so since c is continuous it has every number in the interval so it would be that one?
#help-33
north prism
cunning jackal
Essentially it just means the value of the function should tend to the same thing if you approach the x value from the right or the left.
If they're saying it's NOT continuous at 1, it means that if you plug in a number slightly bigger than 1, and then a number slightly lesser than 1 in the function, and then exactly 1.
These 3 values won't be equal
north prism
ah i get u
cunning jackal
That's the basic meaning of saying it's 'discontinuous '
Now if we look at C, or any polynomial
It doesn't really matter whether you're approaching a number from its right or its left
It will tend to the same value
cunning jackal
so since c is continuous it has every number in the interval so it would be that...
Yes c would be the answer since it is continuous at 1 and 3 as well
faint grove
What is the value of x when x² = 3 * (x + 9)
marsh citrusBOT
What is the value of x when x² = 3 * (x + 9)
desert socket
!status
marsh citrusBOT
What step are you on?
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2. I have begun but got stuck midway
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6. None of the above
1
desert socket
Have you ever solved any similar problem?
faint grove
Same thing with that but with different numbers and without bracket
faint grove
I can't solve when there's bracket
desert socket
Just open the bracket
leaden monolith
I can't solve when there's bracket
What can you do with the bracket
faint grove
The bracket keeps making the answer wrong
desert socket
Show your work
marsh citrusBOT
@faint grove Has your question been resolved?
faint grove
I can't even start because if I try to divide x² to x then you have to multiply x again
The thing is
$x = \sqrt{3 \ast (x+9)}$
elfin berryBOT
아리스킨충∪
stoic saddle
this is legal but will not help you
you didn't answer Frosst's question of "what can you do with the bracket?"
oak dock
@faint grove 한국어가 더 편하시면 말씀하세요. 제가 도와줄 수 있어요.
oak dock
근데 괄호가 있어서
전개 대해 배운적 있으세요?
marsh citrusBOT
@faint grove Has your question been resolved?
marsh citrusBOT
lyric bay
47분.. ㄷㄷ
marsh citrusBOT
Available help channel!
frank orbit
How do I start?
My mind is all over the place
Do I just 10/5 + 12/8 = 3.5 hrs ? Or is there more to it?
marsh citrusBOT
How do I start?
My mind is all over the place
Do I just 10/5 + 12/8 = 3.5 hrs ? ...
@frank orbit Has your question been resolved?
vale jackal
ig u write an expression for time in terms of x
so the running distance is 8-x and the 5km/h swim is $sqrt(100+x^2)$
elfin berryBOT
Ryаn
frank orbit
so the running distance is 8-x and the 5km/h swim is $sqrt(100+x^2)$
Ye I got sqrt(100+x^2)
But isn't the "x" for km?
8 is the speed
vale jackal
which part are u talking about
vale jackal
yeah my bad
frank orbit
j turn it into t(x) = sqrt(100+x^2)/5 + (12-x)/8
Oh lmao holy shit I just need to follow the line-
Cuz this chap is abt differentiation
frank orbit
Hmm where do I apply differentiation?
vale jackal
j turn it into t(x) = sqrt(100+x^2)/5 + (12-x)/8
to this
and then find the zero point
frank orbit
Aight gimme a sec
11 p.m. brain not working well rn
frank orbit
and then find the zero point
$$ t' (x) = \frac {x}{5 \sqrt{100 + x^{2}}}$$
elfin berryBOT
Kai Funaba
marsh citrusBOT
marsh citrusBOT
Available help channel!
fading raptor
Not sure what is being asked of me. Given that the second root has the form 5 + ci, find the other root of the equation.
marsh citrusBOT
Not sure what is being asked of me. Given that the second root has the form 5 + ...
lofty gyro
did you try using Vieta's formula?
marsh citrusBOT
@fading raptor Has your question been resolved?
marsh citrusBOT
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Available help channel!
faint grove
Why ∫^{x}_{0}\frac{k}{k}dk = x
marsh citrusBOT
Why ∫^{x}_{0}\frac{k}{k}dk = x
grizzled bobcat
Why ∫^{x}_{0}\frac{k}{k}dk = x
can you redo this
marsh citrusBOT
@faint grove Has your question been resolved?
faint grove
Is it wrong
marsh citrusBOT
@faint grove Has your question been resolved?
marsh citrusBOT
@faint grove Has your question been resolved?
elfin berryBOT
nebula40
midnight obsidian
faint grove
It results x always
elfin berryBOT
nebula40
faint grove
Yes
versed birch
still temple
Yes
You agree every line has infinite number of points?
Now the distance between very very very close points can be assumes to be dx
Now when we add these very small distances between points one after the another, we get the whole line again
This adding is called , integration
And that's why integral of dx {a very very very small section of x} = x
marsh citrusBOT
@faint grove Has your question been resolved?
marsh citrusBOT
@faint grove Has your question been resolved?
open relic
But it is divided by k
Yes, so k/k = 1. Integral of 1 dk is just k, then you evaluate the limits of the integral
Notice that this works as long as k≠0
open relic
You agree every line has infinite number of points?
