#help-27
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Did u find q?
I did 90 + 90 + 40 - 360 and got 130 then q was 65.
Since tangent and raidus
Mb 210 from 360 and got 150
Is this enough for u?@acoustic plank
It is possible:
angle B = 180β° - Angle q - (Angle O)Γ·2
Ig
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Write in mathematical terms what it means
Form a system of two equations
Omds i posted the wrong image i meant this so srry
@ adoniszgks reference??
Assign a variable for each person
Let's call the income of adonis D
Short for drakes kid
The income of apollo P
And aphrodite L
Please somenoe help
I'm tryna do that cuhπ
So we know that Mr adonis earns 7200 more than apollo
We can make an equation from this
D-P=7200
As we know the difference between D and P is 7200
yea
@deep marsh try making two other equations in a similar manner with other info in the qn
yea
Lastly form an equation based on their total income
what would the euation be
P+L+D=303 200
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hello, does someone mind verifying a solution i have for a calc 3 question
Send
also it is on R = [0,1] x [0,1]
thank you again, for any clarification or help π
I believe your counterexample works yes
an even easier counterexample : f(x,y) = (x-y)^2
perfect! i also have another similar question i am a bit more unsure of, can i send it here too?
you mean D = ... instead of R = ... right?
yes haha my bad
thanks
my intuition was kind of along the lines of 2^0 = 1 so can it be less than 1?
since it's continuous along D and the area of D is 1 can it be below 1?
some people i know thought it was false, so i was a little confused on the proof
would my intuition be correct?
it's growth of the integral
the answer is it cannot
but try the proof
again it's all here
yeah it seems pretty straightforward
$\iint_D 2^{|f(x,y)|}dA \geq \iint_D 1dA = 1$
rafilou is not not born in 2003
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is my reasoning for part (d) valid? if not how can I fix it?
You should also mention R(t) - W(t) = 0
Because youre technically using IVT on R(t) - W(t)
Either works
You must also state that W(0) - R(0) < 0 and W(8) - R(8) > 0
is that a must?
then why?
why is that condition necessary?
to show that it exists in that interval?
where R(t) - W(t) = 0 -> R(t) = W(t)
Bc the IVT only guarantees stuff in between f(a) and f(b)
It seems obvious to you, but the scorers will be strict about this
should i mathematically find the value to state it cause otherwise I wouldn't know...?
I dont think you have to mathematically find the value, you can literally just write those statements
is it cause that must be true for the IVT to do the stuff...?
but the question says "explain why or why not" so that means it could either exist or not exist...?
Well youre given W(t) and R(t) at t = 0, 8
Ig you can solve for W(t) - R(t)
Also its very likely you can use a graphing calculator
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The nature of W(t) makes it exceedingly likely so
yeah its kinda impossible to calculate with hands...
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(a) part is easy, the rank of A is 3 and Ax=0 has solutions s(2,3,1,0) where s is a parameter
not sure how I would go about doing b
I know that the rref has only one non-zero row
and hence only one pivot
no wait, we have three pivots
shouldnt this give it away
it says "reduced row echelon" not "row echelon"
yeah
there'll be three pivots, not one
there's one free variable
which is x3
so the pivots are in the first, second and fourth column
what's a pivot
the first non-zero number in a row after elimination
kheerii
oh
kheerii
should be this then
Fourth row
Oh mb
For (c) it's just that rank = no. of rows so the range is all of R^3
Then yes
it's full row rank
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yo chat how do i solve a system of two quadratic one in terms of x and another in terms of y
but both have x and y terms
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(ignore the blue stuff)
4 or 5?
ok lemme go through it
Ay thank you so much
Yeah I got my maths exam tomorrow and I cannot understand anything related to this sin,cos,tan inverse blah blah thing I am so cooked π
How did you do it without inverse trig
Bro π
Which question π
yeah because i approximated the values at last
Bru π so is the answer right or naw?
