that makes more sense thanks
#help-17
clear sundial
lucid bane
you're subtracting the sum from 1...(k-1)
clear sundial
.close
ionic dagger
vocal sleetBOT
cyan shadow
the formula is $\sum_A = (n-2)\cdot180$
cyan shadow
A is total angle, n is number of sides
twin meteorBOT
cyan shadow
this totation is still wrong but close enough to be understandable
ionic dagger
o
cyan shadow
"the sum of the measure of the interior angles of an n-gon is 180(n-2)"
ionic dagger
so is the last one 42
i think it goes 5, 8, 12, 10, 13, 20, 16, and 42
i think i got it right
but idk
cyan shadow
no
the last one is not 42
i didnt go through and check the rest though
they start being wrong after 5, 8, 12
ionic dagger
oh
cyan shadow
I showed you the formula to calculate them
cyan shadow
"the sum of the measure of the interior angles of an n-gon is 180(n-2)"
you just need to use this fact
can you see how to apply it knowing the sum of the internal angles is 1620?
ionic dagger
yes
cyan shadow
triangles have 3 sides
ionic dagger
yeah
ionic dagger
oh
ionic dagger
can you apply the formula to figure out number of sides?
360(n-4)
ionic dagger
ohh
so its 360(4-2)
so it would be 720
wait
but whats the # for that
i would puti n the answer box
cyan shadow
i have no idea what you are doing
cyan shadow
its total angle = 180(#sides - 2)
given this formula, knowing that the total angle is 360, how many sides are there on the shape?
ionic dagger
4
cyan shadow
how did you do that
cyan shadow
I do not want you to use the square fact
i want you to use the formula
lets say we have 540 degrees internal angles
ionic dagger
ok
cyan shadow
its total angle = 180(#sides - 2)
how many sides, using this formula?
ionic dagger
180(3-2)
cyan shadow
geometry is after algebra, no?
ionic dagger
yes
cyan shadow
180(3-2) = 180(1) = 180
ionic dagger
im doing both at the same time but im further in algebra
cyan shadow
that is not equal to 540
ionic dagger
is the last one 44
cyan shadow
you have $540 = 180(n-2)$ and you need to solve for n
twin meteorBOT
cyan shadow
is the last one 44
yes, but a bunch of them in the middle are wrong as well
ionic dagger
i did
cyan shadow
but if you figured out that mistake, maybe you can fix the rest
cyan shadow
yeah thats right
cyan shadow
whats the angle?
cyan shadow
ya 18 is right
cyan shadow
no
cyan shadow
yes
vocal sleetBOT
vocal sleetBOT
Available help channel!
lean wyvern
vocal sleetBOT
lean wyvern
How to do this, I everytime try to make the triangle, it felt like it doesnt fit
someone help
vocal sleetBOT
@lean wyvern Has your question been resolved?
lean wyvern
<@&286206848099549185>
copper axle
that's a learning activity to learn what you need to make similar triangle for example:
when the sides of all 3 sides of a triangle is known
copper axle
- when 2/3 angle are congruent to 2 other angles of an other triangle
lean wyvern
then
copper axle
lean wyvern
did you read the question uhhhhhh
this feels different but
Triangle ABC must be congruent to Triangle XYZ
copper axle
yes not the best picture
lean wyvern
in which Angle B is congruent to Angle X, C congruent to Z, and A congruent to Y
try to draw it
lean wyvern
show me the figure
copper axle
it's the same triangle but bigger
lean wyvern
of what have you drawn so I can really get it
lean wyvern
im sorry
but I do not still get it
but What seems to be true when three angles of one triangle are congruent to
two angles of another triangle?
the thing is
I do not kno
since it aint even somewhere in the triangle postulates
lean wyvern
are you saying that segments are not gonna be congruent?
lean wyvern
I dont get it still unfortunately since only three angles are supposed to be congruent to two angles
either the segments are not congruent
copper axle
the segments are not congruent but they are proportionnal to each other
since the sum of all angles inside a triangle is 180° then if 2 angles are congruent, the 3rd also is
lean wyvern
congruent
copper axle
yes
copper axle
using THOSE? really ? good luck...
copper axle
i'm here to explain what you have to do, not do it for you.
copper axle
decide what angles will be the triangle, decide what mAB is gonna be then draw it
then using those sames angles, draw an other triange with a different mAB
lean wyvern
ok
copper axle
also since they used that symbol, the angles can be approximately the same no need to be exact
vocal sleetBOT
@lean wyvern Has your question been resolved?
knotty shell
im getting an infinite number of solutions, wondering what im doing wrong
knotty shell
ye my b i close other
i got f=y^2e^y -x^2e^y
i got -2xe^y=0
y^2e^y +e*ye^y -x^2e^y =0
i j added the two equations
got infinite number
vocal sleetBOT
@knotty shell Has your question been resolved?