Now the distance between ve...
You get integral of 1 dk, here it's explained why that is k
or the length of the segment
etc
marsh citrusBOT
@faint grove Has your question been resolved?
faint grove
Then isn't it \frac{x}{k}
lofty gyro
Why ∫^{x}_{0}\frac{k}{k}dk = x
lofty gyro
Then isn't it \frac{x}{k}
hi, do you still need help with your question?
for your question, you can first simplify the expression inside the integral, i.e. k/k
let's assume k≠0, or else it cannot be calculated.
lofty gyro
**nebula40**
now ,that we have k/k=1 and
int 1 dk = k +C, we have
this definite integral equal to
(x)-(0), = x
any question so far? @faint grove
faint grove
$but_ \frac{k}{k} \neq 1_so_it_is_not$
lofty gyro
$but\_ \frac{k}{k} \neq 1\_so\_it\_is\_not$
why?
elfin berryBOT
아리스킨충∪
elfin berryBOT
lofty gyro
even though k can be different values throughout the integration
for each and every case
k/k is k/k
we won't do (k+1)/k for the same value k
lofty gyro
@faint grove all good till here?
faint grove
Yeah
lofty gyro
then any more question?
faint grove
But it doesn't multiply by x
lofty gyro
But it doesn't multiply by x
which doesn't mutliply by x
lyric bay
Why is this channel open for more than a day
undone bane
Why is this channel open for more than a day
try to say that in #help-8 and you'll get sullied.
lyric bay
bruh
pastel linden
so what’s the question
marsh citrusBOT
marsh citrusBOT
Available help channel!
rapid mulch
Find the y-coordinates of the points (if any) where the circle intersects the y-axis. 9x^2+54x+9y^2-6y+64=0
marsh citrusBOT
Find the y-coordinates of the points (if any) where the circle intersects the y-...
tall pewter
!status
marsh citrusBOT
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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
stark trail
@calm anvil
calm anvil
its kind of like double complete the square to factor that equation
thats what I like to think of it as
crystal lintel
@stark trail
rapid mulch
what do you mean by discrimination of the quadratic in terms of y?
rapid mulch
its kind of like double complete the square to factor that equation
let me try this
calm anvil
discriminate
does a function in terms of y even have a discriminate
,w graph 9y^2-6y+64
elfin berryBOT
calm anvil
z axis tf
rapid mulch
also, i needed to find the center and radius of the equation, yet i keep getting sqrt 23/12 as the radius but the answer is sqrt 2
calm anvil
also, i needed to find the center and radius of the equation, yet i keep getting...
send your workings
I dont think so
rapid mulch
send your workings
calm anvil
,w graph 9x^2+54x+9y^2-6y+64=0
elfin berryBOT
calm anvil
you have to use the constant to make the 9 and 1/36
you can't just make numbers out of nothing
rapid mulch
you have to use the constant to make the 9 and 1/36
what do you mean by constant
calm anvil
64/9
its a constant cause its value doesn't change
x and x^2 change with different values of x
rapid mulch
ohmygod wait, so its not +1/36 itd be +1/9 right?
because for completing the square id take -2/3 half it to get -1/3 and then square it to get 1/9
rapid mulch
HOYL SHIT I GOT IT
calm anvil
or you add and subtract the same value
rapid mulch
i realized i was subtracting 1/9 from -64/9 rather than adding it
calm anvil
oh nvm I realized you did just not in the right line
rapid mulch
yeah i now got it, because i was getting like 7.11 something when adding everything (obviously incorrectly) so i was getting weird numbers as a result
okay yeah im set then, thank you so much for the help
.close
woven quarry
marsh citrusBOT
@woven quarry Has your question been resolved?
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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above
1
marsh citrusBOT
@woven quarry Has your question been resolved?
cunning fiber
$\lim_{x \to 2^{-}} f(x)$ represents the value $f$ approaches as $x$ approaches $2$ from the left \ \ Similarly, $\lim_{x \to 2^{+}} f(x)$ represents the value $f$ approaches as $x$ approaches $2$ from the right
elfin berryBOT
Civil Service Pigeon
cunning fiber
$\lim_{x \to 2} f(x)$ represents the value $f$ approaches as $x$ approaches $2$. We say that if $$\lim_{x \to 2^{-}}=\lim_{x \to 2^{+}} f(x)$$ then $\lim_{x \to 2} f(x)$ is equal to both of these limits. \ \ However, if these two one-sided limits are unequal, then $\lim_{x \to 2} f(x)$ does not exist
elfin berryBOT
Civil Service Pigeon
cunning fiber
$f(-2)$ just represents the value of $f$ when $x=-2$, whichc am also be read off the graph
elfin berryBOT
Civil Service Pigeon
strong yacht
cannot find out how to do this problem
marsh citrusBOT
cannot find out how to do this problem
@strong yacht Has your question been resolved?
