#help-33
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H8 to break it to you
You still have 5 mins remaining but that one is wrong
@upper anchor hey listen
I actually messed up
So sorry
We have $\frac{AD}{DC}= \frac{1}{3}$
Lelouch
uh huh
Surely u understand this
yea
With this we have $\frac{CD}{DB}=\frac{1}{3}$
Lelouch
yea
Now we take this equation again
uh huh
We multiply both sides by $(DC)$
y
Lelouch
There is the correction
ok
We multiply both sides by DC
uh huh
So we have $(AD)=\frac{(DC)^2}{DB}$
Lelouch
Lelouch
Now divide both sides by $(DB)^2$
Lelouch
No wait
ok
$\frac{AD}{DB}= \left ( \frac{CD}{DB} \right )^2$
Lelouch
We know that $\frac{CD}{DB}$ is 1/3
Lelouch
im 13
Japan or sum shit
georgia
Tf
Georgia the country? Or r u the typical American with no geo knowledge
yea
I did this when i was 15 but anyways
No 13 year old should be given to do this wtf
remember my parents are asian too
I thought so , this isn't from your school, it's from ur parents
Tell me, do you really like learning math? Or r u being forced?
yea
I think you will learn this in 10th grade
but im tryna get everything done
so i can take a break
and wont have to worry bout this
That's a good mindset but remember to not force urself to learn , that's a grave mistake and is the worst way of learning cuz u will eventually understand nothing
:3 srsly tho
Well if the sides are in ratio 1 is to 3 then what about the areas?
also 1:3?
im in 8th grd so according to my mom im gonna learn it soon
No more hints after this, AOD is similar to COB
yep
No
@upper anchor if i give u any more info , then there is no point so...
I think you should play with equations in this one , for a good 30 minutes
Like don't stop untill u solve
Without help
I am going to eat dinner now
I will take a look at this after that
I used to be stuck on problems for a good week , and still not look at the solution or any hints , cuz it would hurt my ego , and that will build skill in math
oh
i understand
thank you for ur wise words great Lelouch
@digital spindle
GREAT LELOUCH
I GOT IT
ITS 9
TYSM FOR UR GREAT WORDS
Well done !
:3
bye lelouch :))
As u should
Bye,
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I need help to prepare for my exams
Please don't occupy multiple help channels.
can you just send all of your question/doubts in here so that everyone can help?
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Condition 3 should say f(x) > x
instead of f(x) < y
does this transformation exist?
if so what is it?
if not why?
@radiant abyss Has your question been resolved?
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Did you try anything?
yup
what did you try
Are you posting your work?
9^(1/2) is √9
They did that
oh
They didn't show that part
is my working out incorrect
so is it wrong then
Assuming that's a multiplication sign, then everything looks correct
hmmmm i somehow messed up
im kinda confused what you needed help with
surely you didn't mess up 3 * 16
did you multiply 3x16 correctly?
now I did but before i did it wrong
what did you get?
,calc 3*6
Result:
18
👍
lol haha all good
,calc 3*18
Result:
54
✅
....
.close
:\
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, calculate 3"16
The following error occured while calculating:
Error: Unexpected operator " (char 8)
💀
do they not teach multiplication table anymore these days?
Why use brain cells when there is bot calculator
fr
good question, ponder that one for a while 😄
Not the 16 times table
I only know up to 12
3x10 + 3x6, surely you can add those together : )
,w \frac{9^{\frac{1}{2}}}{4^{-2}}
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Perhaps try rephrasing your question?
Click here to refine your query online
Don't think it likes the latex part
there we go
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I'm trying to understand Markov chains, is this a correct Markov chain for the sentence "hello how are hello you"? (p means probability)
@plucky anvil Has your question been resolved?
<@&286206848099549185> maybe someone who knows markov chains?
Looks right, though I don't know if "you" loops back to "hello," but I haven't done Markov chains in forever
if it wouldn't, then what would happen if the current state is "you"?
Terminate? Idk, you're probably better off ignoring me in terms of the ending bit. I'm fairly certain everything else is correct though
I think I'll keep it like that to take into account this "edge case". Thank you for your help!