Atp I gotta accept fate and understand I am cooked tomorrow
<@&286206848099549185> anyone else vro ππ
A
Check the messages above
alright
To find angles sin inverse cos inverse and tan inverse is used
But like I kinda forgor what to do if I got three values like let's assume
ABC
AB is hypotenuse
BC is perpendicular
BA is base
So like now let's assume I got Value of all three of em
If I had to find out like <BCA or something what inverse would I use sin cos or tan and then those questions above make it more confusing so now I don't understand anything atp π
mhm
Bro what is "mhm" ππ
Which question π
Vro I already got the answers I needa see the working
relax
BRO I AM SO COOKED FOR TOMORROW πππ
since LNM is 90 degrees
Hm
Wait how is mln 90Β°
Lm is a curved line π
it had a 90 degree angle
Lnm
lmk if u dont so i can expaain deeper
From 90Β° do you mean angle L or angle N?
From 90 i mean MNL
Actually go on
see how its a right triangle
all right triangles are 90 degrees
the square in the conor is a sign its 90
Only angle <N is 90Β°
N woudlent be a angle by itself
Bro I know that π
that would be considerd a point
Okay okay go on
39?
Yeah oki
Hm
Hm
Isn't that only used on right angles triangles but that another triangle ain't right angled
no
even if the other parts dont form a triangle we only applyed the tan fuction in the right triangle
Oh
Hm
yes
The answer is 8.04 tho behind the book
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Help differentiate these and put in the simplest form
hi all , how are you?
Do you know chain rule?
Apply that, take as much time as you can. Don't rush
you can bring them to a same base
So
e^y=e^2t^2+1
can u check?
First one is correct
2nd is correct
Ayo what u did in the third question
How did u cancel those two with the red pen?
Explain pls
Ye
Else is correct, also one more thing
What you are doing is quite long process, it's not wrong
But you do these directly let's talk about 3rd example
Just differentiate it with respect to t
Ye
Take your time understanding the chain rule
Ur not helping
Correct, but a slight mistake in sign and also I forgot to add brackets
The final result should have -27
@kindred mauve Has your question been resolved?
try rechecking $\frac{dy}{dx}(5x^3 + 4x^2 - 1)$
BuilderDolphin
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On the last step, why did they make it tan(u) instead of keeping it as tan(t)?
The guy in the video said, "Obviously change the t to a u," but it's not very obvious to me. If someone could explain it as simply as possible, I'd greatly appreciate it π
One possible justification is that, since u is defined, its technically "clear" that the bottom expression is correct
I just remembered this is the equation/formula that went along with the example:
Could I say they changed the "t" to "u" so that it would turn into tan(u), which then becomes tan(x^2), matching the final answer of f(x)?
They technically didnt "change" the t to u though
yea ik the "u" came up as a substitution for x^2 but visually, that's what it looks like to me, and it's an easier way for me to put it into words π
but im sticking w the "matching the final answer of f(x)" explanation cuz i think it makes sense
tyvm
Your explanation works if youre justifying it to yourself. Your explanation does not work if youre justifying it to an exam scorer
Lmao got it
But for now, it's just for myself
I'm having my teacher explain this topic next class, so I'll have her do the extra explaining and justifying.
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test
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I need somebody to take a look at this so-called 'solution' to check whether i've made a mistake, because I dont seem to notice any
nvm i found sth
ok i think I got a prettier answer now
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wanted to clarify smthn
so the limit where x approaches 1, both from the left and right side would approach 1/3 right
even tho theres no point there and itll never reach it, they both approach it
so the limit does exist
yes
ty one more thing
very true
if i had the same rgaph but there is a point at x=1 up ther
itd be the same answer right
yes
im just confused rn bc i learnt this whole section on my own before it came in class but then
the teacher said like some other bs and i didnt have time to ask
he said smthn like LS would approach 0.99 and rs would approach 1.01 here
and he said theres no limit
i think hes wrong but he also knows his stuff
the limit deefinitely exists
weird ill ask him tmrw then
thats what im saying but like i also just learnt this topic yesterday on my own
ty then i apprecaite it alot
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np
ps I hate how this implies 0/0 whoopee limit exists, because thatβs not entirely guaranteed
whoopee?