strong yacht
<@&286206848099549185>
still temple
cannot find out how to do this problem
Distance formula
strong yacht
how would that work for the points provided?
still temple
correctly
strong yacht
so it would be like this?
still temple
yes
btw the distance formula is just pythagorean theorem
but instead of $$c^2=a^2+b^2$$ it's $$c= \sqrt{a^2+b^2}$$
elfin berryBOT
Free Geoffrey
still temple
$(x_2-x_1)$ the horizontal leg and $(y_2-y_1)$ being the vertical leg
elfin berryBOT
Free Geoffrey
strong yacht
and how would i find the midpoint?
tender mantle
marsh citrusBOT
tender mantle
Any short method?
plush elk
Multiply by x both above and below
stoic saddle
substitute x = e^t maybe?
oh but then you get something nasty anyway nvm
the shortest method would be to take the derivative of each of these things tbh
and see which one matches
plush elk
It doesn’t look very nasty. I looked at it and think it can be proceed by integration by part
tender mantle
Multiply by x both above and below
What next?
plush elk
It becomes solving integral of (e^t)(t-1)/(t+1)
Oh you mean what next after that, dx/x=d(log(x)), exactly what Ann said
tender mantle
Nope. I am trying your method
stoic saddle
C is the one option that definitely doesnt match
tender mantle
Should i apply L'Hospital?
plush elk
Oh I forgot. If you want you don’t have to solve it
stoic saddle
Should i apply L'Hospital?
to what??
plush elk
You can check each options …
stoic saddle
where do you see any place where l'hop even COULD be applied...
there are no limits
there is an integral
tender mantle
Let me check all the options derivative
I got option B
Simplyyyyyyy
Thank you all
I guess D?
plush elk
It’s solvable though, in the end it becomes solving integral of e^t/t
tender mantle
Should I first apply x up down?
plush elk
That is the same as what Ann said
tender mantle
It becomes solving integral of (e^t)(t-1)/(t+1)
how u got this?
plush elk
That’s after using integration by part
tender mantle
On t/(1+t^2)d(e^t)
d(e^t) comes from?
plush elk
(t/(1+t)^2)d(e^t)=(t/(1+t)^2)e^t- e^t(d(t/(1+t)^2))
tender mantle
$(t/(1+t)^2)d(e^t)=(t/(1+t)^2)e^t- e^t(d(t/(1+t)^2))$
elfin berryBOT
! Arjunn
tender mantle
ohh A+B partition
plush elk
Okay
plush elk
ohh A+B partition
On second thought
marsh citrusBOT
On second thought
plush elk
Maybe this is more effective:
tender mantle
.reopen
plush elk
You are looking for the antiderivative of (te^t)/(1+t)^2, we can just set it be g(t)e^t/(1+t)^2. So
(1+t)^2g’(t)-2(1+t)g(t)+(1+t)^2g(t)=t(1+t)^2
(1+t)g’(t)+(t-1)g(t)=t^2+t
So we set g(t)=a+bt
We can solve that a=b=1
Just like that it is solved
@tender mantle
tender mantle
-x?
plush elk
Because sin(x)dx=-d(cos(x))
plush elk
Yeah
plush elk
Oh wait I have a better idea
tender mantle
Sure sure
plush elk
Call this integral A
Now substitute x with π-y
It becomes integral of y from 0 to π, with any x replaced by π-y
In the end, y is just a symbol you can replace it with x again
So 2A=A+A=π (integral of tan(x)/(sec(x)+cos(x)) dx)
This x is canceled
So A=(π/2) integral of (tan(x)/(sec(x)+cos(x)) dx
tender mantle
hmm it is a litle but complicated and lengthy
is there no short way to solvve by options?
from a book
plush elk
Therefore A=π integral of tan(x)/(sec(x)+cos(x)) dx from 0 to π/2.
This last integral by our previous discussion
Is -1/(1+cos^2(x)) dcos(x)
So difference of -arctan(cos(x)) , at x=π/2 and x=0
=arctan(1)-arctan(0)=π/4
Multiplied by π, answer is π^2/4
@tender mantle
tender mantle
is there any other method?