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i already did the first part. how can i get the second part?
wait i think it's 10654.50 is that right?
yup
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For 14 do I just put e-3 in my calculator
Or do I use ln
Fuck math finals on the last day of school
Help
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Guys, I need to prove that
∃x ∈ N ∀y ∈ N, x ≤ y
I know x=0
Since every number is greater than 0
But I have no idea how to out it into words
Put*
you just did
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<@&286206848099549185>
i need help
on this question
i need to make a algebra equation
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Can someone explain 1d
well what'd you get for 1a,b,c?
you sure that (2,0,2)x(-4,0,-4)=16?
You are right, I ignored those bars. But anyways if you noticed the 2 normal vectors you get from axb and bxc are parallel which tells us that all three lie on the same plane since they share a normal vector.
i dont understand how you know they are parralel
is it because they are both on the xz plane
if you took the cross product of the normal vectors you'd get the 0 vector further telling you that they are parallel. But also vectors are parallel if they are just a scalar multiple and -2(2,0,2)=(-4,0,-4)
ahhh okok
and it can be logically see that a,b,c are coplanar because if axb forms a plane and bxc forms a parallel plane then both planes must share vector b. So, they're coplanar
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can someone give the solution for this the red text is the answer
Why do you have answers next to the question
anyways
you should be aware of what sin, cos, and tan do
@snow sun
i do but i can't make the solution cause i have class in 5 mins and im supposed to be doing a presentation
soh cah toa
the opposite adjacent side
what
omg i have to go
i have the other ones i just need 1 more but ill just do them on the bus
THANKS FOR HELPING THO
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And we are looking for opposite and know adjacent
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What is 98 in z13
either do the division to find the remainder or subtract 13 until you get to a value in Z13
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Observe $$\left(1 + \frac{1}{x^2}\right)^{3x-4} = \frac{\left(1 + \frac{1}{x^2}\right)^{3x}}{\left(1+ \frac{1}{x^2}\right)^4}$$
BenJr
Yea
"good morning world" what a generic ass name
Now for the top, how should we approach it?
"BenJr" is your father Benjamin Franklin?
We know lim((1+1/x)^x) = e.
Yes
Observe $$\lim((1+\frac{1}{x^2})^{3x}) = \lim(e^{3x\log(1+\frac{1}{x^2})})$$.
BenJr
Okay
To evalue the right side lim, we just need to e^lim(…) instead.
Where … is the stuff on top of e.
We have infinitely times 0 when evaluating the limit.
Do you see that?
I switch over to laptop to make this easier.
Alright tks
I know
that is log to the base e
what do you have to find here?
What will happen when” infinity” times 0
0
So it will be e^0
Which is 1
So 1/1
=1
But it is weird seeing infinity times 0
$$\lim_{x\to\infty}\left(1+\frac{1}{x^2}\right)^{3x} = \lim_{x\to\infty}e^{3x\log\left(1+\frac{1}{x^2}\right)$$.
the question arises that the log here is really ln or not
It's not always, different fields of maths take different conventions
if not then the question is simply wrong as log requires base
BenJr
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the right side follows from power rule of log.
They are using log for ln, nothing more going on here
$$\lim_{x\to\infty}e^{3x\log\left(1+\frac{1}{x^2}\right) = e^{\lim_{x\to\infty}3x\log\left(1+\frac{1}{x^2}\right)$$.
BenJr
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3x goes to infity and log(1+1/x^2) goes to 0 as x goes to infinity. So we have an indeterminate form.
Applying lhopital rule we have.
$$e^{\lim{x\to\infty}3x\log\left(1+\frac{1}{x^2}\right) = e^{\lim_{x\to\infty}\frac{\log\left(1+\frac{1}{x^2}\right)}{\frac{1}{x^2}}$$
BenJr
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Sorry about my latex.
when you take the derivative of top and bottom of right side and we evalute the limit we have e^lim(...) = e.
so $\lim_{x\to\infty}\left(1+\frac{1}{x^2}\right)^{3x} = e$.
BenJr
@ashen pivot
yep.
It comes out with this
for which one?