knotty shell
<@&286206848099549185>
tribal moss
im getting an infinite number of solutions, wondering what im doing wrong
$f'{x}=-2xe^{y}=0\Leftrightarrow x=0\\f'{y}=e^{y}\left( y^{2}-x^{2} \right)+2ye^{y}=0\Leftrightarrow y^{2}+2y=0\\\Leftrightarrow y=0\vee y=-2$
twin meteorBOT
Joanna Angel
tribal moss
so you get two points: (0,0) and (0, -2)
vocal sleetBOT
@knotty shell Has your question been resolved?
knotty shell
$f'_{x}=-2xe^{y}=0\Leftrightarrow x=0\\\\f'_{y}=e^{y}\left( y^{2}-x^{2} \right)+...
u cant add the two equations?
then wouldnt u get y^2e^y-x^2e^y-2xe^y+2ye^y
so whenever x=y its valid?
why is that wrong
tribal moss
u cant add the two equations?
you add equations, only sometimes
knotty shell
wait why am i being stupid
knotty shell
you add equations, only sometimes
when cant you
cuz now when i look at it obv x has to be 0
tribal moss
when cant you
you should ask , when you may, you can add only if addition helps you to solve it, i did not need to add them at all
tribal moss
sure, since you have only two critical points
floral gyro
vocal sleetBOT
floral gyro
I used FTC to do the integral
but now i dont know what to do
i also expanded 2+h cubed
but theres no way to cancel out the h
surreal bone
what level of calculus is this?
floral gyro
calc 1
surreal bone
huh weird I don't remember doing anything similar to this
outer dune
i feel like hospital is useful
surreal bone
yeah definitely
surreal bone
I thought that was a given lol oops
outer dune
Ok your thing from the first to second line is not true
floral gyro
how so?
outer dune
You don’t need to integrate that function
floral gyro
oh ok
outer dune
Ohhh i see
Yea so what you have confused is that it’s not f(b)-f(a), it’s F(b)-F(a) where F is the integral
(the theorem)
And so yes the derivative cancels them out
floral gyro
sorry FTC has two parts tho
i used the one where its a constant number
to a function b
im not sure what you mean sorry
outer dune
Ok so you know the integral of f(x) from a to b is F(b)-F(a)
Or in this example, it would be F(2+h)-F(2)
floral gyro
but h is a function
outer dune
Well h is a variable
outer dune
Yes
floral gyro
so can't i just substitute 2+h to get it
outer dune
Yes you can
floral gyro
was subbing two in my mistake then?
outer dune
I was just trying to explain where this came from
outer dune
Well I don’t know why you substituted
floral gyro
what should i do instead
i can sub h in to cancel them out after lhopitals
so i thought it was the way
outer dune
I think you can keep it like root(1+ (2+h)^3
Yes and also the h should go away after hopital
floral gyro
ok ill try
outer dune
Im also not sure how you get from the second to last step to the last step
outer dune
Ok that seems very complicated
I think the easiest way is the write the initial problem and immediately use lhopital, and with FTC 1 you can figure it out
outer dune
Wait wait wait
After your first step, apply hospital
So it’s derivative of that integral / derivative of h
And use FTC 1
floral gyro
u were saying that i would have to integrate to do F(b) - F(a)
outer dune
And you didn’t even need to do the integral
outer dune
Ofc
floral gyro
.close
quiet vector
hey i was just wondering im doing integration by integrating factor. how come sometimes there is no absolute value in integrating?
vocal sleetBOT
hey i was just wondering im doing integration by integrating factor. how come so...
west raptor
shouldnt these both be absolute value
Both ln |cos x| and ln(cos x) are antiderivatives of - tan x, but they are antiderivatives of - tan x on different intervals
Here's the original definition of an antiderivative:
vocal sleetBOT
@quiet vector Has your question been resolved?
quiet vector
Here's the original definition of an antiderivative:
i still dont really understand tbh, so because for a specific interval in ln(cosx) (ie. where cosx >= 0) yes it is the antiderivative, but it is not equal to ln(abs cosx), so these are both valid antiderivatives?
and could you also solve the problem using the absolute value of cos x?
why did they choose to limit the domain then
and is it the same for the 2ond problem
river kettle
I don’t get why this type of fraction split won’t work
vocal sleetBOT
I don’t get why this type of fraction split won’t work
west raptor
Shouldn't the A/(u^2 + 1) term instead be (Au + B)/(u^2 + 1)?
dull bear
Shouldn't the A/(u^2 + 1) term instead be (Au + B)/(u^2 + 1)?
That's what @river kettle is asking about - why it can't be like that [as in the original post]
dull bear
[to which, I can suggest, "try it and see" 
]
river kettle
That's what <@722683408190406667> is asking about - why it can't be like that [a...
