in \bZ$ and $m\in \bN$
derivada.schwarziana
this is a proper subset of the rationals that is dense on $\bR$
derivada.schwarziana
you can check that if $a_n$ is $\varepsilon$-dense in an interval $[a,b]$ for all $\varepsilon>0$, then the set ${a_n: \ n\in\bN}$ is dense in $[a,b]$
derivada.schwarziana
so indeed every point of the interval is a limit point of the sequence
but i still dont understand
how can a sequence have uncountably infinite sublimits
when there are only countably infinite subsequences
the idea of "sublimits" of a sequence, or limit points of a set, is simply that they can be approximated to an arbitrary degree of error by elements of the sequence or set respectively
a priori there's no reason why a countable set cannot approximate an uncountable one
and in fact that's a thing we often desire in maths, e.g. the https://en.wikipedia.org/wiki/StoneβWeierstrass_theorem gives you a way to approximate any continuous function by polynomials, and hence particularly by polynomials with rational coefficients which form a countable set
I guess another way to think about the density of $\bQ$ in $\bR$ is by looking at decimal expansions, suppose we wanted to approximate [x=\pi=3.14159265\dots]
derivada.schwarziana
one way to do this by rational numbers is by truncating the decimal expansion[x_1=3, \quad x_2=3.1, \quad x_3=3.14,]and so on
derivada.schwarziana
and it turns out you can do this for all real numbers
but isnt this the oppsite?
... and that the rational numbers are countable, so we can write $\bQ=(q_n)_{n\geq 1}$ for some sequence $q_n$ (called an \textit{enumeration}). So this is an example of a sequence which has any point of $\bR$ as a sublimit
derivada.schwarziana
how so?
this says that you can map from an uncountably infinite set to a countably infinite one
no?
not quite, it's saying that any real number is a limit point of the set of rational numbers which is what we want to show
it's true that in some sense we're mapping the limit (the real number) to a sequence that approximates it -- we kinda need to show how the sequence can be constructed
basically we want to define a sequence q_1, q_2, q_3, ... that counts all rational numbers right
what youre saying is, for any real number you can create a sequence that approaches it
yeah
yes, a sequence of rational numbers
right
yeah no stil
its true that you can create a sequence of rational numbers that approaches any real number
but you still have uncountably many sequences like that
yes, your original question was how is it possible for one sequence to have uncountably many sublimits
yeah i still dont understand that
which is why I brought up this as an example: you can actually define this sequence q_n explicitly as shown in this diagram, from https://en.wikipedia.org/wiki/Rational_number#Countability
so q_1=1, q_2=2, q_3=1/2 and so on, following the arrows drawn in the diagram
yeah we used that to prove that Q is countably infinite a while back
it can be shown that this sequence passes through all of the positive rationals
right
and from there I think doing an approximation scheme of this sort is not hard to see
take a positive real number, suppose we know its (possibly infinite) decimal expansion
wdym approximation scheme
oh just a way to define a sequence of rationals that converges to it
so we can do this, and then define a subsequence q_n_k by taking the corresponding fractions of denominator 1, 10, 100, and so on, that describe the decimal expansion truncated to k digits
and since this can be done for any positive real number, the sequence q_n has uncountably many sublimits right
I haven't been too rigorous explaining this but hopefully that gives an idea of how this kind of thing can happen
so for example
for the number 0.12345678...
our subsequence would be 0.1, 0.02, 0.003, 0.0004...?
I'd say 0.1, 0.12, 0.123, 0.1234, etc.
since you want the sequence to actually converge to 0.12345678...
but what if the number can be truncated to an irrational number?
and this can be written as 1/10, 12/100, 123/1000 which respects the order of the sequence q_n but that's not too important -- just shows that it is a valid subsequence since q_n goes first through all fractions of denominator 1, then 2, and so on
that doesn't really matter I think, you can always take rational truncations like this and have it converge

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i know that when g(x) translates 1 unit to the left that it becomes g(x+1) but how would i use that to create an equation of the new curve?
do exactly what you said
replace x with x+1
then expand and simplify if you want to do so
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oh correction S1 wont be there
for obvious reasons π₯²
,tex
$S_k$ is the sum of the infinite G.P. where first term $a_k = k$ and common ratio $r_k = \frac{1}{k}$ $(k\in\mathbb{N})$. Find the value of $S_2^2+S_3^2+...+S_{2n-1}^2$.