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In the answer part it says that 2 comes out, I tried with artificies but I ended up confused halfway through the solution
what does artificies mean
try u-sub
probably better to just show the solution
It's like a method of restoring and adding the same amount (I don't speak English, I guess it's another pre-university method), for example +1-1
i see
u can find omega directly with the quadratic formula and be able to solve the equation below but that seems to tedius to be the method they would want you to use
uhmm Okay, thank you. It makes sense to me, I finished high school recently and I donΒ΄t know how to use integrals, which makes me excited because the exam I want to pass is in 4 months.
I just understood why the answer is 2, thank you again.
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how do i determine if g(x) <=0 from looking at g'(x)=f(x) graph...? whats the rule(?) for this? (part d)
You can try marking the area of the triangles
Or, in other words, finding the area under the curve line segments
so integrate it to go a step backwards and find if its <=0?
Yes, you might need to use this rule:
$$\int_a^b f(x) \dd{x}= -\int_b^a f(x) \dd{x}$$
King Leo
how do i select the bounds (a and b)...?
a is already defined as a = 2
b will change based on the value of x
how do i select b...? like which x could make g(x)<=0?
Why are you trying to make g(x) <= 0
cause they want to find where g(x)<= from g'(x)=f(x)
Oh the absolute minimum
Can you label the area of each triangle
Ok, now find g at every value where f crosses the t-axis
t axis...?
Same thing as the x axis in this case
does touching x-axis count or you have to cross it...?
For now, touching
And then you just see which point is the absolute minimum
Although later, you might need to see where it crosses the x-axis
so 2,10?
Also make sure to find g(-4) and g(12)
int from 2 to 2 is 0...? and int from 2 to 10 is -4...?
is it like the the interval where g(x)<=0 is where in g' graph the area below the line that is negative or 0?
@dull thunder Has your question been resolved?
@dull thunder Has your question been resolved?
@dull thunder Has your question been resolved?
@dull thunder
Wait
still need help?
How did you get -4
brother came back right away wtf
Nvm thats right
I was here for ~15s prior reading what they said
you mean 2 hours
The other 2hrs were bc i forgot to look at this channel
:)
whatβs PWR
Stands for Ping When Replying
Cause there have been too many times ive left someone hanging bc i forgot to check the channel again
what happened...?
yea
send it again
how do i determine if g(x) <=0 from looking at g'(x)=f(x) graph? whats the rule(?) for this? (part d)
since g(x) is defined in terms of an integral we simply want to find all intervals in which the resulting integral yields a non positive result
you could do this two ways
either find this directly
or find the intervals in which itβs positive then exclude those parts
and hence construct the remaining intervals
should i try integrating all intervals or...
if you do it this way
you donβt need to do much direct computation
how?
i dont get it
like are we finding the areas and then...?
well we are shown [-4,12] right
yea
now it certainly is true that g(x) is either <= 0 or > 0 yes?
there is no other possibility
yea
ok so if we found the intervals where g was positive wouldnβt that mean the other parts of the interval [-4,12] would have to therefore be not positive (<= 0)
yes
ok so letβs use this strategy
first letβs consider x > 2
since our lower limit is 2
for the integral
ok
clearly the integral is positive for (2,6] right
because we are integrating f
and f > 0 there
we get positive area from that triangle
yes?
no?
questions?
yes
exactly
now
letβs continue further right
and see how far we can go until g is no longer positive
i.e find where g is zero for x > 2
so letβs try subtracting the area of the next triangle
since itβs below the x axis
itβll have negative area
whatβs the area of that triangle and thus what is the integral of f from 2 to 10
8+(-8)+(-4) = -4?
wdym?
you said its 2 to 10 but why should it be to 12...? is that cause that triangle extends to 12...?
i asked you to stop at 10
because if you notice that g = 0 when x = 10 since the positive and negative area there cancels
itβs negative thereafter
so weβd include the 10 to 12 bit
oh so at 10 its g(x)=0
yeah
we donβt need the numerical value
so continue to move one...?
ok so weβve established now that (2,10) will not be in our answer
10 to 12?