Hewo
river kettle
I’ll try it
@dull bear I think it could work?
but I’m not sure if I did something wrong
dull bear
<@788085606483361802> I think it could work?
let's work through it together 
river kettle
🙏
dull bear
So we basically want to find $A$ and $B$ such that $u^2 - 3 = A(u -1) + B(u^2 + 1)$, right? (after rearranging the fractions and stuff)
river kettle
Is that from getting common denominators?
dull bear
Yep you common denominator and then clear both denominators!
twin meteorBOT
@dull bear
river kettle
Okay yeah
dull bear
Cool, let's try to solve this one
So if we set u=1, then we would nicely get that -2 = B(2), right?
dull bear
Like this?
Yep, that’s pretty much what we start off with 
river kettle
So if we set u=1, then we would nicely get that -2 = B(2), right?
Okay yeah i get this
dull bear
Cool cool you have your A’s and B’s the other way round to me haha
river kettle
wai wdym
dull bear
Oh no they’re the same way nvm 
river kettle
Hehe
river kettle
Ye
dull bear
So we get that $A(u - 1) -(u^2 + 1) = u^2 - 3$
twin meteorBOT
@dull bear
river kettle
Yep
dull bear
Now, we can rearrange that a bit more, say, to $A(u - 1) = 2u^2 - 2$
twin meteorBOT
@dull bear
dull bear
Do you see any problems yet? 
river kettle
Hm. Let me try and play around with it
river kettle
Do you see any problems yet? <:Hehe:669001177605603328>
so I think there’s nothing we can do with the just A and u because we don’t want the A to disappear
dull bear
Yea, there are also a number of other things we could spot wrong
river kettle
there are too many solutions?
dull bear
<@788085606483361802>
For example, notice how there’s a squared term on the right, but none of the left
river kettle
Ohh
dull bear
2=0 confirmed 
river kettle
Hehe
dull bear
Also, for example, you could, say, factor the right hand side as 2(u-1)(u+1)
And when u is not 1, you then would be saying that A = 2(u + 1)
…in other words, a constant is equal to something that can vary 
Long story short is that if you try something like that, you run into problems 
dull bear
that’s why you want it to have that general form of the numerator like it is, and when you do it like that, life becomes so much smoother 
dull bear
Generally the idea is that you want the degree of the numerator to be “one less than that of the denominator”, so you’d have it in that general form
Of course when it comes to linear factors, that’s basically just saying that you want the numerator constant, as you’re familiar with already
river kettle
Generally the idea is that you want the degree of the numerator to be “one less ...
If it isn’t then couldn’t you do something like long division
dull bear
For irreducible quadratics, you’d want it in that form Ax + B (the numerator to have degree 1, to be linear)
dull bear
If it isn’t then couldn’t you do something like long division
Well that’s kind of the idea, what you’re trying to do partial fractions on should hopefully already be like that, so when you try and split it into the factors you should hopefully not have the degree of the “new” numerators bigger than that of the correspond denominator (and if you left it in that form, you’d quickly find the other constants to be zero anyway!)
Like with your one, if you chose something like (Au^2 + Bu + C)/(u^2 + 1), when you do similar work, you’d find A to be zero
Which you can do by e.g. comparing coefficients and stuff!
dull bear
That should hopefully go much smoother! Let me know how it goes of course
river kettle
Yep 
vast shale
<@788085606483361802> I think it could work?
bro i think you should factorize your denominator
then apply partial fractions
river kettle
That should hopefully go much smoother! Let me know how it goes of course <:catl...
So if u = 1 I can cancel the left fraction to get C
dull bear
Might be a good idea to multiply out that u^2 + 1 as well, clear all denominators first
vast shale
c will not be 0 actually
dull bear
Well while the left fraction is zero, when you cancel down common factors on the right, you may find something…
(Be careful of hidden divisions by zero and all…)
dull bear
The right is 0/0
Yep hence my comment to clear all denominators first!
river kettle
Like this?
dull bear
Like this?
Yep
river kettle
Then if u = 1 I get c = -1
dull bear
Then if u = 1 I get c = -1
Seems good to me! 
river kettle
I’m kinda stuck on what to do
dull bear
I’m kinda stuck on what to do
Either equate coefficients, or choose other “nice” values of u
If you do the latter, you should get simultaneous equations which you could then solve 
river kettle
Either equate coefficients, or choose other “nice” values of u
But if I do that the Au+B might get cancelled 
dull bear
But if I do that the Au+B might get cancelled <:sad:532803514892812288>
It won’t, the only way to have it cancel was u=1, any other choice won’t disappear!
river kettle
but how do I know what I should choose?
dull bear
You can pick anything you want, apart from 1
Usually it’s a good idea to choose “nice” choices - maybe try u=0? 
river kettle
ooh
dull bear
I’m kinda stuck on what to do
Ohh also hang on 
river kettle
I think there’s a problem
I don’t think there’s a safe value of u I can choose to isolate A
dull bear
Also the simplification you did 
river kettle
Ohhh
dull bear
Happens 
river kettle
dull bear
But otherwise you get B=2 anyway, same idea and all 
dull bear
I don’t think there’s a safe value of u I can choose to isolate A
But anything else (that isn’t 0 or 1, we used those already) then a matter of rearranging 
river kettle
Oh so I can choose like 2?
dull bear
Yep, 2 can do!