Sorry, the whole latex mess was me trying to this ^.
I look for a link online that explains it clearly.
Here, the idea I used is basically what the 2nd person answered.
You have to scroll down to see it. Not talking about the comments to the questions.
I see
I found the answer of the question should be 1
Anyway thanks for your help
.close
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Type .help <command name> for more info on a command.
.help me
No command called "me" found.
.reopen
Why is sinx times 1+cosx equal to -1? And why is 1-cosx times 1+cosx equal to sinx
Not sure how they arrive at that. lol
where do you see that
the red box
yh
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We're playing a game, if I flip heads you give me $30 if I flip tails I give the coin to you, then you go. How much would you pay to go?
i said 15$ because the P(H) = 1/2, and the probability of P(T) = 1/2. the outcome of flipping H is $30, so I multiplied that with 1/2 and the outcome of flipping tails is 0 since i am giving the coin to my opponent so i multiplied that by 0. i than add both events together (30*.5) + (0*.5) and got $15.
The person who flips heads wins on the spot. If tails is flipped, the game continues.
Scenarios in which opponent wins: H, TTH, TTTTH, TTTTTTH, etc
Scenarios in which I win: TH, TTTH, TTTTTH, etc
What is the probability that opponent wins? Sum of probabilities of all those cases. 1/2 + 1/8 + 1/32 +... = 2/3 (infinite geometric series formula)
So there's a 2/3 chance I lose $30
And similarly, a 1/3 chance I gain 30$
So on the whole I expect to obtain 2/3 x (-30) + 1/3 x (30) = -10$
So you'd have to pay me $10 to play
the question is asking how much would you pay?
not me paying you
are those scenarios
where the opponet flips?
Yeah, I assumed the opponent flips first
Ohhh wait
"pay to go" means I get to go first right?
yeah
Okay sorry my bad
I misunderstood that part haha
Then the roles of 'me' and 'opponent' are flipped in my analysis, and I expect to win $10
So I'd pay $10 to go
If I flip heads I win. That's 1/2 chance
If I flip tails and you flip tails and I flip heads I win. That's a 1/8 chance
You can see where this is going?
how 1/8?
TailsxTailsxHeads, 1/2 x 1/2 x 1/2
ah
as long as you flip H at the end you win right?
This forms an infinite geometric progression, there's a nice formula to find the sum for that
Yeah, and as long as the number of tails until then has been an even number
even number because it would mean that the opponent was the last person to give you the coin right?
Yep
is this the formula?
And since r is smaller than 1, r^n basically gets smaller and smaller until it vanishes. So the formula reduces to a/1-r if n is infinity
r is the ratio between successive terms
The series goes like 1/2, 1/8, 1/32, etc
Each time we're multiplying by 1/4, so r=1/4
And the first term a=1/2
i can only have 1 H not 2 H
since 1 H is enough for me to win right?
how did you get the 1/3 than?
a1 right?
Right
Yeah
First I calculated probability I win by doing 0.5/(1-0.25) = 2/3
Then the probability I lose is just 1 minus that
which equation is that?
r still the same
So the first term a1=0.25. R is same
Yeah
why did you multiply 1/3 by 30?
we would pay the amount we are expected to lose so we cant be in debt?
Yeah, that makes sense
if i pay $10
and i lose
i lose all my money but wont be in debt
but if i win
i gain the money i initally spent so i wont really gain money either
Yeah, it's like a break even
why do we do this?
Well um
So there's something called the expectation value
Also known as the mean
If I played the game a million times, how much in average would I expect to win?
That sounds like a fair idea of how much I should pay to go
(ideally I'd like to pay less but yeah)
im asking why do we pay an amount for me to gain profit every time?
If I pay more than that, I expect to lose
If I pay that much or less, I expect to gain
So I'm willing to pay up to 10$
ah
so i would either pay up to $10 since i wouldnt lose any money right?
either no gain or gain
no loss
Yeah
I mean if you only play a few times you could be unlucky
But with probability it's about hypothetically doing it lots of times and seeing the average
Np
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$1+sin(x)=\sqrt{3}cos(x)$
Mohad
any ideas?