plush elk
You better get used to it
tender mantle
it is a little bit complicated
plush elk
Because I view it as perfectly normal
That’s how they normally look like, no reason to feel complicated
tender mantle
it is not normal for me in the exam to instinct of pi-y
plush elk
I used it several times, I think it’s common technique
tender mantle
right, i'll need to work out with it
tender mantle
when do we use this trick mostly i meant in which form
plush elk
Whenever you have a feeling of some kind of symmetry going on
Like this case
I sensed a feeling of symmetry
For this question I can’t even think of any other way. If the author is going to give an answer as a standard one, it probably also goes like this
No reason not to use x+π-x to cancel x
Anyway I am going to ping you one more time so that you can find this dialogue. I will close the channel I think. @tender mantle
.close
marsh citrusBOT
stoic saddle
completing the square
tender mantle
sure
tender mantle
second term is 23x so here we know our square will have 2ab so a=x b=3
(a+b)^2= a^2+2ab+b^2
compare it with that equation
(x+3)^2 so they need extra 3^2 so they give it and minus it for making it perfect square
fine
do it 2 and 3 times
and you will understand it more
marsh citrusBOT
@still temple Has your question been resolved?
marsh citrusBOT
@still temple Has your question been resolved?
elfin berryBOT
tender mantle
because it has two circle surface up and down which area is pi r^2 so they are two so 2pi r^2
but i guess it should be different
2 pi (R-r)^2
marsh citrusBOT
marsh citrusBOT
Available help channel!
sleek crater
someone
marsh citrusBOT
marsh citrusBOT
Available help channel!
tame cedar
What in the world does b, c, d, and e mean exactly?
marsh citrusBOT
What in the world does b, c, d, and e mean exactly?
tame cedar
marsh citrusBOT
@tame cedar Has your question been resolved?
marsh citrusBOT
@tame cedar Has your question been resolved?
marsh citrusBOT
@tame cedar Has your question been resolved?
strong pecan
why when u do vectors u ignore the b in mx+b in 2 dimensional vectors
strong pecan
say we have 5x-3=y
and 7-7x=y2
how come the vector would be <1,-7> and <1,5>
not <1,2> and <1,0>
dont vectors give directions to where a line is going
hazy lion
dont vectors give directions to where a line is going
its maybe more useful to think of vectors as describing like
a displacement
so your vectors here are describing the pair of like
given some change in x, how much would you move in y
also, conventionally vectors arent thought of living anywhere at all, so the +b would be extraneous
sometimes said like "vectors are arrows you can move around"
or whatever
strong pecan
also, conventionally vectors arent thought of living anywhere at all, so the +b ...
ah i c
a displacement
in a point at some time
we describe the displacement
but how come the +b isnt included in that displacement
hazy lion
but how come the +b isnt included in that displacement
maybe going back to that vectors are arrows that can move is helpful
maybe you can think of idk
think of like a speed
if i say he was going 80 mph for 10 minutes
does it matter if he went from bobs house to james house
or james house into the desert
or any other direction
strong pecan
right
hazy lion
were talking about a different collection of information
strong pecan
so it'
it's less information on a specific line
and moreso just
how the line changes from time to time?
cuz isnt that what vectors are
the components of something
or thats what they represent kind of right
hazy lion
you can write lines as vectors
if you want
and encode the intercept in a vector
idk that its really helpful here though
hazy lion
how the line changes from time to time?
yea, thats one thing
i mean you already know its just encoding m, right?
the vector holds the slope
its describing how some point on the line moves given a change in x
hazy lion
if you really really feel like encoding the intercept too
strong pecan
the vectors just describe
hazy lion
you can write it like
strong pecan
a change in something
just because we start somewhere
doesnt mean it effects how it'll be changed
or the rate something will change at
elfin berryBOT
jan Niku
hazy lion
the way this works is you start somewhere
(x_0, y_0)
the m vector is the vector you found in your original post
it just encodes how y changes as you change x
t here is the variable
strong pecan
right
hazy lion
so you can think of this as like
since you can move vectors around
put the (x_0, y_0) vector at the origin
pointing to the place where you start
then put the mt vector at the end of the first vector
pointing to where you land
after t time elapses
wonder if theres a desmos
anyways
hazy lion
doesnt mean it effects how it'll be changed
you got the idea here i think
ah
@strong pecan https://teacher.desmos.com/activitybuilder/custom/5ecc2e10d9e6692d1366bc5e#preview/ffc1ff10-4283-42e7-b1ee-a3fca766c04b
if you'd like
strong pecan
ty!