You could also e.g. pick -1, but any number you go for that isn’t either of the ones we used is good!
river kettle
Yay
dull bear
,w partial fractions (u^2 - 3)/(u^3 - u^2 + u - 1)
twin meteorBOT
river kettle
the hardest part is not even the integration but the fracrtion stuff 😭
dull bear
the hardest part is not even the integration but the fracrtion stuff 😭
It can be yea 
esp making sure you’re doing it all right
river kettle
lmao i was integrating it
and realized i was using u as the variable to sub
so i was doing du/du
lmao
dull bear
Heheh 
river kettle
Anyways thank u for the help
dull bear
Anyways thank u for the help <:catlove:920698762932920370> <:catlove:92069876293...
Always a pleasure 
river kettle
.close
rain hollow
Need help with 5
vocal sleetBOT
Need help with 5
patent nymph
Is the question to evaluate or simplify it?
You can start by using a formula that involves tangent, like its definition
carmine hinge
Mhmm
paper depot
@rain hollow yes, and also rewrite the cotangent as cos(α)/sin(α).
you might find it less time-consuming if you abbreviate $\cos(\alpha)$ and $\sin(\alpha)$ as $c$ and $s$ respectively, with three things worth noting:
\begin{enumerate}
\item the reason this is sensible to do at all is because all trig functions in your expression are of the same argument ($\alpha$ in this case -- there are no $2\alpha$, $\alpha + \pi$ or whatever else as inputs to trig functions)
\item if you're going to submit your work, then you have to note these abbreviations explicitly at the beginning
\item whether or not this is getting submitted to your teacher, you're gonna have to convert back from these abbreviations at the end
\end{enumerate}
twin meteorBOT
Ann
vocal sleetBOT
@rain hollow Has your question been resolved?
rain hollow
Did that now what
rain hollow
paper depot
and also, it looks like you've chosen NOT to go with my "make your life easier" idea
was this on purpose?
rain hollow
Yea my teacher said i should do it this way
paper depot
so your teacher specifically forbade you from doing any abbreviations like i wrote, yes?
rain hollow
No
paper depot
then what did your teacher mean when they said you should "do it this way"?
rain hollow
Like with identitys
paper depot
i didn't say we wouldn't be using identities.
the usage of identities is independent of whether or not you choose to allow yourself to save ink, time and effort
i was simply trying to set you up on a path which i think will make your life easier
rain hollow
So i replace cosa as c and sina as s?
paper depot
yes, but do you understand WHY im telling you to do it?
i do NOT want to have you do something because you see it as dogma.
rain hollow
It’s shorter
rain hollow
So do i rewrite the whole equation
paper depot
expression*
rain hollow
Dooing this?
paper depot
but yes, the whole thing.
rain hollow
Yes
paper depot
replace all instances of cos(a) with the letter c and all instances of sin(a) with the letter s
send me the result
rain hollow
Ok
paper depot
also get a piece of paper and note down c := cos(a), s := sin(a) on it
so that you do not forget what the letters mean
rain hollow
paper depot
why is there a?
paper depot
there is still an a you forgot to fix.
paper depot
ok, cool.
now look at this expression as simply an algebraic expression. forget about trig for the time being.
do you see any simplifications?
answer me only "yes, i see some simplifications" or "no, i do not see any simplifications."
do NOT elaborate on what simplifications you see until i ask for it.
rain hollow
I do but not sure if they are correct
paper depot
yes, cs/c simplifies to just s.
that is correct
do the simplifications you see, and send me the resut.
rain hollow
rain hollow
S/s^2 is 1 /s no?
patent nymph
Yes
paper depot
S/s^2 is 1 /s no?
yes
rain hollow
Ok now what
patent nymph
Then, what is the whole expression now equal to
paper depot
well show what you've got right now
paper depot
one sec, image still loading
rain hollow
Wich is just sin/sin no?
paper depot
(1-c^2)/s, you mean.
rain hollow
2
paper depot
"I forgot a 2. I meant sin^2(a)."
patent nymph
So just sin it the answer
yes, the answer is just sin a for this question
rain hollow
Ok ill rewrite this in my notebook and i have 1 more problem i need help with
paper depot
So just sin it the answer
sin(a), but yes
rain hollow
Yes
patent nymph
again, you can expand it all out, and then use the same method as before
by letting c := cos alpha, s := sin alpha
rain hollow
Expand both?