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Is the center of symmetric group and alternating group S4 and A4 only the identity?
@thorny jungle Has your question been resolved?
@thorny jungle Has your question been resolved?
Yes
If you take a non identity permutation σ in S4, then σ(i)=j for some i,j in {1,2,3,4}. (We could assume without loss of generality and by symmetry that i=1 and j=2, but no problem). But i≠j so σ(j)=k for some k≠i≠j, k≠j.
Now, if we consider (ij) we have that (ij)σ(i)=i but σ(ij)(i)=k, so σ and (ij) does not commute.
I think that suffices
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iam at the radical rn
i need help on what to do next <@&286206848099549185>
what is the equation for the radical
so i have (90,50) and (130,30)
i just need help with the function now
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sorry
but can u help me @proper zodiac
just what is the radiacl function
and where can u substitute my points
so will it be 90^(1/a)=50
-(90^(1/a))=50
yup
wait
ergh
forgot that every new function you add
needs translation from origin
so
it will be
y = -((x-90)^(1/a))+50
hopefuly you get a back that is positive
even better when integer
what is this
btw radical function means that a should equal 2
so it will wind up into
y = -sqrt(x-90)+50
that didnt work
ok
there are 2 options from here to get it to work
thats still wrong its not touching y=30
either you add parameter a before sqrt
as in
y = -a*sqrt(x-90) + 50
and the other
you change degree of a root
what are you saying i dont understand
so
y = -((×-90)^(1/a))+50
but I dunno if this qualifies as radical function
I need to check literaturr
one moment
<@&286206848099549185> ?
what is the *
so it became 50=sqrt(90)
if you plug x-a instead of x
here why did u do this
you just move it to the right by value of a
so that it would atart from point 90
then I knew for value 90 it will be just 90-90
so 0
so it needs a little push to the up
by a 50
then for the final I stretched it a bit by just multiplicating term with x
and solved for second point
it was suppossed to be
y = -a*sqrt(x-90)+50
30 = -a*sqrt(130-90)+50
simplifying
a*sqrt(40) = 20
a = 20/(2*sqrt(10)) = sqrt(10)
then I knew that
sqrt(m)sqrt(n) = sqrt(mn)
so final formula was made
ping if you need any more help
can we go on a vc @glad topaz
wait vc doesnt work for me could u joine this instead https://zoom.us/j/96266401935?pwd=NTJXYnFabGYxNzZEaDgvUmZGZXgxUT09
Zoom is the leader in modern enterprise video communications, with an easy, reliable cloud platform for video and audio conferencing, chat, and webinars across mobile, desktop, and room systems. Zoom Rooms is the original software-based conference room solution used around the world in board, conference, huddle, and training rooms, as well as ex...
@glad topaz
I'll try
@rough plank Has your question been resolved?
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in this khan academy limits course it says this
i don't get how it knows what the function looks like when approaching from the right?
could you please explain this too me?
<@&286206848099549185>
When it approaches -6 from the right, we're looking at the values of x -5.9, -5.99, -5.999, and the value that h(x) gets close to. h(-5.999)=-1.02, that's close to -1.
i still don't get it? it said to also see what it is on the right side?
-5.999 is on the right side of 6, and is close. h(-5.999)=-1.02 is close to -1
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thanks
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Shouldnt it be -a/1+a^2?
One, don't ping specific people, and two read #❓how-to-get-help
And open your own channel
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Greetings!
Can someone tell me if I did the right thing? A negative answer seems wrong to me so I'm wondering if I solved it correctly or not
You can graph the function on Desmos to see if the function is below the x-axis
It actually is in this case
Although if you don't want signed area, then positive is the way to go
So it's fine?
If you'd like you can check the integral in Desmos
You can just give Desmos the integral and it'll give you the signed area
yup!
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How would I do this
can you quote your nth term for me please
Write the formula for Tn
Then equal it to the options i think, see for which one it works🤔
the nth term is a + (n-1)d
8n-5
they will appear in teh sequence if n is an integer
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nope.