hazy lion
a short activity that goes over this sort of way of writing a line with some nice visuals
slide 7 is really nice
strong pecan
ahhh
that makes sense
so like when we have a slope or 3/4
we can say <3,4>
ok ok that makes a lot of sense
tysm
hazy lion
np 
strong pecan
np <:happy:908558103992225822>
can u help me w/ a quick q
i have this q
Find two unit vectors that make an angle of 60° with<6,8>
ik thats j
lets call it vector v
vdot some vector lets say y
vdoty=cos60*10
5=vdoty
5=6x+8y
but highkey im lost on how to get the y vector
strong pecan
np <:happy:908558103992225822>
r u here
hazy lion
dont laugh im trying to remind myself how to do this without just using geometry lol
hazy lion
well you could draw triangles
usually last resort
heres my thought and it might be dumb
let $u$ be your original vector
elfin berryBOT
jan Niku
hazy lion
let $v$ be the one you want to find
elfin berryBOT
jan Niku
hazy lion
so obviously $v$ is a unit vector
elfin berryBOT
jan Niku
hazy lion
then uhh
strong pecan
udotv=umagnitude*vmagnitude=costheta(angle between vectors)
hazy lion
$v_1 ^2 + v_2^2 = 1$
elfin berryBOT
jan Niku
hazy lion
then
$u \cdot v = |u| |v| \cos \theta = 10 \cos \theta$ giving $3v_1 + 4v_2 = 5\cos \theta$
elfin berryBOT
jan Niku
hazy lion
so theres 2 equations
can we just do like
$3 v_1 + 4 \qty( \frac{5 \cos \theta - 3v_1 }{ 4 } )^2 = 1$
elfin berryBOT
jan Niku
strong pecan
how did u get 3v1+4v2=5costheta
hazy lion
okay so $u \cdot v = |u||v| \cos \theta$
elfin berryBOT
jan Niku
hazy lion
this is just a definition
strong pecan
right
elfin berryBOT
jan Niku
hazy lion
so $u \cdot v = u_1 v_1 + u_2 v_2$
elfin berryBOT
jan Niku
strong pecan
icic
hazy lion
oh im just being dumb 
strong pecan
they give us the angle
hazy lion
woe
no no wait
this is fine
sorry man tbh my sleeping pills are kicking in im pushing through jello
we can finish this
elfin berryBOT
jan Niku
hazy lion
it follows $u \cdot v = 6v_1 + 8 v_2$
elfin berryBOT
jan Niku
hazy lion
thats the LHS
strong pecan
ye
hazy lion
the right hand side is $|u||v|\cos \theta = 1 \cdot 10 \cdot \cos \theta$
elfin berryBOT
jan Niku
strong pecan
ye
hazy lion
then $6 v_1 + 8 v_2 = 10 \cos \theta$
elfin berryBOT
jan Niku
strong pecan
right
hazy lion
so divide by 2
hazy lion
how did u get 3v1+4v2=5costheta
that gets you this
strong pecan
why divide by 2
hazy lion
no reason
hazy lion
$3 v_1 + 4 \qty( \frac{5 \cos \theta - 3v_1 }{ 4 } )^2 = 1$
elfin berryBOT
jan Niku
hazy lion
im curious if we can just do this
it should result in a quadratic? so thatd give you the two solutions
strong pecan
that makes sense
strong pecan
i got to that point but i thought i had to make the vector instead of solve for x and y
hazy lion
i dont get what you mean
strong pecan
like i considered y=5/8 - 3/4x a line, and the vector
<4,-3>
but that doesnt make sense
i like got the unit vector of that then
strong pecan
**jan Niku**
if u plug in pi/3 here
u get 5=6v1+8v2
i j did v2=5/8-3/4v1
but i j like thought that was the vector for some reason
idk
but thats only the y component
ok
i sleep
thank you
hazy lion
**jan Niku**
this is just a quadratic for v1
the first component of v
turn it into a root finding problem
(something)v_1^2 + (something) v_1 + (something) = 0
use the quadratic formula
hazy lion
yea
strong pecan
Or not formula sorry
hazy lion
its just solving a quadratic from here
strong pecan
What we got
hazy lion
thats the whole problem
hazy lion
,w 3x + 4 ( (5pi/3 - 3x)/4 ) ^2 = 1
elfin berryBOT
hazy lion
hazy lion
oh, yea, its wrong
strong pecan
It’s cos
hazy lion
,calc cos(pi/3)
elfin berryBOT
Result:
0.5
bonk
strong pecan
Bru
hazy lion
,w 3x + 4 ( (0.5 - 3x)/4 )^2 = 1
elfin berryBOT
strong pecan
**jan Niku**
How did u get the squared
elfin berryBOT
jan Niku
hazy lion
so $v_2^2 = \qty( \frac{ 5 \cos \theta - 3 v_1 }{ 4 } )^2$
elfin berryBOT
jan Niku
strong pecan
I c i c
Wait wuh
Maybe I’m being dumb
How did we insert v2^2
When we only have v2 in the equation
hazy lion
we squared it
strong pecan
Holy ahit
I’m so uh
Lol
I c what y did
Is there really no easier way to do thi
Very tedious
No way we could do this w/o squaring
hazy lion
i guess you could draw a triangle
hazy lion
marsh citrusBOT
@strong pecan Has your question been resolved?
rotund whale
If a, b, c are real numbers and a + b + c = 0, and a^2 + b^2 + c^2 = 1 and a^2 * b^2 * c^2 < or = m/n then find n - m.
marsh citrusBOT
If a, b, c are real numbers and a + b + c = 0, and a^2 + b^2 + c^2 = 1 and a^2 *...