paper depot
^
exact same method
write the thing in terms of c and s, look at it purely as an algebraic expression, and simplify what you can.
rain hollow
Is this not just a^2+b^2?
paper depot
wdym by "just a^2+b^2"?
rain hollow
Like the formula
paper depot
what formula?
rain hollow
Like this?
paper depot
yeah, keep going
from what i've seen so far you generally don't have trouble doing the algebra from this point on
but send your steps anyway so we can debug it in case you do screw up
rain hollow
So do i just square this up
paper depot
"square this up"?
rain hollow
1-c/s
rain hollow
Yes
paper depot
you don't need my permission to do it lol
if you see something you can do that you think helps you, do it
rain hollow
It would be (s-c)^2/s^2 no?
patent nymph
It would be (s-c)^2/s^2 no?
how did you get there?
paper depot
datoo i think it will be better if you just do it and show your work
and then we tell you your errors
rather than you asking at every step
again you clearly see things and just hesitate to do them. don't hesitate. just do it.
rain hollow
It would be (s-c)^2/s^2 no?
Ok is this right tho?
paper depot
half-right
(1 - c/s)^2 does equal (s-c)^2/s^2 yes
but like
reiterating
just do the whole thing. please.
patent nymph
again you clearly see things and just hesitate to do them. don't hesitate. just ...
but also send steps so we know what you’re doing
patent nymph
S^2+2sc+c^2/s^2
(s^2 + 2sc + c^2)/s^2
paper depot
That should be right
please do not send us bits and pieces at a time
rain hollow
/sin
paper depot
it's very hard to check what you are doing with mere glimpses
paper depot
like here is the full work for your previous problem. this is the format in which i personally would prefer to see your work
rain hollow
My writing is very bad
paper depot
try your best, or write it out line by line in desmos or something.
rain hollow
I have this now
patent nymph
I have this now
you can’t cancel the 2 and the s^2
rain hollow
Why?
patent nymph
you can only cancel things on the numerator and the denominator when they’re all in one term
here, there are 2 terms in the numerator
rain hollow
Dominator is the same tho
patent nymph
Dominator is the same tho
,tex .wrong cancel
twin meteorBOT
rysrobrgldvoelr👻ep>vneae=u
patent nymph
basically the same, but flip the fraction
rain hollow
Oh so thats wrong
Ok so final thing i have left is 2s^2+2c^2/s^2
Thats 2/sin^2a
Yup done thanks
vocal sleetBOT
@rain hollow Has your question been resolved?
sour oasis
if 1+1/4+1/9+1/16+1/25+...=a , then 1/9+1/25+1/49+...=
vocal sleetBOT
if 1+1/4+1/9+1/16+1/25+...=a , then 1/9+1/25+1/49+...=
sour oasis
can someone explain this to me
and please explain it in kinda simple terms, cause im kinda dumb
tidal dock
\begin{align*}
1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\cdots&=a\
\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\cdots&=?
\end{align*}
twin meteorBOT
artemetra
sour oasis
this is the original question btw, its the one i found on the internet. but the one on my book says instead of a its x
tidal dock
**artemetra**
don't they look similar?
sour oasis
don't they look similar?
seems like it?
daring fractal
how bout a/4?
outer warren
series they want skips terms
tidal dock
1+\frac{1}{4}+\underbrace{\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\cdots}_{b}=a
sour oasis
series they want skips terms
series? sorry english is my second language and i never learned english math terms
original question. its the same meaning
daring fractal
isnt this the basel prob??
solemn marten
it is
daring fractal
so doesnt its sum converge to pi^2/6?
sour oasis
🤷♂️ my book says its a algebra substituon method
solemn marten
so doesnt its sum converge to pi^2/6?
yeah
although i don't think that's how ur supposed to solve this one
sour oasis
this is the proof that i got on the internet, but i dont really understand it. specifically line 2 from the answers
daring fractal
pi^2 - (5/4)
sour oasis
this is the proof that i got on the internet, but i dont really understand it. s...
id like an explanation of the answer here, specifically line 2
dawn kestrel
Isn’t it more like the sum of the reciprocal of odd squares
sour oasis
i dont get what you mean by that
can someone explain what happened between line 2 and 3
its still the same question
vocal sleetBOT
@sour oasis Has your question been resolved?
abstract palm
let $B=((1,2)^T,(1,3)^T), C=((1,3)^T,(1,4)^T)$ be basis in $\mathbb{R}^2$. $[f]^B_C= \begin{pmatrix} 1&-1\0&2 \end{pmatrix}$. how could i find $f^{-1}$?
vocal sleetBOT
let $B=((1,2)^T,(1,3)^T), C=((1,3)^T,(1,4)^T)$ be basis in $\mathbb{R}^2$. $[f]^...