NightHaunter
i tried partial fractions to turn S_k into a telescoping series but didnt work
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you could get two equations and solve for the intersection point
the Cartesian equation first, so $y = -\frac{1}{2}x + 2$
south
which becomes $r \sin \theta = -\frac{1}{2} r \cos \theta + 2$
south
solve for r then use (x, y) = (r cos theta, r sin theta)
definition of polar form
jeez
this is section 1 on polar and parametric
basically
cool
Iβm saying how do you know the coefficients are equal
it's a change in coordinates
this applies to any equation when you write it in polar form
How do we know it isnβt an elliptical path?
I guess it makes sense
I can draw a circle of some radius and some point on that circle to get any point on that line
Is that the idea?
Convert a polar equation into a cartesian equation: circle!
Convert r = 6sin(theta)-2cos(theta) into cartesian equation.
Polar equation, polar curve of a circle,
polar to rectangular conversion,
polar to rectangular conversion with completing the square
S. Chow
www.blackpenredpen.com
maybe these examples also help
yeah!
Alright much appreciated
Iβll check that out after this question
like you can describe any point on the plane with
- distance to origin
- angle
that's the idea
Got it
yeah thatβs fair
Alright back to the question
Lemme try and solve what you sent rq
I used a on accident and was too lazy to change so a = r
Is this the right idea?
@smoky sparrow
Top equations are seperate
Kind of mushed
right idea
Wrong answer?
but you can just use (x, y) = (r cos theta, r sin theta) after you have found r
anyway I'm sure they're equivalent parametric forms
take r cos(t) = $\frac{2 \cos t}{\sin t + (1/2) \cos t}$ for example
south
if you multiply top and bottom by 2/(cos t)....
also you can realise
$\frac{y}{x} = \frac{r \sin \theta}{r \cos \theta} = \tan \theta$
south
so $x \cdot \tan \theta = y$
south
it's no coincidence
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so
PC=PD then triangle PCD isosceles
M midpoint DC then PM perpendicular on DC
OC=OD then triangle OCD is isosceles
M midpoint DC then MO perpendicular on DC
so P-M-O are colliniars
here i get stuck
<@&286206848099549185>
<@&268886789983436800>
Pls no ping mods for math help
have you drawn a diagram?
yes
could you post it here?
@errant hamlet Has your question been resolved?
AB is radial axis for C(O,OC) and C(O', O'O)
we can proof
so AMOB inscriptible
thats what we need to proof
<@&286206848099549185>
btw D is not always the diameter unless im missing smth
@errant hamlet Has your question been resolved?
@errant hamlet Has your question been resolved?
@errant hamlet Has your question been resolved?
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"show an example of at least one set A containing numbers such that
|A| = 4
for any x,y in A ,x^y and y^x are also in A where x !=y"
idk if this is valid but Z_5 = {1,2,3,4} works
im curious if theres some other solution
I guess it depends on what x and y are
is the operation ^ supposed to be the one on reals, or can you define them however you want?
{1,2,3,4} is consistent with the rules given if we have freedom to choose what x and y are
huh?
oh wait its supposed to be "for any x,y"
but its modulo 5
The way it was initially phrased states "if x,y are in A" which doesn't imply that they are in A
stuff got lost in translation, sorry
ok
x,y could have been elements outside of A, but OP has now clarified that they are definitively in A
uhhhh but the question is talking about exponentiation, it doesn't mention exponentiation modulo 5
yeah it doesnt
just invent your own number system where you define x^y = 0 always
if this is from some textbook, can you ss the original problem?
its in thai hold on
its from a competition
maybe google translate sucks ill have gpt translate it for me
is it "x^y in A or y^x in A"?
yes this is what i'm suspecting too
but i don't think google translate is that dogshit it can't distinguish between "and" and "or"
i got that dollar store translate π
okay, so firstly, unless the question allows you so, you shouldnt just invent your own number systems to solve the question
oh alr
mb
i tried for a while with some numbers and it didnt work
okay thanks
seems more possible now
now try experimenting with it a bit
its quite easy now
keep in mind that you dont need both x^y and y^x be in A, you only need one of them
yeah, i had this one in mind
that works
i didnt fully verify it but i think it works
any pair with 1 obviously works
because x^1 = x which must be in that set
question was just hard because i translated wrong lmao
i think 0 should also be too
Yep, any pair with 0 works as well
and that only leaves (-1, 2)
which works since (-1)^2 = 1 is in the set
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hello
i just left my discrete math ezam and everyone is asking about this
i got 4^2n
is it the correct answer?