note that the interval is open since if we included the endpoints weβd get g = 0 which would be part of our answer
10<=x<=12 is one interval
this will be part of our answer yes
mhm
now letβs look towards x < 2
so for x < 2 we have to realize that whatever the resulting area would be, we have to negate it because of how the limits of integration are
anything below 2 is gonna be negative unlesss its already negative cause if we flip the bounds we stick the - sign? right?
well, we need to analyze carefully here
letβs start by rephrasing the problem
ok
we have $g(x) = \int_2^x f(t) \dd{t}$
knief
but for x < 2 the integral will be negative so the areas weβd get will be misleading
so letβs flip it
$g(x) = -\int_x^2 f(t) \dd{t}$
knief
now
reading from left to right, if the area is negative we have that g is > 0
and aha
yeah
notice that the triangle between -2 and 2 has an area of 8 but we must negate it so -8
and whatβs the area of the other triangle?
the two triangles eat each other up so the rest is 4 and we stick that negative --> -4?
hmm well tbe triangle from -4 to -2 had an area of -4 which gets negated to 4
so what can we say about g for x < 2?
the two triangle before zero is gone
and after zero is 4 and we put the sign so -4?
its -4?
that would be the integral from -4 to 2
i see what youβre doing
and itβs correct
but
what can we then say
about g for all of x < 2
i dont know?
weβre not so concerned with the particular value
itβs always negative isnβt it?
because thereβs more "positive area" for x < 2 which then gets negated
yeah
notice that no matter how far left we go we still have to go through that larger triangle which has more area
so itβs always negative
yea
-4<=x<=2 and 10<=x<=12
yea
its literally that easy if you understand it
thanks for helping π
π
have a good night
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can someone double check my work
basically since inequality signs flip when we take both limits there must be a value $\pi ^*$ that exists where B passes A
Faduzzle
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iv'e been stuck on this for a while, especially b, any help?
i dont really understand interest because i wasnt there for any of the lessons
@vocal pebble Has your question been resolved?
<@&286206848099549185> ?
im unsure how to start it, (both a and b) and when i try it it is incorrect
So like when you deposit money in a bank the bank usually gives you an interest which is a percentage of the amount you deposited
And the formula of simple interest is P x R x T /100
Kk
So the principal is 4000 pounds
okay
so 4000 x 3.5 x 1 year?
We would get 4000 x 1 x 3.5/100
/ 100 okay
Divided by 100
Now see what you get as the interest
Wait
@vocal pebble
The time here is 6 years
its compound though?
Is it?
yeah
no
i kinda know simple interest but not compound
okay
So if u make the denominators same
It would 100+R all over 100
And the rest of the formula I have given u
and r = 3.5 ?
Yes
This formula is for the total amount
To find ci
First u have to find total and then subtract principal from it
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Hi
Can I ask help for science :cc
sure
wht even is that
I'm using a compound microscope
are those cells
Cardiac Muscle Section
reminds me of robert hooke looking at onion peels
Yes
well i know nothing about biology
Tissues*
Yes
I'm dying
anyways i hope you get someone to help you soon! :D
I hope so too
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How do you do this
I first took 31C10-21C10
Stripped away the combinations between the distinct ones themselves since they arent affected
And then (31C10-21C10)/10 since each combination is repeated 10 times
Then i did (31C10-21C10)/10 + 21C10 to get the total
@toxic spire Has your question been resolved?
im assuming you're confused because this seems like a pain to calculate?
No its just why my method aint working
The combinations that contain the identical numbers
so every single combination that consists solely of identical ones? or all combinatios that include at least one of the identical ones
im assuming the latter
this includes all combinations where one of the balls is from the identical pool, ones where two of the balls, three of them, etc.
for the case where only one of them are identical, then your divide by 10 makes sense, as you ball could be any of the 10
but what if two are? then you have 10c2 ways of selecting those two balls
and if three are? then 10c3 and etc
you can't just divide by 10
its better to try and do casework with the # of identical balls
with 0 identical balls, you have 21c10 ways of choosing the rest
with 1 identical ball, you have 21c9
with 2, 21c8
...
with 10, 21c0
this way you dont even have to consider how many ways there are to choose the identical balls from the 10
this has an interesting way of adding them up
theres a somewhat well known sequence where nC0 + nC1 + ... + nCn = 2^n
if you notice where we stop, (21c10), we have actually traversed exactly half this sequence
and the choose function is 'symmetric' in the sense that like 21c2 = 21c19
Hmmm
Answer is that yes
anyway i think this is the main problem with your approach
Do you mean like
If theres AAAA then dividing by 10 won't enough?