It should get deleted any moment
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,rotate
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im given (x1, x2, ..... , xn) thats a set of Rn dimensions, where n is the set of all positive integers, that satisfy (cosx1 + cosx2 + ... + cosxn = 0)
@merry plaza Has your question been resolved?
I assume you're familiar with the notion of continuity -- what it means for a function to be continuous -- right? What definition for continuity are you using?
I think there's something confused in how you phrased the question -- probably you mean that for any fixed positive integer n, you must show that the points (x1, ..., xn) in R^n satisfying (cos x1 + ... + cos xn) = 0 are closed. Is that right?
(And non-compact)
It’s clear for n = 1 that the set for this n is closed. Suppose it’s closed for some n >= 1, we want to show for n+1 the set is closed. Take any fixed points (x_1,…,x_n) such that cos(x_1) + … + cos(x_n) = 0. Let S be the set of real number x such that cos(x_1) + … + cos(x_n) + cos(x) = 0. It follows S is just the set of real numbers x such that cos(x) = 0. Clearly S is closed. The set for n is closed, so the product for the set for n and S is the set for n+1, which is closed. Since the product of closed set is closed. I think this is a way to do it.
It might be wrong, never done a problem like this before.
This is one attempt.
I will take it up with #real-complex-analysis for confirmation.
That doesn't quite work -- cos(x_1) + ... cos(x_n) + cos(x) = 0 has solutions where cos(x) is not zero. So S is a bit more complicated than you've said.
x_1,…,x_n is fixed. That is point (x_1,…,x_n) for the set n.
We have cos(x_1) + … + cos(x_n) = 0.
So we just left with dealing with cos(x) = 0 for S.
Again, if @merry plaza is still around, I'd start with what definition of continuity you're using. If it's that the preimage of any open set is open, your life is very easy for this question.
@normal oracle i need to show the set (x1…xn) is closed
Oh yea
This
My bad
Ok. And you know cos(x) is a continuous function from R to R?
Ok. If you can show cos(x_1) + ... + cos(x_n) is a continuous function from R^n to R, then you can appeal to the fact that {0} is a closed set of R, so its preimage in R^n will also be closed.
(So for example -- you must know or show that adding finitely many continuous functions together results in a continuous function)
Would I do that through the limit definition?
If that's your definition of continuity, then yea. There are many definitions of continuity though -- using limits, epsilon-delta stuff, or open/closed sets.
Do you define a closed set as one that contains all its limit points?
Wait I think this might be referring to being closed under different linear combinations
Hmm, that's an entirely different question, if so. But odd to ask about non-compactness in that context.
It’s supposed to be linear alg and multivar calc
Are you familiar with the notion of a closed set (in the multivariable calculus sense?)
Not rlly
It's a famous fact, for example, that any compact set in R^n must be both closed (in this sense) and bounded (meaning doesn't contain points that get arbitrarily far away from the origin). So if a set is non-compact, it's either not-closed (but your set is closed!) or else not-bounded (so that's what you'd have to show in this problem).
Oh cause the cosines of all the elements converge?
Ok -- lemme take a stab; see if this clicks. A set of points $S$ is closed if, whenever you have a sequence of points $s_1, s_2, \ldots$ in $S$ that converge, meaning that $\lim_{i\to\infty} s_i = s$ for some $s$, then the limit point $s$ is also in the set $S$. Does that ring any bells?
daveamayombo
So what you'd have to show is that, if $S$ is the set of points $(x_1,\ldots,x_n)$ for which $\cos(x_1)+\cdots+\cos(x_n)=0$, and you have a sequence of points $s^1, s^2, ...$ in $S$ that converge to a point $s$, then $s$ is also in $S$.
daveamayombo
(The superscript isn't meant to be exponentiation, by the way -- just an index into the sequence of points)
I see
If this is seeming like a bit too much -- I'd say maybe see if your teacher can point you in the right direction. The issue I'm facing is just that there are many different ways of defining all these things ("closed", "continuous" etc) that all turn out to be equivalent. In an intro class you might only see one definition, and I'm not sure which ones you're using.
Also, the baby's waking up so I gotta run!
Alright ty for the help
Yah, sorry couldn't get you across the finish line this time.