rotund whale
<@&286206848099549185>
marsh citrusBOT
@rotund whale Has your question been resolved?
rotund whale
Ok.
elfin berryBOT
Fossil
rotund whale
Uh.
rotund whale
Except zero.
eternal sand
u should say that eariler
so m is not equal to n
m / n > 1
then there's infinite solutions???
m = 3, n = 2
2 - 3 = -1
m = 6. n = 5
infinite solutions
elfin berryBOT
AnotherColdWind
rotund whale
We have,
$$ \begin{center} a^2 + b^2 + c^2 = 1 \text{ and } a + b + c = 0 \end{center} $$
elfin berryBOT
AnotherColdWind
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
eternal sand
ys
rotund whale
I forgor how to use it 💀
elfin berryBOT
AnotherColdWind
eternal sand
ys u may do that
rotund whale
Also.
eternal sand
(-(b+c))^2 + b^2 + c^2 = 1
elfin berryBOT
AnotherColdWind
eternal sand
umm
rotund whale
It's a property.
rotund whale
If you want the proof here.
marsh citrusBOT
@rotund whale Has your question been resolved?
rotund whale
<@&286206848099549185>
crystal sand
<@&286206848099549185>
What's your progress as of now?
marsh citrusBOT
@rotund whale Has your question been resolved?
rotund whale
What's your progress as of now?
None as of now.
marsh citrusBOT
@rotund whale Has your question been resolved?
plush elk
x^3-s_1x^2+s_2x-s_3=0
s_1=0
s_1^2=t_2+2s_2, t_2=1
Therefore s_2=-1/2
x^3+(-1/2)x-s_3=0
The discriminant of this cubic should >=0
So -4(-1/2)^3-27(s_3)^2>=0
s_3^2<=1/54
m=1, n=54, n-m=53
s_1=a+b+c, s_2=ab+bc+ac, s_3=abc, t_k=a^k+b^k+c^k in my steps
@rotund whale
rotund whale
I didn't get it.
Also I haven't learnt about the discriminant of cubic equations.
I guess I'll learn it now.
rotund whale
s_1^2=t_2+2s_2, t_2=1
I didn't get it.
plush elk
I guess I'll learn it now.
Good
rotund whale
But what's t_2 and the other stuff?
plush elk
s_1=a+b+c, s_2=ab+bc+ac, s_3=abc, t_k=a^k+b^k+c^k in my steps
Notation explained here
plush elk
Np
rotund whale
Is k = 2?
rotund whale
Oh ok.
plush elk
About discriminant, simply this:
Cubic x^3+px+q
Discriminant of it is defined to be:
-4p^3-27q^2
Discriminant >0 <-> three real roots (not the same)
=0 <-> three same real roots
<0 <-> a real root and a pair of conjugate complex roots
rotund whale
So, here we have one real root that is x - 54 or something like that.
rotund whale
x^3-s_1x^2+s_2x-s_3=0
I have a small doubt (I know the general form of cubic equation and signs of co-efficients) but how did you get it in this form or did you assume it to be?
plush elk
x^3+ax^2+bx+c
Let
y=(x+a/3)
Then it becomes y^3+py+q for some p,q
power 2 term canceled
plush elk
x^3+ax^2+bx+c
Now replace x with y-a/3 in this, try
rotund whale
Ok.
elfin berryBOT
AnotherColdWind
rotund whale
Now replace x with y-a/3 in this, try
You there?
rotund whale
How?
rotund whale
y^2 cancelled
How though?
rotund whale
Ah ok.
sacred loom
1
sacred loom
30a+24b
olive compass
Thanks
sacred loom
:)
sacred loom
Bro u smarter than me and I’m in college 😭
wait this is not college maths
cunning basin
olive compass
wait this is not college maths
It is I’m just bad at math so they put me in it
sacred loom
It is I’m just bad at math so they put me in it
what lol its like 8 grade maths
olive compass
I know 💀
still temple
wtf kinda college are you in bro
marsh citrusBOT
@rotund whale Has your question been resolved?
plush elk
You can close it. It’s perfectly solved using discriminant. If you don’t know the proof of that statement about discriminant you can’t get explanations in one of these channels. You need to read, on net or on some textbooks. Algebra textbooks will have it. it’s direct result from solving roots of cubic.
spiral wraith
How
spiral wraith
Just putting it into sin x over cos x form
And im stuck there
My help sheet doesnt help 💀
brave marsh
Multiply top and bottom by cos(x).
Don't distribute the bottom, do it for the top though.
marsh citrusBOT
@spiral wraith Has your question been resolved?
spiral wraith
Where is there 1/cos x
plush elk
Denominator
sin-cos=cos(tan-1)
(1-tan)/(sin-cos)=(1-tan)/cos(tan-1)=-(1/cos)
Since tan doesn’t equal 1 near π/4
marsh citrusBOT
@spiral wraith Has your question been resolved?
plush elk
? I almost told you the answer
You don’t know cos(π/4)?