Aka more than one identical
yeah, if i get what ur saying
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hey guys what does the y hat mean here?
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I guess C is correct?
just a guess π
you should not guess
you are not so lucky as to hit that 1/4 chance enough times to pass the exam or whatever
Please provide some hints
(nobody is)
if there are enough people taking the exam then someone is
try to think about a similar question but with A^3 = 0 instead. it will not be the exact same, but you might have some ideas there.
as in: suppose A^3 = 0; can we say anything about whether either of I Β± A are invertible?
i need to think of that nilpotent actually
nilpotence will be crucial, yes
you did not say anything yet so you are neither on the right nor the wrong way at this moment
are you confused about the word "crucial" itself?
I only get hint that a^n=0 looks like nilpotent
it doesn't just look like nilpotence.
I have no idea rest
A^5 = 0 literally is nilpotence (with a particular index of it, too)
why am i saying what?
we are getting nowhere with this. you pick my words apart until you confuse yourself, but you have made no progress on this question (nor on the simpler one i suggested).
try calculating (I-A)(I+A+A^2). maybe you will get an idea.
That is much better
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how do i do this? i don't know how to do induction
would it be enough to just give an example?
like |u^5| = |uuuuu| = 5 which is the same as 5|u| = 5 * 1 = 5
no an example wont be sufficient
@idle tinsel Has your question been resolved?
Do you know how I would do it?
Do you remember induction?
i did induction many semesters ago but i dont remember it at all
you need to prove it for some base case
so show that |u^n| = n|u|, when n= 1
and then you assume it holds for |u^n| and show it holds for u^n+1
okay, so we can show that for n = 1
n = 1
|u ^ n| = |u| = 1
n|u| = 1 * 1 = 1
then we do
|u ^ (n + 1)| = |uu| = 2
n|u| = 2 * 1 = 2
oh i thought that was still assuming n was 1
thats for the base case
you need to show for any arbitrary value of n, where your equation works
it'll also work for the next one
ahh
and then you have your recursive/inductive proof
since it works for 1 hence it'll work for the next one (2)
but since it works for two it must work for 3
and so on
|u ^ (n + 1)| = (n+1) * |u|
this is what im getting now, im not sure how to break that down further
tahts what you need to show yeah
but you need to show that you can get the left hand side to turn into the right hand side
i'm not 100% certain on my definitions with strings
oops
u^k means string u repeated k times
yeah so u^k * u is u^(k+1)
so like hello^2 = hellohello
right?
yes i believe thats right
$|u^{k+1}| = |u^k * u| = |u^k| + |u| = $
oh man idk how to latex in this
you have an expression for u^k (you assumed it worked for u^k)
so you have k |u| + 1 |u|
(k+1) |u| or something
i am a bit confused on how we went from |u^k| + |u| to k|u| + 1|u|
by your induction hypothesis
aren't we still trying to prove that |u^k| = k|u|
you proved that it holds for some value, 1
you now assume it holds for some value n
and that for some value n where it holds it works for the next one
right, but in this case k could be any number other than 1
correct
it's just any condition where it holds
but like
if we assume some n where it holds and show it works where
there*
you could say ok it works from 1 to 2
but since it holds for 2, (as it holds for n+1)
then we can start again ok it holds for 2 (n) so it must hold for n+1, which is 3
and so on
so instead of proving it holds for 1,2,3,4,5,6 you just have to show it works for the next value from some arbitrary value, where your expression works
that makes sense, I just thought we weren't able to use our own hypothesis that |u^k| = k|u| in our proof, i thought it would be like circular reasoning
thats why induction relies on two parts
a base case and the inductive step
you need to prove that there exists some k where it works
you can use your expression in the inductive step
on u^k that is
okay I think i understand it now, thank you
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yo how can i solve this
4x-3y+2z = 4
6x-6y+3z = 3
12x-11y+8z = 14
How would you do it with 2?