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Considering the determinant of a 2x2 matrix - What is the signature of the quadratic form representing the determinant? I want to say (1,1) because we have ad-bc -> xˆ2-yˆ2 but i'm not sure if this reasoning is correct. Can someone help?
@astral oyster Has your question been resolved?
i'm not sure, but this might help https://math.stackexchange.com/questions/3193710/on-the-signature-of-a-quadratic-form
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integrate (lnx)^2, why is the method using u = (lnx)^2 and dv = dx ? isnt first logical attempt to use u = lnx and dv = lnx ? also what is the theory behind why you are able to use dx as one of the variables when integrating by parts such as the first method in this example, and why do you not need to include dx in either u or dv in other integration by parts questions?
not really theory, just something that works by finding patterns in the product rule
ah ok. but why then do you use dx in one of either u or dv in some questions but not in others? what makes you able to dismiss it?
at this point for you, just trial and error
just so i know, is there a way to know and you are saying that for my level i should just trial and error or is that what everybody does?
this is good, thanks
it's what everybody should do yes
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number 4 please. I understand that revenue=(sales)(price) so I get 52000= (2.00+0.05x)(6500-5x)
but I dont know where to go from here to find the new price and the #sold
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What are the solutions of this equation?
can someone please confirm?
if my reasoning is correct then its A the correct answer
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This equation has 2 real solutions when...
i believe the correct answer is E
because if we take 2 then we will get two results that are equal
can someone please confirm?
Do you know how to use the discriminant
its b^2-4ac
And for what values of the discriminant are there 2 solutions?
when its positive
because if its equal to zero, then we would have two equal solutions
Yes, but i wanted to be sure that the answer was indeed E and not D
I belive its not D because if we take -2 then the statement becomes not true
the discriminant would become equal to 0
Correct
Perfect, thank you 🙂
I have another question if you dont mind
can i post it here in the same thread?
Sure
Two concentric circles delimit a circular circle whose area is half the area of the larger circle. The ratio of the radius of the larger circle to the radius of the smaller circle is...
Could you draw it out?
this is what i tried to do
I used r for the radius of the small circle, while R for the bigger one
The problem tells us that the area of the bigger circle is equal to half the area of the "circular crown" (dont know if this term exist in English)
from there i tried to something but i didnt work out in the end
@echo kernel Has your question been resolved?
<@&286206848099549185>
<@&286206848099549185>
@echo kernel Has your question been resolved?
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can you divide by 0?
no
correct
it is undefined so h(t) won't have a value when the denominator is 0
that's why we find the t values for when the denominator is 0 so we can say it is not continous for those values of t
so i would always have to equal it to zero
if you have a function of t on the denominator, yes
so equal it to 0 then factor it
why cant i just factor it instead of equaling it to zero
because you need to check when the factors = 0?
yes
you have your factors?
yes
ohhh ok i get it
the instructions were werid
so just factor it out first then equal it to zero
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In Figure 8, are represented, in Cartesian reference,
of origin at point O, the lines defined by the equations
Point I is the point of intersection of two of these lines.
What is the system of equations that makes it possible to determine the
point I coordinates?
i have already ruled out option A
because the slopes are opposite of eachother
so there needs to be a negative and positive slope
So you know that the general equation for a line is y = mx + b
You already identified that one line as a positive slope and the other has a negative slope
For the line that has a positive slope, what is the y intercept?
which one
The line with the positive slope
there are 2
You're looking for the lines that goes through point I, aren't you?
oh ye
So would the line that does not go through point I make sense to use?
but they dont associate the equations with the lines
so i dont know which equation represents that line
Well, I'm asking, there are two lines that goes through point I
As you stated, one line as a positive slope and the other has a negative slope
yes
For the line that has a positive slope, what is the y intercept?
-2
Can you use what you know about the slope and the y intercept to determine the proper choice?
yes
👍
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does the identity permutation of $S_n$ have length 0 or length 1 ?
Rœmer
(and why ?)
The identity is even
e.g e = (1,2)(1,2) [a product of 2 transpositions, i.e an even permutation].
so it cannot be written as a cycle of length 1 (if that's what you meant)
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