twin meteorBOT
Slowaq
hasty pulsar
Calculate f^C_B or simply take the inverse of the basis transform
twin meteorBOT
Slowaq
hasty pulsar
yes
abstract palm
but this is not correct
this yields this
and this is correct result
how could i find f^C_B? @hasty pulsar
hasty pulsar
,w inverse of {{0,1},{-1,2}}
twin meteorBOT
hasty pulsar
write every element of B as a LC of the basiselements of C
(1,2) = x * (1,3) + y * (1,4)
dull bear
Do you have a picture of the original question? Presumably you may want f^{-1} in terms of e.g. standard basis?
abstract palm
Do you have a picture of the original question? Presumably you may want f^{-1} i...
yes exaclty sorry i forgot this
dull bear
**Slowaq**
In that case, if so, remember to multiply this (on the appropriate sides) by the correct change of basis matrices
abstract palm
In that case, if so, remember to multiply this (on the appropriate sides) by the...
Hm what did i do wrong here?
dull bear
Hm what did i do wrong here?
You’re doing f rather than f^{-1} presumably
abstract palm
what do you mean?
dull bear
and this is correct result
As in are you wondering why you don’t get this?
abstract palm
yes i guess so
dull bear
**Slowaq**
Well this matrix represents $[f^{-1}]_B^C$, and if you multiply by the appropriate matrices, you should get I think
twin meteorBOT
@dull bear
abstract palm
which matrix?
abstract palm
this?
dull bear
,w [[1, 1], [1, 4]]^(-1) * [[1, -1], [0, 2]]^(-1) * [[1, 2], [1, 3]]
twin meteorBOT
dull bear
Hmm 
dull bear
**Slowaq**
And this one here
dull bear
At least f is invertible I guess 
Ahhhh wait 
,w [[1, 1], [1, 4]]* [[1, -1], [0, 2]]^(-1) * [[1, 2], [1, 3]]^(-1)
twin meteorBOT
dull bear
f is in those other bases, did it the wrong way around
abstract palm
but could you please describe what are you tring to achieve?
dull bear
,w [[1, 1], [3, 4]]* [[1, -1], [0, 2]]^(-1) * [[1, 1], [2, 3]]^(-1)
twin meteorBOT
abstract palm
yes nice finally
dull bear
and this is correct result
Finally 
got mixed up so much 
abstract palm
and could youo please explaint why you did this?
dull bear
but could you please describe what are you tring to achieve?
The idea, which is better explained, is that $[f^{-1}]_B^C = ([f]_C^B)^{-1}$
twin meteorBOT
@dull bear
dull bear
But then you want f^{-1} in standard basis, so first convert your standard basis vector to C basis, then apply f^{-1}, which by multiplying by the inverse matrix we found gets you into B basis, then convert that B basis back to standard basis
abstract palm
and how i i convert standard basis vector to C basis?
dull bear
You know that e.g. the matrix that converts you from C basis to standard basis is $\pmqty{ 1 & 1 \ 3 & 4}$
twin meteorBOT
@dull bear
dull bear
So the matrix that converts you from standard to C basis is the inverse of that
abstract palm
ugh yes maybe i am supposed to know that but it does not ring th bell
abstract palm
,w [[1, 1], [3, 4]]* [[1, -1], [0, 2]]^(-1) * [[1, 1], [2, 3]]^(-1)
and why then you did not use inverce of that here?
dull bear
Good question 
arghhh always don’t like actually doing linalg out 
abstract palm
xddd i feel you
dull bear
,w [[1, 1],[3, 4]] * [[1, -1], [0, 2]] * [[1, 1], [2, 3]]^(-1)
twin meteorBOT
dull bear
That should be f in terms of standard basis, if I haven’t gotten myself mixed up again 
,w [[1, 0], [-1, 2]]^(-1)
twin meteorBOT
dull bear
What did I do then
$[f]_C^B$ is where the initial basis is $B$ and the final one is $C$ right?
twin meteorBOT
@dull bear
abstract palm
i think so i always get this mixed up
dull bear
Yeaaa it’s painful, had to find that one 
dull bear
But then you want f^{-1} in standard basis, so first convert your standard basis...
Anyways, the idea’s something like that 
but without getting mixed up and stuff 
abstract palm
xd alright ill try to absorb it thx for help
dull bear
No worries, hopefully that was actually useful I’m always better with saying what to do rather than actual work when it comes to linalg 
vocal sleetBOT
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heady spade
why is the theta from 0 to pi/2 and not 0 to pi?
vocal sleetBOT
why is the theta from 0 to pi/2 and not 0 to pi?
visual oracle
why would it be 0 to pi
heady spade
uhh because of the y2=1-x2 which makes it so y is greater than or equal to 0
so its only the top part of the circle
idk why its the top right part instead tho
visual oracle
i dont even know cylindrical coordinates 🤣
vocal sleetBOT
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desert hornet
Find the number of ways to select k balls from 3 green, 3 blue, 3 white and 3 gold balls.