Please answer my question
like
multiply the other, and so on
and subtract it
and solve it so that one equiation has only one unknown variable
but yk
Take a look at the coefficients of the x terms. Those all look like they'll play nicely with each other
when there is 2 unknown variables but 3 equations
okay wait
yeah but like
waiz omg
i can do the first times 3
and the second times 2
Indeed
Use any 2 of your equations to find the other 2
ohh
you mean use y
and form 2 new equations?
@neon wagon Has your question been resolved?
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does this needs an integration by parts :
$\int \frac{t}{t+1}dt$
<rajel />
SWR
then i'll have $\int 1 - \frac{1}{t+ 1} dt$
t - log(t+1)
<rajel />
<rajel />
.close
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don't forget your +C
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I have to solve for x by completing the square, how do i square root both sides if the 1/9 is negative?
also for this question how do i show how i got the x-intercepts can i just factor the original equation
,rotate
I think they want u to graph x^2 + 4x and -2x -5
And find where they intercept
,w 3x^2 + 4x + 5/3
Continue with this
Sqrt of -1/9 is 1/3i
wait i thought u couldnt square root negative numbers
You can, just need to introduce another dimension (imaginary numbers)
oh i didnt learn that yetπ i dont think im supposed to use it
Iβm guessing the equation was written wrong and 5/3 should be negative
i had a similar question on another assignment and they took marks off bc i square rooted the negative number
but idk what else im supposed to do
Just ignore it for now and focus on the other questions
Your teacher will probably omit it from the final mark since you havenβt learnt complex numbers yet
@merry plume Has your question been resolved?
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how ddi they go from the (extS)^c cap (S^o)^c?
did they just use lots of distribtuive laws?
and then like some are contradiction/just false so they get rid of those?
@velvet coral Has your question been resolved?
it's by definition, interpreting the factors of the union as sets
a point x is in ext(S)^c if and only if for all positive radii (e) the ball B(x, e) is not entirely contained in S^c (in set theoretic language, B(x, e) intersect S is non-empty)
a point x is in S^o^c if and only if for all positive radii (e) the ball B(x, e) is not entirely contained in S (in set theoretic language, B(x, e) intersect S^c is non-empty)
combining these two conditions gives the result in the image you posted at the start
more explicitly, $x \in \left((S^c)^{\circ}\right)^c \Longleftrightarrow \forall \epsilon > 0, B(x, \varepsilon) \nsubseteq S^c \Longleftrightarrow \forall \varepsilon > 0, B(x, \varepsilon) \cap S \neq \varnothing$
Mqnic_
^ for the first factor, and likewise for the second
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Is f^-1(x) a common expression of the inverse of function f
very much so
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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
its just Dg(f(0,1,0)) Df(0,1,0) right
I see but isn't the set Ext(S) = ${ x \in S^c : \exists \epsilon > 0 \text {such that} B(x, \epsilon) \subseteq S^c }$. Therfore shouldn't $(Ext(S))^c = {x \in E : x \in S \text{ OR } \forall \epsilon > 0, B(x, \epsilon) \nsubseteq S^c }$?
LXDL
wrong question buddy
this is the wrong channel
<@&268886789983436800>
yea basically
sorry for late rpely i was thinking about it and it kinda made sense but then i not really
yea h sorry ill move
@flat ibex Has your question been resolved?
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How do i find C in this
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you didnβt need {} around the x^n dx part
Oh
$\int x^n \dd{x} = \frac{x^{n+1}}{n+1} + c$
knief
Ok, but how do i know what C is?
you don't
youd need more information
Say you input a value for n, how do you calculate C?
the antiderivative is not a function, it's a class of functions
β¦
any constant C differentiated yields a 0
so you just
can't "find out" C
still not enough information
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what
you donβt
you still can't π
the n isnβt the issue
So how would you calculate this?
you just did
you can't π’
π
the + c serves as a vertical shift which doesnβt affect the derivative
What does that mean?