It’s 1/sqrt(2)
So -1/cos(π/4)=-sqrt(2)
-1/cos is continuous near π/4
plush elk
Np
cloud dawn
marsh citrusBOT
cloud dawn
Okay I just wanna check
Am I supposed to be getting the fraction as the answer
or does this thing just want me to convert it to 3/4 and 3/2
maiden thistle
they probably want the answer in fraction form
cloud dawn
Gotcha so I'm essentially right just need to convert the decimals
maiden thistle
yeah try it, see if that works
uneven torrent
marsh citrusBOT
uneven torrent
I’m stuck on what my graph for 2a should look like, this is what I have but I don’t know it
uneven torrent
this is the question
marsh citrusBOT
@uneven torrent Has your question been resolved?
marsh citrusBOT
@uneven torrent Has your question been resolved?
sonic elm
marsh citrusBOT
sonic elm
hello! does anyone know how to do this? i have no idea where to begin with this.
i believe my teacher wants it in IVT sentence form, whatever that means
nova totem
Find h(1) and h(3)
sonic elm
what would I do after?
sonic elm
Do you even know how IVT works
i do not, no
well, I was introduced to it but haven't been taught it
nova totem
I'll write it down
nova totem
Basically, if:
- f(x) is continuous for some interval, x in [a, b]
Then there exists a value, c, such that:
- a < c < b
- f(c) is a value between f(a) and f(b)
Aka either f(a) < f(c) < f(b) or f(b) < f(c) < f(a)
sonic elm
got it. i think i understood that part, but i am not quite sure as to how to put it into a sentence form
for example, given values i’m not sure how to write it like
“Yes there is a value of c such that #<c< # such that……”
that’s about all i know
nova totem
You'd first need to state or show that h(x) is continuous for x in (1, 3)
And then you'd need to find and state the values of h(1) and h(3)
You then need to see if 8 lies in between h(1) and h(3)
If it does, IVT guarantees that there is a value 1 < r < 3 such that h(r) = 8
Otherwise it doesn't guarantee
marsh citrusBOT
@sonic elm Has your question been resolved?
sonic elm
got it, thank you so much
marsh citrusBOT
marsh citrusBOT
Available help channel!
viscid sky
marsh citrusBOT
viscid sky
I tried converting into cos/sin form but still unable to get ans
plush elk
Write cot(π/9) as x
viscid sky
ok thn
viscid sky
didnt understood
marsh citrusBOT
@viscid sky Has your question been resolved?
viscid sky
<@&286206848099549185> ?
long viper
I believe this might be what was intended by Cogwheels' answer:
We have the equation:
$$ \cot^2(\pi / 9) + \cot^2(2\pi / 9) + cot^2(4\pi / 9) $$
Using the following trigonometric identity, let's try and get the inside of all $\cot$ expressions to be $\pi / 9$:
$$ \cot (2u) = \frac{\cot^2(u) - 1}{2 \cdot \cot(u)} $$
(since once they're all the same, we can do some work to simplify)
elfin berryBOT
kiako
viscid sky
i dont know the identity cot(2u)
long viper
Ah
viscid sky
no other way than that?
long viper
Would you happen to have some sort of list of trig identities?
viscid sky
I have list of formula of trig
oh w8 i know this identity
i have it in terms of Tan so converted it to cot
long viper
Ah, that makes sense.
viscid sky
continue
long viper
Well, what should we use for u in our first round of substitutions?
viscid sky
pi/9
long viper
alright, in that case I presume we're going to substitute into the first one?
Just for simplicity let's define these:
We want to find $A + B + C$, where:
\begin{align*}
A &= \cot^2 (\pi / 9) \
B &= \cot^2 (2\pi / 9) \
C &= \cot^2 (4\pi / 9)
\end{align*}
viscid sky
ok
elfin berryBOT
kiako
long viper
Okay, using the identity on $B$ with $u = \pi/9$, what do we get?
elfin berryBOT
kiako
viscid sky
[ cot^2(pi/9) -1 / 2cot(pi/9) ]^2
long viper
just about! remember that B is cot^2, so we're going to have to remember to bring back the square
\begin{align*}
\cot^2(2u) &= \left( \frac{cot^2{u} - 1}{2 \cdot \cot (u)} \right)^2 \
&= \left( \frac{cot^2(\pi / 9) - 1}{2 \cdot \cot (\pi / 9)} \right)^2
\end{align*}
elfin berryBOT
kiako
viscid sky
got it
long viper
alright, let's look at that last one
since it's 4 times pi/9, we're going to need to do a bit more work
long viper
yep, with u being what in this case?
viscid sky
sry?
long viper
I agree that we can say cot^2(4 pi/9) = cot^2(2u), but that won't work if we do u = pi/9, so what should we use in this case?
viscid sky
u=2pi/9?
long viper
yep!