vocal sleetBOT
Find the number of ways to select k balls from 3 green, 3 blue, 3 white and 3 go...
desert hornet
Using generating functions I get that this is equivalent to the coefficient of $x^k$ in the expansion of $(1+x+x^2+x^3)^4$
twin meteorBOT
kheerii
desert hornet
I get that $(1+x+x^2+x^3)$ gives us the generating function for one of the color of balls (since there is exactly one way to choose 0, 1, 2 or 3 balls from a set of 3 identical balls)
twin meteorBOT
kheerii
desert hornet
but why, by raising this to the power of 4, does it give us the generating function for the original problem?
livid horizon
desert hornet
how?
the colored balls are identical, I forgot to mention
also this is more a question about generating functions and less about the actual question
ancient knoll
but why, by raising this to the power of 4, does it give us the generating funct...
4 colors
desert hornet
can I just multiply generating functions like that?
ancient knoll
you can, and it represents the combinations to pick everything that’s being multiplied
ancient knoll
I mean it doesn't really make sense to me how that works
probably easier to first think of every bracket as a polynomial in different variables
say (1+x+x²+x³)(1+y+y²+y³)
first bracket is number of green balls and second bracket is number of blue balls
where every coefficient in both polynomials is 1, so for either color you only have 1 distinct way to pick some number of balls up to 4
multiplying gets you
1+x+x²+x³ +
y+xy+x²y+x³y+
y²+xy²+x²y²+x³y²+
y³+xy³+x²y³+x³y³
something with 16 terms because you have 4 different ways to choose the number of green balls and 4 different ways to choose the number of blue balls
and coefficients of x^n y^m tell you how many ways exact to choose a combination with n blue balls and m green balls (here since the brackets are simple the answer is always 1)
desert hornet
ahh yeah that makes sense
so the total sum of all terms of x^n y^m such that m+n = k will be the answer
so we can just replace y with x and check the coefficient of x^(m+n)
this wouldn't work for all generating functions though, right?
only for these counting type problems
ancient knoll
but here you only care about the total number of balls, so to make life easier you can reuse the same variable
(note that here it’s analogous to summing up a diagonal in the block of expression i gave)
ancient knoll
only for these counting type problems
for a specific kind of counting problems yes
where you don’t really care about the number of any specific “subspecies” you might call, just the total count
you don’t really need the number of balls per color to be identical either
if you have 2 greens and 7 whites you’d just have (1+x+x²)(1+x+x²+x³)² (1+…+x^7)
desert hornet
yeah that makes sense '
how would I stipulate that in a competition environment though?
like it makes sense but how would I show that mathematically
vocal sleetBOT
@desert hornet Has your question been resolved?
vocal sleetBOT
@desert hornet Has your question been resolved?
vocal sleetBOT
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fringe mortar
is this correct?
vocal sleetBOT
is this correct?
jade anvil
yes I think so
jade anvil
np
pale lion
i have no idea where to start with this problem
vocal sleetBOT
i have no idea where to start with this problem
lucid bane
complete the square for x^2+12x and 4y^2+20y
our goal is to try to parameterize the equation for this ellipse, and to set up an expression for the arclength.
vocal sleetBOT
@pale lion Has your question been resolved?
pale lion
so would it be
(x+6)^2 - 36 = x^2 +12x +36 - 36 = x^2
(2y+5)^2 - 25 = 4y^2 + 20y + 25 - 25 = 4y^2 + 20y
and we take away 36 and 25 from 400 so it ends up being
(x+6)^2 + (2y+5)^2 = 339
not gonna lie im not too good with parametric equations atm
sinful magnet
is there a mistake yet
vocal sleetBOT
is there a mistake yet
jade anvil
where did the 1/2 go?
jade anvil
yup seems correct
sinful magnet
kk
now this part im stuck
idk what to do
should i just say its divergent since 1/0 = inf
@jade anvil
jade anvil
no so what you do is instead of using a, put the equation in variable n by using u=n2+1 and limits infinity to 0 after solving the integral
vocal sleetBOT
@sinful magnet Has your question been resolved?
toxic osprey
why is y(2)=0 ? the only initial condition given is that y(0)=0, i don’t see where they got y(2) from
vocal sleetBOT
why is y(2)=0 ? the only initial condition given is that y(0)=0, i don’t see whe...
toxic osprey
why’s that solution valid at t=2 though? doesn’t the function shift at that point
oh is it because it’s a continuous function so the two have to be the same
cyan talon
if you want the derivative to make sense you need the function to be continuous yes
vocal sleetBOT
vocal sleetBOT
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rain hollow
I have a hexagan all the sides are eaquel
vocal sleetBOT
I have a hexagan all the sides are eaquel
rain hollow
CF line makes 2 trapezoids
AC diagnal is 13 AE diagnal is 10
Need to find the sides of the hexagram
pliant gust
If sides ar equal, then AC=AE, but its not true for your example
pliant gust
I have a hexagan all the sides are eaquel
.
ancient knoll
exactly, didn’t say it’s regular
vocal sleetBOT
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twin meteorBOT
Moaen Ism
rain hollow
?
ancient knoll
find a right angle triangle
ancient knoll
i think so yes
and with the right triangle you can relate it to the side lengths given and get an answer
rain hollow
So the side would by 12
Of the right triangle
Sides are 5 13 12
Now what how do i find the sides from that?