i double checked multiple times
,w \sum_{i=1}^n \binom{n}{i} 3^i 4^n
not sure what you mean with "always"
but after you move out the 4^n then the i=0 term would be 1, yes
the -1
oh
fair enough
thanks for answering
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3, 5, 11, 17, ?
what's the next term?
we define a polynomial $a+bx^2+cx^3+dx^4= p(x)$
Ζ( wai ina teacup)= I don't know
no its not. it has a pattern
the increments are +2 +6 +6
i dont see any oattern
ok chatgpt says its alternate prime sequence
3 is the 2nd prime.
5 is the 3rd prime.
11 is the 5th prime.
17 is the 7th prime.
Therefore, the next term is the 11th prime, which is 31.
HAHA GOTCHA
k chatgpt is stupid
did u like this pattern? it's pretty smart yk
did u ask this question to test us
umm i just wanted to ask.. if that was a nice question.....
that's my question.
i made that up.. and its my school project
coming up with random four term that has a pattern
right. ty for the feedback. i really appreciate :)). if u know any interesting patterns, please tell me.
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What is your question?
i need help with the integral
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am i supposed to send it again?
Make a new post in a different available channel
thank you
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hello can someone please explain the weak and weak * topologies ? i wanna know why and what are they . i already have some backgorund but i cant just wrap my head around them
@zinc star Has your question been resolved?
Not heard of them myself, but start at #point-set-topology. Someone there may have better luck helping you.
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A frog is travelling from point A(0,0) to point B(4,6) but each step can only be 1 unit up or 1 unit to the right. Additionally, the frog refuses to move three steps in the same direction consecutively.
Compute the number of ways the frog can move from A to B.
So I boiled it down to arranging RRRRUUUUUU in a way such that that three Rs and three Us arenβt next to each other
No really sure how to proceed
@glossy wave Has your question been resolved?
<@&286206848099549185>
Would it be simpler to calculate the number of ways to arrange it so that there are 3 Rs next to each other, and the number of ways to arrange it so that there are 3 Us next to each other, then subtract them from the total arrangements?
Idk
Still seems like a lot of combinations to go through
I got that the number of ways of having exactly atleast 3 Rs next to each other is 48
Not sure how to go about the Us without double counting tbh
@glossy wave Has your question been resolved?
you have 5 buckets with U
e,g, 1 1 2 1 1 for URURUURURU
0 is allowed
one 4, one 5, one 6 are disjoint cases
4 2 0 0 0: 20 ways
4 1 1 0 0: 30 ways
5 1 0 0 0: 20 ways
6 0 0 0 0: 5 ways
ok 3 is also not allowed
and 2 0 0 2 2 would have 2 consecutive 0 which is bad for the R
but 0 0 2 2 2 is fine because 0s aren't in the middle
yeah that's a lot of stuff
i get the feeling there's no neat way at all lol
so like, 2 2 2 0 0: 5β2 = 3 ways
2 2 1 1 0: 30 ways
2 1 1 1 1: 5 ways
something's missing
oh it's 10β2
yeah
43 is right
yeah
ok i'll explain my approach, maybe we can figure out where i went wrong
U's can be partitioned as 2+2+1 (3c2=3 distinct ways) or 2+1+1+1+1 (5c1=5 distinct ways)
in the former case, two R's must separate the U's, and the other R's have 8 arrangements
in the latter case, all four R's are used to separate the U's
so i got 3*8+5*1
oh i messed up the partitions
they should be:
2+2+2
2+2+1+1
2+1+1+1+1
so you get 1*8+6*5+5*1=43
it works
that's identical to what i said btw
ah maybe not exactly
it's fine
the difference is probably just that i included 0s
so it was clear that Rs usually get spread out, there's only 2 cases with two consecutive 0 in the middle
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How to approach this Q? The answer is 661
<@&268886789983436800>
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
for three-digit numbers, i think it will be a pyramidal number
i mean a tetrahedral number
I just searched about pyramidal no.s and tetrahedral but i think that's a different concept
It's a Q on PnC
Permutations and combinations, Q-question
oh