And that leaves us with
\begin{align*}
C &= \cot^2(2u) \ &= \left( \frac{\cot^2(u) - 1}{2 \cdot \cot (u)} \right)^2 \
&= \left( \frac{cot^2(2 \pi / 9) - 1}{2 \cdot \cot (2 \pi / 9)} \right)^2
\end{align*}
viscid sky
k
elfin berryBOT
kiako
long viper
we're still going to have to use another round of identites, though, since we still have 2pi/9 inside a cot
viscid sky
yes
long viper
are you alright with me just pulling that in from what we computed with B?
viscid sky
what if we just substitute tha value we found of cot^2 2u in this ans
long viper
we can do that, yeah! we'd also want to replace the value in the denominator with cot(2u) (which is just cot^2(2u) without the square at the end)
long viper
Sure! Just for completeness, this is what it looks like:
$$ \left( \frac{\left( \frac{\cot^2(\pi / 9) - 1}{2 \cdot \cot (\pi / 9)} \right)^2 - 1}{2 \cdot \frac{\cot^2(\pi / 9) - 1}{2 \cdot \cot (\pi / 9)}} \right)^2 $$
viscid sky
yup
elfin berryBOT
kiako
viscid sky
Lets assume Cot pi/9 as Y?
long viper
Alright, so our total goal is to compute:
$$ \cot^2(\pi / 9) + \left( \frac{cot^2(\pi / 9) - 1}{2 \cdot \cot (\pi / 9)} \right)^2 + \left( \frac{\left( \frac{\cot^2(\pi / 9) - 1}{2 \cdot \cot (\pi / 9)} \right)^2 - 1}{2 \cdot \frac{\cot^2(\pi / 9) - 1}{2 \cdot \cot (\pi / 9)}} \right)^2 $$
elfin berryBOT
kiako
long viper
Ah, yes, that's a good idea.
$$ Y^2 + \left( \frac{Y^2 - 1}{2 \cdot Y} \right)^2 + \left( \frac{\left( \frac{Y^2 - 1}{2 \cdot Y} \right)^2 - 1}{2 \cdot \frac{Y^2 - 1}{2 \cdot Y}} \right)^2 $$
elfin berryBOT
kiako
long viper
I believe this is what we're looking at?
long viper
Do you think you'll be able to work through simplifying this on your own or would you like to work through it together?
viscid sky
ill tell if i get stuck
long viper
Alright!
viscid sky
If im opening the brackets
its getting to the power 4
biquadratic eq forms in last bracket without opening whole squre
i simplifies and got a eq
long viper
I should ask; are you allowed to use a calculator on this question?
viscid sky
Ofc no
I just typed the eqn to send you here
Didn't use calculator I did it in rough work
plush elk
As I said in the beginning he can express everything in terms of $x=\cot(\frac{\pi}{9})$ and he can solve x by $\frac{\frac{x^{2}-1}{2}-1}{x+\frac{x^{2}-1}{2x}}=\frac{1}{\sqrt{3}}$, quadratic
elfin berryBOT
Cogwheels of the mind
plush elk
Or cubic, but solvable
plush elk
Didn’t read what you have discussed
So there should be only one thing to solve, what is it?
Anyway, that one thing whatever it is, can be solved by cubic
If you know how to solve roots of cubic polynomials then done
viscid sky
I got biquadratic eqn
marsh citrusBOT
@viscid sky Has your question been resolved?
marsh citrusBOT
@viscid sky Has your question been resolved?
marsh citrusBOT
@viscid sky Has your question been resolved?
marsh citrusBOT
marsh citrusBOT
Available help channel!
still temple
need help with algebra
marsh citrusBOT
need help with algebra
cobalt sentinel
That’s definitely integral calculator right?
deft apex
You can play around with factoring
still temple
You can play around with factoring
i thought multiplying both top and bottom by 16 was the way to go but i got the answer wrong by a factor of 1/4
deft apex
Hmm
smoky plover
i thought multiplying both top and bottom by 16 was the way to go but i got the ...
It should
deft apex
Just looking at x^2/4 - 1 is there not something you can take out to get your goal?
still temple
Just looking at x^2/4 - 1 is there not something you can take out to get your go...
i thought that multiplying by 16 would work because sqrt of 16 is 4 so inside it would be like doing x4?
idk
i will show my work for whole problem
smoky plover
No multiply by 2, since sqrt(4)=2
smoky plover
Yeah easy mistake to make
still temple
also
what to do about absolute value? textbook answer is diff than integral calc answer
deft apex
Just looking at x^2/4 - 1 is there not something you can take out to get your go...
All im getting at here is 1/4(x^2 - 4) gets your goal asap and sqrt(1/4) can just be (1/2) , think it should lead to nice stuff
still temple
what does this mean
the textbook answer missing the absolute value x on bottom and the x after arcsec on top
.close
marsh citrusBOT