@ancient knoll
ancient knoll
im trying to think of a way to get you to the next step without directly telling you the answer and it feels kinda hard
ancient knoll
no, then your diagonals are a lot longer than possible
ok giving a hint for this is too hard
with the 5, 12, 13 triangle in mind, rearrange the hexagon into a rectangle
rain hollow
Ok wait is it ez to realise what to do next?
I will just do it later if thats the case
ancient knoll
should be yeah
rain hollow
Can you just tell me
vocal sleetBOT
@rain hollow Has your question been resolved?
warped cliff
Get the maximum and minimum values of the function
vocal sleetBOT
Get the maximum and minimum values of the function
willow bridge
do you mean ln(x) ?
willow bridge
okay
strange crater
do you know the product rule?
warped cliff
I differentiated
willow bridge
what did you get?
warped cliff
Now it's ln(x) + 1/x * (x+1)
I assign this by 0
So that it's ln(x) + 1/x * (x+1) = 0
willow bridge
yes
willow bridge
So 1/x + ln(x) = 0
how? no
willow bridge
$\frac{1}{x} \cdot (x+1) = 1 + \frac{1}{x}$
twin meteorBOT
LF
willow bridge
but yes
warped cliff
Ohhh
Wait
I'm dumb
I forgot to assign the value of x/x by 1
I just forgot about it lol
willow bridge
lol
warped cliff
So ln(x) = -1 -1/x
warped cliff
Yeah it's no solution for some reason
solemn marten
You're getting no solution or some answer key says no solution
warped cliff
Do iiii
willow bridge
what does that mean about the function
warped cliff
Differentiate again?
willow bridge
what does it mean if the derivative is never =0
warped cliff
Uhhhm
solemn marten
Ah right yeah no solution my bad
warped cliff
I don't know
The only hint given in this question is the value of the minimum value which is f(1) = e²
willow bridge
i dont think you gave us the right equation
$f(x) = (x+1) \ln(x) \implies f(1) = 2 \cdot 0 = 0 \neq e^2$
twin meteorBOT
LF
willow bridge
and also if the derivative is never equal to 0 there is no minimum or maximum
willow bridge
good idea
vocal sleetBOT
Available help channel!
sleek escarp
vocal sleetBOT
burnt coral
ok
strange crater
just multiply
burnt coral
write the thickness of a dollar bill in scientific notation
sleek escarp
write the thickness of a dollar bill in scientific notation
1 x 10 to the power of -4 right?
strange crater
yes
burnt coral
yep
broken nimbus
Hi there I'm having some difficulty understanding what exactly a cross section is. I googled the definition and watched a YT vid and am still confused. I think where my confusion is stemming from is I know that singular semi circular there of those functions rotated around the x-axis is the cross section, but I find that to be confusing cus what I have inferred so far from cross sections is that its a cut we make in 3-dimensions making a plane. But this singular semi circle has 3 dimensions here, so how can it be a cross section?
vocal sleetBOT
Hi there I'm having some difficulty understanding what exactly a cross section i...
broken nimbus
For context that singular semi circle represents what happens to a Riemann sum when rotated as its super helpful in conceptualizing applications of integrals problems.
heavy yoke
the semicircle itself is a cross-section (2D), but when we take a riemann sum we multiply it by a small thickness (dx, 1D) to make it a volume (3D). similar to how we multiply a height (1D) by the same thickness (dx, 1D) to obtain an area (2D)
vocal sleetBOT
@broken nimbus Has your question been resolved?
regal bane
A cross-section is a 2D face that results when you cut a 3D shape
For example if you cut a sphere, you'll get a circle. Any cross section of a sphere is a circle.
From your picture, that is not a cross section.
However it does relate heavily to a cross section, and the volume of it is important to your problem
idle perch
Find the volume of the tetrahedron bounded by the planes x + 2y + z = 2, x = 2y, x = 0, y = 0
vocal sleetBOT
Find the volume of the tetrahedron bounded by the planes x + 2y + z = 2, x = 2y,...
idle perch
this is what I've tried so far
rotated version
correct answer should be 1/3 but I don't think I set it up right
vocal sleetBOT
@idle perch Has your question been resolved?
vocal sleetBOT
@idle perch Has your question been resolved?
onyx flax
@idle perch I don't believe that these planes actually define a tetrahedron